Chemical, Biochemical, and Engineering Thermodynamics [5 ed.] 9780470504796, 047050479X

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CHEMICAL, BIOCHEMICAL, AND ENGINEERING

THERMODYNAMICS FIFT H EDIT IO N

STANLEY I. SANDLER

Chemical, Biochemical, and Engineering Thermodynamics Fifth Edition

Stanley I. Sandler University of Delaware

VP AND EDITORIAL DIRECTOR SENIOR DIRECTOR ACQUISITIONS EDITOR EDITORIAL MANAGER CONTENT MANAGEMENT DIRECTOR CONTENT MANAGER SENIOR CONTENT SPECIALIST PRODUCTION EDITOR

Laurie Rosatone Don Fowley Linda Ratts Gladys Soto Lisa Wojcik Nichole Urban Nicole Repasky Ameer Basha

This book was set in 10/12 TimesLTStd by SPi Global and printed and bound by LSC Communications, Inc. The cover was printed by LSC Communications, Inc. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. c 2017, 2006 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be Copyright  reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923 (Web site: www.copyright.com). Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008, or online at: www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at: www.wiley.com/go/returnlabel. If you have chosen to adopt this textbook for use in your course, please accept this book as your complimentary desk copy. Outside of the United States, please contact your local sales representative. ISBN: 978-0-470-50479-6 (PBK) Library of Congress Cataloging-in-Publication Data: Names: Sandler, Stanley I., 1940- author. Title: Chemical, biochemical and engineering thermodynamics / Stanley I. Sandler, University of Delaware. Other titles: Chemical and engineering thermodynamics Description: Fifth edition. | Hoboken, NJ : John Wiley & Sons, Inc., [2016] | Revised edition of: Chemical and engineering thermodynamics. | Includes index. Identifiers: LCCN 2016044996 (print) | LCCN 2016048589 (ebook) | ISBN 9780470504796 (pbk.) | ISBN 9781119343776 (pdf) | ISBN 9781119321286 (epub) Subjects: LCSH: Thermodynamics—Textbooks. | Chemical engineering—Textbooks. | Biochemical engineering—Textbooks. Classification: LCC QD504 .S25 2016 (print) | LCC QD504 (ebook) | DDC 541/.369—dc23 LC record available at https://lccn.loc.gov/2016044996 The inside back cover will contain printing identification and country of origin if omitted from this page. In addition, if the ISBN on the back cover differs from the ISBN on this page, the one on the back cover is correct.

To Judith, Catherine, Joel, And Michael

About the Author STANLEY I. SANDLER earned the B.Ch.E. degree in 1962 from the City College of New York, and the Ph.D. in chemical engineering from the University of Minnesota in 1966. He was then a National Science Foundation Postdoctoral Fellow at the Institute for Molecular Physics at the University of Maryland for the 1966–67 academic year. He joined the faculty of the University of Delaware in 1967 as an assistant professor, and was promoted to associate professor in 1970, professor in 1973 and Henry Belin du Pont Professor of Chemical Engineering in 1982. He was department chairman from 1982 to 1986. He currently is also professor of chemistry and biochemistry at the University of Delaware and founding director of its Center for Molecular and Engineering Thermodynamics. He has been a visiting professor at Imperial College (London), the Technical University of Berlin, the University of Queensland (Australia), the University of California, Berkeley and the University of Melbourne (Australia). In addition to this book, Professor Sandler is the author of over 400 research papers and a monographs, and he is the editor of a book on thermodynamic modeling and five conference proceedings. His most recent book is “Using Aspen Plus(R) in Thermodynamics Instruction: A Step-by-Step Guide” published by AIChE/Wiley in 2015. He was also the editor of the AIChE Journal. Among his many awards and honors are a Faculty Scholar Award (1971) from the Camille and Henry Dreyfus Foundation, a Research Fellowship (1980) and U.S. Senior Scientist Award (1988) from the Alexander von Humboldt Foundation (Germany), the 3M Chemical Engineering Lectureship Award (1988) from the American Society for Engineering Education, the Professional Progress (1984), Warren K. Lewis (1996) and Founders (2004) Awards from the American Institute of Chemical Engineers, the E. V. Murphree Award (1996) from the American Chemical Society, the Rossini Lectureship Award (1997) from the International Union of Pure and Applied Chemistry, and election to the U.S. National Academy of Engineering (1996). He is a Fellow of the American Institute of Chemical Engineers and the Institution of Chemical Engineers (Britian and Australia), and a Chartered Engineer.

Preface

PREFACE FOR INSTRUCTORS This book is intended as the text for a course in thermodynamics for undergraduate and graduate students in chemical engineering and also for practicing engineers. Its previous four editions have served this purpose at the University of Delaware for almost forty years. In writing the first edition of this book I had two objectives that have been retained in the succeeding editions. The first was to develop a modern applied thermodynamics text, especially for chemical engineers, relevant to other parts of the curriculum–specifically to courses in separations processes, chemical reactor analysis, and process design. The other objective was to develop, organize and present material in sufficient detail for students to obtain a good understanding of the basic principles of thermodynamics and a proficiency in applying these principles to the solution of a large variety of energy flow and equilibrium problems. Since the earlier editions largely met these goals, and since the principles of thermodynamics have not changed over the past decade, this edition is similar in structure to the earlier ones. During this time, however, important changes in engineering education have taken place. The first is the increasing availability of powerful desktop computers and computational software, along with well-developed and easy-to-use process simulation software. Another is the increasing application of chemical engineering thermodynamics principles and models to new areas of technology such as polymers, biotechnology, solid-state processing, and the environment. The current edition of this text includes applications that address each of these changes. The availability of desktop computers and equation-solving software has now made it possible to closely align engineering science, industrial practice, and undergraduate education. In their dormitory rooms or at home, students can now perform sophisticated thermodynamics and phase equilibrium calculations similar to those they will encounter in industry. In this fifth edition, I provide several different methods for making such calculations. The first is to utilize the set of programs I have developed for making specific types of calculations included in the fourth edition. These programs enable (1) the calculation of thermodynamic properties and vapor-liquid equilibrium of a pure fluid described by a cubic equation of state; (2) the calculation of the thermodynamic properties and phase equilibria for a multicomponent mixture described by a cubic equation of state; and (3) the prediction of activity coefficients in a mixture using the UNIFAC group-contribution activity coefficient model. These programs are available on the website for this book as both program-code and stand-alone executable modules; they are unchanged from the previous edition of this book. However, I suggest instead the use of the thermodynamics package in Aspen Plus(R), which is continually updated and has an easy-to-use interface. The second is to employ the computer algebra/calculus programs for MATHCAD on the website that provides solutions to many illustrations and homework problems in this edition. Alternatively, students and instructors could use similar programs such as MATHEMATICA, MAPLE, and MATLAB. Students who develop their own codes for such computer programs can achieve a thorough understanding of the methods required (and the computational difficulties involved) in solving complex problems without having to become experts in computer programming and numerical analysis. Students who use my prepared codes will be able to solve interesting problems and concentrate on the subject matter at hand, namely, thermodynamics, without being distracted by computational methods, algorithms, and programming languages. These equation-solving programs are, in my view, valuable educational tools; but there is no material in this textbook that requires their use. Whether to implement them or not is left to the discretion of the instructor. iv

Preface for Instructors v More recently in engineering practice, these one-off thermodynamics programs written by textbook authors have been replaced by suites of programs, process simulators, that make it possible to quickly model a whole chemical plant using current unit operations and thermodynamics models, as well as to access enormous databanks of pure fluid and mixture thermodynamic data. A number of such simulators are available, such as ASPEN, HYSIS, PROSIM, and CAPE-OPEN. In this fifth edition, I have incorporated the ASPEN process simulator by adding thermodynamics illustrations and homework problems that use ASPEN. I recognize, however, that there is no universal agreement on the use of a process simulator in, especially, an undergraduate thermodynamics course. Indeed, there are those in my own department who argue against it. The argument against the use of prepared computer programs in general, and process simulators in particular, is that students will treat them as “black boxes” without understanding the fundamentals of thermodynamics or the methods for choosing the thermodynamic models most appropriate to the problem at hand. My argument for using process simulators in undergraduate instructional courses is two-fold. First, it allows students to solve with great efficiency more interesting and practical problems than they could, within a reasonable time-frame, solve by hand; and it provides them an opportunity to ask and answer “what-if” questions. For example, what happens to the vapor-liquid split and the compositions of each of the co-existing phases in a multi-component Joule-Thomson expansion if the inlet temperature or pressure is changed? Answering such what-if questions allows students to quickly develop an intuitive sense of the way processes behave, an understanding that otherwise might only be attained by repeated, tedious hand calculations. Second, using a process simulator introduces students to a tool they are likely to employ in their professional career. Moreover, modern process-simulation software is generally bug-free, providing an easy-to-use interface that is the same for all problems. In this argument I have taken the middle road. By means of some of the illustrations and problems provided in this text, students will initially develop an understanding of the basic applications and methods of thermodynamics by doing hand calculations. Then, once they understand the basic principles and methods, I encourage them to use process simulators (rather than my previous programs) to explore many additional, and more complicated, applications of thermodynamic principles. Whereas nothing in this new edition requires students or the instructor to use a process simulator, the illustrations do contain examples of the results of using a process simulator. In addition, many opportunities for using process simulator software are provided in the numerous end-of-chapter problems. Furthermore, by using a process simulator the instructor can easily change the input parameters of a homework problem and obtain the solution, thereby providing unlimited opportunities for creating new problems. On the designated website for this new edition, I have, therefore, provided the ASPEN 8.6 input files for numerous illustrations and problems presented in the textbook. I have chosen ASPEN because it appears to be the process simulator most widely used in industry and at colleges and universities, in the United States at least. Clearly, any other process simulation software could be employed, but in these cases users will need to develop their own input files. Since I am introducing ASPEN in this fifth edition, I have not updated the thermodynamics programs included in previous editions of this textbook, and they remain available on the website. Still, I encourage the use of Aspen or other process-simulation software rather than these more primitive programs. (For assistance in employing the thermodynamics packages in Aspen, I suggest consulting my recent book, Using Aspen Plus in Thermodynamics Instruction, published by Wiley/AIChE in 2015.) In an effort to make the subject of thermodynamics more accessible to students, the format of this book provides space for marginal notes. The notes I have added are meant to emphasize important ideas and concepts, as well as to make it easier for students to locate these concepts at a later time. Since I frequently write notes in the margins of books I own, I wanted to provide a place for students to add notes of their own. Also, I continue to enclose important equations in boxes, so that readers can easily identify the equations that are the end results of often detailed analysis. I hope this will enable students to quickly identify the central tree in what seems like a forest of equations. I have also provided a short title or description for each illustration to indicate the primary concept that is to be learned or grasped. Readers familiar with earlier editions of this book will notice that while the basic structure remains the same, it contains many internal changes. For example, there are many new illustrative and homework problems. Illustrations have been added not only to demonstrate new concepts, but also to provide breaks among pages of mathematical derivations or thermodynamic philosophy. These should make thermodynamics and phase

vi Preface equilibria more relevant to the interests of students. There are additional sections on chemical reactions in biochemical systems and I have included additional material on energy and energy-related processes. Furthermore, the biochemical applications now appear throughout the second half of the book rather than being relegated to the final chapter, as was the case in the previous edition. Some of the idiosyncrasies present in earlier editions remain here. For example, I prefer to use the term energy balance rather than the first law, and to show that the Carnot efficiency easily follows once entropy is defined. Here I depart from the more common procedure of introducing entropy (and the second law) in terms of the Carnot cycle. My experience with the latter method is that students then have difficulty making the necessary generalization if the concept of entropy and the second law are introduced in terms of a specific device. Also, I continue to prefer the partial molar Gibbs energy, which describes the function precisely, to the term chemical potential. In most other areas, I employ traditional thermodynamic notation. It has been a decade since the appearance of the fourth edition of this book. During this time many people have encouraged me to prepare a new edition and have graciously contributed their views, ideas, and advice. The most important contributors have been the undergraduate and graduate students I have taught at the University of Delaware. I have benefited greatly from their inquisitive minds and penetrating questions. I have also benefited from the helpful comments of colleagues at the University of Delaware and elsewhere who have used earlier editions of this book, and from the questions and comments of students around the world who have corresponded with me by email. I do refuse, however, to provide these students with solutions to homework problems assigned by their instructors, a not infrequent request. I wish to thank the administration and my colleagues at the University of Delaware, who have provided the unencumbered time of a sabbatical leave necessary for the completion of this new edition. And I am grateful, as always, to my family for their support. Stanley I. Sandler Newark, Delaware January 25, 2016

PREFACE FOR STUDENTS Thermodynamics is essential to the practice of chemical engineering. A major part of the equipment and operating costs of processes developed by chemical engineers is based on design methods that apply the principles of thermodynamics. In courses you will take later in the chemical engineering program–on mass transfer, reaction engineering, and process design you will discover just how important a foundation thermodynamics provides. At this point in your education, you have probably been exposed to some aspects of thermodynamics in courses in general chemistry, physical chemistry, and physics. My recommendation is that you set aside what you have learned about thermodynamics in those courses and start with a fresh mind. To begin with, the notation in this book is different from that employed in those courses and more like the notation used in other chemical engineering courses. In non-engineering courses, thermodynamics is usually applied only to a closed system (for example, a fixed mass of a substance), while engineering applications generally involve open systems–that is, those with mass flows into and/or out of the system. Moreover, you may have been introduced to entropy using a device such as a Carnot cycle. Please expunge from your mind the connection between entropy and such devices. Entropy, like energy, is a very general concept, independent of any such device. Entropy is different from energy (and mass and momentum) in that it is not a conserved property. Indeed that is one of its most important characteristics and allows us to explain why processes go in one direction and not in the reverse. As you will see (in Chapter 4), even though it is a non-conserved property, entropy is very important. For example, if two metal blocks, one hot and the other cold, are put into contact with each other, the concept of entropy leads us to the conclusion that heat will be transferred from the hot block to the cold one, and not the reverse, and that after a while, the two blocks will be at the same temperature. Both of these conclusions are in agreement with our experience. Note that the principle of energy conservation tells us only that the total energy of the system will ultimately equal the total initial energy, but not that the blocks need to be at the same temperature. This is one illustration of the fact that we frequently have to employ the concepts of both energy conservation and entropy to solve problems in thermodynamics.

Preface for Students vii Thermodynamics is applied in two central ways. One is to calculate heat and work (or more generally energy) flows: for example, to determine the conversion of heat to work in various types of engines; and to determine the heat flows accompanying chemical reactions or changes from one state of system to another. The second important type of thermodynamic calculation is to determine the equilibrium state: for example, to calculate the equilibrium compositions of the vapor and liquid of a complex mixture needed to design a method, such as distillation, for purifying the components; or to determine the equilibrium composition of a chemically reacting system. After completing your study of this textbook, you should be able to do all such calculations, as well as some computations relating to biochemical processing, safety, and the distribution of chemicals in the environment. Chemical engineering, and the applied thermodynamics presented in this book deals with real substances; and therein lie two of the difficulties. The first is that the properties of real substances may not be completely known from an experiment or available in tables at all temperatures and pressures (and for mixtures at all compositions). These may need to be approximately described by model equations: for example, a volumetric equation of state that interrelates pressure, volume, and temperature (the ideal gas equation of state applies only to gases at very low pressures and not to conditions generally of interest to chemical engineers); or equations that relate activity coefficients to composition. Any one of several different models may be used to describe a pure substance or mixture, and each will result in slightly different answers when solving a problem. Within the accuracy of the underlying equations, however, all the solutions are likely to be correct if the appropriate models are used. This may be disconcerting to you, as in other courses–especially in mathematics and physics–you may be used to solving problems that have only a single correct answer. The situation here is one that is continually faced by practicing engineers. They must solve a problem, even though the description of the properties is imperfect, and choose which equation of state or activity coefficient model to employ. (Some guidance in making such choices for mixtures is provided in Section 9.11.) The second problem that arises is that the equations of state and activity coefficient models used in thermodynamics are not linear algebraic equations, which can make the computations difficult. It is for this reason that I provide a collection of computer aids on the website for this book. Included are MATHCAD worksheets, VISUAL BASIC programs (as code and stand-alone executable modules), MATLAB programs (as code and essentially stand-alone programs), and older DOS BASIC programs (as code and stand-alone executable programs). These computer aids are described in Appendix B. What I highly recommend, however, is that you use the thermodynamics packages in process simulators such as Aspen. These have the following advantages: they have been developed over many years by experts, so that they are free of the bugs in programs written by professors; they have a nice interface that can be used in solving many different problems; they include the most recent thermodynamic models; and they have large databases with substance-specific parameters for these models. Perhaps most important, these programs are the ones you are most likely to use in later classes in the chemical engineering program and throughout your career. I have also provided several instructional aides to help you in your study of thermodynamics. First, every chapter of this book begins with a list of Instructional Objectives, indicating important items to be learned. I suggest reading these objectives before starting a chapter and then reviewing them while preparing for examinations. Second, important equations are displayed in boxes, and some very important ones within those boxes are indicated by name or description in the margins. Third, at the end of each chapter (and in the case of Chapters 10, 11, and 12, also at the end of each section) you will find many problems to work on to hone your problem-solving skills. Finally, Appendix C provides answers to selected problems. Only the final answers appear, however, not a complete solution containing the steps required to arrive at that answer. Keep in mind that you may be solving a problem correctly but may get a slightly different numerical answer than the one I have provided either because you read a graph of thermodynamic properties slightly differently than I or because you used a correct but different equation of state or activity coefficient model. If your answer and mine differ only slightly, it is likely that both are correct. Good luck in your study of thermodynamics. Stanley I. Sandler Newark, Delaware January 25, 2016

viii Preface

PROPOSED SYLLABI Two semester undergraduate chemical engineering thermodynamics course Cover as much of the book as possible. If necessary, I would omit following material: Sections 2.4 and 3.6 Section 6.9 Section 9.9 Sections 12.3, 4 and 5

Chapter 14 (as this material may be covered in the reaction engineering course) and perhaps Chapter 15 if there is no interest in biochemical engineering

Two quarter undergraduate chemical engineering thermodynamics course I suggest omitting the following chapters and sections: Sections 2.4 and 3.6 Section 5.4 Sections 6.6, 6.9 and 10 Section 7.8 Section 9.9

Sections 12.3, 4 and 5 Chapter 14 (as this material may be covered in the reaction engineering course) and perhaps Chapter 15 if there is no interest in biochemical engineering

One semester undergraduate chemical engineering thermodynamics course following a one semester general or mechanical engineering thermodynamics course I suggest quickly reviewing the notation in Chapters 2, 3 and 4, then start with Chapter 8. With the limited time available, I suggest omitting the following chapters and sections: Section 9.9 Sections 12.3, 4 and 5

Chapter 14 (as this material may be covered in the reaction engineering course) and Chapter 15 if there is no interest in biochemical engineering

One quarter undergraduate chemical engineering thermodynamics course following a general or mechanical engineering thermodynamics course Quickly review the notation in chapters 2, 3 and 4, then go directly to Chapter 8. With the very limited time available, I suggest omitting the following chapters and sections: Section 9.9 Sections 11.3, 4 and 5 Chapter 12

Chapter 14 (as this material may be covered in the reaction engineering course) and Chapter 15 if there is no interest in biochemical engineering

Contents

CHAPTER 1

INTRODUCTION Instructional Objectives for Chapter 1 3 Important Notation Introduced in This Chapter 4 1.1 The Central Problems of Thermodynamics 4 1.2 A System of Units 5 1.3 The Equilibrium State 7 1.4 Pressure, Temperature, and Equilibrium 10 1.5 Heat, Work, and the Conservation of Energy 15 1.6 Specification of the Equilibrium State; Intensive and Extensive Variables; Equations of State 18 1.7 A Summary of Important Experimental Observations 21 1.8 A Comment on the Development of Thermodynamics 23 Problems 23

1

CHAPTER 2

CONSERVATION OF MASS Instructional Objectives for Chapter 2 25 Important Notation Introduced in This Chapter 26 2.1 A General Balance Equation and Conserved Quantities 26 2.2 Conservation of Mass for a Pure Fluid 30 2.3 The Mass Balance Equations for a Multicomponent System with a Chemical Reaction 35 2.4 The Microscopic Mass Balance Equations in Thermodynamics and Fluid Mechanics (Optional - only on the website for this book) 43 Problems 44

25

CHAPTER 3

CONSERVATION OF ENERGY Instructional Objectives for Chapter 3 46 Notation Introduced in This Chapter 46 3.1 Conservation of Energy 47 3.2 Several Examples of Using the Energy Balance 54 3.3 The Thermodynamic Properties of Matter 59 3.4 Applications of the Mass and Energy Balances 69 3.5 Conservation of Momentum 93 3.6 The Microscopic Energy Balance (Optional - only on website for this book) 93 Problems 93

45

ix

x Contents

CHAPTER 4

ENTROPY: AN ADDITIONAL BALANCE EQUATION Instructional Objectives for Chapter 4 99 Notation Introduced in This Chapter 100 4.1 Entropy: A New Concept 100 4.2 The Entropy Balance and Reversibility 108 4.3 Heat, Work, Engines, and Entropy 114 4.4 Entropy Changes of Matter 125 4.5 Applications of the Entropy Balance 128 4.6 Availability and the Maximum Useful Shaft Work that can be obtained In a Change of State 140 4.7 The Microscopic Entropy Balance (Optional - only on website for this book) 145 Problems 145

99

CHAPTER 5

LIQUEFACTION, POWER CYCLES, AND EXPLOSIONS Instructional Objectives for Chapter 5 152 Notation Introduced in this Chapter 152 5.1 Liquefaction 153 5.2 Power Generation and Refrigeration Cycles 158 5.3 Thermodynamic Efficiencies 181 5.4 The Thermodynamics of Mechanical Explosions 185 Problems 194

152

CHAPTER 6

THE THERMODYNAMIC PROPERTIES OF REAL SUBSTANCES Instructional Objectives for Chapter 6 200 Notation Introduced in this Chapter 201 6.1 Some Mathematical Preliminaries 201 6.2 The Evaluation of Thermodynamic Partial Derivatives 205 6.3 The Ideal Gas and Absolute Temperature Scales 219 6.4 The Evaluation of Changes in the Thermodynamic Properties of Real Substances Accompanying a Change of State 220 6.5 An Example Involving the Change of State of a Real Gas 245 6.6 The Principle of Corresponding States 250 6.7 Generalized Equations of State 263 6.8 The Third Law of Thermodynamics 267 6.9 Estimation Methods for Critical and Other Properties 268 6.10 Sonic Velocity 272 6.11 More About Thermodynamic Partial Derivatives (Optional - only on website for this book) 275 Problems 275

200

CHAPTER 7

EQUILIBRIUM AND STABILITY IN ONE-COMPONENT SYSTEMS 285 Instructional Objectives for Chapter 7 285 Notation Introduced in This Chapter 285 7.1 The Criteria for Equilibrium 286 7.2 Stability of Thermodynamic Systems 293 7.3 Phase Equilibria: Application of the Equilibrium and Stability Criteria to the Equation of State 300

Contents xi 7.4 7.5

The Molar Gibbs Energy and Fugacity of a Pure Component 307 The Calculation of Pure Fluid-Phase Equilibrium: The Computation of Vapor Pressure from an Equation of State 322 7.6 Specification of the Equilibrium Thermodynamic State of a System of Several Phases: The Gibbs Phase Rule for a One-Component System 330 7.7 Thermodynamic Properties of Phase Transitions 334 7.8 Thermodynamic Properties of Small Systems, or Why Subcooling and Superheating Occur 341 Problems 344

CHAPTER 8

THE THERMODYNAMICS OF MULTICOMPONENT MIXTURES Instructional Objectives for Chapter 8 353 Notation Introduced in this chapter 353 8.1 The Thermodynamic Description of Mixtures 354 8.2 The Partial Molar Gibbs Energy and the Generalized Gibbs-Duhem Equation 363 8.3 A Notation for Chemical Reactions 367 8.4 The Equations of Change for a Multicomponent System 370 8.5 The Heat of Reaction and a Convention for the Thermodynamic Properties of Reacting Mixtures 378 8.6 The Experimental Determination of the Partial Molar Volume and Enthalpy 385 8.7 Criteria for Phase Equilibrium in Multicomponent Systems 396 8.8 Criteria for Chemical Equilibrium, and Combined Chemical and Phase Equilibrium 399 8.9 Specification of the Equilibrium Thermodynamic State of a Multicomponent, Multiphase System; the Gibbs Phase Rule 404 8.10 A Concluding Remark 408 Problems 408

CHAPTER 9

ESTIMATION OF THE GIBBS ENERGY AND FUGACITY OF A COMPONENT IN A MIXTURE Instructional Objectives for Chapter 9 416 Notation Introduced in this Chapter 417 9.1 The Ideal Gas Mixture 417 9.2 The Partial Molar Gibbs Energy and Fugacity 421 9.3 Ideal Mixture and Excess Mixture Properties 425 9.4 Fugacity of Species in Gaseous, Liquid, and Solid Mixtures 436 9.5 Several Correlative Liquid Mixture Activity Coefficient Models 446 9.6 Two Predictive Activity Coefficient Models 460 9.7 Fugacity of Species in Nonsimple Mixtures 468 9.8 Some Comments on Reference and Standard States 478 9.9 Combined Equation-of-State and Excess Gibbs Energy Model 479 9.10 Electrolyte Solutions 482 9.11 Choosing the Appropriate Thermodynamic Model 490 Appendix A9.1 A Statistical Mechanical Interpretation of the Entropy of Mixing in an Ideal Mixture (Optional – only on the website for this book) 493

353

416

xii Contents Appendix A9.2 Appendix A9.3 Problems

Multicomponent Excess Gibbs Energy (Activity Coefficient) Models 493 The Activity Coefficient of a Solvent in an Electrolyte Solution 495

499

CHAPTER 10

VAPOR-LIQUID EQUILIBRIUM IN MIXTURES Instructional Objectives for Chapter 10 507 Notation Introduced in this Chapter 508 10.0 Introduction to Vapor-Liquid Equilibrium 508 10.1 Vapor-Liquid Equilibrium in Ideal Mixtures 510 Problems for Section 10.1 536 10.2 Low-Pressure Vapor-Liquid Equilibrium in Nonideal Mixtures 538 Problems for Section 10.2 568 10.3 High-Pressure Vapor-Liquid Equilibria Using Equations of State (φ-φ Method) 578 Problems for Section 10.3 595

507

CHAPTER 11

OTHER TYPES OF PHASE EQUILIBRIA IN FLUID MIXTURES Instructional Objectives for Chapter 11 599 Notation Introduced in this Chapter 600 11.1 The Solubility of a Gas in a Liquid 600 Problems for Section 11.1 615 11.2 Liquid-Liquid Equilibrium 617 Problems for Section 11.2 646 11.3 Vapor-Liquid-Liquid Equilibrium 652 Problems for Section 11.3 661 11.4 The Partitioning of a Solute Among Two Coexisting Liquid Phases; The Distribution Coefficient 665 Problems for Section 11.4 675 11.5 Osmotic Equilibrium and Osmotic Pressure 677 Problems for Section 11.5 684

599

CHAPTER 12

MIXTURE PHASE EQUILIBRIA INVOLVING SOLIDS Instructional Objectives for Chapter 12 688 Notation Introduced in this Chapter 688 12.1 The Solubility of a Solid in a Liquid, Gas, or Supercritical Fluid 689 Problems for Section 12.1 699 12.2 Partitioning of a Solid Solute Between Two Liquid Phases 701 Problems for Section 12.2 703 12.3 Freezing-Point Depression of a Solvent Due to the Presence of a Solute; the Freezing Point of Liquid Mixtures 704 Problems for Section 12.3 709 12.4 Phase Behavior of Solid Mixtures 710 Problems for Section 12.4 718 12.5 The Phase Behavior Modeling of Chemicals in the Environment 720 Problems for Section 12.5 726

688

Contents xiii 12.6 Process Design and Product Design 726 Problems for Section 12.6 732 12.7 Concluding Remarks on Phase Equilibria 732

CHAPTER 13

CHEMICAL EQUILIBRIUM 734 Instructional Objectives for Chapter 13 734 Important Notation Introduced in This Chapter 734 13.1 Chemical Equilibrium in a Single-Phase System 735 13.2 Heterogeneous Chemical Reactions 768 13.3 Chemical Equilibrium When Several Reactions Occur in a Single Phase 781 13.4 Combined Chemical and Phase Equilibrium 791 13.5 Ionization and the Acidity of Solutions 799 13.6 Ionization of Biochemicals 817 13.7 Partitioning of Amino Acids and Proteins Between Two Liquids 831 Problems 834

CHAPTER 14

THE BALANCE EQUATIONS FOR CHEMICAL REACTORS, AVAILABILITY, AND ELECTROCHEMISTRY 848 Instructional Objectives for Chapter 14 848 Notation Introduced in this Chapter 849 14.1 The Balance Equations for a Tank-Type Chemical Reactor 849 14.2 The Balance Equations for a Tubular Reactor 857 14.3 Overall Reactor Balance Equations and the Adiabatic Reaction Temperature 860 14.4 Thermodynamics of Chemical Explosions 869 14.5 Maximum Useful Work and Availability in Chemically Reacting Systems 875 14.6 Introduction to Electrochemical Processes 882 14.7 Fuel Cells and Batteries 891 Problems 897

CHAPTER 15

SOME ADDITIONAL BIOCHEMICAL APPLICATIONS OF THERMODYNAMICS Instructional Objectives for Chapter 15 900 Notation Introduced in this Chapter 901 15.1 Solubilities of Weak Acids, Weak Bases, and Amino Acids as a Function of pH 901 15.2 The Solubility of Amino Acids and Proteins as a funciton of Ionic Strength and Temperature 911 15.3 Binding of a Ligand to a Substrate 917 15.4 Some Other Examples of Biochemical Reactions 922 15.5 The Denaturation of Proteins 925 15.6 Coupled Biochemical Reactions: The ATP-ADP Energy Storage and Delivery Mechanism 932 15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 937 15.8 Gibbs-Donnan Equilibrium and Membrane Potentials 960 15.9 Protein Concentration in an Ultracentrifuge 967 Problems 970

900

xiv Contents

APPENDIX A

THERMODYNAMIC DATA Appendix A.I Conversion Factors for SI Units 973 Appendix A.II The Molar Heat Capacities of Gases in the Ideal Gas (Zero Pressure) State 974 Appendix A.III The Thermodynamic Properties of Water and Steam 977 Appendix A.IV Enthalpies and Free Energies of Formation 987 Appendix A.V Heats of Combustion 990

APPENDIX B

BRIEF DESCRIPTIONS OF COMPUTER AIDS FOR USE WITH THIS BOOK

973

992

APPENDIX B on Website only

DESCRIPTIONS OF COMPUTER PROGRAMS AND COMPUTER AIDS FOR USE WITH THIS BOOK Appendix B.I Windows-based Visual Basic Programs B1 Appendix B.II DOS-based Basic Programs B9 Appendix B.III MATHCAD Worksheets B12 Appendix B.IV MATLAB Programs B14

APPENDIX C

ASPEN ILLUSTRATION INPUT FILES. THESE ARE ON THE WEBSITE FOR THIS BOOK

994

ANSWERS TO SELECTED PROBLEMS

995

APPENDIX D INDEX

B1

998

Chapter

1 Introduction A major objective of any field of pure or applied science is to summarize a large amount of experimental information with a few basic principles. The hope is that any new experimental measurement or phenomenon can be easily understood in terms of the established principles, and that predictions based on these principles will be accurate. This book demonstrates how a collection of general experimental observations can be used to establish the principles of an area of science called thermodynamics, and then shows how these principles can be used to study a wide variety of physical, chemical, and biochemical phenomena. Questions the reader of this book might ask include what is thermodynamics and why should one study it? The word thermodynamics consists of two parts: the prefix thermo, referring to heat and temperature, and dynamics, meaning motion. Initially, thermodynamics had to do with the flow of heat to produce mechanical energy that could be used for industrial processes and locomotion. This was the study of heat engines, devices used to operate mechanical equipment, drive trains and cars, and perform many other functions that accelerated progress in the Industrial Age. These started with steam engines and progressed to internal combustion engines, turbines, heat pumps, air conditioners, and other devices. This part of thermodynamics is largely the realm of mechanical engineers. However, because such equipment is also used in chemical processing plants, it is important for chemical engineers to have an understanding of the fundamentals of this equipment. Therefore, such equipment is considered briefly in Chapters 4 and 5 of this book. These applications of thermodynamics generally require an understanding the properties of pure fluids, such as steam and various refrigerants, and gases such as oxygen and nitrogen. More central to chemical engineering is the study of mixtures. The production of chemicals, polymers, pharmaceuticals and other biological materials, and oil and gas processing, all involve chemical or biochemical reactions (frequently in a solvent) that produce a mixture of reaction products. These must be separated from the mixture and purified to result in products of societal, commercial, or medicinal value. It is in these areas that thermodynamics plays a central role in chemical engineering. Separation processes, of which distillation is the most commonly used in the chemical industry, are designed based on information from thermodynamics. Of particular interest in the design of separation and purification processes is the compositions of two phases that are in equilibrium. For example, when a liquid mixture boils, the vapor coming off can be of a quite different composition than the liquid from which it was obtained. This is the basis for distillation, and the design of a distillation column is based on 1

2 Chapter 1: Introduction predictions from thermodynamics. Similarly, when partially miscible components are brought together, two (or more) liquid phases of very different composition will form, and other components added to this two-phase mixture will partition differently between the phases. This phenomenon is the basis for liquid-liquid extraction, another commonly used separation process, especially for chemicals and biochemicals that cannot be distilled because they do not vaporize appreciably or because they break down on heating. The design of such processes is also based on predictions from thermodynamics. Thus, thermodynamics plays a central role in chemical process design. Although this subject is properly considered in other courses in the chemical engineering curriculum, we will provide very brief introductions to distillation, air stripping, liquid-liquid extraction, and other processes so that the student can appreciate why the study of thermodynamics is central to chemical engineering. Other applications of thermodynamics considered in this book include the distribution of chemicals when released to the environment, determining safety by estimating the possible impact (or energy release) of mechanical and chemical explosions, analyzing biochemical processes, and product design, that is, identifying a chemical or mixture that has the properties needed for a specific application. A generally important feature of engineering design is making estimates when specific information on a fluid or fluid mixture is not available, which is almost always the case. To understand why this is so, consider the fact that there are several hundred chemicals commonly used in industry, either as final products or intermediates. If this number were, say, 200, there would be about 20,000 possible binary mixtures, 1.3 million possible ternary mixtures, 67 million possible four-component mixtures, and so on. However, in the history of mankind the vapor-liquid equilibria of considerably fewer than 10,000 different mixtures have been measured. Further, even if we were interested in one of the mixtures for which data exist, it is unlikely that the measurements were done at exactly the temperature and pressure in which we are interested. Therefore, many times engineers have to make estimates by extrapolating the limited data available to the conditions (temperature, pressure, and composition) of interest to them, or predict the behavior of multicomponent mixtures based only on sets of twocomponent mixture data. In other cases predictions may have to be made for mixtures in which the chemical identity of one or more of the components is not known. One example of this is petroleum or crude oil; another is the result of a polymerization reaction or biochemical process. In these cases, many components of different molecular weights are present that will not, and perhaps cannot, be identified by chemical analytic methods, and yet purification methods have to be designed so approximations are made. Although the estimation of thermodynamic properties, especially of mixtures, is not part of the theoretical foundation of chemical engineering thermodynamics, it is necessary for its application to real problems. Therefore, various estimation methods are interspersed with the basic theory, especially in Chapters 6, 8, and 11, so that the theory can be applied. This book can be considered as consisting of two parts. The first is the study of pure fluids, which begins after this introductory chapter. In Chapter 2 is a review of the use of mass balance, largely for pure fluids, but with a digression to reacting mixtures in order to explain the idea of nonconserved variables. Although mass balances should be familiar to a chemical engineering student from a course on stoichiometry or chemical process principles, it is reviewed here to introduce the different forms of the mass balance that will be used, the rate-of-change and difference forms (as well as the microscopic form for the advanced student), and some of the subtleties in applying the mass balance to systems in which flow occurs. The mass balance is the simplest of

Chapter 1: Introduction 3 the balance equations we will use, and it is important to understand its application before proceeding to the use of other balance equations. We then move on to the development of the framework of thermodynamics and its application to power cycles and other processes involving only pure fluids, thereby avoiding the problems of estimating the properties of mixtures. However, in the second part of this book, which begins in Chapter 8 and continues to the end of the book, the thermodynamic theory of mixtures, the properties of mixtures, and many different types of phase equilibria necessary for process design are considered, as are chemical reaction equilibria. It is this part of the book that is the essential background for chemical engineering courses in equipment and process design. We end the book with a chapter on the application of thermodynamics to biological and biochemical processes, though other such examples have been included in several of the preceding chapters. Before proceeding, it is worthwhile to introduce a few of the terms used in applying the balance equations; other, more specific thermodynamic terms and definitions appear elsewhere in this book. Glossary

Adiabatic system: A well-insulated system in which there are no heat flows in or out. Closed system: A system in which there are no mass flows in or out. Isolated system: A system that is closed to the flow of mass and energy in the form of work flows and heat flows (i.e., is adiabatic). Steady-state system: A system in which flows of mass, heat, and work may be present but in such a way that the system properties do not change over time. Cyclic process: A process that follows a periodic path so that the system has the same properties at any point in the cycle as it did at that point in any preceding or succeeding cycle. The chapters in this book are all organized in a similar manner. First, there is a paragraph or two describing the contents of the chapter and where it fits in to the general subject of thermodynamics. This introduction is followed by some specific instructional objectives or desired educational outcomes that the student is expected to develop from the chapter. Next, is a brief list of the new terms or nomenclature introduced within the chapter. After these preliminaries, the real work starts.

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 1 The goals of this chapter are for the student to: • Know the basic terminology of thermodynamics, such as internal energy, potential • • • • • •

energy, and kinetic energy; system, phase, and thermal and mechanical contact; adiabatic and isolated systems; and the difference between a system and a phase Be able to use the SI unit system which is used in this book and throughout the world Understand the concepts of absolute temperature and pressure Understand the difference between heat and work, and between mechanical and thermal energies Understand the general concept of equilibrium, which is very important in the application of thermodynamics in chemical engineering Understand the difference between intensive and extensive variables Understand that total mass and total energy are conserved in any process

4 Chapter 1: Introduction

IMPORTANT NOTATION INTRODUCED IN THIS CHAPTER M N P R T U ˆ U U V Vˆ V

Mass (g) Number of moles (mol) Absolute pressure (kPa or bar) Gas constant (J/mol K) Absolute temperature (K) Internal energy (J) Internal energy per unit mass (J/g) Internal energy per mole (J/mol) Volume (m3 ) Specific volume, volume per unit mass (m3 /g) Volume per mole (m3 /mol)

1.1 THE CENTRAL PROBLEMS OF THERMODYNAMICS Thermodynamics is the study of the changes in the state or condition of a substance when changes in its temperature, state of aggregation, or internal energy are important. By internal energy we mean the energy of a substance associated with the motions, interactions, and bonding of its constituent molecules, as opposed to the external energy associated with the velocity and location of its center of mass, which is of primary interest in mechanics. Thermodynamics is a macroscopic science; it deals with the average changes that occur among large numbers of molecules rather than the detailed changes that occur in a single molecule. Consequently, this book will quantitatively relate the internal energy of a substance not to its molecular motions and interaction, but to other, macroscopic variables such as temperature, which is primarily related to the extent of molecular motions, and density, which is a measure of how closely the molecules are packed and thus largely determines the extent of molecular interactions. The total energy of any substance is the sum of its internal energy and its bulk potential and kinetic energy; that is, it is the sum of the internal and external energies. Our interest in thermodynamics is mainly in changes that occur in some small part of the universe, for example, within a steam engine, a laboratory beaker, or a chemical or biochemical reactor. The region under study, which may be a specified volume in space or a quantity of matter, is called the system; the rest of the universe is its surroundings. Throughout this book the term state refers to the thermodynamic state of a system as characterized by its density, refractive index, composition, pressure, temperature, or other variables to be introduced later. The state of agglomeration of the system (whether it is a gas, liquid, or solid) is called its phase. A system is said to be in contact with its surroundings if a change in the surroundings can produce a change in the system. Thus, a thermodynamic system is in mechanical contact with its surroundings if a change in pressure in the surroundings results in a pressure change in the system. Similarly, a system is in thermal contact with its surroundings if a temperature change in the surroundings can produce a change in the system. If a system does not change as a result of changes in its surroundings, the system is said to be isolated. Systems may be partially isolated from their surroundings. An adiabatic system is one that is thermally isolated from its surroundings; that is, it is a system that is not in thermal contact, but may be in mechanical contact, with its surroundings. If mass can flow into or out of a thermodynamic system, the system is said to be open; if not, the system is closed. Similarly, if heat can be added to the system or work done on it, we say the system is open to heat or work flows, respectively.1 1 Both

heat and work will be defined shortly.

1.2 A System of Units 5 An important concept in thermodynamics is the equilibrium state, which will be discussed in detail in the following sections. Here we merely note that if a system is not subjected to a continual forced flow of mass, heat, or work, the system will eventually evolve to a time-invariant state in which there are no internal or external flows of heat or mass and no change in composition as a result of chemical or biochemical reactions. This state of the system is the equilibrium state. The precise nature of the equilibrium state depends on both the character of the system and the constraints imposed on the system by its immediate surroundings and its container (e.g., a constant-volume container fixes the system volume, and a thermostatic bath fixes the system temperature; see Problem 1.1). Using these definitions, we can identify the two general classes of problems that are of interest in thermodynamics. In the first class are problems concerned with computing the amount of work or the flow of heat either required or released to accomplish a specified change of state in a system or, alternatively, the prediction of the change in thermodynamic state that occurs for given heat or work flows. We refer to these problems as energy flow problems. The second class of thermodynamic problems are those involving equilibrium. Of particular interest here is the identification or prediction of the equilibrium state of a system that initially is not in equilibrium. The most common problem of this type is the prediction of the new equilibrium state of a system that has undergone a change in the constraints that had been maintaining it in a previous state. For example, we will want to predict whether a single liquid mixture or two partially miscible liquid phases will be the equilibrium state when two pure liquids (the initial equilibrium state) are mixed (the change of constraint; see Chapter 11). Similarly, we will be interested in predicting the final temperatures and pressures in two gas cylinders after opening the connecting valve (change of constraint) between a cylinder that was initially filled and another that was empty (see Chapters 3 and 4). It is useful to mention another class of problems related to those referred to in the previous paragraphs, but that is not considered here. We do not try to answer the question of how fast a system will respond to a change in constraints; that is, we do not try to study system dynamics. The answers to such problems, depending on the system and its constraints, may involve chemical kinetics, heat or mass transfer, and fluid mechanics, all of which are studied elsewhere. Thus, in the example above, we are interested in the final state of the gas in each cylinder, but not in computing how long a valve of given size must be held open to allow the necessary amount of gas to pass from one cylinder to the other. Similarly, when, in Chapters 10, 11, and 12, we study phase equilibrium and, in Chapter 13, chemical equilibrium, our interest is in the prediction of the equilibrium state, not in how long it will take to achieve this equilibrium state. Shortly we will start the formal development of the principles of thermodynamics, first qualitatively and then, in the following chapters, in a quantitative manner. First, however, we make a short digression to discuss the system of units used in this text.

1.2 A SYSTEM OF UNITS The study of thermodynamics involves mechanical variables such as force, pressure, and work, and thermal variables such as temperature and energy. Over the years many definitions and units for each of these variables have been proposed; for example, there are several values of the calorie, British thermal unit, and horsepower. Also, whole

6 Chapter 1: Introduction Table 1.2-1 The SI Unit System Unit

Name

Abbreviation

Basis of Definition

Length

meter

m

Mass

kilogram

kg

Time

second

s

Electric current

ampere

A

Temperature

kelvin

K

Amount of substance

mole

mol

Luminous intensity

candela

cd

The distance light travels in a vacuum in 1/299 792 458 second Platinum-iridium prototype at the International Bureau of Weights and Measures, S`evres, France Proportional to the period of one cesium-133 radiative transition Current that would produce a specified force between two parallel conductors in a specified geometry 1/273.16 of the thermodynamic temperature (to be defined shortly) of water at its triple point (see Chapter 7) Amount of a substance that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12 (6.022 × 1023 , which is Avogadro’s number) Related to the black-body radiation from freezing platinum (2045 K)

systems of units, such as the English and cgs systems, have been used. The problem of standardizing units was studied, and the Syst`eme International d’Unit´es (abbreviated SI units) was agreed on at the Eleventh General Conference on Weights and Measures in 1960. This conference was one of a series convened periodically to obtain international agreement on questions of metrology, so important in international trade. The SI unit system is used throughout this book, with some lapses to the use of common units such as volume in liters and frequently pressure in bar. In the SI system the seven basic units listed in Table 1.2-1 are identified and their values are assigned. From these seven basic well-defined units, the units of other quantities can be derived. Also, certain quantities appear so frequently that they have been given special names and symbols in the SI system. Those of interest here are listed in Table 1.2-2. Some other derived units acceptable in the SI system are given in Table 1.2-3, and Table 1.2-4 lists the acceptable scaling prefixes. [It should be pointed Table 1.2-2 Derived Units with Special Names and Symbols Acceptable in SI Units Expression in Quantity Force Energy, work, or quantity of heat Pressure or stress Power Frequency

Name newton joule pascal watt hertz

Symbol N J Pa W Hz

SI Units −2

m kg s m2 kg s−2 m−1 kg s−2 m2 kg s−3 s−1

Derived Units J m−1 Nm N/m2 J/s

1.3 The Equilibrium State 7 Table 1.2-3 Other Derived Units in Terms of Acceptable SI Units Quantity

Expression in SI Units

Symbol

Concentration of substance Mass density (ρ = m/V ) Heat capacity or entropy ˙ Heat flow rate (Q) Molar energy Specific energy Specific heat capacity or specific entropy Specific volume Viscosity (absolute or dynamic) Volume Work, energy (W )

mol m−3 kg m−3 m2 kg s−1 K−1 m2 kg s−3 m2 kg s−2 mol−1 m2 s − 2 m2 s−2 K−1 m3 kg−1 m−1 kg s−1 m3 m2 kg s−2

mol/m3 kg/m3 J/K W or J/s J/mol J/kg J/(kg K) m3 /kg Pa s m3 J or N m

Table 1.2-4 Prefixes for SI Units Multiplication Factor 12

10 109 106 103 102 10 10−1 10−2 10−3 10−6 10−9 10−12 10−15

Prefix

Symbol

tera giga mega kilo hecto deka deci centi milli micro nano pico femto

T G M k (e.g., kilogram) h da d c (e.g., centimeter) m μ

n p f

out that, except at the end of a sentence, a period is never used after the symbol for an SI unit, and the degree symbol is not used. Also, capital letters are not used in units that are written out (e.g. pascals, joules, or meters) except at the beginning of a sentence. When the units are expressed in symbols, the first letter is capitalized only when the unit name is that of a person (e.g. Pa and J, but m).] Appendix A.I presents approximate factors to convert from various common units to acceptable SI units. In the SI unit system, energy is expressed in joules, J, with 1 joule being the energy required to move an object 1 meter when it is opposed by a force of 1 newton. Thus, 1 J = 1 N m = 1 kg m2 s−2 . A pulse of the human heart, or lifting this book 0.1 meters, requires approximately 1 joule. Since this is such a small unit of energy, kilojoules (kJ = 1000 J) are frequently used. Similarly, we frequently use bar = 105 Pa = 0.987 atm as the unit of pressure.

1.3 THE EQUILIBRIUM STATE As indicated in Section 1.1, the equilibrium state plays a central role in thermodynamics. The general characteristics of the equilibrium state are that (1) it does not vary with time; (2) the system is uniform (there are no internal temperature, pressure, velocity,

8 Chapter 1: Introduction or concentration gradients) or is composed of subsystems each of which is uniform; (3) all flows of heat, mass, or work between the system and its surroundings are zero; and (4) the net rate of all chemical reactions is zero. At first it might appear that the characteristics of the equilibrium state are so restrictive that such states rarely occur. In fact, the opposite is true. The equilibrium state will always occur, given sufficient time, as the terminal state of a system closed to the flow of mass, heat, or work across its boundaries. In addition, systems open to such flows, depending on the nature of the interaction between the system and its surroundings, may also evolve to an equilibrium state. If the surroundings merely impose a value of temperature, pressure, or volume on the system, the system will evolve to an equilibrium state. If, on the other hand, the surroundings impose a mass flow into and out of the system (as a result of a pumping mechanism) or a heat flow (as would occur if one part of the system were exposed to one temperature and another part of the system to a different temperature), the system may evolve to a time-invariant state only if the flows are steady. The time-invariant states of these driven systems are not equilibrium states in that the systems may or may not be uniform (this will become clear when the continuous-flow stirred tank and plug-flow chemical reactors are considered in Chapter 14) and certainly do not satisfy part or all of criterion (3). Such time-invariant states are called steady states and occur frequently in continuous chemical and physical processing. Steady-state processes are of only minor interest in this book. Nondriven systems reach equilibrium because all spontaneous flows that occur in nature tend to dissipate the driving forces that cause them. Thus, the flow of heat that arises in response to a temperature difference occurs in the direction that dissipates the temperature difference, the mass diffusion flux that arises in response to a concentration gradient occurs in such a way that a state of uniform concentration develops, and the flux of momentum that occurs when a velocity gradient is present in a fluid tends to dissipate that gradient. Similarly, chemical reactions occur in a direction that drives the system toward equilibrium (Chapter 13). At various points throughout this book it will be useful to distinguish between the flows that arise naturally and drive the system to equilibrium, which we will call natural flows, and flows imposed on the system by its surroundings, which we term forced flows. An important experimental observation in thermodynamics is that any system free from forced flows will, given sufficient time, evolve to an equilibrium state. This empirical fact is used repeatedly in our discussion. It is useful to distinguish between two types of equilibrium states according to their response to small disturbances. To be specific, suppose a system in equilibrium is subjected to a small disturbance that is then removed (e.g., temperature fluctuation or pressure pulse). If the system returns to its initial equilibrium state, this state of the system is said to be stable with respect to small disturbances. If, however, the system does not return to the initial state, that state is said to have been unstable. There is a simple mechanical analogy, shown in Fig. 1.3-1, that can be used to illustrate the concept of stability. Figure 1.3-1a, b, and c represent equilibrium positions of

(a)

(b)

(c)

Figure 1.3-1 Blocks in states (a) and (b) are stable to small mechanical disturbances; the delicately balanced block in state (c) is not.

1.3 The Equilibrium State 9 a block on a horizontal surface. The configuration in Fig. 1.3-1c is, however, precarious; an infinitesimal movement of the block in any direction (so that its center of gravity is not directly over the pivotal point) would cause the block to revert to the configuration of either Fig. 1.3-1a or b. Thus, Fig. l.3-1c represents an unstable equilibrium position. The configurations of Figs. 1.3-1a and b are not affected by small disturbances, and these states are stable. Intuition suggests that the configuration of Fig. 1.3-1a is the most stable; clearly, it has the lowest center of gravity and hence the lowest potential energy. To go from the configuration of Fig. 1.3-1b to that of Fig. 1.3-1a, the block must pass through the still higher potential energy state indicated in Fig. 1.3-1c. If we use Δε to represent the potential energy difference between the configurations of Figs. 1.3-1b and c, we can say that the equilibrium state of Fig. 1.3-1b is stable to energy disturbances less than Δε in magnitude and is unstable to larger disturbances. Certain equilibrium states of thermodynamic systems are stable to small fluctuations; others are not. For example, the equilibrium state of a simple gas is stable to all fluctuations, as are most of the equilibrium states we will be concerned with. It is possible, however, to carefully prepare a subcooled liquid, that is, a liquid below its normal solidification temperature, that satisfies the equilibrium criteria. This is an unstable equilibrium state because the slightest disturbance, such as tapping on the side of the containing vessel, will cause the liquid to freeze. One sometimes encounters mixtures that, by the chemical reaction equilibrium criterion (see Chapter 13), should react; however, the chemical reaction rate is so small as to be immeasurable at the temperature of interest. Such a mixture can achieve a state of thermal equilibrium that is stable with respect to small fluctuations of temperature and pressure. If, however, there is a sufficiently large, but temporary, increase in temperature (so that the rate of the chemical reaction is appreciable for some period of time) and then the system is quickly cooled, a new thermal equilibrium state with a chemical composition that differs from the initial state will be obtained. The initial equilibrium state, like the mechanical state in Fig. 1.3-1b, is then said to be stable with respect to small disturbances, but not to large disturbances. Unstable equilibrium states are rarely encountered in nature unless they have been specially prepared (e.g. the subcooled liquid mentioned earlier). The reason for this is that during the approach to equilibrium, temperature gradients, density gradients, or other nonuniformities that exist within a system are of a sufficient magnitude to act as disturbances to unstable states and prevent their natural occurrence. In fact, the natural occurrence of an unstable thermodynamic equilibrium state is about as likely as the natural occurrence of the unstable mechanical equilibrium state of Fig. 1.3-1c. Consequently, our concern in this book is mainly with stable equilibrium states. If an equilibrium state is stable with respect to all disturbances, the properties of this state cannot depend on the past history of the system or, to be more specific, on the path followed during the approach to equilibrium. Similarly, if an equilibrium state is stable with respect to small disturbances, its properties do not depend on the path followed in the immediate vicinity of the equilibrium state. We can establish the validity of the latter statement by the following thought experiment (the validity of the first statement follows from a simple generalization of the argument). Suppose a system in a stable equilibrium state is subjected to a small temporary disturbance of a completely arbitrary nature. Since the initial state was one of stable equilibrium, the system will return to precisely that state after the removal of the disturbance. However, since any type of small disturbance is permitted, the return to the equilibrium state may be along a path that is different from the path followed in initially achieving the stable

10 Chapter 1: Introduction equilibrium state. The fact that the system is in exactly the same state as before means that all the properties of the system that characterize the equilibrium state must have their previous values; the fact that different paths were followed in obtaining this equilibrium state implies that none of these properties can depend on the path followed. Another important experimental observation for the development of thermodynamics is that a system in a stable equilibrium state will never spontaneously evolve to a state of nonequilibrium. For example, any temperature gradients in a thermally conducting material free from a forced flow of heat will eventually dissipate so that a state of uniform temperature is achieved. Once this equilibrium state has been achieved, a measurable temperature gradient will never spontaneously occur in the material. The two observations that (1) a system free from forced flows will evolve to an equilibrium state and (2) once in equilibrium a system will never spontaneously evolve to a nonequilibrium state, are evidence for a unidirectional character of natural processes. Thus we can take as a general principle that the direction of natural processes is such that systems evolve toward an equilibrium state, not away from it.

1.4 PRESSURE, TEMPERATURE, AND EQUILIBRIUM Most people have at least a primitive understanding of the notions of temperature, pressure, heat, and work, and we have, perhaps unfairly, relied on this understanding in previous sections. Since these concepts are important for the development of thermodynamics, each will be discussed in somewhat more detail here and in the following sections. The concept of pressure as the total force exerted on an element of surface divided by the surface area should be familiar from courses in physics and chemistry. Pressure— or, equivalently, force—is important in both mechanics and thermodynamics because it is closely related to the concept of mechanical equilibrium. This is simply illustrated by considering the two piston-and-cylinder devices shown in Figs. 1.4-1a and b. In each case we assume that the piston and cylinder have been carefully machined so that there is no friction between them. From elementary physics we know that for the systems in these figures to be in mechanical equilibrium (as recognized by the absence

A

B

(a)

B

A

(b)

Figure 1.4-1 The piston separating gases A and B and the cylinder containing them have been carefully machined so that the piston moves freely in the cylinder.

1.4 Pressure, Temperature, and Equilibrium 11 of movement of the piston), there must be no unbalanced forces; the pressure of gas A must equal that of gas B in the system of Fig. 1.4-1a, and in the system of Fig. 1.4-1b it must be equal to the sum of the pressure of gas B and the force of gravity on the piston divided by its surface area. Thus, the requirement that a state of mechanical equilibrium exists is really a restriction on the pressure of the system. Since pressure is a force per unit area, the direction of the pressure scale is evident; the greater the force per unit area, the greater the pressure. To measure pressure, one uses a pressure gauge. A pressure gauge is a device that produces a change in some indicator, such as the position of a pointer, the height of a column of liquid, or the electrical properties of a specially designed circuit, in response to a change in pressure. Pressure gauges are calibrated using devices such as that shown in Fig. 1.4-2. There, known pressures are created by placing weights on a frictionless piston of known weight. The pressure at the gauge Pg due to the metal weight and the piston is

M w + Mp (1.4-1) g A Here g is the local acceleration of gravity on an element of mass; the standard value is 9.80665 m/s2 . The position of the indicator at several known pressures is recorded, and the scale of the pressure gauge is completed by interpolation. There is, however, a complication with this calibration procedure. It arises because the weight of the air of the earth’s atmosphere produces an average pressure of 14.696 lbs force per sq in, or 101.3 kPa, at sea level. Since atmospheric pressure acts equally in all directions, we are not usually aware of its presence, so that in most nonscientific uses of pressure the zero of the pressure scale is the sea-level atmospheric pressure (i.e., the pressure of the atmosphere is neglected in the pressure gauge calibration). Thus, when the recommended inflation pressure of an automobile tire is 200 kPa, what is really meant is 200 kPa above atmospheric pressure. We refer to pressures on such a scale as gauge pressures. Note that gauge pressures may be negative (in partially or completely evacuated systems), zero, or positive, and errors in pressure measurement result from changes in atmospheric pressure from the Pg =

Weight (mass Mw) Piston (mass Mp)

Gauge being tested or calibrated Oil reservoir Valve Piston area A

Figure 1.4-2 A simple deadweight pressure tester. (The purpose of the oil reservoir and the system volume adjustment is to maintain equal heights of the oil column in the cylinder and gauge sections, so that no corrections for the height of the liquid column need be made in the pressure calibration.)

12 Chapter 1: Introduction gauge calibration conditions (e.g. using a gauge calibrated in New York for pressure measurements in Denver). We define the total pressure P to be equal to the sum of the gauge pressure Pg and the ambient atmospheric pressure Patm . By accounting for the atmospheric pressure in this way we have developed an absolute pressure scale, which is a pressure scale with zero as the lowest pressure attainable (the pressure in a completely evacuated region of space). One advantage of such a scale is its simplicity; the pressure is always a positive quantity, and measurements do not have to be corrected for either fluctuations in atmospheric pressure or its change with height above sea level. We will frequently be concerned with interrelationships between the temperature, pressure, and specific volume of fluids. These interrelationships are simplest if the absolute pressure is used. Consequently, unless otherwise indicated, the term pressure in this book refers to absolute pressure. Although pressure arises naturally from mechanics, the concept of temperature is more abstract. To the nonscientist, temperature is a measure of hotness or coldness and as such is not carefully defined, but rather is a quantity related to such things as physical comfort, cooking conditions, or the level of mercury or colored alcohol in a thermometer. To the scientist, temperature is a precisely defined quantity, deeply rooted in the concept of equilibrium and related to the energy content of a substance. The origin of the formal definition of temperature lies in the concept of thermal equilibrium. Consider a thermodynamic system composed of two subsystems that are in thermal contact but that do not interchange mass (e.g. the two subsystems may be two solids in contact, or liquids or gases separated by a thin, impenetrable barrier or membrane) and are isolated from their surroundings. When this composite system achieves a state of equilibrium (detected by observing that the properties of each system are time invariant), it is found that the property measured by the height of fluid in a given thermometer is the same in each system, although the other properties of the subsystems, such as their density and chemical composition, may be different. In accord with this observation, temperature is defined to be that system property which, if it has the same value for any two systems, indicates that these systems are in thermal equilibrium if they are in contact, or would be in thermal equilibrium if they were placed in thermal contact. Although this definition provides the link between temperature and thermal equilibrium, it does not suggest a scale for temperature. If temperature is used only as an indicator of thermal equilibrium, any quantification or scale of temperature is satisfactory provided that it is generally understood and reproducible, though the accepted convention is that increasing hotness of a substance should correspond to increasing values of temperature. An important consideration in developing a thermodynamic scale of temperature is that it, like all other aspects of thermodynamics, should be general and not depend on the properties of any one fluid (such as the specific volume of liquid mercury). Experimental evidence indicates that it should be possible to formulate a completely universal temperature scale. The first indication came from the study of gases at densities so low that intermolecular interactions are unimportant (such gases are called ideal gases), where it was found that the product of the absolute pressure P and the molar volume V of any low-density gas away from its condensation line (see Chapter 7) increases with increasing hotness. This observation has been used as the basis for a temperature scale by defining the temperature T to be linearly proportional to the product of P V for a particular low-density gas, that is, by choosing T so that

P V = A + RT

(1.4-2)

1.4 Pressure, Temperature, and Equilibrium 13 where A and R are constants. In fact, without any loss of generality one can define a new temperature T = T + (A/R), which differs from the choice of Eq. 1.4-2 only by an additive constant, to obtain

P V = RT

(1.4-3)

Since neither the absolute pressure nor the molar volume of a gas can ever be negative, the temperature defined in this way must always be positive, and therefore the ideal gas temperature scale of Eq. 1.4-3 is an absolute scale (i.e., T ≥ 0). To complete this low-density gas temperature scale, it remains to specify the constant R, or equivalently the size of a unit of temperature. This can be done in two equivalent ways. The first is to specify the value of T for a given value of P V and thus determine the constant R; the second is to choose two reproducible points on a hotness scale and to decide arbitrarily how many units of T correspond to the difference in the P V products at these two fixed points. In fact, it is the latter procedure that is used; the ice point temperature of water2 and the boiling temperature of water at standard atmospheric pressure (101.3 kPa) provide the two reproducible fixed-point temperatures. What is done, then, is to allow a low-density gas to achieve thermal equilibrium with water at its ice point and measure the product P V , and then repeat the process at the boiling temperature. One then decides how many units of temperature correspond to this measured difference in the product P V ; the choice of 100 units or degrees leads to the Kelvin temperature scale, whereas the use of 180 degrees leads to the Rankine scale. With either of these choices, the constant R can be evaluated for a given lowdensity gas. The important fact for the formulation of a universal temperature scale is that the constant R and hence the temperature scales determined in this way are the same for all low-density gases! Values of the gas constant R in SI units are given in Table 1.4-1. For the present we assume this low-density or ideal gas Kelvin (denoted by K) temperature scale is equivalent to an absolute universal thermodynamic temperature scale; this is proven in Chapter 6. More common than the Kelvin temperature scale for nonscientific uses of temperature are the closely related Fahrenheit and Celsius scales. The size of the degree is the same in both the Celsius (◦ C) and Kelvin temperature scales. However, the zero point of the Celsius scale is arbitrarily chosen to be the ice point temperature of water. Consequently, it is found that

T (K) = T (◦ C) + 273.15

(1.4-4a)

Table 1.4-1 The Gas Constant R = 8.314 J/mol K = 8.314 N m/mol K = 8.314 × 10−3 kPa m3 /mol K = 8.314 × 10−5 bar m3 /mol K = 8.314 × 10−2 bar m3 /kmol K = 8.314 × 10−6 MPa m3 /mol K

2 The freezing temperature of water saturated with air at 101.3 kPa. On this scale the triple point of water is 0.01◦ C.

14 Chapter 1: Introduction In the Fahrenheit (◦ F) temperature scale the ice point and boiling point of water (at 101.3 kPa) are 32◦ F and 212◦ F, respectively. Thus

T (K) =

T (◦ F) + 459.67 1.8

(1.4-4b)

Since we are assuming, for the present, that only the ideal gas Kelvin temperature scale has a firm thermodynamic basis, we will use it, rather than the Fahrenheit and Celsius scales, in all thermodynamic calculations.3 (Another justification for the use of an absolute-temperature scale is that the interrelation between pressure, volume, and temperature for fluids is simplest when absolute temperature is used.) Consequently, if the data for a thermodynamic calculation are not given in terms of absolute temperature, it will generally be necessary to convert these data to absolute temperatures using Eqs. 1.4-4. The product of P V for a low-density gas is said to be a thermometric property in that to each value of P V there corresponds only a single value of temperature. The ideal gas thermometer is not convenient to use, however, because of both its mechanical construction (see Fig. 1.4-3) and the manipulation required to make a measurement. Therefore, common thermometers make use of thermometric properties of other materials—for example, the single-valued relation between temperature and the specific volume of liquid mercury (Problem 1.2) or the electrical resistance of platinum wire. There are two steps in the construction of thermometers based on these other thermometric properties: first, fabrication of the device, such as sealing liquid mercury in an Open to the atmosphere

h Indicial point Gas bulb

Flexible hose

Figure 1.4-3 A simplified diagram of a constant-volume ideal gas thermometer. In this thermometer the product P V for a gas at various temperatures is found by measuring the pressure P at constant volume. For each measurement the mercury reservoir is raised or lowered until the mercury column at the left touches an index mark. The pressure of the gas in the bulb is then equal to the atmospheric pressure plus the pressure due to the height of the mercury column.

3 Of course, for calculations involving only temperature differences, any convenient temperature scale may be used,

since a temperature difference is independent of the zero of the scale.

1.5 Heat, Work, and the Conservation of Energy 15 otherwise evacuated tube; and second, the calibration of the thermometric indicator with a known temperature scale. To calibrate a thermometer, its readings (e.g., the height of a mercury column) are determined at a collection of known temperatures, and its scale is completed by interpolating between these fixed points. The calibration procedure for a common mercury thermometer is usually far simpler. The height of the mercury column is determined at only two fixed points (e.g., the temperature of an ice-water bath and the temperature of boiling water at atmospheric pressure), and the distance between these two heights is divided into equal units; the number of units depends on whether the Rankine or Kelvin degree is used, and whether a unit is to represent a fraction of a degree, a degree, or several degrees. Since only two fixed points are used in the calibration, intermediate temperatures recorded on such a thermometer may be different from those that would be obtained using an ideal gas thermometer because (1) the specific volume of liquid mercury has a slightly nonlinear dependence on temperature, and (2) the diameter of the capillary tube may not be completely uniform (so that the volume of mercury will not be simply related to its height in the tube; see Problem 1.2).

1.5 HEAT, WORK, AND THE CONSERVATION OF ENERGY As we have already indicated, two systems in thermal contact but otherwise isolated from their surroundings will eventually reach an equilibrium state in which the systems have the same temperature. During the approach to this equilibrium state the temperature of the initially low-temperature system increases while the temperature of the initially high-temperature system decreases. We know that the temperature of a substance is directly related to its internal energy, especially the energy of molecular motion. Thus, in the approach to equilibrium, energy has been transferred from the high-temperature system to the one of lower temperature. This transfer of energy as a result of only a temperature difference is called a flow of heat. It is also possible to increase the total energy (internal, potential, and kinetic) of a system by mechanical processes involving motion. In particular, the kinetic or potential energy of a system can change as a result of motion without deformation of the system boundaries, as in the movement of a solid body acted on by an external force, whereas the internal energy and temperature of a system may change when external forces result in the deformation of the system boundaries, as in the compression of a gas. Energy transfer by mechanical motion also occurs as a result of the motion of a drive shaft, push rod, or similar device across the system boundaries. For example, mechanical stirring of a fluid first results in fluid motion (evidence of an increase in fluid kinetic energy) and then, as this motion is damped by the action of the fluid viscosity, in an increase in the temperature (and internal energy) of the fluid. Energy transfer by any mechanism that involves mechanical motion of, or across, the system boundaries is called work. Finally, it is possible to increase the energy of a system by supplying it with electrical energy in the form of an electrical current driven by a potential difference. This electrical energy can be converted to mechanical energy if the system contains an electric motor, it can increase the temperature of the system if it is dissipated through a resistor (resistive heating), or it can be used to cause an electrochemical change in the system (e.g. recharging a lead storage battery). Throughout this book we consider the flow of electrical energy to be a form of work. The reason for this choice will become clear shortly.

16 Chapter 1: Introduction The amount of mechanical work is, from mechanics, equal to the product of the force exerted times the distance moved in the direction of the applied force, or, alternatively, to the product of the applied pressure and the displaced volume. Similarly, the electrical work is equal to the product of the current flow through the system, the potential difference across the system, and the time interval over which the current flow takes place. Therefore, the total amount of work supplied to a system is frequently easy to calculate. An important experimental observation, initially made by James Prescott Joule between 1837 and 1847, is that a specified amount of energy can always be used in such a way as to produce the same temperature rise in a given mass of water, regardless of the precise mechanism or device used to supply the energy, and regardless of whether this energy is in the form of mechanical work, electrical work, or heat. Rather than describe Joule’s experiments, consider how this hypothesis could be proved in the laboratory. Suppose a sample of water at temperature T1 is placed in a well-insulated container (e.g. a Dewar flask) and, by the series of experiments in Table 1.5-1, the amount of energy expended in producing a final equilibrium temperature T2 is measured. Based on the experiments of Joule and others, we would expect to find that this Table 1.5-1 Experiments Designed to Prove the Energy Equivalence of Heat and Work Form in Which Energy Is Transferred to Water

Mechanism Used

Form of Energy Supplied to Mechanism

Method of Measuring Energy Input

Corrections That Must Be Made to Energy Input Data Electrical energy loss in motor and circuit, temperature rise of paddlewheel Temperature rise of paddlewheel

(1) Mechanical energy

Stirring: Paddlewheel driven by electric motor

Electrical energy

Product of voltage, current, and time

(2) Mechanical energy

Stirring: Paddlewheel driven by pulley and falling weight

Mechanical energy

(3) Heat flow

Electrical energy converted to heat in a resistor

Electrical energy

Change in potential energy of weight: product of mass of weight, change in height, and the gravitational constant g Product of voltage, current, and time

(4) Heat flow

Mechanical energy of falling weight is converted to heat through friction of rubbing two surfaces together, as with a brake on the axle of a pulley

Mechanical energy

Change in potential energy of weight: product of mass of weight, change in height, and g

Temperature rise of resistor and electrical losses in circuit Temperature rise of mechanical brakes, etc.

1.5 Heat, Work, and the Conservation of Energy 17 energy (determined by correcting the measurements of column 4 for the temperature rise of the container and the effects of column 5) is precisely the same in all cases. By comparing the first two experiments with the third and fourth, we conclude that there is an equivalence between mechanical energy (or work) and heat, in that precisely the same amount of energy was required to produce a given temperature rise, independent of whether this energy was delivered as heat or work. Furthermore, since both mechanical and electrical energy sources have been used (see column 3), there is a similar equivalence between mechanical and electrical energy, and hence among all three energy forms. This conclusion is not specific to the experiments in Table 1.5-1 but is, in fact, a special case of a more general experimental observation; that is, any change of state in a system that occurs solely as a result of the addition of heat can also be produced by adding the same amount of energy as work, electrical energy, or a combination of heat, work, and electrical energy. Returning to the experiments of Table 1.5-1, we can now ask what has happened to the energy that was supplied to the water. The answer, of course, is that at the end of the experiment the temperature, and hence the molecular energy, of the water has increased. Consequently, the energy added to the water is now present as increased internal energy. It is possible to extract this increased internal energy by processes that return the water to its original temperature. One could, for example, use the warm water to heat a metal bar. The important experimental observation here is that if you measured the temperature rise in the metal, which occurred in returning the water to its initial state, and compared it with the electrical or mechanical energy required to cause the same temperature rise in the metal, you would find that all the energy added to the water in raising its temperature could be recovered as heat by returning the water to its initial state. Thus total energy has been conserved in the process. The observation that energy has been conserved in this experiment is only one example of a general energy conservation principle that is based on a much wider range of experiments. The more general principle is that in any change of state, the total energy, which is the sum of the internal, kinetic, and potential energies of the system, heat, and electrical and mechanical work, is conserved. A more succinct statement is that energy is neither created nor destroyed, but may change in form. Although heat, mechanical work, and electrical work are equivalent in that a given energy input, in any form, can be made to produce the same internal energy increase in a system, there is an equally important difference among the various energy forms. To see this, suppose that the internal energy of some system (perhaps the water in the experiments just considered) has been increased by increasing its temperature from T1 to a higher temperature T2 , and we now wish to recover the added energy by returning the system to its initial state at temperature T1 . It is clear that we can recover the added energy completely as a heat flow merely by putting the system in contact with another system at a lower temperature. There is, however, no process or device by which it is possible to convert all the added internal energy of the system to mechanical energy and restore both the system and the surroundings to their initial states, even though the increased internal energy may have resulted from adding only mechanical energy to the system. In general, only a portion of the increased internal energy can be recovered as mechanical energy, the remainder appearing as heat. This situation is not specific to the experiments discussed here; it occurs in all similar efforts to convert both heat and internal energy to work or mechanical energy. We use the term thermal energy to designate energy in the form of internal energy and heat, and mechanical energy to designate mechanical and electrical work and

18 Chapter 1: Introduction the external energy of a system. This distinction is based on the general experimental observation that while, in principle, any form of mechanical energy can be completely converted to other forms of mechanical energy or thermal energy, only a fraction of the thermal energy can be converted to mechanical energy in any cyclic process (a process that at the end of the cycle restores the system and surroundings to their states at the beginning of the cycle), or any process in which the only change in the universe (system and surroundings) is the conversion of thermal energy to mechanical energy. The units of mechanical work arise naturally from its definition as the product of a force and a distance. Typical units of mechanical work are foot–pound force, dynecentimeter, and erg, though we will use the newton-meter (N m), which is equal to one joule; and from the formulation of work as pressure times displaced volume, pascalmeter3 , which is also equal to one joule. The unit of electrical work is the volt-amperesecond or, equivalently, the watt-second (again equal to one joule). Heat, however, not having a mechanical definition, has traditionally been defined experimentally. Thus, the heat unit calorie was defined as the amount of heat required to raise the temperature of 1 gram of water from 14.5◦ C to 15.5◦ C, and the British thermal unit (BTU) was defined to be the amount of heat required to raise 1 lb of water from 59◦ F to 60◦ F. These experimental definitions of heat units have proved unsatisfactory because the amount of energy in both the calorie and BTU have been subject to continual change as measurement techniques improved. Consequently, there are several different definitions of the Btu and calorie (e.g. the thermochemical calorie, the mean calorie, and the International Table calorie) that differ by less than one and one-half parts in a thousand. Current practice is to recognize the energy equivalence of heat and work and to use a common energy unit for both. We will use only the joule, which is equal to 0.2390 calorie (thermochemical) or 0.9485 × 10−3 Btu (thermochemical).

1.6 SPECIFICATION OF THE EQUILIBRIUM STATE; INTENSIVE AND EXTENSIVE VARIABLES; EQUATIONS OF STATE Since our main interest throughout this book is with stable equilibrium states, it is important to consider how to characterize the equilibrium state and, especially, what is the minimum number of properties of a system at equilibrium that must be specified to fix the values of all its remaining properties completely.4 To be specific, suppose we had 1 kg of a pure gas, say oxygen, at equilibrium whose temperature is some value T , pressure some value P , volume V , refractive index R, electrical permitivity ε, and so on, and we wanted to adjust some of the equilibrium properties of a second sample of oxygen so that all the properties of the two samples would be identical. The questions we are asking, then, are what sorts of properties, and how many properties, must correspond if all of the properties of the two systems are to be identical? The fact that we are interested only in stable equilibrium states is sufficient to decide the types of properties needed to specify the equilibrium state. First, since gradients in velocity, pressure, and temperature cannot be present in the equilibrium state, they do not enter into its characterization. Next, since, as we saw in Sec. 1.3, the properties of a stable equilibrium state do not depend on the history of the system or its approach to equilibrium, the stable equilibrium state is characterized only by equilibrium properties of the system. 4 Throughout

this book, we implicitly assume that a system contains a large number of molecules (at least several tens of thousands), so that the surface effects present in small systems are unimportant. See, however, Sec. 7.8.

1.6 Specification of the Equilibrium State; Intensive and Extensive Variables; Equations of State 19 The remaining question, that of how many equilibrium properties are necessary to specify the equilibrium state of the system, can be answered only by experiment. The important experimental observation here is that an equilibrium state of a single-phase, one-component system in the absence of external electric and magnetic fields is completely specified if its mass and two other thermodynamic properties are given. Thus, going back to our example, if the second oxygen sample also weighs 1 kg, and if it were made to have the same temperature and pressure as the first sample, it would also be found to have the same volume, refractive index, and so forth. If, however, only the temperature of the second 1-kg sample was set equal to that of the first sample, neither its pressure nor any other physical property would necessarily be the same as that of the first sample. Consequently, the values of the density, refractive index, and, more generally, all thermodynamic properties of an equilibrium single-component, singlephase fluid are completely fixed once the mass of the system and the values of at least two other system parameters are given. (The specification of the equilibrium state of multiphase and multicomponent systems is considered in Chapters 7 and 8.) The specification of an equilibrium system can be made slightly simpler by recognizing that the variables used in thermodynamic descriptions are of two different types. To see this, consider a gas of mass M that is at a temperature T and pressure P and is confined to a glass bulb of volume V . Suppose that an identical glass bulb is also filled with mass M of the same gas and heated to the same temperature T . Based on the previous discussion, since the values of T , V , and M are the same, the pressure in the second glass bulb is also P . If these two bulbs are now connected to form a new system, the temperature and pressure of this composite system are unchanged from those of the separated systems, although the volume and mass of this new system are clearly twice those of the original single glass bulb. The pressure and temperature, because of their size-independent property, are called intensive variables, whereas the mass, volume, and total energy are extensive variables, or variables dependent on the size or amount of the system. Extensive variables can be transformed into intensive variables by dividing by the total mass or total number of moles so that a specific volume (volume per unit mass or volume per mole), a specific energy (energy per unit mass or per mole), and so forth are obtained. By definition, the term state variable refers to any of the intensive variables of an equilibrium system: temperature, pressure, specific volume, specific internal energy, refractive index, and other variables introduced in the following chapters. Clearly, from the previous discussion, the value of any state variable depends only on the equilibrium state of the system, not on the path by which the equilibrium state was reached. With the distinction now made between intensive and extensive variables, it is possible to rephrase the requirement for the complete specification of a thermodynamic state in a more coherent manner. The experimental observation is that the specification of two state variables uniquely determines the values of all other state variables of an equilibrium, single-component, single-phase system. [Remember, however, that to determine the size of the system, that is, its mass or total volume, one must also specify the mass of the system, or the value of one other extensive parameter (total volume, total energy, etc.).] The implication of this statement is that for each substance there exist, in principle, equations relating each state variable to two others. For example,

P = P (T, Vˆ ) ˆ =U ˆ (T, Vˆ ) U

(1.6-1)

20 Chapter 1: Introduction

ˆ the internal energy per unit mass, and Vˆ Here P is the pressure, T the temperature, U 5 the volume per unit mass. The first equation indicates that the pressure is a function of the temperature and volume per unit mass; the second indicates that the internal energy is a function of temperature and volume. Also, there are relations of the form ˆ =U ˆ (T, P ) U ˆ ˆ (P, Vˆ ) U =U ˆ , Vˆ ) P = P (U (1.6-2) Similar equations are valid for the additional thermodynamic properties to be introduced later. The interrelations of the form of Eqs 1.6-1 and 1.6-2 are always obeyed in nature, though we may not have been sufficiently accurate in our experiments, or clever enough in other ways to have discovered them. In particular, Eq. 1.6-1 indicates that if we prepare a fluid such that it has specified values T and Vˆ , it will always have the same pressure P . What is this value of the pressure P ? To know this we would have either done the experiment sometime in the past or know the exact functional relationship between T , Vˆ , and P for the fluid being considered. What is frequently done for fluids of scientific or engineering interest is to make a large number of measurements of P , Vˆ , and T and then to develop a volumetric equation of state for the fluid, that is, a matheˆ, matical relationship between the variables P , Vˆ , and T . Similarly, measurements of U Vˆ , and T are made to develop a thermal equation of state for the fluid. Alternatively, the data that have been obtained may be presented directly in graphical or tabular form. (In fact, as will be shown later in this book, it is more convenient to formulate volumetric equations of state in terms of P , V , and T than in terms of P , Vˆ , and T , since in this case the same gas constant of Eq. 1.4-3 can be used for all substances. If volume on a per-mass basis Vˆ was used, the constant in the ideal gas equation of state would be R divided by the molecular weight of the substance.) There are some complications in the description of thermodynamic states of systems. For certain idealized fluids, such as the ideal gas and the incompressible liquid (both discussed in Sec. 3.3), the specification of any two state variables may not be sufficient to fix the thermodynamic state of the system. To be specific, the internal energy of the ideal gas is a function only of its temperature, and not of its pressure or density. Thus, the specification of the internal energy and temperature of an ideal gas contains no more information than specifying only its temperature and therefore is insufficient to determine its pressure. Similarly, if a liquid is incompressible, its molar volume will depend on temperature but not on the pressure exerted on it. Consequently, specifying the temperature and the specific volume of an incompressible liquid contains no more information than specifying only its temperature. The ideal gas and the incompressible liquid are limiting cases of the behavior of real fluids, so that although the internal energy of a real gas depends on density and temperature, the density dependence may be weak; also the densities of most liquids are only weakly dependent on their pressure.

5 In this book we use letters with carets to indicate properties per unit mass, such as U ˆ and Vˆ , and letters with underbars, such as U and V , to indicate properties per mole, which are referred to as molar properties. When, in later chapters, we consider mixtures and have to distinguish between species, the notation will become a bit more complicated in that U i and V i will be used to designate the molar internal energy and volume, respectively, of pure species i. Also, when necessary, within parentheses we can indicate the temperature and/or pressure (and in later chapters the composition) of the substance. In these cases, notation such as U i (T, P ) and V i (T, P ) will be used.

1.7 A Summary of Important Experimental Observations 21 Therefore, although in principle any two state variables may be used to describe the ˆ and T as the independent state thermodynamic state of a system, one should not use U variables for gases or Vˆ and T as the independent state variables for liquids and solids. As was pointed out in the previous paragraphs, two state variables are needed to fix the thermodynamic state of an equilibrium system. The obvious next question is, how does one specify the thermodynamic state of a nonequilibrium system? This is clearly a much more complicated question, and the detailed answer would involve a discussion of the relative time scales for changes imposed on the system and the changes that occur within the system (as a result of chemical reaction, internal energy flows, and fluid motion). Such a discussion is beyond the scope of this book. The important observation is that if we do not consider very fast system changes (as occur within a shock wave), or systems that relax at a very slow but perceptible rate (e.g., molten polymers), the equilibrium relationships between the fluid properties, such as the volumetric and thermal equations of state, are also satisfied in nonequilibrium flows on a point-by-point basis. That is, even though the temperature and pressure may vary in a flowing fluid, as long as the changes are not as sharp as in a shock wave and the fluid internal relaxation times are rapid,6 the properties at each point in the fluid are interrelated by the same equations of state as for the equilibrium fluid. This situation is referred to as local equilibrium. This is an important concept since it allows us to consider not only equilibrium phenomena in thermodynamics but also many flow problems involving distinctly nonequilibrium processes.

1.7 A SUMMARY OF IMPORTANT EXPERIMENTAL OBSERVATIONS An objective of this book is to present the subject of thermodynamics in a logical, coherent manner. We do this by demonstrating how the complete structure of thermodynamics can be built from a number of important experimental observations, some of which have been introduced in this chapter, some of which are familiar from mechanics, and some of which are introduced in the following chapters. For convenience, the most important of these observations are listed here. From classical mechanics and chemistry we have the following two observations. Experimental observation 1. In any change of state (except one involving a nuclear reaction, which is not considered in this book) total mass is conserved. Experimental observation 2. In any change of state total momentum is a conserved quantity. In this chapter the following eight experimental facts have been mentioned. Experimental observation 3 (Sec. 1.5). In any change of state the total energy (which includes internal, potential, and kinetic energy, heat, and work) is a conserved quantity. Experimental observation 4 (Sec. 1.5). A flow of heat and a flow of work are equivalent in that supplying a given amount of energy to a system in either of these forms can be made to result in the same increase in its internal energy. Heat and work, or more generally, thermal and mechanical energy, are not equivalent in the sense

6 The

situation being considered here is not as restrictive as it appears. In fact, it is by far the most common case in engineering. It is the only case that is considered in this book.

22 Chapter 1: Introduction that mechanical energy can be completely converted to thermal energy, but thermal energy can be only partially converted to mechanical energy in a cyclic process. Experimental observation 5 (Sec. 1.3). A system that is not subject to forced flows of mass or energy from its surroundings will evolve to a time-invariant state that is uniform or composed of uniform subsystems. This is the equilibrium state. Experimental observation 6 (Sec. 1.3). A system in equilibrium with its surroundings will never spontaneously revert to a nonequilibrium state. Experimental observation 7 (Sec. 1.3). Equilibrium states that arise naturally are stable to small disturbances. Experimental observation 8 (Secs. 1.3 and 1.6). The stable equilibrium state of a system is completely characterized by values of only equilibrium properties (and not properties that describe the approach to equilibrium). For a single-component, singlephase system the values of only two intensive, independent state variables are needed to fix the thermodynamic state of the equilibrium system completely; the further specification of one extensive variable of the system fixes its size. Experimental observation 9 (Sec. 1.6). The interrelationships between the thermodynamic state variables for a fluid in equilibrium also apply locally (i.e., at each point) for a fluid not in equilibrium, provided the internal relaxation processes are rapid with respect to the rate at which changes are imposed on the system. For fluids of interest in this book, this condition is satisfied. Although we are not, in general, interested in the detailed description of nonequilibrium systems, it is useful to note that the rates at which natural relaxation processes (i.e. heat fluxes, mass fluxes, etc.) occur are directly proportional to the magnitude of the driving forces (i.e., temperature gradients, concentration gradients, etc.) necessary for their occurrence. Experimental observation 10. The flow of heat Q˙ (units of J/s or W) that arises because of a temperature difference ΔT is linearly proportional to the magnitude of the temperature difference:7

Q˙ = −hΔT

(1.7-1)

Here h is a positive constant, and the minus sign in the equation indicates that the heat flow is in the opposite direction to the temperature difference; that is, the flow of heat is from a region of high temperature to a region of low temperature. Similarly, on a microscopic scale, the heat flux in the x-coordinate direction, denoted by qx (with units of J/m2 s), is linearly related to the temperature gradient in that direction:

∂T (1.7-2) ∂x The mass flux of species A in the x direction, jA |x (kg/m2 s), relative to the fluid mass average velocity is linearly related to its concentration gradient, qx = −k

jA |x = −ρD

∂wA ∂x

(1.7-3)

7 Throughout this book we use a dot, as on Q, ˙ to indicate a flow term. Thus, Q˙ is a flow of heat with units of J/s, ˙ is a flow of mass with units of kg/s. Also, radiative heat transfer is more complicated, and is not considered and M here.

1.8 A Comment on the Development of Thermodynamics 23 and for many fluids, the flux of the x-component of momentum in the y -coordinate direction is ∂vx τyx = −μ (1.7-4) ∂y In these equations T is the temperature, ρ the mass density, wA the mass fraction of species A, and vx the x-component of the fluid velocity vector. The parameter k is the thermal conductivity, D the diffusion coefficient for species A, and μ the fluid viscosity; from experiment the values of these parameters are all greater than or equal to zero (this is, in fact, a requirement for the system to evolve toward equilibrium). Equation 1.7-2 is known as Fourier’s law of heat conduction, Eq. 1.7-3 is called Fick’s first law of diffusion, and Eq. 1.7-4 is Newton’s law of viscosity.

1.8 A COMMENT ON THE DEVELOPMENT OF THERMODYNAMICS The formulation of the principles of thermodynamics that is used in this book is a reflection of the author’s preference and experience, and is not an indication of the historical development of the subject. This is the case in most textbooks, as a good textbook should present its subject in an orderly, coherent fashion, even though most branches of science have developed in a disordered manner marked by both brilliant, and frequently unfounded, generalizations and, in retrospect, equally amazing blunders. It would serve little purpose to relate here the caloric or other theories of heat that have been proposed in the past, or to describe all the futile efforts that went into the construction of perpetual motion machines. Similarly, the energy equivalence of heat and work seems obvious now, though it was accepted by the scientific community only after 10 years of work by J. P. Joule. Historically, this equivalence was first pointed out by a medical doctor, J. R. Mayer. However, it would be foolish to reproduce in a textbook the stages of his discovery, which started with the observation that the venous blood of sailors being bled in Java was unusually red, made use of a theory of Lavoisier relating the rate of oxidation in animals to their heat losses, and ultimately led to the conclusion that heat and work were energetically equivalent. The science of thermodynamics as we now know it is basically the work of the experimentalist, in that each of its principles represents the generalization of a large amount of varied experimental data and the life’s work of many. We have tried to keep this flavor by basing our development of thermodynamics on a number of key experimental observations. However, the presentation of thermodynamics in this book, and especially in the introduction of entropy in Chapter 4, certainly does not parallel its historical development. PROBLEMS 1.1 For each of the cases that follow, list as many properties of the equilibrium state as you can, especially the constraints placed on the equilibrium state of the system by its surroundings and/or its container. a. The system is placed in thermal contact with a thermostatic bath maintained at temperature T .

b. The system is contained in a constant-volume container and thermally and mechanically isolated from its surroundings. c. The system is contained in a frictionless piston and cylinder exposed to an atmosphere at pressure P and thermally isolated from its surroundings.

24 Chapter 1: Introduction d. The system is contained in a frictionless piston and cylinder exposed to an atmosphere at pressure P and is in thermal contact with a thermostatic bath maintained at temperature T . e. The system consists of two tanks of gas connected by tubing. A valve between the two tanks is fully opened for a short time and then closed. 1.2 The following table lists the volumes of 1 gram of water and 1 gram of mercury as functions of temperature. a. Discuss why water would not be an appropriate thermometer fluid between 0◦ C and 10◦ C. b. Because of the slightly nonlinear temperature dependence of the specific volume of liquid mercury, there is an inherent error in using a mercury-filled thermometer that has been calibrated against an ideal gas thermometer at only 0◦ C and 100◦ C. Using the data in the table, prepare a graph of the error, ΔT , as a function of temperature. c. Why does a common mercury thermometer consist of a large-volume mercury-filled bulb attached to a capillary tube?

T (◦ C)

Volume of 1 gram of H2 O (cm3 )

Volume of 1 gram of Hg (cm3 )

0 1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 90 100

1.0001329 1.0000733 1.0000321 1.0000078 1.0000000 1.0000081 1.0000318 1.0000704 1.0001236 1.0001909 1.0002719 1.0015678 1.0043408 1.0078108 1.012074 1.017046 1.022694 1.028987 1.035904 1.043427

0.0735560 0.0735694 0.0735828 0.0735961 0.0736095 0.0736228 0.0736362 0.0736496 0.0736629 0.0736763 0.0736893 0.0738233 0.0739572 0.0740910 0.0742250 0.0743592 0.0744936 0.0746282 0.0747631 0.0748981

*Based on data in R. H. Perry and D. Green, eds., Chemical Engineers’ Handbook, 6th ed., McGraw-Hill, New York, 1984, pp. 3-75–3-77.

Chapter

2 Conservation of Mass In this chapter we start the quantitative development of thermodynamics using one of the qualitative observations of the previous chapter, that mass is conserved. Here we begin by developing the balance equations for the total mass of a system (a piece of equipment, a defined volume in space, or whatever is convenient for the problem at hand) by considering systems of only a single component. In this case, the mass of a single species being considered is also the total mass, which is conserved, and we can write the balance equation either based on mass or by dividing by the molecular weight, on the number of moles. We develop two forms of these mass balance equations—the first for computing the rate at which mass in a system changes with time, and the second set, obtained by integrating these rate equations over an interval of time, to compute only the change in mass (or number of moles) in that time interval. We next consider the mass balances for a mixture. In this case while total mass is conserved, there will be a change in mass of some or all species if one or more chemical reactions occur. For this case, it is more convenient to develop the mass balance for mixtures on a molar basis, as chemical reaction stoichiomentry is much easier to write on a molar basis than on a mass basis. In this chapter we will consider only the case of a single chemical reaction; in later chapters the more general case of several chemical reactions occuring simultaneously will be considered. The most important goals of this chapter are for the student to understand when to use the rate-of-change form of the mass balance equation and when to use the difference form, and how to use these equations to solve problems. Mastering the use of the mass balance equations here will make it easier to use the more complicated energy and other balance equations that will be introduced in the following two chapters.

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 2 The goals of this chapter are for the student to: • Be able to use the rate-of-change form of the pure component mass balance in

problem solving (Sec. 2.2)

• Be able to use the difference form of the pure component mass balance in problem

solving (Sec. 2.2)

• Be able to use the rate-of-change form of the multicomponent mass balance in

problem solving (Sec. 2.3)

• Be able to use the difference form of the multicomponent mass balance in problem

solving (Sec. 2.3)

• Be able to solve mass balance problems involving a single chemical reaction

(Sec. 2.3) 25

26 Chapter 2: Conservation of Mass

IMPORTANT NOTATION INTRODUCED IN THIS CHAPTER Mi M˙ k ˙ (Mi )k Ni N˙ k (N˙ i )k t x X νi

Mass of species i (g) Mass flow rate at location k (g/s) Mass flow rate of species i at location k (g/s) Moles of species i (mol) Molar flow rate at location k (g/s) Molar flow rate of species i at location k (g/s) Time (s) Set of mole fractions of all species x1 , x2 , x3 , . . . Molar extent of reaction (mol) Stoichiometric coefficient of species i

2.1 A GENERAL BALANCE EQUATION AND CONSERVED QUANTITIES The balance equations used in thermodynamics are conceptually simple. Each is obtained by choosing a system, either a quantity of mass or a region of space (e.g., the contents of a tank), and equating the change of some property of this system to the amounts of the property that have entered and left the system and that have been produced within it. We are interested both in the change of a system property over a time interval and in its instantaneous rate of change; therefore, we will formulate equations of change for both. Determining which formulation of the equations of change is used for the description of a particular physical situation will largely depend on the type of information desired or available. To illustrate the two types of descriptions and the relationship between them, as well as the idea of using balance equations, consider the problem of studying the total mass of water in Lake Mead (the lake behind Hoover Dam on the Colorado River). If you were interested in determining, at any moment, whether the water level in this lake was rising or falling, you would have to ascertain whether the water flows into the lake were greater or less than the flows of water out of the lake. That is, at some instant you would determine the rates at which water was entering the lake (due to the flow of the Colorado River and rainfall) and leaving it (due to flow across the dam, evaporation from the lake surface, and seepage through the canyon walls), and then use the equation

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ Rate of change of Rate at which Rate at which ⎝ amount of water ⎠ = ⎝ water flows ⎠ − ⎝water flows out⎠ in the lake into the lake of the lake

(2.1-1)

to determine the precise rate of change of the amount of water in the lake. If, on the other hand, you were interested in determining the change in the amount of water for some period of time, say the month of January, you could use a balance equation in terms of the total amounts of water that entered and left the lake during this time:

2.1 A General Balance Equation and Conserved Quantities 27

⎛

⎞ ⎛ ⎞ ⎛ ⎞ Change in amount Amount of water that Amount of water that ⎜ of water in the ⎟ ⎜ flowed into the lake ⎟ ⎜ ⎟ flowed out of ⎝ lake during the ⎠ = ⎝ during the month of ⎠ − ⎝ the lake during the ⎠ month of January January month of January (2.1-2) Notice that Eq. 2.1-1 is concerned with an instantaneous rate of change, and it requires data on the rates at which flows occur. Equation 2.1-2, on the other hand, is for computing the total change that has occurred and requires data only on the total flows over the time interval. These two equations, one for the instantaneous rate of change of a system property (here the amount of water) and the other for the change over an interval of time, illustrate the two types of change-of-state problems that are of interest in this book and the forms of the balance equations that are used in their solution. There is, of course, an interrelationship between the two balance equations. If you had information on each water flow rate at each instant of time for the whole month of January, you could integrate Eq. 2.1-1 over that period of time to obtain the same answer for the total change in the amount of water as would be obtained directly from Eq. 2.1-2 using the much less detailed information on the total flows for the month. The example used here to illustrate the balance equation concept is artificial in that although water flows into and out of a lake are difficult to measure, the amount of water in the lake can be determined directly from the water level. Thus, Eqs. 2.1-1 and 2.1-2 are not likely to be used. However, the system properties of interest in thermodynamics and, indeed, in most areas of engineering are frequently much more difficult to measure than flow rates of mass and energy. Therefore, the balance equation approach may be the only practical way to proceed. While our interest here is specifically in the mass balance, to avoid having to repeat the analysis leading to equation 2.1-4, which follows, for other properties, such as energy (see next chapter), we will develop a general balance equation for any extensive property θ. We will then replace θ with the total mass. In the next section θ will be the mass of only one of the species (which may undergo a chemical reaction), and in the next chapter θ will be replaced by the total energy. In the remainder of this section, the balance equations for an unspecified extensive property θ of a thermodynamic system are developed. With the balance equations formulated in a general manner, they will be applicable (by appropriate simplification) to all systems studied in this book. In this way it will not be necessary to rederive the balance equations for each new problem; we will merely simplify the general equations. Specific choices for θ, such as total mass, mass (or number of moles) of a single species, and energy, are considered in Sec. 2.2, 2.3, and 3.2, respectively. We consider a general system that may be moving or stationary, in which mass and energy may flow across its boundaries at one or more places, and the boundaries of which may distort. Since we are concerned with equating the total change within the system to flows across its boundaries, the details of the internal structure of the system will be left unspecified. This “black-box” system is illustrated in Fig. 2.1-1, and characteristics of this system are 1. Mass may flow into one, several, all, or none of the K entry ports labeled 1, 2, . . . , K (i.e., the system may be either open or closed to the flow of mass). Since we are concerned with pure fluids here, only one molecular species will be involved, although its temperature and pressure may be different at each entry port. The

28 Chapter 2: Conservation of Mass M2 M3

M1

Mk

Figure 2.1-1 A single-component system with several mass flows.

˙ k , so that M˙ k > 0 mass flow rate into the system at the k th entry port will be M ˙ for flow into the system, and Mk < 0 for flow out of the system. 2. The boundaries of the black-box system may be stationary or moving. If the system boundaries are moving, it can be either because the system is expanding or contracting, or because the system as a whole is moving, or both. The following two characteristics are important for the energy balance. 3. Energy in the form of heat may enter or leave the system across the system boundaries. 4. Energy in the form of work (mechanical shaft motion, electrical energy, etc.) may enter or leave the system across the system boundaries. Throughout this book we will use the convention that a flow into the system, whether it be a mass flow or an energy flow, is positive and a flow out of the system is negative. The balance equation for the total amount of any extensive quantity θ in this system is obtained by equating the change in the amount of θ in the system between times t and t + Δt to the flows of θ into and out of the system, and the generation of θ within the system, in the time interval Δt. Thus,    Amount of θ in the Amount of θ in the Amount of θ that entered the system across = − system at time t + Δt system at time t system boundaries between t and t + Δt  Amount of θ that left the system across − system boundaries between t and t + Δt  Amount of θ generated within the + system between t and t + Δt (2.1-3) The meaning of the first two terms on the right side of this equation is clear, but the last term deserves some discussion. If the extensive property θ is equal to the total mass, total energy, or total momentum (quantities that are conserved, see experimental observations 1 to 3 of Sec. 1.7), then the internal generation of θ is equal to zero. This is easily seen as follows for the special case of a system isolated from its environment (so that the flow terms across the system boundaries vanish); here Eq. 2.1-3 reduces to ⎛ ⎞ ⎞ ⎛  Amount of θ in the Amount of θ generated ⎝ system at time ⎠ − Amount of θ in the = ⎝ within the system ⎠ (2.1-3a) system at time t t + Δt between t and t + Δt Since neither total mass, total momentum, nor total energy can be spontaneously produced, if θ is any of these quantities, the internal generation term must be zero. If,

2.1 A General Balance Equation and Conserved Quantities 29 however, θ is some other quantity, the internal generation term may be positive (if θ is produced within the system), negative (if θ is consumed within the system), or zero. For example, suppose the black-box system in Fig. 2.1-1 is a closed (batch) chemical reactor in which cyclohexane is partially dehydrogenated to benzene and hydrogen according to the reaction C6 H12 → C6 H6 + 3H2 If θ is set equal to the total mass, then, by the principle of conservation of mass, the internal generation term in Eq. 2.1-3a would be zero. If, however, θ is taken to be the mass of benzene in the system, the internal generation term for benzene would be positive, since benzene is produced by the chemical reaction. Conversely, if θ is taken to be the mass of cyclohexane in the system, the internal generation term would be negative. In either case the magnitude of the internal generation term would depend on the rate of reaction. The balance equation (Eq. 2.1-3) is useful for computing the change in the extensive property θ over the time interval Δt. We can also obtain an equation for computing the instantaneous rate of change of θ by letting the time interval Δt go to zero. This is done as follows. First, we use the symbol θ(t) to represent the amount of θ in the system at time t , and we recognize that for a very small time interval Δt (over which the flows into and out of the system are constant) we can write ⎞ ⎛ ⎛ ⎞ Amount of θ that enters the Rate at which θ enters ⎝system across system boundaries⎠ as ⎝the system across system⎠ Δt between t and t + Δt boundaries with similar expressions for the outflow and generation terms. Next we rewrite Eq. 2.1-3 as  θ(t + Δt) − θ(t) Rate at which θ enters the system = across system boundaries Δt  Rate at which θ leaves the system − across system boundaries  Rate at which θ is generated + within the system Finally, taking the limit as Δt → 0 and using the definition of the derivative from calculus,

dθ θ(t + Δt) − θ(t) = lim dt Δt→0 Δt we obtain

dθ = dt

  Rate of change of Rate at which θ enters the = system across system boundaries θ in the system  Rate at which θ leaves the − system across system boundaries  Rate at which θ is generated + within the system (2.1-4)

30 Chapter 2: Conservation of Mass Balance Eq. 2.1-4 is general and applicable to conserved and nonconserved quantities. There is, however, the important advantage in dealing with conserved quantities that the internal generation term is zero. For example, to use the total mass balance to compute the rate of change of mass in the system, we need know only the mass flows into and out of the system. On the other hand, to compute the rate of change of the mass of cyclohexane undergoing a dehydrogenation reaction in a chemical reactor, we also need data on the rate of reaction in the system, which may be a function of concentration, temperature, catalyst activity, and internal characteristics of the system. Thus, additional information may be needed to use the balance equation for the mass of cyclohexane, and, more generally, for any nonconserved quantity. The applications of thermodynamics sometimes require the use of balance equations for nonconserved quantities.

2.2 CONSERVATION OF MASS FOR A PURE FLUID The first balance equation of interest in thermodynamics is the conservation equation for total mass. If θ is taken to be the total mass in the system, designated by the symbol M , we have, from Eq. 2.1-3 ⎛ ⎞ ⎛ ⎞ Amount of mass that Amount of mass ⎜ entered the system ⎟ ⎜ that left the system ⎟ ⎜ ⎟ ⎜ ⎟ M (t+Δt) − M (t) = ⎜ across the system ⎟ − ⎜ across the system ⎟ (2.2-1a) ⎝ boundaries between ⎠ ⎝boundaries between ⎠ t and t + Δt t and t + Δt where we have recognized that the total mass is a conserved quantity and that the only ˙k mechanism by which mass enters or leaves a system is by a mass flow. Using M to represent the mass flow rate into the system at the k th entry point, we have, from Eq. 2.1-4, the equation for the instantaneous rate of change of mass in the system: Rate-of-change mass balance

dM M˙ k = dt K

(2.2-1b)

k=1

Equations 2.2-1a and b are general and valid regardless of the details of the system and whether the system is stationary or moving. Since we are interested only in pure fluids here, we can divide Eqs. 2.2-1a and b by the molecular weight of the fluid and use the fact that N , the number of moles in the system, is equal to M/mw, where mw is the molecular weight, and N˙ k , the molar flow rate into ˙ k /mw, to obtain instead of Eq. 2.2-1a a similar the system at the k th entry port, is M equation in which the term moles replaces the word mass and, instead of Eq. 2.2-1b, Rate-of-change mass balance: molar basis

dN N˙ k = dt K

(2.2-2)

k=1

We introduce this equation here because it is frequently convenient to do calculations on a molar rather than on a mass basis.

2.2 Conservation of Mass for a Pure Fluid 31 In Sec. 2.1 it was indicated that the equation for the change of an extensive state variable of a system in the time interval Δt could be obtained by integration over the time interval of the equation for the rate of change of that variable. Here we demonstrate how this integration is accomplished. For convenience, t1 represents the beginning of the time interval and t2 represents the end of the time interval, so that Δt = t2 − t1 . Integrating Eq. 2.2-1b between t1 and t2 yields



t2

t1

dM dt = dt K



k=1

t2

M˙ k dt

(2.2-3)

t1

The left side of the equation is treated as follows:



t2 t1

dM dt = dt



⎛

M (t2 )

M (t1 )

⎞ Change in total mass dM = M (t2 ) − M (t1 ) = ⎝ of system between ⎠ t1 and t2

where M (t) is the mass in the system at time t. The term on the right side of the equation may be simplified by observing that

 t2 Mass that entered the M˙ k dt = system at the k th entry ≡ ΔMk t1 port between t1 and t2 Thus Integral mass balance

M (t2 ) − M (t1 ) =

K

ΔMk

(2.2-4)

k=1

This is the symbolic form of Eq. 2.2-1a. Equation 2.2-4 may be written in a simpler form when the mass flow rates are steady, that is, independent of time. For this case  t2  t2 ˙ ˙ Mk dt = Mk dt = M˙ k Δt t1

t1

so that

M (t2 ) − M (t1 ) =

K

M˙ k Δt (steady flows)

(2.2-5)

k=1

The equations in this section that will be used throughout this book are listed in Table 2.2-1. Illustration 2.2-1 Use of the Difference Form of the Mass Balance A tank of volume 25 m3 contains 1.5 × 104 kg of water. Over a two-day period the inlet to the tank delivers 2.0 × 103 kg, 1.3 × 103 kg leaves the tank through the outlet port, and 50 kg of

32 Chapter 2: Conservation of Mass Table 2.2-1 The Mass Conservation Equation Mass Basis

Molar Basis

Rate-of-change form of the mass balance  dM ˙k = M dt k=1

 dN = N˙ k dt k=1

dM =0 dt M = constant

dN =0 dt N = constant

K

General equation

K

Special case: Closed system

Difference form of the mass balance* General equation

M2 − M1 =

K 

ΔMk

N 2 − N1 =

k=1

Special cases: Closed system Steady flow

M2 − M1 =

ΔNk

k=1

M2 = M1 K 

K 

N 2 = N1 M˙ k Δt

N 2 − N1 =

k=1

K 

N˙ k Δt

k=1

*Here we have used the abbreviated notation Mi = M (ti ) and Ni = N (ti ).

water leaves the tank by evaporation. How much water is in the tank at the end of the two-day period?

Solution Since we are interested only in the change in the mass of water in the tank over the two-day period, and not in the rate of change, we will use the difference form of the mass balance over the period from the initial time (which we take to be t = 0) until two days later (t = 2 days). We use Eq. 2.2-4, recognizing that we have three flow terms: M1 (flow into the tank) = +2.0 × 103 kg, M2 (flow from the tank) = −1.3 × 103 kg, and M3 (evaporation) = −50 kg. (Remember, in our notation the + sign is for flow into the system, the tank, and the − sign is for flow out of the system.) Therefore, M (t = 2 days) − M (t = 0) = M1 + M2 + M3 M (t = 2 days) − 1.5 × 104 kg = 2.0 × 103 − 1.3 × 103 − 50 M (t = 2 days) = 1.5 × 104 + 2.0 × 103 − 1.3 × 103 − 50 = 1.565 × 104 kg 1

Illustration 2.2-2 Use of the Rate-of-Change Form of the Mass Balance A storage tank is being used in a chemical plant to dampen fluctuations in the flow to a downstream chemical reactor. The exit flow from this tank will be kept constant at 1.5 kg/s; if the instantaneous flow into the tank exceeds this, the level in the tank will rise, while if the instantaneous flow is less, the level in the tank will drop. If the instantaneous flow into the storage tank is 1.2 kg/s, what is the rate of change of mass in the tank?

1 Throughout

this text the symbol

will be used to indicate the end of an illustration.

2.2 Conservation of Mass for a Pure Fluid 33 Solution Since we are interested in the rate of change of mass, here we use the rate-of-change form of the mass balance (Eq. 2.2-1b):  dM M˙ k = dt k

or, in this case, kg dM = −1.5 + 1.2 = −0.3 dt s

Thus, at the moment the measurements were made, the amount of liquid in the tank was decreasing by 0.3 kg/s or 300 g/s.

Comment Remember, if we are interested in the rate of change of mass, as we are here, we use the rate-ofchange form of the mass balance, Eq. 2.2-1b. However, if we are interested only in the change of total mass over a period of time, we use Eq. 2.2-4.

Illustration 2.2-3 Use of the Rate-of-Change Form of the Mass Balance Gas is being removed from a high-pressure storage tank through a device that removes 1 percent of the current contents of the tank each minute. If the tank initially contains 1000 mols of gas, how much will remain at the end of 20 minutes?

Solution Since 1 percent of the gas is removed at any time, the rate at which gas leaves the tank will change with time. For example, initially gas is leaving at the rate of 0.01 × 1000 mol/min = 10 mol/min. However, later when only 900 mol of gas remain in the tank, the exiting flow rate will be 0.01 × 900 mol/min = 9 mol/min. In fact, the exiting flow rate is continuously changing with time. Therefore, we have to use the rate-of-change or differential form of the mass (mole) balance. Starting from the rate-of-change form of the mass balance (Eq. 2.2-1b) around the tank that has only a single flow term, we have dN = N˙ dt

where

N˙ = −0.01 × N

so that

d ln N = −0.01 dt

The solution to this first-order differential equation is 

N (t) ln N (t = 0)

 = −0.01t

or N (t) = N (t = 0)e−0.01t

Therefore, N (t = 20) = N (t = 0)e−0.01×20 = 1000e−0.2 = 818.7 mol

Note that if we had merely (and incorrectly) used the initial rate of 10 mol/min we would have obtained the incorrect answer of 800 mol remaining in the tank.

34 Chapter 2: Conservation of Mass Comment To solve any problem in which the mass (or molar) flow rate changes with time, we need to use the differential or rate-of-change form of the mass balance. For problems in which all of the flow terms are constant, we can use the general difference form of the mass balance (which has been obtained from the rate-of-change form by integration over time), or we can use the rate-ofchange form and then integrate over time. However, it is important to emphasize that if one (or more) flow rates are changing with time, the rate-of-change form must be used.

Illustration 2.2-4 Another Problem Using the Rate-of-Change Form of the Mass Balance An open cylindrical tank with a base area of 1 m2 and a height of 10 m contains 5 m3 of water. As a result of corrosion, the tank develops a leak at its bottom. The rate at which water leaves the tank through the leak is 

Leak rate

m3 s



√

= 0.5 ΔP

where ΔP is the pressure difference in bar between the fluid at the base of the tank and the atmosphere. (You will learn about the origin of this equation in a course dealing with fluid flow.) Determine the amount of water in the tank at any time.

Solution Note that the pressure at the bottom of the tank is equal to the atmosphere pressure plus the hydrostatic pressure due to the water above the leak; that is, P = 1.013 bar + ρh, where ρ is the density of water and h is the height of water above the leak. Therefore, ΔP = (1.013 + ρh) − 1.013 = ρh and ΔP = 103

m Pa m s kg bar × h m × 9.807 2 × 1 × 105 = 0.09807h bar m3 s kg Pa

Since the height of fluid in the tank is changing with time, the flow rate of the leak will change with time. Therefore, to solve the problem, we must use the rate-of-change form of the mass balance. The mass of water in the tank at any time is M (t) = ρAh(t) = 103

kg · 1 m2 · h(t) m = 103 h(t) kg m3

The mass balance on the contents of the tank at any time is   dM (t) dh(t) = 103 = M˙ = −0.5 0.098 07h(t) = −0.1566 h(t) dt dt

where the negative sign arises because the flow is out of the tank. Integrating this equation between t = 0 and any later time t yields    √ 2 h(t) − 2 h(0) = 2 h(t) − 2 5 = −0.1566 × 10−3 t or  √ h(t) = 5 −

which can be rearranged to



0.1566 2



× 10−3 t

2.3 The Mass Balance Equations for a Multicomponent System with a Chemical Reaction 35 h(t) =

√

5 − 0.7829 × 10−4 t

2

and M (t) =

√

5 − 0.7829 × 10−4 t

2

× 103 kg

From this equation, one finds that the tank will be completely drained in 28 580 s or 7.938 hr.

Comment Since the rate of flow of water out of the tank depends on the hydrostatic pressure due to the water column above the leak, and since the height of this column changes with time, again we must use the rate-of-change form of the mass balance to solve the problem. It is important in any problem to be able to recognize whether the flows are steady, in which case the difference form of the mass balance can be used, or the flows vary with time, as is the case here, in which case the rate-of-change form of the mass balance must be used.

2.3 THE MASS BALANCE EQUATIONS FOR A MULTICOMPONENT SYSTEM WITH A CHEMICAL REACTION When chemical reactions occur, the mass (or mole) balance for each species is somewhat more complicated since the amount of the species can increase or decrease as a result of the reactions. Here we will consider mass balances when there is only a single chemical reaction; in Chapter 8 and later chapters the more general case of several chemical reactions occurring simultaneously is considered. Also, we will write the mass balances using only the number of moles since the stoichiometry of chemical reactions is usually written in terms of the number of moles of each species that undergoes chemical   reaction rather than the mass of each species that reacts. Using the notation N˙ i for the rate at which moles of species i enter (if positive) or leave (if negative) k in flow stream k, we have the differential or rate-of-change form of the mass balance on species i as

dNi ˙ (Ni )k + = dt K

Rate-of-change mass balance with chemical reaction on a molar basis

k=1



dNi dt

(2.3-1) rxn

where the last term is new and describes the rate at which species i is produced (if positive) or consumed (if negative) within the system by chemical reaction. The difference form of this equation, obtained by integrating over the time period from t1 to t2 , is Difference form of the mass balance

Ni (t2 ) − Ni (t1 ) =

K ⎧ K



t2 ⎭ (Ni )k dt + (ΔNi )rxn = ΔNk + (ΔNi )rxn k=1

t1

k=1

(2.3-2)

If a flow rate is steady

where the summation terms after the equal signs are the changes in the number of moles of the species due to the flow streams, and the second terms are the result  of  the ˙ chemical reaction. Note that only if the flow rate of a stream is steady (i.e., Ni is k constant), then   (ΔNi )k = N˙ i t k

36 Chapter 2: Conservation of Mass Now consider the mass (mole) balances for a reactor in which the following chemical reaction occurs

C2 H4 + Cl2 → C2 H4 Cl2 but in which neither ethylene nor chlorine is completely consumed. The mass balances for these species, in which stream 1 is pure ethylene and stream 2 is pure chlorine, are



     dNC2 H4 dN C H 2 4 C2 H4 : = N˙ C2 H4 + N˙ C2 H4 + dt dt 1 3 rxn       dNCl2 dNCl2 = N˙ Cl2 + N˙ Cl2 + Cl2 : (2.3-3) dt dt 2 3 rxn     dNC2 H4 Cl2 dNC2 H4 Cl2 ˙ C2 H4 Cl2 : = NC2 H4 Cl2 + dt dt 3 rxn where all the exit streams [i.e., (...)3 ] will be negative in value. From the stoichiometry of this reaction, all the reaction rate terms are interrelated. In this case, the rate at which ethylene chloride is created (the number of moles per second) is equal to the rate at which ethylene is consumed, which is also equal to the rate at which chlorine is consumed. That is,



dNC2 H4 Cl2 dt



 =−

rxn

dNC2 H4 dt



 =−

rxn

dNCl2 dt

rxn

so we can simplify Eqs. 2.3-3 by replacing the three different reaction rates with a single one. We can generalize this discussion of the interrelationships between reaction rates by introducing the following convenient notation for the description of chemical reactions. Throughout this book the chemical reaction

αA + βB + · · · −→ ←− ρR + · · · where α, β, . . . , are the molar stoichiometric coefficients, will be written as

ρR + · · · − αA − βB − · · · = 0 or



νi I = 0

(2.3-4)

i

Here νi is the stoichiometric coefficient of species I, defined so that νi is positive for reaction products, negative for reactants, and equal to zero for inert species. In this notation the electrolytic dissociation reaction H2 O = H2 + 12 O2 is written as H2 + 1 O − H2 O = 0, so that νH2 O = −1, νH2 = +1, and νO2 = + 12 . 2 2 We will use Ni to represent the number of moles of species i in the system at any time t and Ni,0 to be the initial number of moles of species i. For a closed system, Ni and Ni,0 are related through the reaction variable X , called the molar extent of reaction, and

2.3 The Mass Balance Equations for a Multicomponent System with a Chemical Reaction 37 the stoichiometric coefficient νi by the equation Molar extent of reaction

Ni = Ni,0 + νi X

(2.3-5a)

or

X=

Ni − Ni,0 νi

(2.3-5b)

An important characteristic of the reaction variable X defined in this way is that it has the same value for each molecular species involved in a reaction; this is illustrated in the following example. Thus, given the initial mole numbers of all species and X (or the number of moles of one species from which the molar extent of reaction X can be calculated) at time t , one can easily compute all other mole numbers in the system. In this way the complete progress of a chemical reaction (i.e., the change in mole numbers of all the species involved in the reaction) is given by the value of the single variable X . Illustration 2.3-1 Using the Molar Extent of Reaction Notation The electrolytic decomposition of water to form hydrogen and oxygen occurs as follows: H2 O −→ H2 + 12 O2 . Initially, only 3.0 mol of water are present in a closed system. At some later time it is found that 1.2 mol of H2 and 1.8 mol of H2 O are present. a. Show that the molar extents of reaction based on H2 and H2 O are equal. b. Compute the number of moles of O2 in the system.

Solution a. The reaction H2 O → H2 + 12 O2 is rewritten as H2 + 12 O2 − H2 O = 0

so that νH2 O = −1

νH2 = +1

and

νO2 = + 12

From the H2 data, X=

1.2 − 0.0 = +1.2 mol +1

X=

1.8 − 3.0 = +1.2 mol −1

From the H2 O data,

b. Starting from Ni = Ni,0 + νi X , we have NO2 = 0 + (+ 12 )(1.2) = 0.6 mol

Comment Note that the molar extent of reaction is not a fractional conversion variable; therefore, its value is not restricted to lie between 0 and 1. As defined here X , which has units of number of moles,

38 Chapter 2: Conservation of Mass is the number of moles of a species that has reacted divided by the stoichiometric coefficient for the species. In fact, X may be negative if the reaction proceeds in the reverse direction to that indicated (e.g., if hydrogen and oxygen react to form water).

The rate of change of the number of the moles of species i resulting from a chemical reaction is  dNi = νi X˙ (2.3-6) dt rxn where the subscript rxn indicates that this is the rate of change of the number of moles of species i due to chemical reaction alone, and X˙ is the rate of change of the molar extent of reaction. Using this notation, the balance equation for species i is

dNi ˙ dX (Ni )k + νi = dt dt K

Rate-of-change form of the mass balance with chemical reaction: molar basis

(2.3-7)

k=1

The difference form of this equation, obtained by integrating over the time period from t1 to t2 , is Difference form of mass balance with chemical reaction

K ⎧ K



t2 ˙ ⎭ Ni (t2 ) − Ni (t1 ) = (Ni )k dt + (ΔNi )rxn = (ΔNi )k + νi ΔX k=1

t1

k=1

(2.3-8) Using this notation for the description of the production of ethylene dichloride considered earlier, we have      dNC2 H4 dX C2 H4 : = N˙ C2 H4 + N˙ C2 H4 − dt dt 1 3      dX dNCl2 = N˙ Cl2 + N˙ Cl2 − (2.3-9) Cl2 : dt dt 2 3    dX dNC2 H4 Cl2 = N˙ C2 H4 Cl2 + C2 H4 Cl2 : dt dt 3 Illustration 2.3-2 Mass Balance for a Mixture with Chemical Reaction At high temperatures acetaldehyde (CH3 CHO) dissociates into methane and carbon monoxide by the following reaction CH3 CHO−→CH4 + CO

At 520o C the rate at which acetaldehyde dissociates is dCCH3 CHO m3 2 = −0.48CCH 3 CHO dt kmol s

2.3 The Mass Balance Equations for a Multicomponent System with a Chemical Reaction 39 where C is concentration in kmol/m3 . The reaction occurs in a constant-volume, 1-L vessel, and the initial concentration of acetaldehyde is 10 kmol/m3 . a. If 5 mols of the acetaldehyde reacts, how much methane and carbon monoxide is produced? b. Develop expressions for the amounts of acetaldehyde, methane, and carbon monoxide present at any time, and determine how long it would take for 5 mol of acetaldehyde to have reacted.

Solution First, we must determine the initial amount of acetaldehyde present. Since the initial concentration is CCH3 CHO = 10

mol kmol = 10 L m3

it follows that initially NCH3 CHO = 10

mol × 1 L = 10 mol L

Next, we write the stoichiometry for the reaction in terms of the molar extent of reaction X as follows: NCH3 CHO = 10 − X

NCH4 = X

and

NCO = X

(a)

a. To determine the amounts of each species after a given amount of acetaldehyde has reacted, we can use the difference form of the mass balance for this system with no flows of species into or out of the reactor: Ni (t) − Ni (t = 0) = (ΔNi )rxn = νi ΔX

Therefore, for acetaldehyde NCH3 CHO (t) − NCH3 CHO (t = 0) = 5 − 10 mol = −5 mol = −X

so that X = 5 mol. Then amounts of the other species are NCH4 (t) = X = 5 mol and NCO (t) = X = 5 mol

b. To determine the amount of each species as a function of time is more difficult and must be done using the rate-of-change form of the mass balance since the rate of reaction and therefore the value of X change with time. However, because the amounts of the species are always related by the stoichiometry of Eq. a, we can use the mass balance for one of the species to determine the time variation of X , and then can use the expression for X(t) to obtain the compositions of all species in the reaction as a function of time. Since the rate expression is written for acetaldehyde, we will use this substance to determine the time dependence of X . Since there are no flows into or out of the reactor, Eq. 2.3-7 is dNCH3 CHO = dt



dNCH3 CHO dt

 = ν CH3 CHO rxn

dX dX =− dt dt

Next the reaction rate expression can be written as  d

NCH3 CHO V

dt

 =

2 NCH 1 dNCH3 CHO m3 2 3 CHO = −0.48CCH = −0.48 3 CHO V dt V2 kmol s

40 Chapter 2: Conservation of Mass which, after using NCH3 CHO = 10 − X , becomes 0.48(10 − X)2 mol dX mol = = 0.48(10 − X)2 dt V Ls s Now integrating from t = 0, at which X = 0, to some other time t gives 1 1 − = 0.48t 10 − X(t) 10

or X(t) = 10 −

1 mol 0.48t + 0.1

Therefore,

NCH3 CHO (t) =

1 0.48t + 0.1

and NCH4 (t) = NCO (t) = 10 −

1 0.48t + 0.1

Finally, solving this equation for X(t) = 5 mol gives t = 0.208 s; that is, half of the acetaldehyde dissociates within approximately two-tenths of a second.

Comment Notice again that solving the rate-of-change form of the mass balance requires more information (here the rate of reaction) and more effort than solving the difference form of the mass balance. However, we also get more information—the amount of each species present as a function of time. In Sec. 2.4, which is optional and more difficult, we consider another, even deeper level of description, where not only is time allowed to vary, but the system is not spatially homogeneous; that is, the composition in the reactor varies from point to point. However, this section is not for the faint-hearted and is best considered after a course in fluid mechanics.

Illustration 2.3-3 Mass Balance for a Liquid Mixture with a Reversible Reaction The ester ethyl acetate is produced by the reversible reaction k

CH3 COOH + C2 H5 OH −→ ←− CH3 COOC2 H5 + H2 O k

in the presence of a catalyst such as sulfuric or hydrochloric acid. The rate of ethyl acetate production has been found, from the analysis of chemical kinetics data, to be given by the following equation: dCEA = kCA CE − k CEA CW dt

where the subscripts EA, A, E, and W denote ethyl acetate, acetic acid, ethanol, and water, respectively, and the concentration of each species in units of kmol/m3 . The values of the reaction rate constants at 100◦ C and the catalyst concentration of interest are k = 4.76 × 10−4 m3 /kmol min

and k = 1.63 × 10−4 m3 /kmol min

Develop expressions for the number of moles of each species as a function of time if the feed to the reactor is 1 m3 of an aqueous solution that initially contains 250 kg of acetic acid and 500 kg of ethyl alcohol. The density of the solution may be assumed to be constant and equal to 1040 kg/m3 , and the reactor will be operated at a sufficiently high pressure that negligible amounts

2.3 The Mass Balance Equations for a Multicomponent System with a Chemical Reaction 41 of reactants or products vaporize. Compute the number of moles of each species present 100 minutes after the reaction has started, and at infinite time when the reaction will have stopped and the system is at equilibrium.

Solution Since the reaction rate expression is a function of the compositions, which are changing as a function of time, the mass balance for each species must be written in the rate-of-change form. Since the species mole numbers and concentrations are functions of the molar extent of reaction, X, we first determine how X varies with time by solving the mass balance for one species. We will use ethyl acetate since the reaction rate is given for that species. Once the amount of ethyl acetate is known, the other species mole numbers are easily computed as shown below. The initial concentration of each species is 250 kg/m3 = 4.17 kmol/m3 60 g/mol 500 kg/m3 CE = = 10.9 kmol/m3 46 g/mol (1040 − 250 − 500) kg/m3 CW = = 16.1 kmol/m3 18 g/mol CA =

Since there is 1 m3 of solution, the initial amount of each species is NA NE NW NEA

= 4.17 kmol = 10.9 kmol = 16.1 kmol = 0 kmol

and by the reaction stoichiometry, the amount of each species present at any time (in kmol) and its concentration (since 1 m3 of volume is being considered) is NA = 4.17 − X

CA = 4.17 − X

NE = 10.9 − X

CE = 10.9 − X

NW = 16.1 + X

CW = 16.1 + X

and NEA = X

CEA = X

Because the concentration of a species is equal to the number of moles N divided by the volume V, the chemical reaction rate equation can be written as d dt



NEA V

 =k

N A NE NEA NW − k V V V V

Now using V = 1 m3 and the mole numbers, we have d X = k(4.17 − X)(10.9 − X) − k X(16.1 + X) dt

or dX = 4.76 × 10−4 (4.17 − X)(10.9 − X) − 1.63 × 10−4 (16.1 + X)X dt = 2.163 × 10−2 (1 − 0.4528X − 0.01447X 2 ) kmol/m3 min

42 Chapter 2: Conservation of Mass which can be rearranged to 2.163 × 10−2 dt =

dX 1 − 0.4528X + 0.01447X 2

Integrating this equation between t = 0 and time t yields2 2.163 × 10−2



t

dt = 2.163 × 10−2 × t =



0

X

dX 1 − 0.4528X + 0.01447X 2

0

or  ln

0.02894X − 0.8364 0.02894X − 0.0692





− ln

0.8364 0.0692



= 0.8297 × 10−2 t

(c)

and on rearrangement X (t) = 2.3911

e0.008297t − 1 − 0.08274

e0.008297t

for t in minutes and X in kmol. Therefore, NA (t) = CA (t) = 4.17 − X = 4.17 − 2.3911

e0.008297t − 1 e0.008297t − 0.08274

NE (t) = CE (t) = 10.9 − X = 10.9 − 2.3911

e0.008297t − 1 e0.008297t − 0.08274

NW (t) = CW (t) = 16.1 + X = 16.1 + 2.3911

e0.008297t − 1 e0.008297t − 0.08274

and NEA (t) = CEA (t) = X = 2.3911

e0.008297t − 1 − 0.08274

e0.008297t

At 100 minutes, X = 1.40 kmol so that NA = 4.17 − X = 4.17 − 1.4 = 2.77

NE = 9.5,

NW = 17.5,

NEA = 1.4

Also, at infinite time X = 2.39 (actually 2.3911) and NA = 4.17 − X = 4.17 − 2.39 = 1.78,

NE = 8.51,

NW = 18.49,

NEA = 2.39

Illustration 2.3-4 Mass Balance Modeling of a Simple Environmental Problem Water in a lake initially contains a pollutant at a parts-per-million concentration. This pollutant is no longer present in the water entering the lake. The rate of inflow of water to the lake from a creek is constant and equal to the rate of outflow, so the lake volume does not change.

2 Note

that 

where q = 4ac − b2 .

 √  2cX + b − −q dX 1 √ √ ln = a + bX + cX 2 2cX + b + −q 2

2.4 The Microscopic Mass Balance Equations in Thermodynamics and Fluid Mechanics 43 a. Assuming the water in the lake is well mixed, so its composition is uniform and the pollutant concentration in the exit stream is the same as in the lake, estimate the number of lake volumes of water that must be added to the lake and then leave in order for the concentration of the pollutant in the water to decrease to one-half of its initial concentration. b. How many lake volumes would it take for the concentration of the pollutant in the lake to decrease to one-tenth of its initial concentration? c. If the volume of water in the lake is equal to the inflow for a one-year period, assuming the inflow of water is uniform in time, how long would it take for the concentration of the pollutant in the lake to decrease to one-half and one-tenth of its initial concentration?

Solution a. In writing the overall mass balance for the lake, which we take to be the system, we use that the flow rates into and out of the lake are constant and equal, and that the concentration of the pollutant is so low that its change has a negligible effect on the total mass of the water in the lake. With these simplifications the mass balance is dM ˙ )2 , ˙ )1 + (M = 0 = (M dt

so that (M˙ )1 = −(M˙ )2 = M˙

That is, the rate of mass flow out of the lake is equal in magnitude and opposite in sign to the rate of mass flow into the lake. The mass balance on the pollutant is

d (Cp M ) d (Cp ) dMp = =M = M˙ p 2 = (M˙ )2 Cp = −(M˙ )1 Cp dt dt dt

where we have used that the total amount of pollutant is equal to the product of its concentration per unit mass Cp and the total mass M of water in the lake. Therefore, M˙ dCp = − dt Cp M

which has the solution Cp (t) = Cp (t = 0) e−(M /M )t

 ˙t M Cp (t) = 0.5 = exp − Cp (t = 0) M

˙

or M˙ t = 0.693M

Now M˙ t = M , which is the amount of water that entered the lake over the time interval from 0 to t. Therefore, when the amount of fresh water that has entered the lake M t = M equals 69.3 percent of the initial (polluted) water in the lake, the concentration of the pollutant in the lake will have decreased to half its initial value. b. We proceed as in part (a), except that we now have  ˙t M Cp (t) = 0.1 = exp − Cp (t = 0) M

or M˙ t = 2.303M

so that when an amount of water that enters is equal to 2.303 times the initial volume of water in the lake, the concentration of pollutant will have decreased to one-tenth its initial value. c. It will take 0.693 years (253 days) and 2.303 years (840 days) for the concentration of the pollutant to decrease to 50 percent and 10 percent of its inital concentration, respectively.

2.4 THE MICROSCOPIC MASS BALANCE EQUATIONS IN THERMODYNAMICS AND FLUID MECHANICS3 This section appears on the website that accompanies this text. 3 This

section is optional—only for graduate and advanced undergraduate students.

44 Chapter 2: Conservation of Mass PROBLEMS 2.1 As a result of a chemical spill, benzene is evaporating at the rate of 1 gram per minute into a room that is 6 m × 6 m × 3 m in size and has a ventilation rate of 10 m3 /min. a. Compute the steady-state concentration of benzene in the room. b. Assuming the air in the room is initially free of benzene, compute the time necessary for benzene to reach 95 percent of the steady-state concentration. 2.2 The insecticide DDT has a half-life in the human body of approximately 7 years. That is, in 7 years its concentration decreases to half its initial concentration. Although DDT is no longer in general use in the United States, it was estimated that 25 years ago the average farm worker had a body DDT concentration of 22 ppm (parts per million by weight). Estimate what the farm worker’s present concentration would be. 2.3 At high temperatures phosphine (PH3 ) dissociates into phosphorus and hydrogen by the following reaction: 4PH3 −→ P4 + 6H2

At 800◦ C the rate at which phosphine dissociates is dCPH3 = −3.715 × 10−6 CPH3 dt

for t in seconds. The reaction occurs in a constantvolume, 2-L vessel, and the initial concentration of phosphine is 5 kmol/m3 a. If 3 mol of the phosphine reacts, how much phosphorus and hydrogen are produced? b. Develop expressions for the number of moles of phosphine, phosphorus, and hydrogen present at any time, and determine how long it would take for 3 mol of phosphine to have reacted. 2.4 The following reaction occurs in air: 2NO + O2 −→ 2NO2

At 20◦ C the rate of this reaction is dCNO 2 CO2 = −1.4 × 10−4 CNO dt

for t in seconds and concentrations in kmol/m3 . The reaction occurs in a constant-volume, 2-L vessel, and the initial concentration of NO is 1 kmol/m3 and that of O2 is 3 kmol/m3 a. If 0.5 mol of NO reacts, how much NO2 is produced? b. Determine how long it would take for 0.5 mol of NO to have reacted.

Chapter

3 Conservation of Energy In this chapter we continue the quantitative development of thermodynamics by deriving the energy balance, the second of the three balance equations that will be used in the thermodynamic description of physical, chemical, and (later) biochemical processes. The mass and energy balance equations (and the third balance equation, to be developed in the following chapter), together with experimental data and information about the process, will then be used to relate the change in a system’s properties to a change in its thermodynamic state. In this context, physics, fluid mechanics, thermodynamics, and other physical sciences are all similar in that the tools of each are the same: a set of balance equations, a collection of experimental observations (equationof-state data in thermodynamics, viscosity data in fluid mechanics, etc.), and the initial and boundary conditions for each problem. The real distinction between these different subject areas is the class of problems, and in some cases the portion of a particular problem, that each deals with. One important difference between thermodynamics and, say, fluid mechanics and chemical reactor analysis is the level of description used. In fluid mechanics one is usually interested in a very detailed microscopic description of flow phenomena and may try to determine, for example, the fluid velocity profile for flow in a pipe. Similarly, in chemical reactor analysis one is interested in determining the concentrations and rates of chemical reaction everywhere in the reactor. In thermodynamics the description is usually more primitive in that we choose either a region of space or an element of mass as the system and merely try to balance the change in the system with what is entering and leaving it. The advantage of such a description is that we can frequently make important predictions about certain types of processes for which a more detailed description might not be possible. The compromise is that the thermodynamic description yields information only about certain overall properties of the system, though with relatively little labor and simple initial information. In this chapter we are concerned with developing the equations of energy conservation to be used in the thermodynamic analysis of systems of pure substances. (The thermodynamics of mixtures is more complicated and will be considered in later chapters.) To emphasize both the generality of these equations and the lack of detail necessary, we write these energy balance equations for a general black-box system. For contrast, and also because a more detailed description will be useful in Chapter 4, the rudiments of the more detailed microscopic description are provided in the final, optional section of this chapter. This microscopic description is not central to our development of thermodynamic principles, is suitable only for advanced students, and may be omitted. 45

46 Chapter 3: Conservation of Energy To use the energy balances, we will need to relate the energy to more easily measurable properties, such as temperature and pressure (and in later chapters, when we consider mixtures, to composition as well). The interrelationships between energy, temperature, pressure, and composition can be complicated, and we will develop this in stages. In this chapter and in Chapters 4, 5, and 6 we will consider only pure fluids, so composition is not a variable. Then, in Chapters 8 to 15, mixtures will be considered. Also, here and in Chapters 4 and 5 we will consider only the simple ideal gas and incompressible liquids and solids for which the equations relating the energy, temperature, and pressure are simple, or fluids for which charts and tables interrelating these properties are available. Then, in Chapter 6, we will discuss how such tables and charts are prepared.

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 3 The goals of this chapter are for the student to: • Be able to use the differential form of the pure component energy balance in prob-

lem solving (Secs. 3.1 and 2)

• Be able to use the difference form of the pure component energy balance in prob-

lem solving (Secs. 3.1 and 2)

• Be able to compute changes in energy with changes in temperature and pressure

for the ideal gas (Sec. 3.3)

• Be able to compute changes in energy with changes in temperature and pressure

of real fluids using tables and charts of thermodynamic properties (Sec. 3.3)

NOTATION INTRODUCED IN THIS CHAPTER CP CP∗ CV CV∗ H H ˆ H fus H sub H vap H ˆ vap H Q˙ Q TR U ˆ U V ˙ W W Ws ˙s W ψ

Constant-pressure molar heat capacity (J/mol K) Ideal gas constant-pressure molar heat capacity (J/mol K) Constant-volume molar heat capacity (J/mol K) Ideal gas constant-volume molar heat capacity (J/mol K) Enthalpy (J) Enthalpy per mole, or molar enthalpy (J/mol) Enthalpy per unit mass, or specific enthalpy (J/g) Molar enthalpy of melting or fusion (J/mol) Molar enthalpy of sublimation (J/mol) Molar enthalpy of vaporization (J/mol) Enthalpy of vaporization per unit mass (J/g) Rate of flow of heat into the system (J/s) Heat that has flowed into the system (J) Reference temperature for internal energy of enthalpy (K) Internal energy per mole, or molar internal energy (J/mol) Internal energy per unit mass, or specific internal energy (J/g) Volume per mole, or molar volume (m3 /mol) Rate at which work is being done on the system (J/s) Work that has been done on the system (J) Shaft work that has been done on the system (J) Rate at which shaft work is being done on the system (J/s) Potential energy per unit of mass (J/g)

3.1 Conservation of Energy 47

ω I Mass fraction of phase I (quality for steam) ψ Potential energy per unit of mass (J/g)

3.1 CONSERVATION OF ENERGY To derive the energy conservation equation for a single-component system, we again use the black-box system of Figure 2.1-1 and start from the general balance equation, Eq. 2.1-4. Taking θ to be the sum of the internal, kinetic, and potential energy of the system,   2 v +ψ θ =U +M 2 Here U is the total internal energy, v 2 /2 is the kinetic energy per unit mass (where v is the center of mass velocity), and ψ is the potential energy per unit mass.1 If gravity is the only force field present, then ψ = gh, where h is the height of the center of mass with respect to some reference, and g is the force of gravity per unit mass. Since energy is a conserved quantity, we can write

  2      v Rate at which energy Rate at which energy U +M − +ψ = enters the system leaves the system 2 (3.1-1) To complete the balance it remains only to identify the various mechanisms by which energy can enter and leave the system. These are as follows. Energy flow accompanying mass flow. As a fluid element enters or leaves the system, it carries its internal, potential, and kinetic energy. This energy flow accompanying the mass flow is simply the product of a mass flow and the energy per unit mass, d dt

K  k=1

 M˙ k

2 ˆ + v +ψ U 2

 (3.1-2) k

ˆk is the internal energy per unit mass of the k th flow stream, and M˙ k is its mass where U flow rate. Heat. We use Q˙ to denote the total rate of flow of heat into the system, by both conduction and radiation, so that Q˙ is positive if energy in the form of heat flows into the system and negative if heat flows from the system to its surroundings. If heat flows occur at several different places, the total rate of heat flow into the system is  Q˙ = Q˙ j where Q˙ j is the heat flow at the j th heat flow port. Work. The total energy flow into the system due to work will be divided into several parts. The first part, called shaft work and denoted by the symbol Ws , is the mechanical energy flow that occurs without a deformation of the system boundaries. For example,

1 In

writing this form for the energy term, it has been assumed that the system consists of only one phase, that is, a gas, a liquid, or a solid. If the system consists of several distinct parts—for example, gas and a liquid, or a gas and the piston and cylinder containing it—the total energy, which is an extensive property, is the sum of the energies of the constituent parts.

48 Chapter 3: Conservation of Energy if the system under consideration is a steam turbine or internal combustion engine, the ˙ s is equal to the rate at which energy is transferred across the rate of shaft work W stationary system boundaries by the drive shaft or push rod. Following the convention ˙ s is positive if the surroundings do work that energy flow into the system is positive, W on the system and negative if the system does work on its surroundings. For convenience, the flow of electrical energy into or out of the system will be in˙ s = ±EI , where E is the electrical cluded in the shaft work term. In this case, W potential difference across the system and I is the current flow through the system. The positive sign applies if electrical energy is being supplied to the system, and the negative sign applies if the system is the source of electrical energy. Work also results from the movement of the system boundaries. The rate at which work is done when a force F is moved through a distance in the direction of the applied force dL in the time interval dt is

˙ = F dL W dt Here we recognize that pressure is a force per unit area and write ˙ = −P dV W (3.1-3) dt where P is the pressure exerted by the system at its boundaries.2 The negative sign in this equation arises from the convention that work done on a system in compression (for which dV /dt is negative) is positive, and work done by the system on its surroundings in an expansion (for which dV /dt is positive) is negative. The pressure at the boundaries of a nonstationary system will be opposed by (1) the pressure of the environment surrounding the system, (2) inertial forces if the expansion or compression of the system results in an acceleration or deceleration of the surroundings, and (3) other external forces, such as gravity. As we will see in Illustration 3.4-7, the contribution to the energy balance of the first of these forces is a term corresponding to the work done against the atmosphere, the second is a work term corresponding to the change in kinetic energy of the surroundings, and the last is the work done that changes the potential energy of the surroundings. Work of a flowing fluid against pressure. One additional flow of energy for systems open to the flow of mass must be included in the energy balance equation; it is more subtle than the energy flows just considered. This is the energy flow that arises from the fact that as an element of fluid moves, it does work on the fluid ahead of it, and the fluid behind it does work on it. Clearly, each of these work terms is of the P ΔV type. To evaluate this energy flow term, which occurs only in systems open to the flow of mass, we will compute the net work done as one fluid element of mass (M )1 enters a system, such as the valve in Fig. 3.1-1, and another fluid element of

Pressure P1

^

Volume = V1⌬M1

2 In

Pressure P2

Valve

^

Volume = V2⌬M2

Figure 3.1-1 A schematic representation of flow through a valve.

writing this form for the work term, we have assumed the pressure to be uniform at the system boundary. If this is not the case, Eq. 3.1-3 is to be replaced with an integral over the surface of the system.

3.1 Conservation of Energy 49 mass (ΔM )2 leaves the system. The pressure of the fluid at the inlet side of the valve is P1 , and the fluid pressure at the outlet side is P2 , so that we have ⎛ ⎞ Work done by surrounding fluid in ⎝pushing fluid element of mass (M )1 ⎠ = P1 Vˆ1 ΔM1 into the valve ⎛ ⎞ Work done on surrounding fluid ⎜ by movement of fluid element of ⎟ ⎜ ⎟ ⎜mass (ΔM )2 out of the valve (since⎟ = −P2 Vˆ2 ΔM2 ⎝ this fluid element is pushing the ⎠ surrounding fluid)   Net work done on the system due to = P1 Vˆ1 ΔM1 − P2 Vˆ2 ΔM2 movement of fluid For a more general system, with several mass flow ports, we have ⎛ ⎞ Net work done on the system due K ⎜to the pressure forces acting on ⎟  ΔMk P Vˆk ⎝the fluids moving into and out of ⎠ = k=1 the system Finally, to obtain the net rate at which work is done, we replace each mass flow Mk ˙ k , so that with a mass flow rate M

⎛

⎞ Net rate at which work is done on K ⎜ the system due to pressure forces ⎟  ˙ = Mk (P Vˆ )k ⎝acting on fluids moving into and out⎠ k=1 of the system ˙ k. where the sign of each term of this energy flow is the same as that of M One important application of this pressure-induced energy flow accompanying a mass flow is hydroelectric power generation, schematically indicated in Fig. 3.1-2. Here a water turbine is being used to obtain mechanical energy from the flow of water through the base of a dam. Since the water velocity, height, and temperature are approximately the same at both sides of the turbine (even though there are large velocity changes within the turbine), the mechanical (or electrical) energy obtained is a result of only the mass flow across the pressure difference at the turbine. Collecting all the energy terms discussed gives

d dt

  2     K v v2 ˙ ˆ U +M Mk U + +ψ = + ψ + Q˙ 2 2 k k=1

˙ s − P dV + +W dt

K 

(3.1-4)

M˙ k (P Vˆ )k

k=1

This equation can be written in a more compact form by combining the first and last terms on the right side and introducing the notation

50 Chapter 3: Conservation of Energy

Reservoir

Water turbine Water flow Dam Hydroelectric power generating station

Figure 3.1-2 A hydroelectric power generating station: a device for obtaining work from a fluid flowing across a large pressure drop.

H = U + PV ˙ to represent where the function H is called the enthalpy, and by using the symbol W ˙ s and expansion work −P (dV /dt), that is, W ˙ = the combination of shaft work W ˙ Ws − P (dV /dt). Thus we have Complete energy balance, frequently referred to as the first law of thermodynamics

d dt

  2     K v v2 ˙ ˙ ˆ U +M +ψ = + ψ + Q˙ + W Mk H + 2 2 k

(3.1-4a)

k=1

It is also convenient to have the energy balance on a molar rather than a mass basis. ˙ kH ˆ k can equally well be This change is easily accomplished by recognizing that M ˙ written as Nk H k , where H is the enthalpy per mole or molar enthalpy,3 and M (v 2 /2+ ψ) = N m (v 2 /2 + ψ), where m is the molecular weight. Therefore, we can write the energy balance as

d dt

  2   2    K v v ˙ ˙ U + Nm Nk H + m + Q˙ + W +ψ = +ψ 2 2 k k=1

(3.1-4b) Several special cases of Eqs. 3.1-4a and b are listed in Table 3.1-1. The changes in energy associated with either the kinetic energy or potential energy terms, especially for gases, are usually very small compared with those for the thermal (internal) energy terms, unless the fluid velocity is near the velocity of sound, the change in height is very large, or the system temperature is nearly constant. This point will become evident in some of the illustrations and problems (see particularly Illustration 3.4-2). Therefore, it is frequently possible to approximate Eqs. 3.1-4a and b by 3H

= U + P V , where U and V are the molar internal energy and volume, respectively.

3.1 Conservation of Energy 51

 dU ˆ k + Q˙ + W ˙ (M˙ H) = dt

(mass basis)

(3.1-5a)

 dU ˙ (N˙ H)k + Q˙ + W = dt

(molar basis)

(3.1-5b)

K

Commonly used forms of energy balance

k=1

K

k=1

As with the mass balance, it is useful to have a form of the energy balance applicable to a change from state 1 to state 2. This is easily obtained by integrating Eq. 3.1-4a over the time interval t1 to t2 , the time required for the system to go from state 1 to state 2.

Table 3.1-1 Differential Form of the Energy Balance General equation d dt

   2    K v2 v ˙ ˆ ˙ Mk H + + Q˙ + W U +M +ψ = +ψ 2 2 k k=1

(a)

Special cases: (i) Closed system dM =0 dt

M˙ k = 0,

so d dU +M dt dt



v2 +ψ 2

 ˙ = Q˙ + W

(b)

(ii) Adiabatic process in Eqs. a, b, and d Q˙ = 0

(iii) Open and steady-state system dV = 0, dt

dM = 0, dt

so 0=

K 

d dt

(c)

  2  v U +M +ψ =0 2

  2 ˆ + v +ψ ˙s H + Q˙ + W 2 k

˙k M

k=1

(d)

(iv) Uniform system In Eqs. a and b ˆ U = MU

(e)

Note: To obtain the energy balance on a molar basis, make the following substitutions: Replace

 M  ˙k M

v2 +ψ 2

with



2 ˆ + v +ψ H 2 ˆ MU



 v2 +ψ 2  2   v N˙ k H + m +ψ 2 k Nm

 k

NU

52 Chapter 3: Conservation of Energy The result is

  2   2   v v U +M − U +M +ψ +ψ 2 2 t2 t1  

K  t2 v2 ˙ ˆ +ψ Mk H + = dt + Q + W 2 t1 k

(3.1-6)

k=1

where

t2

Q= t1

Q˙ dt

t2

Ws =

˙ s dt W

V (t2 )

t2

P dV = V (t1 )

t1

P t1

dV dt dt

and

W = Ws −

V (t2 )

P dV V (t1 )

The first term on the right side of Eq. 3.1-6 is usually the most troublesome to evaluate because the mass flow rate and/or the thermodynamic properties of the flowing fluid may change with time. However, if the thermodynamic properties of the fluids entering and leaving the system are independent of time (even though the mass flow rate may depend on time), we have K  k=1

t2

t1

   t2 K  2 2  v v ˆ+ ˆ+ dt = +ψ +ψ M˙ k H M˙ k dt H 2 2 k k t1 k=1

=

K  k=1

  2 ˆ + v +ψ ΔMk H 2 k

(3.1-7)

If, on the other hand, the thermodynamic properties of the flow streams change with time in some arbitrary way, the energy balance of Eq. 3.1-6 may not be useful since it may not be possible to evaluate the integral. The usual procedure, then, is to try to choose a new system (or subsystem) for the description of the process in which these time-dependent flows do not occur or are more easily handled (see Illustration 3.4-5). Table 3.1-2 lists various special cases of Eq. 3.1-6 that will be useful in solving thermodynamic problems. For the study of thermodynamics it will be useful to have equations that relate the differential change in certain thermodynamic variables of the system to differential changes in other system properties. Such equations can be obtained from the differential form of the mass and energy balances. For processes in which kinetic and potential energy terms are unimportant, there is no shaft work, and there is only a single mass flow stream, these equations reduce to

dM = M˙ dt

3.1 Conservation of Energy 53 and

dU ˆ + Q˙ − P dV = M˙ H dt dt which can be combined to give

dU ˆ dM + Q˙ − P dV (3.1-8) =H dt dt dt ˆ is the enthalpy per unit mass entering or leaving the system. (Note that for a where H ˙ = 0.) Defining Q = Q˙ dt to be equal system closed to the flow of mass, dM/dt = M ˙ dt = to the heat flow into the system in the differential time interval dt, and dM = M (dM/dt) dt to be equal to the mass flow in that time interval, we obtain the following expression for the change of the internal energy in the time interval dt: ˆ dM + Q − P dV dU = H

(3.1-9a)

Table 3.1-2 Difference Form of the Energy Balance General equation

  2   2     K  t2  v v v2 ˙ ˆ Mk H + − U +M = dt + Q + W U +M +ψ +ψ +ψ 2 2 2 t2 t1 k k=1 t1

Special cases: (i) Closed system

  2   2   v v U +M +ψ +ψ − U +M =Q+W 2 2 t2 t1

(a)

(b)

and M (t1 ) = M (t2 )

(ii) Adiabatic process In Eqs. a and b (c)

Q=0

(iii) Open system, flow of fluids of constant thermodynamic properties K  

t2

M˙ k

t1

k=1

    K 2 2  ˆ + v +ψ ˆ + v +ψ H dt = ΔMk H 2 2 k k k=1

(d)

in Eq. a (iv) Uniform system

     2 2 v ˆ + v +ψ +ψ =M U U +M 2 2

in Eqs. a and b Note: To obtain the energy balance on a molar basis, make the following substitutions: Replace

 M 

t2 t1

2

v +ψ 2

  2 ˆ + v +ψ H dt 2 k   2 ˆ + v +ψ M U 2

˙k M

with





 v2 +ψ 2  2   v N˙ k H + m dt +ψ 2 k  2   v N U +m +ψ 2 Nm



t2 t1

(e)

54 Chapter 3: Conservation of Energy For a closed system this equation reduces to

dU = Q − P dV

(3.1-9b)

Since the time derivative operator d/dt is mathematically well defined, and the operator d is not, it is important to remember in using Eqs. 3.1-9a and b that they are abbreviations of Eq. 3.1-8. It is part of the traditional notation of thermodynamics to use dθ to indicate a differential change in the property θ, rather than the mathematically more correct dθ/dt.

3.2 SEVERAL EXAMPLES OF USING THE ENERGY BALANCE The energy balance equations developed so far in this chapter can be used for the description of any process. As the first step in using these equations, it is necessary to choose the system for which the mass and energy balances are to be written. The important fact for the student of thermodynamics to recognize is that processes occurring in nature are in no way influenced by our mathematical description of them. Therefore, if our descriptions are correct, they must lead to the same final result for the system and its surroundings regardless of which system choice is made. This is demonstrated in the following example, where the same result is obtained by choosing for the system first a given mass of material and then a specified region in space. Since the first system choice is closed and the second open, this illustration also establishes the way in ˙ is related to the closed-system work term which the open-system energy flow P V M P (dV /dt). Illustration 3.2-1 Showing That the Final Result Should Not Depend on the Choice of System A compressor is operating in a continuous, steady-state manner to produce a gas at temperature T2 and pressure P2 from one at T1 and P1 . Show that for the time interval Δt ˆ 1 ) ΔM ˆ2 − H Q + Ws = (H

where ΔM is the mass of gas that has flowed into or out of the system in the time Δt. Establish this result by (a) first writing the balance equations for a closed system consisting of some convenient element of mass, and then (b) by writing the balance equations for the compressor and its contents, which is an open system.

Solution a. The closed-system analysis Here we take as the system the gas in the compressor and the mass of gas ΔM , that will enter the compressor in the time interval Δt. The system is enclosed by dotted lines in the figure.

P1, T1

Compressor

Ws

Q

P2, T2

3.2 Several Examples of Using the Energy Balance 55 At the later time t + Δt, the mass of gas we have chosen as the system is as shown below.

P1, T1

P2, T2

Compressor

Ws

Q

We use the subscript c to denote the characteristics of the fluid in the compressor, the subscript 1 for the gas contained in the system that is in the inlet pipe at time t, and the subscript 2 for the gas in the system that is in the exit pipe at time t + Δt. With this notation the mass balance for the closed system is M2 (t + Δt) + Mc (t + Δt) = M1 (t) + Mc (t)

Since the compressor is in steady-state operation, the amount of gas contained within it and the properties of this gas are constant. Thus, Mc (t + Δt) = Mc (t) and M2 (t + Δt) = M1 (t) = ΔM

The energy balance for this system, neglecting the potential and kinetic energy terms (which, if retained, would largely cancel), is ˆ2 |t+Δt + Mc U ˆc |t+Δt − M1 U ˆ1 |t − Mc U ˆc |t = Ws + Q + P1 Vˆ1 M1 − P2 Vˆ2 M2 M2 U

(a)

In writing this equation we have recognized that the flow terms vanish for the closed system and  that there are two contributions of the P dV type, one due to the deformation of the system boundary at the compressor inlet and another at the compressor outlet. Since the inlet and exit pressures are constant at P1 and P2 , these terms are 

−

 P dV = −P1

 dV |inlet

− P2

dV |outlet

= −P1 {V1 (t + Δt) − V1 (t)} − P2 {V2 (t + Δt) − V2 (t)}

However, V1 (t + Δt) = 0 and V2 (t) = 0, so that 

−

P dV = +P1 V1 − P2 V2 = P1 Vˆ1 M1 − P2 Vˆ2 M2

Now, using the energy balance and Eq. a above, and recognizing that since the compressor is in steady-state operation, ˆc |t+Δt = Mc U ˆc |t Mc U

we obtain ˆ1 ) = Ws + Q + P1 Vˆ1 ΔM − P2 Vˆ2 ΔM ˆ2 − U ΔM (U

or ˆ1 − P1 Vˆ1 ) = ΔM (H ˆ 1 ) = Ws + Q ˆ2 + P2 Vˆ2 − U ˆ2 − H ΔM (U

56 Chapter 3: Conservation of Energy b. The open-system analysis Here we take the contents of the compressor at any time to be the system. The mass balance for this system over the time interval Δt is  M (t + Δt) − M (t) =



t+Δt

t+Δt

M˙ 1 dt + t

M˙ 2 dt = (M )1 + (M )2 t

and the energy balance is   2   2   v v U +M − U +M +ψ +ψ 2 2 t+Δt t  t+Δt  ˆ 1 + v 2 /2 + ψ1 ) dt + M˙ 1 (H = 1 t

t+Δt

˙ 2 (H ˆ 2 + v 2 /2 + ψ2 ) dt + Q + W M 2

t

These equations may be simplified as follows: 1. Since the compressor is operating continuously in a steady-state manner, its contents must, by definition, have the same mass and thermodynamic properties at all times. Therefore, M (t + Δt) = M (t)

and   2   2   v v U +M = U +M +ψ +ψ 2 2 t+Δt t

2. Since the thermodynamic properties of the fluids entering and leaving the turbine do not change in time, we can write 



t+Δt

t+Δt

ˆ 1 + v 2 /2 + ψ1 ) dt = (H ˆ 1 + v 2 /2 + ψ1 ) M˙ 1 (H 1 1 t

˙ 1 dt M t

ˆ 1 + v 2 /2 + ψ1 ) ΔM1 = (H 1

with a similar expression for the compressor exit stream. 3. Since the volume of the system here, the contents of the compressor, is constant 

V2

P dV = 0 V1

so that W = Ws

4. Finally, we will neglect the potential and kinetic energy changes of the entering and exiting fluids. With these simplifications, we have 0 = ΔM1 + ΔM2

or

ΔM1 = −ΔM2 = ΔM

and ˆ 1 + ΔM2 H ˆ 2 + Q + Ws 0 = ΔM1 H

Combining these two equations, we obtain ˆ 1 ) ΔM ˆ2 − H Q + Ws = (H

This is the same result as in part (a).

3.2 Several Examples of Using the Energy Balance 57 Comment Notice that in the closed-system analysis the surroundings are doing work on the system (the mass element) at the inlet to the compressor,  while the system is doing work on its surroundings at the outlet pipe. Each of these terms is a P dV –type work term. For the open system this work term has been included in the energy balance as a P Vˆ ΔM term, so that it is the enthalpy, rather than the internal energy, of the flow streams that appears in the equation. The explicit P dV term that does appear in the open-system energy balance represents only the work done if the system boundaries deform; for the choice of the compressor and its contents as the system here this term is zero unless the compressor (the boundary of our system) explodes.

This illustration demonstrates that the sum Q+Ws is the same for a fluid undergoing some change in a continuous process regardless of whether we choose to compute this sum from the closed-system analysis on a mass of gas or from an open-system analysis on a given volume in space. In Illustration 3.2-2 we consider another problem, the compression of a gas by two different processes, the first being a closed-system pistonand-cylinder process and the second being a flow compressor process. Here we will find that the sum Q + W is different in the two processes, but the origin of this difference is easily understood. Illustration 3.2-2 Showing that Processes in Closed and Open Systems Are Different A mass M of gas is to be compressed from a temperature T1 and a pressure P1 to T2 and P2 in (a) a one-step process in a frictionless piston and cylinder,4 and (b) a continuous process in which the mass M of gas is part of the feed stream to the compressor of the previous illustration. Compute the sum Q + W for each process.

Solution a. The piston-and-cylinder process

P1, T1

P2, T2

Here we take the gas within the piston and cylinder as the system. The energy balance for this closed system is ˆ1 ) = Q + W (piston-cylinder process) ˆ2 − U M (U  It is useful to note that Ws = 0 and W = − P dV .

b. The flow compressor process (see the figures in Illustration 3.2-1) If we take the contents of the compressor as the system and follow the analysis of the previous illustration, we obtain 

ˆ1) = Q + W ˆ2 − H M (H

(flow compressor)

where, since P dV = 0, W = Ws .

Comment From these results it is evident that the sum Q + W is different in the two cases, since the two processes are different. The origin of the difference in the flow and nonflow energy changes

4 Since

the piston is frictionless, the pressure of the gas is equal to the pressure applied by the piston.

58 Chapter 3: Conservation of Energy accompanying a change of state is easily identified by considering two different ways of compressing a mass M of gas in a piston and cylinder from (T1 , P1 ) to (T2 , P2 ). The first way is merely to compress the gas in situ. The sum of heat and work flows needed to accomplish the change of state is, from the preceding computations, ˆ1 ) ˆ2 − U Q + W = M (U

A second way to accomplish the compression is to open a valve at the side of the cylinder and use the piston movement (at constant pressure P1 ) to inject the gas into the compressor inlet stream, use the compressor to compress the gas, and then withdraw the gas from the compressor exit stream by moving the piston against a constant external pressure P2 . The energy required in the compressor stage is the same as that found above: ˆ1) ˆ2 − H (Q + W )c = M (H

To this we must add the work done in using the piston movement to pump the fluid into the compressor inlet stream,  W1 =

P dV = P1 V1 = P1 Vˆ1 M

(this is the work done by the system on the gas in the inlet pipe to the compressor), and subtract the work obtained as a result of the piston movement as the cylinder is refilled,  W2 = − P dV = −P2 V2 = −P2 Vˆ2 M

(this is the work done on the system by the gas in the compressor exit stream). Thus the total energy change in the process is Q + W = (Q + W )c + W1 + W2 ˆ 1 ) + P1 Vˆ1 M − P2 Vˆ2 M = M (U ˆ1 ) ˆ2 − H ˆ2 − U = M (H

which is what we found in part (a). Here, however, it results from the sum of an energy requireˆ 1 ) in the flow compressor and the two pumping terms. ˆ2 − H ment of M (H

Consider now the problem of relating the downstream temperature and pressure of a gas in steady flow across a flow constriction (e.g., a valve, orifice, or porous plug) to its upstream temperature and pressure. Illustration 3.2-3 A Joule-Thomson or Isenthalpic Expansion A gas at pressure P1 and temperature T1 is steadily exhausted to the atmosphere at pressure P2 through a pressure-reducing valve. Find an expression relating the downstream gas temperature T2 to P1 , P2 , and T1 . Since the gas flows through the valve rapidly, one can assume that there is no heat transfer to the gas. Also, the potential and kinetic energy terms can be neglected.

T1, P1

T2 = ?, P2

Solution The flow process is schematically shown in the figure. We will consider the region of space that includes the flow obstruction (indicated by the dashed line) to be the system, although, as in Illustration 3.2-1, a fixed mass of gas could have been chosen as well. The pressure of the

3.3 The Thermodynamic Properties of Matter 59 gas exiting the reducing valve will be P2 , the pressure of the surrounding atmosphere. (It is not completely obvious that these two pressures should be the same. However, in the laboratory we find that the velocity of the flowing fluid will always adjust in such a way that the fluid exit pressure and the pressure of the surroundings are equal.) Now recognizing that our system (the valve and its contents) is of constant volume, that the flow is steady, and that there are no heat or work flows and negligible kinetic and potential energy changes, the mass and energy balances (on a molar basis) yield 0 = N˙ 1 + N˙ 2

or

N˙ 2 = −N˙ 1

and 0 = N˙ 1 H 1 + N˙ 2 H 2 = N˙ 1 (H 1 − H 2 )

Thus Isenthalpic or Joule-Thomson expansion

H1 = H2

or, to be explicit, H(T1 , P1 ) = H(T2 , P2 )

or

ˆ 1 , P1 ) = H(T ˆ 2 , P2 ) H(T

so that the initial and final states of the gas have the same enthalpy. Consequently, if we knew how the enthalpy of the gas depended on its temperature and pressure, we could use the known values of T1 , P1 , and P2 to determine the unknown downstream temperature T2 .

Comments 1. The equality of enthalpies in the upstream and downstream states is the only information we get from the thermodynamic balance equations. To proceed further we need constitutive information, that is, an equation of state or experimental data interrelating H , T , and P. Equations of state are discussed in the following section and in much of Chapter 6. 2. The experiment discussed in this illustration was devised by William Thomson (later Lord Kelvin) and performed by J. P. Joule to study departures from ideal gas behavior. The Joule-Thomson expansion, as it is called, is used in the liquefaction of gases and in refrigeration processes (see Chapter 5).

3.3 THE THERMODYNAMIC PROPERTIES OF MATTER The balance equations of this chapter allow one to relate the mass, work, and heat flows of a system to the change in its thermodynamic state. From the experimental observations discussed in Chapter 1, the change of state for a single-component, single-phase system can be described by specifying the initial and final values of any two independent intensive variables. However, certain intensive variables, especially temperature and pressure, are far easier to measure than others. Consequently, for most problems we will want to specify the state of a system by its temperature and pressure rather than by its specific volume, internal energy, and enthalpy, which appear in the energy balance. What are needed, then, are interrelations between the fluid properties that allow one to eliminate some thermodynamic variables in terms of other, more easily measured ones. Of particular interest is the volumetric equation of state, which is a relation between temperature, pressure, and specific volume, and the thermal equation of state, which is usually either in the form of a relationship between internal energy, temperature, and specific (or molar) volume, or between specific or molar enthalpy,

60 Chapter 3: Conservation of Energy temperature, and pressure. Such information may be available in either of two forms. First, there are analytic equations of state, which provide an algebraic relation between the thermodynamic state variables. Second, experimental data, usually in graphical or tabular form, may be available to provide the needed interrelationships between the fluid properties. Equations of state for fluids are considered in detail in Chapter 6. To illustrate the use of the mass and energy balance equations in a simple form, we briefly consider here the equation of state for the ideal gas and the graphical and tabular display of the thermodynamic properties of several real fluids. An ideal gas is a gas at such a low pressure that there are no interactions among its molecules. For such gases it is possible to show, either experimentally or by the methods of statistical mechanics, that at all absolute temperatures and pressures the volumetric equation of state is

P V = RT (3.3-1) (as indicated in Sec. 1.4) and that the enthalpy and internal energy are functions of temperature only (and not pressure or specific volume). We denote this latter fact by H = H(T ) and U = U (T ). This simple behavior is to be compared with the enthalpy for a real fluid, which is a function of temperature and pressure [i.e., H = H(T, P )] and the internal energy, which is usually written as a function of temperature and specific volume [U = U (T, V )], as will be discussed in Chapter 6. The temperature dependence of the internal energy and enthalpy of all substances (not merely ideal gases) can be found by measuring the temperature rise that accompanies a heat flow into a closed stationary system. If a sufficiently small quantity of heat is added to such a system, it is observed that the temperature rise produced, ΔT , is linearly related to the heat added and inversely proportional to N , the number of moles in the system: Q = CΔT = C {T (t2 ) − T (t1 )} N where C is a parameter and Q is the heat added to the system between the times t1 and t2 . The object of the experiment is to accurately measure the parameter C for a very small temperature rise, since C generally is also a function of temperature. If the measurement is made at constant volume and with Ws = 0, we have, from the energy balance and the foregoing equation, U (t2 ) − U (t1 ) = Q = N CV {T (t2 ) − T (t1 )} Thus

U (t2 ) − U (t1 ) U (t2 ) − U (t1 ) = N {T (t2 ) − T (t1 )} T (t2 ) − T (t1 ) where the subscript V has been introduced to remind us that the parameter C was determined in a constant-volume experiment. In the limit of a very small temperature difference, we have CV =

Constant-volume heat capacity definition

U (t2 ) − U (t1 ) CV (T, V ) = lim = T (t2 )−T (t1 )→0 T (t2 ) − T (t1 )



∂U ∂T



 = V

∂U (T, V ) ∂T

 V

(3.3-2)

3.3 The Thermodynamic Properties of Matter 61 so that the measured parameter CV is, in fact, equal to the temperature derivative of the internal energy at constant volume. Similarly, if the parameter C is determined in a constant-pressure experiment, we have

Q = U (t2 ) − U (t1 ) + P {V (t2 ) − V (t1 )} = U (t2 ) + P (t2 )V (t2 ) − U (t1 ) − P (t1 )V (t1 ) = H(t2 ) − H(t1 ) = N CP {T (t2 ) − T (t1 )} where we have used the fact that since pressure is constant, P = P (t2 ) = P (t1 ). Then

 Constant-pressure heat capacity definition

CP (T, P ) =

∂H ∂T



 = P

∂H(T, P ) ∂T

 (3.3-3) P

so that the measured parameter here is equal to the temperature derivative of the enthalpy at constant pressure. The quantity CV is called the constant-volume heat capacity, and CP is the constant-pressure heat capacity; both appear frequently throughout this book. Partial derivatives have been used in Eqs. 3.3-2 and 3.3-3 to indicate that although the internal energy is a function of temperature and density or specific (or molar) volume, CV has been measured along a path of constant volume; and although the enthalpy is a function of temperature and pressure, CP has been evaluated in an experiment in which the pressure was held constant. For the special case of the ideal gas, the enthalpy and internal energy of the fluid are functions only of temperature. In this case the partial derivatives above become total derivatives, and dH dU CP∗ (T ) = (3.3-4) and CV∗ (T ) = dT dT so that the ideal gas heat capacities, which we denote using asterisks as CP∗ (T ) and CV∗ (T ), are only functions of T as well. The temperature dependence of the ideal gas heat capacity can be measured or, in some cases, computed using the methods of statistical mechanics and detailed information about molecular structure, bond lengths, vibrational frequencies, and so forth. For our purposes CP∗ (T ) will either be considered to be a constant or be written as a function of temperature in the form Ideal gas heat capacity

CP∗ (T ) = a + bT + cT 2 + dT 3 + · · ·

(3.3-5)

Since H = U + P V , and for the ideal gas P V = RT , we have H = U + RT and

CP∗ (T ) =

dH d(U + RT ) = = CV∗ (T ) + R dT dT

so that CV∗ (T ) = CP∗ (T ) − R = (a − R) + bT + cT 2 + dT 3 + · · · . The constants in Eq. 3.3-5 for various gases are given in Appendix A.II. The enthalpy and internal energy of an ideal gas at a temperature T2 can be related to their values at T1 by integration of Eqs. 3.3-4 to obtain

H IG (T2 ) = H IG (T1 ) +

T2

T1

CP∗ (T ) dT

62 Chapter 3: Conservation of Energy and

U IG (T2 ) = U IG (T1 ) +

T2

CV∗ (T ) dT

(3.3-6)

T1

where the superscript IG has been introduced to remind us that these equations are valid only for ideal gases. Our interest in the first part of this book is with energy flow problems in singlecomponent systems. Since the only energy information needed in solving these problems is the change in internal energy and/or enthalpy of a substance between two states, and since determination of the absolute energy of a substance is not possible, what is done is to choose an easily accessible state of a substance to be the reference state, for which H IG is arbitrarily set equal to zero, and then report the enthalpy and internal energy of all other states relative to this reference state. (That there is a state for each substance for which the enthalpy has been arbitrarily set to zero does lead to difficulties when chemical reactions occur. Consequently, another energy convention is introduced in Chapter 8.) If, for the ideal gas, the temperature TR is chosen as the reference temperature (i.e., H IG at TR is set equal to zero), the enthalpy at temperature T is then

T IG H (T ) = CP∗ (T ) dT (3.3-7) TR

Similarly, the internal energy at T is

T

∗

U (T ) = U (TR ) + CV (T ) dT = {H (TR ) − RTR } + TR

T CV∗ (T ) dT − RTR = IG

IG

T

IG

CV∗ (T ) dT

TR

TR

(3.3-8) One possible choice for the reference temperature TR is absolute zero. In this case,

T

T IG IG ∗ H (T ) = CP (T ) dT and U (T ) = CV∗ (T ) dT 0

0

However, to use these equations, heat capacity data are needed from absolute zero to the temperature of interest. These data are not likely to be available, so a more convenient reference temperature, such as 0◦ C, is frequently used. For the special case in which the constant-pressure and constant-volume heat capacities are independent of temperature, we have, from Eqs. 3.3-7 and 3.3-8,

H IG (T ) = CP∗ (T − TR )

(3.3-7  )

U IG (T ) = CV∗ (T − TR ) − RTR = CV∗ T − CP∗ TR

(3.3-8  )

and

which, when TR is taken to be absolute zero, simplify further to

H IG (T ) = CP∗ T

and U IG (T ) = CV∗ T

3.3 The Thermodynamic Properties of Matter 63 As we will see in Chapters 6 and 7, very few fluids are ideal gases, and the mathematics of relating the enthalpy and internal energy to the temperature and pressure of a real fluid is much more complicated than indicated here. Therefore, for fluids of industrial and scientific importance, detailed experimental thermodynamic data have been collected. These data can be presented in tabular form (see Appendix A.III for a table of the thermodynamic properties of steam) or in graphical form, as in Figs. 3.3-1 to 3.3-4 for steam, methane, nitrogen, and the environmentally friendly refrigerant HFC134a. (Can you identify the reference state for the construction of the steam tables in Appendix A.III?) With these detailed data one can, given values of temperature and pressure, easily find the enthalpy, specific volume, and entropy (a thermodynamic quantity that is introduced in the next chapter). More generally, given any two intensive variables of a single-component, single-phase system, the remaining properties can be found. Notice that different choices have been made for the independent variables in these figures. Although the independent variables may be chosen arbitrarily,5 some choices are especially convenient for solving certain types of problems. Thus, as we will see, an enthalpy-entropy (H -S ) or Mollier diagram,6 such as Fig. 3.3-1a, is useful for problems involving turbines and compressors; enthalpy-pressure (H -P ) diagrams (for example Figs. 3.3-2 to 3.3-4) are useful in solving refrigeration problems; and temperatureentropy (T -S ) diagrams, of which Fig. 3.3-1b is an example, are used in the analysis of engines and power and refrigeration cycles (see Sec. 3.4 and Chapter 5). An important characteristic of real fluids is that at sufficiently low temperatures they condense to form liquids and solids. Also, many applications of thermodynamics of interest to engineers involve either a range of thermodynamic states for which the fluid of interest undergoes a phase change, or equilibrium multiphase mixtures (e.g., steam and water at 100◦ C and 101.3 kPa). Since the energy balance equation is expressed in terms of the internal energy and enthalpy per unit mass of the system, this equation is valid regardless of which phase or mixture of phases is present. Consequently, there is no difficulty, in principle, in using the energy balance (or other equations to be introduced later) for multiphase or phase-change problems, provided thermodynamic information is available for each of the phases present. Figures 3.3-1 to 3.3-4 and the steam tables in Appendix A.III provide such information for the vapor and liquid phases and, within the dome-shaped region, for vapor-liquid mixtures. Similar information for many other fluids is also available. Thus, you should not hesitate to apply the equations of thermodynamics to the solution of problems involving gases, liquids, solids, and mixtures thereof. There is a simple relationship between the thermodynamic properties of a two-phase mixture (e.g., a mixture of water and steam), the properties of the individual phases, and the mass distribution between the phases. If θˆ is any intensive property, such as internal energy per unit mass or volume per unit mass, its value in an equilibrium two-phase mixture is Properties of a two-phase mixture: the lever rule

θˆ = ω I θˆI + ω II θˆII = ω I θˆI + (1 − ω I )θˆII

(3.3-9)

Here ω I is the mass fraction of the system that is in phase I, and θˆI is the value of the variable in that phase. Also, by definition of the mass fraction, ω I + ω II = 1. 5 See,

however, the comments made in Sec. 1.6 concerning the use of the combinations U and T , and V and T as the independent thermodynamic variables. 6 Entropy, denoted by the letter S, is a thermodynamic variable to be introduced in Chapter 4.

64 Chapter 3: Conservation of Energy

(a)

Figure 3.3-1 (a) Enthalpy-entropy of Mollier diagram for steam. [Source: ASME Steam Tables in SI (Metric) Units for Instructional Use, American Society of Mechanical Engineers, New York, 1967. Used with permission.] (This figure appears as an Adobe PDF file on the website for this book, and may be enlarged and printed for easier reading and for use in solving problems.)

3.3 The Thermodynamic Properties of Matter 65

(b)

Figure 3.3-1(b) Temperature-entropy diagram for steam. [Source: J. H. Keenan, F. G. Keyes, P. G. Hill, and J. G. Moore, Steam Tables (International Edition—Metric Units). Copyright 1969. John Wiley & Sons, Inc., New York. Used with permission.] (This figure appears as an Adobe PDF file on the website for this book, and may be enlarged and printed for easier reading and for use in solving problems.)

66 Chapter 3: Conservation of Energy

Figure 3.3-2 Pressure-enthalpy diagram for methane. (Source: W. C. Reynolds, Thermodynamic Properties in SI, Department of Mechanical Engineering, Stanford University, Stanford, CA, 1979. Used with permission.) (This figure appears as an Adobe PDF file on the website for this book, and may be enlarged and printed for easier reading and for use in solving problems.)

(For mixtures of steam and water, the mass fraction of steam is termed the quality and is frequently expressed as a percentage, rather than as a fraction; for example, a steam-water mixture containing 0.02 kg of water for each kg of mixture is referred to as steam of 98 percent quality.) Note that Eq. 3.3-9 gives the property of a two-phase mixture as a linear combination of the properties of each phase weighted by its mass fraction. Consequently, if charts such as Figures 3.3-1 to 3.3-4 are used to obtain the thermodynamic properties, the properties of the two-phase mixture will fall along a line connecting the properties of the individual phases, which gives rise to referring to Eq. 3.3-9 as the lever rule. If a mixture consists of two phases (i.e., vapor and liquid, liquid and solid, or solid and vapor), the two phases will be at the same temperature and pressure; however, other properties of the two phases will be different. For example, the specific volume of the vapor and liquid phases can be very different, as will be their internal energy and enthalpy, and this must be taken into account in energy balance calculations. The notation that will be used in this book is as follows:

ˆ =H ˆV − H ˆ L = enthalpy of vaporization per unit mass or on a molar basis, vap H vap H = H V − H L = molar enthalpy of vaporization Also (but for brevity, only on a molar basis), fus H = H L − H S = molar enthalpy of melting or fusion sub H = H V − H S = molar enthalpy of sublimation

3.3 The Thermodynamic Properties of Matter 67

Figure 3.3-3 Pressure-enthalpy diagram for nitrogen. (Source: W. C. Reynolds, Thermodynamic Properties in SI, Department of Mechanical Engineering, Stanford University, Stanford, CA, 1979. Used with permission.) (This figure appears as an Adobe PDF file on the website for this book, and may be enlarged and printed for easier reading and for use in solving problems.)

Similar expressions can be written for the volume changes and internal energy changes on a phase change. It is also useful to note that several simplifications can be made in computing the thermodynamic properties of solids and liquids. First, because the molar volumes of condensed phases are small, the product P V can be neglected unless the pressure is high. Thus, for solids and liquids, Solids or liquids at low pressure

H ≈U

A further simplification commonly made for liquids and solids is to assume that they are also incompressible; that is, their volume is only a function of temperature, so that

 Idealized incompressible fluid or solid

(3.3-10)

∂V ∂P

 =0

(3.3-11)

T

In Chapter 6 we show that for incompressible fluids, the thermodynamic properties U , CP , and CV are functions of temperature only. Since, in fact, solids, and most liquids away from their critical point (see Chapter 7) are relatively incompressible, Eqs. 3.3-10 and 3.3-11, together with the assumption that these properties depend only on temperature, are reasonably accurate and often used in thermodynamic studies involving liquids and solids. Thus, for example, the internal energy of liquid water at a temperature T1 and pressure P1 is, to a very good approximation, equal to the internal

68 Chapter 3: Conservation of Energy

Figure 3.3-4 Pressure–enthalpy diagram for HFC-134a. (Used with permission of DuPont Fluoroproducts.) (This figure appears as an Adobe PDF file on the website for this book, and may be enlarged and printed for easier reading and for use in solving problems.)

energy of liquid water at the temperature T1 and any other pressure. Consequently, the entries for the internal energy of liquid water for a variety of temperatures (at pressures corresponding to the vapor-liquid coexistence or saturation pressures at each temperature) given in the saturation steam tables of Appendix A.III can also be used for the internal energy of liquid water at these same temperatures and higher pressures. Although we will not try to quantitatively relate the interactions between molecules to their properties—that is the role of statistical mechanics, not thermodynamics—it is useful to make some qualitative observations. The starting point is that the interactions between a pair of simple molecules (for example, argon or methane) depend on the separation distance between their centers of mass, as shown in Fig. 3.3-5. There we see that if the molecules are far apart, the interaction energy is very low, and it vanishes at infinite separation. As the molecules are brought closer together, they attract each other, which decreases the energy of the system. However, if the molecules are brought too close together (so that their electrons overlap), the molecules repel each other, resulting in a positive energy that increases rapidly as one attempts to bring the molecules closer. We can make the following observations from this simple picture of molecular interactions. First, if the molecules are widely separated, as occurs in a dilute gas, there will be no energy of interaction between the molecules; this is the case of an ideal gas. Next, as the density increases, and the molecules are somewhat closer together, molecular attractions become more important, and the energy of the system decreases. Next, at liquid densities, the average distance between the molecules will be near the deepest (most attractive) part of the interaction energy curve shown in Fig. 3.3-5. (Note that the molecules will not be at the very lowest value of the interaction energy curve as a result of their thermal motion, and because the behavior of a large collection of

Interaction energy

3.4 Applications of the Mass and Energy Balances 69

+

0 −

Separation distance

Figure 3.3-5 The interaction energy between two molecules as a function of their separation distance. Since the molecules cannot overlap, there is a strong repulsion (positive interaction energy) at small separation distances. At larger separation distances the interactions between the electrons result in an attraction between the molecules (negative interaction energy), which vanishes at very large separations.

molecules is more complicated than can be inferred by examining the interaction between only a pair of molecules.) Consequently, a liquid has considerably less internal energy than a gas. The energy that must be added to a liquid to cause its molecules to move farther apart and vaporize is the heat of vaporization Δvap H . In solids, the molecules generally are located very close to the minimum in the interaction energy function in an ordered lattice, so that a solid has even less internal energy than a liquid. The amount of energy required to slightly increase the separation distances between the molecules in a solid and form a liquid is the heat of melting or the heat of fusion Δfus H . As mentioned earlier, the constant-pressure heat capacity of solids is a function of temperature; in fact, CP goes to zero at the absolute zero of temperature and approaches a constant at high temperatures. An approximate estimate for CP of solids for temperatures of interest to chemical engineers comes from the empirical law (or observation) of DuLong and Petit that

CP = 3N R = 24.942N

J mol K

(3.3-12)

where N is the number of atoms in the formula unit. For comparison, the constantpressure heat capacities of lead, gold, and aluminum at 25◦ C are 26.8, 25.2, and 24.4 J/(mol K), respectively. Similarly, Eq. 3.3-12 gives a prediction of 49.9 for gallium arsenide (used in the electronics industry), which is close to the measured value of 47.0 J/mol K. For Fe3 C the prediction is 99.8 J/mol K and the measured value is 105.9. So we see that the DuLong-Petit law gives reasonable though not exact values for the heat capacities of solids.

3.4 APPLICATIONS OF THE MASS AND ENERGY BALANCES In many thermodynamics problems one is given some information about the initial equilibrium state of a substance and asked to find the final state if the heat and work flows are specified, or to find the heat or work flows accompanying the change to a specified final state. Since we use thermodynamic balance equations to get the information

70 Chapter 3: Conservation of Energy needed to solve this sort of problem, the starting point is always the same: the identification of a convenient thermodynamic system. The main restriction on the choice of a system is that the flow terms into and out of the system must be of a simple form—for example, not varying with time, or perhaps even zero. Next, the forms of the mass and energy balance equations appropriate to the system choice are written, and any information about the initial and final states of the system and the flow terms is used. Finally, the thermal equation of state is used to replace the internal energy and enthalpy in the balance equations with temperature, pressure, and volume; the volumetric equation of state may then be used to eliminate volume in terms of temperature and pressure. In this way equations are obtained that contain temperature and pressure as the only state variables. The volumetric equation of state may also provide another relationship between the temperature, pressure, mass, and volume when the information about the final state of the system is presented in terms of total volume, rather than volume per unit mass or molar volume (see Illustration 3.4-5). By using the balance equations and the equation-of-state information, we will frequently be left with equations that contain only temperature, pressure, mass, shaft work (Ws ), and heat flow (Q). If the number of equations equals the number of unknowns, the problem can be solved. The mass and energy balance equations, together with equationof-state information, are sufficient to solve many, but not all, energy flow problems. In some situations we are left with more unknowns than equations. In fact, we can readily identify a class of problems of this sort. The mass and energy balance equations together can, at most, yield new information about only one intensive variable of the system (the internal energy or enthalpy per unit mass) or about the sum of the heat and work flows if only the state variables are specified. Therefore, we are not, at present, able to solve problems in which (1) there is no information about any intensive variable of the final state of the system, (2) both the heat flow (Q) and the shaft work (Ws ) are unspecified, or (3) one intensive variable of the final state and either Q or Ws are unknown, as in Illustration 3.4-4. To solve these problems, an additional balance equation is needed; such an equation is developed in Chapter 4. The seemingly most arbitrary step in thermodynamic problem solving is the choice of the system. Since the mass and energy balances were formulated with great generality, they apply to any choice of system, and, as was demonstrated in Illustration 3.2-1, the solution of a problem is independent of the system chosen in obtaining the solution. However, some system choices may result in less effort being required to obtain a solution. This is demonstrated here and again in Chapter 4. Illustration 3.4-1 Joule-Thomson Calculation Using a Mollier Diagram and Steam Tables Steam at 400 bar and 500◦ C undergoes a Joule-Thomson expansion to 1 bar. Determine the temperature of the steam after the expansion using a. Fig. 3.3-1a b. Fig. 3.3-1b c. The steam tables in Appendix A.III

Solution (Since only one thermodynamic state variable—here the final temperature—is unknown, from the discussion that precedes this illustration we can expect to be able to obtain a solution to this problem.)

3.4 Applications of the Mass and Energy Balances 71 We start from Illustration 3.2-3, where it was shown that ˆ 1 = H(T ˆ 1 , P1 ) = H(T ˆ 2 , P2 ) = H ˆ2 H ˆ 1 can be found from either for a Joule-Thomson expansion. Since T1 and P1 are known, H ˆ 2 (= H ˆ 1 , from the foregoing) and P2 are known, T2 Fig. 3.3-1 or the steam tables. Then, since H can be found.

a. Using Fig. 3.3-1a, the Mollier diagram, we first locate the point P = 400 bar = 40 000 kPa ˆ 1 = 2900 kJ/kg. Following a line of constant and T = 500◦ C, which corresponds to H enthalpy (a horizontal line on this diagram) to P = 1 bar = 100 kPa, we find that the final temperature is about 214◦ C. b. Using Fig. 3.3-1b, we locate the point P = 400 bar and T = 500◦ C (which is somewhat easier to do than it was using Fig. 3.3-1a) and follow the curved line of constant enthalpy to a pressure of 1 bar to see that T2 = 214◦ C. c. Using the steam tables of Appendix A.III, we have that at P = 400 bar = 40 MPa and ˆ = 2903.3 kJ/kg. At P = 1 bar = 0.1 MPa, H ˆ = 2875.3 kJ/kg at T = 200◦ C T = 500◦ C, H ◦ ˆ and H = 2974.3 kJ/kg at T = 250 C. Assuming that the enthalpy varies linearly with temperature between 200 and 250◦ C at P = 1 bar, we have by interpolation T = 200 + (250 − 200) ×

2903.3 − 2875.3 = 214.1◦ C 2974.3 − 2875.3



R The Aspen Plus simulation for this illustration available on the Wiley website for this book in the folder Aspen Illustrations>Chapter 3>3.4-1. The results using the IAPWS-95 model (equations on which the steam tables are based) give the following results:

exit temperature = 488.85 K = 215.7◦ C Using the more approximate Peng-Robinson equation of state (discussed in Chapter 5), the result is exit temperature = 484.87 K = 211.7◦ C This result

is in reasonable agreement with that obtained with the more accurate IAPWS-95 model.

Comment For many problems a graphical representation of thermodynamic data, such as Figure 3.3-1, is easiest to use, although the answers obtained are approximate and certain parts of the graphs may be difficult to read accurately. The use of tables of thermodynamic data, such as the steam tables, generally leads to the most accurate answers; however, one or more interpolations may be required. For example, if the initial conditions of the steam had been 475 bar and 530◦ C instead of 400 bar and 500◦ C, the method of solution using Fig. 3.3-1 would be unchanged; however, using the steam tables, we would have to interpolate with respect to both temperature and pressure to get the initial enthalpy of the steam. One way to do this is first, by interpolation between temperatures, to obtain the enthalpy of steam at 530◦ C at both 400 bar = 40 MPa and 500 bar. Then, by interpolation with respect to pressure between these two values, we obtain the enthalpy at 475 bar. That is, from ˆ (40 MPa, 500◦ C) = 2903.3 kJ/kg H ˆ (40 MPa, 550◦ C) = 3149.1 kJ/kg H

ˆ (50 MPa, 500◦ C) = 2720.1 kJ/kg H ˆ (50 MPa, 550◦ C) = 3019.5 kJ/kg H

and the interpolation formula Θ(x + Δ) = Θ(x) + Δ

Θ(y) − Θ(x) y−x

72 Chapter 3: Conservation of Energy where Θ is any tabulated function, and x and y are two adjacent values at which Θ is available, we have ◦ ◦ ˆ ˆ ˆ (40 MPa, 530◦ C) = H ˆ (40 MPa, 500◦ C) + 30 × H (40 MPa, 550 C) − H (40 MPa, 500 C) H 550 − 500 3149.1 − 2903.3 = 2903.3 + 147.5 = 2903.3 + 30 × 50 = 3050.8 kJ/kg

and ˆ (50 MPa, 530◦ C) = 2720.1 + 30 × 3019.5 − 2720.1 = 2899.7 kJ/kg H 50

Then ◦ ◦ ˆ ˆ ˆ (47.5 MPa, 530◦ C) = H ˆ (40 MPa, 530◦ C) + 7.5 × H (50 MPa, 530 C) − H (40 MPa, 530 C) H 50 − 40 7.5 × (2881.7 − 3050.8) = 2924.0 kJ/kg = 3050.8 + 10

Illustration 3.4-2 Application of the Complete Energy Balance Using the Steam Tables An adiabatic steady-state turbine is being designed to serve as an energy source for a small electrical generator. The inlet to the turbine is steam at 600◦ C and 10 bar, with a mass flow rate of 2.5 kg/s through an inlet pipe that is 10 cm in diameter. The conditions at the turbine exit are T = 400◦ C and P = 1 bar. Since the steam expands through the turbine, the outlet pipe is 25 cm in diameter. Estimate the rate at which work can be obtained from this turbine.

T1 = 600°C P1 = 10 bar

T2 = 400°C P2 = 1 bar

Ws

Solution (This is another problem in which there is only a single thermodynamic unknown, the rate at which work is obtained, so we can expect to be able to solve this problem.) The first step in solving any energy flow problem is to choose the thermodynamic system; the second step is to write the balance equations for the system. Here we take the turbine and its contents to be the system. The mass and energy balance equations for this adiabatic, steady-state system are dM = 0 = M˙ 1 + M˙ 2 dt

(a)

and d dt

    2  2 2 v ˆ 1 + v1 + M ˆ 2 + v2 + W ˙2 H ˙s U +M + gh = 0 = M˙ 1 H 2 2 2

(b)

3.4 Applications of the Mass and Energy Balances 73 In writing these equations we have set the rate of change of mass and energy equal to zero because the turbine is in steady-state operation; Q is equal to zero because the process is adiabatic, and P (dV /dt) is equal to zero because the volume of the system is constant (unless the turbine explodes). Finally, since the schematic diagram indicates that the turbine is positioned horizontally, we have assumed there is no potential energy change in the flowing steam. ˆ1, H ˆ2, W ˙ s , v1 , and v2 —in Eqs. a and b. However, both veThere are six unknowns—M˙ 2 , H locities will be found from the mass flow rates, pipe diameters, and volumetric equation-of-state information (here the steam tables in Appendix A.III). Also, thermal equation-of-state information (again the steam tables in Appendix A.III) relates the enthalpies to temperature and pressure, ˆ 2 are the only real unknowns, and these may be found both of which are known. Thus M˙ 2 and H from the balance equations above. From the mass balance equation, we have M˙ 2 = −M˙ 1 = −2.5 kg/s

From the steam tables or, less accurately from Fig. 3.3-1, we have ˆ 1 = 3697.9 kJ/kg H Vˆ1 = 0.4011 m3 /kg

ˆ 2 = 3278.2 kJ/kg H Vˆ2 = 3.103 m3 /kg

The velocities at the inlet and outlet to the turbine are calculated from Volumetric flow rate = M˙ Vˆ =

πd2 v 4

where d is the pipe diameter. Therefore, kg m3 · 0.4011 m s kg = 127.7 2 3.14159 · (0.1 m) s 4M˙ 2 Vˆ2 m v2 = = 158.0 πd2out s

˙ 1 Vˆ1 4M v1 = = πd2in

4 · 2.5

Therefore, the energy balance yields   2 2 ˙ s = −M˙ 2 H ˙1 H ˆ 2 + v2 − M ˆ 1 + v1 W 2 2   kg 1 ˆ 2 ) + (v 2 − v 2 ) kJ ˆ1 − H (H = −2.5 1 2 s 2 kg ⎧ ⎫ J ⎪ ⎪ ⎪ ⎪ 1 ⎨ 2 kJ kg 1 1 kJ ⎬ kg 2 2 m 419.7 · + (127.7 − 158.0 ) 2 · = −2.5 m2 1000 J ⎪ s ⎪ kg 2 s ⎪ ⎪ ⎩ ⎭ s2 kg kJ kJ = −2.5 = −1038.5 (= −1329 hp) {419.7 − 4.3} s kg s

R The Aspen Plus simulation for this illustration available on the Wiley website for this book in the folder Aspen Illustrations>Chapter 3>3.4-2. The items to notice in that solution are:

1. The isentropic efficiency of the turbine had to be adjusted by trial-and-error to 57.85% to achieve the exit conditions of 400◦ C and 1 bar R 2. Kinetic and potential energy changes are not easily taken into account using Aspen Plus . However, they are small for this case, as already shown and can be neglected

74 Chapter 3: Conservation of Energy R result of 1049.85 kJ/s is in quite good agreement with the result of 3. The Aspen Plus 1038.5 kJ/s this illustration. The difference of 1% is a result of the differences in the values of the thermodynamic properties in the steam tables and the properties equations used in  R Aspen Plus .

Comment If we had completely neglected the kinetic energy terms in this calculation, the error in the work term would be 4.3 kJ/kg, or about 1%. Generally, the contribution of kinetic and potential energy terms can be neglected when there is a significant change in the fluid temperature, as was suggested in Sec. 3.2.

Illustration 3.4-3 Use of Mass and Energy Balances with an Ideal Gas A compressed-air tank is to be repressurized to 40 bar by being connected to a high-pressure line containing air at 50 bar and 20◦ C. The repressurization of the tank occurs so quickly that the process can be assumed to be adiabatic; also, there is no heat transfer from the air to the tank. For this illustration, assume air to be an ideal gas with CV∗ = 21 J/(mol K). a. If the tank initially contains air at 1 bar and 20◦ C, what will be the temperature of the air in the tank at the end of the filling process? b. After a sufficiently long period of time, the gas in the tank is found to be at room temperature (20◦ C) because of heat exchange with the tank and the atmosphere. What is the new pressure of air in the tank? 50 bar 20°C

Solution (Each of these problems contains only a single unknown thermodynamic property, so solutions should be possible.) a. We will take the contents of the tank to be the system. The difference form of the mass (or rather mole) and energy balances for this open system are N2 − N1 = ΔN N2 U 2 − N1 U 1 = (ΔN )H in

(a) (b)

In writing the energy balance we have made the following observations: 1. The kinetic and potential energy terms are small and can be neglected. 2. Since the tank is connected to a source of gas at constant temperature and pressure, H in is constant.

3.4 Applications of the Mass and Energy Balances 75 3. The initial process is adiabatic, so Q = 0, and the system (the contents of the tank) is of constant volume, so ΔV = 0. Substituting Eq. a in Eq. b, and recognizing that for the ideal gas H(T ) = CP∗ (T − TR ) and U (T ) = CV∗ (T − TR ) − RTR , yields ∗ ∗ N2 {CV (T2 − TR ) − RTR } − N1 {CV (T1 − TR ) − RTR } = (N2 − N1 )CP∗ (Tin − TR )

or ∗ ∗ N2 CV T2 − N1 CV T1 = (N2 − N1 )CP∗ Tin

(Note that the reference temperature TR cancels out of the equation, as it must, since the final result cannot depend on the arbitrarily chosen reference temperature.) Finally, using the ideal gas equation of state to eliminate N1 and N2 , and recognizing that V1 = V2 , yields P1 C∗ P2 = + V∗ T2 T1 CP



P2 − P1 Tin

 or

T2 =

C∗ P1 + V∗ T1 CP

P2   P2 − P1 Tin

The only unknown in this equation is T2 , so, formally, the problem is solved. The answer is T = 405.2 K = 132.05◦ C. Before proceeding to the second part of the problem, it is interesting to consider the case in which the tank is initially evacuated. Here P1 = 0, and so T2 =

CP∗ ∗ Tin CV

independent of the final pressure. Since CP is always greater than CV , the temperature of the gas in the tank at the end of the filling process will be greater than the temperature of gas in the line. Why is this so? b. To find the pressure in the tank after the heat transfer process, we use the mass balance and the equation of state. Again, choosing the contents of the tank as the system, the mass (mole) balance is N2 = N1 , since there is no transfer of mass into or out of the system during the heat transfer process (unless, of course, the tank is leaking; we do not consider this complication here). Now using the ideal gas equation of state, we have P1 P2 = T2 T1

or

P2 = P1

T2 T1

Thus, P2 = 28.94 bar.

Illustration 3.4-4 Example of a Thermodynamics Problem that Cannot Be Solved with Only the Mass and Energy Balances7 A compressor is a gas pumping device that takes in gas at low pressure and discharges it at a higher pressure. Since this process occurs quickly compared with heat transfer, it is usually assumed to be adiabatic; that is, there is no heat transfer to or from the gas during its compression. Assuming that the inlet to the compressor is air [which we will take to be an ideal gas with CP∗ = 29.3 J/(mol K)] at 1 bar and 290 K and that the discharge is at a pressure of 10 bar, estimate the temperature of the exit gas and the rate at which work is done on the gas (i.e., the power requirement) for a gas flow of 2.5 mol/s.

7 We

return to this problem in the next chapter after formulating the balance equation for an additional thermodynamic variable, the entropy.

76 Chapter 3: Conservation of Energy Solution (Since there are two unknown thermodynamic quantities, the final temperature and the rate at which work is being done, we can anticipate that the mass and energy balances will not be sufficient to solve this problem.) The system will be taken to be the gas contained in the compressor. The differential form of the molar mass and energy balances for this open system are dN = N˙ 1 + N˙ 2 dt dU ˙ = N˙ 1 H 1 + N˙ 2 H 2 + Q˙ + W dt

where we have used the subscript 1 to indicate the flow stream into the compressor and 2 to indicate the flow stream out of the compressor. Since the compressor operates continuously, the process may be assumed to be in a steady state, dN =0 dt dU =0 dt

or

N˙ 1 = −N˙ 2

that is, the time variations of the mass of the gas contained in the compressor and of the energy content of this gas are both zero. Also, Q˙ = 0 since there is no heat transfer to the gas, and ˙ = W ˙ s since the system boundaries (the compressor) are not changing with time. Thus we W have ˙ s = N˙ 1 H 2 − N˙ 1 H 1 = N˙ 1 C ∗ (T2 − T1 ) W P

or W s = CP∗ (T2 − T1 ) ˙ s /N˙ 1 is the work done per mole of gas. Therefore, the power necessary to drive where W s = W the compressor can be computed once the outlet temperature of the gas is known, or the outlet temperature can be determined if the power input is known. We are at an impasse; we need more information before a solution can be obtained. It is clear by comparison with the previous examples why we cannot obtain a solution here. In the previous cases, the mass balance and the energy balance, together with the equation of state of the fluid and the problem statement, provided the information necessary to determine the final state of the system. However, here we have a situation where the energy balance contains two unknowns, the final temperature and W s . Since neither is specified, we need additional information about the system or process before we can solve the problem. This additional information will be obtained using an additional balance equation developed in the next chapter.

Illustration 3.4-5 Use of Mass and Energy Balances to Solve an Ideal Gas Problem8 A gas cylinder of 1 m3 volume containing nitrogen initially at a pressure of 40 bar and a temperature of 200 K is connected to another cylinder of 1 m3 volume that is evacuated. A valve between the two cylinders is opened until the pressures in the cylinders equalize. Find the final temperature and pressure in each cylinder if there is no heat flow into or out of the cylinders or between the gas and the cylinder. You may assume that the gas is ideal with a constant-pressure heat capacity of 29.3 J/(mol K).

8 An

alternative solution to this problem giving the same answer is given in the next chapter.

3.4 Applications of the Mass and Energy Balances 77 Solution This problem is more complicated than the previous ones because we are interested in changes that occur in two separate cylinders. We can try to obtain a solution to this problem in two different ways. First, we could consider each tank to be a separate system, and so obtain two mass balance equations and two energy balance equations, which are coupled by the fact that the mass flow rate and enthalpy of the gas leaving the first cylinder are equal to the like quantities entering the second cylinder.9 Alternatively, we could obtain an equivalent set of equations by choosing a composite system of the two interconnected gas cylinders to be the first system and the second system to be either one of the cylinders. In this way the first (composite) system is closed and the second system is open. We will use the second system choice here; you are encouraged to explore the first system choice independently and to verify that the same solution is obtained. The difference form of the mass and energy balance equations (on a molar basis) for the twocylinder composite system are N1i = N1f + N2f

(a)

N1i U i1 = N1f U f1 + N2f U f2

(b)

and

Here the subscripts 1 and 2 refer to the cylinders, and the superscripts i and f refer to the initial and final states. In writing the energy balance equation we have recognized that for the system consisting of both cylinders there is no mass flow, heat flow, or change in volume. Now using, in Eq. a, the ideal gas equation of state written as N = P V /RT and the fact that the volumes of both cylinders are equal yields P1f P2f P1i = + T1i T1f T2f

(a  )

Using the same observations in Eq. b and further recognizing that for a constant heat capacity gas we have, from Eq. 3.3-8, that ∗ U (T ) = CV T − CP∗ TR

yields P1f P2f P1i ∗ i ∗ ∗ f ∗ { C T − C T } = { C T − C T } + {CV∗ T2f − CP∗ TR } R R P V 1 P T1i V 1 T1f T2f

which, on rearrangement, gives 

−

P1f P2f P1i − − T1i T1f T2f

 ∗ CP∗ TR + CV {P1i − P1f − P2f } = 0

Since the bracketed quantity in the first term is identically zero (see Eq. a ), we obtain P1i = P1f + P2f

9 That

(c)

the enthalpy of the gas leaving the first cylinder is equal to that entering the second, even though the two cylinders are at different pressures, follows from the fact that the plumbing between the two can be thought of as a flow constriction, as in the Joule-Thomson expansion. Thus the analysis of Illustration 3.2-3 applies to this part of the total process.

78 Chapter 3: Conservation of Energy (Note that the properties of the reference state have canceled. This is to be expected, since the solution to a change-of-state problem must be independent of the arbitrarily chosen reference state. This is an important point. In nature, the process will result in the same final state independent of our arbitrary choice of reference temperature, TR . Therefore, if our analysis is correct, TR must not appear in the final answer.) Next, we observe that from the problem statement P1f = P2f ; thus P1f = P2f = 12 P1i = 20 bar

and from Eq. a 1 T1f

+

1 T2f

=

2 T1i

(c  )

Thus we have one equation for the two unknowns, the two final temperatures. We cannot assume that the final gas temperatures in the two cylinders are the same because nothing in the problem statement indicates that a transfer of heat between the cylinders necessary to equalize the gas temperatures has occurred. To get the additional information necessary to solve this problem, we write the mass and energy balance equations for the initially filled cylinder. The rate-of-change form of these equations for this system are

and

dN1 = N˙ dt

(d)

d(N1 U 1 ) = N˙ H 1 dt

(e)

˙ s , and dV /dt In writing the energy balance equation, we have made use of the fact that Q˙ , W are all zero. Also, we have assumed that while the gas temperature is changing with time, it is spatially uniform within the cylinder, so that at any instant the temperature and pressure of the gas leaving the cylinder are identical with those properties of the gas in the cylinder. Thus, the molar enthalpy of the gas leaving the cylinder is H = H(T1 , P1 ) = H 1

Since our interest is in the change in temperature of the gas that occurs as its pressure drops from 40 bar to 20 bar due to the escaping gas, you may ask why the balance equations here have been written in the rate-of-change form rather than in terms of the change over a time interval. The answer is that since the properties of the gas within the cylinder (i.e., its temperature and pressure) are changing with time, so is H 1 , the enthalpy of the exiting gas. Thus, if we were to use the form of Eq. e integrated over a time interval (i.e., the difference form of the energy balance equation),  N1f U f1 − N1i U i1 =

N˙ H 1 dt

we would have no way of evaluating the integral on the right side. Consequently, the difference equation provides no useful information for the solution of the problem. However, by starting with Eqs. d and e, it is possible to obtain a solution, as will be evident shortly. To proceed with the solution, we first combine and rearrange the mass and energy balances to obtain d(N1 U 1 ) dU dN1 dN1 ≡ N1 1 + U 1 = N˙ H 1 = H 1 dt dt dt dt

3.4 Applications of the Mass and Energy Balances 79 so that we have N1

dN1 dU 1 = (H 1 − U 1 ) dt dt

Now we use the following properties of the ideal gas (see Eqs. 3.3-7 and 3.3-8) H = CP∗ (T − TR )

N = P V /RT

and ∗ U = CV (T − TR ) − RTR

to obtain 

P1 V ∗ dT1 d C = RT1 RT1 V dt dt



P1 V RT1

Simplifying this equation yields ∗ 1 dT1 T1 d CV = R T1 dt P1 dt





P1 T1

or ∗ d ln T1 d CV = ln R dt dt



P1 T1



Now integrating between the initial and final states, we obtain 

T1f T1i

CV∗ /R

 =

P1f P1i



T1i



T1f

or 

T1f T1i

CP∗ /R

 =

P1f P1i



(f)

where we have used the fact that for the ideal gas CP∗ = CV∗ + R. Equation f provides the means to compute T1f , and T2f can then be found from Eq. c . Finally, using the ideal gas equation of state we can compute the final number of moles of gas in each cylinder using the relation NJf =

VcylJ PJf RTJf

(g)

where the subscript J refers to the cylinder number. The answers are T1f = 164.3 K

N1f = 1.464 kmol

T2f

N2f = 0.941 kmol

= 255.6 K

Comments The solution of this problem for real fluids is considerably more complicated than for the ideal gas. The starting points are again P1f = P2f

(h)

80 Chapter 3: Conservation of Energy and N1f + N2f = N1i

(i)

and Eqs. d and e. However, instead of Eq. g we now have N1f =

Vcyl1

N2f =

Vcyl2

V f1

(j)

and V f2

(k)

where V f1 and V f2 are related to (T1f , P1f ) and (T2f , P2f ), respectively, through the equation of state or tabular P V T data of the form V f1 = V f1 (T1f , P1f ) V

f 2

=V

f f f 2 (T2 , P2 )

(l) (m)

The energy balance for the two-cylinder composite system is N1i U i1 = N1f U f1 + N2f U f2

(n)

Since a thermal equation of state or tabular data of the form U = U (T, V ) are presumed available, Eq. n introduces no new variables. Thus we have seven equations among eight unknowns (N1f , N2f , T1f , T2f , P1f , P2f , V f1 , and f V 2 ). The final equation needed to solve this problem can, in principle, be obtained by the manipulation and integration of Eq. e, as in the ideal gas case, but now using the real fluid equation of state or tabular data and numerical integration techniques. Since this analysis is difficult, and a simpler method of solution (discussed in Chapter 6) is available, the solution of this problem for the real fluid case is postponed until Sec. 6.5.

Illustration 3.4-6 Showing That the Change in State Variables between Fixed Initial and Final States Is Independent of the Path Followed It is possible to go from a given initial equilibrium state of a system to a given final equilibrium state by a number of different paths, involving different intermediate states and different amounts of heat and work. Since the internal energy of a system is a state property, its change between any two states must be independent of the path chosen (see Sec. 1.3). The heat and work flows are, however, path-dependent quantities and can differ on different paths between given initial and final states. This assertion is established here by example. One mole of a gas at a temperature of 25◦ C and a pressure of 1 bar (the initial state) is to be heated and compressed in a frictionless piston and cylinder to 300◦ C and 10 bar (the final state). Compute the heat and work required along each of the following paths. Path A. Isothermal (constant temperature) compression to 10 bar, and then isobaric (constant pressure) heating to 300◦ C Path B. Isobaric heating to 300◦ C followed by isothermal compression to 10 bar Path C. A compression in which P V γ = constant, where γ = CP∗ /CV∗ , followed by an isobaric cooling or heating, if necessary, to 300◦ C.

3.4 Applications of the Mass and Energy Balances 81 For simplicity, the gas is assumed to be ideal with CP∗ = 38 J/(mol K). 350 Final state 300

Path B

T (°C)

250 200 th

150

Pa

C

100 50 0

Path A Initial state 0

2

4

6 P (bar)

8

10

12

Solution The 1-mol sample of gas will be taken as the thermodynamic system. The difference form of the mass balance for this closed, deforming volume of gas is N = constant = 1 mol

and the difference form of the energy balance is  ΔU = Q −

P dV = Q + W

Path A i. Isothermal compression  V 2

 V 2

 V 2 dV dV V P2 = −RT = −RT ln 2 = RT ln V V V P V1 V1 V1 1 1 10 = 8.314 J/(mol K) × 298.15 K × ln = 5707.7 J/mol 1

Wi = −

P dV = −

RT

Since 

T2

ΔU =

∗ ∗ CV dT = CV (T2 − T1 )

and

T2 = T1 = 25◦ C

T1

we have ΔU = 0

and

Qi = −Wi = −5707.7 J/mol

82 Chapter 3: Conservation of Energy ii. Isobaric heating

Wii = −

 V 3 V2

P2 dV = −P2 

 V 3 V2 T3

ΔU =

dV = −P2 (V 3 − V 2 ) = −R(T3 − T2 )

∗ ∗ CV dT = CV (T3 − T2 )

T2

and ∗ ∗ Qii = ΔU − Wii = CV (T3 − T2 ) + R(T3 − T2 ) = (CV + R)(T3 − T2 ) = CP∗ (T3 − T2 )

[This is, in fact,  a special case of the general result that at constant pressure for a closed system, Q = CP∗ dT . This is easily proved by starting with dU dV Q˙ = +P dt dt

and using the fact that P is constant to obtain dU d dH d dT Q˙ = + (P V ) = (U + P V ) = = CP∗ dt dt dt dt dt   ∗ Now setting Q = Q˙ dt yields Q = CP dT .]

Therefore, Wii = −8.314 J/(mol K) × 275 K = −2286.3 J/mol Qii = 38 J/(mol K) × 275 K = 10 450 J/mol Q = Qi + Qii = −5707.7 + 10 450 = 4742.3 J/mol W = Wi + Wii = 5707.7 − 2286.3 = 3421.4 J/mol

Path B i. Isobaric heating Qi = CP∗ (T2 − T1 ) = 10 450 J/mol Wi = −R(T2 − T1 ) = −2286.3 J/mol

ii. Isothermal compression

  P2 10 = 8.314 × 573.15 ln = 10 972.2 J/mol P1 1 Qii = −Wii = −10 972.2 J/mol Q = 10 450 − 10 972.2 = −522.2 J/mol W = −2286.3 + 10 972.2 = 8685.9 J/mol

Wii = RT ln

Path C i. Compression with P V γ = constant  V 2 constant constant 1−γ dV = − (V 2 − V 11−γ ) Vγ 1−γ V1 V1 1 −R(T2 − T1 ) −R(T2 − T1 ) ∗ (P2 V 2 − P1 V 1 ) = = =− ∗ = CV (T2 − T1 ) 1−γ 1−γ 1 − (CP∗ /CV )

Wi = −

 V 2

P dV = −

3.4 Applications of the Mass and Energy Balances 83 where T2 can be computed from  P1 V γ1 = P1

RT1 P1

γ

 = P2 V γ2 = P2

RT2 P2

γ

or T2 = T1



P2 P1

(γ −1)/γ

Now γ=

CP∗ 38 = 1.280 ∗ = CV 38 − 8.314

so that T2 = 298.15 K (10)0.280/1.280 = 493.38 K

and ∗ Wi = CV (T2 − T1 ) = (38 − 8.314) J/(mol K) × (493.38 − 298.15) K

= 5795.6 J/mol ∗ (T2 − T1 ) = 5795.6 J/mol ΔUi = CV

Qi = ΔUi − Wi = 0

ii. Isobaric heating Qii = CP∗ (T3 − T2 ) = 38 J/(mol K) × (573.15 − 493.38) K = 3031.3 J/mol Wii = −R(T3 − T2 ) = −8.314 J/(mol K) × (573.15 − 493.38) K = −663.2 J/mol

and Q = 0 + 3031.3 = 3031.3 J/mol W = 5795.6 − 663.2 = 5132.4 J/mol

Solution Path

Q (J/mol)

W (J/mol)

Q + W = ΔU (J/mol)

A B C

4742.3

−522.2

3421.4 8685.9 5132.4

8163.7 8163.7 8163.7

3031.3

Comment Notice that along each of the three paths considered (and, in fact, any other path between the initial and final states), the sum of Q and W, which is equal to ΔU , is 8163.7 J/mol, even though Q and W separately are different along the different paths. This illustrates that whereas the internal energy is a state property and is path independent (i.e., its change in going from state 1 to state 2 depends only on these states and not on the path between them), the heat and work flows depend on the path and are therefore path functions.

84 Chapter 3: Conservation of Energy Illustration 3.4-7 Showing That More Work Is Obtained If a Process Occurs without Friction An initial pressure of 2.043 bar is maintained on 1 mol of air contained in a piston-and-cylinder system by a set of weights W, the weight of the piston, and the surrounding atmosphere. Work is obtained by removing some of the weights and allowing the air to isothermally expand at 25◦ C, thus lifting the piston and the remaining weights. The process is repeated until all the weights have been removed. The piston has a mass of ω = 5 kg and an area of 0.01 m2 . For simplicity, the air can be considered to be an ideal gas. Assume that, as a result of sliding friction between the piston and the cylinder wall, all oscillatory motions of the piston after the removal of a weight will eventually be damped. Compute the work obtained from the isothermal expansion and the heat required from external sources for each of the following: a. b. c. d.

The weight W is taken off in one step. The weight is taken off in two steps, with W/2 removed each time. The weight is taken off in four steps, with W/4 removed each time. The weight is replaced by a pile of sand (of total weight W), and the grains of sand are removed one at a time.

Processes b and d are illustrated in the following figure.10

Solution I. Analysis of the problem. Choosing the air in the cylinder to be the system, recognizing that for an ideal gas at constant temperature U is constant so that ΔU = 0, and neglecting the kinetic and potential energy terms for the gas (since the mass of 1 mol of air is only 29 g), we obtain the following energy balance equation: 

0=Q−

(a)

P dV

The total work done by the gas in lifting and accelerating the piston and the weights against the frictionalforces, and in expanding the system volume against atmospheric pressure is contained in the − P dV term. To see this we recognize that the laws of classical mechanics apply to the piston and weights, and equate, at each instant, all the forces on the piston and weights to their acceleration, 

Forces on piston and weights =



Mass of piston × Acceleration and weights

and obtain [P × A − Patm × A − (W + ω)g + Ffr ] = (W + ω)

dv dt

(b)

Here we have taken the vertical upward (+z) direction as being positive and used P and Patm to represent the pressure of the gas and atmosphere, respectively; A the piston area; ω its mass; W the mass of the weights on the piston at any time; v the piston velocity; and Ffr the frictional force, which is proportional to the piston velocity. Recognizing that the piston velocity v is equal to the rate of change of the piston height h or the gas volume V , we have v=

dh 1 dV = dt A dt

10 From H. C. Van Ness, Understanding Thermodynamics, McGraw-Hill, New York, 1969. Used with permission of the McGraw-Hill Book Co.

3.4 Applications of the Mass and Energy Balances 85

ᐃ/2

ᐃ/2

ᐃ/2 ᐃ/2

ᐃ/2

ᐃ/2

Process b

Process d

Also we can solve Eq. b for the gas pressure: P = Patm +

(W + ω) (W + ω) dv Ffr g− + A A A dt

(c)

At mechanical and thermodynamic equilibrium (i.e., when dv/dt = 0 and v = 0), we have P = Patm +

(W + ω) g A

(d)

With these results, the total work done by the gas can be computed. In particular,  



 (W + ω) dv Ffr + dV A A A dt     (W + ω) (W + ω) 1 dv g ΔV − Ffr dV + dV = Patm + A A A dt



P dV =

Patm +

W +ω



g−

This equation can be simplified by rewriting the last integral as follows: 1 A



1 dv dV = dt A



dv dV dt = dt dt



1 dv v dt = dt 2



dv 2 dt = Δ dt



1 2 v 2



86 Chapter 3: Conservation of Energy where the symbol Δ indicates the change between the initial and final states. Next, we recall from mechanics that the force due to sliding friction, here Ffr , is in the direction opposite to the relative velocity of the moving surfaces and can be written as Ffr = −kfr v

where kfr is the coefficient of sliding friction. Thus, the remaining integral can be written as 1 A

 Ffr dV = −

1 A

 kfr v

dV dt = −kfr dt

 v 2 dt

The energy balance, Eq. a, then becomes 

 Q=

P dV = Patm ΔV + (W + ω)g Δh + kfr

v 2 dt + (W + ω) Δ

1 2

v2



(e)

This equation relates the heat flow into the gas (to keep its temperature constant) to the work the gas does against the atmosphere in lifting the piston and weights (hence increasing their potential energy) against friction and in accelerating the piston and weights (thus increasing their kinetic energy). The work done against frictional forces is dissipated into thermal energy, resulting in a higher temperature at the piston and cylinder wall. This thermal energy is then absorbed by the gas and appears as part of Q. Consequently, the net flow of heat from a temperature bath to the gas is  QNET = Q − kfr

v 2 dt

(f)

Since heat will be transferred to the gas, and the integral is always positive, this equation establishes that less heat will be needed to keep the gas at a constant temperature if the expansion occurs with friction than in a frictionless process. Also, since the expansion occurs isothermally, the total heat flow to the gas is, from Eq. a, 



Vf

Q=

Vf

P dV = Vi

Vi

Vf N RT dV = N RT ln V Vi

(g)

Combining Eqs. e and f, and recognizing that our interest here will be in computing the heat and work flows between states for which the piston has come to rest (v = 0), yields  QNET = Q − kfr

v 2 dt = Patm ΔV + (W + ω)g Δh = P ΔV = −W NET

(h)

where P is the equilibrium final pressure given by Eq. d. Also from Eqs. f, g, and h, we have  QNET = −W NET = N RT ln

V2 V1





− kfr

v 2 dt

(i)

Here W NET represents the net work obtained by the expansion of the gas (i.e., the work obtained in raising the piston and weights and in doing work against the atmosphere). The foregoing equations can now be used in the solution of the problem. In particular, as a weight is removed, the new equilibrium gas pressure is computed from Eq. d, the resulting volume change from the ideal gas law, W NET and QNET from Eq. h, and the work against friction from Eq. i. There is, however, one point that should be mentioned before we proceed with this calculation. If there were no mechanism for the dissipation of kinetic energy to thermal energy (that is, sliding friction between the piston and cylinder wall, and possibly also viscous dissipation on expansion and compression of the gas due to its bulk viscosity), then when a weight was removed the piston would be put into a perpetual oscillatory motion. The presence

3.4 Applications of the Mass and Energy Balances 87 of a dissipative mechanism will damp the oscillatory motion. (As will be seen, the value of the coefficient of sliding friction, kfr , does not affect the amount of kinetic energy ultimately dissipated as heat. Its value does, however, affect the dynamics of the system and thus determine how quickly the oscillatory motion is damped.) II. The numerical solution. First, the mass W of the weights is computed using Eq. d and the fact that the initial pressure is 2.043 bar. Thus,

P = 2.043 bar = 1.013 bar +

(5 + W ) kg m 1 Pa 1 bar × 9.807 2 × × 5 0.01 m2 s kg/(m s2 ) 10 Pa

or (5 + W ) kg = 105.0 kg

so that W = 100 kg. The ideal gas equation of state for 1 mol of air at 25◦ C is P V = N RT = 1 mol × 8.314 × 10−5 −2

= 2.479 × 10

bar m3 × (25 + 273.15) K mol K

(j)

3

bar m = 2479 J

and the initial volume of the gas is V =

2.479 × 10−2 bar m3 = 1.213 × 10−2 m3 2.043 bar

Process a The 100-kg weight is removed. The equilibrium pressure of the gas (after the piston has stopped oscillating) is P1 = 1.013 bar +

5 kg × 9.807 m/s2 10−5 bar × = 1.062 bar 2 0.01 m kg/(m s2 )

and the gas volume is V1 =

2.479 × 10−2 bar m3 = 2.334 × 10−2 m3 1.062 bar

Thus ΔV = (2.334 − 1.213) × 10−2 m3 = 1.121 × 10−2 m3 J − W NET = 1.062 bar × 1.121 × 10−2 m3 × 105 bar m3 = 1190.5 J = QNET

and Q = N RT ln

V1 2.334 × 10−2 = 2479 J ln = 1622.5 J V0 1.213 × 10−2

(since N RT = 2479 J from Eq. j). Consequently, the work done against frictional forces (and converted to thermal energy), which we denote by Wfr , is −Wfr = Q − QNET = (1622.5 − 1190.5) J = 432 J The total useful work obtained, the net heat supplied, and the work against frictional forces are given in Table 1. Also, the net work, P ΔV , is shown as the shaded area in the accompanying figure, together with the line representing the isothermal equation of state, Eq. j. Note that in this case the net work is that of raising the piston and pushing back the atmosphere.

88 Chapter 3: Conservation of Energy Table 1 Process

−W NET = QNET (J)

(J)

−Wfr (J)

1190.5 1378.7 1493.0 1622.5

1622.5 1622.5 1622.5 1622.5

432.0 243.8 129.5 0

a b c d

Q

Process b The situation here is similar to that of process a, except that the weight is removed in two 50-kg increments. The pressure, volume, work, and heat flows for each step of the process are given in Table 2, and −WiNET = Pi (ΔV )i , the net work for each step, is given in the figure. Process c Here the weight is removed in four 25-kg increments. The pressure, volume, work, and heat flows for each step are given in Table 2 and summarized in Table 1. Also, the net work for each step is given in the figure. 2.5

2.5 2

0

2

0

P, bar

P, bar

1 1 1

0 0.01

0.02

2 1

0 0.01

0.025

0.02

V, m3

V, m3

Process a

Process b

0.025

Table 2 Process b P

−WiNET = Pi (ΔV )i

Stage

(bar)

V (m3 )

0 1 2 Total

2.043 1.552 1.062

1.213 × 10−2 1.597 × 10−2 2.334 × 10−2

Q = N RT ln

(J)

(J)

596.0 782.7 1378.7

681.8 940.7 1622.5

85.8 158.0 243.8

Stage 0 1 2 3 4 Total

(bar) 2.043 1.798 1.552 1.307 1.062

−WiNET = Pi (ΔV )i

V (m3 )

−Wfr

(J)

Process c P

Vi Vi−1

Q = N RT ln

Vi Vi−1

−Wfr

(J)

(J)

(J)

298.5 338.3 392.1 464.1 1493.0

318.0 363.8 426.8 513.9 1622.5

19.5 25.5 34.7 49.8 129.5

−2

1.213 × 10 1.379 × 10−2 1.597 × 10−2 1.897 × 10−2 2.334 × 10−2

3.4 Applications of the Mass and Energy Balances 89 Process d The computation here is somewhat more difficult since the number of stages to the calculation is almost infinite. However, recognizing that in the limit of the mass of a grain of sand going to zero there is only a differential change in the pressure and volume of the gas and negligible velocity or acceleration of the piston, we have −W

NET

=



 Pi (ΔVi ) →

P dV = NRT ln

i

Vf = QNET Vi

and Wfr = 0, since the piston velocity is essentially zero at all times. Thus, − W NET = QNET = Q = 1 mol × 8.314 = 1622.5 J

J 2.334 × 10−2 × 298.15 K × ln mol K 1.213 × 10−2

This result is given in Table 1 and the figure. 2.5

2.5 0

0 2

2

1 P, bar

P, bar

2 3 4 1

0 0.01

0.02 V, m3

0.025

1

0 0.01

0.02

0.025

V, m3

Comments Several points are worth noting about this illustration. First, although the initial and final states of the gas are the same in all three processes, the useful or net work obtained and the net heat required differ. Of course, by the energy conservation principle, it is true that −W NET = QNET for each process. It is important to note that the most useful work is obtained for a given change of state if the change is carried out in differential steps, so that there is no frictional dissipation of mechanical energy to thermal energy (compare process d with processes a, b, and c). Also, in this case, if we were to reverse the process and compress the gas, it would be found that the minimum work required for the compression is obtained when weights are added to the piston in differential (rather than finite) steps. (See Problem 3.29.) It should also be pointed out that in each of the four processes considered here the gas did 1622.5 J of work on its surroundings (the piston, the weights, and the atmosphere) and absorbed 1622.5 J of heat (from the thermostatic bath maintaining the system temperature constant and from the piston and cylinder as a result of their increased temperature due to frictional heating). We can see this from Table 1, since −(W NET + Wfr ) = Q = 1622.5 J for all four processes. However, the fraction of the total work of the gas obtained as useful work versus work against friction varies among the different processes.

At first glance it might appear that in the process in illustration 3.4.7 the proscription in Chapter 1 that thermal energy (here heat) cannot be completely converted to mechanical energy (here work) has been violated. However, that statement included the requirement that such a complete conversion was not possible in a cyclic process or in

90 Chapter 3: Conservation of Energy a process in which there were no changes in the universe (system plus surroundings). In the illustration here heat has been completely converted to work, but this can only be done once since the system, the gas contained within the cylinder, has not been restored to its initial pressure. Consequently, we cannot continue to add heat and extract the same amount of energy as work.

Illustration 3.4-8 Computing the Pressure, and Heat and Work Flows in a Complicated Ideal Gas Problem Consider the cylinder filled with air and with the piston and spring arrangement shown below. The external atmospheric pressure is 1 bar, the initial temperature of the air is 25◦ C, the no-load length of the spring is 50 cm, the spring constant is 40 000 N/m, the frictionless piston weighs 500 kg, and the constant-volume heat capacity of air can be taken to be constant at 20.3 J/(mol K). Assume air is an ideal gas. Compute a. The initial pressure of the gas in the cylinder b. How much heat must be added to the gas in the cylinder for the spring to compress 2 cm

55 cm

15 cm

30 cm

Solution a. The pressure of the gas inside the cylinder is a result of the atmospheric pressure (1 bar), the force exerted on the gas as a result of the weight of the piston and the force of the spring. The contribution from the piston is

PPiston

F = = A =

M kg × 9.807

m 1 Pa 1 bar × × 5 s2 kg/(m s2 ) 10 Pa π(d2 /4) m2

M 500 × 1.2487 × 10−4 bar = × 1.2487 × 10−4 bar = 0.6937 bar d2 (0.3)2

The contribution due to the spring is PSpring =

F N −k(x − x0 ) = = −40 000 × (55 − 50) × 10−2 m A π(d2 /4) m

−40 000 =

N × (55 − 50) × 10−2 m m kg Pa 1 bar m ×1 2 ×1 × 5 2 2 2 π[(0.3) /4] m s N kg/(m s ) 10 Pa

= −0.2829 bar

Therefore, the pressure of the air in the cylinder is P = PAtmosphere + PPiston + PSpring = 1 + 0.6937 − 0.2829 bar = 1.4108 bar

3.4 Applications of the Mass and Energy Balances 91 For later reference we note that the number of moles of gas in the cylinder is, from the ideal gas law, π 1.4108 bar × (0.3)2 (0.15) m3 PV 4 N = = bar m3 RT 298.15 K × 8.314 × 10−5 mol K = 0.6035 mol

which is equal to 17.5 g = 0.0175 kg

b. The contribution to the total pressure due to the spring after heating is computed as above, except that now the spring extension is 53 cm (55 cm − 2 cm). Thus −40 000 PSpring =

N × (53 − 50) × 10−2 m m = −0.1697 bar (0.3)2 m2 π 4

So the final pressure of the air in the cylinder is P = 1 + 0.6937 − 0.1697 bar = 1.524 bar

[Note that the pressure at any extension of the spring, x, is P = 1.6937 − 5.658(x − 0.50) bar

where x is the extension of spring (m)]

The volume change on expansion of the gas (to raise the piston by 2 cm) is ΔV =

π (0.3)2 m2 × 0.02 m = 1.414 × 10−3 m3 4

Also, the final temperature of the gas, again from the ideal gas law, is PV = T = NR

π 2 2 0.3 m × 0.17 m 4 3 = 365.0 K −5 bar m 0.6035 mol × 8.314 × 10 mol K 1.524 bar ×

The work done can be computed in two ways, as shown below. For simplicity, we first compute the individual contributions, and then see how these terms are combined. Work done by the gas against the atmosphere is WAtmosphere = −PAtmosphere × ΔV = −1 bar × 1.414 × 10−3 m3 × 105

J = −141.4 J bar m3

(The minus sign indicates that the gas did work on the surrounding atmosphere.) Work done by the gas in compressing the spring is k WSpring = − [(x2 − x0 )2 − (x1 − x0 )2 ] 2 40 000 N 2 [3 × 10−4 m2 − 52 × 10−4 m2 ] =− 2 m J = 32 J = −2 × (9 − 25) = 32 N m × 1 Nm

Work done by the gas in raising the 500-kg piston is WPiston = −500 kg × 0.02 m × 9.807

m 1J × 2 = −98.07 J s2 m kg/s2

92 Chapter 3: Conservation of Energy Work done by the gas in raising its center of mass by 1 cm (why does the center of mass of the gas increase by only 1 cm if the piston is raised 2 cm?) is Wgas = −17.5 g × 0.01 m × 9.807

m 1J 1 kg × 2 × = −0.0017 J s2 m kg/s2 1000 g

which is negligible compared with the other work terms. If we consider the gas in the cylinder to be the system, the work done by the gas is W = Work to raise piston + Work against atmosphere + Work to compress spring = −98.1 J − 141.4 J + 32.0 J = −207.5 J

[Here we have recognized that the gas is doing work by raising the piston and by expansion against the atmosphere, but since the spring is extended beyond its no-load point, it is doing work on the gas as it contracts. (The opposite would be true if the spring were initially compressed to a distance less than its no-load point.)] An alternative method of computing this work is as follows. The work done by the gas on expanding (considering only the gas in the system) is  W =−

 P dV = −A

0.17 m

 P dy = −A

0.15 m

0.17 m

[1.6937 − 5.658(x − 0.5)] dy

0.15 m

where y is the height of the bottom of the piston at any time. The difficulty with the integral above is that two different coordinate systems are involved, since y is the height of the piston and x is the extension of the spring. Therefore, we need to make a coordinate transformation. To do this we note that when y = 0.15, x = 0.55, and when y = 0.17, x = 0.53. Consequently, x = 0.7 − y and  W = −A

0.17 m

[1.6937 − 5.658(0.7 − y − 0.5)] dy   0.17 m (0.2 − y) dy = −A 1.6937 × 0.02 − 5.658 0.15 m  

= −A 1.6937 × 0.02 − 5.658 0.2 × 0.02 − 12 (0.172 − 0.152 ) 

0.15 m

= −7.068 × 10−2 [0.03387 − 0.02263 + 0.01811] J = −0.2075 × 10−2 bar m3 × 105 = −207.5 J bar m3

This is identical to the result obtained earlier. Finally, we can compute the heat that must be added to raise the piston. The difference form of the energy balance on the closed system consisting of the gas is U(final state) − U(initial state) = Q + W = N CV (Tf − Ti )

so that Q = N CV (Tf − Ti ) − W J × (365 − 298.15) K + 207.5 J mol K = 819.0 J + 207.5 J = 1026.5 J

= 0.6035 mol × 20.3

Therefore, to accomplish the desired change, 1026.5 J of heat must be added to the gas. Of this amount, 819 J are used to heat the gas, 98 J to raise the 500-kg piston, and 141.4 J to push back the atmosphere; during the process, the spring supplies 32 J.

Problems 93

3.5 CONSERVATION OF MOMENTUM Based on the discussion of Sec. 3.4 and Illustration 3.4-4, we can conclude that the equations of mass and energy conservation are not sufficient to obtain the solution to all the problems of thermodynamics in which we might be interested. What is needed is a balance equation for an additional thermodynamic state variable. The one conservation principle that has not yet been used is the conservation of momentum. If, in Eq. 2.1-4, θ is taken to be the momentum of a black-box system, we have ⎞ ⎛ ⎞ ⎛ Rate at which momentum Rate at which momentum d (M v) = ⎝ enters system by all ⎠ − ⎝ leaves system by all ⎠ (3.5-1) dt mechanisms mechanisms where v is the center-of-mass velocity vector of the system, and M its total mass. We could now continue the derivation by evaluating all the momentum flows; however, it is clear by looking at the left side of this equation that we will get an equation for the rate of change of the center-of-mass velocity of the system, not an equation of change for a thermodynamic state variable. Consequently, the conservation-of-momentum equation will not lead to the additional balance equation we need, so this derivation will not be completed. The development of an additional, useful balance equation is not a straightforward task, and will be delayed until Chapter 4.

3.6 THE MICROSCOPIC ENERGY BALANCE (OPTIONAL) This section appears on the website for this text.

PROBLEMS 3.1 a. A bicyclist is traveling at 20 km/hr when he encounters a hill 70 m in height. Neglecting road and wind resistance (a poor assumption), what is the maximum vertical elevation gain the bicyclist could achieve without pedaling? b. If the hill is a down hill, what speed would the bicyclist achieve without pedaling, again neglecting road and wind resistance? 3.2 Water in the Niagara River approaches the falls at a velocity of 8 km/hr. Niagara Falls is approximately 55 m high. Neglecting air resistance, estimate the velocity of the water at the bottom of the falls. 3.3 a. One kilogram of steam contained in a horizontal frictionless piston and cylinder is heated at a constant pressure of 1.013 bar from 125◦ C to such a temperature that its volume doubles. Calculate the amount of heat that must be added to accomplish this change, the final temperature of the steam, the work the steam does against its surroundings, and the internal energy and enthalpy changes of the steam for this process. b. Repeat the calculation of part (a) if the heating occurs at a constant volume to a pressure that is twice the initial pressure.

c. Repeat the calculation of part (a) assuming that steam is an ideal gas with a constant-pressure heat capacity of 34.4 J/mol K. d. Repeat the calculation of part (b) assuming steam is an ideal gas as in part (c). 3.4 In Joule’s experiments, the slow lowering of a weight (through a pulley and cable arrangement) turned a stirrer in an insulated container of water. As a result of viscosity, the kinetic energy transferred from the stirrer to the water eventually dissipated. In this process the potential energy of the weight was first converted to kinetic energy of the stirrer and the water, and then as a result of viscous forces, the kinetic energy of the water was converted to thermal energy apparent as a rise in temperature. Assuming no friction in the pulleys and no heat losses, how large a temperature rise would be found in 1 kg of water as a result of a 1-kg weight being lowered 1 m? 3.5 Steam at 500 bar and 600◦ C is to undergo a JouleThomson expansion to atmospheric pressure. What will be the temperature of the steam after the expansion? What would be the downstream temperature if the steam were replaced by an ideal gas?

94 Chapter 3: Conservation of Energy 3.6 Water in an open metal drum is to be heated from room temperature (25◦ C) to 80◦ C by adding steam slowly enough that all the steam condenses. The drum initially contains 100 kg of water, and steam is supplied at 3.0 bar and 300◦ C. How many kilograms of steam should be added so that the final temperature of the water in the tank is exactly 80◦ C? Neglect all heat losses from the water in this calculation. 3.7 Consider the following statement: “The adiabatic work necessary to cause a given change of state in a closed system is independent of the path by which that change occurs.” a. Is this statement true or false? Why? Does this statement contradict Illustration 3.4-6, which establishes the path dependence of work? b. Show that if the statement is false, it would be possible to construct a machine that would generate energy. 3.8 A nonconducting tank of negligible heat capacity and 1 m3 volume is connected to a pipeline containing steam at 5 bar and 370◦ C, filled with steam to a pressure of 5 bar, and disconnected from the pipeline. a. If the tank is initially evacuated, how much steam is in the tank at the end of the filling process, and what is its temperature? b. If the tank initially contains steam at 1 bar and 150◦ C, how much steam is in the tank at the end of the filling process, and what is its temperature? 3.9 a. A 1-kg iron block is to be accelerated through a process that supplies it with 1 kJ of energy. Assuming all this energy appears as kinetic energy, what is the final velocity of the block? b. If the heat capacity of iron is 25.10 J/(mol K) and the molecular weight of iron is 55.85, how large a temperature rise would result from 1 kJ of energy supplied as heat? 3.10 The voltage drop across an electrical resistor is 10 volts and the current through it is 1 ampere. The total heat capacity of the resistor is 20 J/K, and heat is dissipated from the resistor to the surrounding air according to the relation Q˙ = −h(T − Tam )

where Tam is the ambient air temperature, 25◦ C; T is the temperature of the resistor; and h, the heat transfer coefficient, is equal to 0.2 J/(K s). Compute the steadystate temperature of the resistor, that is, the temperature of the resistor when the energy loss from the resistor is just equal to the electrical energy input. 3.11 The frictionless piston-and-cylinder system shown here is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of 0.15 m2 , and the initial volume of the entrapped ideal gas is 0.12 m3 . The piston and cylinder do not conduct heat, but heat can be added to the gas by a heating coil. The gas has

a constant-volume heat capacity of 30.1 J/(mol K) and an initial temperature of 298 K, and 10.5 kJ of energy are to be supplied to the gas through the heating coil. a. If stops placed at the initial equilibrium position of the piston prevent it from rising, what will be the final temperature and pressure of the gas? b. If the piston is allowed to move freely, what will be the final temperature and volume of the gas?

Heating coil

3.12 As an energy conservation measure in a chemical plant, a 40-m3 tank will be used for temporary storage of exhaust process steam. This steam is then used in a later stage of the processing. The storage tank is well insulated and initially contains 0.02 m3 of liquid water at 50◦ C; the remainder of the tank contains water vapor in equilibrium with this liquid. Process steam at 1.013 bar and 90 percent quality enters the storage tank until the pressure in the tank is 1.013 bar. How many kilograms of wet steam enter the tank during the filling process, and how much liquid water is present at the end of the process? Assume that there is no heat transfer between the steam or water and the tank walls. 3.13 The mixing tank shown here initially contains 50 kg of water at 25◦ C. Suddenly the two inlet valves and the single outlet valve are opened, so that two water streams, each with a flow rate of 5 kg/min, flow into the tank, and a single exit stream with a flow rate of 10 kg/min leaves the tank. The temperature of one inlet stream is 80◦ C, and that of the other is 50◦ C. The tank is well mixed, so that the temperature of the outlet stream is always the same as the temperature of the water in the tank.

Problems 95

3.14

3.15

3.16

3.17

a. Compute the steady-state temperature that will finally be obtained in the tank. b. Develop an expression for the temperature of the fluid in the tank at any time. A well-insulated storage tank of 60 m3 contains 200 L of liquid water at 75◦ C. The rest of the tank contains steam in equilibrium with the water. Spent process steam at 2 bar and 90 percent quality enters the storage tank until the pressure in the tank reaches 2 bar. Assuming that the heat losses from the system to the tank and the environment are negligible, calculate the total amount of steam that enters the tank during the filling process and the fraction of liquid water present at the end of the process. An isolated chamber with rigid walls is divided into two equal compartments, one containing gas and the other evacuated. The partition between the compartments ruptures. After the passage of a sufficiently long period of time, the temperature and pressure are found to be uniform throughout the chamber. a. If the filled compartment initially contains an ideal gas of constant heat capacity at 1 MPa and 500 K, what are the final temperature and pressure in the chamber? b. If the filled compartment initially contains steam at 1 MPa and 500 K, what are the final temperature and pressure in the compartment? c. Repeat part (a) if the second compartment initially contains an ideal gas, but at half the pressure and 100 K higher temperature. d. Repeat part (b) if the second compartment initially contains steam, but at half the pressure and 100 K higher temperature. a. An adiabatic turbine expands steam from 500◦ C and 3.5 MPa to 200◦ C and 0.3 MPa. If the turbine generates 750 kW, what is the flow rate of steam through the turbine? b. If a breakdown of the thermal insulation around the turbine allows a heat loss of 60 kJ per kg of steam, and the exiting steam is at 150◦ C and 0.3 MPa, what will be the power developed by the turbine if the inlet steam conditions and flow rate are unchanged? Intermolecular forces play an important role in determining the thermodynamic properties of fluids. To see this, consider the vaporization (boiling) of a liquid such as water in the frictionless piston-and-cylinder device shown. a. Compute the work obtained from the piston when 1 kg of water is vaporized to steam at 100◦ C (the vapor and liquid volumes of steam at the boiling point can be found in the steam tables). b. Show that the heat required for the vaporization of the steam is considerably greater than the work

done. (Note that the enthalpy change for the vaporization is given as 2257 kJ/kg in the steam tables in Appendix A.III.)

Steam

Water

3.18 It is sometimes necessary to produce saturated steam from superheated steam (steam at a temperature higher than the vapor-liquid coexistence temperature at the given pressure). This change can be accomplished in a desuperheater, a device in which just the right amount of water is sprayed into superheated steam to produce dry saturated steam. If superheated steam at 3.0 MPa and 500◦ C enters the desuperheater at a rate of 500 kg/hr, at what rate should liquid water at 2.5 MPa and 25◦ C be added to the desuperheater to produce saturated steam at 2.25 MPa? 3.19 Nitrogen gas leaves a compressor at 2.0 MPa and 120◦ C and is collected in three different cylinders, each of volume 0.3 m3 . In each case the cylinder is to be filled to a pressure of 2.0 MPa. Cylinder 1 is initially evacuated, cylinder 2 contains nitrogen gas at 0.1 MPa and 20◦ C, and cylinder 3 contains nitrogen at 1 MPa and 20◦ C. Find the final temperature of nitrogen in each of the cylinders, assuming nitrogen to be an ideal gas with CP∗ = 29.3 J/mol K. In each case assume the gas does not exchange heat with the cylinder walls. 3.20 A clever chemical engineer has devised the thermally operated elevator shown in the accompanying diagram. The elevator compartment is made to rise by electrically heating the air contained in the pistonand-cylinder drive mechanism, and the elevator is lowered by opening a valve at the side of the cylinder, allowing the air in the cylinder to slowly escape. Once the elevator compartment is back to the lower level, a small pump forces out the air remaining in the cylinder and replaces it with air at 20◦ C and a pressure just sufficient to support the elevator compartment. The cycle can then be repeated. There is no heat transfer between the piston, cylinder, and the gas; the weight of the piston, elevator, and elevator

96 Chapter 3: Conservation of Energy ∗ γ = CP∗ /CV . Show that such a pressure-volume re-

3.24 2nd floor

Elevator compartment

3m 1st floor

Heating coil

3.25 Centrifugal pump

contents is 4000 kg; the piston has a surface area of 2.5 m2 ; and the volume contained in the cylinder when the elevator is at its lowest level is 25 m3 . There is no friction between the piston and the cylinder, and the air in the cylinder is assumed to be an ideal gas with CP∗ = 30 J/(mol K). a. What is the pressure in the cylinder throughout the process? b. How much heat must be added to the air during the process of raising the elevator 3 m, and what is the final temperature of the gas? c. What fraction of the heat added is used in doing work, and what fraction is used in raising the temperature of the gas? d. How many moles of air must be allowed to escape in order for the elevator to return to the lowest level? 3.21 The elevator in the previous problem is to be designed to ascend and descend at the rate of 0.2 m/s, and to rise a total of 3 m. a. At what rate should heat be added to the cylinder during the ascent? b. How many kilomoles per second of air should be removed from the cylinder during the descent? 3.22 Nitrogen gas is being withdrawn at the rate of 4.5 g/s from a 0.15-m3 cylinder, initially containing the gas at a pressure of 10 bar and 320 K. The cylinder does not conduct heat, nor does its temperature change during the emptying process. What will be the temperature and pressure of the gas in the cylinder after 5 minutes? What will be the rate of change of the gas temperature at this time? Nitrogen can be considered to be an ideal gas with CP∗ = 30 J/(mol K). 3.23 In Illustration 3.4-6 we considered the compression of an ideal gas in which P V γ = constant, where

3.26

3.27

3.28

3.29

lationship is obtained in the adiabatic compression of an ideal gas of constant heat capacity. Air in a 0.3-m3 cylinder is initially at a pressure of 10 bar and a temperature of 330 K. The cylinder is to be emptied by opening a valve and letting the pressure drop to that of the atmosphere. What will be the temperature and mass of gas in the cylinder if this is accomplished? a. In a manner that maintains the temperature of the gas at 330 K? b. In a well-insulated cylinder? For simplicity assume, in part (b), that the process occurs sufficiently rapidly that there is no heat transfer between the cylinder walls and the gas. The gas is ideal, and CP∗ = 29 (J/mol K). A 0.01-m3 cylinder containing nitrogen gas initially at a pressure of 200 bar and 250 K is connected to another cylinder 0.005 m3 in volume, which is initially evacuated. A valve between the two cylinders is opened until the pressures in the cylinders equalize. Find the final temperature and pressure in each cylinder if there is no heat flow into or out of the cylinder. You may assume that there is no heat transfer between the gas and the cylinder walls and that the gas is ideal with a constantpressure heat capacity of 30 J/(mol K). Repeat the calculation of Problem 3.25, but now assume that sufficient heat transfer occurs between the gas in the two cylinders that both final temperatures and both final pressures are the same. Repeat the calculation in Problem 3.25, but now assume that the second cylinder, instead of being evacuated, is filled with nitrogen gas at 20 bar and 160 K. A 1.5 kW heater is to be used to heat a room with dimensions 3.5 m × 5.0 m × 3.0 m. There are no heat losses from the room, but the room is not airtight, so the pressure in the room is always atmospheric. Consider the air in the room to be an ideal gas with CP∗ = 29 J/(mol K), and its initial temperature is 10◦ C. a. Assuming that the rate of heat transfer from the air to the walls is low, what will be the rate of increase of the temperature in the room when the heater is turned on? b. What would be the rate of increase in the room temperature if the room were hermetically sealed? The piston-and-cylinder device of Illustration 3.4-7 is to be operated in reverse to isothermally compress the 1 mol of air. Assume that the weights in the illustration have been left at the heights they were at when they were removed from the piston (i.e., in process b the first 50-kg weight is at the initial piston height and the second is at Δh = ΔV /A = 0.384 m above the initial piston height). Compute the minimum work that must be done by the surroundings and the net heat that must

Problems 97 be withdrawn to return the gas, piston, and weights to their initial states. Also compute the total heat and the total work for each of the four expansion and compression cycles and comment on the results. 3.30 The piston-and-cylinder device shown here contains an ideal gas at 20 bar and 25◦ C. The piston has a mass of 300 kg and a cross-sectional area of 0.05 m2 . The initial volume of the gas in the cylinder is 0.03 m3 , the piston is initially held in place by a pin, and the external pressure on the piston and cylinder is 1 bar. The pin suddenly breaks, and the piston moves 0.6 m farther up the cylinder, where it is stopped by another pin. Assuming that the gas is ideal with a constant-pressure heat capacity of 30 J/(mol K), and that there is no heat transfer between the gas and the cylinder walls or piston, estimate the piston velocity, and the temperature and pressure of the gas just before the piston hits the second pin. Do this calculation assuming a. No friction between the piston and the cylinder b. Friction between the piston and the cylinder List and defend all assumptions you make in solving this problem.

Pin 2

0.6 m Pin 1

3.31 A 0.6 m diameter gas pipeline is being used for the long-distance transport of natural gas. Just past a pumping station, the gas is found to be at a temperature of 25◦ C and a pressure of 3.0 MPa. The mass flow rate is 125 kg/s, and the gas flow is adiabatic. Forty miles down the pipeline is another pumping station. At this point the pressure is found to be 2.0 MPa. At the pumping station the gas is first adiabatically compressed to a pressure of 3.0 MPa and then isobarically (i.e., at constant pressure) cooled to 25◦ C. a. Find the temperature and velocity of the gas just before it enters the pumping station. b. Find the rate at which the gas compressor in the pumping station does work on the gas, the temperature of the gas leaving the compressor, and the heat load on the gas cooler. You may assume that the compressor exhaust is also a 0.6-m pipe. (Explain why you cannot solve this problem. You will have another chance in Chapter 4.)

Natural gas can be assumed to be pure methane [molecular weight = 16, CP∗ = 36.8 J/(mol K)], and an ideal gas at the conditions being considered here. Note that the mass flow rate M is ρvA, where ρ is the mass density of the gas, v is the average gas velocity, and A is the area of the pipe. 3.32 Nitrogen can be liquefied using a Joule-Thomson expansion process. This is done by rapidly and adiabatically expanding cold nitrogen gas from high pressure to a low pressure. If nitrogen at 135 K and 20 MPa undergoes a Joule-Thomson expansion to 0.4 MPa, a. Estimate the fraction of vapor and liquid present after the expansion, and the temperature of this mixture using the pressure-enthalpy diagram for nitrogen. b. Repeat the calculation assuming nitrogen to be an ideal gas with CP∗ = 29.3 J/(mol K). 3.33 A very large mass M of hot porous rock equal to 1012 kg is to be utilized to generate electricity by injecting water and using the resulting hot steam to drive a turbine. As a result of heat extraction, the temperature of the rock drops according to Q˙ = −M CP dT /dt, where CP is the specific heat of the rock which is assumed to be independent of temperature. If the plant produces 1.36 × 109 kW hr of energy per year, and only 25 percent of the heat extracted from the rock can be converted to work, how long will it take for the temperature of the rock to drop from 600◦ C to 110◦ C? Assume that for the rock CP = 1 J/(g K). 3.34 The human body generates heat by the metabolism of carbohydrates and other food materials. Metabolism provides energy for all biological activities (e.g., muscle contractions). The metabolic processes also generate heat, and there are special cells in the body whose main function is heat generation. Now let us assume that our friend Joe BlueHen ingests 1 L of ice, which he allows to melt in his mouth before swallowing. a. How much energy is required to melt the ice and warm the water to the body temperature? b. If 1 g of fat when metabolized releases approximately 42 kJ of energy, how much fat will Joe burn by ingesting the water? c. Suppose that, instead of ice, Joe drank 1 L of water at 0◦ C. How would the answers to parts (a) and (b) change? Several years ago there was a story circulating the Internet that a good way to lose weight is to drink a lot of very cold water, since considerable energy would be expended within the body in heating up the cold water. Based on your calculations above; is that a reasonable method of weight loss? (The reason this claim was widely circulated is a result of the sloppy use of units. Some countries report the biologically

98 Chapter 3: Conservation of Energy accessible energy in food as being in units of calories, when in fact the number reported is kilocalories. For example, in the United States a teaspoon of sugar is reported ito contain 16 calories, when it is actually 16 kilocalories. Calories is incorrectly being used as an abbreviation for kilocalories.) 3.35 Water is to be heated from its pipeline temperature of 20◦ C to 90◦ C using superheated steam at 450◦ C and 2.5 MPa in a steady-state process to produce 10 kg/s of heated water. In each of the processes below, assume there is no heat loss. a. The heating is to be done in a mixing tank by direct injection of the steam, all of which condenses. Determine the two inlet mass flows needed to meet the desired hot water flow rate. b. Instead of direct mixing, a heat exchanger will be used in which the water to be heated will flow inside copper tubes and the steam will partially condense on the outside of the tubes. In this case heat will flow from the steam to the water, but the two streams are not mixed. Calculate the steam flow rate if the steam leaves the heat exchanger at 50 percent quality at 100◦ C.

3.36 People partially cool themselves by sweating, which releases water that evaporates. If during exercise a human “burns” 1000 kcal (4184 kJ) in one hour of exercise, how many grams of water must evaporate at a body temperature of 37◦ C? Assuming only 75 percent of the sweat evaporates (the rest being retained by the exercise clothes), how many grams of sweat must actually be produced? 3.37 It is thought that people develop respiratory infections during air travel because much of the airplane cabin air is recirculated. Airlines claim that using only fresh air in the cabins is too costly since at an altitude of 30 000 feet the outside conditions are −50◦ C and 0.1 bar, so that the air would have to be compressed and heated before being introduced into the cabin. The airplane cabin has a volume of 100 m3 with air at the in-flight conditions of 25◦ C and 0.8 bar. What would be the cost of completely refreshing the air every minute if air has a heat capacity of CP∗ = 30 J/(mol K) and energy costs $0.2 per kW hr? R 3.38 Redo Problem 3.5 using Aspen Plus . R 3.39 Redo Problem 3.16a using Aspen Plus . R  3.40 Redo Problem 3.32 using Aspen Plus .

R Aspen Plus problem solutions are available for Problems 3.5, 3.16a and 3.32

Solution R and the IAPWS-95 model, see the file Illustration 3.5 in the Aspen 3.5 Using Aspen Plus folder on the website gives the results of 387.4K = 114.25◦ C R 3.16 Using Aspen Plus and the IAPWS-95 equations for steam, to obtain the exit condi◦ tions of 200 C an 0.3 MPa, the isentropic efficiency, by trial-and-error, was found to be 89.33%. This results in a work flow of 0.1627 kW/kg-hr. Therefore, the steam flow rate is 750 kW/0.1627 kW/kg-hr or 4609 kg/hr

Chapter

4 Entropy: An Additional Balance Equation Several of the illustrations and problems in Chapter 3 show that the equations of mass and energy conservation are not sufficient to solve all the thermodynamic energy flow problems in which we might be interested. To be more specific, these two equations are not always sufficient to determine the final values of two state variables, or the heat and work flows for a system undergoing a change of state. What is needed is a balance equation for an additional state variable. As we have seen, the principle of momentum conservation does not provide this additional equation. Although a large number of additional state variables could be defined and could serve as the basis for a new balance equation, these variables would have the common feature that they are not conserved quantities. Thus the internal generation term for each of these variables would, in general, be nonzero and would have to be evaluated for the balance equation to be of use. Clearly the most useful variable to introduce as the basis for a new balance equation is one that has an internal generation rate that can be specified and has some physical significance. Another defect in our present development of thermodynamics has to do with the unidirectional character of natural processes that was considered in Sec. 1.3. There it was pointed out that all spontaneous or natural processes proceed only in the direction that tends to dissipate the gradients in the system and thus lead to a state of equilibrium, and never in the reverse direction. This characteristic of natural processes has not yet been included in our thermodynamic description. To complete our thermodynamic description of pure component systems, it is therefore necessary that we (1) develop an additional balance equation for a state variable and (2) incorporate into our description the unidirectional character of natural processes. In Sec. 4.1 we show that both these objectives can be accomplished by introducing a single new thermodynamic function, the entropy. The remaining sections of this chapter are concerned with illustrating the properties and utility of this new variable and its balance equation.

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 4 The goals of this chapter are for the student to: • Be able to use the rate-of-change form of the pure component entropy balance in

problem solving 99

100 Chapter 4: Entropy: An Additional Balance Equation • Be able to use the difference form of the pure component entropy balance in prob-

lem solving

• Be able to calculate the entropy change between two states of an ideal gas • Be able to calculate the entropy change of a real fluid using thermodynamic prop-

erties charts and tables

• To understand the concept of availability and be able to compute the maximum

useful shaft work that can be obtained in a change of state

NOTATION INTRODUCED IN THIS CHAPTER A A ˆ A, Bˆ G G Q˙ R S S Sˆ ˙ Sgen Sgen TR W rev γ

Helmholtz energy = U + P V (J) Molar Helmholtz energy (J/mol) Availability Gibbs energy = H + P V (J) Molar Gibbs energy (J/mol) Radiant heat flux (J/s) Entropy (J/K) Entropy per mole [J/(mol K)] Entropy per unit mass [J/(kg K)] Rate at which entropy is generated within the system [J/(mol K s)] Entropy generated within the system [J/(mol K)] Temperature of body emitting radiation (K) Work in a reversible process (J) = CP /CV

4.1 ENTROPY: A NEW CONCEPT We take as the starting point for the identification of an additional thermodynamic variable the experimental observation that all spontaneous processes occurring in an isolated constant-volume system result in the evolution of the system to a state of equilibrium (this is a special case of experimental observation 5, Sec. 1.7). The problem is to quantify this qualitative observation. We can obtain some insight into how to do this by considering the general balance equation (Eq. 2.1-4) for any extensive variable θ of a closed, isolated, constant-volume system     dθ Rate of change of Rate at which θ is generated = (4.1-1) = within the system θ in the system dt Alternatively, we can write Eq. 4.1-1 as

dθ = θ˙gen dt

(4.1-2)

where θ˙gen is the rate of internal generation of the yet-unspecified state variable θ. Now, if the system under consideration were in a true time-invariant equilibrium state, dθ/dt = 0 (since, by definition of a time-invariant state, no state variable can change with time). Thus θ˙gen = 0 at equilibrium (4.1-3)

4.1 Entropy: A New Concept 101 Equations 4.1-2 and 4.1-3 suggest a way of quantifying the qualitative observation of the unidirectional evolution of an isolated system to an equilibrium state. In particular, suppose we could identify a thermodynamic variable θ whose rate of internal generation θ˙gen was positive,1 except at equilibrium, where θ˙gen = 0. For this variable

dθ away from equilibrium >0 dt ⎫ dθ ⎬ =0 dt at equilibrium ⎭ or θ = constant

(4.1-4)

Furthermore, since the function θ is increasing in the approach to equilibrium, θ must be a maximum at equilibrium subject to the constraints of constant mass, energy, and volume for the isolated constant-volume system.2 Thus, if we could find a thermodynamic function with the properties given in Eq. 4.1-4, the experimental observation of unidirectional evolution to the equilibrium state would be built into the thermodynamic description through the properties of the function θ. That is, the unidirectional evolution to the equilibrium state would be mathematically described by the continually increasing value of the function θ and the occurrence of equilibrium in an isolated, constant-volume system to the attainment of a maximum value of the function θ. The problem, then, is to identify a thermodynamic state function θ with a rate of internal generation, θ˙gen , that is always greater than or equal to zero. Before searching for the variable θ, it should be noted that the property we are looking for is θ˙gen ≥ 0; this is clearly not as strong a statement as θ˙gen = 0 always, which occurs if θ is a conserved variable such as total mass or total energy, but it is as strong a general statement as we can expect for a nonconserved variable. We could now institute an extensive search of possible thermodynamic functions in the hope of finding a function that is a state variable and also has the property that its rate of internal generation is a positive quantity. Instead, we will just introduce this new thermodynamic property by its definition and then show that the property so defined has the desired characteristics. Definition The entropy (denoted by the symbol S ) is a state function. In a system in which ˙ s and P (dV /dt)] there are flows of both heat by conduction (Q˙ ) and work [W across the system boundaries, the conductive heat flow, but not the work flow, causes a change in the entropy of the system; this rate of entropy change is ˙ , where T is the absolute thermodynamic temperature of the system at Q/T we were to choose the other possibility, θ˙gen being less than zero, the discussion here would still be valid except that θ would monotonically decrease to a minimum, rather than increase to a maximum, in the evolution to the equilibrium state. The positive choice is made here in agreement with standard thermodynamic convention. 2 Since an isolated constant-volume system has fixed mass, internal energy, and volume, you might ask how θ can vary if M , U , and V or alternatively, the two state variables U and V are fixed. The answer is that the discussion of Secs. 1.3 and 1.6 established that two state variables completely fix the state of a uniform one-component, one-phase system. Consequently, θ (or any other state variable) can vary for fixed U and V in (1) a nonuniform system, (2) a multicomponent system, or (3) a multiphase system. The first case is of importance in the approach to equilibrium in the presence of internal relaxation processes, and the second and third cases for chemical reaction equilibrium and phase equilibrium, which are discussed later in this book. 1 If

102 Chapter 4: Entropy: An Additional Balance Equation the point of the heat flow. If, in addition, there are mass flows across the system boundaries, the total entropy of the system will also change due to this convected flow. That is, each element of mass entering or leaving the system carries with it its entropy (as well as internal energy, enthalpy, etc.). Using this definition and Eq. 2.1-4, we have the following as the balance equation for entropy:3 Entropy balance, also called the second law of thermodynamics, for an open system

K Q˙ dS  ˙ ˆ = Mk Sk + + S˙ gen dt T

(4.1-5a)

k=1

where K 

M˙ k Sˆk = net rate of entropy flow due to the flows of mass into and out of the k=1 system (Sˆ = entropy per unit mass) ˙ Q = rate of entropy flow due to the flow of heat across the system boundary T S˙ gen = rate of internal generation of entropy within the system.

˙ k = 0), we have For a system closed to the flow of mass (i.e., all M dS Q˙ = + S˙ gen dt T

Entropy balance for a closed system

(4.1-5b)

Based on the discussion above, we then have (from Eqs. 4.1-3 and 4.1-4)

Sgen ≥ 0,

and

dS = 0 at equilibrium dt

(4.1-5c)

Equations 4.1-5 are usually referred to as the second law of thermodynamics. Before we consider how the entropy balance, Eq. 4.1-5a, will be used in problem solving, we should establish (1) that the entropy function is a state variable, and (2) that it has a positive rate of internal generation, that is, S˙ gen ≥ 0. Notice that Eq. 4.1-5 cannot provide general information about the internal generation of entropy since it is an equation for the black-box description of a system, whereas S˙ gen depends on the detailed internal relaxation processes that occur within the system. In certain special cases, however, one can use Eq. 4.1-5 to get some insight into the form of S˙ gen . To see this, consider the thermodynamic system of Fig. 4.1-1, which is a composite of two subsystems, A and B . These subsystems are well insulated except at their interface, so that the only heat transfer that occurs is a flow of heat from the high-temperature subsystem A to the low-temperature subsystem B . We assume that the resistance to heat transfer at this interface is large relative to the internal resistances of the subsystems (which would occur if, for example, the subsystems were well-mixed liquids or highly conducting solids), so that the temperature of each subsystem is uniform, but varying with time. 3 For simplicity, we have assumed that there is only a single heat flow into the system. If there are multiple heat flows

 ˙ ˙ by conduction, the term Q/T is to be replaced by a Qj /Tj , where Q˙ j is the heat flow and Tj the temperature at the jth heat flow port into the system.

4.1 Entropy: A New Concept 103

A

B

Figure 4.1-1 Systems A and B are free to interchange energy, but the composite system (A + B ) is isolated from the environment.

In this situation the heat transfer process occurs in such a way that, at any instant, each subsystem is in a state of internal thermal equilibrium (i.e., if the two subsystems were suddenly separated, they would be of uniform but different temperatures and would not change with time); however, the composite system consisting of both subsystems (at different temperatures) is not in thermal equilibrium, as evidenced by the fact that though the composite system is isolated from the environment, its properties are changing with time (as heat is transferred from A to B and the temperatures of these subsystems are changing). The rate-of-change form of the entropy balances for subsystems A and B , which are passing through a succession of equilibrium states,4 and therefore have no internal generation of entropy, are   dSA TA − T B Q˙ A = −h (4.1-6a) = dt TA TA and

  dSB TA − TB Q˙ B = +h = dt TB TB

(4.1-6b)

In writing these equations we have recognized that the amount of heat that leaves subsystem A enters subsystem B , and that the heat flow from A to B is proportional to the temperature difference between the two systems, that is,

Q˙ A = −Q˙ B = −h (TA − TB ) where h is the heat transfer coefficient (experimental observation 10 of Sec. 1.7). The entropy balance for the isolated (Q˙ = 0), nonequilibrium composite system composed of subsystems A and B is

dS (4.1-7) = S˙ gen dt Since the total entropy is an extensive property, S = SA +SB , and Eqs. 4.1-6 and 4.1-7 can be combined to yield     h(TA − TB )2 h(ΔT )2 TA − TB T A − TB +h = = (4.1-8) S˙ gen = −h TA TB TA TB TA TB Since h, TA , and TB are positive, Eq. 4.1-8 establishes that for this simple example the entropy generation term is positive. It is also important to note that S˙ gen is proportional to the second power of the system nonuniformity, here (ΔT )2 . Thus, S˙ gen is positive away from equilibrium (when TA = TB ) and equal to zero at equilibrium. 4A

process in which a system goes through a succession of equilibrium states is termed a quasistatic process.

104 Chapter 4: Entropy: An Additional Balance Equation This expression for the rate of entropy generation was obtained by partitioning a black-box system into two subsystems, resulting in a limited amount of information about processes internal to the overall system. In Sec. 4.6 a more general derivation of the entropy generation term is given, based on the detailed microscopic description introduced in Sec. 2.4, and it is shown that the entropy is indeed a state function and that S˙ gen is positive, except in the equilibrium state, where it is equal to zero. It is also established that S˙ gen is proportional to the second power of the gradients of temperature and velocity; thus the rate of generation of entropy is related to the square of the departure from the equilibrium state. Table 4.1-1 gives several special cases of the entropy balance equation, on both a mass and a molar basis, for situations similar to those considered for the mass and energy balance equations in Tables 2.2-1, 3.1-1, and 3.1-2. Frequently, one is interested in the change in entropy of a system in going from state 1 to state 2, rather than the rate of change of entropy with time. This entropy change can be determined by integrating Eq. 4.1-5 over the time interval t1 to t2 , where (t2 − t1 ) is the (perhaps unknown) time required to go between the two states. The result is Difference form of entropy balance

S2 − S1 =

 k

t2

 M˙ k Sˆk dt +

t1

Q˙ dt + Sgen T

t2

t1

(4.1-9)

where



t2

Sgen = total entropy generated =

S˙ gen dt

t1

Since S˙ gen is always greater than or equal to zero in any process, it follows that its integral Sgen must also be greater than or equal to zero. There are two important simplifications of Eq. 4.1-9. First, if the entropy per unit mass of each stream entering and leaving the system is constant in time (even though Table 4.1-1 Rate-of-Change Form of the Entropy Balance General equation:

K ˙  dS ˙ k Sˆk + Q + S˙ gen M = dt T k=1

Special cases: (i) Closed system

set M˙ k = 0 Q˙ dS = + S˙ gen dt T set Q˙ = 0 set S˙ gen = 0

so (ii) Adiabatic process (iii) Reversible process (iv) Open steady-state system

(a)

(b) in Eqs. a, b, and e in Eqs. a, b, and e

(c) (d)

dS =0 dt

so (v) Uniform system

0=

K 

˙ ˙ k Sˆk + Q + S˙ gen M T k=1 S = M Sˆ

(e) in Eqs. a and b

(f)

Note: To obtain the entropy balance on a molar basis, replace M˙ k Sˆk by N˙ k S k , and M Sˆ by N S , where S is the entropy per mole of fluid.

4.1 Entropy: A New Concept 105 the flow rates may vary), we have     t2 ˙ ˆ Mk Sk dt = Sˆk t1

k

t2

M˙ k dt =



t1

k

ΔMk Sˆk

k

t2

˙ k dt is the total mass that has entered the system from the kth where (ΔM )k = t1 M stream. Next, if the temperature is constant at the location where the heat flow occurs, then   t2 ˙ 1 t2 ˙ Q Q dt = Q dt = T t1 T t1 T t where Q = t12 Q˙ dt is the total heat flow into the system between t1 and t2 . If either of these simplifications is not valid, the respective integrals must be evaluated if Eq. 4.1-9 is to be used. This may be a difficult or impossible task, so that, as with the energy balance, the system for which the entropy balance is to be written must be chosen with care, as illustrated later in this chapter. Table 4.1-2 summarizes various forms of the integrated entropy balance. It should be pointed out that we have introduced the entropy function in an axiomatic and mathematical fashion. In the history of thermodynamics, entropy has been presented in many different ways, and it is interesting to read about these alternative approaches. One interesting source is The Second Law by P. W. Atkins (W. H. Freeman, New York, 1984). The axiom that is used here, Sgen ≥ 0, is a statement of what has historically been called the second law of thermodynamics. The principle of conservation of energy discussed previously is referred to as the first law. It is interesting to compare the form of the second law used here with two other forms that have been proposed in the history of thermodynamics, both of which deal with the transformation of heat to work. Table 4.1-2 Difference Form of the Entropy Balance S2 − S 1 =

General equation

K   k=1

Special cases: (i) Closed system

t2

 ˙ k Sˆk dt + M

t1

t2 t1

set M˙ k = 0  t2 ˙ Q S2 − S1 = dt + Sgen t1 T

so (ii) Adiabatic process



t2

set t1

Q˙ dt + Sgen T

(a)

in Eq. a (b)

Q˙ dt = 0 T

in Eq. a

(c)

(iii) Reversible process set Sgen = 0 in Eqs. a and b (iv) Open system: Flow of fluids of constant thermodynamic properties set

K  

k=1

(v) Uniform system

t2

˙ k Sˆk dt = M

t1

K 

ΔMk Sˆk

in Eq. a

(d)

(e)

k=1

S = M Sˆ

Note: To obtain the entropy balance on a molar basis, replace M Sˆ by N S , and ΔMk Sˆk with ΔNk S k .

in Eq. a  t2 t1

˙ k Sˆk dt by M

(f)  t2 t1

N˙ k S k dt,

106 Chapter 4: Entropy: An Additional Balance Equation Illustration 4.1-1 Clausius Statement of the Second Law The second-law statement of Rudolf Clausius (1822–1888) is that it is not possible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a colder body to a hotter body. Show from the axiom Sgen ≥ 0 that the process below is impossible, so that the Clausius statement of the second law is consistent with what has been presented in this book. T2 Q2 Clausius device

T2 > T1

Q1 T1

Solution The energy balance over one complete cycle, so that the device is in the same state at the beginning and at the end of the process, is −→

Uf − Ui = 0 = Q1 + Q2 + W

0

since there is no work produced or absorbed in the device. Therefore, the energy balance yields Q2 = −Q1 . Thus the energy balance tells us that the process may be possible as long as the heat flows in and out balance, that is, if energy is conserved. The entropy balance over one complete cycle is Sf − Si = 0 =

Q1 Q2 + + Sgen T1 T2

or Sgen = −

Q1 Q2 Q1 Q1 − = − = Q1 T2 T1 T2 T1



1 1 − T2 T1



The second-law statement we have used is Sgen > 0. However, since T2 > T1 , if Q1 were positive (heat flow into device from cold body), Sgen < 0, which is a violation of our axiom. If Q1 were negative, so that the heat flow was from high temperature to low temperature (or equivalently T1 > T2 ), the process would be possible. Consequently, our statement of the second law, Sgen ≥ 0, is consistent with the Clausius version, but much more general.

Illustration 4.1-2 Kelvin-Planck Statement of the Second Law An alternative statement of the second law, due to Lord Kelvin (William Thomson, 1824–1907) and Max Planck (1858–1947), is that it is not possible to construct a device operating in a cycle that results in no effect other than the production of work by transferring heat from a single body. A schematic diagram of a Kelvin-Planck device is shown below. Kelvin-Planck device

W

Q T

Show from our axiom Sgen ≥ 0 that this process is indeed impossible.

4.1 Entropy: A New Concept 107 Solution The energy balance over one complete cycle is Uf − Ui = 0 = Q + W

or

W = −Q

As long as the work produced equals the heat absorbed, this device satisfies the principle of energy conservation, and is possible by the first law of thermodynamics. The entropy balance over one complete cycle is Sf − Si = 0 =

Q + Sgen T

or Sgen = −

Q T

Since T is positive (remember, we are using absolute temperature), for the entropy generation to be positive, Q must be negative. That is, to be consistent with our statement of the second law, the device cannot absorb heat and convert all of it to work. However, the reverse process, in which the device receives work and converts all that work to heat, is possible. Therefore, our statement of the second law is consistent with that of Kelvin and Planck, but again more general.

Heat flow by radiation

An alternative wording of the Kelvin-Planck statement is that heat cannot be completely converted to work in a cyclic process. However, it is possible, as shown above, to do the converse and completely convert work to heat. Since heat cannot be completely converted to work, heat is sometimes considered a less useful form of energy than work. When work or mechanical energy is converted to heat, for example, by friction, it is said to be degraded. Finally, notice the very different roles of the first law (energy conservation) and second law (Sgen ≥ 0) in the analyses above. The first law did not provide a definitive result as to whether a process was possible or not, but merely set some constraints (i.e., Q2 = −Q1 for the Clausius device and W = −Q for the Kelvin-Planck device). However, the second law is definitive in that it establishes that it is impossible to construct devices that would operate in the manners proposed. So far we have considered the heat flow Q˙ to be due only to conduction across the system boundaries. However, heat can also be transmitted to a system by radiation—for example, as the sun heats the earth. If there is a radiative heat flow, it can be incorporated into the energy balance as an additional energy flow that can be combined with the conductive heat flow. The change in entropy accompanying radiative heat transfer is more complicated than for conductive heat transfer, and must be treated separately. In particular, the entropy flow depends on the distribution of energy in the radiation, which in turn depends on the source and temperature of the radiation. Radiation can be monochromatic, that is, of a single frequency, when produced by a laser, or it can have a broad spectrum of energies, for example, following the Stefan-Boltzmann law for typical radiating bodies, such as the sun, a red-hot metal, or even our bodies (which do radiate heat). Further, the frequency distribution (or energy spectrum) emitted from fluorescent and incandescent lamps is different. Here we will consider only the entropic contribution from the most common form of radiation, that which follows the Stefan-Boltzmann law. In this case, for a radiant energy transfer to a system of Q˙ R the rate of entropy change is Entropy flow accompanying radiation =

4Q˙ R 3TR

108 Chapter 4: Entropy: An Additional Balance Equation where TR is the temperature of the body emitting the radiation, not the temperature of the receiving body. Therefore, the entropy balance for a system, including heat transfer by radiation (from a body or device emitting radiation with a Stefan-Boltzmann energy distribution), is, on a mass basis K Q˙ 4Q˙ R dS  ˙ ˆ + S˙ gen = Mk Sk + + dt T 3TR

(4.1-10a)

K 4Q˙ R dS  ˙ Q˙ = + S˙ gen Nk S k + + dt T 3TR

(4.1-10b)

k=1

and on a molar basis

k=1

Since radiative heat transfer is not generally important in chemical engineering applications of thermodynamics, Eqs. 4.1-10a and b will not be used much in this book.

4.2 THE ENTROPY BALANCE AND REVERSIBILITY Reversible processes

An important class of processes is that for which the rate of generation of entropy is always zero. Such processes are called reversible processes and are of special interest in thermodynamics. Since S˙ gen is proportional to the square of the temperature gradients and velocity gradients in the system, such gradients must vanish in a process in which S˙ gen is zero. Notice, however, that although the rate of entropy generation is second order in the system gradients, the internal relaxation processes that occur in the approach to equilibrium are linearly proportional to these gradients (i.e., the heat flux q is proportional to the temperature gradient ∇T , the stress tensor is proportional to the velocity gradient, etc. as shown in Sec. 1.7). Therefore, if there is a very small temperature gradient in the system, the heat flux q, which depends on ∇T , will be small, and S˙ gen , which depends on (∇T )2 , may be so small as to be negligible. Similarly, the rate of entropy generation S˙ gen may be negligible for very small velocity gradients. Processes that occur with such small gradients in temperature and velocity that S˙ gen is essentially zero can also be considered to be reversible. The designation reversible arises from the following observation. Consider the change in state of a general system open to the flow of mass, heat, and work, between two equal time intervals, 0 to t1 and t1 to t2 , where t2 = 2t1 . The mass, energy, and entropy balances for this system are, from Eqs. 2.2-4, 3.1-6, and 4.1-9,

M2 − M0 =



U2 − U0 =

M˙ k dt +

0

k



t1

t1



k



0

k



t2

+ t1





t2

M˙ k dt

t1

 dV ˙ s + Q˙ dt ˆk − P +W M˙ k H dt

 k

 dV ˙ s + Q˙ dt ˆk − P +W M˙ k H dt

4.2 The Entropy Balance and Reversibility 109 and

   t2  ˙ ˙ Q Q S2 − S0 = dt + dt M˙ k Sˆk + M˙ k Sˆk + T T 0 t 1 k k  t2  t1 S˙ gen dt + S˙ gen dt + 

t1





0

t1

Now, suppose that all of the mass, heat, and work flows are just reversed between t1 and t2 from what they had been between 0 and t1 , so that  t1  t1  t2  t2 ˙ ˙ ˙ Mk dt = − Mk dt Q dt = − Q˙ dt 0



t1

t1



ˆ k dt = − M˙ k H

0

t2

 ˆ k dt M˙ k H

t



t1

M˙ k Sˆk dt = −

0



 M˙ k Sˆk dt

t

t1

P 0

 1t2

dV dt = − dt

t1

t1

0 t1

Q˙ dt = − T

˙ s dt = − W



P

t2

t

 1t2

0

 1t2 t1

0

Q˙ dt T

(4.2-1)

˙ s dt W

t1

dV dt dt

In this case the equations reduce to

M 2 = M0 U2 = U0  S2 = S0 + 0

t1

 S˙ gen dt +

t2

(4.2-2a) (4.2-2b)

S˙ gen dt

(4.2-2c)

t1

In general, S˙ gen ≥ 0, so the two integrals in Eq. 4.2-2c will be positive, and the entropy of the initial and final states will differ. Thus the initial and final states of the system will have different entropies and therefore must be different. If, however, the changes were accomplished in such a manner that the gradients in the system over the whole time interval are infinitesimal, then S˙ gen = 0 in each time interval, and S2 = S0 . In this case the system has been returned to its initial state from its state at t1 by a process in which the work and each of the flows were reversed. Such a process is said to be reversible. If S˙ gen had not been equal to zero, then S2 > S0 , and the system would not have been returned to its initial state by simply reversing the work and other flows; the process is then said to be irreversible. The main characteristic of a reversible process is that it proceeds with infinitesimal gradients within the system. Since transport processes are linearly related to the gradients in the system, this requirement implies that a reversible change occurs slowly on the time scale of macroscopic relaxation times. Changes of state in real systems can be approximated as being reversible if there is no appreciable internal heat flow or viscous dissipation; they are irreversible if such processes occur. Consequently, expansions and compressions that occur uniformly throughout a fluid, or in well-designed turbines, compressors, and nozzles for which viscous dissipation and internal heat flows are unimportant, can generally be considered to occur reversibly (i.e., S˙ gen = 0). Flows through pipes, through flow constrictions (e.g., a valve or a porous plug), and through shock waves all involve velocity gradients and viscous dissipation and hence are irreversible. Table 4.2-1 contains some examples of reversible and irreversible processes.

110 Chapter 4: Entropy: An Additional Balance Equation Table 4.2-1 Examples of Reversible and Irreversible Processes Reversible process

Reversible process: A process that occurs with no (appreciable) internal temperature, pressure, or velocity gradients, and therefore no internal flows or viscous dissipation Examples: Fluid flow in a well-designed turbine, compressor, or nozzle Uniform and slow expansion or compression of a fluid Many processes in which changes occur sufficiently slowly that gradients do not appear in the system

Irreversible process

Irreversible process: A process that occurs with internal temperature, pressure, and/or velocity gradients so that there are internal flows and/or viscous dissipation Examples: Flow of fluid in a pipe or duct in which viscous forces are present Flow of fluid through a constriction such as a partially open valve, a small orifice, or a porous plug (i.e., the Joule-Thomson expansion) Flow of a fluid through a sharp gradient such as a shock wave Heat conduction process in which a temperature gradient exists Any process in which friction is important Mixing of fluids of different temperatures, pressures, or compositions

Another characteristic of a reversible process is that if the surroundings are extracting work from the system, the maximum amount of work is obtained for a given change of state if the process is carried out reversibly (i.e., so that S˙ gen = 0). A corollary to this statement is that if the surroundings are doing work on the system, a minimum amount of work is needed for a given change of state if the change occurs reversibly. The first of these statements is evident from Illustration 3.4-7, where it was found that the maximum work (W NET ) was extracted from the expansion of a gas in a piston-andcylinder device between given initial and final states if the expansion was carried out reversibly (by removing grains of sand one at a time), so that only differential changes were occurring, and there was no frictional dissipation of kinetic energy to thermal energy in the work-producing device. It will be shown, in Illustration 4.5-8, that such a process is also one for which Sgen = 0. Although few processes are truly reversible, it is sometimes useful to model them to be so. When this is done, it is clear that any computations made based on Eqs. 4.1-5 or 4.1-9, with S˙ gen = Sgen = 0, will only be approximate. However, these approximate results may be very useful, since the term neglected (the entropy generation) is of known sign, so we will know whether our estimate for the heat, work, or any state variable is an upper or lower bound to the true value. To see this, consider the energy and entropy balances for a closed, isothermal, constant-volume system:

U2 = U1 + Q + Ws

(4.2-3)

Q + Sgen T

(4.2-4)

and

S2 = S1 +

4.2 The Entropy Balance and Reversibility 111 Eliminating Q between these two equations and using T1 = T2 = T gives

Ws = (U2 − T2 S2 ) − (U1 − T1 S1 ) + T Sgen = A2 − A1 + T Sgen

(4.2-5)

Here we have defined a new thermodynamic state variable, the Helmholtz energy, as

A = U − TS

Helmholtz energy

(4.2-6)

(Note that A must be a state variable since it is a combination of state variables.) The work required to bring the system from state 1 to state 2 by a reversible (i.e., Sgen = 0), isothermal, constant-volume process is Reversible work at constant N , V and T

Wsrev = A2 − A1

(4.2-7)

while the work in an irreversible (i.e., Sgen > 0), isothermal, constant-volume process between the same initial and final states is

Ws = A2 − A1 + T Sgen = Wsrev + T Sgen Since T Sgen > 0, this equation establishes that more work is needed to drive the system from state 1 to state 2 if the process is carried out irreversibly than if it were carried out reversibly. Conversely, if we are interested in the amount of work the system can do on its surroundings at constant temperature and volume in going from state 1 to state 2 (so that Ws is negative), we find, by the same argument, that more work is obtained if the process is carried out reversibly than if it is carried out irreversibly. One should not conclude from Eq. 4.2-7 that the reversible work for any process is equal to the change in Helmholtz energy, since this result was derived only for an isothermal, constant-volume process. The value of Wsrev , and the thermodynamic functions to which it is related, depends on the constraints placed on the system during the change of state (see Problem 4.3). For example, consider a process occurring in a closed system at fixed temperature and pressure. Here we have

U2 = U1 + Q + Ws − (P2 V2 − P1 V1 ) where P2 = P1 , T2 = T1 , and

S2 = S1 +

Q + Sgen T

Thus Reversible work at constant N , P and T

Wsrev = G2 − G1 where

G ≡ U + PV − TS = H − TS Gibbs energy

(4.2-8)

G is called the Gibbs energy. Therefore, for the case of a closed system change at constant temperature and pressure, we have Ws = G2 − G1 + T Sgen = Wsrev + T Sgen

112 Chapter 4: Entropy: An Additional Balance Equation The quantity T Sgen in either process can be interpreted as the amount of mechanical energy that has been converted to thermal energy by viscous dissipation and other system irreversibilities. To see this, consider a reversible process (Sgen = 0) in a closed system. The energy balance is

Qrev = U2 − U1 − W rev (here W is the sum of the shaft work and P ΔV work). If, on the other hand, the process was carried out between the same initial and final states so that macroscopic gradients arose in the system (that is, irreversibly, so Sgen > 0), then

Q = U2 − U1 − W = U2 − U1 − W rev − T Sgen or

Q = Qrev − T Sgen

(4.2-9a)

W = W rev + T Sgen

(4.2-9b)

and Since T Sgen is greater than zero, less heat and more work are required to accomplish a given change of state in the second (irreversible) process than in the first.5 This is because the additional mechanical energy supplied in the second case has been converted to thermal energy. It is also generally true that system gradients that lead to heat flows, mass flows, and viscous dissipation result in the conversion of mechanical energy to thermal energy and in a decrease in the amount of work that can be obtained from a work-producing device between fixed initial and final states. Finally, since the evaluation of the changes in the thermodynamic properties of a system accompanying its change of state are important in thermodynamics, it is useful to have an expression relating the entropy change to changes in other state variables. To obtain this equation we start with the rate-of-change form of the mass, energy, and entropy balances for a system in which the kinetic and potential energy terms are unimportant, there is only one mass flow stream, and the mass and heat flows occur at the common temperature T :

dM = M˙ dt dU ˆ + Q˙ − P dV + W ˙s = M˙ H dt dt and

Q˙ dS = M˙ Sˆ + + S˙ gen dt T ˆ ˆ where H and S are the enthalpy per unit mass and entropy per unit mass of the fluid ˙ between these equations and multiplying entering or leaving the system. Eliminating M the third equation by T yields dU ˆ dM + Q˙ − P dV + W ˙s =H dt dt dt

5 The

argument used here assumes that work is required to drive the system from state 1 to state 2. You should verify that the same conclusions would be reached if work were obtained in going from state 1 to state 2.

4.2 The Entropy Balance and Reversibility 113 and

dM dS = T Sˆ + Q˙ + T S˙ gen dt dt Of particular interest is the form of these equations applicable to the change of state occurring in the differential time interval dt. As in Sec. 3.1, we write T

Q = Q˙ dt = heat flow into the system in the time interval dt ˙ s dt = shaft work into the system in the time interval dt Ws = W ˙ Sgen = Sgen dt = entropy generated in the system in the time interval dt and obtain the following balance equations for the time interval dt:

ˆ dM + Q − P dV + Ws dU = H

(4.2-10a)

T dS = T Sˆ dM + Q + T Sgen

(4.2-10b)

and

Equations 4.2-10 can be used to interrelate the differential changes in internal energy, entropy, and volume that occur between fixed initial and final states that are only slightly different. This is accomplished by first solving Eq. 4.2-10b for Q,

Q = T dS − T Sgen − T Sˆ dM and using this result to eliminate the heat flow from Eq. 4.2-10a, to obtain

ˆ − T S) ˆ dM dU = T dS − T Sgen − P dV + Ws + (H ˆ = T dS − T Sgen − P dV + Ws + G dM

(4.2-11a) 6

(4.2-11b)

For a reversible process (Sgen = 0), these equations reduce to

Qrev = T dS − T Sˆ dM

(4.2-12a)

ˆ dM dU = T dS + (−P dV + Ws )rev + G

(4.2-12b)

and

In writing these equations we have recognized that since the initial and final states of the system are fixed, the changes in the path-independent functions are the same for reversible and irreversible processes. The heat and work terms that depend on the path followed are denoted by the superscript rev. Equating the changes in the (state variables) internal energy and entropy for the reversible process (Eqs. 4.2-12) and the irreversible process (Eqs. 4.2-11) yields

Qrev = Qirrev + T Sgen and

(−P dV + Ws )rev = (−P dV + Ws )irrev − T Sgen

6 Alternatively,

this equation and those that follow can be written in terms of the molar Gibbs energy and the mole ˆ dM by G dN . number change by replacing G

114 Chapter 4: Entropy: An Additional Balance Equation Substituting these results into Eqs. 4.2-11 again gives Eqs. 4.2-12, which are now seen to apply for any process (reversible or irreversible) for which kinetic and potential changes are negligible. There is a subtle point here. That is, Eq. 4.2-12b can be used to interrelate the internal energy and entropy changes for any process, reversible or irreversible. However, to relate the heat flow Q and the entropy, Eq. 4.2-11a is always valid, whereas Eq. 4.2-12a can be used only for reversible processes. Although both Eqs. 4.2-11b and 4-2.12b provide relationships between the changes in internal energy, entropy, mass, and the work flow for any real process, we will, in fact use only Eq. 4.2-12b to interrelate these changes since, in many cases, this simplifies the computation. Finally, we note that for a system with no shaft work7 Differential entropy change for an open system

ˆ dM dU = T dS − P dV + G

(4.2-13a)

and, further, if the system is closed to the flow of mass,

dU = T dS − P dV or Differential entropy change for a closed system

dU = T dS − P dV

(4.2-13b)

Since we are always free to choose the system for a given change of state such that there is no shaft work (only P ΔV work), Eq. 4.2-13a can generally be used to compute entropy changes in open systems. Similarly, Eq. 4.2-13b can generally be used to compute entropy changes in closed systems. This last point needs to be emphasized. In many cases it is necessary to compute the change in thermodynamic properties between two states of a substance. Since, for this computation, we can choose the system in such a way that it is closed and shaft work is excluded, Eq. 4.2-13b can be used in the calculation of the change in thermodynamic properties of the substance between the given initial and final states, regardless of the device or path used to accomplish this change.

4.3 HEAT, WORK, ENGINES, AND ENTROPY The discussion of Sec. 4.2, and especially Eqs. 4.2-9, reveals an interesting distinction between mechanical energy (and work) and heat or thermal energy. In particular, while both heat and work are forms of energy, the relaxation processes (i.e., heat flow and viscous dissipation) that act naturally to reduce any temperature and velocity gradients in the system result in the conversion of mechanical energy in the form of work or the potential to do work, to heat or thermal energy. Friction in any moving object, which reduces its speed and increases its temperature, is the most common example of this phenomenon. A problem of great concern to scientists and engineers since the late eighteenth century has been the development of devices (engines) to accomplish the reverse transformation, the conversion of heat (from the combustion of wood, coal, oil, natural gas, and other fuels) to mechanical energy or work. Much effort has been spent on developing engines of high efficiency, that is, engines that convert as large a fraction of the heat supplied to useful work as is possible. 7 Equivalently,

on a molar basis, we have dU = T dS − P dV + G dN .

4.3 Heat, Work, Engines, and Entropy 115

Q1 T1

Heat engine

Q2 T2

N1

Fluid flow engine (turbine)

W

Q, Ws

(a)

(b)

N2

Figure 4.3-1 (a) Schematic diagram of a simple heat engine. (b) Schematic diagram of a fluid flow engine.

Such engines are schematically represented in Fig. 4.3-1a, where Q˙ 1 is the heat flow rate into the engine from the surroundings at temperature T1 , Q˙ 2 is the heat flow rate ˙ is the rate at which work from the engine to its surroundings at temperature T2 , and W is done by the engine. Remember that from the Clausius and Kelvin-Planck statements discussed earlier that Q˙ 2 cannot be zero; that is, some of the heat entering the engine cannot be converted to work and must be expelled at a lower temperature. The engine in Fig. 4.3-1a may operate either in a steady-state fashion, in which case ˙ are independent of time, or cyclically. If the energy and entropy balQ˙ 1 , Q˙ 2 , and W ances for the engine are integrated over a time interval Δt, which is the period of one cycle of the cyclic engine, or any convenient time interval for the steady-state engine, one obtains8 0 = Q1 + Q2 + W (4.3-1)

Q1 Q2 0= + + Sgen (4.3-2) T1 T2 t t ˙ s − P (dV /dt)) dt. The left sides of Eqs. 4.3-1 where Q = t12 Q˙ dt and W = t12 (W and 4.3-2 are zero for the cyclic and steady-state engines, as both engines are in the same state at time t2 as they were at t1 . Eliminating Q2 between Eqs. 4.3-1 and 4.3-2 yields   T1 − T2 − T2 Sgen (4.3-3) Work done by the engine = −W = Q1 T1 Based on the discussion of the previous section, to obtain the maximum work from an engine operating between fixed temperatures T1 and T2 it is necessary that all processes be carried out reversibly, so that Sgen = 0. In this case   T 1 − T2 (4.3-4) Maximum work done by the engine = −W = Q1 T1 ⎛ ⎞ and Fraction of heat −W T1 − T 2 = (4.3-5) Engine efficiency = ⎝ supplied that is ⎠ = Q T1 1 converted to work

8 Independent

of the direction the arrows have been drawn in these figures, we still use the sign convention that a flow of heat, work, or mass into the system or device is positive, and a flow out is negative. The arrows in this and the following figures are drawn to remind the reader of the expected directions of the flows. Consequently, Q˙ 2 in Fig. 4.3-1a will be negative in value.

116 Chapter 4: Entropy: An Additional Balance Equation This is a surprising result since it establishes that, independent of the engine design, there is a maximum engine efficiency that depends only on the temperature levels between which the engine operates. Less than ideal design or irreversible operation of the engine will, of course, result in the efficiency of an engine being less than the maximum efficiency given in Eq. 4.3-5. Industrial heat engines, due to design and operating limitations, heat losses, and friction, typically operate at about half this efficiency. Another aspect of the conversion of heat to work can be illustrated by solving for Q2 , the heat leaving the engine, in Eqs. 4.3-1 and 4.3-3 to get T2 + T2 Sgen (4.3-6) −Q2 = Q1 T1 This equation establishes that it is impossible to convert all the heat supplied to an engine to work (that is, for Q2 to equal zero) in a continuous or cyclic process, unless the engine has a lower operating temperature (T2 ) of absolute zero. In contrast, the inverse process, that of converting work or mechanical energy completely to heat, can be accomplished completely at any temperature, and unfortunately occurs frequently in natural processes. For example, friction can convert mechanical energy completely to heat, as when stopping an automobile by applying the brake, which converts mechanical energy (here kinetic energy) to heat in the brake drum. Therefore, it is clear that although heat and work are equivalent in the sense that both are forms of energy (experimental observations 3 and 4 of Sec. 1.7), there is a real distinction between them in that work or mechanical energy can spontaneously (naturally) be converted completely to heat or thermal energy, but thermal energy can, with some effort, be only partially converted to mechanical energy. It is in this sense that mechanical energy is regarded as a higher form of energy than thermal energy. At this point one might ask if it is possible to construct an engine having the efficiency given by Eq. 4.3-5. Nicolas L´eonard Sadi Carnot (1796–1832) described such a cyclic engine in 1824. The Carnot engine consists of a fluid enclosed in a frictionless piston-and-cylinder device, schematically shown in Fig. 4.3-2. Work is extracted from this engine by the movement of the piston. In the first step of the four-part cycle, the fluid is isothermally and reversibly expanded from volume Va to volume Vb at a constant temperature T1 by adding an amount of heat Q1 from the first heat source. The V mechanical work obtained in this expansion is Vab P dV . The next part of the cycle is a reversible adiabatic expansion of the fluid from the state (Pb , Vb , T1 ) to the state V (Pc , Vc , T2 ). The work obtained in this expansion is Vbc P dV and is gotten by directly converting the internal energy of the fluid to work. The next step in the cycle is to reversibly and isothermally compress the fluid to the state (Pd , Vd , T2 ). V The work done on the fluid in this process is Vcd P dV , and the heat removed is Q2 . The final step in the cycle is a reversible adiabatic compression to the initial state V (Pa , Va , T1 ). The work done on the fluid in this part of the process is Vda P dV . The complete work cycle is summarized in the following table. The energy and entropy balances for one complete cycle are 0 = W + Q1 + Q2 (4.3-7) Q1 Q2 0= + (4.3-8) T1 T2 where Carnot cycle work

 W =−

Vb

Va

 P dV −

Vc Vb

 P dV −

Vd Vc

 P dV −

Va

P dV Vd

4.3 Heat, Work, Engines, and Entropy 117 Heat bath at temperature T1 Heat bath at temperature T2 (a) a T1

b

P

a

b

d

c

T d T2

c

V

S

(b)

(c)

a

b

T

T c

0 0

S

b

d

c

T d

0

a

0 S

0

S

0

Qa→b

Qc→d

W NET

(d)

(e)

(f)

Figure 4.3-2 The Carnot cycle. (a) Schematic diagram of a Carnot engine. (b) The Carnot cycle on a pressure-volume plot. (c) The Carnot cycle on a temperature-entropy plot. (d) Heat flow into the cycle going from point a to b. (e) Heat flow into the cycle going from point c to point d. (f ) Net work flow.

Work Done on the Fluid

Path reversible isothermal

(Pa , Va , T1 ) −−−−−→ (Pb , Vb , T1 ) expansion

reversible

(Pb , Vb , T1 ) −−−−−→ (Pc , Vc , T2 ) adiabatic expansion

reversible isothermal

(Pc , Vc , T2 ) −−−−−→ (Pd , Vd , T2 ) compression reversible

(Pd , Vd , T2 ) −−−−−→ (Pa , Va , T1 ) adiabatic compression



− 

−

Heat Added to the Fluid

Vb

P dV

Q1

P dV

0

P dV

Q2

P dV

0

Va Vc Vb



− 

−

Vd Vc Va

Vd

118 Chapter 4: Entropy: An Additional Balance Equation Now using Eq. 4.3-8 to eliminate Q2 from Eq. 4.3-7 yields

 −W = Q1 −

T2 Q1 T1

 =

T1 − T 2 Q1 T1

(4.3-9)

and −W T1 − T 2 = Q1 T1

Carnot cycle efficiency

which is exactly the result of Eq. 4.3-4. Equations 4.3-5 and 4.3-6 then follow directly. Thus we conclude that the Carnot engine is the most efficient possible in the sense of extracting the most work from a given flow of heat between temperature baths at T1 and T2 in a cyclic or continuous manner. Perhaps the most surprising aspect of the Carnot cycle engine (or Eq. 4.3-4, for that matter) is that the work obtained depends only on T1 , T2 , and the heat flow Q1 , and does not depend on the working fluid, that is, which fluid is used in the piston-and-cylinder device. Consequently, the efficiency, −W/Q1 , of the Carnot cycle depends only on the temperatures T1 and T2 , and is the same for all fluids. Since the work supplied or obtained in each step of the Carnot cycle is expressible in the form − P dV , the enclosed area on the P-V diagram of Fig. 4.3-2b is equal to the total work supplied by the Carnot engine to its surroundings in one complete cycle. (You should verify that if the Carnot engine is driven in reverse, so that the cycle in Fig. 4.3-2b is traversed counterclockwise, the enclosed area is equal to the work absorbed by the engine from its surroundings in one cycle.9 ) Since the differential entropy change dS and the heat flow Q for a reversible process in a closed system are related as Q dS = or Q = T dS T the heat flows (and the work produced) can be related to areas on the T-S diagram for the process as follows. The heat flow into the Carnot cycle in going from point a to point b is equal to  Qa→b = T dS = T1 ΔSa→b = T1 · (Sb − Sa ) since the temperature is constant This heat flow is given by the area shown in Fig. 4.3-2d . Similarly, the heat flow from point c to point d is equal in magnitude to the area shown in Fig. 4.3-2e, but negative in value (since Sc is larger than Sd ). Finally, since from Eq. 4.3-7 the net work flow (work produced less work used in the isothermal compression step) is equal to the difference between the two heat flows into the Carnot cycle, this work flow is given by the rectangular area in Fig. 4.3-2f. A similar graphical analysis of the heat and work flows can be used for other cycles, as will be shown in Sec. 5.2. Thus, for reversible cycles, the P-V diagram directly supplies information about the net work flow, and the T-S diagram provides information about the net heat flow. 9 Note

that if the Carnot heat engine is operated as shown in Fig. 4.3-2 it absorbs heat from the high-temperature bath, exhausts heat to the low-temperature bath, and produces work. However, if the engine is operated in reverse, it accepts work, absorbs heat from the low-temperature bath, and exhausts heat to the high-temperature bath. In this mode it is operating as a refrigerator, air conditioner, or heat pump.

4.3 Heat, Work, Engines, and Entropy 119 Rotor Stator

Stationary blades

Rotating blades

Figure 4.3-3 (a) Sketch of an axial flow steam turbine. High pressure steam enters, is expanded against the rotating turbine blades and low pressure steam leaves going to a condenser. The expanding steam drives the turbine blades producing work that is captured through the shaft that may drive an electric generator. (b) Sketch of an axial flow compressor in which low pressure gas enters, is compressed by the rotating blades that are driven by a motor, and high pressure gas leaves the compressor. {Note that these two devices are the inverse of each other. The turbine starts with fluid at high pressure and produces work and a low pressure fluid, while the compressor starts with a low pressure fluid and requires work to produce a fluid of higher pressure. Adapted from “Fundamentals of Engineering Thermodynamics”, 5th ed. by M. J. Morgan and H. N. Shapiro, J. Wiley & Sons, Inc., 2004. Used with permission.}

For irreversible processes (i.e., processes for which Sgen = 0), the heat flow and entropy change are not simply related as above, and the area on a T-S diagram is not directly related to the heat flow. In addition to the Carnot heat engine, other cycles and devices may be used for the conversion of thermal energy to mechanical energy or work, although the conversion efficiencies for these other cycles, because of to the paths followed, are less than that of the Carnot engine. Despite their decreased efficiency, these other engines offer certain design and operating advantages over the Carnot cycle, and hence are more widely used. The efficiencies of some other cycles are considered in Sec. 5.2. Another class of work-producing devices are engines that convert the thermal energy of a flowing fluid to mechanical energy. Examples of this type of engine are the nozzleturbine systems of Fig. 4.3-3. Here a high-pressure, high-temperature fluid, frequently steam, is expanded through a nozzle to obtain a low-pressure, high-velocity gas. This gas then impinges on turbine blades, where the kinetic energy of the gas is transferred to the turbine rotor, and thus is available as shaft work. The resulting low-pressure, lowvelocity gas leaves the turbine. Of course, many other devices can be used to accomplish the same energy transformation. All of these devices can be schematically represented as shown in Fig. 4.3-1b. The steady-state mass, energy, and entropy balances on a molar basis for such heat engines are 0 = N˙ 1 + N˙ 2 or N˙ 2 = −N˙ 1 (4.3-10) ˙s 0 = (H − H )N˙ 1 + Q˙ + W 1

2

and

Q˙ (4.3-11) + S˙ gen T Here we have assumed that the kinetic and potential energy changes of the fluid entering and leaving the device cancel or are negligible (regardless of what happens internally) and that the heat flow Q˙ can be identified as occurring at a single temperature T .10 0 = (S 1 − S 2 )N˙ 1 +

˙ this is not the case, a sum of Q/T terms or an integral of the heat flux divided by the temperature over the surface of the system is needed.

10 If

120 Chapter 4: Entropy: An Additional Balance Equation

˙ s yields Solving these equations for the heat flow Q˙ and the work flow W Q˙ = −T (S 1 − S 2 )N˙ 1 − T S˙ gen

(4.3-12)

˙ s = −N˙ 1 [(H 1 − T S 1 ) − (H 2 − T S 2 )] + T S˙ gen W

(4.3-13)

and

Note that the quantity (H 1 − T S 1 ) is not equal to the Gibbs energy unless the temperature T at which heat transfer occurs is equal to the inlet fluid temperature (that is, G1 = H 1 − T1 S 1 ); a similar comment applies to the term (H 2 − T S 2 ). Several special cases of these equations are important. First, for the isothermal flow engine (i.e., for an engine in which the inlet temperature T1 , the outlet temperature T2 , and the operating temperature T are all equal), Eq. 4.3-13 reduces to

˙ s = −N˙ 1 (G1 − G2 ) + T S˙ gen W The maximum rate at which work can be obtained from such an engine for fixed inlet and exit pressures and fixed temperature occurs when the engine is reversible; in this case

˙ srev = −N˙ 1 (G1 − G2 ) W and the heat load for reversible, isothermal operation is

Q˙ rev = −T N˙ 1 (S 1 − S 2 ) Second, for adiabatic operation (Q˙ = 0) of a flow engine, we have from Eq. 4.3-10 that

˙ s = −N˙ 1 (H 1 − H 2 ) W

(4.3-14)

0 = N˙ 1 (S 1 − S 2 ) + S˙ gen

(4.3-15)

and

so that the work flow is proportional to the difference in the enthalpies of the inlet and exiting fluids. Usually, the inlet temperature and pressure and the exit pressure of the adiabatic flow engine can be specified by the design engineer; the exit temperature cannot be specified, but instead adjusts so that Eq. 4.3-15 is satisfied. Thus, although the entropy generation term does not explicitly appear in the work term of Eq. 4.3-14, it is contained implicitly through the exit temperature and therefore the exit enthalpy H 2 . By example (Problems 4.9 and 4.24), one can establish that a reversible adiabatic engine has the lowest exit temperature and enthalpy for fixed inlet and exit pressures, and thereby achieves the best conversion of fluid thermal energy to work. Finally, we want to develop an expression in terms of the pressure and volume for the maximum rate at which work is obtained, or the minimum rate at which work must be added, to accomplish a given change of state in continuous-flow systems such as turbines and compressors. Figure 4.3-4 is a generic diagram of a device through which fluid is flowing continuously. The volume element in the figure contained within the dashed lines is a very small region of length ΔL in which the temperature and pressure of the fluid can be taken to be approximately constant (in fact, shortly we will consider the limit in which ΔL → 0). The mass, energy, and entropy balances for this steadystate system are

4.3 Heat, Work, Engines, and Entropy 121 ωs

q

NL + ΔL

NL

L

Figure 4.3-4 Device with fluid, heat, and work flows.

L + ΔL

dN = 0 = N˙ |L − N˙ |L+ΔL dt dU = 0 = N˙ H |L − N˙ H |L+ΔL + q˙ ΔL + ω˙ s ΔL dt

(4.3-16a) (4.3-16b)

q˙ dS (4.3-16c) = 0 = N˙ S |L − N˙ S |L+ΔL + ΔL + σ˙ gen ΔL dt T In these equations, q˙, ω˙ s , and σ˙ gen are, respectively, the heat and work flows and the rate of entropy generation per unit length of the device. Dividing by ΔL, taking the limit as ΔL → 0, and using the definition of the total derivative from calculus gives N˙ |L+ΔL − N˙ |L dN˙ = =0 or N˙ = constant ΔL→0 ΔL dL   H |L+ΔL − H |L dH ˙ = N˙ = q˙ + ω˙ s N lim ΔL→0 ΔL dL   q˙ S |L+ΔL − S |L dS = + σ˙ gen N˙ lim = N˙ ΔL→0 ΔL dL T lim

(4.3-17a)

(4.3-17b) (4.3-17c)

From the discussion of the previous section, the maximum work that can be obtained, or the minimum work required, in a given change of state occurs in a reversible process. Setting σ˙ gen = 0 yields

q˙ dS = N˙ dL T   dH dS ˙ −T ω˙ s = N dL dL Now, from H = U + P V ,

dH = dU + d(P V ) = dU + P dV + V dP and from Eq. 4.2-13b,

dU = T dS − P dV we have

dH − T dS = T dS − P dV + P dV + V dP − T dS = V dP

122 Chapter 4: Entropy: An Additional Balance Equation or

dS dH dP =V −T dL dL dL and

dP ω˙ s = N˙ V dL Further, integrating over the length of the device gives   dP ˙ ˙ dL Ws = ω˙ s dL = N V dL or

˙s  W Ws = = V dP N˙

(4.3-18)

Equation 4.3-18 is the desired result. It is useful to consider several applications of Eq. 4.3-18. For the ideal gas undergoing an isothermal change, so that P V = RT = constant,

 Ws =

 V dP =

RT dP = RT P



P2 dP = RT ln P P1

(4.3-19)

For an expansion or compression for which

P1 V γ1 = P2 V γ2 = constant 

 Ws =

=

V dP = (constant)1/γ

Work in a polytropic process

dP P 1/γ 

 (constant)1/γ P21−1/γ − P11−1/γ 1−

=

(4.3-20)

1 γ

(P2 V γ2 )1/γ P21−1/γ − (P1 V γ1 )1/γ P11−1/γ 1 1− γ Ws =

γ(P2 V 2 − P1 V 1 ) γ−1

(4.3-21)

A process that obeys Eq. 4.3-20 is referred to as a polytropic process. For an ideal gas it is easy to show that

γ = 0 for an isobaric process γ = 1 for an isothermal process γ = ∞ for a constant-volume (isochoric) process

4.3 Heat, Work, Engines, and Entropy 123 Also, we will show later that

γ = CP∗ /CV∗

for a constant-entropy (isentropic) process in an ideal gas of constant heat capacity

Note that since entropy is a state property, once two properties of a one-phase system, such as temperature and pressure, are fixed, the value of the entropy is also fixed. Consequently, the entropy of steam can be found in the steam tables or the Mollier diagram, and that of methane, nitrogen, and HFC-134a in the appropriate figures in Chapter 3. In the next section we consider entropy changes for an ideal gas, and in Chapter 6 we develop the equations to be used to compute entropy changes for nonideal fluids. In preparation for the discussion of how entropy changes accompanying a change of state can be computed, it is useful to consider the thermodynamic balance equations for a change of state of a closed system. The difference form of the energy balance is

Uf − Ui = Q + W Consequently we see that the change in the state variable U , the internal energy, is related to two path variables, the heat and the work. For a change between a given initial and final state, the internal energy change will always be the same, even though, as we have shown in the previous chapter, the heat and work flows along different paths will be different. Similarly, the entropy balance is

Sf − Si =

Use of a reversible path to calculate the change in state variables

Q + Sgen T

and we have a similar situation as above, where a change in a state function between two states is related to two path functions, here Q/T and the entropy generation. This discussion brings up a subtle, sometimes confusing, but very important point. Since properties such as the internal energy, enthalpy, and entropy are state functions, the changes in their values with a change of state depend only on the initial and final states, not on the path used to go between these two states. Therefore, in calculating a change in a state property, such as the internal energy or entropy, between fixed initial and final states, any convenient path can be used; to some degree this was demonstrated in Illustration 3.4-6. Frequently, a reversible path will be the most convenient for computing the change in the state variables between given initial and final states even though the actual system change is not reversible. However, for path variables such as heat flows, work flows, and entropy generation, the path is important, and different values for these quantities will be obtained along different paths. Therefore, to accurately compute the heat flows, the work flows, and the entropy generation, the actual path of the system change must be followed. To conclude this section, we consider a brief introduction to the thermodynamic limits on the conversion of sunlight to electrical (or mechanical) energy.

Illustration 4.3-1 Conversion of Radiant Energy to Mechanical or Electrical Energy Show that a solar or photovoltaic cell that converts solar energy to mechanical or electrical energy must emit some of the energy of the incident radiation as heat.

124 Chapter 4: Entropy: An Additional Balance Equation Solution The way we will prove that a solar cell must emit heat is to assume that it does not, and show that this would be a violation of the entropy generation principle (that is, the second law of thermodynamics). The steady-state energy balance on a solar cell absorbing radiation but not releasing any heat is ˙ 0 = Q˙ R + W

or ˙ = Q˙ R −W

and the entropy balance is 0=

4 Q˙ R + S˙ gen 3 TR

so that 4 Q˙ R S˙ gen = − 3 TR

The only way that S˙ gen can be greater than or equal to zero, which is necessary to satisfy the second law of thermodynamics, is if Q˙ R ≤ 0; that is, radiant energy must be released rather than absorbed. Therefore, we see that a solar cell also obeys the Kelvin-Planck statement of the second law (see Illustration 4.1-2). However, note that the cell can operate in the reverse manner in that it can receive electrical energy and completely convert it to radiant energy. A light-emitting diode (LED) or a metal wire electrically heated until it is red hot are two examples of complete conversion of electrical energy to radiant energy.

Illustration 4.3-2 Maximum Conversion of Solar Energy to Mechanical or Electrical Energy Based on analysis of the frequency distribution of radiation from the sun, it can be considered to be emitting radiant energy with a Stefan-Boltzmann distribution at a temperature of 6000 K. Estimate the maximum efficiency with which this radiant energy can be converted to electrical (or mechanical) energy using solar cells (commonly called photovoltaic cells). For this analysis, assume that the solar cell is operating in steady state and is receiving radiant energy, that its surface temperature is 300 K, and that it is losing heat by conduction to the environment.

Solution The energy balance on this solar cell is ˙ = Q˙ R + Q˙ 2 + W ˙ 0 = Q˙ 1 + Q˙ 2 + W

or ˙ = Q˙ R + Q˙ 2 −W

and the entropy balance is 0=

4 Q˙ R Q˙ 2 + + S˙ gen 3 TR T

4.4 Entropy Changes of Matter 125 For maximum conversion efficiency, S˙ gen = 0. Therefore, 4 Q˙ R Q˙ 2 = − T 3 TR

and ˙ = Q˙ R − −W

  4 Q˙ R 4 T T = Q˙ R 1 − 3 TR 3 TR

and the maximum efficiency is ˙ −W Q˙ R

=1−

4 T 3 TR

With TR = 6000 K and T = 300 K, the maximum efficiency for solar energy conversion is 0.9333, or 93.33 percent.

Comments The efficiency of commercial solar cells is generally 10 percent or less, rather the 93.33 percent from the calculation above. The most important reason for this is that is solar cells (and also biological cells in photosynthesis) can use radiant energy in just a small portion of the radiation frequency spectrum. Consequently, most of the radiation received by a solar cell is merely absorbed as heat, which is why many solar systems combine photovoltaic cells for the production of electricity with panels for heating water. Another loss factor is that some of the radiant energy received is reradiated to the sky (which can be considered a black body at a temperature of 0 K.) Also, one should note that the efficiency calculated above is somewhat less than the Carnot efficiency, η =1−

300 = 0.95 6000

operating between the same two temperatures. However, as there is no way to transfer heat at 6000 K from the the sun to the earth by conduction, the Carnot efficiency is not applicable.

4.4 ENTROPY CHANGES OF MATTER Equation 4.2-13b provides the basis for computing entropy changes for real fluids, and it will be used in that manner in Chapter 6. However, to illustrate the use of the entropy balance here in a simple way, we consider the calculation of the entropy change accompanying a change of state for 1 mol of an ideal gas, and for incompressible liquids and solids. From the discussion of Sec. 3.3 the internal energy change and pressure of an ideal gas are

dU = CV∗ dT

and P = RT /V

respectively, so that for 1 mol of an ideal gas we have

dS =

1 P C∗ R dU + dV = V dT + dV T T T V

(4.4-1)

If CV∗ is independent of temperature, we can immediately integrate this equation to obtain

126 Chapter 4: Entropy: An Additional Balance Equation



 V2 dT dV S(T2 , V 2 ) − S(T1 , V 1 ) = CV +R T T1 V1 V     T2 V2 ∗ = CV ln + R ln T1 V1

Ideal gas entropy change with T and V as independent variables if C∗V is a constant

∗

T2

(4.4-2)

Using the ideal gas law, we can eliminate either the temperature or the volume in this equation and obtain expressions for the change in entropy with changes in temperature and pressure,   RT2 /P2 T2 + R ln S(T2 , P2 ) − S(T1 , P1 ) = CV∗ ln T1 RT1 /P1     T2 P2 ∗ = (CV + R) ln − R ln T1 P1

 Entropy change of an ideal gas with T and P as independent variables if C∗P is a constant

∗

S(T2 , P2 ) − S(T1 , P1 ) = CP ln

T2 T1



 − R ln

P2 P1

 (4.4-3)

and pressure and volume:

  V2 P2 V 2 /R + R ln S(P2 , V 2 ) − S(P1 , V 1 ) = CV ln P1 V 1 /R V1     V2 P2 ∗ ∗ + CV ln = (CV + R) ln V1 P1     V2 P2 + CV∗ ln = CP∗ ln V1 P1 ∗

(4.4-4)

The evaluation of the entropy change for an ideal gas in which the heat capacity is a function of temperature (see Eq. 3.3-5) leads to more complicated equations than those given here. It is left to the reader to develop the appropriate expressions (Problem 4.13). For liquids or solids we can generally write

1 P 1 (4.4-5) dU + dV ≈ dU T T T since the molar volume is very weakly dependent on either temperature or pressure (i.e., dV is generally small). Furthermore, since for a liquid or a solid CV ≈ CP , we have dU = CV dT ≈ CP dT dS =

for these substances, so that

dS = CP

dT T

and Entropy change for a solid or liquid

 S(T2 ) − S(T1 ) =

T2

CP T1

dT T

(4.4-6)

4.4 Entropy Changes of Matter 127 Finally, if thermodynamic tables and charts that include entropy have been prepared for real fluids, the entropy changes accompanying a change in state can easily be calculated. In this way the entropy changes on a change of state for methane, nitrogen, HFC-134a, and steam can be calculated using the figures of Chapter 3 and the tables in Appendix A.III for steam. Illustration 4.4-1 Calculation of Entropy Generation for a Process Compute the entropy generated on mixing 1 kg of steam at 1 bar and 200◦ C (state 1) with 1 kg of steam at 1 bar and 300◦ C (state 2).

Solution Considering the 2 kg of steam to be a closed system, the mass balance is Mf − (M1 + M2 ) = 0

or

Mf = M1 + M2 = 2 kg

The energy balance is ˆf − M1 U ˆ1 − M2 U ˆ2 = −P (Mf Vˆf − M1 Vˆ1 − M2 Vˆ2 ) Mf U

Since the pressure is constant, the U and P V terms can be combined to give ˆ f − M1 H ˆ 1 − M2 H ˆ2 = 0 Mf H

or ˆ f =?, P = 1 bar) 2 kg · H(T ˆ ˆ = 200◦ C, P = 1 bar) + 1 kg · H(T = 300◦ C, P = 1 bar) = 1 kg · H(T

Therefore, ˆ f =?, P = 1 bar) = 1 [H(T ˆ ˆ H(T = 200◦ C, P = 1 bar) + H(T = 300◦ C, P = 1 bar)] 2 kJ kJ = 12 [2875.3 + 3074.3] = 2974.8 kg kg

and from the steam tables we find for this value of the enthalpy at a pressure of 1 bar that Tf = 250◦ C. Now from the entropy balance, we have Mf Sˆf − M1 Sˆ1 − M2 Sˆ2 = Sgen

or ˆ = 250◦ C, P = 1 bar) − 1 kg · S(T ˆ = 200◦ C, P = 1 bar) Sgen = 2 kg · S(T ◦ ˆ − 1 kg · S(T = 300 C, P = 1 bar) kJ kJ kJ = 2 kg · 8.0333 − 1 kg · 7.8343 − 1 kg · 8.2158 kg K kg K kg K kJ = 0.0165 K

So we see that mixing two fluids of the same pressure but different temperatures generates entropy and therefore is an irreversible process.

128 Chapter 4: Entropy: An Additional Balance Equation Illustration 4.4-2 Illustration 3.4-1 Continued Compute the entropy generated by the flow of 1 kg/s of steam at 400 bar and 500◦ C undergoing a Joule-Thomson expansion to 1 bar.

Solution In Illustration 3.4-1, from the energy balance, we found that ˆ 1 = H(T ˆ 1 = 500◦ C, P1 = 400 bar) = H ˆ 2 = H(T ˆ 2 =?, P2 = 1 bar) H

and then by interpolation of the information in the steam tables that T2 = 214.1◦ C. From the entropy balance on this steady-state system, we have dS ˙ 2 Sˆ2 + S˙ gen = M ˙ 1 (Sˆ1 − Sˆ2 ) + S˙ gen ˙ 1 Sˆ1 + M =0=M dt

so that S˙ gen = M˙ 1 (Sˆ2 − Sˆ1 )

From the steam tables (using interpolation to obtain the entropy of steam at 214.1◦ C and 1 bar), we have ˆ = 500◦ C, P = 400 bar) = 5.4700 kJ Sˆ1 = S(T kg K

and ˆ = 214.1◦ C, P = 1 bar) = 7.8904 kJ Sˆ2 = S(T kg K

Therefore ˙ 1 (Sˆ2 − Sˆ1 ) = 1 kg · (7.8904 − 5.4700) kJ = 2.4202 kJ S˙ gen = M s kg K Ks

Since S˙ gen > 0, the Joule-Thomson expansion is also an irreversible process.

4.5 APPLICATIONS OF THE ENTROPY BALANCE In this section we show, by example, that the entropy balance provides a useful additional equation for the analysis of thermodynamic problems. In fact, some of the examples considered here are continuations of the illustrations of the previous chapter, to emphasize that the entropy balance can provide the information needed to solve problems that were unsolvable using only the mass and energy balance equations, or, in some cases, to develop a simpler solution method for problems that were solvable (see Illustration 4.5-2). Illustration 4.5-1 Illustration 3.4-4 Continued, Using the Entropy Balance In Illustration 3.4-4 we tried to estimate the exit temperature and power requirements for a gas compressor. From the steady-state mass balance we found that N˙ 1 = −N˙ 2 = N˙

(a)

and from the steady-state energy balance we had ˙ s = N˙ C ∗ (T2 − T1 ) W P

(b)

4.5 Applications of the Entropy Balance 129 ˙ s and T2 . Now, writing a molar which resulted in one equation (Eq. b) with two unknowns, W entropy balance for the same system yields 0 = (S 1 − S 2 )N˙ + S˙ gen

(c)

To obtain an estimate of the exit temperature and the power requirements, we assume that the compressor is well designed and operates reversibly, that is S˙ gen = 0

(d)

S1 = S2

(e)

Thus, we have which is the additional relation for a state variable needed to solve the problem. Now using Eq. 4.4-3, S(P2 , T2 ) − S(P1 , T1 ) = CP∗ ln



T2 T1





− R ln

P2 P1



and recognizing that S(P2 , T2 ) = S(P1 , T1 ) yields 

T2 T1



 =

P2 P1

R/CP∗

(f)

or  T2 = T1

P2 P1

R/CP∗

 = 290 K

10 1

8.314/29.3 = 557.4 K

or

284.2◦ C

(g)

Thus T2 is known, and hence W s can be computed: W s = CP∗ (T2 − T1 ) = 29.3 × (557.4 − 290) = 7834.8 J/mol

and ˙ s = N˙ W s = 2.5 mol × 7834.8 J = 19.59 kJ W s mol s

Before considering the problem to be solved, we should try to assess the validity of the assumption S˙ gen = 0. However, this can be done only by experiment. One method is to measure the inlet and exit temperatures and pressures for an adiabatic turbine and see if Eq. e is satisfied. Experiments of this type indicate that Eq. e is reasonably accurate, so that reversible operation is a reasonable approximation for a gas compressor. 

R The Aspen Plus simulation to solve this illustration is available on the Wiley website for this book in the folder Aspen Illustrations>Chapter 4>4.5-1 (also available there are folders for using the Peng-Robinson equation of state, and also varying compressor efficiencies). The simulation using the ideal gas equation of state and assuming 100% isentropic efficiency leads to an exit temperature of 557.25 K or 284.2◦ C. The compressor work needed is 19.62 kW = 19.62 kJ/s. Repeating the calculation with the Peng-Robinson equation of state (see Chapter 6), the results are an exit temperature of 557.67 K and compressor work of 19.64 kW. (Aspen Illustrations> Chapter 4>4.5-1 PREOS) Finally, repeating the calculation using the Peng-Robinson equation of state and an isentropic R efficiency of 0.72 (Aspen Plus default value) results in an exit temperature of 658.0 K and compressor work of 27.27 kW (Aspen Illustrations>Chapter 4>4.5-1 PREOS Comp eff).

130 Chapter 4: Entropy: An Additional Balance Equation Illustration 4.5-2 An Alternative Way to Solve One Problem Sometimes it is possible to solve a thermodynamic problem several ways, based on different choices of the system. To see this, we consider Illustration 3.4-5, which was concerned with the partial evacuation of a compressed gas cylinder into an evacuated cylinder of equal volume. Suppose we now choose for the system of interest only that portion of the contents of the first cylinder that remains in the cylinder when the pressures have equalized (see Fig. 4.5-1, where the thermodynamic system of interest is within the dashed lines). Note that with this choice the system is closed, but of changing volume. Furthermore, since the gas on one side of the imaginary boundary has precisely the same temperature as the gas at the other side, we can assume there is no heat transfer across the boundary, so that the system is adiabatic. Also, with the exception of the region near the valve (which is outside what we have taken to be the

Initial state

Figure 4.5-1 The dashed lines enclose a system consisting of gas initially in the first cylinder that remains in that cylinder at the end of the process.

Final state

system), the gas in the cylinder is undergoing a uniform expansion so there will be no pressure, velocity, or temperature gradients in the cylinder. Therefore, we can assume that the changes taking place in the system occur reversibly. The mass, energy, and entropy balances (on a molar basis) for this system are N1f = N1i

(a)

 N1f U f1

=

N1i U i1

−

f

V1

(b)

P dV V1i

and N1f S f1 = N1i S i1

(c)

Now the important observation is that by combining Eqs. a and c, we obtain S f1 = S i1

(d)

so the process is isentropic (i.e., occurs at constant entropy) for the system we have chosen. Using Eq. 4.4-3, 

∗

S 1 (P , T ) − S 1 (P , T ) = CP ln f

and Eq. d yields

f

i



T1f T1i

i

CP∗ /R

 =

T1f T1i

P1f P1i





− R ln

P1f P1i





This is precisely the result obtained in Eq. f of Illustration 3.4-5 using the energy balance on the open system consisting of the total contents of cylinder 1. The remainder of the problem can now be solved in exactly the same manner used in Illustration 3.4-5. Although the system choice used in this illustration is an unusual one, it is one that leads quickly to a useful result. This demonstrates that sometimes a clever choice for the thermodynamic system can be the key to solving a thermodynamic problem with minimum effort.

4.5 Applications of the Entropy Balance 131 However, one also has to be careful about the assumptions in unusual system choices. For example, consider the two cylinders connected as in Fig. 4.5-2, where the second cylinder is not initially evacuated. Here we have chosen to treat that part of the initial contents of cylinder 1 that will be in that cylinder at the end of the process as one system and the total initial contents of cylinder 2 as a second system. The change that occurs in the first system, as we already discussed, is adiabatic and reversible, so that 

1

T1f T1i

CP∗ /R



=

2

P1f P1i



(e)

1

Initial state

2

Final state

Figure 4.5-2 Incorrect system choice for gas contained in cylinder 2. One might expect that a similar relation would hold for the system shown in cylinder 2. This is not the case, however, since the gas entering cylinder 2 is not necessarily at the same temperature as the gas already there (hydrodynamics will ensure that the pressures are the same). Therefore, temperature gradients will exist within the cylinder, and our system, the partial contents of cylinder 2, will not be adiabatic. Consequently, Eq. e will not apply.

Illustration 4.5-3 Illustration 3.4-5 Continued, Using the Entropy Balance Instead of the Energy Balance A gas cylinder of 1 m3 volume containing nitrogen initially at a pressure of 40 bar and a temperature of 200 K is connected to another cylinder of 1 m3 volume, which is evacuated. A valve between the two cylinders is opened only until the pressure in both cylinders equalizes and then is closed. Find the final temperature and pressure in each cylinder if there is no heat flow into or out of the cylinders, or between the gas and the cylinder walls. The properties of nitrogen gas are given in Fig. 3.3-3.

Solution Except for the fact that nitrogen is now being considered to be a real, rather than an ideal, gas, the problem here is the same as in Illustration 3.4-5. In fact, Eqs. h–n in the comments to that illustration apply here. The additional equation needed to solve this problem is obtained in the same manner as in Illustration 4.5-2. Thus, for the cylinder initially filled we have S i1 = S f1

(o)

For an ideal gas of constant heat capacity (which is not the case here) this reduces to 

P1f P1i



 =

T1f T1i

CP∗ /R

by Eq. 4.4-3. For real nitrogen gas, Eq. o requires that the initial and final states in cylinder 1 be connected by a line of constant entropy in Fig. 3.3-3. With eight equations (Eqs. h–o) and eight unknowns, this problem can be solved, though the solution is a trial-and-error process. In general, a reasonable first guess for the pressure in the nonideal gas problem is the ideal gas solution, which was P1f = P2f = 20 bar

132 Chapter 4: Entropy: An Additional Balance Equation Locating the initial conditions (P = 40 bar = 4 MPa and T = 200 K) in Fig. 3.3-3 yields11 ˆ i ≈ 337 kJ/kg and Vˆ i ≈ 0.0136 m3 /kg, so that H 1 1 M1i =

V1 = Vˆ1i

1 m3 = 73.5 kg m3 0.136 kg

Now using Eq. o (in the form Sˆ1i = Sˆ1f ) and Fig. 3.3-3, we find T1f ≈ 165 K, H1f = 310 kJ/kg, and Vˆ1f ≈ 0.0224 m3 /kg, so that M1f =

V1 = Vˆ1f

1 m3 = 44.6 kg m3 0.0224 kg

and M2f = M1i − M1f = 28.9 kg, which implies 1 m3 m3 Vˆ2f = = 0.0346 28.9 kg kg

Locating P = 20 bar (2 MPa) and Vˆ = 0.0346 m3 /kg on Fig. 3.3-3 gives a value for T2f of ˆ 2f = 392 kJ/kg. about 240 K and H Finally, we must check whether the energy balance is satisfied for the conditions computed here based on the final pressure we have assumed. To do this we must first compute the internal energies of the initial and final states as follows: ˆi = H ˆ i − P i Vˆ i U 1 1 1 1 kJ kJ m3 Pa 1J 1 kJ = 282.6 = 337 − 40 bar × 0.0136 × 105 × 3 × kg kg bar m Pa 1000 J kg

Similarly, ˆ1f = 265.2 kJ/kg U

and ˆ2f = 322.8 kJ/kg U

The energy balance (Eq. n of Illustration 3.4-5) on a mass basis is ˆ i = M1f U ˆ1f + M2f U ˆ2f M1i U 1

or

73.5 × 282.6 kJ = 44.6 × 265.2 + 28.9 × 322.8 kJ 2.0771 × 104 kJ ≈ 2.1157 × 104 kJ

Thus, to the accuracy of our calculations, the energy balance can be considered to be satisfied and the problem solved. Had the energy balance not been satisfied, it would have been necessary to make another guess for the final pressures and repeat the calculation. It is interesting to note that the solution obtained here is essentially the same as that for the ideal gas case. This is not generally true, but occurs here because the initial and final pressures are sufficiently low, and the temperature sufficiently high, that nitrogen behaves as an ideal gas. Had we chosen the initial pressure to be higher, say several hundred bars, the ideal gas and real gas solutions would have been significantly different (see Problem 4.22).

Illustration 4.5-4 Illustration 3.4-6 Continued, Showing That Entropy Is a State Function Show that the entropy S is a state function by computing ΔS for each of the three paths of Illustration 3.4-6. 11 Since

the thermodynamic properties in Fig. 3.3-3 are on a mass basis, all calculations here will be on a mass, rather than a molar, basis.

4.5 Applications of the Entropy Balance 133 Solution Since the piston-and-cylinder device is frictionless (see Illustration 3.4-6), each of the expansion processes will be reversible (see also Illustration 4.5-8). Thus, the entropy balance for the gas within the piston and cylinder reduces to dS Q˙ = dt T

Path A i. Isothermal compression. Since T is constant, ΔSA =

QA 5707.7 J/mol =− = −19.14 J/(mol K) T 298.15 K

ii. Isobaric heating dT Q˙ = CP∗ dt

so Q˙ C ∗ dT dS = = P dt T T dt

and ΔS B = CP∗ ln

T2 J 573.15 = 38 × ln = 24.83 J/(mol K) T1 mol K 298.15

ΔS = ΔS A + ΔS B = −19.14 + 24.83 = 5.69 J/(mol K)

Path B i. Isobaric heating ΔS A = CP∗ ln

T2 = 24.83 J/(mol K) T1

ii. Isothermal compression ΔS B =

Q 10 972.2 =− = −19.14 J/(mol K) T 573.15

ΔS = 24.83 − 19.14 = 5.69 J/(mol K)

Path C i. Compression with P V γ = constant ΔS A =

Q =0 T

ii. Isobaric heating ΔS B = CP∗ ln

T3 J 573.15 = 38 × ln = 5.69 J/(mol K) T2 mol K 493.38 ΔS = 5.69 J/(mol K)

134 Chapter 4: Entropy: An Additional Balance Equation

1

2

Figure 4.5-3 A well-insulated box divided into two equal compartments.

Comment This example verifies, at least for the paths considered here, that the entropy is a state function. ˙ are For reversible processes in closed systems, the rate of change of entropy and the ratio Q/T ˙ equal. Thus, for reversible changes, Q/T is also a state function, even though the total heat flow Q˙ is a path function.

Illustration 4.5-5 Showing That the Entropy Reaches a Maximum at Equilibrium in a Closed, Isolated System (In Sec. 4.1 we established that the entropy function will be a maximum at equilibrium in an isolated system. This is illustrated by example for the system shown here.) Figure 4.5-3 shows a well-insulated box of volume 6 m3 divided into two equal volumes. The left-hand cell is initially filled with air at 100◦ C and 2 bar, and the right-hand cell is initially evacuated. The valve connecting the two cells will be opened so that gas will slowly pass from cell 1 to cell 2. The wall connecting the two cells conducts heat sufficiently well that the temperature of the gas in the two cells will always be the same. Plot on the same graph (1) the pressure in the second tank versus the pressure in the first tank, and (2) the change in the total entropy of the system versus the pressure in tank 1. At these temperatures and pressures, air can be considered to be an ideal gas of constant heat capacity.

Solution For this system Total mass = N = N1 + N2 Total energy = U = U1 + U2 Total entropy = S = S1 + S2 = N1 S 1 + N2 S 2 From the ideal gas equation of state and the fact that V = N V , we have N1i =

PV = RT

2 bar × 3 m3 = 193.4 mol = 0.1934 kmol bar m3 8.314 × 10−5 × 373.15 K mol K

Now since U = constant, T1 = T2 , and for the ideal gas U is a function of temperature only, we conclude that T1 = T2 = 100◦ C at all times. This result greatly simplifies the computation. Suppose that the pressure in cell 1 is decreased from 2 bar to 1.9 bar by transferring some gas from cell 1 to cell 2. Since the temperature in cell 1 is constant, we have, from the ideal gas law, N1 = 0.95N1i

and by mass conservation, N2 = 0.05N1i . Applying the ideal gas relation, we obtain P2 = 0.1 bar.

For any element of gas, we have, from Eq. 4.4-3, S f − S i = CP∗ ln

Tf Pf Pf − R ln i = −R ln i i T P P

4.5 Applications of the Entropy Balance 135 0.7

3

0.6 0.5

2 P2 (bar)

ΔS N1iR

0.4 0.3 1 0.2 0.1 0

2

1.5

1.0

0.5

0

0

P1 (bar)

Figure 4.5-4 The system entropy change and the pressure in cell 2 as a function of the pressure in cell 1. since temperature is constant. Therefore, to compute the change in entropy of the system, we visualize the process of transferring 0.05N1i moles of gas from cell 1 to cell 2 as having two effects: 1. To decrease the pressure of the 0.95N1i moles of gas remaining in cell 1 from 2 bar to 1.9 bar 2. To decrease the pressure of the 0.05N1i moles of gas that have been transferred from cell 1 to cell 2 from 2 bar to 0.1 bar Thus S f − S i = −0.95N1i R ln(1.9/2) − 0.05N1i R ln(0.1/2)

or

ΔS = −0.95 ln 0.95 − 0.05 ln 0.05 N1i R = 0.199

P1 = 1.9 bar P2 = 0.1 bar

Similarly, if P1 = 1.8 bar, ΔS = −0.9 ln 0.9 − 0.1 ln 0.1 N1i R = 0.325



P1 = 1.8 bar P2 = 0.2 bar

and so forth. The results are plotted in Fig. 4.5-4. From this figure it is clear that ΔS , the change in entropy from the initial state, and therefore the total entropy of the system, reaches a maximum value when P1 = P2 = 1 bar. Consequently, the equilibrium state of the system under consideration is the state in which the pressure in both cells is the same, as one would expect. (Since the use of the entropy function leads to a solution that agrees with one’s intuition, this example should reinforce confidence in the use of the entropy function as a criterion for equilibrium in an isolated constant-volume system.)

Illustration 4.5-6 Showing That the Energy and Entropy Balances Can Be Used to Determine Whether a Process Is Possible

136 Chapter 4: Entropy: An Additional Balance Equation An engineer claims to have invented a steady-flow device that will take air at 4 bar and 20◦ C and separate it into two streams of equal mass, one at 1 bar and −20◦ C and the second at 1 bar and 60◦ C. Furthermore, the inventor states that his device operates adiabatically and does not require (or produce) work. Is such a device possible? [Air can be assumed to be an ideal gas with a constant heat capacity of CP∗ = 29.3 J/(mol K)].

P = 4 bar T = 20°C

N1

N2

P = 1 bar T = –20°C

N3

P = 1 bar T = 60°C

Black box

Solution The three principles of thermodynamics—(1) conservation of mass, (2) conservation of energy, and (3) S˙ gen ≥ 0—must be satisfied for this or any other device. These principles can be used to test whether any device can meet the specifications given here. The steady-state mass balance  equation for the open system consisting of the device and its ˙ ˙ ˙ ˙ contents is dN/dt = 0 = k Nk = N1 + N2 + N3 . Since, from the problem statement, N˙ 2 = N˙ 3 = − 12 N˙ 1 , mass is conserved. The steady-state energy balance for this device is  dU N˙ k H k = N˙ 1 H 1 − 12 N˙ 1 H 2 − 12 N˙ 1 H 3 =0= dt k = N˙ 1 CP∗ (293.15 K −

1 2

× 253.15 K −

1 2

× 333.15 K) = 0

so the energy balance is also satisfied. Finally, the steady-state entropy balance is  dS N˙ k S k + S˙ gen = N˙ 1 S 1 − 12 N˙ 1 S 2 − 12 N˙ 1 S 3 + S˙ gen =0= dt K = 12 N˙ 1 [(S 1 − S 2 ) + (S 1 − S 3 )] + S˙ gen

Now using Eq. 4.4-3, we have   T1 P1 T1 P1 1 S˙ gen = − N˙ 1 CP∗ ln − R ln + CP∗ ln − R ln 2 T2 P2 T3 P3   293.15 × 293.15 4×4 1 J = 11.25N˙ 1 − 8.314 ln = − N˙ 1 29.3 ln 2 253.15 × 333.15 1×1 Ks

Therefore, we conclude, on the basis of thermodynamics, that it is possible to construct a device with the specifications claimed by the inventor. Thermodynamics, however, gives us no insight into how to design such a device. That is an engineering problem. Two possible devices are indicated in Fig. 4.5-5. The first device consists of an air-driven turbine that extracts work from the flowing gas. This work is then used to drive a heat pump (an air conditioner or refrigerator) to cool part of the gas and heat the rest. The second device, the Hilsch-Ranque vortex tube, is somewhat more interesting in that it accomplishes the desired change of state with only a valve and no moving parts. In this device the air expands as it enters the tube, thus gaining kinetic energy at the expense of internal energy (i.e., at the end of the expansion process we have high-velocity air of both lower pressure and lower temperature than the incoming air). Some of this cooled air is withdrawn from the center of the vortex tube. The rest of the air swirls down the tube, where, as a result of viscous dissipation, the kinetic energy is dissipated into heat, which increases the internal energy (temperature) of the air. Thus, the air being withdrawn at the valve is warmer than the incoming air.

4.5 Applications of the Entropy Balance 137 (a) Turbine-heat pump system

Hot air, 1 bar Q

Compressed air at room temperature

Heat pump driven by turbine

Turbine Air at 1 bar

Q Cold air, 1 bar

Work (b) Hilsch-Ranque vortex tube Cold air

Warm air

Compressed air Compressed air (cross section)

Figure 4.5-5 Two devices to separate compressed air into two low-pressure air streams of different temperature.

Illustration 4.5-7 Another Example of Using the Entropy Balance in Problem Solving A steam turbine operates at the following conditions:

Velocity (m/min) T (K) P (MPa) Flow rate (kg/hr) Heat loss (kJ/hr)

Inlet

Outlet

2000 800 3.5

7500 440 0.15 10 000 125 000

a. Compute the horsepower developed by the turbine and the entropy change of the steam. b. Suppose the turbine is replaced with one that is well insulated, so that the heat loss is eliminated, and well designed, so that the expansion is reversible. If the exit pressure and velocity are maintained at the previous values, what are the outlet steam temperature and the horsepower developed by the turbine?

Solution The steady-state mass and energy balances on the turbine and its contents (the system) yield dM ˙2 ˙1 +M =0=M dt

M˙ 2 = −M˙ 1 = −10 000 kg/hr

   2  2 2 v d ˆ 1 + v1 + M˙ 2 H ˆ 2 + v2 + W ˙1 H ˙ s + Q˙ U +M + gh =0=M dt 2 2 2

138 Chapter 4: Entropy: An Additional Balance Equation a. From the Mollier diagram of Fig. 3.3-1 (or Appendix A.III), ˆ 1 3510 J/g H

and Sˆ1 7.23 J/(g K)

Also, 1 v12 = 2 2



2000 m/min2 60 s/min

2

×

1 J/kg 1 kg × = 0.56 J/g m2 /s2 1000 g

so that 2 ˆ 1 + v1 = 3510.6 J/g H 2

Similarly, ˆ 2 2805 J/g H

2 ˆ 2 + v2 = 2805 + 7.8 = 2812.8 J/g H 2

and Sˆ2 7.50 J/(g K)

Therefore, ˙ s = (3510.6 − 2812.8) −W = 6.853 × 109

J kg g J kJ × 10 000 × 1000 − 12.5 × 104 × 1000 g hr kg hr kJ

J 1 kJ 1 hr × × = 1903.6 kJ/s = 1.9036 × 106 W hr 1000 J 3600 s

= 2553 hp

Also, ΔSˆ = (Sˆ2 − Sˆ1 ) = 0.27 J/(g K)

b. The steady-state entropy balance for the turbine and its contents is dS ˙ 1 Sˆ2 + S˙ gen = 0 = M˙ 1 Sˆ1 − M dt since Q˙ = 0, and M˙ 2 = −M˙ 1 . Also, the turbine operates reversibly so that S˙ gen = 0, and Sˆ1 = Sˆ2 ; that is, the expansion is isentropic. We now use Fig. 3.3-1, the entropy-enthalpy

plot (Mollier diagram) for steam, to solve this problem. In particular, we locate the initial steam conditions (T = 800 K, P = 3.5 MPa) on the chart and follow a line of constant entropy (a vertical line on the Mollier diagram) to the exit pressure (0.15 MPa), to obtain ˆ 2 ≈ 2690 J/g) and its final temperature (T ≈ 373 K). the enthalpy of the exiting steam (H Since the exit velocity is known, we can immediately compute the horsepower generated by the turbine:

˙ s = [(3510 + 0.6) − (2690 + 7.8)] −W

J kg 1000 g/kg 1 hr × 10 000 × × g hr 1000 J/kJ 3600 s

= 2257.8 kJ/s = 2.2578 × 106 W = 3028 hp

4.5 Applications of the Entropy Balance 139 Comments 1. Here, as before, the kinetic energy term is of negligible importance compared with the internal energy term. 2. Notice from the Mollier diagram that the turbine exit steam is right at the boundary of a two-phase mixture of vapor and liquid. For the solution of this problem, no difficulties arise if the exit steam is a vapor, a liquid, or a two-phase vapor-liquid mixture since our mass, energy, and entropy balances are of general applicability. In particular, the information required to use these balance equations is the internal energy, enthalpy, and entropy per unit mass of each of the flow streams. Provided we have this information, the balance equations can be used independent of whether the flow streams consist of single or multiple phases, or, in fact, single or multiple components. Here the Mollier diagram provides the necessary thermodynamic information, and the solution of this problem is straightforward. 3. Finally, note that more work is obtained from the turbine by operating it in a reversible and adiabatic manner.

Illustration 4.5-8 Showing That Sgen = 0 for a Reversible Process, and Sgen > 0 for an Irreversible Process a. By considering only the gas contained within the piston-and-cylinder device of Illustration 3.4-7 to be the system, show that the gas undergoes a reversible expansion in each of the four processes considered in that illustration. That is, show that Sgen = 0 for each process. b. By considering the gas, piston, and cylinder to be the system, show that processes a, b, and c of Illustration 3.4-7 are not reversible (i.e., Sgen > 0), and that process d is reversible.

Solution a. The entropy balance for the 1 mol of gas contained in the piston and cylinder is Sf − Si =

Q + Sgen T

where T is the constant temperature of this isothermal system and Q is the total heat flow (from both the thermostatic bath and the cylinder walls) to the gas. From Eq. g of Illustration 3.4-7, we have for the 1 mol of gas Q = RT ln

Vf Vi

and from Eq. 4.4-2, we have S f − S i = R ln

Vf Vi

since the temperature of the gas is constant. Thus Sgen = S f − S i −

Vf Q 1 − = R ln T Vi T

  Vf =0 RT ln Vi

so that the gas undergoes a reversible expansion in all four processes. b. The entropy balance for the isothermal system consisting of 1 mol of gas and the piston and cylinder is Sf − Si =

Q + Sgen T

140 Chapter 4: Entropy: An Additional Balance Equation where Q is the heat flow to the piston, cylinder, and gas (QNET of Illustration 3.4-7) and Sf − Si is the entropy change for that composite system: Sf − Si = (S f − S i )gas + (Sf − Si )piston-cylinder

Since the system is isothermal, (S f − S i )gas = R ln

Vf Vi

(see Eq. 4.4-2)

and (Sf − Si )piston-cylinder = 0

(see Eq. 4.4-6)

Consequently, Sgen = R ln

Vf Vi

−

1622.5 − QNET Q = J/K T 298.15

so we find, using the entries in Table 1 of Illustration 3.4-7, that

Sgen

⎧ 1.4473 J/K ⎪ ⎪ ⎪ ⎨ 0.8177 J/K = ⎪ 0.4343 J/K ⎪ ⎪ ⎩ 0

for process a for process b for process c for process d

Thus, we conclude that for the piston, cylinder, and gas system, processes a, b, and c are not reversible, whereas process d is reversible.

Comment From the results of part (a) we find that for the gas all expansion processes are reversible (i.e., there are no dissipative mechanisms within the gas). However, from part (b), we see that when the piston, cylinder, and gas are taken to be the system, the expansion process is irreversible unless the expansion occurs in differential steps. The conclusion, then, is that the irreversibility, or the dissipation of mechanical energy to thermal energy, occurs between the piston and the cylinder. This is, of course, obvious from the fact that the only source of dissipation in this problem is the friction between the piston and the cylinder wall.

4.6 AVAILABILITY AND THE MAXIMUM USEFUL SHAFT WORK THAT CAN BE OBTAINED IN A CHANGE OF STATE A question that arises in thermodynamics is what is the maximum useful work that can be obtained for a change of state of a substance, fluid or solid that is moving or stationary? This quantity is frequently referred to as the available work that can be obtained as a result of a change of state. To explore this, we first consider a stationary, steady-state system shown below. M1

M2

Figure 4.6-1 Schematic diagram of a flow process

4.6 Availability and the Maximum Useful Shaft Work that can be obtained in a Change of State 141 Since the system is operating in steady-state and is stationary, the mass, energy and entropy balances are M˙ 1  + M˙ 2 = 0 or M˙ 2 = −M˙1   ˆ (T1 , P1 ) + 1 v12 + gh1 + M˙ 2 H ˆ (T2 , P2 ) + 1 v22 + gh2 + Q˙ + W ˙s=0 M˙ 1 H 2 2     ˙s=0 ˆ (T1 , P1 ) + 1 v12 + gh1 − M˙ 1 H ˆ (T2 , P2 ) + 1 v22 + gh2 + Q˙ + W or M˙ 1 H 2 2

Q˙ and M˙ 1 Sˆ (T1 , P1 ) + M˙ 2 Sˆ (T2 , P2 ) + + S˙ gen = T ˙ Q M˙ 1 Sˆ (T1 , P1 ) − M˙ 1 Sˆ (T2 , P2 ) + + S˙ gen = 0 T

(4.6-1) Of special interest is that when the final state of the system is the so-called dead state in which the temperature and pressure are the same as the environment that we indicate as T amb and Pamb . Since, if the temperature of the exiting stream was something other than T amb , additional shaft work could be obtained by inserting a power cycle, for example a Carnot cycle that would operate between the exit stream temperature and the ambient temperature. Similarly, any heat transfer from the system should occur at the ambient temperature; if not a power cycle could be used to obtain work from this heat flow. In a similar fashion the exiting pressure should be Pamb since otherwise a turbine could be added to the system to extract work from the higher pressure of the exit stream. The velocity of the exit stream should be zero, otherwise a turbine could be added to the system to extract work from this stream. Also, to obtain the maximum shaft work, the height of the exit stream should be a ground level, i.e., h2 = 0. Finally, to obtain the maximum useful shaft work, any process should operate reversibly, that is S˙ gen = 0. With these restrictions, we have that   ˆ (Tamb , Pamb ) + Q˙ + W ˆ (T1 , P1 ) + 1 v12 + gh1 − M˙ 1 H ˙ s,max = 0 M˙ 1 H 2 ˙ amb = 0 and M˙ 1 Sˆ (T1 , P1 ) − M˙ 1 Sˆ (Tamb , Pamb ) + Q/T (4.6-2) Now eliminating the heat flow between the energy and entropy balances yields   ˆ (T1 , P1 ) − Tamb Sˆ (T1 , P1 ) + 1 v12 + gh1 − M˙ 1 H 2   (4.6-3) ˙ s,max = 0 ˆ (Tamb , Pamb ) − Tamb Sˆ (Tamb , Pamb ) + W M˙ 1 H Since temperature, enthalpy and entropy are all state properties, it is convenient to define a new state variable, the open system availability B(T, P ) = H(T, P ) − Tamb S(T, P ). This is similar to the Gibbs energy G(T, P ) = H(T, P ) − T S(T, P ) except that the ambient temperature T amb multiplies the entropy instead of the system temperature. With this notation the maximum work that can be obtained from a flowing stream in any device is     ˙ s,max = −M˙ 1 Bˆ (T1 , P1 ) + 1 v12 + gh1 − Bˆ (Tamb , Pamb ) + gh2 or W 2 ˙ s,max W ˙1 M

= Bˆ (Tamb , Pamb ) − Bˆ (T1 , P1 ) − 12 v12 − gh1 (4.6-4)

Illustration 4.6-1 What is the maximum work that can be obtained from steam at 2 MPa and 700◦ C that is flowing continuously at 5 m/sec in a pipe that is 5 m above ground level?

142 Chapter 4: Entropy: An Additional Balance Equation Solution Using the steam tables ˆ = 3071.8 kJ/kg, Sˆ = 7.8926 kJ/kg · K, so that At 2 MPa and 700◦ C:H ˆ B = 3071.8 − 298.15 × 7.8926 = 718.62 kJ/kg and at 0.10135 MPa and 25◦ C ˆ = 104.89 kJ/kg, Sˆ = 0.3674 kJ/kg · K, so that Bˆ = 104.89 − 298.15 × 0.3674 = H −4.65 kJ/kg

So that

˙ s,max W ˙1 M

= Bˆ (Tamb , Pamb ) −Bˆ (T1 , P1 ) − 12 v12 − gh1 = 2

= ((−4.65) − 718.62)

kJ kg

−

J/kg m2 /s2 J 2·1000 kJ

25 m2 × s

− 5m × 9.8 sm2 ×

J/kg m2 /s2 J 1000 kJ

kJ = −723.27 − 0.01 − 0.049 kJ = −723.77 kg kg

where the minus sign indicates work done by the maximum shaft work that could be done by system on the surroundings. We see that the maximum useful shaft work that can be obtained from this flowing stream is 723.77 kJ/kg, and that the kinetic and potential energy terms, that is the terms due to the stream velocity and the initial stream height, contribute very little (0.059 kJ/kg) to the maximum work that can be obtained. This is generally the case if there is a significant difference between the entering fluid and the ambient temperature.

Note that the maximum useful shaft work that can be obtained in a transformation from a state 1 to any state 2 is, by simple extension of the analysis above, ˙ s,max W ˙1 M

= Bˆ (T2 , P2 ) + 12 v22 + gh2 − Bˆ (T1 , P1 ) − 12 v12 − gh1 = Bˆ (T2 , P2 ) − Bˆ (T1 , P1 ) + 12 Δv 2 + gΔh

(4.6-5)

where Δv 2 = v22 − v12 and Δh = h2 − h1 . However, as shown in this example, these two terms are usually negligible if there is a significant temperature change between the two states. Illustration 4.6-2 One continuously flowing stream of steam is at 2 MPa and 800◦ C, and another is at 1 MPa and 900◦ C. Which could, in principle, produce the greatest amount of useful work in a flow process?

Solution ˆ = 4150.3 kJ/kg, Sˆ = 8.1765 kJ/kg · K, so that Using the steam tables at 2 MPa and 800◦ C: H Bˆ = 4150.3 − 298.15 × 8.1765 = 1712.5 kJ/kg and at 1 MPa and 900◦ C ˆ = 4392.9 kJ/kg, Sˆ = 8.7118 kJ/kg · K, so that Bˆ = 4392.9 − 298.15 × 8.7118 H = 1795.5 kJ/kg

Therefore, we see that the lower pressure stream has the greater potential to do useful work due to its higher temperature.

For convenience, changing to a molar basis rather than a mass basis as tables thermodynamic properties such as the steam tables are not available for most fluids that are of interest, and since (see Section 6.2)     ∂H ∂S CP = CP and = ∂T P ∂T P T

4.6 Availability and the Maximum Useful Shaft Work that can be obtained in a Change of State 143 It then follows that         ∂H ∂B ∂S CP Tamb = − Tamb = CP − Tamb = CP 1 − ∂T P ∂T P ∂T P T T (4.6-6) and    T2 Tamb CP 1 − (4.6-7a) dT B (T2 , P1 ) − B (T1 , P1 ) = T T1 If the constant pressure heat capacity independent of temperature     T2  Tamb T2 1− B (T2 , P1 )−B (T1 , P1 ) = CP dT = CP T2 − T1 − Tamb ln T T1 T1 (4.6-7b) Also, since 

∂H ∂P



 =V −T

T

∂V ∂P



 and

T

∂S ∂P



 =−

T

∂V ∂P

 T

then         ∂B ∂V ∂V ∂V = V −T +Tamb = V +(Tamb − T ) (4.6-8) ∂P T ∂P T ∂P T ∂P T and

      ∂V ∂V V −T + Tamb dP ∂P T ∂P T T m P1    P2  ∂V dP V + (Tamb − T ) = ∂P P1 T (4.6-9) For the ideal gas this reduces to     P2  ∂V B (T, P2 ) − B (T, P1 ) = V + (Tamb − T ) dP ∂P T P1   P2  R RT dP = + (Tamb − T ) P P P1 (4.6-10)  P2 dP = RTamb P P 1  P2 = RTamb ln P1 B (T, P2 ) − B (T, P1 ) =



P2

If, instead of a steady flow situation just considered we had a closed system with no mass flows entering of leaving, the analysis of the useful work that can be obtained is somewhat different. The mass, energy and entropy balances in this case are

ˆ ˆ ˆ ˆ ˙ dM U dU dM ˙ s − P dV ; and dM S = M dS = Q + S˙ gen = 0; =M = Q˙ + W dT dT dT dT dT dT T (4.6-11)

144 Chapter 4: Entropy: An Additional Balance Equation Now invoking the same restrictions as previously, that all heat transfer occurs at ambient conditions, and that the process is reversible so that S˙ gen = 0 results in

ˆ ˙ ˆ dU ˙ s,max − Pamb dV ; M dS = Q ; = Q˙ + W dT dT dT Tamb   ˆ − Tamb Sˆ d U ˙ s,max − Pamb dV so that M =W dT dT   ˆ + Pamb Vˆ − Tamb Sˆ d U ˙ s,max or M (4.6-12) =W dT Here it is convenient to define a new function, the closed system availability, as A = U + Pamb V − Tamb S , that is similar to the definition of the Gibbs energy except that Pamb multiples the volume instead of the system pressure, and T amb multiplies the entropy rather than the system temperature T. In this case the maximum useful work that can be obtained from a closed, stationary system is   ˆ + Pamb Vˆ − Tamb Sˆ d U dAˆ ˙ s,max = M (4.6-13a) W =M dT dT which on integration between the initial and ambient states gives   Ws,max =M Aˆ (Tamb , Pamb ) − Aˆ (T1 , P1 ) (4.6-13b) M

Here the useful shaft work has been separated from the work as a result of expansion or contraction of the system volume against the ambient pressure as this latter work cannot be captured and therefore is not included in the useful shaft work. The maximum useful work that can be obtained between any initial state 1 and final state 2 is, by extension of the analysis above,   Ws,max = M Aˆ (T1 , P1 ) −Aˆ (T2 , P2 ) (4.6-14)

Illustration 4.6-3 What is the maximum work that can be obtained from steam at 2 MPa and 700◦ C in a non-flow process?

Solution Using the data in the previous illustration with the following additional information from the steam tables (note that at the ambient conditions the steam has condensed to liquid water). At 2 MPa and 700◦ C: ˆ = 2808.6 kJ/kg, Vˆ = 1.3162 m3 /kg so that U 3 Aˆ = 2808.6 kJ + 1.0 bar × 1.3162 m × 102.67 kg

kg

kJ bar·m3

kJ − 298.15 × 7.8926 kg = 590.55 kJ/kg

and at the ambient conditions 0.10135 MPa and 25◦ C ˆ = 104.88 kJ/kg, Vˆ = 0.001 m3 /kg, so that U Aˆ = 104.88 + 1 × 0.001 × 102.67 − 298.15 × 0.3674 = −4.56 kJ/kg

Problems 145 Therefore, ˙ s,max W kJ = Aˆ (Tamb , Pamb ) − Aˆ (T1 , P1 ) = −4.56 − 590.55 = −595.11 kg M˙ 1

We see that the maximum useful work that can be obtained from this stagnant stream is 595.1 kJ/kg. (As usual, the negative sign indicates that shaft work is being done by the system.)

Another function that is sometimes used in thermodynamics is the exergy, which is the difference in availability between the current state of the system and the availability if the system was at ambient conditions. For the flow system, the exergy per unit mass is

ˆ s,max exergy = Bˆ (T1 , P1 ) − Bˆ (Tamb , Pamb ) = −W

(4.6-15)

which in this case is just the negative of the maximum shaft work per unit mass. The analogous expression for the non-flow system is

ˆ s,max exergy = Aˆ (T1 , P1 ) − Aˆ (Tamb , Pamb ) = −W

(4.6-16)

Finally, there is a similarity between the analysis in this section and that in Section 4.3. There we obtained the maximum shaft work that could be obtained from a thermal engine operating between two temperatures, Eq. 4.3-4, without knowing what the engine would be, or how it would operate. It was left to an inventor, Carnot, to invent the machinery to accomplish the maximum shaft work. Here we have obtained a general expressions for the maximum shaft work that could be obtained for any change of state, Eqs. 4.6-15 and 16, without specifying the device to obtain that maximum shaft work. It is left to the engineer to design the device to accomplish this.

4.7 THE MICROSCOPIC ENTROPY BALANCE (OPTIONAL) This section appears on the website for this book. PROBLEMS 4.1 A 5-kg copper ball at 75◦ C is dropped into 12 kg of water, initially at 5◦ C, in a well-insulated container. a. Find the common temperature of the water and copper ball after the passage of a long period of time. b. What is the entropy change of the water in going from its initial to final state? Of the ball? Of the composite system of water and ball? Data: CP (copper) = 0.5 J/(g K) CP (water) = 4.2 J/(g K)

4.2 In a foundry, metal castings are cooled by quenching in an oil bath. Typically, a casting weighing 20 kg and at a temperature of 450◦ C is cooled by placing it in a 150-kg involatile oil bath initially at 50◦ C. If CP of the metal is 0.5 J/(kg K), and CP of the oil 2.6 J/(kg K), determine the common final temperature of the oil and casting after quenching if there are no heat losses. Also, find the entropy change in this process. 4.3 a. Show that the rate at which shaft work is obtained or required for a reversible change of state in a closed system at constant internal energy and volume is

equal to the negative of the product of the temperature and the rate of change of the entropy for the system. b. Show that the rate at which shaft work is obtained or required for a reversible change of state in a closed system at constant entropy and pressure is equal to the rate of change of enthalpy of the system. 4.4 Steam at 700 bar and 600◦ C is withdrawn from a steam line and adiabatically expanded to 10 bar at a rate of 2 kg/min. What is the temperature of the steam that was expanded, and what is the rate of entropy generation in this process? 4.5 Two metal blocks of equal mass M of the same substance, one at an initial temperature T1i and the other at an initial temperature T2i , are placed in a well-insulated (adiabatic) box of constant volume. A device that can produce work from a flow of heat across a temperature difference (i.e., a heat engine) is connected between the two blocks. Develop expressions for the maximum amount of work that can be obtained from this process and the common final temperature of the blocks when this amount of work is obtained. You may assume that

146 Chapter 4: Entropy: An Additional Balance Equation

4.6

4.7

4.8

4.9

Steam

the heat capacity of the blocks does not vary with temperature. The compressor discussed in Illustrations 3.4-4 and 4.51 is being used to compress air from 1 bar and 290 K to 10 bar. The compression can be assumed to be adiabatic, and the compressed air is found to have an outlet temperature of 575 K. a. What is the value of ΔS for this process? b. How much work, Ws , is needed per mole of air for the compression? c. The temperature of the air leaving the compressor here is higher than in Illustration 4.5-1. How do you account for this? In your calculations you may assume air is an ideal gas with CP∗ = 29.3 J/(mol K). A block of metal of total heat capacity CP is initially at a temperature of T1 , which is higher than the ambient temperature T2 . Determine the maximum amount of work that can be obtained on cooling this block to ambient temperature. The Ocean Thermal Energy Conversion (OTEC) project in Hawaii produces electricity from the temperature difference between water near the surface of the ocean (about 27◦ C) and the 600 m deep water at 5◦ C that surrounds the island. Estimate the maximum net work (total work less the work of pumping the water to the surface) that can be obtained from each kilogram of water brought to the surface, and the overall efficiency of the process. a. A steam turbine in a small electric power plant is designed to accept 4500 kg/hr of steam at 60 bar and 500◦ C and exhaust the steam at 10 bar. Assuming that the turbine is adiabatic and has been well designed (so that S˙ gen = 0), compute the exit temperature of the steam and the power generated by the turbine. b. The efficiency of a turbine is defined to be the ratio of the work actually obtained from the turbine to the work that would be obtained if the turbine operated isentropically between the same inlet and exit pressures. If the turbine in part (a) is adiabatic but only 80 percent efficient, what would be the exit temperature of the steam? At what rate would entropy be generated within the turbine? c. In off-peak hours the power output of the turbine in part (a) (100 percent efficient) is decreased by adjusting a throttling valve that reduces the turbine inlet steam pressure to 30 bar (see diagram) while keeping the flow rate constant. Compute T1 , the

60 bar

30 bar

500°C

T1 = ?

Turbine

W

10 bar T2 = ?

4.10 4.11

4.12

4.13

steam temperature to the turbine, T2 , the steam temperature at the turbine exit, and the power output of the turbine. Complete part (b) of Problem 3.31, assuming the compressor operates reversibly and adiabatically. Steam is produced at 70 bar and some unknown temperature. A small amount of steam is bled off just before entering a turbine and goes through an adiabatic throttling valve to atmospheric pressure. The temperature of the steam exiting the throttling valve is 400◦ C. The unthrottled steam is fed into the turbine, where it is adiabatically expanded to atmospheric pressure. a. What is the temperature of the steam entering the turbine? b. What is the maximum work per kilogram of steam that can be obtained using the turbine in its present mode of operation? c. Tests on the turbine exhaust indicate that the steam leaving is a saturated vapor. What is the efficiency of the turbine and the entropy generated per kilogram of steam? d. If the ambient temperature is 25◦ C and the ambient pressure is 1 bar, what is the maximum possible work that could be obtained per kilogram of steam in any continuous process? A well-insulated, 0.7-m3 gas cylinder containing natural gas (which can be considered to be pure methane) at 70 bar and 300 K is exhausted until the pressure drops to 3.5 bar. This process occurs fast enough that there is no heat transfer between the cylinder walls and the gas, but not so rapidly as to produce large velocity or temperature gradients in the gas within the cylinder. Compute the number of moles of gas withdrawn and the final temperature of the gas in the cylinder if a. Methane gas is assumed to be ideal with CP∗ = 36 J/(mol K). b. Methane is considered to be a real gas with the properties given in Fig. 3.3-2. If the heat capacity of an ideal gas is given by ∗ CV = (a − R) + bT + cT 2 + dT 3 + e/T 2

show that



S(T2 , V 2 ) − S(T1 , V 1 ) = (a − R) ln

T2 T1

 + b(T2 − T1 )

c 2 d (T − T12 ) + (T23 − T13 ) 2 2 3   V2 e −2 −2 − (T2 − T1 ) + R ln 2 V1

+

Also develop expressions for this fluid that replace Eqs. 4.4-3 and 4.4-4. 4.14 a. Steam at 35 bar and 600 K enters a throttling valve that reduces the steam pressure to 7 bar. Assuming there is no heat loss from the valve, what is the exit temperature of the steam and its change in entropy?

Problems 147

4.15

4.16

4.17

4.18

4.19

b. If air [assumed to be an ideal gas with CP∗ = 29.3 J/(mol K)] entered the valve at 35 bar and 600 K and left at 7 bar, what would be its exit temperature and entropy change? A tank contains 20 percent liquid water and 80 percent steam by volume at 200◦ C. Steam is withdrawn from the top of the tank until the fluid remaining in the tank is at a temperature of 150◦ C. Assuming the tank is adiabatic and that only vapor is withdrawn, compute a. The pressure in the tank finally b. The fraction of vapor and liquid in the tank finally c. The fraction of the total water present initially that was withdrawn One mole of carbon dioxide is to be compressed adiabatically from 1 bar and 25◦ C to 10 bar. Because of irreversibilities and poor design of the compressor, the compressor work required is found to be 25 percent greater than that for a well-designed (reversible) compressor. Compute the outlet temperature of the carbon dioxide and the work that must be supplied to the compressor for both the reversible and irreversible compressors for the two cases below. a. Carbon dioxide is an ideal gas with a constantpressure heat capacity of 37.151 J/(mol K). b. Carbon dioxide is an ideal gas with the constantpressure heat capacity given in Appendix A.II. Hydrogen has an auto-ignition temperature of 853 K; that is, hydrogen will ignite spontaneously at that temperature if exposed to oxygen. Hydrogen is to be adiabatically and reversibly compressed from 1 bar and 300 K to a high pressure. To avoid accidental explosions in case of a leak, the maximum allowed exit temperature from the compressor will be 800 K. Compute the maximum pressure that can be obtained in the compressor. You may consider hydrogen to be an ideal gas with the heat capacity given in Appendix A.II. If it is necessary to compress hydrogen to a higher pressure than is possible with the single-compression step above, an alternative is to use two compressors (or a two-stage compressor) with intercooling. In such a process the hydrogen is compressed in the first stage of the compressor, then cooled at constant pressure to a lower temperature, and then compressed further in a second compressor or stage. Although it may not be economical to do so, more than two stages can be used. a. Compute the maximum pressure that can be obtained in a two-stage compression with intercooling to 300 K between the stages, assuming hydrogen to be an ideal gas with the heat capacity given in Appendix A.II. b. Repeat the calculation above for a three-stage compression with intercooling to 300 K. Joe Unidel claims to have invented a steady-state flow device in which the inlet is steam at 300◦ C and 5 bar,

4.20

4.21

4.22

4.23

the outlet is saturated steam at 100◦ C and 1 bar, the device is adiabatic and produces approximately 388 kJ per kilogram of steam passed through the device. Should we believe his claim? Steam at 20 bar and 300◦ C is to be continuously expanded to 1 bar. a. Compute the final temperature, the entropy generated, the heat required, and the work obtained per kilogram of steam if this expansion is done by passing the steam through an adiabatic expansion valve. Will the final state be a vapor, a liquid, or a vaporliquid mixture? b. Compute the final temperature, the entropy generated, the heat required, and the work obtained per kilogram of steam if this expansion is done by passing the steam through a well-designed, adiabatic turbine. Will the final state be a vapor, a liquid, or a vapor-liquid mixture? c. Compute the final temperature, the entropy generated, the heat required, and the work obtained per kilogram of steam if this expansion is done by passing the steam through a well-designed, isothermal turbine. Will the final state be a vapor, a liquid, or a vapor-liquid mixture? In a large refrigeration plant it is necessary to compress a fluid, which we will assume to be an ideal gas with constant heat capacity, from a low pressure P1 to a much higher pressure P2 . a. If the compression is done in a single compressor that operates reversibly and adiabatically, obtain an expression for the work needed for the compression in terms of the mass flow rate, P1 , P2 , and the initial temperature, T1 . b. If the compression is to be done in two stages, first compressing the gas from P1 to P ∗ , then cooling the gas at constant pressure down to the compressor inlet temperature T1 , and then compressing the gas to P2 , develop an expression for the work needed for the compression. What should the value of the intermediate pressure be to accomplish the compression with minimum work? Repeat Problem 3.25, now considering nitrogen to be a real gas with the thermodynamic properties given in Fig. 3.3-3. An isolated chamber with rigid walls is divided into two equal compartments, one containing steam at 10 bar and 370◦ C, and the other evacuated. A valve between the compartments is opened to permit steam to pass from one chamber to the other. a. After the pressures (but not the temperatures) in the two chambers have equalized, the valve is closed, isolating the two systems. What are the temperature and pressure in each cylinder? b. If the valve were left open, an equilibrium state would be obtained in which each chamber has

148 Chapter 4: Entropy: An Additional Balance Equation the same temperature and pressure. What are this temperature and pressure? (Note: Steam is not an ideal gas under the conditions here.) 4.24 An adiabatic turbine is operating with an ideal gas working fluid of fixed inlet temperature and pressure, T1 and P1 , respectively, and a fixed exit pressure, P2 . Show that a. The minimum outlet temperature, T2 , occurs when the turbine operates reversibly, that is, when Sgen = 0. b. The maximum work that can be extracted from the turbine is obtained when Sgen = 0. 4.25 a. Consider the following statement: “Although the entropy of a given system may increase, decrease, or remain constant, the entropy of the universe cannot decrease.” Is this statement true? Why? b. Consider any two states, labeled 1 and 2. Show that if state 1 is accessible from state 2 by a real (irreversible) adiabatic process, then state 2 is inaccessible from state 1 by a real adiabatic process. 4.26 A very simple solar engine absorbs heat through a collector. The collector loses some of the heat it absorbs by convection, and the remainder is passed through a heat engine to produce electricity. The heat engine operates with one-half the Carnot efficiency with its lowtemperature side at ambient temperature Tamb and its high-temperature side at the steady-state temperature of the collector, Tc . The expression for the heat loss from the collector is Rate of heat loss from the collector = k (Tc − Tamb ) where T1 is the temperature at which heat enters the solar collector, and k is the overall heat transfer coefficient; all temperatures are absolute. What collector temperature produces the maximum rate at which work is produced for a given heat flux to the collector? 4.27 It is necessary to estimate how rapidly a piece of equipment can be evacuated. The equipment, which is 0.7 m3 in volume, initially contains carbon dioxide at 340 K and 1 bar pressure. The equipment will be evacuated by connecting it to a reciprocating constantdisplacement vacuum pump that will pump out 0.14 m3 /min of gas at any conditions. At the conditions here carbon dioxide can be considered to be an ideal gas with CP∗ = 39 J/(mol K). a. What will be the temperature and pressure of the carbon dioxide inside the tank after 5 minutes of pumping if there is no exchange of heat between the gas and the process equipment? b. The gas exiting the pump is always at 1 bar pressure, and the pump operates in a reversible adiabatic manner. Compute the temperature of the gas exiting the pump after 5 minutes of operation.

4.28 A 0.2-m3 tank containing helium at 15 bar and 22◦ C will be used to supply 4.5 moles per minute of helium at atmospheric pressure using a controlled adiabatic throttling valve. a. If the tank is well insulated, what will be the pressure in the tank and the temperature of the gas stream leaving the throttling valve at any later time t? b. If the tank is isothermal, what will be the pressure in the tank as a function of time? You may assume helium to be an ideal gas with CP∗ = 22 J/(mol K), and that there is no heat transfer between the tank and the gas. 4.29 A portable engine of nineteenth-century design used a tank of compressed air and an “evacuated” tank as its power source. The first tank had a capacity of 0.3 m3 and was initially filled with air at 14 bar and a temperature of 700◦ C. The “evacuated” tank had a capacity of 0.75 m3 . Unfortunately, nineteenth-century vacuum techniques were not very efficient, so the “evacuated” tank contained air at 0.35 bar and 25◦ C. What is the maximum total work that could be obtained from an airdriven engine connected between the two tanks if the process is adiabatic? What would be the temperature and pressure in each tank at the end of the process? You may assume that air is an ideal gas with CP∗ = 7R/2. 4.30 A heat exchanger is a device in which heat flows between two fluid streams brought into thermal contact through a barrier, such as a pipe wall. Heat exchangers can be operated in either the cocurrent (both fluid streams flowing in the same direction) or countercurrent (streams flowing in opposite direction) configuration; schematic diagrams are given here. Cocurrent flow

Countercurrent flow

The heat flow rate from fluid 1 to fluid 2 per unit ˙ is proportional to the length of the heat exchanger, Q, temperature difference (T1 − T2 ): ⎛

⎞

Heat flow rate from fluid 1 ⎜ ⎟ Q˙ = ⎝to fluid 2 per unit length of⎠ = κ(T1 − T2 ) heat exchanger where κ is a constant of proportionality with units of J/(m s K). The fluids in the two streams are the same and their flow rates are equal. The initial and final temperatures of stream 1 will be 35◦ C and 15◦ C, respectively, and those for stream 2 will be −15◦ C and 5◦ C. a. Write the balance equations for each fluid stream in a portion of the heat exchanger of length

Problems 149 device to see if your firm should buy it. All you are told is that 10 kW of work is obtained under the following operating conditions: Outlet Outlet Inlet Stream 1 Stream 2

dL and obtain differential equations by letting dL → 0.

4.31

4.32

4.33

4.34

b. Integrate the energy balance equations over the length of the exchanger to obtain expressions for the temperature of each stream at any point in the exchanger for each flow configuration. Also compute the length of the exchanger, in units of ˙ is the mass flow rate L0 = M˙ CP /2κ (where M of either stream), needed to accomplish the desired heat transfer. c. Write an expression for the change of entropy of stream 1 with distance for any point in the exchanger. a. Compute the maximum useful work that can be obtained when 1 kg of steam undergoes a closedsystem change of state from 30 bar and 600◦ C to 5 bar and 300◦ C when the atmospheric conditions are 1.013 bar and 298.15 K. b. Compute the maximum useful work that can be obtained when 1 kg of steam undergoes a closedsystem change of state from 30 bar and 600◦ C to 5 bar and 300◦ C when the atmospheric conditions are 1.013 bar and 298.15 K. Two tanks are connected as shown here. Tank 1 initially contains an ideal gas at 10 bar and 20◦ C, and both parts of tank 2 contain the same gas at 1 bar and 20◦ C. The valve connecting the two tanks is opened long enough to allow the pressures in the tanks to equilibrate and is then shut. There is no transfer of heat from the gas to the tanks or the frictionless piston, and the constant-pressure heat capacity of the gas is 4R. Compute the temperature and pressure of the gas in each part of the system at the end of the process and the work done on the gas behind the piston (i.e., the gas in subsystem 3). An important factor in determining the extent of air pollution in urban areas is whether the atmosphere is stable (poor mixing, accumulation of pollutants) or unstable (good mixing and dispersion of pollutants). Whether the atmosphere is stable or unstable depends on how the temperature profile (the so-called lapse rate) in the atmosphere near ground level compares with the “adiabatic lapse rate.” The adiabatic lapse rate is the atmospheric temperature that would result at each elevation if a packet of air (for example, as contained in a balloon) rose from ground level without any heat or mass transfer into or out of the packet. Assuming that air is an ideal gas, with a molecular weight of 29 and a constant-pressure heat capacity of 29.2 J/(mol K), obtain expressions for the pressure profile and the adiabatic lapse rate in the atmosphere. A device is being marketed to your company. The device takes in hot, high-pressure water and generates work by converting it to two outlet streams: steam and low-pressure water. You are asked to evaluate the

Mass flow rate (kg/s) T (◦ C) P (MPa) Steam

4.35

4.36

4.37

4.38

1.0 0.5 300 100 300 5.0 Liquid Quality unknown Sat.liquid

a. What is the quality of the steam in outlet stream 1? b. Is this device thermodynamically feasible? Diesel engines differ from gasoline engines in that the fuel is not ignited by a spark plug. Instead, the air in the cylinder is first compressed to a higher pressure than in a gasoline engine, and the resulting high temperature results in the spontaneous ignition of the fuel when it is injected into the cylinder. a. Assuming that the compression ratio is 25:1 (that is, the final volume of the air is 1/25 times the initial volume), that air can be considered an ideal gas with CV∗ = 21J/(mol K), that the initial conditions of the air are 1 atm and 30◦ C, and that the compression is adiabatic and reversible, determine the temperature of the air in the cylinder before the fuel is injected, and the minimum amount of work that must be done in the compression. b. The compression ratio in a supercharged gasoline engine is usually 10:1 (or less). Repeat the calculations of part (a) for this compression ratio. A rigid, isolated container 300 L in volume is divided into two parts by a partition. One part is 100 L in volume and contains nitrogen at a pressure of 200 kPa and a temperature of 100◦ C. The other part, 200 L in volume, contains nitrogen at a pressure of 2 MPa and a temperature of 200◦ C. If the partition breaks and a sufficiently long time elapses for the temperature and pressure to become uniform within the container, assuming that there is no heat transfer to the container and that nitrogen is an ideal gas with CP∗ = 30 J/(mol K) a. What is the final temperature and pressure of the gas in the container? b. How much entropy is generated in this process? In normal operation, a paper mill generates excess steam at 20 bar and 400◦ C. It is planned to use this steam as the feed to a turbine to generate electricity for the mill. There are 5000 kg/hr of steam available, and it is planned that the exit pressure of the steam will be 2 bar. Assuming that the turbine is well insulated, what is the maximum power that can be generated by the turbine? Atmospheric air is to be compressed and heated as shown in the figure before being fed into a chemical reactor. The ambient air is at 20◦ C and 1 bar, and is first compressed to 5 bar and then heated at constant

150 Chapter 4: Entropy: An Additional Balance Equation pressure to 400◦ C. Assuming air can be treated as a single-component ideal gas with CV∗ = 21 J/(mol K), and that the compressor operates adiabatically and reversibly, a. Determine the amount of work that must be supplied to the compressor and the temperature of the gas leaving the compressor. b. Determine the heat load on the heat exchanger. 20°C, 1 bar

5 bar

Heater

400°C, 5 bar

Ambient air

4.42

4.39 One kilogram of saturated liquid methane at 160 K is placed in an adiabatic pistion-and-cylinder device, and the piston will be moved slowly and reversibly until 25 percent of the liquid has vaporized. Compute the maximum work that can be obtained. 4.40 A sugar mill in Florida has been disposing of the bagasse (used sugar cane) by open-air burning. You, as a new chemical engineer, determine that by using the dried bagasse as boiler fuel in the mill, you can generate 5000 kg/hr of surplus steam at 20 bar and 600◦ C. You propose to your supervisor that the mill invest in a steam turbine–electrical generator assembly to generate electricity that can be sold to the local power company. a. If you purchase a new, state-of-the-art adiabatic and isentropic steam turbine, and the turbine exit pressure is 1 bar, what is the temperature of the exit steam, how much electrical energy (in kW) can be generated, and what fraction of the steam exiting the turbine is vapor? b. Instead, a much cheaper used turbine is purchased, which is adiabatic but not isentropic, that at the same exit pressure of 1 bar produces 90 percent of the work produced by the state-of-the-art turbine. What is the temperature of the exit steam, what fraction of the steam exiting the turbine is vapor, and what is the rate of entropy generation? 4.41 A well-insulated cylinder fitted with a frictionless piston initially contains nitrogen at 15◦ C and 0.1 MPa. The piston is placed such that the volume to the right of the piston is 0.5 m3 , and there is negligible volume to the left of the piston and between the piston and the tank, which is also well insulated. Initially the tank of volume 0.25 m3 contains nitrogen at 400 kPa and 200◦ C. For the calculations below, assume that nitrogen is an ideal gas with CP∗ = 29.3J/(mol K)

Tank

4.43

4.44

Cylinder

4.45 Piston

a. The piston is a perfect thermal insulator, and the valve between the cylinder and the tank is opened only long enough for the pressure in the tank and cylinder to equalize. What will the equalized pressure be, and what will be the temperatures of the nitrogen in the tank, in the cylinder to the left of the piston, and in the cylinder to the right of the piston? b. What will the volumes in the cylinder be to the left and to the right of the piston? c. How much entropy has been generated by the process? If the piston in the previous problem is replaced by one that is a perfect conductor, so that after the valve is closed the temperatures of the gas on the two sides of the piston equalize (clearly, the piston will move in this process), a. What will the equalized pressure in the cylinder be, and what will be the temperature of the nitrogen in the cylinder? b. What will the volumes in the cylinder be to the left and to the right of the piston? c. How much entropy has been generated in going from the initial state (before the valve was opened) to the final state (equalized temperature and pressure on the two sides of the piston)? A tank of 0.1 m3 volume initially containing nitrogen at 25◦ C and 1 bar will be filled with compressed nitrogen at a rate of 20 mol/s. The nitrogen coming from the compressor and into the tank is at an absolute pressure of 110 bar and a temperature of 80◦ C. The filling process occurs sufficiently rapidly that there is negligible heat transfer between the gas and the tank walls, and a valve is closed to stop the filling process when the pressure in the tank reaches 100 bar. Assuming nitrogen is an ideal gas with CP∗ = 29.3 J/(mol K), a. What is the nitrogen temperature immediately after the filling process ends? b. How long did it take for the pressure of the gas in the tank to reach 100 bar? c. After a sufficiently long period of time, due to heat transfer with the surroundings, the temperature of the gas drops to 25◦ C. What is the pressure of the nitrogen in the tank? A stream of hot water at 85◦ C and a rate of 1 kg/s is needed for the pasteurizing unit in a milk-bottling plant. Such a stream is not readily available, and will be produced in a well-insulated mixing tank by directly injecting steam from the boiler plant at 10 bar and 200◦ C into city water available at 1 bar and 20◦ C. a. Calculate the flow rates of city water and steam needed. b. Calculate the rate of entropy production in the mixing tank. Low-density polyethylene is manufactured from ethylene at medium to high pressure in a radical chain

Problems 151

4.46

4.47

4.48

4.49 4.50 4.51

polymerization process. The reaction is exothermic, and on occasion, because of cooling system failure or operator error, there are runaway reactions that raise the pressure and temperature in the reactor to dangerous levels. For safety, there is a relief valve on the reactor that will open when the total pressure reaches 200 bar and discharge the high-pressure ethylene into a holding tank with a volume that is four times larger than that of the reactor, thereby reducing the risk of explosion and discharge into the environment. Assuming the volume occupied by the polymer is very small, that the gas phase is pure ethylene, that in case of a runaway reaction the temperature in the reactor will rise to 400◦ C, and that the holding tank is initially evacuated, what will be the temperature and pressure in the reactor and the holding tank if ethylene can be considered to be an ideal gas with CP∗ = 73.2 J/(mol K)? An inventor has proposed a flow device of secret design for increasing the superheat of steam. He claims that a feed steam at 1 bar and 100◦ C is converted to superheated steam at 1 bar and 250◦ C and saturated water (that is, water at its boiling point) also at 1 bar, and that no additional heat or work is needed. He also claims that 89.9 percent of the outlet product is superheated steam, and the remainder is saturated water. Will the device work as claimed? Two separate experiments are performed on a gas enclosed in a piston-and-cylinder device, both starting from the same initial state. The result of the first experiment is to be used to predict the outcome of the second. a. In the first experiment, the piston is free to move, with the external pressure held constant. A small amount of heat is added to the gas in the cylinder, resulting in the expansion of the gas. The temperature of the gas is found to have increased. b. In the second experiment, the piston is not free to move; instead, its position is adjusted manually. In addition, the device is insulated, so that no heat flows to or from the surroundings. The piston is moved outward slowly, allowing the gas to expand by a small amount. Does the temperature of the gas increase or decrease? Consider the Hilsch-Ranque vortex tube discussed in Illustration 4.5-6 (pages 135–136). Starting with air at 4 bar and 25◦ C, an exhaust pressure of 1.013 bar, and that half the air that enters the tube will be withdrawn at the higher temperature, what is the maximum temperature difference that can be obtained? Treat air as J . an ideal gas with CP∗ = 29.3 mol ·K R Redo Problem 4.4 using Aspen Plus . Redo Problem 4.6 using Aspen Plus R . R Redo Problem 4.9 using Aspen Plus .

4.52 4.53 4.54 4.55 4.56 4.57 4.58 4.59

4.60

4.61

4.62

4.63

4.64

R Redo Problem 4.11 using Aspen Plus . R Redo Problem 4.17 using Aspen Plus . R Redo Problem 4.18 using Aspen Plus . R Redo Problem 4.20 using Aspen Plus . R Redo Problem 4.37 using Aspen Plus . R Redo Problem 4.38 using Aspen Plus . R Redo Problem 4.40 using Aspen Plus . 100 kg of steam is available at 2 MPa and 800◦ C. a. Determine the maximum amount of shaft work that can be obtained from this steam in a non-flow process if the ambient conditions are 25◦ C and 1 bar. b. This steam will be used in a non-flow work producing process that reduces its pressure to 0.6 MPa and its temperature to 400◦ C. Determine the maximum amount of shaft work that could have been obtained in that process. c. Determine the maximum amount of non-flow shaft work that could be obtained from the steam available at the end of the process in part b. Steam is available at 2 MPa and 800◦ C. a. Determine the maximum amount of shaft work that can be obtained from this steam in a flow process if the ambient conditions are 25◦ C and 1 bar. b. This steam will be used in a work producing process that reduces its pressure to 0.6 MPa and its temperature to 400◦ C. Determine the maximum amount of shaft work that could have been obtained in that process. c. Determine the maximum amount of shaft work that could be obtained in a flow from the steam available at the end of the process in part b. From a process, 100 kg of steam is available at 2 MPa and 800◦ C. a. Determine the maximum amount of shaft work that can be obtained from this steam in a non-flow process if the ambient conditions are 25◦ C and 1 bar. b. This steam will be used in a work producing nonflow process that reduces its pressure to 0.6 MPa and its temperature to 400◦ C. Determine the maximum amount of non-flow shaft work that can be obtained in that process. Which has the greater potential to produce more available work steam at 2 MPa and 800◦ C or steam at 1.4 MPa and 900◦ C? A stream of flowing air is available at 25 bar and 1000◦ C or at 40 bar and 900◦ C. Assuming air is an ideal gas with a constant pressure heat capacity of 29.7 J/mol K, which stream has the greater potential produce available work? a. Derive the equations analogous to Eqs. 4.6-6 and 8 for the nonflow available work. b. Derive the integrated forms of these equations for an ideal gas of constant heat capacity.

Chapter

5 Liquefaction, Power Cycles, and Explosions In this chapter we consider several practical applications of the energy and entropy balances that we have developed. These include the liquefaction of a gas, and the analysis of cycles used to convert heat to work and to provide refrigeration and air conditioning. Finally, as engineers we have a social responsibility to consider safety as a paramount issue in anything we design or operate. Therefore, methods to estimate the energy that could be released in different types of nonchemical explosions are presented.

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 5 The goals of this chapter are for the student to: • Be able to solve problems involving the liquefaction of gases (Sec. 5.1) • Be able to compute the work that can be obtained from different types of power

cycles and using different working fluids (Sec. 5.2)

• Be able to compute the work required for the operation of refrigeration cycles

(Sec. 5.2)

• Be able to compute the energy release resulting from the uncontrolled expansion

of a gas (Sec. 5.4)

• Be able to compute the energy release resulting from an explosion that involves a

boiling liquid (Sec. 5.4)

NOTATION INTRODUCED IN THIS CHAPTER C.O.P. ˆ vap H vap H Tb ˆ vap U vap U 152

Coefficient of performance of a refrigeration cycle Enthalpy change (or heat) of vaporization per unit mass (J/kg) Molar enthalpy change on vaporization (J/mol) Boiling temperature of a liquid at atmospheric pressure (K) Internal energy change on vaporization per unit mass (J/kg) Molar internal energy change on vaporization (J/mol)

5.1 Liquefaction 153

5.1 LIQUEFACTION An important industrial process is the liquefaction of gases, such as natural gas (to produce liquid natural gas (LNG)), propane, and refrigerant gases, to name a few. One way to liquefy a gas is to cool it below its boiling-point temperature at the desired pressure. However, this would require refrigeration equipment capable of producing very low temperatures. Therefore, such a direct liquefaction process is not generally used. What is more commonly done is to start with a gas at low pressure, compress it to high pressure (which increases its temperature since work has been done on it), cool this high-temperature gas at the constant high pressure, and then expand it to low pressure and low temperature using a Joule-Thomson expansion, which produces a mixture of liquid and vapor. In this way the cooling is done at a higher temperature (and pressure), so that low-temperature refrigeration is not needed. (Such a process is also used internally in many refrigeration cycles, including your home refrigerator, as discussed in the next section.) The vapor and liquid are then separated in a flash drum (an insulated, constant-pressure container). The process just described is shown schematically in Fig. 5.1-1. 5 Gas 2

4

3

1 Throttling valve Compressor

Flash drum

Cooler

6

Liquefied gas

(single or multistage)

Figure 5.1-1 A simple liquefaction process without recycle.

The efficiency of this process, that is, the amount of liquefied gas produced for each unit of work done in the compressor, can be improved upon by better engineering design. For example, instead of merely discarding the low-temperature, low-pressure gas leaving the flash drum (stream 5), the gas can be used to cool the high-pressure gas upstream of the throttle valve and then returned to the compressor, so that none of the gas is wasted or exhausted to the atmosphere. This process, referred to as the Linde process, is shown in Fig. 5.1-2. In this way the only stream leaving the liquefaction plant is liquefied gas, and, as shown in Illustration 5.1-1, more liquefied gas is produced per unit of energy expended in the compressor.

5' 5 2

4

3'

1

3 1'

Heat exchanger

Throttling valve

Cooler

Compressor (single- or multistage)

Figure 5.1-2 The more efficient Linde liquefaction process.

6 Flash drum

Liquefied gas

154 Chapter 5: Liquefaction, Power Cycles, and Explosions Illustration 5.1-1 Comparing the Efficiency of the Simple and Linde Liquefaction Processes It is desired to produce liquefied natural gas (LNG), which we consider to be pure methane, from that gas at 1 bar and 280 K (conditions at point 1 in Figs. 5.1-1 and 5.1-2). Leaving the cooler, methane is at 100 bar and 210 K (point 3). The flash drum is adiabatic and operates at 1 bar, and the compressor can be assumed to operate reversibly and adiabatically. However, because of the large pressure change, a three-stage compressor with intercooling is used. The first stage compresses the gas from 1 bar to 5 bar, the second stage from 5 bar to 25 bar, and the third stage from 25 bar to 100 bar. Between stages the gas is isobarically cooled to 280 K. a. Calculate the amount of work required for each kilogram of methane that passes through the compressor in the simple liquefaction process. b. Calculate the fractions of vapor and liquid leaving the flash drum in the simple liquefaction process of Fig. 5.1-1 and the amount of compressor work required for each kilogram of LNG produced. c. Assuming that the recycled methane leaving the heat exchanger in the Linde process (Fig. 5.1-2) is at 1 bar and 200 K, calculate the amount of compressor work required for each kilogram of LNG produced. Data: The thermodynamic properties of methane are given in Fig. 3.3-2

Solution a. For each stage of compression, assuming steady-state and reversible adiabatic operation, we have from the mass, energy, and entropy balances, respectively: ˙ in + M ˙ out 0=M ˆ in + M˙ out H ˆ out + W ˙ in H ˙ 0=M ˙ in Sˆin + M ˙ out Sˆout 0=M

or or or

˙ out = −M˙ in = −M˙ M ˙ = M˙ (H ˆ out − H ˆ in ) W Sˆout = Sˆin

So through each compressor (but not intercooler) stage, one follows a line of constant entropy in Fig. 3.3-2 (which is redrawn here with all the stages indicated). For the first compressor stage, we have ˆ in (T = 280 K, P = 1 bar) = 940 kJ/kg H

and

Sˆin (T = 280 K, P = 1 bar) = 7.2 kJ/(kg K) ˆ out (Sˆ = 7.2 kJ/(kg K), P = 5 bar) = 1195 kJ/kg H

and

Tout = 388 K

so that the first-stage work per kilogram of methane flowing through the compressor is ˙ (first stage) = (1195 − 940) kJ/kg = 225 kJ/kg W

After cooling, the temperature of the methane stream is 280 K, so that for the second compressor stage, we have ˆ in (T = 280 K, P = 5 bar) = 938 kJ/kg H

and

Sˆin (T = 280 K, P = 5 bar) = 6.35 kJ/(kg K) ˆ out (Sˆ = 6.35 kJ/(kg K), P = 25 bar) = 1180 kJ/kg H

and

Tout = 386 K

5.1 Liquefaction 155

Figure 5.1-3 Pressure-enthalpy diagram for methane. (Source: W. C. Reynolds, Thermodynamic Properties in SI, Department of Mechanical Engineering, Stanford University, Stanford, CA, 1979. Used with permission.)

Therefore, the second-stage work per kilogram of methane flowing through the compressor is ˙ (second stage) = (1180 − 938) kJ/kg = 242 kJ/kg W

Similarly, after intercooling, the third-stage compressor work is found from ˆ in (T = 280 K, P = 25 bar) = 915 kJ/kg H

and

Sˆin (T = 280 K, P = 25 bar) = 5.5 kJ/(kg K) ˆ out (Sˆ = 5.5 kJ/(kg K), P = 100 bar) = 1140 kJ/kg H

and

Tout = 383 K

Therefore, the third-stage work per kilogram of methane flowing through the compressor is ˙ (third stage) = (1140 − 915) kJ/kg = 225 kJ/kg W

Consequently, the total compressor work through all three stages is ˙ = 255 + 242 + 225 = 722 kJ/kg W

b. The liquefaction process is a Joule-Thomson expansion, and therefore occurs at constant enthalpy. The enthalpy of the methane leaving the cooler at 100 bar and 210 K is 493 kJ/kg.

156 Chapter 5: Liquefaction, Power Cycles, and Explosions At 1 bar the enthalpy of the saturated vapor is 582 kJ/kg, and that of the liquid is 71 kJ/kg. Therefore, from the energy balance on the throttling valve and flash drum, we have1 ˆ out ˆ in = H H

or

ˆ ˆ ˆ H(210 K, 100 bar) = (1 − ω)H(saturated vapor, 1 bar) + ω H(saturated liquid, 1 bar) 493

kJ kJ kJ = (1 − ω) · 71 + ω · 582 kg kg kg

so that ω = 0.826 is the fraction of vapor leaving the flash drum, and (1 − ω) = 0.174 is the fraction of the methane that has been liquefied. Therefore, for each kilogram of methane that enters the simple liquefaction unit, 826 g of methane are lost as vapor, and only 174 g of LNG are produced. Further, since 722 kJ of work are required in the compressor to produce 174 g of LNG, approximately 4149 kJ of compressor work are required for each kilogram of LNG produced. c. While the Linde process looks more complicated, it can be analyzed in a relatively simple manner. First, we choose the system for writing balance equations to be the subsystem consisting of the heat exchanger, throttle valve, and flash drum (though other choices could be made). The mass and energy balances for this subsystem (since there are no heat losses to the outside or any work flows) are M˙ 3 = M˙ 5 + M˙ 6

or, taking M˙ 3 = 1 and letting ω be the mass fraction of vapor, 1 = (1 − ω) + ω

for the mass balance, and ˆ3 = M ˆ 5 + M ˆ6 ˙ 5 H ˙ 6H M˙ 3 H ˆ H(T = 210 K, P = 100 bar) ˆ ˆ = ω · H(T = 200 K, P = 1 bar) + (1 − ω) · H(saturated liquid, P = 1 bar) 493

kJ kJ kJ = ω · 770 + (1 − ω) · 70.7 kg kg kg

for the energy balance. The solution to this equation is ω = 0.604 as the fraction of vapor that is recycled, and 0.396 as the fraction of liquid. The balance equations for mixing the streams immediately before the compressor are ˙ 5 + M ˙1 = M ˙ 1 M

and basing the calculation on 1 kg of flow into the compressor, M˙ 1 = 1

1 In

M˙ 5 = 0.604

and M˙ 1 = 0.396

this equation saturated liquid refers to the fact that the phase is at its equilibrium (boiling) temperature at the specified pressure.

5.1 Liquefaction 157 for the mass balance, and ˙ 1 H ˆ 1 + M ˆ 5 = M˙ 1 H ˆ 1 ˙ 5 H M ˆ = 280 K, P = 1 bar) 0.396 · H(T ˆ ˆ 1 (T = ?, P = 1 bar) + 0.604 · H(T = 200 K, P = 1 bar) = H

0.396 · 940

kJ kJ kJ ˆ + 0.604 · 770 = 837.3 = H(T = ?, P = 1 bar) kg kg kg

for the energy balance. Using the redrawn Fig. 3.3-2 here, we conclude that the temperature of a stream entering the compressor is 233 K. To complete the solution to this problem, we need to compute the compressor work load. Following the solution for part (a), we find for the first stage that ˆ in (T = 233 K, P = 1 bar) = 837 kJ/kg H

and

Sˆin (T = 233 K, P = 1 bar) = 6.8 kJ/(kg K) ˆ out (Sˆ = 6.8 kJ/(kg K), P = 5 bar) = 1020 kJ/kg H

and

Tout = 388 K

Therefore, now the first-stage work per kilogram of methane flowing through the compressor is ˙ (first stage) = 1020 − 837 kJ/kg = 183 kJ/kg W

Since the methane leaving the intercooler is at 280 K and 5 bar, the work in each of the second and third stages of the compressor is the same as in part (a). Therefore, the total compressor work is ˙ = 183 + 242 + 225 = 650 kJ/kg of methane through the compressor W

However, each kilogram of methane through the compressor results in only 0.396 kg of LNG, as the remainder of the methane is recycled. Consequently, the compressor work required per kilogram of LNG produced is (650 kJ/kg)/0.396 kg = 1641 kJ/kg of LNG produced. This is to be compared with 4149 kJ/kg of LNG produced in the simple liquefaction process.  R simulation for this illustration to solve the simple liquefaction process The Aspen Plus with the Peng-Robinson equation of state is available on the Wiley website for this book in the folder Aspen Illustrations>Chapter 5>5.1-1a+b. The results are as follows: Compressor Work

=

281.3(stage 1) + 260.6(stage 2) + 224.0(stage 3)

=

765.8(total) 

Also, the fraction of the liquid leaving the flash unit is 0.164.

Comment By comparing the Linde process with the simple liquefaction process we see the importance and advantages of clever engineering design. In particular, the Linde process uses only about

158 Chapter 5: Liquefaction, Power Cycles, and Explosions 40 percent of the energy required in the simple process to produce a kilogram of LNG. Also, unlike the simple process, the Linde process releases no gaseous methane, which is a greenhouse gas that contributes to global warming, into the atmosphere. Finally, it should be pointed out that since a graph of thermodynamic properties, rather than a more detailed table of values, was used, the numerical values of the properties read from the diagram are approximate.

5.2 POWER GENERATION AND REFRIGERATION CYCLES So far we have considered only the Carnot cycle for converting heat (produced by burning coal, oil, or natural gas) to work (usually electricity). While this cycle is the most efficient possible for converting heat to work, in practice it is rarely used because of the large amount of work that must be supplied during the isothermal compression step. In this section other power generation and refrigeration cycles are introduced. However, we consider only ideal such cycles, that is, those that are the most efficient, theoretical implementations. In actual operation, the efficiency of these cycles will be lower because of irreversibilities due to fluid friction, that there is not a step change in temperature as modelled, but rather a more gradual temperature change in a heat exchanger, that there is a variation in pressure not a step change in a pump or turbine, and that there may be heat conduction in the device between the hot and cold parts. Therefore, while different parts of the cycles to be considered are modelled here to be isentropic, isothermal, adiabatic or at constant pressure, they will not be so in an actual machine. Therefore the efficiencies computed here are upper bounds. The most widely used commercial cycle is the Rankine cycle, shown in Fig. 5.2-1 with water as the operating or so-called working fluid, though other fluids may also be used. The Rankine cycle is used in most electrical power generation plants. The cycle is closed in that the working fluid is continually recirculated without any addition or removal. When the Rankine cycle is operated in reverse, as will be described later, it is used in refrigeration plants, air conditioners, heat pumps and refrigerators. Because of the temperatures involved, the working fluid in such cases is a refrigerant such as a partially halogenated hydrocarbon such as 1,1,1,2 tetrafluoroethane (HFC- 134a). In this cycle the turbine and the pump are considered to operate isentropically, the condenser operates isobarically, and the fluid in the boiler is heated at constant pressure. The properties and path for this cycle are:

QB

Rankine cycle High-pressure steam

Boiler

3

WT

High-pressure water

Pump

Turbine 4 Low-pressure steam

2

Condenser

QC

WP

1 Low-pressure water

Figure 5.2-1 Rankine cycle.

5.2 Power Generation and Refrigeration Cycles 159 Point

Path to Next Point

1

T

Sˆ

P

T1

P1

Sˆ1

ˆ1 H

T2

P2

Sˆ2 = Sˆ1

ˆ2 H

T3

P3 = P2

Sˆ3

ˆ3 H

T4

P4

Sˆ4 = Sˆ3

ˆ4 H

T1

P1 = P4

Sˆ1

ˆ1 H

↓

Isentropic 2 Isobaric 3

↓

↓

Isentropic 4 Isobaric 1

ˆ H

↓

Energy Flow ˙P=M ˙ Vˆ1 (P2 − P1 ) W Q˙ B ˙ (H ˆ4 − H ˙T=M ˆ3) W Q˙ C

The work flows in this process are computed as follows. Step 1 → 2 (pump) The mass and energy balances on the open system consisting of the pump and its contents operating at steady state are

dM = 0 = M˙ 1 + M˙ 2 dt

or M˙ 1 = −M˙ 2 = M˙

and

dU ˙P ˆ 1 + M˙ 2 H ˆ2 + W = 0 = M˙ 1 H dt so that

˙ P = −M˙ 2 H ˆ 2 − M˙ 1 H ˆ 1 = M˙ (H ˆ 1) ˆ2 − H W ˆ1 − P1 Vˆ1 ) ˆ2 + P2 Vˆ2 − U = M˙ 1 (U = M˙ Vˆ1 (P2 − P1 ) since the molar volume and internal energy of the liquid are essentially independent of pressure at constant temperature. Step 3 → 4 (turbine)

dM = 0 = M˙ 3 + M˙ 4 dt

or M˙ 3 = −M˙ 4 = M˙

since the mass flow rate is constant throughout the process, and

dU ˙T ˆ 3 − M˙ 4 H ˆ4 + W = 0 = M˙ 3 H dt

˙ T = M˙ (H ˆ4 − H ˆ 3) or W

The efficiency η of this process (and other cyclic processes as well) is defined to be ˙T+W ˙ P ), where W ˙ T is negative and W ˙P the ratio of the net work obtained, here −(W ˙ is positive in value, to the heat input QB ; that is,

η=

˙T+W ˙ P) −(W Q˙ B

160 Chapter 5: Liquefaction, Power Cycles, and Explosions The plot of this cycle on a temperature–entropy diagram is such that 1 → 2 is a line of constant entropy (vertical line), 2 → 3 is a line of constant pressure, 3 → 4 is another line of constant entropy, and 4 → 1 closes the cycle with a line of constant temperature. Several different locations for these lines, depending largely on the operation of the boiler, are shown in Fig. 5.2-2. Figure 5.2-2a looks like the T -Sˆ diagram for the Carnot cycle (Fig. 4.3-2c), and indeed at these conditions the Rankine cycle has the same efficiency as the Carnot cycle. The practical difficulty is that the high-speed turbine blades will severely erode if impacted by liquid droplets. Therefore, step 3 → 4 should be completely in the vapor

2

T

2

3

4

1

3

T

4

1

^

^

S (a)

S (b) 3 3

T

T

2 4

1

2 4

1

^

^

S (c)

S (d)

3

T

3

T

T

2

2 4

1 0

2 4

1 0

0

3

^

S

4

1 0

0

^

S

0

^

S

Q2→3

Q4→1

W NET

(e)

(f)

(g)

Figure 5.2-2 (a)–(d) T -Sˆ diagrams of various Rankine cycles. (e) Area equal to heat flow into boiler. (f ) Area equal to heat removed in condenser. (g ) Area equal to net work produced in cycle.

5.2 Power Generation and Refrigeration Cycles 161 region, as in Fig. 5.2-2b and d. Also, to minimize the work required for the pump, the fluid passing through it should be all liquid, as in Fig. 5.2-2c and d. Therefore, of the choices in Fig. 5.2-2, the conditions in Fig. 5.2-2d are used for practical operation of a Rankine power cycle. In this case, the efficiency is less than that of the Carnot cycle. Also, the condenser operating temperature is generally determined by the available cooling water. However, lower temperatures and pressures in the condenser improve the efficiency of the cycle (see Problems 5.2 and 5.3). As pointed out in Sec. 4.3, the differential entropy change dS and the heat flow Q for a reversible process in a closed system are related as follows:

dS =

Q T

or Q = T dS

so that the heat flows (and the work produced) are related to areas on the T -S diagram for the Rankine power cycle as follows. The heat flow into the cycle on going from point 2 to 3 is equal to  3 Q2→3 = T dS 2

This heat flow is given by the area shown in Fig. 5.2-2e. Similarly, the heat flow from point 4 to 1 is equal in magnitude to the area shown in Fig 5.2-2f, but negative in value (since S4 is larger than S1 ). Finally, since from the energy balance, the net work flow is equal to the difference of the two heat flows into the cycle, this work flow is given by the area in Fig. 5.2-2g. Similar graphical analyses can be done for the other cycles considered in this section, though we will not do so.

Illustration 5.2-1 Calculating the Efficiency of a Steam Power Cycle A Rankine power generation cycle using steam operates at a temperature of 100◦ C in the condenser, a pressure of 3.0 MPa in the evaporator, and a maximum temperature of 600◦ C. Assuming the pump and turbine operate reversibly, plot the cycle on a T -Sˆ diagram for steam and compute the efficiency of the cycle.

Solution We start by plotting the cycle on the temperature-entropy diagram for steam (Fig. 3.3-1b, repeated below). Then we construct the table of thermodynamic properties (which follows) for each point in the cycle using the known operating conditions and the path from each point in the cycle to the next point. Based on this information, we can then compute the efficiency of the cycle η=

˙T+W ˙ P) −(W Q˙ B

=

−(−947.6 + 3.03) 3260.2

=

944.3 = 0.290 3260.2

or 29.0 percent

162 Chapter 5: Liquefaction, Power Cycles, and Explosions

5.2 Power Generation and Refrigeration Cycles 163

Point

(State) Path to Next Point

T ( C)

P (MPa)

Sˆ (kJ/kgK)

1

(Saturated liquid)

100

0.10135

1.3069 ↓

419.04

3.0 ↑

1.3069

422.07

3.0

7.5085 ↓

3682.3

0.10135 ↑

7.5085

∼ 2734.7

0.10135

1.3069

◦

ˆ H (kJ/kg)

Vˆ (m /kg) 3

0.001044 ˆ P = Vˆ1 (P2 − P1 ) W = 0.001044 m3 /kg × (3.0 − 0.10135) MPa = 3.03 kJ/kg

Isentropic

∼ 103

2

ˆ B = 3682.3 − 422.1 Q = 3260.2 kJ/kg

Isobaric 3

600 Isentropic ∼ 126.2

4

Saturated liquid

100

ˆ T = (H ˆ3) ˆ4 − H W = 2734.7 − 3682.3 = −947.6 kJ/kg ˆC = H ˆ4 ˆ1 − H Q = 419.04 − 2734.7 = −2315.7 kJ/kg

Isobaric

1

Energy Flow

419.04

Note: In preparing this table we have used bold notation to indicate data given in the problem statement (e.g., 100◦ C) or from information on an adjacent point in the cycle (also indicated by arrows connecting two points), italics to indicate numbers found in the steam tables (e.g., 0.10135 MPa), and a tilde to indicate numbers obtained from the steam tables by interpolation (e.g., ∼ 2734.7 kJ/kg).



R simulation with the Peng-Robinson equation of state to solve this Using an Aspen Plus illustration is available Wiley website for this book in the folder Aspen>Illustration 5.2-1. R The Aspen Plus simulation for this illustration available on the Wiley website for this book in the folder Aspen Illustrations>Chapter 5>5.2-1 PR. The results are as follows:

Point 1 2 3 4

Pump↓ Boiler↓ Turbine↓ Condenser

T(◦ C) 100.2 600 129.6 100.0

P(MPa) 3.0 3.0 0.1013

Q(kJ/kg) 3334.4 −2388.1

W(kJ/kg) 3.02 −942.6

R simulation with IAPWS-95 thermodynamic properties of water and Using an Aspen Plus streamto solve this illustration is available Wiley website for this book in the folder Aspen Illustrations>Chapter 5>5.2-1 IAPWS-95. The results are as follows:

Point 1 2 3 4

Pump↓ Boiler↓ Turbine↓ Condenser

T(◦ C) 100.2 600 129.6 100.0

P(MPa) 3.0 3.0 0.1013

Q(kJ/kg) 3260.6 −2316.7

W(kJ/kg) 3.02 −946.9 

Those results are in excellent agreement with the results obtained using the stream tables.

The Rankine power cycle above receives heat at a high temperature, emits it at a lower temperature, and produces work. (Does this cycle violate the Kelvin-Planck statement of the second law? The answer is no since there are two heat flows, one into the cycle

164 Chapter 5: Liquefaction, Power Cycles, and Explosions from a hot body, and the second from the cycle into a cooler body or cooling water. Consequently, not all the heat available at high temperature is converted into work, only a portion of it.) A similar cycle, known as the Rankine refrigeration cycle, operates essentially in reverse by using work to pump heat from a low-temperature region to a high-temperature region. This refrigeration cycle is shown in Fig. 5.2-3a. In a home refrigerator the condenser is the air-cooled coil usually found at the bottom or back of the refrigerator, and the evaporator is the cooling coil located in the freezer section. Here the compressor and the turbine are considered to operate isentropically, and the condenser and evaporator to operate at constant pressure. Frequently, and this is the case we consider here, the condenser receives a two-phase mixture and emits a saturated liquid at the same pressure. The properties and path for this cycle are given in the following table. ˆ diagrams are Schematic diagrams of the path of this cycle on both T -Sˆ and P -H shown in Fig. 5.2-3b. As in the previous case, the mass flow is the same in all parts of the cycle, and we ˙ . The steady-state energy balance around the evaporator and its designate this by M contents yields

ˆ 2 − M˙ H ˆ 3 + Q˙ B 0 = M˙ H

ˆ3 − H ˆ 2) or Q˙ B = M˙ (H

The energy balance on the compressor is

ˆ4 + W ˆ 3 − M˙ H ˙P 0 = M˙ H

ˆ 3) ˙ P = M˙ (H ˆ4 − H or W

QB

Refrigeration cycle Low-pressure vapor 3

2

Evaporator (or boiler)

Expansion turbine

Compressor

WP

Low-pressure vapor-liquid mixture WT

1 Condenser

High-pressure 4 vapor

High-pressure liquid QC (a)

4 1

1 T

4

P

2

3

2

^

3 ^

S

H (b)

Figure 5.2-3 (a) The Rankine refrigeration cycle. (b) The Rankine refrigeration ˆ diagrams. cycle on T -Sˆ and P -H

5.2 Power Generation and Refrigeration Cycles 165 The energy balance on the condenser is

ˆ 1 + Q˙ C ˆ 4 − M˙ H 0 = M˙ H

Point 1 2

3 4

(State) Path to Next Point (Saturated liquid) Isentropic (Vapor-liquid mix.) Isobaric [also isothermal in this case] (Saturated vapor) Isentropic (Superheated vapor) Isobaric

1

ˆ 4) ˆ1 − H or Q˙ C = M˙ (H

T

Sˆ

P

ˆ H

P1

Sˆ1

ˆ1 H

T2

P2

Sˆ2 = Sˆ1

ˆ2 H

T3 = T2

P3 = P2

Sˆ3

ˆ3 H

T4

P4

Sˆ4 = Sˆ3

ˆ4 H

P1 = P4

Sˆ1

ˆ1 H

T1

↓

T1

↓

↓

↓

↓

Energy Flow ˙T W Q˙ B

˙P W Q˙ C

and the energy balance on the turbine is

ˆ 1 − M˙ H ˙T ˆ2 + W 0 = M˙ H

˙ T = M˙ (H ˆ2 − H ˆ 1) or W

In refrigeration it is common to use the term coefficient of performance (C.O.P.), defined as

C.O.P. =

Q˙ B Heat removed from low-temperature region =− ˙P+W ˙T Net work required for heat removal W

rather than efficiency (as was used in the Rankine power generation cycle). For the cycle shown here, we have

C.O.P. =

ˆ2 ˆ3 − H Q˙ B H = ˙T ˙P+W ˆ 3 ) + (H ˆ 1) ˆ4 − H ˆ2 − H W (H

The amount of work recovered in the expansion turbine of the Rankine refrigeration cycle is small. Consequently, to simplify the design and operation of refrigerators and small air-conditioning units, and to reduce the number of moving parts in such equipment, it is common to replace the expansion turbine with an expansion valve, capillary tube, orifice, or other partial obstruction so that the refrigerant undergoes a Joule-Thomson expansion, rather than an isentropic expansion, between points 1 and 2 in the cycle. Such a cycle is shown in Fig. 5.2-4a. The properties and path for this cycle are given in the table below. This cycle is referred to as the vapor-compression refrigeration cycle, and is more commonly used than the Rankine refrigeration cycle. Note that the net process in both the Rankine and vapor-compression refrigeration cycles is to use work to remove heat from a low-temperature source and exhaust it to a reservoir at a higher temperature.

166 Chapter 5: Liquefaction, Power Cycles, and Explosions The compressor in this cycle is considered to operate isentropically, the expansion valve results in a Joule-Thomson (isenthalpic) expansion, and the condenser and evaporator operate at constant pressure. Here we consider the case in which the condenser receives a two-phase mixture and emits a saturated liquid at the same pressure; consequently, the condenser also operates at constant temperature. The properties and path for this cycle are given in the table.

Point 1 2

3 4

(State) Path to Next Point

T

(Saturated liquid) Isenthalpic (Vapor-liquid mixture) Isobaric heating [also isothermal in this case] (Saturated vapor) Isentropic (Superheated vapor) Isobaric

ˆ H

T1

P1

Sˆ1

ˆ1 H

T2

P2

Sˆ2

ˆ2 = H ˆ1 H

T3 = T2

P3 = P2

Sˆ3

ˆ3 H

T4

P4

Sˆ4 = Sˆ3

ˆ4 H

P1 = P4

Sˆ1

ˆ1 H

↓

1

Sˆ

P

↓

↓

T1

Energy Flow

↓

↓

0 Q˙ B

˙P W

− Q˙ C

QB Low-pressure vapor 3

2

Evaporator (or boiler)

Compressor

WP

Low-pressure vapor-liquid mixture Expansion valve 1

High-pressure 4 vapor

Condenser High-pressure liquid QC (a)

4 1

1 T

4

P

2

2

3 ^

3 ^

S

H (b)

Figure 5.2-4 (a) The vapor-compression refrigeration cycle. (b) The ˆ diagrams. vapor-compression refrigeration cycle on T -Sˆ and P -H

5.2 Power Generation and Refrigeration Cycles 167

ˆ diagrams of the path of this cycle are shown in Fig. 5.2-4b. Schematic T -Sˆ and P -H ˙ . The The mass flow is the same in all parts of the cycle, and we designate this by M steady-state energy balance around the evaporator and its contents yields ˆ 2 − M˙ H ˆ 3 + Q˙ B 0 = M˙ H

ˆ3 − H ˆ 2) or Q˙ B = M˙ (H

The energy balance on the compressor is

ˆ4 + W ˆ 3 − M˙ H ˙P 0 = M˙ H

ˆ 3) ˙ P = M˙ (H ˆ4 − H or W

and the energy balance on the condenser is

ˆ 4 − M˙ H ˆ 1 + Q˙ C 0 = M˙ H

ˆ1 − H ˆ 4) or Q˙ C = M˙ (H

The coefficient of performance (C.O.P.) for the vapor-compression refrigeration cycle is

C.O.P. = =

Heat removed from low-temperature region Net work required for that heat removal

ˆ2 ˆ3 − H Q˙ B H = ˙P ˆ4 − H ˆ3 W H

Illustration 5.2-2 Calculating the Coefficient of Performance of an Automobile Air Conditioner An automobile air conditioner uses a vapor-compression refrigeration cycle with the environmentally friendly refrigerant HFC-134a as the working fluid. The following data are available for this cycle. Point 1 2 3 4

Fluid State Saturated liquid Vapor-liquid mixture Saturated vapor Superheated vapor

Temperature 55◦ C 5◦ C

a. Supply the missing temperatures and the pressures in the table. b. Evaluate the coefficient of performance for the refrigeration cycle described in this problem.

Solution The keys to being able to solve this problem are (1) to identify the paths between the various locations and (2) to be able to use the thermodynamic properties chart of Fig. 3.3-4 for HFC-134a to obtain the properties at each of the locations. The following figure shows the path followed in the cycle, and the table provides the thermodynamic properties for each stage of the cycle as read from the figure.

168 Chapter 5: Liquefaction, Power Cycles, and Explosions With this information, the C.O.P. is C.O.P. =

ˆ3 − H ˆ2 402 − 280 H Q˙ B = = = 4.07 ˙ ˆ4 − H ˆ3 432 − 402 W H



Since an equation of state specific to HFC-134a is not available, the Peng-Robinson equation (See Chapter 6) will be used. To proceed, we need to know the equilibrium pressures for HFC-134a as predicted by the Peng-Robinson equation of state. To calculate these use Analysis > Pure, and choose PL as the property we find at 5◦ C the vapor pressure is 3.484 bar = 0.3484 MPa and at 55◦ C the vapor pressure is 14.96 bar = 1.496 MPa. Using these results in the simulation, the following is obtained Point 1 2 3 4

Condenser Valve Boiler Pump

T (◦ C)

P (MPa)

55 5 5 58.9

1.496 0.348 0.348 1.496

Q (Watts)

−152.84

W (Watts)

+122.31 30.53

(Results for Q and W based on 1kg/sec flow.) The coefficient of performance is C.O.P =

122.31 = 4.01 30.53

in good agreement with the hand calculation using the results of using the thermodynamic properties chart for HFC-134a. R Also an Aspen Plus simulation with the Peng-Robinsonequation of state is available Wiley website for this book in the folder Aspen>Illustration 5.2-2.

5.2 Power Generation and Refrigeration Cycles 169 Point

State

Path

1

Saturated liquid

2

Vapor-liquid mixture

3

Saturated vapor

4

Superheated vapor

1

Saturated liquid

T (◦ C)

P (MPa)

ˆ (kJ/kg) H

Sˆ (kJ/kg K)

55

1.493

280

1.262

5

0.350

↓ 280

1.30

5

↑ 0.350

402

1.725

60

1.493

432

↓ 1.725

55

↑ 1.493

280

1.262

Isenthalp to 5◦ C

P constant

Isentrope to 1493 kPa

P constant

In this table the numbers in boldface are properties known from the problem statement or from an adjacent step in the cycle (i.e., a process that is at constant pressure as in steps 2 → 3 and 4 → 1, at constant enthalpy as in step 1 → 2, or at constant entropy as in step 3 → 4). The numbers in italics are found from Fig. 3.3-4 with the state of the refrigerant fixed from the values of the known variables.

Stirling cycle

A large number of other cycles and variations to the standard cycles considered above have been proposed. We consider only a few additional cycles here. The Stirling cycle, shown in Fig. 5.2-5, operates with a vapor-phase working fluid, rather than a two-phase mixture as in the cycles considered above. In Stirling cycle the compressor and turbine, which are on the same shaft, are cooled and heated, respectively, in order to operate isothermally. The heat exchanger operates isochorically (that is, at constant volume). In many implementations, the compressor is replaced by a simple gas-phase pump, and the turbine by any work-receiving device such as a flywheel. The Stirling cycle has the advantages that it is reasonably simple to construct, and any form of heat source can be used, including solar energy. The P -Vˆ and T -Sˆ traces of this cycle are shown in Fig. 5.2-6. The properties and path are shown in the table. Point

Path to Next Point

1 Vˆ = constant

2 T = constant

3 Vˆ = constant

4 T = constant

1

T

Vˆ

P

ˆ H

T1

P1

Vˆ1

T2

P2

Vˆ2 = Vˆ1

ˆ2 H

T3 = T2

P3

Vˆ3

ˆ3 H

T4

P4

Vˆ4 = Vˆ3

ˆ4 H

T1 = T4

P1

Vˆ1

ˆ1 H

↓

↓

↓

↓

ˆ1 H

Energy Flow Q˙ 12 ˙ 23 Q˙ T , − W

−Q˙ 34 = Q˙ 12 ˙ 41 −Q˙ C , − W

3 4 QC

1 Isothermal compressor

Regenerator or heat exchanger

2 Isothermal turbine

QT

Wout

Figure 5.2-5 The Stirling cycle.

170 Chapter 5: Liquefaction, Power Cycles, and Explosions 2 2

3

T = constant ^

V = constant P

3

1

T

^

V = constant

T = constant 4 4

1 ^

^

V

S

Figure 5.2-6 The Stirling cycle on P -Vˆ and T -Sˆ diagrams.

From an energy balance on stream 3 in the regenerator, we have (at steady state)

ˆ 3 − M˙ H ˆ 4 + Q˙ 34 0 = M˙ H and on stream 1

ˆ 2 + Q˙ 12 ˆ 1 − M˙ H 0 = M˙ H Adding these two equations and noting that Q˙ 12 + Q˙ 34 = 0, since there is no net heat flow into or out of the heat exchanger, gives

ˆ 4 ) + (H ˆ 2) = 0 ˆ3 − H ˆ1 − H (H An energy balance on the compressor (4 → 1) gives

˙ 41 + Q˙ C = 0 ˆ4 − H ˆ 1) + W M˙ (H and on the turbine (2 → 3) gives

ˆ2 − H ˆ 3) + W ˙ 23 + Q˙ T = 0 M˙ (H Adding these last two equations together gives

ˆ2 − H ˙ 41 + W ˙ 23 ) + Q˙ C + Q˙ T = 0 ˆ4 − H ˆ 1 ) + M˙ (H ˆ 3 ) + (W M˙ (H From the balance equation on the heat exchanger, the first two terms cancel. Also, ˙ 23 is equal to W ˙ 41 + W ˙ out . Thus we obtain W

˙ out + Q˙ C + Q˙ T = 0 W Therefore, the efficiency of this Stirling cycle, η , is

η= Ericsson cycle

˙ out ˙ 23 − W ˙ 41 ˙ out −W −W W = = ˙ out + Q˙ C Q˙ T W Q˙ T

The Ericsson cycle is similar to the Stirling cycle except that each of the streams in the regenerator is at a constant (but different) pressure rather than at a constant volume. The P -Vˆ and T -Sˆ traces are shown in Fig. 5.2-7.

5.2 Power Generation and Refrigeration Cycles 171

2 1

2

3

P = constant

P

T

P = constant

T = constant

T= constant

4

1 3

4 ^

^

S

V

Figure 5.2-7 The Ericsson cycle on P -Vˆ and T -Sˆ diagrams.

The properties and path are given in the following table. Point

Path to Next Point

1

T

P

T1

P1

ˆ1 H

T2

P2 = P1

ˆ2 H

T3 = T2

P3

ˆ3 H

T4

P4 = P3

ˆ4 H

P1

ˆ1 H

↓

P = constant

2 T = constant

3

↓

P = constant

4 T = constant

1

ˆ H

↓

T1 = T4

↓

Energy Flow Q˙ 12 ˙ 23 Q˙ T , − W

−Q˙ 34 = Q˙ 12 ˙ 41 −Q˙ C , − W

The balance equations are exactly as for the Stirling cycle, so that the efficiency is

η=

˙ out ˙ 23 − W ˙ 41 −W −W −Wout = = ˙ ˙ ˙ QT QT QC + Wout

Indeed, the only computational difference between the Stirling and Ericsson cycles is that the thermodynamic properties at the various points of the cycle are slightly different because of the difference between the constant-pressure and constant-volume paths. Illustration 5.2-3 Calculating the Efficiency of an Ericsson Cycle An Ericsson cycle with air as the working fluid is operating between a low temperature of 70◦ C and a high temperature of 450◦ C, a low pressure of 2 bar, and a pressure compression ratio of 8 (so that the high pressure is 16 bar). Assume that at these conditions air can be considered to be an ideal gas with CP∗ constant. Compute the properties at each point in each of these cycles and the cycle efficiencies.

Solution We first compute the energy flows in each step of the process. In this analysis the process is at steady state so that all the time derivatives are equal to zero and the mass flow rate M˙ (actually we will use the molar flow rate N˙ ) is constant throughout the process. 1 → 2 Through the regenerative heat exchanger Energy balance: 0 = N˙ H 1 − N˙ H 2 + Q˙ 12

or

Q˙ 12 = (H 2 − H 1 ) = CP∗ (T2 − T1 ) N

172 Chapter 5: Liquefaction, Power Cycles, and Explosions 2 → 3 Through the isothermal turbine Energy balance: ˙ 23 0 = N˙ H 2 − N˙ H 3 + Q˙ T + W

Entropy balance: Q˙ T 0 = N˙ S 2 − N˙ S 3 + T

or   Q˙ T T3 P3 P3 = T2 (S 3 − S 2 ) = T2 CP∗ ln = −RT2 ln − R ln ≡ −RT2 ln KT T2 P2 P2 N˙

where KT = P3 /P2 is the compression ratio of the turbine. Consequently, ˙ 23 W Q˙ T = H3 − H2 − = CP∗ (T3 − T2 ) + RT2 ln KT = RT2 ln KT N˙ N˙

3 → 4 Through the regenerative heat exchanger Energy balance: 0 = N˙ H 3 − N˙ H 4 + Q˙ 34

Q˙ 34 = H 4 − H 3 = CP∗ (T4 − T3 ) N˙

or

However, Q˙ 12 = −Q˙ 34

which implies (since CP∗ is constant) that T2 − T1 = T3 − T4 . Also, since both the compressor and turbine are isothermal (so that T2 = T3 and T4 = T1 ), the temperature decrease of stream 1 exactly equals the temperature increase of stream 3 for this case of an ideal gas of constant heat capacity. 4 → 1 Isothermal compressor As for the isothermal turbine, we find Q˙ C P1 = RT1 ln P4 N˙

and

˙ 41 W P1 = RT1 ln P4 N˙

The efficiency of this cycle is  η=

˙ out −W Q˙ T

=

˙ 23 + W ˙ 41 ) −(W Q˙ T

P3 P1 + RT1 ln P2 P4 P3 RT2 ln P2



RT2 ln =

Since each of the streams in the heat exchanger operates at constant pressure, it follows that P3 = P4 and P2 = P1 . Therefore, 

P3 P1 + RT1 ln P2 P3 P3 RT2 ln P2



RT2 ln η=

=

T2 − T1 T2

5.2 Power Generation and Refrigeration Cycles 173 As T2 is the high temperature of the cycle and T1 is the low temperature, this efficiency is exactly equal to the Carnot efficiency. (Indeed, it can be shown that for any cycle, if the two heat transfers to the surroundings occur at different but constant temperatures, as is the case here with an isothermal compressor and an isothermal turbine, the efficiency of the cycle will be that of a Carnot cycle.) Therefore, the efficiency of the Ericsson cycle considered here is η=

(450 + 273.15) − (70 + 273.15) = 0.525 (450 + 273.15)

Note that the efficiency is dependent on the temperatures of the various parts of the cycle, but not on the pressures or the compression ratio. However, the pressure ratio does affect the heat and work flows in each part of the cycle.

Illustration 5.2-4 Calculating the Efficiency of a Stirling Cycle Compute the efficiency of a Stirling cycle operating under the same conditions as the Ericsson cycle above.

Solution The terms for each of the steps in the Stirling cycle are the same as for the Ericsson cycle; however, the properties are slightly different since in this case the heat exchanger (regenerator) operates such that each of its streams is at constant volume, not constant pressure. Therefore, rather than P2 = P1 as in the Ericsson cycle, here we have (by the ideal gas law) P2 = P1 T2 /T1 . Similarly, P4 = P3 T4 /T3 . Therefore, P3 P2 T1 /T2 P3 P1 RT2 ln + RT1 ln + RT1 ln P P P2 P4 2 3 T4 /T3 η= = = ˙ P P 3 3 QT RT2 ln RT2 ln P2 P2 P3 P2 + RT1 ln RT2 ln T2 − T1 P2 P3 = = T2 P3 RT2 ln P2 Since T2 = T3 and T1 = T4 , both the compressor and turbine operate isothermally. Therefore, ˙ out −W

RT2 ln

the efficiency of the Stirling cycle is also equal to that of the Carnot cycle. For the operating conditions here, η = 0.525.

A common method of producing mechanical work (usually for electrical power generation) is to use a gas turbine and the Brayton or gas turbine power cycles. This open cycle consists of a compressor (on the same shaft as the turbine), a combustor in which fuel is added and ignited to heat the gas, and a turbine that extracts work from the high-temperature, high-pressure gas, which is then exhausted to the atmosphere. The open-cycle gas turbine is used, for example, in airplane jet engines and in some trucks. This cycle is shown in Fig. 5.2-8. In this cycle the compressor and turbine are assumed to operate isentropically, and the gas flow through the combustor is at constant pressure. Note that the temperatures of streams 1 and 4 may not be the same, though the pressures are both atmospheric. The closed gas turbine cycle is shown in Fig. 5.2-9. Closed cycles are typically used in nuclear power plants, where the heating fluid in the high-temperature heat exchanger

174 Chapter 5: Liquefaction, Power Cycles, and Explosions 2

Combustion region

3

Compressor

W

Turbine 4

1 Air

Exhaust gas

3 P

2

3

T 4

1

2

4

1

^

^

V

S

Figure 5.2-8 The open Brayton power cycle.

is the reactor cooling fluid, which for safety and environmental reasons must be contained within the plant, and the low-temperature coolant is river water. The properties and path for this cycle are given below. Point

Path to Next Point

1

T

P P1

S1

H1

T2

P2

S2 = S1

H2

T3

P3 = P2

S3

H3

T4

P4

S4 = S3

H4

T1

P1 = P4

S1

H1

Isobaric 3

↓

↓

Isentropic 4 Isobaric 1

H

T1

Isentropic 2

S

↓

↓

Energy Flow ˙ 12 W Q˙ 23 ˙ 34 W Q˙ 41

High-temperature heat exchanger

Closed Brayton cycle

3

2

Compressor

Turbine

1

WT

4

Low-temperature heat exchanger

Figure 5.2-9 The closed Brayton power generation cycle.

5.2 Power Generation and Refrigeration Cycles 175 Illustration 5.2-5 Calculating the Efficiency of a Brayton Cycle Assuming the working fluid in a Brayton power cycle is an ideal gas of constant heat capacity, show that the efficiency of the cycle is η=

˙ 34 − W ˙ 12 ) −(W Q˙ 23

and obtain an explicit expression for the efficiency in terms of the cycle temperatures.

Solution For path 1 → 2, through the compressor, we have from the energy balance that ˙ 12 W = H 2 − H 1 = CP∗ (T2 − T1 ) N˙

Also from S 2 = S 1 , we have 0 = CP∗ ln

T2 P2 − R ln T1 P1

or

T2 = T1



P2 P1

R/CP∗

Similarly, across the gas turbine, 3 → 4, we have ˙ 34 W = H 4 − H 3 = CP∗ (T4 − T3 ) N˙

and T4 = T3



P4 P3

R/CP∗

But P3 = P2 and P4 = P1 , so that T4 = T3



P4 P3



R/CP∗ =

P1 P2

R/CP∗ =

T1 T2

Across the high-temperature heat exchanger, 2 → 3, we have 0 = N˙ H 2 − N˙ H 3 + Q˙ 23

or

Q˙ 23 = H 3 − H 2 = CP∗ (T3 − T2 ) N˙

Similarly, across the low-temperature heat exchanger, 4 → 1, Q˙ 41 = H 1 − H 4 = CP∗ (T1 − T4 ) N˙

Therefore, η=−

˙ 34 + W ˙ 12 ) (W −CP∗ (T4 − T3 ) − CP∗ (T2 − T1 ) = ˙ CP∗ (T3 − T2 ) Q23

(T3 − T4 ) + (T1 − T2 ) (T3 − T2 ) − (T4 − T1 ) (T4 − T1 ) =1− = T3 − T2 (T3 − T2 ) (T3 − T2 )   T4 −1 T1 T1   =1− T2 T3 −1 T2 =

176 Chapter 5: Liquefaction, Power Cycles, and Explosions However, from T4 /T3 = T1 /T2 , we have that T4 /T1 = T3 /T2 , so that η =1−

T1 =1− T2



P1 P2

R/CP∗

∗ −R/CP

= 1 − KT

where KT = P2 /P1 is the compression ratio.

The most ubiquitous implementation of the open Brayton cycle is the jet engine on airplanes. There air is taken in through the compressor at the front of the engine, mixed with fuel and ignited in the combustion region, and the resulting high temperature, high pressure gas is released through the exit turbine that is on the same shaft as the compressor. Here the turbine is designed to produce only the power needed to drive the inlet compressor, while most of the energy in the process goes to produce a high velocity exhaust gas that provides the thrust to drive the aircraft. The final cyclic device we consider is the heat pump, that is, a device used to pump heat from a low-temperature source to a high-temperature sink by expending work. Refrigerators and air conditioners are examples of heat pumps. Heat pumps (with appropriate valving) are now being installed in residential housing so that they can be used for both winter heating (by pumping heat to the house from its surroundings) and summer cooling (by pumping heat from the house to the surroundings). The surroundings may be either the atmosphere, the water in a lake, or, with use of underground coils, the earth. The term pumping is used since in both the winter and summer modes, heat is being taken from a region of low temperature and exhausted to a region of higher temperature. In principle, any power generation or refrigeration cycle that can be made to operate in reverse can serve as a heat pump. A heat pump that uses the vapor-compression refrigeration cycle and two connected (so that they operate together) three-way valves is schematically shown in Fig. 5.2-10. In this way the indoor coil is the condenser during the winter months and the evaporator or boiler during the summer. Similarly, the outdoor coil is the evaporator (boiler) during the winter and the condenser during the summer. The vapor compression refrigeration ˆ plots are given cycle was discussed earlier in this section, and the paths on T-Sˆ and P-H in Fig. 5.2-4b.

Heating cycle

Cooling cycle

Heat pump

4

4 3

Compressor

Outdoor coil

Compressor Indoor coil

Outdoor coil

3

2

Joule-Thomson (isenthalpic) expansion valve or capillary tube

1

1

Indoor coil

Joule-Thomson (isenthalpic) expansion valve or capillary tube

Figure 5.2-10 Heat pump in heating (winter) and cooling (summer) cycles.

2

5.2 Power Generation and Refrigeration Cycles 177 Illustration 5.2-6 Calculating the Coefficients of Performance of a Heat Pump Consider a residential heat pump that uses lake water as a heat source in the winter and as a heat sink in the summer. The house is to be maintained at a winter temperature of 18◦ C and a summer temperature of 25◦ C. To do this efficiently, it is found that the indoor coil temperature should be at 50◦ C in the winter and 5◦ C in the summer. The outdoor coil temperature can be assumed to be 5◦ C during the winter months and 35◦ C during the summer. a. Compute the coefficient of performance of a reverse Carnot cycle (Carnot cycle heat pump) in the winter and summer if it is operating between the temperatures listed above. b. Instead of the reverse Carnot cycle, a vapor-compression cycle will be used for the heat pump with HFC-134a as the working fluid. Compute the winter and summer coefficients of performance for this heat pump. Assume that the only pressure changes in the cycle occur across the compressor and the expansion valve, and that the only heat transfer to and from the refrigerant occurs in the indoor and outdoor heat transfer coils.

Point

Heating Temperature

Cooling Temperature

50◦ C

35◦ C

5◦ C

5◦ C

1 2 3 4

Fluid State Saturated liquid Vapor-liquid mixture Saturated vapor Superheated vapor

Solution a. Carnot cycle coefficients of performance: Q1

Q2

T1

T2 W T1 < T2

The energy balance on the cycle is ˙ 0 = Q˙ 1 + Q˙ 2 + W

The entropy balance is 0=

Q˙ 1 Q˙ 2 + T1 T2

since Sgen = 0 for the Carnot cycle. Therefore, −T2 ˙ Q˙ 2 = Q1 T1

 and

˙ = Q˙ 2 + Q˙ 1 = Q˙ 1 −W

T1 − T2 T1



Winter Operation (Heating Mode) C.O.P. (winter) =

−Q˙ 1 ˙ W

=

T1 T2 − T1

=

(5 + 273.15) = 6.18 45

178 Chapter 5: Liquefaction, Power Cycles, and Explosions Summer Operation (Cooling Mode): C.O.P. (summer) =

Q˙ 1 (5 + 273.15) T1 = = 9.27 = ˙ T2 − T1 30 W

b. Vapor compression cycle coefficients of performance: The following procedure is used to locate the cycles on the pressure-enthalpy diagram of Fig. 3.3-4, a portion of which is repeated here. First, point 1 is identified as corresponding to the saturated liquid at the condenser temperature. (In this illustration the condenser is the indoor coil at 50◦ C in the heating mode, and the outdoor coil at 35◦ C in the cooling mode.) Since the path from point 1 to 2 is an isenthalpic (Joule-Thomson) expansion, a downward vertical line is drawn to the evaporator temperature (corresponding to the outdoor coil temperature of 5◦ C in the heating mode, or in the cooling mode to the indoor coil temperature, which also happens to be 5◦ C in this illustration). This temperature and the enthalpy of point 1 fix the location of point 2. Next, point 3 is found by drawing a horizontal line to the saturated vapor at the evaporator pressure. The path from point 2 to 3 describes the vaporization of the vapor-liquid mixture that resulted from the Joule-Thomson expansion. The evaporator is the outdoor coil in the heating mode and the indoor coil in the cooling mode. The path from point 3 to 4 is through the compressor, which is assumed to be isentropic, and so corresponds to a line of constant entropy on the pressure-enthalpy diagram. Point 4 is located on this line at its intersection with the horizontal line corresponding to the pressure of the evaporator. The fluid at point 4 is a superheated vapor. The path from 4 to 1 is a horizontal line of constant pressure terminating at the saturated liquid at the temperature and pressure of the condenser. The tables below give the values of those properties (read from the pressure-enthalpy diagram) at each point in the cycle needed for the calculation of the coefficient of performance in the heating and cooling modes of the heat pump. In these tables the properties known at each point from the problem statement or from an adjacent point appear in boldface, and the properties found from the chart at each point appear in italics. Heating (winter operation)



kJ kg

Point

T (◦ C)

P (MPa)

1

50

1.319

Saturated liquid

2

5

Vapor-liquid mixture

3

5

0.3499 ↓ 0.3499

271.9 ↓ 271.9

Saturated vapor

401.7

4

58

1.319

Superheated vapor

432

Condition

ˆ H



Therefore, −Q˙ 23

ˆ 2 = (407.1 − 271.9) kJ = 129.8 kJ ˆ3 − H =H ˙ kg kg M ˙ W34 ˆ 3 = (432 − 401.7) kJ = 30.3 kJ ˆ4 − H =H ˙ kg kg M

and C.O.P. =

129.8 = 4.28 30.3

 Sˆ

kJ kg K

1.7252 ↓ 1.7252



5.2 Power Generation and Refrigeration Cycles 179

180 Chapter 5: Liquefaction, Power Cycles, and Explosions Cooling (summer operation)



kJ kg

Point

T (◦ C)

P (MPa)

1

35

0.8879

Saturated liquid

2

5

Vapor-liquid mixture

3

5

0.3499 ↓ 0.3499

249.2 ↓ 249.2

Saturated vapor

401.7

4

45

0.8879

Superheated vapor

436

ˆ H

Condition



 Sˆ

kJ kg K



1.7252 ↓ 1.7252

Consequently, here Q˙ 23 ˆ 2 = (401.7 − 249.2) kJ = 152.5 kJ ˆ3 − H =H ˙ kg kg M ˙ W34 ˆ4 − H ˆ 3 = (436 − 401.7) kJ = 34.3 kJ =H ˙ kg kg M

and C.O.P. =

152.5 = 4.45 34.3

Note that in both the heating and cooling modes, the heat pump in this illustration has a lower coefficient of performance (and therefore lower efficiency) than a reverse Carnot cycle operating between the same temperatures. Finally, it should be mentioned that the thermodynamic properties listed above were obtained from a detailed thermodynamic properties table for HFC-134a, akin to the steam tables for water in Appendix A.III; values cannot be obtained from Fig. 3.3-4 to this level of accuracy. R As in Illustration 5.2-2 Aspen Plus with the Peng-Robinson equation of state will be used. To use the folder Aspen Illustration>Chapter 5>5.2-6 on the Wiley website for this book it is necessary to compute the vapor pressures of HFC-134a as predicted by equation of state at 5◦ C, 35◦ C and 50◦ C. These are computed using Properties > Pure option in R which predicts the following Aspen Plus T (◦ C)

5 35 50 Winter Point 1 2 3 4

P (MPa)

0.3484 0.8864 1.3205

Condenser Valve Boiler Compressor

T (◦ C)

P (MPa)

50 5 5 56.4

1.3205 0.3484 0.3484 1.3205

Q (Watts)

−158276

W (Watts)

130762 2796.4

130762 C.O.P = = 4.676 27964

Summer Point 1 2 3 4

Condenser Valve Boiler Compressor

T (◦ C)

P (MPa)

35 5 5 37.3

0.8864 0.3484 0.3484 0.8864

Q (Watts)

−174316

W (Watts)

154692 19625

5.3 Thermodynamic Efficiencies 181

5.3 THERMODYNAMIC EFFICIENCIES We have so far considered the operations of adiabatic turbines (work-producing devices) and compressors to be completely efficient, that is that there are no energy losses in the process. However, real turbines and compressors are not 100% efficient, and two measures of efficiency are commonly used. The first efficiency measure is the isentropic efficiency that compares the work out of a turbine or the work required in a compressor with that if the process were isentropic. the isentropic work obtained from a turbine expanding a fluid from T1 and P1 to an outlet pressure P2 is obtained by first computing T2,S where the subscript S is used to indicate this temperature is found along a path of constant entropy. The isentropic work, WS , is then ˆ 1 , P1 ) ˆ 2,S , P2 ) − H(T WS = H(T (5.3-1) The actual work obtained is

ˆ 1 , P1 ) ˆ 2 , P2 ) − H(T WA = H(T

(5.3-2)

where T2 is the actual measured temperature. The isentropic efficiency, η , for a turbine is the ˆ 1 , P1 ) ˆ 2 , P2 ) − H(T H(T actual work η= = (5.3-3) ˆ 2,S , P2 ) − H(T ˆ 1 , P1 ) isentropic work H(T So if the isentropic efficiency of a turbine is given, one proceeds as follows to determine the work obtained and fluid exit temperature: a) Determine the exit temperature of the fluid assuming the process is isentropic b) Compute the ideal work obtained using Eq. (5.3-1) c) Compute the actual work obtained using the specified isentropic efficiency from

ˆ 1 , P1 )) ˆ 2,S , P2 ) − H(T WA = ηWS = η(H(T

(5.3-4)

d) the final enthalpy is computed from

ˆ 2,S , P2 ) = W + H(T ˆ 1 , P1 )) H(T

(5.3-5)

and e) the final temperature is computed using the known pressure and the final enthalpy. Illustration 5.3-1 Effect of Turbine Efficiency Steam at 500◦ C and 10 bar is expanded in an adiabatic turbine 1 bar. (a) Compute the final temperature of the steam and work obtained if the process is 100% efficient. (b) Compute the final temperature of the steam and work obtained if the process has an isentropic efficiency of 85%.

Solution a) From the steam tables, the thermodynamic properties of the initial state are ˆ H(T = 500◦ C, 1.0 MPa) = 3478.5 kJ/kg and ˆ = 500◦ C, 1.0 MPa) = 7.7622 kJ/kg K S(T

182 Chapter 5: Liquefaction, Power Cycles, and Explosions The final fluid conditions are P = 0.1 MPa and Sˆ = 7.7622 kJ/kg K

Interpolating from the data in the steam tables, we have T = 183.7◦ C and H = 2843.0 kJ/kg therefore, WS = 2843.0 − 3478.5 kJ/kg = −635.5 kJ/kg

of work produced by the isentropic turbine b) For the non-isentropic turbine the work obtained is WA = 0.85 WS = −540.2 kJ/kg

and the final enthalpy = 3478.5 − 540.2 = 2938.3 kJ/kg. At 0.1 MPa, this enthalpy corresponds to a temperature of 238◦ C.

Comment Note that less work is obtained from the non-isentropic turbine. Since energy is conserved, a greater portion of the energy in the initial steam now appears in the exiting steam in the form of a higher temperature.

Isentropic work WS and the actual work WA required in a compressor are also computed from Eqs. 5.3-1 and 2 respectively. However, the compressor efficiency is defined to be

η=

ˆ 1 , P1 ) ˆ 1,S , P2 ) − H(T isentropic work H(T = ˆ 2 , P2 ) − H(T ˆ 1 , P1 ) actual work H(T

(5.3-6)

Illustration 5.3-2 Analysis of compressor operation An ideal gas with a constant pressure heat capacity of 30 kJ/kg K is to be compressed from 1 bar and 25◦ C to 10 bar. (a) Compute the final temperature of the gas and the work required if the compressor operates isentropically (b) Compute the final temperature of the gas and the work required if the compressor has an isentropic efficiency of 75%.

Solution (a) For an isentropic process, from Eq. 4.4-3 T2 CP ln = R ln T1 ∗



P2 P1



or  T2,S = T1

P2 P1

R/CP∗

5.3 Thermodynamic Efficiencies 183 so that T2,S = 564.38 K and W

=

CP∗ (T2,S − 298.15)

=

30(564.38 − 298.15) = 7986.8 kJ/kg

(b) Here WA = 7986.8/0.75 = 106490.0 kJ/kg. Therefore, T2 = 653.1 K

10649.0 + 298.15 = 30

Comment In this case more work is required to compress the gas than with a isentropic compressor, and this extra work requirement results in a higher gas exit temperature.

Another measure of thermodynamic efficiency is the polytropic efficiency. A polytropic process is one in which

P Vˆ n = constant

or

P1 Vˆ1n = P2 Vˆ2n

(5.3-7)

where n is the polytropic index. If n = 0 the process is at constant pressure (isobaric), if n = 1 the process is isothermal (for an ideal gas), if n = CP /CV the process is isentropic, while if n = ∞ the process is at constant volume (isochoric). The ideal work obtained from, or done on, a fluid can be calculated for a closed system from  state2 Vˆ dP (5.3-8) state1

while the actual work required or obtained between the two states is

ˆ state2 − H ˆ state1 H The polytropic efficiency is defined as

ηP =

ˆ state2 − H ˆ state1 actual change in energy H =  state2 ideal change in energy Vˆ dP

(5.3-9)

state1

In some cases, a compressor is designed to operate approximately isothermally rather than adiabatically. This is done using fins, internally cooled turbine blades or other methods. One reason for doing this is that on compression, especially if there is a large pressure change, the temperature of the gas may get sufficiently high as to be dangerous. For example, a combustible gas may self-ignite if there is oxygen infiltration and the gas temperature is above the auto-ignition temperature (frequently abbreviated as AIT). In this case, the ideal operation of the compressor is a reversible isothermal process operating at the same inlet temperature and pressure and the same exit temperature and pressure, and the efficiency is defined as

ηisothermal compressor =

Wisothermal Wactual

(5.3-10)

So in this case the inlet and exit stream properties are identical, but more work will be required and more heat released in the real isothermal compressor than in the ideal isothermal compressor.

184 Chapter 5: Liquefaction, Power Cycles, and Explosions Illustration 5.3-3 Operation of an isothermal compressor Methane at 260 K is to be isothermally compressed from 0.1 MPa to 1.0 MPa. a) What is the minimum work required, and how much heat must be removed to keep the compression process isothermal? b) If the compressor is only 75% efficient what is the work required, and how much heat must be removed to keep the compression process isothermal?

Solution a) The steady-state energy and entropy balances for this case are 0

=

ˆ ˆ = 260 K, P = 1 MPa) + W + Q H(T = 260 K, P = 0.1 MPa) − H(T

=

882 kJ/kg − 897 kJ/kg + W + Q

or W = 15 kJ/kg − Q and 0

= =

ˆ = 260 K, P = 1 MPa) + Q ˆ = 260 K, P = 0.1 MPa) − S(T S(T T Q (7.1 − 5.9) kJ/kg K + 260 K

so that Q = −312 kJ/kg and then W = 15 + 312 = 327 kJ/kg The Aspen process simulator does not provide the option of an isothermal compressor,

so there are no results to compare with. b) Since the turbine is only 75% efficient, the actual work done by the compressor is 436 kJ/kg. Since the temperature and pressure of the inlet and exit streams are fixed, the enthalpies of the streams remain the same as in part a. So W = 436 kJ/kg = 15 kJ/kg − Q

or

Q = 15 − 436 = −421 kJ/kg

Therefore, the extra work that must be added as a result of the turbine efficiency turns into heat that must be removed.

A less commonly used measure of thermodynamic efficiency is for nozzles. The objective of using a nozzle is to produce a fluid of high velocity and low pressure from a low-velocity, high-pressure inlet stream. This is what is done, for example, inside a steam turbine where a low-velocity, high-pressure stream passes through a nozzle to produce a high-velocity, low-pressure stream that impinges upon the blades to drive the turbine. For nozzles, the thermodynamic efficiency is defined as the ratio of the kinetic energy (or simply the square of the velocity) of the exit stream to that in a reversible adiabatic (that is, isentropic) nozzle at the same exit pressure. Thus,

ν2 ηnozzle = 2 actual νisentropic

(5.3-11)

In the case of the actual nozzle, the velocity of the exit stream will be lower, and its temperature higher than for an isentropic nozzle. In order do accurate thermodynamic calculations in all of the cases above, one has to know the efficiency of the device, be it a turbine, compressor or nozzle. This may be impossible to predict knowing only the design of the device. Therefore, the usual

5.4 The Thermodynamics of Mechanical Explosions 185 procedure is to do one or several measurements on a device with a fluid whose thermodynamic properties are precisely known, and from those data compute the efficiency of the device. A common example of this is when an electrical power generation company orders a turbine, one of the specifications to the manufacturer is the turbine efficiency. After installation, the efficiency is then certified using steam as the working fluid. Failure to meet the specification by even tenths of a percent will result in additional fuel needed for each unit of electrical power generated, and result in a penalty to the turbine manufacturer of millions of dollars. It is for this reason that extremely accurate thermodynamic properties of steam are needed, and there is an international organization, the International Association for the Properties of Water and Steam (IAPWS), that develops critically-evaluated tables of data that are the standard for such certifications. IAPWS-95 is the currently accepted steam tables (also available in equation form for use in computers), and it is one of the thermodynamic model choices in the ASPEN process simulator.

5.4 THE THERMODYNAMICS OF MECHANICAL EXPLOSIONS In this section we are interested in predicting the uncontrolled energy release of an explosion that occurs without a chemical reaction. That is, we are interested in energy released from an explosion that results from the bursting of an overpressurized tank, or the rapid depressurization of a hot liquid that leads to its partially boiling (flashing) to a vapor-liquid mixture. We do not consider chemical explosions, for example, the detonation of TNT or a natural gas explosion, both of which involve multiple chemical reactions and require the estimate of properties of mixtures that we have not yet considered. Chemical explosions are considered in Chapter 13. First we consider the impact of an explosion involving a high-pressure gas, and then we consider the explosion of a flashing liquid due to rapid depressurization. A primary characteristic of an explosion is that it is rapid, indeed so rapid that there is insufficient time for the transfer of heat or mass to or from the exploding material. Instead, the exploding matter undergoes a rapid expansion, pushing away the surrounding air. This is schematically illustrated in Fig. 5.4-1.

Expanding shock wave

Initial system boundary

Uniform expansion

Figure 5.4-1 Explosion—a rapid, uniform expansion. If the shock wave moves at less than the speed of sound, the event is classified as a deflagration. If the shock wave travels at the speed of sound, the event is an explosion.

186 Chapter 5: Liquefaction, Power Cycles, and Explosions During a detonation, a shock wave is produced; outside this shock wave the pressure is ambient, but inside the shock wave the pressure is above ambient. This shock wave continues to travel outward until, as a result of the expansion of the gas behind the shock front, the pressure inside the wave falls to ambient pressure and the temperature is below the initial temperature of the gas. If the initial pressure is sufficiently high, the shock wave will travel at, but not exceed, the speed of sound. This is the strict definition of an explosion. If the shock wave travels at a slower speed, the event is referred to as a deflagration. It is the overpressurization or pressure difference resulting from the passage of a shock wave that causes most of the damage in an explosion. We compute the energy released from an explosion, since the energy release is related to the extent of overpressurization and the amount of damage. We do this by writing the thermodynamic balance equations for a closed system consisting of the initial contents of an exploding tank. This is a system with an expanding boundary (as the explosion occurs) that is closed to the transfer of mass and is adiabatic (i.e., Q = 0) since the expansion occurs too rapidly for heat transfer to occur. We also assume that the expansion occurs uniformly, so that there are no temperature or pressure gradients, except at the boundary of the shock wave. The difference form of the mass balance for this closed system inside the shock wave is Mf − Mi = 0 or Mf = Mi = M (5.4-1) Here the subscript i denotes the initial state, before the explosion has occurred, and f denotes the final state, when the system volume has expanded to such an extent that the pressure inside the shock wave is the ambient pressure. The energy balance for this system is ˆ (Tf , Pf ) − Mi U ˆ (Ti , Pi ) = W Mf U (5.4-2) Using the mass balance and the fact that the final pressure is ambient gives Balance equations for a vapor-phase explosion

ˆ (Ti , Pi )] = W ˆ (Tf , P = ambient) − U M [U

(5.4-3)

where −W is the energy that the exploding system imparts to its surroundings. ˆ (Tf , P = ambient), that we cannot This equation contains two unknowns, W and U evaluate since Tf is unknown. Therefore, we need another equation, which we get from the entropy balance. The entropy balance for this system is

ˆ f , P = ambient) − S(T ˆ i , Pi )] = Sgen M [S(T

(5.4-4)

Here Sgen is the entropy generated during the explosion. Since the expansion is a uniform expansion (that is, there are no spatial gradients except at the shock wave boundary), the only generation of entropy occurs across the shock wave. If we neglect this entropy generation, the work we compute will be somewhat too high. However, in safety problems we prefer to be conservative and err on the side of overpredicting an energy release resulting from an explosion, since we are usually interested in estimating the maximum energy release and the maximum damage that could result. Further, we really do not have a good way of computing the amount of entropy generated during an explosion. Consequently, we will set Sgen = 0, and from Eqs. 5.4-2 and 4 obtain

ˆ i , Pi ) ˆ f , P = ambient) = S(T S(T ˆf . which provides the additional equation necessary to find Tf and U

(5.4-5)

5.4 The Thermodynamics of Mechanical Explosions 187 Equations 5.4-3 and 5.4-5 will be used to compute the work or explosive energy imparted to the surroundings as a result of a gas phase explosion. This work is transferred to the surroundings by the damage-producing overpressurization of the shock wave. The procedure to be used to compute the energy transfer to the surroundings is as follows: 1. Knowing the initial temperature and pressure of the system, compute the initial ˆi , and specific entropy Sˆi . specific volume Vˆi , specific internal energy U 2. If the mass of the system is known, proceed to step 3. If the mass of the system is not known, but the total initial volume of the system VS is known, compute the mass from

M=

VS ˆ Vi (Ti , Pi )

3. Since the initial and final specific entropies of the system are the same (see Eq. 5.4-5), compute the final temperature of the system from the known final specific entropy and the ambient pressure. 4. From the ambient pressure and computed final temperature, calculate the final ˆf . specific internal energy U 5. Finally, use Eq. 5.4-3 to determine how much work the exploding system does on its surroundings. To use the procedure described above, we need to be able to compute the thermodynamic properties of the fluid. We may have access to tables of thermodynamic properties for the fluid. For example, if the fluid is water vapor, we could use the steam tables. Other possibilities include using graphs for the properties of the gas under consideration, the ideal gas law, or if available an accurate equation of state for the fluid (to be discussed in the next chapter).

Illustration 5.4-1 Energy Released on the Explosion of a Steam Tank A tank of volume 0.1 m3 that contains steam at 600◦ C and 1 MPa bursts. Estimate the energy of the blast. For comparison, the blast energy of trinitrotoluene (TNT) is 4600 kJ/kg.

Solution From the steam tables we have at the initial conditions that Vˆ = 0.4011 m3 /kg, Uˆ = 3296.8 kJ/kg, and Sˆ = 8.0290 kJ/(kg K). At P = 0.1 MPa = 1 bar we have that Sˆ = 8.0290 kJ/(kg K) at ˆ ≈ 2731 kJ/kg. Consequently, T = 248◦ C, where U M =

0.1 m3 = 0.2493 kg m3 0.4011 kg

and −W = 0.2493 kg × (3296.8 − 2731) This is equivalent to the energy released by 30.7 g of TNT.

kJ = 141.1 kJ kg

188 Chapter 5: Liquefaction, Power Cycles, and Explosions The simplest and least accurate procedure for calculating the blast energy is to assume that the fluid is an ideal gas with a constant heat capacity. In this case we have (using a molar basis rather than per unit mass)

N = number of moles =

Pi VS RTi

(5.4-6)

where VS is the initial system volume (i.e., the volume of the tank). Also,

S(Tf , P = ambient) − S(Ti , Pi ) = 0 = CP∗ ln or

 Tf = Ti

P = ambient Pi

Tf (P = ambient) − R ln Ti Pi

R/CP∗ (5.4-7)

Then

W = N [U (Tf , P = ambient) − U (Ti , Pi )] = N CV∗ (Tf − Ti )   R/CP∗ Pi VS ∗ P = ambient C Ti −1 = RTi V Pi   R/CP∗ P = ambient Pi VS ∗ C −1 = R V Pi Finally, using γ = CP∗ /CV∗ , we have   (γ −1)/γ  Pi VS P = ambient 1− −W = γ−1 Pi

(5.4-8)

(5.4-9)

Here Pi is the initial pressure, that is, the pressure at which the tank bursts, VS is the initial total volume of the system or tank, P is the ambient pressure, and −W (which is a positive number) is the work or energy the explosion imparts to its surroundings, which is what does the damage. Illustration 5.4-2 Energy Released on the Explosion of a Compressed Air Tank A compressed air tank has a volume of 0.167 m3 and contains air at 25◦ C and 650 bar when it explodes. Estimate the amount of work done on the surroundings in the explosion. Compute the TNT equivalent of the compressed air tank blast. Data: For air CP∗ = 29.3 J/(mol K) and γ = CP∗ /CV∗ = 1.396.

Solution    650 bar × 0.167 m3 1 bar (1.396−1)/1.396 −W = 1− (1.396 − 1) 650 bar J 1 3 Pa 1 kJ m × × 3 = 23 020 kJ = 230.2 bar m × 10 bar Pa 10 J 3

5

5.4 The Thermodynamics of Mechanical Explosions 189 Therefore, the blast energy is equivalent to 23 020 kJ = 5.0 kg of TNT kJ 4600 kg TNT

Clearly, this is a sizable explosion. In fact, a blast of 5 kg of TNT will cause the total destruction of structures not reinforced to withstand blasts within a circle of radius 7 meters from the blast site, substantial damage out to a radius of 14 meters, minor structural damage out to 55 meters, and broken windows out to 130 meters. Also, eardrum ruptures can be expected up to 10 meters from the site of the explosion.

So far we have considered the blast energy when the fluid is a gas. If the fluid contained in the vessel is a liquid, the same equations apply:

ˆ i , Pi ) ˆ f , P = ambient) = S(T S(T

(5.4-5)

ˆ (Ti , Pi )] ˆ (Tf , P = ambient) − U W = M [U

(5.4-3)

and

However, for a liquid the entropy and internal energy are, to an excellent approximation, independent of pressure. This implies that there is no entropy or internal energy change from the failure of a vessel containing a liquid (as long as the liquid does not vaporize). Consequently, there is no blast energy resulting from the change in thermodynamic properties of the fluid. However, there will be some release of energy that had been stored as elastic or strain energy in the walls of the container. Since the evaluation of this contribution is based on solid mechanics and not thermodynamics, it is not discussed here. The important observation is that compressed gases can result in devastating explosions; however, compressed liquids, as long as they do not vaporize, contain little energy to release on an explosion.2 It is for this reason that liquids, not gases, are used to test high-pressure vessels before they are put into use. A vaporizing liquid is considered next. If a tank initially under high pressure ruptures, it is possible that as a result of the sudden depressurization the liquid or liquid-vapor mixture initially in the tank may partially or completely vaporize. Such explosions are known as boiling liquid–evaporating vapor explosions (BLEVEs). Here we estimate the energy released on a BLEVE. For generality we assume that both the initial and final states include two phases, and we use ωi and ωf to represent the fraction of vapor in the initial and final states, respectively. The mass balance for this system is

Mi [ωi + (1 − ωi )] = Mf [ωf + (1 − ωf )]

(5.4-10a)

M i = Mf = M

(5.4-10b)

or The energy balance is

ˆ L (Tf , P = ambient)] ˆ V (Tf , P = ambient) + (1 − ωf )U M [ωf U ˆ V (Ti , Pi ) + (1 − ωi )U ˆ L (Ti , Pi )] = W − M [ωi U 2 It

(5.4-11)

should be noted that we have considered only the internal energy change of an explosion. For air or steam this is all we need to consider. However, if the tank contents are combustible, the result can be more devastating. There are numerous examples of a tank of combustible material rupturing and producing a flammable vapor cloud. At some point distant from the initial explosion, this vapor cloud comes in contact with sufficient oxygen and an ignition source, which results in a second, chemical explosion. In chemical plants such secondary explosions are usually more devastating than the initial explosion.

190 Chapter 5: Liquefaction, Power Cycles, and Explosions and the entropy balance is Balance equations for a vapor-liquid explosion

ωf SˆV (Tf , P = ambient) + (1 − ωf )SˆL (Tf , P = ambient) = ωi SˆV (Ti , Pi ) + (1 − ωi )SˆL (Ti , Pi )

(5.4-12)

Here the superscripts V and L indicate the vapor and liquid phases, respectively. Also, it should be remembered that if two phases are present, the temperature must be the saturation temperature of the fluid at the specified pressure. For example, if the fluid is water and the final condition is ambient pressure (1 atm or 1.013 bar), then Tf must be 100◦ C. Illustration 5.4-3 Energy Released from a Steam-Water Explosion Joe Udel decides to install his own 52-gallon (0.197 m3 ) hot-water heater. However, being cheap, he ignores safety codes and neglects to add safety devices such as a thermostat and a rupture disk or pressure relief valve. In operation, the hot-water heater is almost completely filled with liquid water and the pressure in the heater tank is the waterline pressure of 1.8 bar or the water saturation pressure at the heater temperature, whichever is greater. The water heater tank will rupture at a pressure of 20 bar. Several hours after Joe completes the installation of the water heater, it explodes. a. What was the temperature of the water when the tank exploded? b. Estimate the energy released in the blast.

Solution a. From Appendix A.III the thermodynamic properties of saturated liquid water at 20 bar (2 MPa) are Ti = 212.42◦ C

and

m3 VˆiL = 0.001177 kg

ˆ L = 906.44 kJ U i kg

kJ SˆiL = 2.4474 kg K

Therefore, the water temperature when the tank explodes is 212.42◦ C. Also, as indicated in the problem statement, the tank contains only liquid, so ωi = 0 and M =

VT = VˆiL

0.197 m3 = 167.24 kg m3 0.001177 kg

b. After the explosion we expect to have a vapor-liquid mixture. Since the pressure is 1 atm (actually, we use 1 bar), the temperature is 100◦ C and the other thermodynamic properties are ˆ L = 417.36 kJ U kg kJ SˆL = 1.3026 kg K

ˆ V = 2506.1 kJ U kg kJ SˆV = 7.3594 kg K

We first use the entropy balance to determine the fractions of vapor and liquid present SˆiL = ωf SˆfV + (1 − ωf )SˆfL

5.4 The Thermodynamics of Mechanical Explosions 191 or

2.4474 = ωf 7.3594 + (1 − ωf )1.3026

which gives ωf = 0.1890

Next, to calculate the blast energy we use the energy balance, Eq. 5.4-11: − W = M [UˆiL − Uˆf ] = M [UˆiL − {ωf UˆfV + (1 − ωf )UˆfL }] = 167.24 kg [906.44 − {0.1890 × 2506.1 + 0.8110 × 417.36}] kJ/kg = 15 772 kJ

This is equivalent to 3.43 kg of TNT.

Comment In January 1982 a large hot-water tank exploded in an Oklahoma school, killing 7 people and injuring 33 others. The tank was found 40 meters from its original location, and part of the school cafeteria was destroyed. It was estimated that the tank failed at a pressure of only 7 bar.

If thermodynamic tables for the fluid in the explosion are not available, it may still be possible to make an estimate of the fraction of vapor and liquid present after the explosion and the energy released knowing just the heat of vaporization of the fluid and its liquid heat capacity. To do this, we first write the unsteady-state mass and energy balances on the open system consisting of the liquid, shown in Fig. 5.4-2. In this figure M˙ is the molar flow rate of the liquid being vaporized. The mass and energy balances for this open, constant-volume system are as follows: Mass balance

dM = −M˙ dt

(5.4-13)

d ˆ L ) = −M˙ H ˆV (M U dt

(5.4-14)

Energy balance

Here we have used the superscripts L and V to indicate the vapor and liquid phases. ˆL ≈ H ˆ L , as discussed earlier. Also, we will use that for liquids at moderate pressure U Therefore, using this result and combining the two above equations, we have

M

ˆL dH ˆ L dM = dM H ˆV +H dt dt dt

or

M CPL

dM dT ˆ = Δvap H dt dt

(5.4-15)

Vapor

Liquid

Figure 5.4-2 Boiling liquid–evaporating vapor event. The dashed line indicates the system for which the mass and energy balances are being written.

192 Chapter 5: Liquefaction, Power Cycles, and Explosions

ˆ L = C L dT . Also, we have defined the heat of vaporization of Here we have used dH P ˆ a liquid, Δvap H , to be the difference between the enthalpies of the vapor and liquid ˆ =H ˆV − H ˆ L ). when both are at the boiling temperature (i.e., Δvap H Integrating this equation between the initial and final states, we have 

Mf Mi

1 dM = M



Tf

Ti

L CPL ∼ CP dT = ˆ ˆ Δvap H Δvap H



Tf

dT

(5.4-16)

Ti

where, in writing the last of these equations, we have assumed that both the liquid heat capacity and the heat of vaporization are independent of temperature. We then obtain

ln

Mf CPL (T − Ti ) = ˆ f Mi Δvap H

(5.4-17)

Note that the final temperature, Tf , is the normal boiling temperature of the fluid, Tb . Also, the initial and final masses in this equation are those of the liquid. Further, if the pre-explosion state is only liquid (ωi = 0), then Mf /Mi is the fraction of liquid initially present that is not vaporized in the BLEVE. For this case we have   CPL 1 − ωf = exp (T − Ti ) ˆ b Δvap H or



 CPL (T − Ti ) ωf = 1 − exp ˆ b Δvap H

(5.4-18)

To proceed further, we need to compute the change in internal energy between the initial and final states. For simplicity in this calculation, we choose the reference state to be one in which the internal energy is zero for the liquid at its normal boiling point, and compute the internal energies of the other states relative to this reference state since our interest is only with changes in internal energy. (Note that this choice of reference state is just a matter of convenience and not a necessity. Can you show that any state can be chosen as the reference state for the calculation of the energy differences without affecting the final result?) Consequently, we have

ˆ L (Tb , P = 1 bar) = 0 U (5.4-19a) ˆ L (Tb , P = 1 bar) + Δvap U ˆ V (Tb , P = 1 bar) = U ˆ U   (5.4-19b) RTb RTb ˆ ˆ = Δvap H − = 0 + Δvap H − MW MW and

 ˆ L (Ti , Pi ) = U

Ti Tb

∼ C L (Ti − Tb ) CPL dT = P

(5.4-19c)

5.4 The Thermodynamics of Mechanical Explosions 193 Therefore, the energy released in an explosion is Simplified equation for estimating the energy released on a two-phase explosion

ˆ L (Ti , Pi ) − ωf U ˆ V (Tb , P = ambient) − − W = M [U ˆ L (Tb , P = ambient)] (1 − ωf )U    RTb L ˆ = M CP (Ti − Tb ) − ωf Δvap H − MW

(5.4-20)

where ωf is given by Eq. 5.4-18.

Illustration 5.4-4 Use of the Simplified Equation to Calculate the Energy Released in a Two-Phase Explosion (BLEVE) Rework the previous illustration using only the fact that the density of water at 212.4◦ C is about 0.85 g/cm3 , and that over the temperature range CPL = 4.184

J gK

and

ˆ = 2250 J = 2250 kJ Δvap H g kg

Solution Given the density of water above, we have M = 0.197 m3 × 0.85

g cc kg × 106 3 × 10−3 = 167.5 kg cc m g

Next, we use  ωf = 1 − exp

CPL ˆ Δvap H

(Tb − Ti )



 4.184 (100 − 212.42) = 0.189 = 1 − exp 2250

which is equal to the value found in the previous illustration using accurate thermodynamic tables. Next we have 

  8.314 × 373.15 − W = M 4.184(212.42 − 100) − 0.189 2250 − 18 = 167.5 kg [470.37 − 392.68]

kJ = 13 013 kJ kg

compared with 15 772 kJ from the more accurate calculation of the previous example. Since the total mass and the fraction vaporized have been correctly calculated, all the error is a result of the approximate calculation of the internal energies. Nonetheless, such approximate calculations are useful when detailed thermodynamic data are not available.

194 Chapter 5: Liquefaction, Power Cycles, and Explosions PROBLEMS 5.1 a. An automobile air conditioner uses the vaporcompression refrigeration cycle with HFC-134a as the refrigerant. The operational temperature of the evaporator is 7◦ C and that of the condenser is 45◦ C. Determine the coefficient of performance of this airconditioning system and the amount of work needed for each kilojoule of cooling provided by the air conditioner. b. For service in high-temperature areas, the condenser temperature may go up to 65◦ C. How would the answers to part (a) change in this case? 5.2 It is desired to improve the thermal efficiency of the Rankine power generation cycle. Two possibilities have been suggested. One is to increase the evaporator temperature, and the second is to decrease the condenser temperature (and consequently the pressure) of the low-pressure part of the cycle. a. Draw a T -Sˆ diagram for the Rankine cycle similar to that in Fig. 5.2-2d, but with a higher evaporator temperature. Show from a comparison of those two diagrams that the efficiency of the Rankine power generation cycle increases with increasing evaporator temperature. b. Repeat part (a) for a lower condenser temperature. 5.3 Using a T -Sˆ diagram, discuss the effect of subcooling in the condenser and superheating in the evaporator on the efficiency of a Rankine (or other) power generation cycle. 5.4 A power plant using a Rankine power generation cycle and steam operates at a temperature of 80◦ C in the condenser, a pressure of 2.5 MPa in the evaporator, and a maximum evaporator temperature of 700◦ C. Draw the two cycles described below on a temperatureentropy diagram for steam, and answer the following questions. a. What is the efficiency of this power plant, assuming the pump and turbine operate adiabatically and reversibly? What is the temperature of the steam leaving the turbine? b. If the turbine is found to be only 85 percent efficient, what is the overall efficiency of the cycle? What is the temperature of the steam leaving the turbine in this case? 5.5 Forest cabins, remote mobile homes, Amish farms, and residential structures in locations where electricity is not available are often equipped with absorption refrigerators that rely on changes from absorption at low temperatures to desorption at high temperatures to produce pressure changes in a refrigeration system instead of a compressor. The energy source for such refrigeration systems is a flame, typically produced by propane or another fuel. The most common absorption refrigeration working fluid is the ammonia-water system.

The only patent ever awarded to Albert Einstein was for an absorption refrigeration design. The simplest representation of an absorption refrigeration is given in the figure below. The energy flows in such a device are a high-temperature (TH ) heat flow (Q˙ H ) that supplies the energy for the refrigerator, a lowtemperature (TL ) heat flow (Q˙ L ) into the condenser of the refrigeration cycle extracted from the cold box of the refrigerator, and a moderate-temperature (TM ) heat flow (Q˙ M ) out of the refrigerator. TH QH

TM QM Absorption refrigerator QL TL

a. Compute the coefficient of performance for an absorption refrigerator defined as C.O.P. =

Q˙ L Q˙ H

assuming that the absorption refrigeration cycle is reversible. b. Calculate the maximum coefficient of performance that can be achieved if heat transfer from the flame occurs at 750◦ C, the ambient temperature near the refrigerator is 27◦ C, and the temperature inside the refrigerator is −3◦ C. 5.6 a. Nitrogen can be liquefied using a simple JouleThomson expansion process. This is done by rapidly and adiabatically expanding cold nitrogen from a high-pressure gas to a low-temperature, lowpressure vapor-liquid mixture. To produce the high pressure, nitrogen initially available at 0.1 MPa and 135 K is reversibly and adiabatically compressed to 2 MPa, isobarically cooled to 135 K, recompressed to 20 MPa, and again isobarically cooled before undergoing the Joule-Thomson expansion to 0.1 MPa. What is the temperature of the liquid nitrogen, and how much compressor work is required per kilogram of liquid nitrogen produced? b. If, to improve efficiency, the Linde process is used with the same two-stage compressor as in part (a) and with nitrogen vapor leaving the heat exchanger at 0.1 MPa and 125 K, how much compressor work is required per kilogram of liquid nitrogen produced?

Problems 195 5.7 A Rankine steam cycle has been proposed to generate work from burning fuel. The temperature of the burning fuel is 1100◦ C, and cooling water is available at 15◦ C. The steam leaving the boiler is at 20 bar and 700◦ C, and the condenser produces a saturated liquid at 0.2 bar. The steam lines are well insulated, the turbine and pump operate reversibly and adiabatically, and some of the mechanical work generated by the turbine is used to drive the pump. a. What is the net work obtained in the cycle per kilogram of steam generated in the boiler? b. How much heat is discarded in the condenser per kilogram of steam generated in the boiler? c. What fraction of the work generated by the turbine is used to operate the pump? d. How much heat is absorbed in the boiler per kilogram of steam generated? e. Calculate the engine efficiency and compare it with the efficiency of a Carnot cycle receiving heat at 1100◦ C and discharging heat at 15◦ C. 5.8 As in Illustration 5.1-1 it is desired to produce liquefied methane; however, the conditions are now changed so that the gas is initially available at 1 bar and 200 K, and methane leaving the cooler will be at 100 bar and 200 K. The flash drum is adiabatic and operates at 1 bar, and each compressor stage can be assumed to operate reversibly and adiabatically. A three-stage compressor will be used, with the first stage compressing the gas from 1 bar to 5 bar, the second stage from 5 bar to 25 bar, and the third stage from 25 bar to 100 bar. Between stages the gas will be isobarically cooled to 200 K. a. Calculate the amount of work required for each kilogram of methane that passes through the compressor in the simple liquefaction process. b. Calculate the fractions of vapor and liquid leaving the flash drum in the simple liquefaction process of Fig. 5.1-1 and the amount of compressor work required for each kilogram of LNG produced. c. Assuming that the recycled methane leaving the heat exchanger in the Linde process (Fig. 5.1-2) is at 1 bar and 200 K, calculate the amount of compressor work required per kilogram of LNG produced. 5.9 High-pressure helium is available from gas producers in 0.045-m3 cylinders at 400 bar and 298 K. Calculate the explosion equivalent of a tank of compressed helium in terms of kilograms of TNT. Assume helium is an ideal gas. 5.10 The “Quick Fill” bicycle tire filling system consists of a small (2 cm diameter, 6.5 cm long) cylinder filled with nitrogen to a pressure of 140 bar. Estimate the explosion equivalent of the gas contained in the cylinder in grams of TNT. Assume nitrogen is an ideal gas.

5.11 A tank containing liquid water in equilibrium with a small amount of vapor at 25 bar suddenly ruptures. Estimate the fraction of liquid water in the tank that flash vaporizes, and the explosive energy released per kilogram of water initially in the tank. 5.12 Electrical power is to be produced from a steam turbine connected to a nuclear reactor. Steam is obtained from the reactor at 540 K and 36 bar, the turbine exit pressure is 1.0 bar, and the turbine is adiabatic. a. Compute the maximum work per kilogram of steam that can be obtained from the turbine. A clever chemical engineer has suggested that the single-stage turbine considered here be replaced by a two-stage adiabatic turbine, and that the steam exiting from the first stage be returned to the reactor

Steam 540 K 36 bar

Part (a)

Turbine

1 bar Ws

Nuclear reactor

540 K 36 bar

Turbine

Ws

540 K P∗

P∗ Parts (b) and (c)

Turbine

1 bar Ws

and reheated, at constant pressure, to 540 K, and then fed to the second stage of the turbine. (See the figure.) b. Compute the maximum work obtained per kilogram of steam if the two-stage turbine is used and the exhaust pressure of the first stage is P ∗ = 12 (36 + 1.0) = 18.5 bar. c. Compute the maximum work obtained per kilogram of steam if the two-stage turbine is used √ and the exhaust pressure of the first stage is P ∗ = 36 × 1 = 6.0 bar. d. Compute the heat absorbed per kilogram of steam in the reheating steps in parts (b) and (c). 5.13 A coal-fired power plant had been operating using a standard Rankine cycle to produce power. The operating conditions are as given in Illustration 5.2-1. However, the boiler is aging and will need to be replaced. While waiting for the replacement, it has been

196 Chapter 5: Liquefaction, Power Cycles, and Explosions suggested that for safety the operating temperature be reduced from 600◦ C to 400◦ C. The plant operates with steam, with a condenser temperature of 100◦ C, and in this emergency mode the boiler would operate at 3.0 MPa and 400◦ C. a. Can the plant function in this mode? Why or why not? b. A clever operator suggests that a Joule-Thompson valve could be placed after the boiler and before the turbine. This valve is to be designed such that the exhaust from the turbine is at the same conditions as in Illustration 5.2-1, 0.10135 MPa and approximately 126◦ C. Assuming that the pump and turbine operate adiabatically and reversibly, fill in the missing thermodynamic properties in the table below. [Values that are unchanged from Illustration 5.2-1 are given in bold.] c. Determine the heat and work flows per kg of steam in the pump, boiler, turbine and condenser. d. What is the efficiency of this new cycle? How does it compare with the efficiency of the cycle in Illustration 5.2-1? 5.14 During methane liquefaction, about 1000 kg of methane are stored at a pressure of 10 MPa and 180 K. The plant manager is worried about the possibility of explosion. Determine the energy released by a sudden rupture of this storage tank and the temperature and physical state of the methane immediately after the rupture. 5.15 A Rankine power generation cycle is operated with water as the working fluid. It is found that 100 MW of power is produced in the turbine by 89 kg/s of steam that enters the turbine at 700◦ C and 5 MPa and leaves at 0.10135 MPa. Saturated liquid water exits the condenser and is pumped back to 5 MPa and fed to the boiler, which operates isobarically. a. The turbine operates adiabatically, but not reversibly. Find the temperature of the steam exiting the turbine. b. Determine the rate of entropy generation in the turbine and compute the efficiency of the turbine. c. How much work must be supplied to the pump?

QB

Boiler

WP

Pump

QC

5.16 The United States produces about 2700 megawatts (MW) of electricity from geothermal energy, which is comparable to burning 60 million barrels of oil each year. Worldwide about 7000 MW of geothermal electricity are produced. The process is that naturally occurring steam or hot water that is not far below the earth’s surface (especially in places such as Yellowstone National Park and other volcanic and geothermal areas) is brought to the surface and used to heat a working fluid in a binary fluid power generation cycle, such as that shown in Fig 5.2-9. (Geothermal steam and water are not directly injected into a turbine, as the dissolved salts and minerals would precipitate and quickly damage the equipment.) For geothermal water at temperatures less than 200◦ C, isobutane is used as the working fluid. Isobutane is vaporized and superheated to 480 K and 10 MPa in the heat exchanger by the geothermal water, and is then passed through a turbine (which we will assume to be adiabatic and isentropic) connected to an electrical generator. The isobutane next passes through an isobaric condenser that produces a saturated liquid at 320 K. A pressure-enthalpy diagram for isobutane follows. a. At what pressure does the condenser operate? b. What are the temperature and pressure of the isobutane leaving the turbine?

T (◦ C)

P (MPa)

Sˆ [kJ/(kg K)]

ˆ (kJ/kg) H

1 Exit from condenser entry to pump 2 Exit from pump entry to boiler 3 Exit from boiler entry to J-T valve 4 Exit from J-T valve entry to turbine 5 Exit from turbine entry to condenser

100 100

0.10135 0.10135

1.3069 1.3069

419.04 419.04

103

3.0

1.3069

422.07

3.0

0.10135

WT

Condenser

Point

126.2

Turbine

7.5085

2734.7

Problems 197 withdrawn at the higher temperature, what is the maximum temperature difference that can be obtained? Treat

c. Determine the work produced by the turbine per kilogram of isobutane circulating, and the flow rate of isobutane necessary to produce 3 MW of electricity. d. Draw the cycle of the process on the isobutane pressure-enthalpy diagram. e. Obtain the heat or work requirements for the four units of the cycle in the table below.

Unit

Heat flow (kJ/kg)

Work flow (kJ/kg)

0 ? 0 ?

? 0 ? 0

Pump Boiler Turbine Condenser

J

air as an ideal gas with CP∗ = 29.3 . mol · K 5.19 An Automobile engine can be modelled as an idealized four-stroke Otto cycle, although it actually consists of 6 steps: Step 0: a fuel-air mixture is drawn into a cylinder at constant pressure (admission stroke). Step 1: adiabatic and reversible (isentropic) compression of the air as the piston moves from the bottom (V2 ) of the cylinder to the top (V1 )(compression stroke). Step 2: isochoric (constant volume) heat transfer to the working gas from an external source while the piston is at V1 . (This represents an instantaneous combustion of the fuel-air mixture.) Step 3: adiabatic and reversible (isentropic) expansion of the cylinder contents (power stroke).

f. What is the efficiency of the proposed cycle?

©Department of Mechanical Engineering, Stanford University

20 580

00

3

022

0.

0.0

0.0018

500

T= 440

K) kg · kJ/(

2.6

2.4

2.8 s=

3.2

0.3

0.2

3.4

d Satu

rate

500

0.4

0.5

280

400

0.6

0.2

3

line

vapor

0.9 Satura ted

0.8

0.7

0.1

300

300

1 0.8

0.07

0.14

320

0.5 x =0 .6

0.4

0.3

2

0.05

2.2

2

1.8

1.6

1.4

1.2

e lin 0.2

0.03

K

id

4

0.02

0.7

600 h, kJ/kg

P, MPa

s =1 kJ/(kg · K)

480

4

400 360 340

0.1

6

0.01

420 380

320

liqu

540

7

0.00

460

360 340

200

10 8

520

400

0 0.0

0.01

280

100

3 /kg

5m

v=

300

0

560

ISOBUTANE

600

THERMODYNAMIC PROPERTIES OF

700

800

900

1000

1100

0.1 1200

(Used with permission of the Department of Mechanical Engineering, Stanford University.) 5.17 Steam enters a turbine 350◦ C and 0.8 MPa and exits at 0.1 MPa.Compute the work exiting temperature and work obtained if: a. the turbine operates isentropically; and b. the turbine has an isentropic efficiency of 0.8. 5.18 Consider the Hilsch-Ranque vortex tube discussed in Illustration 4.5-6 (pages 135-137). Starting with air at 4 bar and 25◦ C, and exhaust pressure of 1.013 bar, and that half the air that enters the tube will be

Step 4: isochoric heat transfer in which heat is released by the gas to the atmosphere through the cylinder walls while the piston is at V2 . Step 5: isobaric release of exhaust gases to the atmosphere (exhaust stroke). Here steps 0 and 5 can be ignored as no appreciable work is obtained or consumed. Steps 1 to 4 produce the net shaft work to power the automobile. The heat source for the engine is the burning of the fuel;

198 Chapter 5: Liquefaction, Power Cycles, and Explosions for simplicity this heat will be considered as coming from an external heat source. a. Draw this cycle on a pressure-entropy diagram. b. Assuming that the working fluid behaves like an ideal gas of constant heat capacity, calculate the thermal efficiency of the cycle (net work divided by heat flow in) as a function of the specific heat ratio k = CP /CV and the compression ratio γ = V1 /V2 , where V1 and V2 are the minimum an maximum volumes occupied by the gas, respectively. 5.20 n-butane is to be liquefied to make liquid petroleum gas (LPG). The butane is available at 25◦ C and 1 bar, it will be compressed to 15 bar in a compressor that has an isentropic efficiency of 85%, cooled to 0◦ C in a heat exchanger, and then expanded adiabatically to 1 bar, with the vapor recycled to the compressor. Determine the work required per kg of LPG produced. 5.21 One suggestion that has been made to conserve energy is that all new electrical power generation plants should be co-generation facilities. In a typical power plant the combustion of coal or natural gas is used to produce steam that is run through a turbine and the only useful energy that results is electricity. In such cases the pressure at the downstream end of the turbine is kept as low as possible to produce the most work (electricity). This is done by having a condenser after the turbine cooled by (frequently river) water or air. Another alternative is a co-generation power plant in which the temperature of the exiting steam is kept higher so that the steam leaving the turbine can be used for heating purposes (as process stream in a chemical plant or for residential heating as in New York City). In a cogeneration plant the useful energy obtained is the sum of the electrical energy and energy that can be used for heating. Calculate the useful energy and overall energy efficiency obtained from: a. a standard power generation plant, and b. a co-generation plant. The following data are available: i The heat combustion of the fuel is used to produce steam at 900◦ C and 25 bar from water at 1 bar and 25◦ C. This is accomplished with 80% efficiency (that is 20% of the heating value of the fuel is lost in the process); ii The steam turbines are adiabatic and reversible; iii The condensed saturated steam leaving the standard power plant is at 45◦ C which sets the pressure at the exit of the turbine and iv The steam exit pressure in the co-generation plant is 1 bar, and this steam in then used as a heat transfer fluid until is condensed and its temperature is 40◦ C. 5.22 Isobutane is to be liquefied to make liquid petroleum gas (LPG). The butane is available at 25◦ C and 1 bar,

5.23 5.24 5.25 5.26

5.27

5.28

5.29

5.30

it will be compressed to 15 bar, cooled to 0◦ C in a heat exchanger, and expanded and flashed to 1 bar in an adiabatic valve, and the vapor and liquid separated. Determine the work required per kg of LPG produced if the compressor has an isentropic efficiency of 100%, and none of the vapor is recycled. Repeat the calculation of problem 5.22 with the vapor being recycled to the compressor. Repeat the calculation of problem 5.22 if the compressor has an isentropic efficiency of 72%. Repeat the calculation of problem 5.23 if the compressor has an isentropic efficiency of 87%. The inlet to an adiabatic turbine is steam at 0.8 MPa and 350◦ C, and the turbine exit pressure is 0.1 MPa. a. Determine the maximum work that can be obtained from each kg of steam and the exit temperature of the steam. b. If the turbine has an isentropic efficiency of 80%, determine the work obtained per kg of steam and the exit temperature. Methane at 260 K is to be isothermally compressed from 0.1 MPa to 1.0 MPa. a. What is the minimum work required, and how much heat must be removed to keep the compression process isothermal? b. If the compressor is only 75% isentropically efficient, what work is required, and how much heat must be removed to keep the compression process isothermal? The inlet to an adiabatic turbine is steam at 1.3 MPa and 385◦ C, and the turbine exit pressure is 0.1 MPa. a. Determine the maximum work that can be obtained from each kg of steam and the exit temperature of the steam. b. If the turbine has an isentropic efficiency of 80%, determine the work obtained per kg of steam and the exit temperature. The inlet to an adiabatic compressor is nitrogen at 1 bar (0.1 MPa) and 150 K, and the exit pressure is 10 bar (1 MPa). a. Determine the minimum work required for each kg of nitrogen compressed and the exit temperature of the nitrogen. b. If the compressor is found to require 20% more work than the minimum, determine efficiency of the compressor, the work obtained per kg of nitrogen and the exit temperature. A gas is continuously passed through an adiabatic turbine at the rate of 2 mol/s. Its initial temperature is 600 K, its initial pressure is 5 bar and its exiting pressure is 1 bar. Determine the maximum rate at which work can be obtained in this process. The gas is described by an augmented Clausius equation of state

Problems 199 P (V − b) = RT b0 = 3 × 10−5

m3 mol

with

and

b = b0 +

b1 ; T

b1 = 3 × 10−5

m3 · K mol

CP∗ = CP,0 + CP,1 ∗ T ; CP,0 = 25

J mol · K

and

CP,1 = 1.8 × 10−3

J mol · K2

5.31 In a continuous manufacturing process, chlorodifluoromethane (CHClF2 ) initially at 10 bar and 420 K, passes through an adiabatic pressure reducing valve so that its pressure is 0.1 bar (this last pressure is low enough that CHClF2 can be considered to be an ideal gas). At these operating conditions, the gas can be represented by a one-term virial equation of state:

5.33

B(T ) PV =1+ RT V

The following data are available for this gas: 256, 100 cm3 T mol and the ideal gas heat capacity is CP = 27.93 + J 0.093T mol · K B(T ) = 297.6 −

What is the temperature of the chlorodifluoromethane exiting valve? How much entropy is generated in the process per mole of chlorodifluoromethane that flows through the valve? 5.32 Two separate experiments are performed on a gas enclosed in a piston-and-cylinder device, both starting from the same initial state. The result of the first experiment is to be used to predict the outcome of the second.

5.34 5.35 5.36 5.37 5.38 5.39 5.40 5.41 5.42 5.43 5.44 5.45

a. In the first experiment, the piston is free to move, with the external pressure held constant. A small amount of heat is added to the gas in the cylinder, resulting in expansion of the gas. The temperature of the gas is found to have increased. b. In the second experiment, the piston is not free to move; instead, its position is adjusted manually. In addition, the device is insulated, so that no heat flows to or from the surroundings. The piston is moved outward slowly, allowing the gas to expand by a small amount. Does the temperature of the gas increase or decrease? R Use Aspen Plus to compute the coefficient of performance of a Rankine cycle using water as the working fluid (described by the IAPWS-95 method) and the following state conditions: Condenser: saturated liquid exiting at 1 bar Pump: exit pressure is 30 bar Boiler: exit temperature is 600◦ C at 30 bar Turbine: operates isentropically with an exit pressure of 1 bar. (Hint: This is a closed system, so you must specify the flowrate of one stream.) R Redo Problem 5.1 using Aspen Plus . R Redo Problem 5.2 using Aspen Plus . R Redo Problem 5.4 using Aspen Plus . R . Redo Problem 5.7 using Aspen Plus R Redo Problem 5.20 using Aspen Plus . R Redo Problem 5.22 using Aspen Plus . R . Redo Problem 5.23 using Aspen Plus R Redo Problem 5.24 using Aspen Plus . R Redo Problem 5.25 using Aspen Plus . R . Redo Problem 5.27 using Aspen Plus R Redo Problem 5.28 using Aspen Plus . R Redo Problem 5.29 using Aspen Plus .

Chapter

6 The Thermodynamic Properties of Real Substances In Chapters 2, 3, and 4 we derived a general set of balance equations for mass, energy, and entropy that can be used to compute energy changes, and heat or work requirements, for the changes of state of any substance. However, these balance equations are in terms of the internal energy, enthalpy, and entropy rather than the pressure and temperature, the variables most easily measured and thus most often used to specify the thermodynamic state of the system. To illustrate the use of the balance equations in the simplest manner, examples were given using either ideal gases or fluids whose thermodynamic properties were available in graphical and tabular form. Unfortunately, no gas is ideal over the whole range of pressure and temperature, and thermodynamic properties tables are not always available, so a necessary ingredient of many thermodynamic computations is the calculation of the thermodynamic properties of real substances in any state. The main topic of this chapter is establishing how to solve thermodynamic problems for real substances given heat capacity data and the relationship between pressure, volume, and temperature. The problem of constructing a thermodynamic properties chart from such data is also considered. The discussion of the relationship between the ideal gas and absolute temperature scales, which began in Chapter 1, is completed here. Finally, the principle of corresponding states and generalized equations of state are considered, as is their application.

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 6 The goals of this chapter are for the student to: • Be able to evaluate the partial derivative of a thermodynamic variable with respect

to one variable (e.g., temperature) while holding a second variable constant (e.g., pressure) (Sec. 6.2) • Be able to interrelate the partial derivatives that arise in thermodynamics (Sec. 6.2) • Be able to obtain volumetric equation of state parameters from critical properties (Sec. 6.4) 200

6.1 Some Mathematical Preliminaries 201 • Be able to solve problems for real fluids using volumetric equations of state (e.g.,

van der Waals or Peng-Robinson) (Secs. 6.4 and 6.7)

• Be able to construct tables and charts of thermodynamic properties (Sec. 6.4).

NOTATION INTRODUCED IN THIS CHAPTER a(T ) A b(T ) B B(T ) C(T ) Pc Pr Tc Tr Vc Vr

Equation-of-state parameter (dimensions depend on equation) Dimensionless form of equation-of-state parameter a Equation-of-state parameter (m3 /mol) Dimensionless form of equation-of-state parameter b Second virial coefficient (m3 /mol) Third virial coefficient (m3 /mol)2 Pressure at the critical point (kPa) Reduced pressure = P/Pc Temperature at the critical point (K) reduced temperature = T /Tc Molar volume at the critical point (m3 /mol) Reduced volume = V /V c PV Z Compressibility factor = RT Zc Compressibility factor at the critical point   1 ∂V α coefficient of thermal expansion = (K−1 ) V ∂T P α(T ) Temperature-dependent term in equation of state   1 ∂V κT isothermal compressibility = − (kPa−1 ) V ∂P T κ Parameter in temperature dependence  of a(T ) in equation of state  ∂T μ Joule-Thomson coefficient = (K kPa−1 ) ∂P H ω Acentric factor

6.1 SOME MATHEMATICAL PRELIMINARIES In the previous chapters eight thermodynamic state variables (P , T , V , S , U , H , A, and G), which frequently appear in thermodynamic calculations, were introduced. If the values of any two of these variables are given, the thermodynamic state of a pure, single-phase system is fixed, as are the values of the remaining six variables (experimental observation 8 of Sec. 1.7). Mathematically we describe this situation by saying that any two variables may be chosen as the independent variables for the singlecomponent, one-phase system, and the remaining six variables are dependent variables. If, for example, T and V are taken as the independent variables, then any other variable, such as the internal energy U , is a dependent variable; this is denoted by U = U (T, V ) to indicate that the internal energy is a function of temperature and specific volume. The change in internal energy dU , which results from differential changes in T and V , can be computed using the chain rule of differentiation:     ∂U ∂U dU = dT + dV ∂T V ∂V T where the subscript on each derivative indicates the variable being held constant; that is, (∂U /∂T )V denotes the differential change in molar internal energy accompanying

202 Chapter 6: The Thermodynamic Properties of Real Substances a differential change in temperature in a process in which the molar volume is constant. Note that both (∂U /∂T )V and (∂U /∂V )T are partial derivatives of the type   ∂X (6.1-1) ∂Y Z where X , Y , and Z are used here to denote the state variables P , T , V , S , U , H , A, and G. Since derivatives of this type occur whenever one tries to relate a change in one thermodynamic function (here U ) to changes in two others (here T and V ), one of the goals of this chapter is to develop methods for computing numerical values for these derivatives, as these will be needed in many thermodynamic calculations. In open systems extensive properties, such as the total internal energy U , the total enthalpy H , and the total entropy S , are functions of three variables, usually taken to be two thermodynamic variables (either intensive or extensive) and the total mass (M ) or mole number (N ). Thus, for example, the total internal energy can be considered to be a function of temperature, total volume, and number of moles, so that for a differential change in T , V , and N       ∂U ∂U ∂U dU = dT + dV + dN ∂T V,N ∂V T,N ∂N T,V where two subscripts are now needed to indicate the variables being held constant for each differential change. The derivatives of extensive properties at constant mole number or mass are simply related to the analogous derivatives among the state variables, that is, to derivatives of the form of Eq. 6.1-1. To see this, we note that since any extensive property X can be written as N X , where X is an intensive (or molar) property, the derivative of an extensive property with respect to an intensive property (e.g., temperature, pressure, or specific volume) at constant mole number is       ∂X ∂(N X) ∂X = =N ∂Y Z,N ∂Y ∂Y Z Z,N Thus,



∂U ∂T



 =N V,N

∂U ∂T

 = N CV V

where CV is the constant-volume heat capacity defined in Chapter 3. Similarly, the derivative of an extensive property with respect to an extensive property at a constant mole number is found from the observation that         ∂(N X) ∂X ∂X N ∂X = = = ∂Y Z,N ∂(N Y ) Z,N N ∂Y Z ∂Y Z so that, for example,



∂U ∂V



 = T,N

∂U ∂V

 T

Consequently, once a method is developed to obtain numerical values for derivatives in the form of Eq. 6.1-1,

6.1 Some Mathematical Preliminaries 203



∂X ∂Y

 Z

it can also be used to evaluate derivatives of the form     ∂X ∂X and ∂Y Z,N ∂Y Z,N The derivative of an extensive property with respect to mole number [e.g., the derivative (∂U/∂N )T,V in this discussion] is not of the form of Eq. 6.1-1. Such derivatives are considered in the following sections and in Sec. 6.10. We start the analysis of partial derivatives of the type indicated in Eq. 6.1-1 by listing several of their important properties. First, their numerical value depends on the path followed, that is, on which variable is being held constant. Thus,     ∂U ∂U = ∂T V ∂T P as will be verified shortly (Illustration 6.2-1). If two intensive variables are held constant in a one-component, single-phase system, all derivatives, such as   ∂U ∂T V ,P must equal zero, since by fixing the values of two intensive variables, one also fixes the values of all the remaining variables. We will assume that all the thermodynamic variables in which we are interested exist and are well behaved in some mathematical sense that we will leave unspecified. In this case, the derivative of Eq. 6.1-1 has the properties that   1 ∂X = (6.1-2) ∂Y Z (∂Y /∂X)Z and

∂   ∂Z Y



∂X ∂Y

 Z

  ∂  ∂X =  ∂Y Z ∂Z Y

(6.1-3)

The last equation, the commutative property, states that in a mixed second derivative the order of taking derivatives is unimportant. Also,



∂X ∂Y

 =0

(6.1-4a)

=1

(6.1-4b)

X

and



∂X ∂X

 Z

since the first derivative is the change in X along a path of constant X , and the second derivative is the change in X in response to an imposed change in X .

204 Chapter 6: The Thermodynamic Properties of Real Substances If X is any dependent variable, and Y and Z the two independent variables, by the chain rule of differentiation we can write X = X(Y , Z), and     ∂X ∂X dX = dY + dZ (6.1-5) ∂Y Z ∂Z Y Now from Eqs. 6.1-4a and 6.1-5, we have           ∂X ∂Y ∂Z ∂X ∂X =0= + ∂Y X ∂Y Z ∂Y X ∂Z Y ∂Y X and using Eq. 6.1-4b we obtain



∂X ∂Y

  Z

∂Z ∂X

  Y

∂Y ∂Z

 = −1

(6.1-6a)

X

or

 Triple-product rule

∂X ∂Y

  Z

∂Z ∂X



 =−

Y

∂Z ∂Y

 (6.1-6b) X

Equations 6.1-6a and b are known as the triple product rule and will be used frequently in this book; this rule is easily remembered by noting the symmetric form of Eq. 6.1-6a in that each variable appears in each derivative position once and only once. There are two other important results to be derived from Eq. 6.1-5. The first is the expansion rule, obtained by introducing any two additional thermodynamic properties K and L,           ∂X ∂Y ∂Z ∂X ∂X = + (6.1-7) ∂K L ∂Y Z ∂K L ∂Z Y ∂K L and the second is a special case of the first in which L = Z , so that, by Eq. 6.1-4a,

 Chain rule

∂X ∂K



 =

Z

∂X ∂Y

  Z

∂Y ∂K

 (6.1-8) Z

This equation is known as the chain rule. (You should compare Eq. 6.1-6b with Eq. 6.1-8 and note the difference between them.) Thus, if for some reason it is convenient, we can use this equation interpose a new variable in evaluating a partial derivative, as is the case with the variable Y in Eq. 6.1-8. Finally, we note that for an open system it is usually convenient to use the mass M or mole number N , and two variables from among T , P , and the extensive variables U , V , S , G, H , and A, as the independent variables. Letting X , Y , and Z represent variables from among the set (U , V , S , G, H , A, T , and P ), we have X = X(Y, Z, N ) and       ∂X ∂X ∂X dX = dY + dZ + dN (6.1-9) ∂Y Z,N ∂Z Y,N ∂N Y,Z

6.2 The Evaluation of Thermodynamic Partial Derivatives 205

6.2 THE EVALUATION OF THERMODYNAMIC PARTIAL DERIVATIVES Whenever the change in a thermodynamic property is related to changes in two others, derivatives of the type of Eq. 6.1-1 occur. Four of these partial derivatives occur so frequently in both experiment and calculation that they have been given special designations:   ∂U constant-volume heat capacity (6.2-1) = CV = (see Sec. 3.3) ∂T V   ∂H constant-pressure heat capacity = CP = (6.2-2) (see Sec. 3.3) ∂T P

1 V

Definitions



1 − V



∂V ∂T ∂V ∂P

 = α = coefficient of thermal expansion

(6.2-3)

= κT = isothermal compressibility

(6.2-4)

P



T

The starting point of the analysis of other thermodynamic derivatives is Eq. 4.2-13a for open systems,

dU = T dS − P dV + G dN

(6.2-5a)

and Eq. 4.2-13b for closed systems:

 dU = T dS − P dV =  which leads to

∂U ∂S

 T

∂U ∂S



 dS + V

∂U ∂V

 dV

(6.2-5b)

S

  = T and ∂U = −P . Alternatively, these equations can ∂V S

be rearranged to

dS =

1 P G dU + dV − dN T T T

(6.2-5c)

1 P dU + dV T T

(6.2-5d)

and

dS = By definition, H = U + P V , so that

dH = dU + V dP + P dV = T dS − P dV + G dN + V dP + P dV = T dS + V dP + G dN

(6.2-6a)

and for the closed system

dH = T dS + V dP

(6.2-6b)

Similarly, from Eq. 4.2-6 we have A = U − T S , so that dA = dU − S dT − T dS . Using Eq. 6.2-5 yields

dA = −P dV − S dT + G dN

(6.2-7a)

206 Chapter 6: The Thermodynamic Properties of Real Substances which for the closed system becomes

dA = −P dV − S dT

(6.2-7b)

Finally, from Eq. 4.2-8, we have G = H − T S , so that

dG = V dP − S dT + G dN and

(6.2-8a)

dG = V dP − S dT (6.2-8b) Next, we note the analogy between Eqs. 6.1-5 and 6.1-9 and Eqs. 6.2-5 through 6.2-8 and obtain the following relations:     ∂U ∂U = =T (6.2-9a) ∂S V,N ∂S V     ∂U ∂U = = −P (6.2-9b) ∂V S,N ∂V S   ∂U =G (6.2-9c) ∂N S,V   G ∂S =− (6.2-9d) ∂N U,V T     ∂H ∂H = =T (6.2-10a) ∂S P,N ∂S P     1 ∂H ∂H = =V (6.2-10b) N ∂P S,N ∂P S   ∂H =G (6.2-10c) ∂N P,S     ∂A ∂A = = −P (6.2-11a) ∂V T,N ∂V T     ∂A 1 ∂A = = −S (6.2-11b) N ∂T V,N ∂T V   ∂A =G (6.2-11c) ∂N T,V     1 ∂G ∂G = =V (6.2-12a) N ∂P T,N ∂P T     1 ∂G ∂G = = −S (6.2-12b) N ∂T P,N ∂T P and   ∂G =G (6.2-12c1 ) ∂N T,P

1 Note

that comparing Eqs. 6.2-9c and 6.2-9d, 6.2-10c, 6.2-11c, and 6.2-12c we have           ∂U ∂H ∂A ∂G ∂S = = = = −T =G ∂N S,V ∂N P,S ∂N T,V ∂N T,P ∂N U,V

The multicomponent analogues of these equations are given in Sec. 8.2.

6.2 The Evaluation of Thermodynamic Partial Derivatives 207 To get expressions for several additional thermodynamic derivatives we next use Eq. 6.1-3, the commutative property of mixed second derivatives. In this way, starting with Eq. 6.2-9 we obtain     ∂  ∂T ∂U =  ∂V S ∂S V ∂V S     ∂U ∂P ∂  =−  ∂S V ∂V S ∂S V so that from Eq. 6.1-3 we have  Maxwell relations

∂T ∂V



 =−

S

∂P ∂S

 (6.2-13) V

Similarly, from Eqs. 6.2-10 through 6.2-12 we obtain     ∂T ∂V = ∂P S ∂S P     ∂P ∂S = ∂T V ∂V T and     ∂V ∂S =− ∂T P ∂P T

(6.2-14) (6.2-15)

(6.2-16)

Equations 6.2-13 through 6.2-16 are known as the Maxwell relations. (It is left to you to derive the Maxwell relations for open systems; see Problem 6.27.) Equation 6.2-5d relates the change in entropy to changes in internal energy and volume. Since temperature and pressure, or temperature and volume, are, because of the ease with which they can be measured, more common choices for the independent variables than U and V , it would be useful to have expressions relating dS to dT and dV , or to dT and dP . We can derive such expressions by first writing S as a function of T and V , S = S(T, V ) and then using the chain rule of partial differentiation to obtain     ∂S ∂S dS = dT + dV ∂T V ∂V T

(6.2-17)

From the application of first Eq. 6.1-8, then Eqs. 6.1-2a and b, and finally Eqs. 6.2-1 and 6.2-9a, we obtain



∂S ∂T



 =

V

∂S ∂U

  V

∂U ∂T



 =

V

∂U ∂T

  V

∂U ∂S

and from Eq. 6.2-15, we have



∂S ∂V



 =

T

∂P ∂T

 V

 −1 = CV /T V

(6.2-18)

208 Chapter 6: The Thermodynamic Properties of Real Substances Thus

CV dT + dS = T



∂P ∂T

 dV

(6.2-19)

V

Consequently, given heat capacity data as a function of T and P or T and V , volumetric equation-of-state information, that is, a relationship between P , V , and T , and a value of the entropy at some value of T and V , it is possible to compute the entropy at any other value of T and V by integration of Eq. 6.2-19. Similarly, starting from S = S(T, P ), one can easily show that       CP ∂S ∂V ∂S = =− , ∂T P T ∂P T ∂T P and

CP dS = dT − T



∂V ∂T

 dP

(6.2-20)

P

Equations 6.2-19 and 6.2-20 can now be used in Eqs. 6.2-5 and 6.2-6 to get

dU = T dS − P dV     CV ∂P =T dV − P dV dT + T ∂T V     ∂P = CV dT + T − P dV ∂T V

(6.2-21)



 ∂V dP dH = CP dT + V − T ∂T P

(6.2-22)

and

From these last two equations we obtain     ∂P ∂U =T −P ∂V T ∂T V and



∂H ∂P



 =V −T T

∂V ∂T

(6.2-23)

 (6.2-24) P

Table 6.2-1 summarizes the definitions used and some of the thermodynamic identities developed so far in this chapter. The equations in this table can be useful in obtaining information about some thermodynamic derivatives, as indicated in Illustration 6.2-1. Illustration 6.2-1 Showing That Similar Partial Derivatives with Different State Variables Held Constant Are Not Equal Obtain expressions for the two derivatives (∂U /∂T )V and (∂U /∂T )P , and show that they are not equal.

6.2 The Evaluation of Thermodynamic Partial Derivatives 209 Important table

Table 6.2-1 Some Useful Definitions and Thermodynamic Identities Definitions



   ∂U ∂S Constant-volume heat capacity = CV = =T ∂T V ∂T V     ∂H ∂S Constant-pressure heat capacity = CP = =T ∂T P ∂T P   1 ∂V Isothermal compressibility = κT = − V ∂P T   1 ∂V Coefficient of thermal expansion = α = V ∂T P     ∂V V −T ∂T P Joule-Thomson coefficient = μ = − CP

Maxwell relations 



∂T ∂V ∂P ∂T







 ∂P ∂S V   ∂S = ∂V T =−



S

V



∂T ∂P ∂V ∂T

Thermodynamic identities

= 

   ∂H ∂U = =T ∂S P ∂S V     ∂U ∂A = = −P ∂V S ∂V T

S

P





 ∂V ∂S P   ∂S =− ∂P T 





∂G ∂P ∂A ∂T



 = 

T

 =

V

∂H ∂P ∂G ∂T

 =V 

S

= −S P

Thermodynamic functions

    ∂P dU = T dS − P dV = CV dT + T − P dV ∂T V     ∂V dP dH = T dS + V dP = CP dT + V − T ∂T P dA = −P dV − S dT dG = V dP − S dT

Miscellaneous  

∂U ∂V ∂H ∂P



   G ∂  H =− 2 ∂T P T T    A ∂  U =− 2  ∂T V T T



∂P Tα −P = −P ∂T V κT   ∂V =V −T = V (1 − T α) ∂T P =T

T



 G=



T

∂G ∂N



 = T,P

∂A ∂N



 = T,V

∂H ∂N



 = P,S

∂U ∂N



 = −T S,V

∂S ∂N

∂ 1 ∂ T ∂ 1 ∂  T U,V

    G  =H  T P     A  =U  T V

210 Chapter 6: The Thermodynamic Properties of Real Substances Solution Starting from Eq. 6.2-21,     ∂P dU = CV dT + T − P dV ∂T V

Using Eq. 6.1-7 yields 

∂U ∂T



 = CV V

∂T ∂T

     ∂P ∂V + T −P ∂T V ∂T V

 V

which, with Eq. 6.1-4, reduces to 

∂U ∂T

 = CV V

Again starting with Eq. 6.2-21, we obtain 

∂U ∂T

 P

     ∂P ∂V = CV + T −P ∂T V ∂T P

(6.2-25)

Clearly, then, 

∂U ∂T





= V

∂U ∂T

 P

(see Problem 6.3).

The form of Eqs. 6.2-19 through 6.2-22 is nice for two reasons. First the equations relate the change in entropy, internal energy, and enthalpy to changes in only P , V , and T . Next, the right sides of these equations contain only CP , CV , P , V , and T , and partial derivatives involving P , V , and T . Thus, given heat capacity data and the volumetric equation of state for the fluid, the changes in S , U , and H accompanying a change in system temperature, pressure, or volume can be computed. Ideally, we would like to develop equations similar to Eqs. 6.2-19 through 6.2-22 for all the thermodynamic variables of interest and, more generally, to be able to relate numerically the change in any thermodynamic property to the changes in any two others. To do this we must be able to obtain a numerical value for any derivative of the form (∂X/∂Y )Z . Since engineers generally use two variables from among pressure, temperature, and volume as the independent variables, and they also have most information about the interrelationship between these variables, the discussion that follows centers on reducing all partial derivatives to functions of P , V , and T ; their mutual derivatives; and the heat capacity, as in the equations already derived. Unfortunately, it is not possible to reduce all partial derivatives to functions of only these variables because certain partial derivatives introduce the entropy (see Eqs. 6.2-11 and 6.2-12). Since, using Eqs. 6.2-19 and 6.2-20, entropy can be evaluated from heat capacity and volumetric equation-of-state data, its inclusion introduces no real difficulty. Thus, we will be satisfied if we can reduce any partial derivative to a form containing P , V , T , S , and CP or CV , and derivatives containing only P , V , and T . In fact, as we shall see later in this section (Eq. 6.2-30), there are only three independent partial derivatives from among the four in Eqs. 6.2-1 through 6.2-4, for example, CP , α, and κT , so that is possible to reduce the partial derivatives encountered in this chapter to functions of only P , V , T , S , α, κT , and CP or CV .

6.2 The Evaluation of Thermodynamic Partial Derivatives 211 Since eight different variables (T , P , V , U , H , S , A, and G) may be used in the thermodynamic description of a one-component system, there are 8 × 7 × 6 = 336 possible nonzero derivatives of the form of Eq. 6.1-1 to be considered. (In a binary mixture there is a ninth variable, composition, and three independent variables—temperature, specific volume, and composition—so that the thermodynamic partial derivatives of possible interest number in the thousands.) Therefore, it is necessary that a systematic procedure be developed for reducing any such derivative. Although not needed for the discussion that follows, such a procedure is introduced in Sec. 6.10. As mentioned earlier, the Joule-Thomson expansion is a process that occurs at constant enthalpy. The Joule-Thomson coefficient μ is defined to be the change in temperature accompanying a differential change in pressure at constant enthalpy; that is,   ∂T μ= (6.2-26) ∂P H From Eq. 6.2-22 we have that at constant enthalpy 

 ∂V dP |H 0 = CP dT |H + V − T ∂T P or 

 ∂V   V −T ∂T P ∂T μ= =− ∂P H CP

(6.2-27)

(Note: The procedure used above can also be used with Eqs. 6.2-5 through 6.2-8, 6.2-19 through 6.2-22, and elsewhere in the evaluation of thermodynamic partial derivatives.) For later reference, we also note   ∂  G G S (H − T S) H 1 ∂G = − 2 =− − =− 2 (6.2-28)  ∂T P T T ∂T T T T2 T and

∂  A 1 =  ∂T V T T



∂A ∂T

 − V

A S (U − T S) U =− − =− 2 T2 T T2 T

(6.2-29)

Also, since

∂ ∂ ∂   =− = −T 2 2 (1/T )∂T ∂T 1 ∂ T we have

 ∂(G/T )  =H ∂(1/T ) P

and

 ∂(A/T )  =U ∂(1/T ) V

(6.2-30)

Illustration 6.2-2 Showing Again That Similar Partial Derivatives with Different State Variables Held Constant Have Different Values

212 Chapter 6: The Thermodynamic Properties of Real Substances For the discussion of the difference between the constant-pressure heat capacity CP and the constant-volume heat capacity CV , it is useful to have an expression for the derivative (∂S/∂T )P in which T and V are the independent variables. Derive such an expression.

Solution Starting from Eq. 6.2-19, CV dT + T

dS =

we have



dS ∂T





∂P ∂T

 dV V

     ∂T ∂V ∂P + ∂T P ∂T V ∂T P     ∂V ∂P CV + = T ∂T V ∂T P CV T

= P



(6.2-31)

Now using the triple-product rule, 

we get



 

∂P ∂T

∂S ∂T

V

 P



∂S ∂T

  T



CV − T

=



CV − T

=

and finally

∂V ∂P

 = P

∂T ∂V

∂P ∂V ∂V ∂P

 = −1

(6.2-32)

P

  T

∂V ∂T

T

∂P ∂T

 

CV V α2 + T κT

2

(6.2-33a) P

2

(6.2-33b) V

(6.2-34)

Compare this with 

∂S ∂T

 = V

CV T

In Eqs. 6.2-1 through 6.2-4, four partial derivatives that frequently occur were introduced. As has already been indicated, only three of these derivatives are independent in that given the values of three derivatives, we can easily compute the value of the fourth. We establish this here by deriving an equation that relates CP to CV , α, and κT . The starting point is the relation CP = T (∂S/∂T )P and the results developed in Illustration 6.2-2, which yield       ∂V ∂S ∂P CP = T = CV + T ∂T P ∂T V ∂T P 2    ∂V ∂P = CV − T ∂V T ∂T P (6.2-35)   2  ∂P ∂V = CV − T ∂P T ∂T V

= CV + T V α2 /κT establishing that CP , CV , α, and κT are all interrelated.

6.2 The Evaluation of Thermodynamic Partial Derivatives 213 For the discussion of the following section, we need to know the dependence of the constant-volume heat capacity on specific volume (or density) at constant temperature. To obtain (∂CV /∂V )T , we start with     ∂P dU = CV dT + T − P dV ∂T V and note that



∂U ∂T



 = CV

and

V

∂U ∂V



 =T T

∂P ∂T

 −P V

Use of the commutative property, Eq. 6.1-3, yields the desired result:

   2      ∂  ∂CV ∂ P ∂  ∂U ∂U = =T =   ∂V T ∂T V ∂V T ∂T 2 V ∂T V ∂V T In a similar fashion, starting with Eq. 6.2-22, one obtains    2  ∂CP ∂ V = −T ∂P T ∂T 2 P

(6.2-36)

(6.2-37)

Illustration 6.2-3 Use of Partial Derivative Interrelations to Obtain Useful Results Develop expressions for the coefficient of thermal expansion α, the isothermal compressibility κT , the Joule-Thomson coefficient μ, and the difference CP − CV for (a) the ideal gas and (b) the gas that obeys the volumetric equation of state   a P + 2 (V − b) = RT V

(6.2-38a)

where a and b are constants. (This equation of state was developed by J. D. van der Waals in 1873, and fluids that obey this equation of state are called van der Waals fluids.)

Solution a. For the ideal gas P V = RT ; thus 

∂V ∂T

 = P

R V = P T

so that α =

1 V



∂V ∂T

 =

1 T

=

1 P

P

and 

∂V ∂P

 =− T

V P

so that κT = −

1 V



∂V ∂P

 T

From Eq. 6.2-27 we have  

∂T ∂P

=μ=− H

 V −T



∂V ∂T

CP

  P

=−

V [1 − T α] CP

214 Chapter 6: The Thermodynamic Properties of Real Substances For the ideal gas,

 μ=

∂T ∂P

 =− H

  V 1 =0 1 − T · CP∗ T

and ∗ CP∗ = CV +

T V α2 PV 1 ∗ ∗ ∗ = CV + T V 2 P = CV + +R = CV κT T T

b. For the van der Waals fluid, we first rewrite the equation of state as

so that

T =

PV a Pb ab − − + R R VR RV 2





∂T ∂V

= P

P 2ab a − 2 + R RV RV 3

and (α)−1 = V



∂T ∂V

 = P

PV 2ab a − + R RV RV 2

Now rewriting the van der Waals equation as P =

RT a − 2 V −b V

(6.2-38b)

allows us to eliminate P from the expression for α to obtain (α)−1 =

TV 2ab TV 2a 2a − + = − (V − b) (V − b) RV RV 2 (V − b) RV 2

An expression for κT is obtained as follows: 

∂P ∂V

 =− T

2a RT + 3 (V − b)2 V

or (κT )−1 =

  TV RT V R 2a 2a − = − (V − b) (V − b)2 V2 (V − b) (V − b) RV 2 =

R α−1 (V − b)

Consequently,  μ=

∂T ∂P

 =− H

V V ⎡ (1 − T α) = − 1− CP CP ⎢ ⎣

and T V α2 = CV + T V α2 κT R = CV + 2a (V − b)2 1− RT V3

CP = CV +

⎤ 1   V 2a V − b ⎥ ⎦ − V −b RT V2



R (V − b)α



6.2 The Evaluation of Thermodynamic Partial Derivatives 215 Finally, it is useful to note that although the molar internal energy can be considered to be a function of any two state variables, an equation of state that gives the internal energy as a function of entropy and volume is, in principle, more useful than an equation of state for the internal energy in terms of temperature and volume or any other pair of state variables. To see that this is so, suppose we had thermal equations of state of the form U = U (S, V ) and U = U (T, V ). From Eq. 6.2-5b it is evident that we could differentiate the first equation of state to get other thermodynamic functions directly, for example,     ∂U ∂U T = and P = − ∂S V ∂V S However, using the second equation of state U (T, V ) and Eq. 6.2-21, we obtain       ∂U ∂P ∂U = CV and =T −P ∂T V ∂V T ∂T V In this case, on differentiation, we do not obtain thermodynamic state functions directly, but rather derivatives of state functions or combinations of state functions and their derivatives. In a similar fashion, it is possible to show that an equation of state that relates H , S , and P , or A, V , and T , or G, P , and T is more useful than other equations of state. Equations of state relating (S , U , and V ), (H , S , and P ), (A, V , and T ), or (G, T , and P ) are called fundamental equations of state, a term first used by the American physicist Josiah Willard Gibbs in 1873. Unfortunately, in general we do not have the information to obtain or construct a fundamental equation of state. More commonly, we have only a volumetric equation of state, that is, an equation relating P , V , and T . Illustration 6.2-4 Showing That a Fundamental Equation of State Contains All the Information about a Fluid Show that from an equation of state relating the Gibbs energy, temperature, and pressure, equations of state for all other state functions (and their derivatives as well) can be obtained by appropriate differentiation.

Solution Suppose we had an equation of state of the form G = G(T, P ). The entropy and volume, as a function of temperature and pressure, are then immediately obtained using Eqs. 6.2-12a, b, and c:  S(T, P ) = −

∂G ∂T



 and

V (T, P ) =

P

∂G ∂P

 T

Since we have G as a function of T and P , these derivatives can be evaluated. Next, the enthalpy and internal energy can be found as follows:  H(T, P ) = G(T, P ) + T S(T, P ) = G(T, P ) − T

∂G ∂T

 P

and  U (T, P ) = G(T, P ) + T S(T, P ) − P V (T, P ) = G(T, P ) − T

∂G ∂T





−P P

∂G ∂P

 T

216 Chapter 6: The Thermodynamic Properties of Real Substances The Hemholtz energy is obtained from  A(T, P ) = G(T, P ) − P V = G(T, P ) − P

∂G ∂P

 T

and the constant-pressure and constant-volume heat capacities can then be found as follows:  CP (T, P ) =

∂H ∂T

 P

  2     ∂  ∂G ∂ G = = −T  G(T, P ) − T ∂T P ∂T P ∂T 2 P

and CV (T, P ) = CP +

T (∂V /∂T )2P = −T (∂V /∂P )T



∂2G ∂T 2





−T P

∂2G ∂T ∂P

2 

∂2G ∂P 2

−1 T

Finally, the isothermal compressibility κT and coefficient of thermal expansion α are found as κT = −

1 V

and α=

1 V





∂V ∂P

∂V ∂T

 =− T

 = P

(∂ 2 G/∂P 2 )T (∂G/∂P )T

(∂ 2 G/∂P ∂T ) (∂G/∂P )T

Thus, if the fundamental equation of state for a substance in the form G = G(T, P ) were available, we could, using these relations, obtain explicit equations relating all other state variables for this substance to temperature and pressure by taking the appropriate derivatives of the fundamental equation.

Questions 1. Why do we need two equations, a volumetric equation of state P = P (T, V ) and a thermal equation of state U = U (T, V ), to define an ideal gas? 2. Can you develop a single equation of state that would completely specify all the properties of an ideal gas? (See Problem 7.7.)

Although fundamental equations of state are, in principle, the most useful thermodynamic descriptions of any substance, it is unlikely that such equations will be available for all fluids of interest to engineers. In fact, frequently only heat capacity and P V T data are available; in Sec. 6.4 we consider how these more limited data are used in thermodynamic problem solving. Illustration 6.2-5 Calculation of the Joule-Thomson Coefficient of Steam from Data in the Steam Tables Use the steam tables in Appendix A.III to evaluate the Joule-Thomson coefficient of steam at 600◦ C and 0.8 MPa.

Solution From Eq. 6.2-27,   μ=

∂T ∂P

V −T

 =− H



∂V ∂T

CP



  P

Vˆ − T =−



∂ Vˆ ∂T

ˆP C

  P

6.2 The Evaluation of Thermodynamic Partial Derivatives 217 

and ˆP = C

ˆ ∂H ∂T

 P

From the superheated vapor section of the steam tables we have the following entries at P = 0.8 MPa: T (◦ C)

Vˆ (m3 /kg)

ˆ (kJ/kg) H

500 600 700

0.4433 0.5018 0.5601

3480.6 3699.4 3924.2

So  ˆP = C

ˆ ∂H ∂T

 P

kJ (3924.2 − 3480.6) ◦ ◦ ˆ ˆ H(700 kJ C) − H(500 C) kg ≈ = = 2.218 (700 − 500)◦ C 200 K kg K

and 

∂ Vˆ ∂T

 P

m3 (0.5601 − 0.4433) ◦ ◦ ˆ ˆ V (700 C) − V (500 C) m3 kg ≈ = = 5.84 × 10−4 ◦ (700 − 500) C 200 K kg K

Therefore,

μ=−

[0.5018 − (600 + 273.15) × 5.84 × 10−4 ] 2.218

m3 kg

kJ kg K

= 3.66 × 10−3

m3 K kJ

Since 1 kJ = 10−3 MPa m3 , and 1 MPa = 10 bar, μ = 3.66

K K = 0.366 MPa bar

Since μ is positive, the temperature of steam at these conditions will decrease if the pressure is reduced, that is, if the steam is isenthalpically expanded. Note: An alternative method of doing this calculation is to use the definition of the Joule-Thomson coefficient, (∂T /∂P )H , directly. For example, at 0.6 MPa we have T (◦ C)

ˆ (kJ/kg) H 500 3482.8 600 3700.9 ˆ 0.6 MPa) = By interpolation we can find the temperature at 0.6 MPa at which: H(T, ◦ ˆ 3699.4 kJ/kg = H(600 C, 0.8 MPa). ◦ ˆ ˆ H(600 0.6 MPa) 3700.9 − 3699.4 600 − T C, 0.6 MPa) − H(T, = = ◦ ◦ C, 0.6 MPa) ˆ ˆ 3700.9 − 3482.8 600 − 500 H(600 C, 0.6 MPa) − H(500

which gives T = 599.3◦ C or ΔT = −0.6977 K. Therefore,  μ=

∂T ∂P

 = H

K K −0.6977 K = 3.49 = 0.349 −0.2 MPa MPa bar

which is in reasonably good agreement with the value obtained above.

218 Chapter 6: The Thermodynamic Properties of Real Substances Illustration 6.2-6 Calculation of CP for Water The constant-pressure heat capacity of liquid water at 25◦ C at 1 bar is 4.18 J/(g K). Estimate the heat capacity of liquid water at 25◦ C at 100 bar. Data: 1 α= V



∂V ∂T



1 = Vˆ

P





∂ Vˆ ∂T

−4

= 2.56 × 10

−1

K

 ,

P

∂α ∂T



= 9.6 × 10−6 K−2

P

and Vˆ = 1.003 cm3 /g.

Solution From Eq. 6.2-37, we have 

∂CP ∂P



 = −T T

∂2V ∂T 2





or equivalently P

ˆP ∂C ∂P





∂ 2 Vˆ ∂T 2

= −T T

 P

Now 

∂ Vˆ ∂T



 = Vˆ α

so

P

∂ 2 Vˆ ∂T 2



 =

P

∂ Vˆ ∂T



 α + Vˆ

P

∂α ∂T

 P

and 

ˆP ∂C ∂P



  = −T α T

∂ Vˆ ∂T



 + Vˆ P

= −298.15 K × 1.003 = −2.890 × 10−3

∂α ∂T

  P

    ∂α = −T α2 Vˆ + Vˆ ∂T P

cm3 [6.5536 × 10−8 + 9.6 × 10−6 ] K−2 g

cm3 1J J × = −2.890 × 10−4 gK 10 bar cm3 g K bar

Assuming that (∂ CˆP /∂P )T is reasonably constant with pressure, we have  ˆP (T, P + ΔP ) − C ˆP (T, P ) = C

ˆP ∂C ∂P

 ΔP T

and ˆP (25◦ C, 100 bar) − C ˆP (25◦ C, 1 bar) = −2.890 × 10−4 C = −0.0286

J × 99 bar g K bar

J gK

Therefore, ˆP (25◦ C, 100 bar) = 4.18 − 0.0286 = 4.1514 J C gK

and we see that the heat capacity of water at these conditions (and indeed that of most liquids and solids) is only very weakly dependent on pressure.

6.3 The Ideal Gas and Absolute Temperature Scales 219 Finally, there are two other equations based on Eq. 6.2-12b that will be useful in analyzing phase behavior. The first of these equations is

∂   ∂T P,N



G T

 =−

H T2

(6.2-39)

The second equation, which is related to the one above, is

⎡  ⎤ G ∂ ⎥ ⎢ ⎢ T  ⎥ =H ⎣ 1 ⎦ ∂ T P,N

(6.2-40)

Various forms of Eqs. 6.2-28 and 6.2-30 are used later in this book.

6.3 THE IDEAL GAS AND ABSOLUTE TEMPERATURE SCALES In Chapters 1 and 3, we introduced the concept of the ideal gas and suggested, without proof, that if the ideal gas were used to establish a scale of temperature, an absolute and universal, or thermodynamic scale would be obtained. We now have developed sufficient thermodynamic theory to prove this to be the case. We take as the starting point for this discussion the facts that the product P V and the internal energy U of an ideal gas are both unspecified but increasing functions of the absolute temperature T and are independent of density, pressure, or specific volume (see Sec. 3.3). To be perfectly general at this point we denote these characteristics by

P V = RT IG = Θ1 (T )

(6.3-1)

U = Θ2 (T )

(6.3-2)

and

IG

where T is the temperature on the ideal gas temperature scale. To prove the equality of the ideal gas and thermodynamic temperature scales it is necessary to establish that Θ1 (T ) is a linear function of T as was suggested in Eqs. 1.4-2 and 3.3-1. It is important to note that in the balance equations the thermodynamic temperature ˙ first appears in the introduction of the entropy function—in particular, in the Q/T term in Eq. 4.1-5. Thus it is the thermodynamic temperature T that appears in all equations derived from Eq. 4.1-5, and therefore in the equations of Sec. 6.2. From Eq. 6.2-21 we have, in general,     ∂P dU = CV dT + T − P dV ∂T V whereas from Eq. 6.3-2 we have, for the ideal gas,

dU =

dΘ2 (T ) dT dT

(6.3-3)

220 Chapter 6: The Thermodynamic Properties of Real Substances The only way to reconcile these two equations is for the coefficient of the dV term in Eq. 6.2-21 to be zero for the ideal gas, that is,   ∂P P =T ∂T V This implies that for changes at constant volume, dT  dP   =  P V T V or

T2 P2 = P1 T1

(6.3-4)

for any two states 1 and 2 with the same specific volume. However, from Eq. 6.3-1 we have, that under these circumstances,

P2 Θ1 (T2 ) = P1 Θ1 (T1 )

(6.3-5)

To satisfy Eqs. 6.3-4 and 6.3-5, Θ1 (T ) must be a linear function of temperature, that is, Θ1 (T ) = RT (6.3-6) where R is a constant related to the unit of a degree (see Sec. 1.4). Using Eq. 6.3-6 in Eq. 6.3-1 yields P V = RT which establishes that the ideal gas temperature scale is also a thermodynamic temperature scale.

6.4 THE EVALUATION OF CHANGES IN THE THERMODYNAMIC PROPERTIES OF REAL SUBSTANCES ACCOMPANYING A CHANGE OF STATE The Necessary Data In order to use the energy and entropy balances for any real substance for which thermodynamic tables are not available, we must be able to compute the changes in its internal energy, enthalpy, and entropy for any change of state. The equations of Sec. 6.2 provide the basis for such computations. However, before we discuss these calculations it is worthwhile to consider the minimum amount of information needed and the form in which this information is likely to be available. Volumetric Equation-of-State Information Clearly, to use Eqs. 6.2-19 through 6.2-22 we need volumetric equation-of-state data, that is, information on the interrelationship between P , V , and T . This information may be available as tables of experimental data, or, more frequently, as approximate equations with parameters that have been fitted to experimental data. Hundreds of analytic equations of state have been suggested for the correlation of P V T data.

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 221 The equation



a P+ 2 V

 (V − b) = RT

(6.2-38a)

or equivalently, The van der Waals equation of state

P =

RT a − 2 V −b V

(6.2-38b)

with a and b constants, was proposed by J. D. van der Waals in 1873 to describe the volumetric or P V T behavior of both vapors and liquids, work for which he was awarded the Nobel Prize in Physics in 1910. The constants a and b can be determined either by fitting this equation to experimental data or, more commonly, from critical-point data as will be described in Sec. 6.6. Values for the parameters for several gases appear in Table 6.4-1. These parameters were computed from critical-point data as described in Sec. 6.6. The van der Waals equation of state is not very accurate and is mainly of historic interest in that it was the first equation capable of predicting the transition between vapor and liquid; this will be discussed in Sec. 7.3. It also is the prototype for modern, more accurate equations of state, such as those of Redlich-Kwong (1949);2

P =

RT a − 1/2 V − b T V (V + b)

(6.4-1)

Soave (1972),3 in which the a/T 1/2 term in Eq. 6.4-1 is replaced with a(T ), a function of temperature; and Peng and Robinson (1976),4 The Peng-Robinson equation of state

P =

RT a(T ) − V − b V (V + b) + b(V − b)

Table 6.4-1 Parameters for the van der Waals Equation of State

2 O.

Gas

a (Pa m6 /mol2 )

b [(m3 /mol) × 105 ]

O2 N2 H2 O CH4 CO CO2 NH3 H2 He

0.1381 0.1368 0.5542 0.2303 0.1473 0.3658 0.4253 0.0248 0.00346

3.184 3.864 3.051 4.306 3.951 4.286 3.737 2.660 2.376

Redlich and J. N. S. Kwong, Chem. Rev. 44, 233 (1949). Soave, Chem. Eng. Sci. 27, 1197 (1972). 4 D.-Y. Peng and D. B. Robinson, IEC Fundam. 15, 59 (1976). 3 G.

(6.4-2)

222 Chapter 6: The Thermodynamic Properties of Real Substances Equations 6.2-38b, 6.4-1, and 6.4-2 are special cases of the general class of equations of state

P =

RT (V − η)θ − V − b (V − b)(V 2 + δV + ε)

(6.4-3)

where each of the five parameters b, θ, δ , ε, and η can depend on temperature. In practice, however, generally only θ is taken to be a function of temperature, and it is adjusted to give the correct boiling temperature as a function of pressure; this will be discussed in Sec. 7.5. Table 6.4-2 gives the parameters of Eq. 6.4-3 for some common equations of state from among the hundreds of this class that have been published. Clearly, many other choices are possible. Numeric values for equation-of-state parameters are commonly obtained in one of two ways. First, parameters can be obtained by fitting the equation to P V T and other data for the fluid of interest; this leads to the most accurate values, but is very tedious. Second, as will be discussed in Sec. 6.7, general relations can be obtained between the equation-of-state parameters and critical-point properties. From these equations, somewhat less accurate parameter values are easily obtained from only critical-point properties. Each of the equations of state discussed here can be written in the form

Z 3 + αZ 2 + βZ + γ = 0

Cubic form of the above equations

Table 6.4-2 Parameters for Cubic Equations of State P =

(6.4-4)

RT −Δ V −b

Author

Year

θ

η

δ

ε

Δ

van der Waals

1873

a

b

0

0

a V2

Clausius

1880

a/T

b

2c

c2

a/T (V + c)2

Berthelot

1899

a/T

b

0

0

a /T V2

Redlich-Kwong

1949

a/ T

b

b

0

a/ T V (V + b)

Soave

1972

θS (T )

b

b

0

θS (T ) V (V + b)

Lee-Erbar-Edmister

1973

θL (T )

η(T )

b

0

θL (T )[V − η(T )] (V − b)(V + b)

Peng-Robinson

1976

θP R (T )

b

2b

− b2

Patel-Teja

1981

θP T (T )

b

b+c

− cb

Note: If η = b, Eq. 6.4-3 reduces to P =

θ RT − 2 V −b V + δV + ε

√

√

θP R (T ) V (V + b) + b(V − b) θP T (T ) V (V + b) + c(V − b)

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 223 Table 6.4-3 Parameters in Eq. 6.4-4 for the Three Equations of State van der Waals −1−B

α β γ

A

− AB

Redlich-Kwong and Soave

Peng-Robinson

−1 A − B − B2 − AB

−1+B A − 3B 2 − 2B − AB + B 2 + B 3

where Z = P V /RT B = bP/RT

and A=

⎧ aP/(RT )2 ⎪ ⎪ ⎨ 2 2.5 ⎪ ⎪ ⎩aP/R T

in the van der Waals, Soave, and Peng-Robinson equations of state in the Redlich-Kwong equation of state

where Z = P V /RT is the compressibility factor, and the parameters α, β , and γ for some representative equations of state are given in Table 6.4-3. Consequently, these equations are said to be cubic equations of state. Many such equations have been suggested in the scientific literature. One should remember that all cubic equations of state are approximate; generally, they provide a reasonable description of the P V T behavior in both the vapor and liquid regions for hydrocarbons, and of the vapor region only for many other pure fluids. The Soave–Redlich-Kwong and the Peng-Robinson equations are, at present, the most commonly used cubic equations of state. A different type of equation of state is the virial equation

B(T ) C(T ) PV + + ··· =1+ RT V V 2

Virial equation of state

(6.4-5)

where B(T ) and C(T ) are the temperature-dependent second and third virial coefficients. Although higher-order terms can be defined in a similar fashion, data generally are available only for the second virial coefficient.5 The virial equation was first used by H. Kamerlingh Onnes in 1901 and is of theoretical interest since it can be derived from statistical mechanics, with explicit expressions obtained for the virial coefficients in terms of the interaction energy between molecules. The virial equation of state is a power series expansion in density (that is, reciprocal volume) about the ideal gas result (P V /RT = 1). With a sufficient number of coefficients, the virial equation can give excellent vapor-phase predictions, but it is not applicable to the liquid phase. When truncated at the B(T ) term, as is usually the case because of a lack of higher virial coefficient data, the virial equation of state can be used only at low densities; as a general rule, it should not be used at pressures above 10 bar for most fluids. There are other, more complicated equations of state that accurately predict the P V T behavior in most of the vapor and liquid regions; such equations contain the reciprocal

5A

recent tabulation, The Virial Coefficients of Pure Gases and Mixtures: A Critical Evaluation, by J. H. Dymond and E. B. Smith, Clarendon Press, Oxford, 1980, provides second virial coefficient data for more than 250 compounds.

224 Chapter 6: The Thermodynamic Properties of Real Substances volume in both integral powers (like the virial equation) and exponential functions. One example is the equation of Benedict, Webb, and Rubin (1940)6 ,   PV A C a  1 1  − + b − =1+ B− RT RT RT 3 V RT V 2   (6.4-6a) γ aα β 2 1 + 2 exp(−γ/V ) + + RT V 5 RT 3 V V where the eight constants a, b, A, B , C , α, β , and γ are specific to each fluid and are obtained by fitting the equation of state to a variety of experimental data. The exponential term in this equation (and others in this class of equations) is meant to compensate for the truncation of the virial series since, if expanded in a Taylor series around V = ∞, the exponential function generates terms of higher order in reciprocal volume. The 20-constant Bender equation (1970)7 ,  

B 1 T C D E F H 2 R+ + 2 + 3 + 4 + 5 + G+ 2 P = exp(−a20 /V ) V V V V V V V V2 (6.4-6b) with B = a1 − a2 /T − a3 /T 2 − a4 /T 3 − a5 /T 4 C = a6 + a7 /T + a8 /T 2 D = a9 + a10 /T E = a11 + a12 /T F = a13 /T G = a14 /T 3 + a15 /T 4 + a16 /T 5 H = a17 /T 3 + a18 /T 4 + a19 /T 5 is another example of an equation of this type. Although such equations provide more accurate descriptions of fluid behavior, including the vapor-liquid phase transition, than simple equations of state, they are useful only with digital computers. Furthermore, because of the amount of data needed the coefficients that appear in these equations are known only for light hydrocarbons and a few other substances. Some recent equations of state are expressed in the form of the Helmholtz energy as a function of temperature and specific volume. The advantage of such an interrelationship is that it is a fundamental equation of state, as discussed in Sec. 4.2. An example of this is the 1995 International Association for the Properties of Water and Steam (IAPWS) formulation8 which contains 56 parameters to very accurately describe the properties of water and steam. Such high accuracy in the thermodynamic properties is needed in a number of applications, including evaluating the efficiency of newly installed steam turbines, which is frequently a contract specification between the power company and the turbine manufacturer. Few other fluids are of sufficient industrial interest to justify the expense of measuring the very large amount of data needed to develop such highaccuracy equations. For an evaluation of volumetric equations of state important in engineering, refer to J. M. Prausnitz, B. E. Poling, and J. P. O’Connell, The Properties of Gases and 6 M.

Benedict, G. B. Webb, and L. C. Rubin, J. Chem. Phys. 8. 334 (1940), and later papers by the same authors. Bender, 5th Symposium on Thermophysical Properties, ASME, New York (1970), p. 227, and later papers by the same author. 8 W. Wagner and A. Pruss, J. Phys. Chem. Ref. Data, 31, 387 (2002) 7 E.

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 225 Liquids, 5th ed. (McGraw-Hill, New York, 2001). In the discussion that follows, we will generally assume that a volumetric equation of state in analytic form is available. Heat Capacity Data It is also evident from Eqs. 6.2-19 through 6.2-22 that data for CP and CV are needed to compute changes in thermodynamic properties. At first glance it might appear that we need data for CV as a function of T and V , and CP as a function of T and P for each fluid over the complete range of conditions of interest. However, from Eqs. 6.2-36 and 6.2-37 it is clear that our need for heat capacity data is much more modest once we have volumetric equation-of-state information. To see this, consider the situation in which we have data for CP as a function of temperature at a pressure P1 , and want CP as a function of temperature at another pressure, P2 . At each temperature we can integrate Eq. 6.2-37 to obtain the desired result:  P2 ,T  P2 ,T  2  ∂ V dCP = CP (P2 , T ) − CP (P1 , T ) = −T dP (6.4-7) ∂T 2 P P1 ,T P1 ,T or

 CP (P2 , T ) = CP (P1 , T ) − T

P2 ,T



P1 ,T



∂2V ∂T 2

dP

(6.4-8)

P

Here we have included T in the limits of integration to stress that the integration is carried out over pressure at a fixed value of temperature. Similarly, for the constantvolume heat capacity, one obtains (from Eq. 6.2-36)  V 2 ,T  2  ∂ P CV (V 2 , T ) = CV (V 1 , T ) + T dV (6.4-9) ∂T 2 V V 1 ,T Therefore, given the volumetric equation of state (or, equivalently, a numerical tabulation of the volumetric data for a fluid) and heat capacity data as a function of temperature at a single pressure or volume, the value of the heat capacity in any other state can be computed. In practice, heat capacity data are tabulated only for states of very low pressure or, equivalently, large specific volume, where all fluids are ideal gases.9 Therefore, if P1 and V 1 are taken as 0 and ∞, respectively, in Eqs. 6.4-8 and 6.4-9, we obtain ∗



P,T

CP (P, T ) = CP (T ) − T P =0,T

and ∗

CV (V , T ) = CV (T ) + T



V ,T

V =∞,T



∂2V ∂T 2





∂2P ∂T 2

dP

(6.4-10)

dV

(6.4-11)

P

 V

where we have used the notation

CP∗ (T ) = CP (P = 0, T ) 9 That

all fluids become ideal gases at large specific volumes is easily verified by observing that all volumetric equations of state (e.g., Eqs. 6.2-33, 6.4-1, 6.4-2, and 6.4-3) reduce to P V = RT in the limit of V → ∞.

226 Chapter 6: The Thermodynamic Properties of Real Substances and

CV∗ (T ) = CV (V = ∞, T )

where the asterisk denotes the ideal gas heat capacity, as in Chapter 3. Data for CP∗ and CV∗ are available in most data reference books, such as the Chemical Engineers’ Handbook10 or The Handbook of Chemistry and Physics.11 This information is frequently presented in the form

CP∗ (T ) = a + bT + cT 2 + dT 3 + · · ·

Ideal gas heat capacity

See Appendix A.II for CP∗ data for some compounds.

The Evaluation of Δ H, Δ U, and Δ S To compute the change in enthalpy in going from the state (T1 , P1 ) to the state (T2 , P2 ), we start from  T2 ,P2 ΔH = H(T2 , P2 ) − H(T1 , P1 ) = dH (6.4-12) T1 ,P1

and note that since enthalpy is a state function, we can compute its change between two states by evaluating the integral along any convenient path. In particular, if the path indicated by the solid line in Fig. 6.4-1 is used (isothermal expansion followed by isobaric heating and isothermal compression), we have, from Eq. 6.2-22,

 

 T2 ,P =0 ∂V CP∗ dT V −T dP + ∂T T1 ,P =0 P1 ,T1 P 

  P2 ,T2 ∂V V −T dP + ∂T P P =0,T2 

Enthalpy change between two states

ΔH =

P =0,T1

(6.4-13)

Alternatively, we could compute the enthalpy change using the path indicated by the dashed line in Fig. 6.4-1—isobaric heating followed by isothermal compression. For

P2

P1

0

10 R.

T1

T2

Figure 6.4-1 Two paths for the integration of Eq. 6.4-12.

H. Perry and D. Green, eds., Chemical Engineers’ Handbook, 6th ed., McGraw-Hill, New York (1984). C. Weast, ed., The Handbook of Chemistry and Physics, Cleveland Chemical Rubber Co. This handbook is updated annually.

11 R.

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 227 this path



  T2 ∂V V −T dP = ΔH = CP dT + CP∗ dT ∂T P P1 ,T1 P1 ,T2 T1 

   T2  P1  2   P2 ,T2 ∂ V ∂V V −T dP − T dP dT + ∂T 2 P ∂T P T1 0 P1 ,T2 



P1 ,T2

P2 ,T2

(6.4-14a) where in going from the first to the second of these equations we have used Eq. 6.4-10. The equality of Eqs. 6.4-13 and 6.4-14 is easily established as follows. First we note that





T2

T1



P1

∂2V ∂T 2 0  P1  T2

T



 dP

dT

P

  ∂2V = T dT dP ∂T 2 P 0 T1  

  P1  T2 ∂V ∂  = − V dT dP  T ∂T P P =0 T1 ∂T P



   P1 ,T1   P1 ,T2  ∂V ∂V = T T − V dP − − V dP ∂T P ∂T P P =0,T2 P =0,T1 

(6.4-15) where we have used the fact that the order of integration with respect to T and P can be interchanged, and then recognized that T (∂ 2 V /∂T 2 ) has an exact differential. Next, substituting Eq. 6.4-15 into Eq. 6.4-14 yields Eq. 6.4-13, verifying that the enthalpy change between given initial and final states is independent of the path used in its computation. Using the solid-line path in Fig. 6.4-1 and Eq. 6.2-20, we obtain Entropy change between two states

 ΔS = −

P =0,T1



P1 ,T1

∂V ∂T





T2

dP + T1

P

CP∗ dT − T



P2 ,T2



P =0,T2

∂V ∂T

 dP P

(6.4-16a) By following the same argument, we can show that the entropy function is also path independent. The path in the V -T plane analogous to the solid-line path in the P -T plane of Fig. 6.4-1 is shown in Fig. 6.4-2. Here the gas is first isothermally expanded to zero pressure (and, hence, infinite volume), then heated (at V = ∞) from T1 to T2 , and finally compressed to a specific volume V 2 . The entropy and internal energy changes from Eqs. 6.2-19 and 6.2-21 are



V =∞,T1

ΔS = V 1 ,T1



∂P ∂T





T2

dV + V

T1

CV∗ dT + T



V 2 ,T2

V =∞,T2



∂P ∂T

 dV (6.4-17a) V

228 Chapter 6: The Thermodynamic Properties of Real Substances _V = ∞

V _1

V _2

T1

and

Figure 6.4-2 Integration path in the V -T plane.

T2

    ∂P T ΔU = − P dV ∂T V V 1 ,T1    T2  V 2 ,T2   ∂P T + CV∗ dT + − P dV ∂T V T1 V =∞,T2 

V =∞,T1

(6.4-18)

It can be easily shown that alternative paths lead to identical results for ΔS and ΔU . Note that only CP∗ , CV∗ , and terms related to the volumetric equation of state appear in Eqs. 6.4-13, 6.4-14a, 6.4-16a, 6.4-17a, and 6.4-18. Thus, as has already been pointed out, we do not need heat capacity data at all densities, but merely in the low-density (ideal gas) limit and volumetric equation-of-state information. Given CP∗ or CV∗ data and volumetric equation-of-state data (in either analytic or tabular form), it is possible to compute ΔH , ΔU , and ΔS for any two states of a fluid; given the value of S in any one state, it is also possible to compute ΔG and ΔA. Thus, we now have the equations necessary to construct complete tables or charts interrelating H , U , S , T , P , and V , such as those in Chapter 3. Although the process of constructing thermodynamic properties tables and charts is tedious (see Illustration 6.4-1), their availability, as we saw in Chapters 3, 4 and 5, makes it possible to use the balance equations to solve thermodynamic problems for real fluids quickly and with good accuracy. It is interesting to compare the equations just derived with the analogous results for an ideal gas. Since

 Ideal gas results

∂V ∂T

IG

 =

P

∂V ∂T

IG P

V R = = T P

 so V

IG

−T

∂V ∂T

IG =0 P

and



∂P ∂T

IG = V

R P = V T

we have

 H IG (T2 , P2 ) − H IG (T1 , P1 ) =

T2 ,P =0 T1 ,P =0

CP∗ dT

(6.4-14b)

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 229



T1 ,P =0

S (T2 , P2 ) − S (T1 , P1 ) = − IG

IG

 =

T1 ,P1 T2 CP∗

T

T1

 S (T2 , V 2 ) − S (T1 , V 1 ) = + IG



T1 ,V 1 T2

= T1



T2

T1

CP∗ dT − T



T2 ,P2

T2 ,P =0

R dP P

P2 (6.4-16b) P1  T2 ,V 2  T2 ∗ R CV R dT + dV + dV V T V T1 T2 ,V =∞

dT − R ln

T1 ,V =∞

IG

R dP + P

V CV∗ dT + R ln 2 T V1

(6.4-17b)

Thus, comparing Eqs. 6.4-14a and b, Eqs. 6.4-17a and b, and Eqs. 6.4-16a and b, we have for the real fluid Enthalpy and entropy changes in terms of departure functions

H(T2 , P2 ) − H(T1 , P1 ) = H IG (T2 , P2 ) − H IG (T1 , P1 ) 



   T2 ,P2  T1 ,P =0 ∂V ∂V V −T dP + V −T dP + ∂T P ∂T P T1 ,P1 T2 ,P =0 = H IG (T2 , P2 ) − H IG (T1 , P1 ) + (H − H IG )T2 ,P2 − (H − H IG )T1 ,P1 (6.4-19)

S(T2 , P2 ) − S(T1 , P1 ) = S IG (T2 , P2 ) − S IG (T1 , P1 ) 



 T1 ,P =0   T2 ,P2  R R ∂V ∂V dP − dP − − − ∂T P P ∂T P P T1 ,P1 T2 ,P =0 = S IG (T2 , P2 ) − S IG (T1 , P1 ) + (S − S IG )T2 ,P2 − (S − S IG )T1 ,P1 (6.4-20) and

S(T2 , V 2 ) − S(T1 , V 1 ) = S IG (T2 , V 2 ) − S IG (T1 , V 1 )      T2 ,V 2   T1 ,V 1 =∞  R R ∂P ∂P − − dV + dV + ∂T V V ∂T V V T1 ,V 1 T2 ,V =∞ = S IG (T2 , V 2 ) − S IG (T1 , V 1 ) + (S − S IG )T2 ,V 2 − (S − S IG )T1 ,V 1 (6.4-21) where Departure functions



 ∂V V −T dP = ∂T P T,P =0 

(H − H )T,P IG

T,P

(6.4-22)

230 Chapter 6: The Thermodynamic Properties of Real Substances 

 T,P  R ∂V IG (S − S )T,P = − − dP ∂T P P T,P =0 and

(6.4-23)

(S − S IG )T,V = S(T, V ) − S IG (T, V )    T,V  R ∂P dV − = ∂T V V T,V = ∞

(6.4-24)

The interpretation of Eqs. 6.4-19, 6.4-20, and 6.4-21 is clear: The changes in enthalpy and entropy of a real fluid are equal to those for an ideal gas undergoing the same change of state plus the departure of the fluid from ideal gas behavior at the end state less the departure from ideal gas behavior of the initial state. These departure functions, given by Eqs. 6.4-22, 6.4-23, and 6.4-24, can be computed once the fluid equation of state is known. Before leaving this subject, we note that although Eqs. 6.4-22, 6.4-23, and 6.4-24 are useful for calculating the enthalpy and entropy departures from ideal gas behavior for some equations of state, their form is less helpful for the van der Waals, PengRobinson and other equations of state considered in this section in which V and T are the convenient independent variables.12 In such cases it is useful to have alternative expressions for the departure functions at fixed temperature and pressure. To obtain such expressions, we start with Eqs. 6.4-22 and 6.4-23 and use 1 P dP = d(P V ) − dV (6.4-25) V V and the triple product rule (Eq. 6.1-6a) in the form             ∂V ∂V ∂P ∂T ∂P   = −1 or dP  = − dV  ∂T P ∂V T ∂P V ∂T P ∂T V T T (6.4-26) to obtain  

 P V (T,P )  P ∂V V −T dP = d(P V ) ∂T P P =0 P V = RT    V = V (T,P )

T

+ V =∞

 = (P V − RT ) +

∂P ∂T

−P V

V =V (T,P ) V =∞





dV

    ∂P − P dV T ∂T V

therefore Useful form of enthalpy departure equation

 H(T, P ) − H (T, P ) = RT (Z − 1) +

V = V (T,P )

IG

V =∞

    ∂P − P dV T ∂T V (6.4-27)

12 That

is, with such equations of state, it is easier to solve for P given V and T than for V given P and T . Consequently, derivatives of P with respect to V or T are more easily found in terms of V and T than are derivatives of V with respect to P and T (try it!).

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 231 where Z = P V /RT . Similarly, from

dV d(P V ) R R −R = Rd ln (P V ) − dV dP = R P PV V V and Eq. 6.4-23 we obtain 

  P  P V (T,P ) R ∂V − d ln (P V ) dP = R ∂T P P =0 P P V = RT



V = V (T,P )



+ V =∞

 = R ln and

PV RT



S(T, P ) − S (T, P ) = R ln Z +





∂P ∂T

 V

R − V



V = V (T,P )

+ V =∞

V = V (T,P )

IG

V =∞



∂P ∂T



 V

dV

∂P ∂T

R − V

 V

R − V

 dV

 dV

(6.4-28)

The desired equations, Eqs. 6.4-27 and 6.4-28, are general and can be used with any equation of state. Using, for example, the Peng-Robinson equation of state, one obtains (see Problem 6.2) Enthalpy and entropy changes for the Peng-Robinson equation of state

 T H(T, P ) − H IG (T, P ) = RT (Z − 1) +

 da √

−a dT Z + (1 + 2)B √ √ ln 2 2b Z + (1 − 2)B (6.4-29)

and

da √

Z + (1 + 2)B dT √ S(T, P ) − S (T, P ) = R ln(Z − B) + √ ln 2 2b Z + (1 − 2)B IG

(6.4-30) Demonstrating how an equation of state is used in the construction of a thermodynamics properties chart

where

Z = P V /RT

and B = P b/RT

Illustration 6.4-1 Making of a Thermodynamic Properties Chart As an introduction to the problem of constructing a chart or table of the thermodynamic properties of a real fluid, develop a thermodynamic properties chart for oxygen over the temperature range of −100◦ C to +150◦ C and a pressure range of 1 to 100 bar. (A larger temperature range, including the vapor-liquid two-phase region, will be considered in Chapter 7.) In particular,

232 Chapter 6: The Thermodynamic Properties of Real Substances calculate the compressibility factor, specific volume, molar enthalpy, and molar entropy as a function of temperature and pressure. Also, prepare a pressure-volume plot, a pressure-enthalpy plot (see Fig. 3.4-2), and a temperature-entropy plot (see Fig. 3.4-3) for oxygen. Data: For simplicity we will assume oxygen obeys the Peng-Robinson equation of state and has an ideal gas heat capacity given by CP∗



J mol K



= 25.46 + 1.519 × 10−2 T − 0.7151 × 10−5 T 2 + 1.311 × 10−9 T 3

We will choose the reference state of oxygen to be the ideal gas state at 25◦ C and 1 bar: H IG (T = 25◦ C, P = 1 bar) = 0

and

S IG (T = 25◦ C, P = 1 bar) = 0

As will be explained shortly, the Peng-Robinson parameters for oxygen are a(T ) = 0.45724

R2 Tc2 α(T ) Pc

RTc Pc  

b(T ) = 0.07780

1/2

[α(T )]

=1+κ 1−

T Tc



where κ = 0.4069, Tc = 154.6 K is the critical temperature of oxygen, and Pc = 5.046 MPa is its critical pressure.

Solution I. Volume The volume is the easiest of the properties to calculate. For each value of temperature we compute values for a(T ) and b, and then at each pressure solve the equation P =

RT a(T ) − V −b V (V + b) + b(V − b)

for the volume, or equivalently (and preferably), solve the equation Z 3 + (−1 + B)Z 2 + (A − 3B 2 − 2B)Z + (−AB + B 2 + B 3 ) = 0

with B = bP/RT and A = aP/(RT )2 (see Eq. 6.4-4 and Table 6.4-3) for the compressibility factor Z = P V /RT , from which the volume is easily calculated as V = ZRT /P . Repeating the calculation a number of times leads to the entries in Table 6.4-4. These calculations can be done using the Visual Basic computer program described in Appendix B.I-2, the DOS-based program PR1 described in Appendix B.II-1, the MATHCAD worksheet described in Appendix B.III, or the MATLAB program described in Appendix B.IV and included on the web site for this book. Since the pressure and specific volume vary by several orders of magnitude over the range of interest, these results have been plotted in Fig. 6.4-3 as ln P versus ln V . Remember that for the ideal gas P V = RT , so that an ideal gas isotherm on a log-log plot is a straight line. (Note that Fig. 6.4-3 also contains isotherms for temperatures below −100◦ C. The calculation of these will be discussed in Sec. 7.5.)

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 233 100 Critical point

50 T = –200°C

T = 150°C T = 100°C T = 50°C T = 0°C T = –50°C T = –100°C

T = –125°C

T = –150°C 10

d

te

ra

tu

5

Sa

Saturate d liquid

P, bar

20

r

po

va

T = –175°C 2

1.0 0.01

0.02

0.05

0.1

0.2

0.5 1.0 V _ , m3/kmol

2

5

10

20

50

Figure 6.4-3 Pressure-volume diagram for oxygen calculated using the Peng-Robinson equation of state.

II. Enthalpy To compute the difference in the enthalpy of oxygen between a state of temperature T and pressure P and the ideal gas reference state at 25◦ C and 1 bar, we use H(T, P ) − H IG (T = 25◦ C, P = 1 bar) = H(T, P ) − H IG (T, P ) + H IG (T, P ) − H IG (T = 25◦ C, P = 1 bar)  T = (H − H IG )T,P + CP∗ dT T =298.15K

= (H − H IG )T,P + 25.46(T − 298.15) +

−

1.519 × 10−2 2 (T − 298.152 ) 2

1.311 × 10−9 4 0.7151 × 10−5 3 (T − 298.153 ) + (T − 298.154 ) 3 4

The quantity (H − H IG )T,P is computed using Eq. 6.4-29 and the value of the compressibility factor Z found earlier at each set of (T, P ) values. The values of enthalpy computed in this manner appear in Table 6.4-4 and have been plotted in Fig. 6.4-4 as a pressure and enthalpy diagram. The Peng-Robinson program (Appendices B.I-3 and B.II-1) was used for this calculation as well. [Note that at T = 25◦ C and P = 1 bar, oxygen is not quite an ideal gas; Z = 0.9991, not unity. Consequently, (H − H IG )T =25◦ C,P =1 bar = −9.44 J/mol, so that the enthalpy of oxygen at T = 25◦ C, P = 1 bar is −9.44 J/mol, whereas if it were an ideal gas at these conditions its enthalpy would be zero.] Once again, lower-temperature results, including the two-phase region, appear in Fig 6.4-6. The basis for those calculations will be discussed in Sec. 7.5.

234 P = 1 bar Z V H S P = 2 bar Z V H S P = 5 bar Z V H S P = 10 bar Z V H S P = 20 bar Z V H S P = 30 bar Z V H S

T (◦ C) 0.9963 16.4200 −2898.69 −11.81 0.9925 8.1760 −2917.04 −17.63 0.9813 3.2330 −2972.47 −25.44 0.9626 1.5860 −3066.16 −31.53 0.9252 0.7621 −3258.71 −37.95 0.8878 0.4876 −3458.44 −42.01

0.9889 7.1190 −3626.18 −21.38 0.9722 2.7990 −3694.40 −29.33 0.9439 1.3590 −3811.36 −35.55 0.8856 0.6375 −4059.21 −42.27 0.8245 0.3957 −4329.71 −46.72

−75

0.9945 14.3200 −3603.75 −15.59

−100

0.9246 0.5718 −2638.59 −38.11

0.9491 0.8804 −2480.87 −34.25

0.9743 1.8080 −2325.11 −28.00

0.9871 3.6630 −2248.04 −22.00

0.9948 9.2290 −2202.08 −14.23

0.9974 18.5100 −2186.80 −8.43

−50

0.9479 0.6519 −1842.18 −34.73

0.9646 0.9951 −1713.17 −30.99

0.9820 2.0260 −1584.04 −24.85

0.9909 4.0890 −1519.52 −18.90

0.9963 10.2800 −1480.03 −11.17

0.9982 20.5900 −1480.83 −5.38

−25

0.9636 0.7295 −1057.37 −31.72

0.9751 1.1070 −949.56 −28.06

0.9872 2.2420 −840.75 −21.99

0.9936 4.5130 −786.05 −16.09

0.9974 11.3300 −753.13 −8.38

0.9987 22.6800 −742.14 −2.59

0

0.9745 0.8053 −278.07 −28.89

0.9825 1.2180 −186.68 −25.39

0.9910 2.4570 −93.94 −19.38

0.9954 4.9350 −47.12 −13.50

0.9982 12.3700 −18.88 −5.81

0.9991 24.7700 −9.44 −0.02

25

0.9824 0.8798 499.24 −26.48

0.9878 1.3270 577.48 −22.92

0.9937 2.6700 657.16 −16.97

0.9968 5.3560 697.51 −11.10

0.9987 13.4200 721.89 −3.43

0.9994 26.8500 730.04 2.36

50

Table 6.4-4 Thermodynamic Properties of Oxygen Calculated Using the Peng-Robinson Equation of State

0.9882 0.9535 1276.72 −24.17

0.9918 1.4350 1344.15 −20.64

0.9957 2.8820 1413.02 −14.71

0.9978 5.7760 1447.97 −8.86

0.9991 14.4600 1469.10 −1.19

0.9996 28.9300 1476.18 4.58

75

0.9925 1.0260 2055.76 −22.00

0.9947 1.5430 2114.16 −18.50

0.9972 3.0940 2173.92 −12.60

0.9986 6.1960 2204.29 −6.77

0.9994 15.5000 2222.68 0.87

0.9997 31.0200 2228.83 6.67

100

0.9958 1.0990 2837.29 −20.00

0.9969 1.6500 2888.07 −16.50

0.9983 3.3050 2940.01 −10.61

0.9991 6.6150 2966.37 −4.79

0.9996 16.5500 2982.49 2.87

0.9998 33.1000 2987.86 8.64

125

0.9983 1.1710 3621.95 −18.07

0.9987 1.7570 3666.08 −14.60

0.9992 3.5150 3711.36 −8.73

0.9996 7.0330 3734.44 −2.92

0.9998 17.5900 3748.42 4.73

0.9999 35.1800 3753.10 10.50

150

235

0.8508 0.3504 −3665.63 −45.14 0.8143 0.2683 −3880.25 −47.77 0.7788 0.2139 −4101.78 −50.09 0.7450 0.1753 −4328.89 −52.21 0.7135 0.1469 −4559.08 −54.19 0.6852 0.1254 −4788.49 −56.03 0.6611 0.1089 −5012.01 −57.75

0.7602 0.2736 −4629.51 −50.35 0.6922 0.1993 −4968.26 −53.66 0.6202 0.1448 −5358.88 −56.90 0.5471 0.1125 −5878.46 −60.27 0.4833 0.0870 −6296.72 −63.65 0.4444 0.0711 −6719.52 −66.53 0.4297 0.0619 −7026.48 −68.69

0.7851 0.1457 −3750.23 −51.74

0.8003 0.1650 −3596.54 −50.35

0.8173 0.1896 −3439.53 −48.86

0.8361 0.2216 −3280.08 −47.23

0.8564 0.2648 −3119.39 −45.42

0.8781 0.3258 −2958.41 −43.39

0.9008 0.4178 −2797.93 −41.01

0.8571 0.1768 −2711.04 −47.32

0.8669 0.1987 −2593.54 −46.09

0.8778 0.2264 −2473.08 −44.75

0.8898 0.2623 −2350.13 −43.27

0.9029 0.3105 −2225.13 −41.62

0.9170 0.3784 −2098.53 −39.72

0.9320 0.4807 −1970.75 −37.50

V [=] m3 /kmol; H [=] J/mol = kJ/kmol; S [=] J/(mol K) = kJ/(kmol K).

P = 40 bar Z V H S P = 50 bar Z V H S P = 60 bar Z V H S P = 70 bar Z V H S P = 80 bar Z V H S P = 90 bar Z V H S P = 100 bar Z V H S 0.9034 0.2052 −1765.61 −43.69

0.9096 0.2295 −1670.95 −42.54

0.9167 0.2602 −1573.76 −41.29

0.9246 0.3000 −1472.25 −39.90

0.9332 0.3532 −1372.64 −38.34

0.9426 0.4282 −1269.13 −36.54

0.9528 0.5409 −1163.97 −34.40

0.9348 0.2317 −869.40 −40.55

0.9387 0.2586 −790.91 −39.46

0.9433 0.2923 −710.26 −38.27

0.9484 0.3358 −627.54 −36.94

0.9541 0.3942 −542.81 −35.44

0.9603 0.4761 −456.28 −33.70

0.9671 0.5994 −367.98 −31.61

0.9570 0.2571 −1.92 −37.75

0.9593 0.2864 64.27 −36.70

0.9621 0.3231 132.35 −35.55

0.9652 0.3705 202.27 −34.26

0.9698 0.4338 273.98 −32.80

0.9729 0.5228 347.92 −31.11

0.9774 0.6565 422.53 −29.06

0.9732 0.2817 847.77 −35.22

0.9743 0.3134 904.24 −34.20

0.9757 0.3530 962.34 −33.08

0.9775 0.4042 1022.07 −31.82

0.9796 0.4726 1083.39 −30.39

0.9821 0.5686 1146.30 −28.73

0.9850 0.7128 1210.75 −26.72

0.9851 0.3056 1686.27 −32.89

0.9854 0.3397 1734.77 −31.90

0.9859 0.3823 1784.71 −30.80

0.9866 0.4373 1836.08 −29.57

0.9877 0.5107 1888.88 −28.16

0.9890 0.6137 1943.10 −26.52

0.9906 0.7683 1998.73 −24.54

0.9941 0.3291 2517.77 −30.74

0.9937 0.3655 2559.61 −29.76

0.9935 0.4111 2602.71 −28.68

0.9935 0.4698 2647.08 −27.46

0.9937 0.5483 2692.73 −26.07

0.9942 0.6582 2739.64 −24.45

0.9949 0.8233 2787.83 −22.50

1.0010 0.3522 3345.07 −28.72

1.0001 0.3909 3381.24 −37.76

0.9993 0.4395 3418.53 −26.69

0.9987 0.5020 3456.93 −25.49

0.9983 0.5854 3496.47 −24.12

0.9981 0.7023 3537.15 −22.51

0.9981 0.8779 3578.97 −20.56

236 Chapter 6: The Thermodynamic Properties of Real Substances 100

T = –120°C

T =–125°C T = –130°C

P, bar

T = –140°C

–125°C

T = –150°C

10

–100°C –75°C –50°C

T = –160°C

–25°C 0°C 25°C 50°C

T = –170°C

75°C 100°C

T = –175°C

125°C

T = –180°C

150°C

T = –183°C 1 –16

–14

–12

–10

–8

–6

–4

–2

0

2

4

6

H, kJ/mol —

Figure 6.4-4 Pressure-enthalpy diagram for oxygen calculated using the Peng-Robinson equation of state. III. Entropy To compute the difference in the entropy of oxygen between the state (T , P ) and the ideal gas reference state at 25◦ C and 1 bar, we use S(T, P ) − S IG (T = 25◦ C, P = 1 bar) = S(T, P ) − S IG (T, P ) + S IG (T, P ) − S IG (T = 25◦ C, P = 1 bar)    T P CP∗ IG = (S − S )T,P + dT − R ln 1 bar T =298.15K T = (S − S IG )T,P + 25.46 ln

T + 1.519 × 10−2 (T − 298.15) 298.15

1.311 × 10−9 3 0.7151 × 10−5 2 (T − 298.152 ) + (T − 298.153 ) 2 3   P − R ln 1 bar

−

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 237 150

100

P=1 00 ba r P=5 0 bar P = 20 bar P = 10 bar P=5b ar P=2 bar P=1 bar

50

T (°C)

0

–50

Critical point –100

P = 20 bar P = 10 bar P = 5 bar P = 2 bar P = 1 bar

–150

–200 –120

–100

–80

–60

–40

–20

0

20

_S, J/(mol K)

Figure 6.4-5 Temperature-entropy diagram for oxygen calculated using the Peng-Robinson equation of state. The quantity (S − S IG )T,P is computed using Eq. 6.4-30, and the value of the compressibility factor Z is found earlier at each T and P . The values of the entropy computed in this manner with the programs in either Appendix B.I or B.II (on the website for this book) appear in Table 6.4-4 and as a temperature-entropy diagram in Fig. 6.4-5.

Comments For illustrative purposes, the Peng-Robinson equation of state was used here. Generally, if one were going to take the time and effort to construct tables or plots such as those developed here, much more complicated equations of state, for example, the Bender equation (Eq. 6.4-6b), would be used to obtain more accurate results. Note that in this illustration we chose the hypothetical ideal gas at 25◦ C and 1 bar as the reference state for the calculation, rather than the real gas at those conditions. Since the actual state of oxygen from the Peng-Robinson equation is not quite ideal, we have Z(1 bar, 25◦ C) = 0.9991 H(1 bar, 25◦ C) = −9.44 kJ/mol S(1 bar, 25◦ C) = −0.02 kJ/(mol K)

At first glance it might appear that to avoid such differences one could choose zero pressure as the reference state, since all gases are then ideal. The reason this is not done is that the entropy of a gas is infinite at zero pressure (see, for example, Eq. 4.4-3).

238 Chapter 6: The Thermodynamic Properties of Real Substances Table 6.4-5 Molar Volume, Coefficient of Thermal Expansion, and Coefficient of Isothermal Compressiblity at 20◦ C for Some Condensed Phases

Solids Copper Graphite Platinum Silver Sodium chloride Liquids Benzene Carbon tetrachloride Ethanol Methanol Water Mercury

V (cc/mol)

α × 104 (K−1 )

7.12 5.31 9.10 10.27 27.02

0.492 0.24 0.265 0.583 1.21

88.87 96.44 58.40 40.47 18.02 14.81

12.4 12.4 11.2 12.0 2.07 1.81

κT × 106 (bar−1 ) 0.77 2.96 0.375 1.0 4.15 92.8 102 109 118 44.7 3.80

In Chapter 2 we made the assumption that some thermodynamic properties of condensed phases, that is, liquids and solids, were independent of pressure. We now have the equations that allow us to examine and improve upon this assumption. We will do this using the data in Table 6.4-5, which includes the molar volume V , the coefficient of thermal expansion α = (1/V )(∂V /∂T )P , and the isothermal compressibility κT = −(1/V )(∂V /∂P )T for a number of substances. Here we are interested in the change in thermodynamic properties with pressure when α and κT are the only information available, as is typically the case for a liquid or a solid. To compute the change in volume of a condensed phase from the pressure P to the pressure P +ΔP , assuming that the isothermal compressibility is independent of pressure, we use   ∂V V (T, P + ΔP ) = V (T, P ) + ΔP ∂P T (6.4-31)

= V (T, P ) − κT V (T, P )ΔP = V (T, P )[1 − κT ΔP ] Note that for an incompressible fluid κT = 0, so that V (T, P + ΔP ) = V (T, P ). Illustration 6.4-2 Effect of Pressure on the Volume of Liquids and Solids Compute the change in the molar volume of copper, sodium chloride, ethanol, and mercury for the very large pressure change of going from 1 bar to 1000 bar at 20◦ C.

Solution Using the equation, we have cc × [1 − 0.77 × 10−6 bar−1 × 999 bar] mol cc cc = 7.12 × [1 − 0.000769] = 7.115 mol mol

V Cu (20◦ C, 1000 bar) = 7.12

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 239 V NaCl (20◦ C, 1000 bar) = 27.02

cc cc × [1 − 4.15 × 10−6 bar−1 × 999 bar] = 26.91 mol mol

V EtOH (20◦ C, 1000 bar) = 58.40

V Hg (20◦ C, 1000 bar) = 14.81

cc cc [1 − 109 × 10−6 bar−1 × 999 bar] = 52.04 mol mol

cc cc [1 − 3.80 × 10−6 bar−1 × 999 bar] = 14.75 mol mol

Comment We see that the change in volume with a factor of 1000 change in pressure is very small for the two solids and the liquid metal considered, but that there is an 8.9 percent change in volume for the more compressible liquid ethanol.

From Eq. 6.2-22 we have that the change in enthalpy with pressure at constant temperature is  

∂V dH = V − T (6.4-32) dP = [V − T V α]dP ∂T P Therefore, assuming that V and α are essentially constant with pressure gives

H(T, P + ΔP ) − H(T, P ) = V [1 − T α]ΔP

(6.4-33)

Illustration 6.4-3 Effect of Pressure on the Enthalpy of Liquids and Solids Compute the change in enthalpy of the four substances considered in the previous example on going from 1 bar to 1000 bar at 20◦ C.

Solution H Cu (20◦ C, 1000 bar) − H Cu (20◦ C, 1 bar)  cc  = 7.12 1 − 293.15 K × 0.492 × 10−4 K−1 × 999 bar mol J cc bar cc [1 − 0.0144] × 999 bar = 7010.3 = 701 = 7.12 mol mol mol H NaCl (20◦ C, 1000 bar) − H NaCl (20◦ C, 1 bar) cc = 27.02 × [1 − 293.15 × 1.21 × 10−4 ] × 999 bar mol J = 2604 mol

240 Chapter 6: The Thermodynamic Properties of Real Substances H EtOH (20◦ C, 1000 bar) − H EtOH (20◦ C, 1 bar) = 58.40

cc × [1 − 293.15 × 11.2 × 10−4 ] × 999 bar mol

= 3919

J mol

and H Hg (20◦ C, 1000 bar) − H Hg (20◦ C, 1 bar) = 14.81

cc × [1 − 293.15 × 1.81 × 10−4 ] × 999 bar mol

= 1401

J mol

Comment Note that in each case the enthalpy change due to the pressure change is large and cannot be neglected. In all cases the contribution from the V ΔP dominates, with the contribution from the isothermal compressibility being only 1.4 percent of the total for copper, 3.6 percent for sodium chloride, and 5.3 percent for liquid mercury. However, this contribution is 32.8 percent for liquid ethanol.

The easiest way to compute the change in internal energy with pressure at constant temperature for a condensed phase is to note that U = H − P V . Therefore, if we can assume the volume of a condensed phase does not change much with pressure, we have

U (T, P + ΔP ) − U (T, P ) = H(T, P + ΔP ) − (P + ΔP )V (T, P ) − [H(T, P ) − P V (T, P )]

= H(T, P + ΔP ) − H(T, P ) − V (T, P )ΔP

(6.4-34)

= V ΔP − V T αΔP − V ΔP = −V T αΔP Illustration 6.4-4 Effect of Pressure on the Internal Energy of Liquids and Solids Compute the change in internal energy of the four substances considered in the previous illustrations on going from 1 bar to 1000 bar at 20◦ C.

Solution U Cu (20◦ C, 1000 bar) − U Cu (20◦ C, 1 bar) = −V T αΔP = −7.12

cc 1J × 293.15 K × 0.492 × 10−4 K−1 × 999 bar × mol 10 cc bar

= −10.3

J mol

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 241 U NaCl (20◦ C, 1000 bar) − U NaCl (20◦ C, 1 bar) = −95.7

J mol

U EtOH (20◦ C, 1000 bar) − U EtOH (20◦ C, 1 bar) = −1915.5 U Hg (20◦ C, 1000 bar) − U Hg (20◦ C, 1 bar) = −78.5

J mol

J mol

Comment With the exception of ethanol, the changes in internal energy are small, much smaller than the changes in enthalpy for these substances. Indeed, for moderate changes in pressure, rather than the large pressure change considered here, the change in internal energy with pressure can be neglected.

The change in entropy with pressure at constant temperature is, from Eq. 6.2-20,   ∂V dS = − dP (6.4-35) ∂T P Assuming the coefficient of thermal expansion is independent of pressure, we obtain

S(T, P + ΔP ) − S(T, P ) = −αV ΔP

(6.4-36)

Illustration 6.4-5 Effect of Pressure on the Entropy of Liquids and Solids Compute the change in entropy of the four substances considered in the previous example on going from 1 bar to 1000 bar.

Solution S Cu (20◦ C, 1000 bar) − S Cu (20◦ C, 1 bar) cc 1J = 7.12 × 0.492 × 10−4 K−1 × 999 bar × mol 10 cc bar J = −0.035 mol K S NaCl (20◦ C, 1000 bar) − S NaCl (20◦ C, 1 bar) = −0.327

J mol K

S EtOH (20◦ C, 1000 bar) − S EtOH (20◦ C, 1 bar) = −6.534

J mol K

S Hg (20◦ C, 1000 bar) − S Hg (20◦ C, 1 bar) = −0.268

J mol K

Comment Again, with the exception of ethanol, the changes in entropy are quite small. For moderate changes in pressure the entropy change of a condensed phase with pressure is negligible.

242 Chapter 6: The Thermodynamic Properties of Real Substances The illustrations above show that unless the pressure change is large, and the condensed phase is quite compressible (as is the case for ethanol here), the changes in volume, internal energy, and entropy with changing pressure are quite small. However, the change in enthalpy can be significant and must be accounted for. Indeed, it is this enthalpy difference that is important in computing the work required to pump a liquid from a lower pressure to a higher pressure.

Illustration 6.4-6 Energy Required to Pressurize a Liquid Compute the minimum work required and the heat that must be removed to isothermally pump liquid ethanol from 20◦ C and 1 bar to 1000 bar in a continuous process.

Solution Consider the pump and its contents to be the system around which balance equations are to be written. N2 N1

The steady-state mass balance (on a molar basis) is dN = 0 = N˙ 1 + N˙ 2 dt

or

N˙ 2 = −N˙ 1 = −N˙

The steady-state energy balance is dU ˙ = 0 = N˙ 1 H 1 + N˙ 2 H 2 + Q˙ + W dt

or ˙ W Q˙ = − + (H 2 − H 1 ) N˙ N˙

The steady-state entropy balance (with S˙ gen = 0 for minimum work) is Q˙ dS = 0 = N˙ 1 S 1 + N˙ 2 S 2 + dt T

or Q˙ = T (S 2 − S 1 ) N˙

From the previous illustrations, we have S 2 − S 1 = S(20◦ C, 1000 bar) − S(20◦ C, 1 bar) = −6.534 J/(mol K)

and H 2 − H 1 = H(20◦ C, 1000 bar) − H(20◦ C, 1 bar) = 3919 J/mol

6.4 Thermodynamic Properties of Real Substances Accompanying a Change of State 243 Therefore, Q˙ J J = −1915 = 293.15 K × (−6.534) mol K mol N˙

and ˙ J W = 1915 + 3919 = 5834 ˙ mol N

Comment If we neglected the effect of thermal expansion, that is, set α = 0, then from Eq. 6.4-33, H(T, P + ΔP ) − H(T, P ) = V [1 − T α]ΔP = V ΔP

and from Eq. 6.4-36, S(T, P + ΔP ) − S(T, P ) = −αV ΔP = 0

Then, for the problem here, H(20◦ C, 1000 bar) − H(20◦ C, 1 bar) = 58.40

J cc 1J = 5834 × 999 bar × mol 10 cc bar mol

and S(20◦ C, 1000 bar) − S(20◦ C, 1 bar) = 0

so that Q˙ =0 N˙

and

˙ W J = 5834 ˙ mol N

Note that the total work is the same as before for this isothermal pressurization (why?), but in this case no heat would have to be removed to keep the liquid at constant temperature. To pressurize liquid mercury, which is closer to being incompressible than ethanol, over the same conditions one obtains (without making the assumption that α = 0)   Q˙ J J = 293.15 K × −0.286 = −78.6 mol K mol N˙

and ˙ J J W = 78.6 + 1401 = 1478.6 mol mol N˙

Much less work is needed for this less compressible fluid since very little work is being done to change the volume of the liquid; the work is largely going to change the fluid pressure.

Illustration 6.4-7 Computing the Difference between CP and CV Compute the difference between the constant-pressure and constant-volume heat capacities at 20◦ C for the four condensed phases considered in the previous illustrations.

244 Chapter 6: The Thermodynamic Properties of Real Substances Solution From Eq. 6.2-35 we have CP − CV = T V (CP − CV )Cu = 293.15 K × 7.12

α2 κT

cc (0.492 × 10−4 )2 bar 1J × × mol K2 0.77 × 10−6 10 bar cc

J mol K J = 2.795 mol K J = 19.70 mol K J = 3.742 mol K

= 0.656 (CP − CV )NaCl (CP − CV )EtOH (CP − CV )Hg

Comment Note that the difference between the constant-pressure and constant-volume heat capacities is significant, and in fact can be quite large if the fluid has a large coefficient of thermal expansion, as is the case for ethanol. For comparison, the difference between CP and CV for an ideal gas is R = 8.314 J/(mol K).

Illustration 6.4-8 Effect of Pressure on the Enthalpy of Liquid Water The coefficient of thermal expansion of liquid water at 25◦ C was given in Illustration 6.2-6 to be 1 α= Vˆ



∂ Vˆ ∂T

 = 2.56 × 10−4 K P

Use this information and the fact that Vˆ (25◦ C, saturated liquid at 3.169 kPa) = 0.001003 m3 /kg ˆ (25◦ C, saturated liquid at 3.169 kPa) = 104.89 kJ/kg to determine the enthalpy of liquid and H water at (a) 25◦ C and 1 bar and (b) 25◦ C and 10 bar.

Solution Using Eq. 6.4-33, on a mass basis we have ◦ ◦ ˆ ˆ C, 1 bar) = H(25 C, 3.169 kPa) + Vˆ [1 − T α](100 − 3.169) kPa H(25

= 104.89

kJ m3 + 1.003 × 10−3 [1 − 298.15 × 2.56 × 10−4 ] × 96.83 kPa kg kg

= 104.89

kJ kJ kJ Pa m3 ×1 = 104.98 + 0.0897 kg kg kPa m3 kg

(a) and ◦ ˆ H(25 C, 10 bar) = 104.89 + 1.003 × 10−3 [1 − 298.15 × 2.56 × 10−4 ](1000 − 3.169) kJ = 104.89 + 0.92 = 105.81 kg

(b)

6.5 An Example Involving the Change of State of a Real Gas 245 Comment Here we see that for modest pressure changes, the liquid enthalpy changes little with pressure. Therefore, little error is incurred if, at fixed temperature, one uses the saturated liquid enthalpy in the steam tables at higher pressures. However, this will lead to errors near the critical point, where the density of the liquid changes rapidly with small changes in both temperature and pressure.

6.5 AN EXAMPLE INVOLVING THE CHANGE OF STATE OF A REAL GAS Given low-density heat capacity data and volumetric equation-of-state information for a fluid, we can use the procedures developed in this chapter to calculate the change in the thermodynamic properties of the fluid accompanying any change in its state. Thus, at least in principle, we can use the mass, energy, and entropy balances to solve energy flow problems for all fluids. The starting point for solving such problems is exactly the same here as that used in Chapters 2, 3, and 4, in that the balance equations are written for a convenient choice of the thermodynamic system. For real fluids, tedious calculations may be necessary to eliminate the internal energy, enthalpy, and entropy, which appear in the balance equations in terms of the pressure, temperature, and specific volume. If the fluid under study is likely to be of continual interest, it is probably worthwhile to construct a complete thermodynamic properties chart, as in Figs. 3.3-1 through 3.3-3 and in Illustration 6.4-1, so that all problems can be solved rapidly, as was demonstrated in Chapters 2, 3, and 4. If, on the other hand, the fluid is of limited interest, it is more logical to use the heat capacity and volumetric equation-of-state data only to solve the problem at hand. This type of calculation is illustrated in the following example which uses the van der Waals equation of state merely as a prototype for the equations of state of real fluids. This example is also considered, and enlarged upon, in the following two sections. Of course, the use of the Benedict-Webb-Rubin and other, more complicated equations of state would lead to predictions of greater accuracy, but with a great deal more computational effort. Illustration 6.5-1 Comparing Solutions to a Problem Assuming the Gas Is Ideal, Using a Thermodynamic Properties Chart, and Assuming That the Gas Can Be Described by the van der Waals Equation of State Nitrogen gas is being withdrawn from a 0.15-m3 cylinder at the rate of 10 mol/min. The cylinder initially contains the gas at a pressure of 100 bar and 170 K. The cylinder is well insulated, and there is a negligible heat transfer between the cylinder walls and the gas. How many moles of gas will be in the cylinder at any time? What will the temperature and pressure of the gas in the cylinder be after 50 minutes? a. Assume that nitrogen is an ideal gas. b. Use the nitrogen properties chart of Fig. 3.3-3. c. Assume that nitrogen is a van der Waals fluid. Data: For parts (a) and (c) use CP∗ [J/(mol K)] = 27.2 + 4.2 × 10−3 T (K)

246 Chapter 6: The Thermodynamic Properties of Real Substances Solution The mass and energy balances for the contents of the cylinder are dN = N˙ dt

N (t) = N (t = 0) + N˙ t

or

(a)

and d(N U ) = N˙ H dt

(b)

where N˙ = −10 mol/min. Following Illustration 4.5-2, the result of writing an entropy balance for that portion of the gas that always remains in the cylinder is dS =0 dt

or

S(t = 0) = S(t)

(c)

Also, from V = N V , we have N (t = 0) =

0.15 m3 Vcyl = V (t = 0) V (t = 0)

Equations a, b, c, and d apply to both ideal and nonideal gases. Computation of N (t) a. Using the ideal gas equation of state V (t = 0) =

8.314 × 10−5 bar m3 /mol K × 170 K RT (t = 0) = P (t = 0) 100 bar

= 1.4134 × 10−4 m3 /mol

so that N (t = 0) = 1061.3 mol

and

N (t) = 1061.3 − 10t mol

b. Using Figure 3.3-3 Vˆ (T = 170 K, P = 100 bar) = Vˆ (T = 170 K, P = 10 MPa) ≈ 0.0035 m3 /kg V (T = 170 K, P = 10 MPa) = 28.014

g m3 1 kg × 0.0035 × mol kg 1000 g

= 9.80 × 10−5 m3 /mol

so that N (t = 0) =

0.15 m3 = 1529.8 mol 9.80 × 10−5 m3 /mol

and N (t) = 1529.8 − 10t mol

c. Using the van der Waals equation of state The van der Waals equation is P =

RT a − 2 V −b V

(d)

6.5 An Example Involving the Change of State of a Real Gas 247 With the constants given in Table 6.4-1, we have 8.314 × 170 0.1368 − V (t = 0) − 3.864 × 10−5 [V (t = 0)]2

100 bar = 1 × 107 Pa =

Solving this cubic equation, we obtain V (t = 0) = 9.435 × 10−5 m3 /mol

so that N (t = 0) = 1589.9 mol

and N (t) = 1589.9 − 10t mol

Note also that since the volume of the cylinder is constant (0.15 m3 ), and as N (t) is known, we can compute the molar volume of the gas at any later time from V (t) =

0.15 m3 N (t)

(e)

In particular, we can use this equation to compute V (t = 50 min). These results, together with other information gathered so far, and some results from the following sections, are listed in Table 6.5-1. Since we know the specific volume after 50 minutes, we need to determine only one further state property to have the final state of the system completely specified. In principle, either the energy or entropy balance could be used to find this property. The entropy balance is more convenient to use, especially for the nonideal gas calculations. Thus, all the calculations here are based on the fact (see Eq. c) that S(t = 0) = S(t = 50 min)

Computation of T (t = 50 min) and P (t = 50 min) a. Using the ideal gas equation of state From Eq. 4.4-1, dS = CV

dV dT +R T V

Now for the ideal gas,

∗ CV = CV = CP∗ − R = (27.2 − 8.314) + 4.2 × 10−3 T

= 18.9 + 4.2 × 10−3 T J/(mol K)

Table 6.5-1 Equation of State

N (t = 0)

N (t)

V (t = 50 min) (m3 /mol)

Ideal gas Fig 3.3-3 van der Waals Corresponding states (Illustration 6.6-2) Peng-Robinson (Illustration 6.7-1)

1061.3 1529.8 1589.9 1432.7

1061.3 − 10t 1529.8 − 10t 1589.9 − 10t 1432.7 − 10t

2.672 × 10−4 1.457 × 10−4 1.376 × 10−4 1.608 × 10−4

1567.9

1567.9 − 10t

1.405 × 10−4

248 Chapter 6: The Thermodynamic Properties of Real Substances Also, since S = constant, or ΔS = 0, we have 

T T =170

(18.9 + 4.2 × 10−3 T ) dT + 8.314 T

 V =2.672×10−4 dV =0 V V =1.413×10−4

or 18.9 ln

T 2.672 × 10−4 =0 + 4.2 × 10−3 (T − 170) + 8.314 ln 170 1.413 × 10−4

The solution to this equation is 130.3 K, so that P (t = 50 min) =

RT (t = 50 min) = 40.5 bar V (t = 50 min)

b. Using Fig. 3.3-3 g m3 1 mol × × 1000 mol 28.014 g kg m3 kg

V (t = 50 min) = 1.457 × 10−4 = 5.20 × 10−3

To find the final temperature and pressure using Fig. 3.3-3, we locate the initial point (T = 170 K, P = 10 MPa) and follow a line of constant entropy (dashed line) through this point to the intersection with a line of constant specific volume Vˆ = 5.2 × 10−3 m3 /kg. This intersection gives the pressure and temperature of the t = 50 min state. We find T (t = 50 min) ≈ 133 K P (t = 50 min) ≈ 3.9 MPa = 39 bar

This construction is shown in the following diagram, which is a portion of Fig. 3.3-3. T = 180 K

^

S = 2 kJ/kg K 20

T= 5K

16

3

^

V

.00

=0

i

^

S = 2.25 kJ/(kg K) 10 ^

V = 0.005 T = 150 K

6

T = 135 K

4

f

200

250 ^

H, kJ/kg

2

P, MPa

8

6.5 An Example Involving the Change of State of a Real Gas 249 c. Using the van der Waals equation of state Here we start with Eq. 6.2-19, dS =

CV dT + T



∂P ∂T

 dV V

and note that for the van der Waals gas 

∂P ∂T

 = V

R V −b

The integration path to be followed is i. T (t = 0), V (t = 0) → T (t = 0), V = ∞ ii. T (t = 0), V = ∞ → T (t = 50 min), V = ∞ (CV = CV∗ for this step) iii. T (t = 50 min), V = ∞ → T (t = 50 min), V (t = 50 min) so that  S(t = 50 min) − S(t = 0) = 0 = R

V =∞

dV V −b

V (t=0)



T (t=50)

+ T =170K

∗ CV dT + R T



V (t=50) V =∞

dV V −b

or 

V (t=50)

0=R V (t=0)

dV V −b



T (t=50)

+ T =170K

18.9 + 4.2 × 10−3 T dT T

Thus, using b = 3.864 × 10−5 m3 /mol, we have  0 = 8.314 ln

 T (13.77 − 3.864) × 10−5 + 18.9 ln + 4.2 × 10−3 (T − 170) (9.437 − 3.864) × 10−5 170

The solution to the equation is T (t = 50 min) = 133.1 K

Now, using the van der Waals equation of state with T = 133.1 K and V = 13.76 × 10−5 m3 /mol, we find P (t = 50 min) = 39.5 bar

Summary Equation of State

V (t = 50 min) (m3 /mol)

T (t = 50 min) (K)

P (t = 50 min) (bar)

Ideal gas Fig. 3.3-3 van der Waals Corresponding states (Illustration 6.6-2) Peng-Robinson (Illustration 6.7-1)

2.672 × 10−4 1.457 × 10−4 1.376 × 10−4 1.608 × 10−4

129.6 133 133.1 136

40.3 39 39.5 41

1.405 × 10−4

134.7

40.6

250 Chapter 6: The Thermodynamic Properties of Real Substances Comments It is clear from the results of this problem that one cannot assume that a high-pressure gas is ideal and expect to make useful predictions. Of the techniques that have been considered for the thermodynamic calculations involving real fluids, the use of a previously prepared thermodynamic properties chart for the fluid is the most rapid way to proceed. The alternatives, the use of a simple volumetric equation of state for the fluid or of corresponding-states correlations (discussed in the following section), are always tedious, and their accuracy is hard to assess. However, if you do not have a thermodynamic properties chart available for the working fluid, you have no choice but to use the more tedious methods.

6.6 THE PRINCIPLE OF CORRESPONDING STATES The analysis presented in Sec. 6.4 makes it possible to solve thermodynamic problems for real substances or to construct tables and charts of their thermodynamic properties given only the low-density (ideal gas) heat capacity and analytic or tabular information on the volumetric equation of state. Unfortunately, these can be tedious tasks, and the necessary volumetric equation-of-state information is not available for all fluids. Thus, we consider here the principle of corresponding states, which allows one to predict some thermodynamic properties of fluids from generalized property correlations based on available experimental data for similar fluids. Before we introduce the concept of generalized fluid properties correlations, it is useful to consider which properties we can hope to get from this sort of correlation scheme, and which we cannot. The two types of information needed in thermodynamic calculations are the volumetric equation of state and the ideal gas heat capacity. The ideal gas heat capacity is determined solely by the intramolecular structure (e.g., bond lengths, vibration frequencies, configuration of constituent atoms) of only a single molecule, as there is no intermolecular interaction energy in an ideal gas. As the structure of each molecular species is sufficiently different from that of others, and since experimental low-density heat capacity data are frequently available, we will not attempt to develop correlations for heat capacity data or for the ideal gas part of the enthalpy, internal energy, or entropy, which can be computed directly from CP∗ and CV∗ . However, the volumetric equation of a state of a fluid is determined solely by the interactions of each molecule with its neighbors. An interesting fact that has emerged from the study of molecular behavior (i.e., statistical mechanics) is that as far as molecular interactions are concerned, molecules can be grouped into classes, such as spherical molecules, nonspherical molecules, molecules with permanent dipole moments, and so forth, and that within any one class, molecular interactions are similar. It has also been found that the volumetric equations of state obeyed by all members of a class are similar in the sense that if a given equation of state (e.g., Peng-Robinson or Benedict-WebbRubin) fits the volumetric data for one member of a class, the same equation of state, with different parameter values, is likely to fit the data for other molecular species in the same class. The fact that a number of different molecular species may be represented by a volumetric equation of state of the same form suggests that it might be possible to construct generalized correlations, that is, correlations applicable to many different molecular species, for both the volumetric equation of state and the density- (or pressure-) dependent contribution to the enthalpy, entropy, or other thermodynamic properties. Historically, the first generalized correlation arose from the study of the van der Waals equation of state. Although present correlations are largely based on experimental data, it is useful to review the van der Waals generalized correlation scheme since, although not very

6.6 The Principle of Corresponding States 251

T > Tc

P

T = Tc T < Tc

V _

Figure 6.6-1 The pressure-volume behavior of the van der Waals equation of state.

accurate, it does indicate both the structure and correlative parameters used in modern fluid property correlations. A plot of P versus V for various temperatures for a van der Waals fluid is given in Fig. 6.6-1. An interesting characteristic of this figure is that isotherms below the one labeled Tc exhibit a local minimum followed by a local maximum with increasing V over part of the specific volume-pressure range. As we will see in Chapter 7, such behavior is associated with a vapor-liquid transition. For T > Tc this structure in the P -V plot has vanished, whereas at T = Tc this minimum and maximum have converged to a single point. At this point the first and second derivatives of P with respect to V at constant T are zero. For the present discussion, this point is taken to be the critical point of the fluid, that is, the point of highest temperature at which a liquid can exist (justification for this identification is given in Chapter 7). A brief list of measured critical temperatures, critical pressures Pc , and critical volumes V c (i.e., the pressure and volume of the highest-temperature liquid) for various fluids is given in Table 6.6-1. To identify the critical point for the van der Waals fluid analytically, we make use of the following mathematical requirements for the occurrence of an inflection point on an isotherm in the P -V plane:

 Critical-point conditions

∂P ∂V



 = 0 and Tc

∂2P ∂V 2

 = 0 at Pc and Vc

(6.6-1)

Tc

(Also, the first nonzero derivative should be odd and negative in value, but this is not used here.) Using Eqs. 6.6-1a and b together with Eq. 6.2-38, we find that at the critical point a RTc Pc = − 2 (6.6-2a) Vc−b Vc

252 Chapter 6: The Thermodynamic Properties of Real Substances     RTc 2a ∂P  =0=− + 3  2 ∂V T evaluated at (V c − b) Vc

(6.6-2b)

Tc ,Pc ,V c

and



∂2P ∂V 2

   2RTc 6a  evaluated at = 0 = (V − b)3 − V 4 c c T

(6.6-2c)

Tc ,Pc ,V c

Thus, three equations interrelate the two unknowns a and b. Using Eqs. 6.6-2b and c to ensure that the critical-point conditions of Eqs. 6.6-1 are satisfied, we obtain (Problem 6.5) 9V c RTc a= (6.6-3a) 8 and V b= c 3 Using these results in Eq. 6.6-2a yields a Pc = (6.6-3b) 27b2 By direct substitution we find that the compressibility at the critical point is For the van der Waals equation only

Zc =

Pc V c 3 = = 0.375 RTc 8

(6.6-3c)

This last equation can be used to obtain expressions for the van der Waals parameters in terms of (1) the critical temperature and pressure, For the van der Waals equation only

27R2 Tc2 64Pc or (2) the critical pressure and volume: a=

a = 3Pc V 2c

and b =

RTc 8Pc

(6.6-4a)

and b =

Vc 3

(6.6-4b)

Substituting either of these sets of parameters into Eq. 6.2-38 yields   2   

P T V Vc 3 +3 −1 =8 Pc V Vc Tc Now defining a dimensionless or reduced temperature Tr , reduced pressure Pr , and reduced volume Vr by

Tr = we obtain

T Tc

Pr =

P Pc

and Vr =



3 Pr + 2 [3Vr − 1] = 8Tr Vr

V Vc

(6.6-5)

(6.6-6)

6.6 The Principle of Corresponding States 253 The form of Eq. 6.6-6 is very interesting since it suggests that for all fluids that obey the van der Waals equation, Vr is the same function of Tr and Pr . That is, at given values of the reduced temperature (Tr = T /Tc ) and reduced pressure (Pr = P/Pc ), all van der Waals fluids will have the same numerical value of reduced volume (Vr = V /V c ). This does not mean that at the same value of T and P all van der Waals fluids have the same value of V ; this is certainly not the case, as can be seen from the illustration that follows. Two fluids that have the same values of reduced temperature and pressure and therefore the same reduced volume, are said to be in corresponding states. Illustration 6.6-1 Simple Example of the Use of Corresponding States Assume that oxygen (Tc = 154.6 K, Pc = 5.046 × 106 Pa, and V c = 7.32 × 10−5 m3 /mol) and water (Tc = 647.3 K, Pc = 2.205 × 107 Pa, and V c = 5.6 × 10−5 m3 /mol) can be considered van der Waals fluids. a. Find the value of the reduced volume both fluids would have at Tr = 3/2 and Pr = 3. b. Find the temperature, pressure, and volume of each gas at Tr = 3/2 and Pr = 3. c. If O2 and H2 O are both at a temperature of 200◦ C and a pressure of 2.5 × 106 Pa, find their specific volumes.

Solution a. Using Tr = 3/2 and Pr = 3 in Eq. 6.6-6, one finds that Vr = 1 for both fluids. b. Since Vr = 1 at these conditions [see part (a)], the specific volume of each fluid is equal to the critical volume of that fluid. So V O2 = 7.32 × 10−5 m3 /mol

V H2 O = 5.6 × 10−5 m3 /mol

at

at

PO2 = 3 × Pc,O2 = 1.514 × 107 Pa

PH2 O = 3 × Pc,H2 O = 6.615 × 107 Pa

and

and

TO2 = 1.5 × Tc,O2 = 232.2 K

TH2 O = 1.5 × Tc,H2 O = 971 K

c. For oxygen we have Pr =

2.5 × 106 = 0.495 5.046 × 106

and Tr = 473.2/154.6 = 3.061. Therefore, Vr = 16.5 and V = 1.208 × 10−3 m3 /mol. Similarly for water, 2.5 × 106 Pr = = 0.113 2.205 × 107 Tr = 473.2/647.3 = 0.731, so that Vr = 15.95 and V = 8.932 × 10−4 m3 /mol.

As has been already indicated, the accuracy of the van der Waals equation is not very good. This may be verified by comparing the results of the previous illustration with experimental data, or by comparing the compressibility factor Z = P V /RT for the van der Waals equation of state with experimental data for a variety of gases. In particular, at the critical point (see Eq. 6.6-3c)  Zc van der Waals = Pc V c /RTc = 0.375 while for most fluids the critical compressibility Zc is in the range 0.23 to 0.31 (Table 6.6-1), so that the van der Waals equation fails to predict accurate critical-point

254 Chapter 6: The Thermodynamic Properties of Real Substances Table 6.6-1 The Critical and Other Constants for Selected Fluids

Substance

Symbol

Acetylene Ammonia Argon Benzene n-Butane Isobutane 1-Butene Carbon dioxide Carbon monoxide Carbon tetrachloride n-Decane n-Dodecane Ethane Ethyl ether Ethylene Helium n-Heptane n-Hexane Hydrogen Hydrogen fluoride Hydrogen sulfide Methane Naphthalene Neon Nitric oxide Nitrogen n-Octane Oxygen n-Pentane Isopentane Propane Propylene Refrigerant R12 Refrigerant HFC-134a Sulfur dioxide Toluene Water Xenon

C2 H2 NH3 Ar C6 H6 C4 H10 C4 H10 C4 H8 CO2 CO CCl4 C10 H22 C12 H26 C2 H6 C4 H10 O C2 H4 He C7 H16 C6 H14 H2 HF H2 S CH4 C10 H8 Ne NO N2 C8 H18 O2 C5 H12 C5 H12 C3 H8 C3 H6 CCl2 F2 CH2 FCF3 SO2 C7 H8 H2 O Xe

Molecular Weight (g mol−1 )

Tc (K)

Pc (MPa)

Vc (m3 /kmol)

Zc

ω

Tboil (K)

26.038 17.031 39.948 78.114 58.124 58.124 56.108 44.010 28.010 153.823 142.286 170.340 30.070 74.123 28.054 4.003 100.205 86.178 2.016 20.006 34.080 16.043 128.174 20.183 30.006 28.013 114.232 31.999 72.151 72.151 44.097 42.081 120.914 102.03 64.063 92.141 18.015 131.300

308.3 405.6 150.8 562.1 425.2 408.1 419.6 304.2 132.9 556.4 617.6 658.3 305.4 466.7 282.4 5.19 540.2 507.4 33.2 461.0 373.2 190.6 748.4 44.4 180.0 126.2 568.8 154.6 469.6 460.4 369.8 365.0 385.0 374.23 430.8 591.7 647.3 289.7

6.140 11.28 4.874 4.894 3.800 3.648 4.023 7.376 3.496 4.560 2.108 1.824 4.884 3.638 5.036 0.227 2.736 2.969 1.297 6.488 8.942 4.600 4.05 2.756 6.485 3.394 2.482 5.046 3.374 3.384 4.246 4.620 4.124 4.060 7.883 4.113 22.048 5.836

0.113 0.0724 0.0749 0.259 0.255 0.263 0.240 0.0940 0.0931 0.276 0.603 0.713 0.148 0.280 0.129 0.0573 0.304 0.370 0.065 0.069 0.0985 0.099 0.410 0.0417 0.058 0.0895 0.492 0.0732 0.304 0.306 0.203 0.181 0.217 0.198 0.122 0.316 0.056 0.118

0.271 0.242 0.291 0.271 0.274 0.283 0.277 0.274 0.295 0.272 0.247 0.24 0.285 0.262 0.276 0.301 0.263 0.260 0.305 0.12 0.284 0.288 0.267 0.311 0.250 0.290 0.259 0.288 0.262 0.271 0.281 0.275 0.280 0.258 0.268 0.264 0.229 0.286

0.184 0.250 −0.004 0.212 0.193 0.176 0.187 0.225 0.049 0.194 0.490 0.562 0.098 0.281 0.085 −0.387 0.351 0.296 −0.22 0.372 0.100 0.008 0.302 0 0.607 0.040 0.394 0.021 0.251 0.227 0.152 0.148 0.176 0.332 0.251 0.257 0.344 0.002

189.2 239.7 87.3 353.3 272.7 261.3 266.9 194.7 81.7 349.7 447.3 489.5 184.5 307.7 169.4 4.21 371.6 341.9 20.4 292.7 212.8 111.7 491.1 27.0 121.4 77.4 398.8 90.2 309.2 301.0 231.1 225.4 243.4 247.1 263 383.8 373.2 165.0

Source: Adapted from R. C. Reid, J. M. Prausnitz, and B. E. Poling, The Properties of Gases and Liquids, 4th ed., McGraw-Hill, New York, 1986, Appendix A and other sources.

behavior. (It is, however, a great improvement over the ideal gas equation of state, which predicts that Z = 1 for all conditions.) The fact that the critical compressibility of the van der Waals fluid is not equal to that for most real fluids also means that different values for the van der Waals parameters are obtained for any one fluid, depending on whether Eq. 6.6-3a, Eq. 6.6-4a, or Eq. 6.6-4b are used to relate these parameters to the critical properties. In practice, the

6.6 The Principle of Corresponding States 255 critical volume of a fluid is known with less experimental accuracy than either the critical temperature or critical pressure, so that Eq. 6.6-4a and critical-point data are most frequently used to obtain the van der Waals parameters. Indeed, the entries in Tables 6.4-1 and 6.6-1 are related in this way. Thus, if the parameters in Table 6.4-1 are used in the van der Waals equation, the critical temperature and pressure will be correctly predicted, but the critical volume will be too high by the factor  Zc van der Waals 3 = Zc 8Zc where Zc is the real fluid critical compressibility. Although the van der Waals equation is not accurate, the idea of a correspondence of states to which it historically led is both appealing and, as we will see, useful. Attempts at using the corresponding-states concept over the last 40 years have been directed toward representing the compressibility factor Z as a function of the reduced pressure and temperature, that is, Two-parameter corresponding states

Z = P V /RT = Z(Pr , Tr )

(6.6-7)

where the functional relationship between Tr , Pr , and Z is determined from experimental data, or from a very accurate equation of state. That such a procedure has some merit is evident from Fig. 6.6-2, where the compressibility data for different fluids have been made to almost superimpose by plotting each as a function of its reduced temperature and pressure. A close study of Fig. 6.6-2 indicates that there are systematic deviations from the simple corresponding-states relation of Eq. 6.6-7. In particular, the compressibility factors for the inorganic fluids are almost always below those for the hydrocarbons. Furthermore, if Eq. 6.6-7 were universally valid, all fluids would have the same value of the critical compressibility Zc = Z(Pr = 1, Tr = 1); however, from Table 6.6-1, it is clear that Zc for most real fluids ranges from 0.23 to 0.31. These failings of Eq. 6.6-7 have led to the development of more complicated corresponding-states principles. The simplest generalization is the suggestion that there should not be a single Z = Z(Pr , Tr ) relationship for all fluids, but rather a family of relationships for different values of Zc . Thus, to find the value of the compressibility factor for given values Tr and Pr , it would be necessary to use a chart of Z = Z(Pr , Tr ) prepared from experimental data for fluids with approximately the same value of Zc . This is equivalent to saying that Eq. 6.6-7 is to be replaced by Z = Z(Tr , Pr , Zc ) Alternatively, fluid characteristics other than Zc can be used as the additional parameter in the generalization of the simple corresponding-states principle. In fact, since for many substances the critical density, and hence Zc , is known with limited accuracy, if at all, there is some advantage in avoiding the use of Zc . Pitzer13 has suggested that for nonspherical molecules the acentric factor ω be used as the third correlative parameter, where ω is defined to be

ω = −1.0 − log10 [P vap (Tr = 0.7)/Pc ]

Acentric factor

13 See

Appendix 1 of K. S. Pitzer, Thermodynamics, 3rd ed., McGraw-Hill, New York, 1995.

256 Chapter 6: The Thermodynamic Properties of Real Substances 1.1

Tr = 2.00

1.0

0.9 Tr = 1.50 0.8 Tr = 1.30

Z=

PV _ RT

0.7

0.6

Tr = 1.20

0.5 Tr = 1.10 0.4

Legend Tr = 1.00

0.3

Average curve based on data for hydrocarbons

0.2

0.1

i-Pentane n-Heptane Nitrogen Carbon dioxide Water

Methane Ethylene Ethane Propane n-Butane

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

7.0

Reduced pressure, Pr

Figure 6.6-2 Compressibility factors for different fluids as a function of the reduced temperature and pressure. [Reprinted with permission from G.-J. Su, Ind. Eng. Chem. 38. 803 (1946). Copyright American Chemical Society.]

Here P vap (Tr = 0.7) is the vapor pressure of the fluid at Tr = 0.7, a temperature near the normal boiling point. In this case the corresponding-states relation would be of the form Three-parameter corresponding states

Z = Z(Tr , Pr , ω) Even these extensions of the corresponding-states concept, which are meant to account for molecular structure, cannot be expected to be applicable to fluids with permanent dipoles and quadrupoles. Since molecules with strong permanent dipoles interact differently than molecules without dipoles, or than molecules with weak dipoles, one would expect the volumetric equation of state for polar fluids to be a function of the dipole moment. In principle, the corresponding-states concept could be further generalized to include this new parameter, but we will not do so here. Instead, we refer you to the book by Reid, Prausnitz, and Poling for a detailed discussion of the correspondingstates correlations commonly used by engineers.14 The last several paragraphs have emphasized the shortcomings of a single corresponding-states principle when dealing with fluids of different molecular classes. However, it is useful to point out that a corresponding-states correlation can be an accurate 14 J.

M. Prausnitz, B. E. Poling, and J. P. O’Connell, Properties of Gases and Liquids, 5th ed., McGraw-Hill, New York, 2001.

6.6 The Principle of Corresponding States 257

5.00

3.00

0.80 1.00

2.00

10 .00

0.50

1.3

0.70

1.4

0.90

0.60

1.5

15

1.2

.00

Generalized compressibility factors (Zc = 0.27)

1.1 Tr

Compressibility factor, Z

1.0

15.00 10.00 5.00 3.00 2.00

2.00

0.9

0

0.8

5

Sat

0.9

0

0.9

5

ura

0.8

1.50 1.30

1.2

0.8

ted

0 1.1

1.80 1.70

0

1.60

1.0

0

1.50

gas

1.40 1.35

0.7

1.30

0.6

1.25 0.75 1.20 0.80 1.15

0.85

1.10

0.90

0.4

1.08 1.06 1.04

0.95

0.3

0.9

1.02

0

0.5

1.00

0.2 0

0.5

0.1

quid ated li

Satur

0 0.1

0.2

0.3 0.4 0.5

1.0

2.0

3.0 4.0 5.0

10

20

30

Reduced pressure, Pr

Figure 6.6-3 (Reprinted with permission from O. A. Hougen, K. M. Watson, and R. A. Ragatz, Chemical Process Principles Charts, 2nd ed., John Wiley & Sons, New York, 1960. This figure appears as an Adobe PDF file on the website for this book, and may be enlarged and printed for easier reading and for use in solving problems.)

representation of the equation-of-state behavior within any one class of similar molecules. Indeed, the volumetric equation-of-state behavior of many simple fluids and most hydrocarbons is approximately represented by the plot in Fig. 6.6-3, which was developed from experimental data for molecules with Zc = 0.27. The existence of an accurate corresponding-states relationship of the type Z = Z(Tr , Pr ) (or perhaps a whole family of such relationships for different values of Zc or ω ) allows one to also develop corresponding-states correlations for that contribution to the thermodynamic properties of the fluid that results from molecular interactions, or nonideal behavior, that is, the departure functions of Sec. 6.4. For example, starting with Eq. 6.4-22, we have 

  P,T ∂V IG V −T dP H(T, P ) − H (T, P ) = ∂T P P =0,T

258 Chapter 6: The Thermodynamic Properties of Real Substances and using the corresponding-states relation, Eq. 6.6-7, yields

RT Z(Tr , Pr ) P     RT 2 ∂Z ∂V V −T =− ∂T P P ∂T P V =

and

so that

H(T, P ) − H IG (T, P )      P,T  Pr ,Tr RT 2 ∂Z Tr2 ∂Z dP = −RTc dPr =− P ∂T P ∂Tr Pr P =0,T Pr =0,Tr Pr or

H(T, P ) − H IG (T, P ) = −RTr2 Tc



Pr ,Tr Pr =0,Tr

1 Pr



∂Z ∂Tr

 dPr

(6.6-8)

Pr

The important thing to notice about this equation is that nothing in the integral depends on the properties of a specific fluid, so that when the integral is evaluated using the corresponding-states equation of state, the result will be applicable to all correspondingstates fluids. Figure 6.6-4 contains in detailed graphical form the corresponding-states prediction for the enthalpy departure from ideal gas behavior computed from Fig. 6.6-3 (for fluids with Zc = 0.27) and Eq. 6.6-8.15 The enthalpy change of a real fluid in going from (T0 , P = 0) to (T, P ) can then be computed using Fig. 6.6-4 as indicated here:

H(T, P ) − H(T0 , P = 0) = H IG (T, P ) − H(T0 , P = 0)

 T + {H(T, P ) − H IG (T, P )} = CP∗ dT

T0 H(T, P ) − H IG (T, P ) + Tc Tc from Fig.6.6−4 (6.6-9) Similarly, the enthalpy change in going from any state (T1 , P1 ) to state (T2 , P2 ) can be computed from the repeated application of Eq. 6.6-9, which yields Enthalpy change from corresponding states





H(T, P ) − H IG (T, P ) H(T2 , P2 ) − H(T1 , P1 ) = CP dT + Tc Tc T1 

H(T, P ) − H IG (T, P ) − Tc Tr ,Pr T2



∗

1

1

Tr2 ,Pr2

from Fig. 6.6-4

(6.6-10) 15 Note that Eqs. 6.6-8, 6.6-9, and 6.6-10 contain the term (H−H IG )/T

c , whereas Fig. 6.6-4 gives (H

IG

−H)/Tc .

6.6 The Principle of Corresponding States 259 15.0

15.0

0.50

14.0

14.0

0.55 0.60

13.0

13.0

Tr 0.70

12.0

0.50

0.65

12.0 0.55

Sa

tu

H cal (1 cal = 4.184 J) _ IG – H _ Tc mol K

11.0

0.80

ra

ted

liq

0.90

uid

10.0

0.94 0.98

9.0

1.02

0.75

11.0

0.85

0.60

0.92 0.96 1.0

0.65

10.0 0.70

9.0

1.08

0.80

8.0

1.15

0.90

1.04

8.0

0.75

1.06

0.80

1.10

0.85 0.90

7.0

7.0

0.92 0.94

6.0

1.20

0.96

1.00 1.25 1.10

6.0

1.20

5.0

1.30

0.98

1.40

5.0

1.30

4.0

1.50

1.60

4.0

1.40

3.0

1.7 0

5

0.9

0.90

1.9 0 2.40

2.0

2.20

1.60

0 1.2 .30 1 0 1.5

2.60

2.80

1.80 2.00

1.0

2.00

0.1

0.2

0.3 0.4 0.5

3.0

0

1.0

0

0

Satu

s

d ga

rate

1.0

1.50

2.00

1.1

2.0

1.80

1.0

3.00

2.20

4.00

3.00

2.0

3.0 4.0 5.0

2.40

10

Reduced pressure, Pr Generalized enthalpy departure from ideal gas behavior (per mole of gas or liquid, Zc = 0.27)

0

2.60 2.80 4.00 3.00

10

20

–1.0 –2.0

30

Figure 6.6-4 (Reprinted with permission from O. A. Hougen, K. M. Watson, and R. A. Ragatz, Chemical Process Principles Charts, 2nd ed., John Wiley & Sons, New York, 1960. This figure appears as an Adobe PDF file on the website for this book, and may be enlarged and printed for easier reading and for use in solving problems.)

The form of Eqs. 6.6-9 and 6.6-10 makes good physical sense in that each consists of two terms with well-defined meanings. The first term depends only on the ideal gas heat capacity, which is a function of the molecular structure and is specific to the molecular species involved. The second term, on the other hand, represents the nonideal behavior of the fluid due to intermolecular interactions that do not exist in the ideal gas, but whose contribution can be estimated from the generalized correlation. In a manner equivalent to that just used, it is also possible to show (see Problem 6.6) that     Pr ,Tr  Z − 1 Tr ∂Z IG dPr (6.6-11) S(T, P ) − S (T, P ) = −R + Pr Pr ∂Tr Pr Pr =0,Tr

260 Chapter 6: The Thermodynamic Properties of Real Substances This equation and Fig. 6.6-3 are the bases for the entropy departure plot given in Fig. 6.6-5.16 The change in entropy between any two states (T1 , P1 ) and (T2 , P2 ) can then be computed from Entropy change from corresponding states



 P2 CP∗ dP dT − R S(T2 , P2 ) − S(T1 , P1 ) = T P T1 P1     + S − S IG Tr ,Pr − S − S IG Tr T2

2



1 ,Pr1

2

from Fig. 6.6-5

(6.6-12) Similarly, corresponding-states plots could be developed for the other thermodynamic properties U , A, and G, though these properties are usually computed from the relations

U = H − PV A = U − TS G = H − TS and the corresponding-states figures already given.

(6.6-13)

Illustration 6.6-2 Using Corresponding States to Solve a Real Gas Problem Rework Illustration 6.5-1, assuming that nitrogen obeys the generalized correlations of Figs. 6.6-3, 6.6-4, and 6.6-5.

Solution From Table 6.6-1 we have for nitrogen Tc = 126.2 K and Pc = 33.94 bar, and from the initial conditions of the problem, Tr =

170 = 1.347 126.2

and

Pr =

100 = 2.946 33.94

From Fig. 6.6-3, Z = 0.741, so V (t = 0) = Z

RT (t = 0) = ZV IG (t = 0) = 1.047 × 10−4 m3 /mol P (t = 0)

Therefore, following Illustration 6.5-1, N (t = 0) = 1432.7 mol N (t) = 1432.7 − 10t mol

and V (t = 50 min) =

0.15 m3 = 1.6082 × 10−4 m3 /mol (1432.7 − 500) mol

To compute the temperature and pressure at the end of the 50 minutes, we use S(t = 0) = S(t = 50 min)

16 Note

that Eqs. 6.6-11 and 6.6-12 contain the term S − S IG , whereas Fig. 6.6-5 gives S IG − S.

6.6 The Principle of Corresponding States 261 21.0 20.0 Generalized enthalpy departure from ideal gas behavior (per mole of gas or liquid, Zc = 0.27)

19.0 18.0 17.0 Tr

16.0

0.50

15.0

0.50

Sa ate

tur

14.0

0.55

0.65

0.60

id

cal (1 cal = 4.184 J) mol K

0.55 0.60

iqu

_S IG – _S

dl

13.0 12.0

0.70

11.0

0.65

10.0 0.75

0.80

0.80

0.85

9.0

0.75 0.70

0.90 0.90

0.92

8.0

0.94

0.92

0.80

0.96

0.90

1.02 1.04

1.00

0.96 0.98

6.0

0.75

0.98

0.94

7.0

0.85

1.00

1.06 1.10

1.08

5.0

1.10

1.15 1.20

1.15

4.0 1.30 1.20

3.0

1.40

1.30

1.50

1.50

1.60 1.80 2.00

1.60

1.80

2.50

2.00 3.00

2.50

2.0 1.40

0

4

as

ed g turat

1.0

Sa

0.1

0.90

0.2

0.95

0 1.0

0.3 0.4 0.5

1.0

1.20

0

1.1

1.40 1.60

1.0

2.0

3.0 4.0 5.0

3.00

10

20

30

Reduced pressure, Pr

Figure 6.6-5 (Reprinted with permission from O. A. Hougen, K. M. Watson, and R. A. Ragatz, Chemical Process Principles Charts, 2nd ed., John Wiley & Sons, New York, 1960. This figure appears as an Adobe PDF file on the website for this book, and may be enlarged and printed for easier reading and for use in solving problems.)

262 Chapter 6: The Thermodynamic Properties of Real Substances and recognize that S(t = 0) = S IG (T, V )t=0 + (S − S IG )t=0

and S(t = 50 min) = S IG (T, V )t=50 min + (S − S IG )t=50 min

so that S(t = 50 min) − S(t = 0) = S IG (T, P )t=50 min − S IG (T, P )t=0 + (S − S IG )t=50 min − (S − S IG )t=0

where S IG (T, P )t=50 min − S IG (T, P )t=0 = 27.2 ln

T (t = 50 min) 170

+ 4.2 × 10−3 [T (t = 50 min) − 170] − 8.314 ln

P (t = 50 min) 100

Both of the (S − S IG ) terms are obtained from the corresponding-states charts. (S − S IG )t=0 is easily evaluated, since the initial state is known; that is, Tr = 1.347 and Pr = 2.946, so that, from Fig. 6.6-5 (S − S IG )t=0 = −2.08 cal/(mol K) = −8.70 J/(mol K). To compute (S − S IG ) at t = 50 minutes is more difficult because neither Tr nor Pr is known. The procedure to be followed is 1. V (t = 50 min) is known, so guess a value of T (t = 50 min). (A reasonable first guess is the ideal gas solution obtained earlier.) 2. Use V (t = 50 min) and T (t = 50 min) to compute, by trial and error, P (t = 50 min) from P =

RT Z V



T P , Tc Pc

 =

RT Z(Tr , Pr ) V

3. Use the values of P and T from steps 1 and 2 to compute (S − S IG )t=50 min . 4. Determine whether S(t = 50 min) = S(t = 0) is satisfied with the trial values of T and P . If not, guess another value of T (t = 50) and go back to step 2. Our solution after a number of trials is T (t = 50 min) = 136 K P (t = 50 min) = 41 bar

Comment Because of the inaccuracy in reading numerical values from the corresponding-states graphs, this solution cannot be considered to be of high accuracy.

It should be pointed out that although the principle of corresponding states and Eqs. 6.6-7, 6.6-8, and 6.6-11 appear simple, the application of these equations can become tedious, as is evident from this illustration. Also, the use of generalized correlations will lead to results that are not as accurate as those obtained using tabulations of the thermodynamic properties for the fluid of interest. Therefore, the correspondingstates principle is used in calculations only when reliable thermodynamic data are not available.

6.7 Generalized Equations of State 263

6.7 GENERALIZED EQUATIONS OF STATE Although the discussion of the previous section focused on the van der Waals equation and corresponding-states charts for both the compressibility factor Z and the thermodynamic departure functions, the modern application of the corresponding states idea is to use generalized equations of state. The concept is most easily demonstrated by again using the van der Waals equation of state. From Eqs. 6.2-38,

P =

a RT − 2 V −b V

(6.2-38b)

and the result of the inflection point analysis of Sec. 6.6, the constants a and b can be obtained from the fluid critical properties using

27R2 Tc2 RTc and b = (6.6-4a) 64Pc 8Pc The combination of Eqs. 6.2-38b and 6.6-4a is an example of a generalized equation of state, since we now have an equation of state that is presumed to be valid for a class of fluids with parameters (a and b) that have not been fitted to a whole collection of experimental data, but rather are obtained only from the fluid critical properties. The important content of these equations is that they permit the calculation of the P V T behavior of a fluid knowing only its critical properties, as was the case in correspondingstates theory. It must be emphasized that the van der Waals equation of state is not very accurate and has been used here merely for demonstration because of its simplicity. It is never used for engineering design predictions, though other cubic equations of state are used. To illustrate the use of generalized equations of state, we will consider only the Peng-Robinson equation, which is commonly used to represent hydrocarbons and inorganic gases such as nitrogen, oxygen, and hydrogen sulfide. The generalized form of the Peng-Robinson equation of state is a=

Complete generalized Peng-Robinson equation of state

P =

RT a(T ) − V − b V (V + b) + b(V − b)

(6.4-2)

with a(T ) = 0.45724

R2 Tc2 α(T ) Pc

RTc b = 0.07780 Pc  

√ α=1+κ 1−

(6.7-1) (6.7-2)

T Tc

 (6.7-3)

and κ = 0.37464 + 1.54226ω − 0.26992ω 2 where ω is the acentric factor defined earlier and given in Table 6.6-1.

(6.7-4)

264 Chapter 6: The Thermodynamic Properties of Real Substances Equations 6.7-1 through 6.7-4 were obtained in the following manner. First, the critical-point restrictions of Eq. 6.6-1 were used, which leads to (see Problem 6.11)

R2 Tc2 RTc and b = 0.07780 Pc Pc Next, to improve the predictions of the boiling pressure as a function of temperature, that is, the vapor pressure (which will be discussed in Sec. 7.5), Peng and Robinson added an additional temperature-dependent term to their equation by setting a(Tc ) = 0.45724

a(T ) = a(Tc )α(T ) Note that to satisfy the critical-point restrictions, α(T = Tc ) must equal unity, as does the form of Eq. 6.7-3. The specific form of α given by Eqs. 6.7-3 and 6.7-4 was chosen by fitting vapor pressure data for many fluids. There are two points to be noted in comparing the generalized van der Waals and Peng-Robinson equations of state. First, although the parameter a is a constant in the van der Waals equation, in the Peng-Robinson equation it is a function of temperature (actually reduced temperature, Tr = T /Tc ) through the temperature dependence of α. Second, the generalized parameters of the Peng-Robinson equation of state are functions of the critical temperature, the critical pressure, and the acentric factor ω of the fluid. Consequently, the Peng-Robinson equation of state, as generalized here, is said to be a three-parameter (Tc , Pc , ω) equation of state, whereas the van der Waals equation contains only two parameters, Tc and Pc . This generalized form of the Peng-Robinson equation of state (or other equations of state) can be used to compute not only the compressibility, but also the departure functions for the other thermodynamic properties. This is done using Eqs. 6.4-29 and 6.4-30. In particular, to obtain numerical values for the enthalpy or entropy departure for a fluid that obeys the Peng-Robinson equation of state, one uses the following procedure: 1. Use the critical properties and acentric factor of the fluid to calculate b, κ, and the temperature-independent part of a using Eqs. 6.7-1, 6.7-2, and 6.7-4. 2. At the temperature of interest, compute numerical values for α and a using Eqs. 6.7-1 and 6.7-3. 3. Solve the equation of state, Eq. 6.4-2, for V and compute Z = P V /RT . Alternatively, solve for Z directly from the equivalent equation

Z 3 − (1 − B)Z 2 + (A − 3B 2 − 2B)Z − (AB − B 2 − B 3 ) = 0

(6.7-5)

where B = P b/RT and A = aP/R2 T 2 . 4. Use the computed value of Z and For Peng-Robinson equation of state

 R2 Tc2 da α κ = −0.45724 dT Pc T Tc to compute [H(T, P ) − H IG (T, P )] and/or [S(T, P ) − S IG (T, P )] as desired, using Eqs. 6.4-29 and 6.4-30. The enthalpy and entropy departures from ideal gas behavior calculated in this way can be used to solve thermodynamic problems in the same manner as the similar functions obtained from the corresponding-states graphs were used in the previous section.

6.7 Generalized Equations of State 265 It is clear that the calculation outlined here using the Peng-Robinson equation of state is, when doing computations by hand, more tedious than merely calculating the reduced temperature and pressure and using the graphs in Sec. 6.6. However, the equations here have some important advantages with the use of digital computers. First, this analytic computation avoids putting the three corresponding-states graphs in a computer memory in numerical form. Second, the values of the compressibility factor and departures from ideal gas properties obtained in the present three-parameter calculation should be more accurate than those obtained from the simple two-parameter (Tc , Pc ) corresponding-states method of the previous section because of the additional fluid parameter (acentric factor) involved. Also, there is an absence of interpolation errors. Finally, if, at some time in the future, it is decided to use a different equation of state, only a few lines of computer code need be changed, as opposed to one’s having to draw a complete new series of corresponding-states graphs. Before there was easy access to electronic calculators and computers, it was common practice to apply the corresponding-states principle by using tables and graphs as illustrated in the previous section. Now, however, the usual industrial practice is to directly incorporate the corresponding-states idea that different fluids obey the same form of the equation of state, by using digital computer programs and generalized equations of state such as the one discussed here. This is demonstrated in the following illustration. Illustration 6.7-1 Using the Peng-Robinson Equation of State to Solve a Real Gas Problem Rework Illustration 6.5-1 assuming that nitrogen can be described using the Peng-Robinson equation of state.

Solution The equations for solving this problem are the same as in Illustration 6.5-1. Specifically, the final temperature and pressure in the tank should be such that the molar entropy of gas finally in the tank is equal the initial molar entropy (Eq. c of Illustration 6.5-1), and the final molar volume should be such that the correct number of moles of gas remains in the tank (Eq. e of Illustration 6.5-1). The main difference here is that the Peng-Robinson equation of state is to be used in the solution. The general procedure used to calculate thermodynamic properties from the Peng-Robinson equation of state, and specifically for this problem, is as follows: 1. Choose values of T and P (these are known for the initial state and here will have to be found by trial and error for the final state). 2. Calculate a and b using Eqs. 6.7-1 through 6.7-4 and then A = aP/R2 T 2 and B = P b/RT . 3. Find the compressibility Z or the molar volume (for the vapor phase in this problem) by solving the cubic equation, Eq. 6.7-5 (here for the largest root). 4. Using the value of Z found above and Eq. 6.4-30 to calculate the entropy departure from ideal gas behavior, S − S IG . (Note that though they are not needed in this problem, the enthalpy departure and other properties can also be computed once the compressibility is known.) The equations to be solved are first the Peng-Robinson equation of Eqs. 6.7-1 to 6.7-4 for the initial molar volume or compressibility, and then the initial number of moles in the tank using Eq. d of Illustration 6.5-2. The results, using the Visual Basic computer program described in Appendix B.I-2, the DOS-based program PR1 described in Appendix B.II-1, the MATHCAD worksheet described in Appendix B.III, or the MATLAB program described in B.IV included

266 Chapter 6: The Thermodynamic Properties of Real Substances on the web site for this book, are V (t = 0) = 0.9567 × 10−4 m3 /mol

Z = 0.6769 N (t = 0) = 1567.9 mol

and

(S − S IG )t=0 = −9.18 J/(mol K)

Consequently, N (t = 50) = 1567.9 − 10 × 50 = 1067.9 mol

and

(a)

V (t = 50) = 0.15 m /1067.9 mol = 1.4046 × 10−4 m3 /mol 3

Also S(t = 50) − S(t = 0) = 0 = S IG (t = 50) − S IG (t = 0) + (S − S IG )t=50 − (S − S IG )t=0    T (t=50) ∗ P (t = 50) CP (T ) = dT − R ln T P (t = 0) T (t=0) + (S − S IG )t=50 − (S − S IG )t=0   T (t = 50) + 0.0042 × (T (t = 50) − 170) = 27.2 ln 170   P (t = 50) − R ln + (S − S IG )t=50 − (S − S IG )t=0 P (t = 0)

(b)

Equations a and b are to be solved together with the Peng-Robinson equation of state and Eq. 6.4-30 for the entropy departure. This can be done in several ways. The simplest is to use an equation-solving program; this is illustrated using the MATHCAD worksheet described in Appendix B III. A somewhat more tedious method is to use one of the other Peng-Robinson equation-of-state programs described in Appendix B in an iterative fashion. That is, one could use the following procedure: 1. Guess the final temperature of the expansion process (the ideal gas result is generally a good initial guess). 2. Using the Peng-Robinson equation of state, iterate on the pressure until the correct value of V (t = 50) is obtained for the guessed value of T (t = 50). 3. With the values of T (t = 50) and P (t = 50) so obtained, check whether Eq. b above is satisfied. If it is, the correct solution has been obtained. If not, adjust the guessed value of T (t = 50) and repeat the calculation. The result, directly from the MATHCAD worksheet or after several iterations with the other programs following the procedure above, is T (t = 50 min) = 134.66 K

and

P (t = 50 min) = 40.56 bar

[Note that at these conditions (S − S )t=50 = −10.19 J/mol.] IG

R Alternatively, Aspen Plus with the Peng-Robinson equation of state can be used to solve this problem.

Though we will usually use the generalized Peng-Robinson equation of state for calculations and illustrations in this text, it is of interest to also list the generalized version of the Soave-Redlich-Kwong equation of state since it is also widely used in industry: Soave-Redlich-Kwong equation of state

P =

a(T ) RT − V − b V (V + b)

(6.4-1b)

6.8 The Third Law of Thermodynamics 267 with

a(T ) = 0.42748 b = 0.08664

R2 Tc2 α(T ) Pc

(6.7-6)

RTc Pc

(6.7-7)

    T α(T ) = 1 + κ 1 − Tc and

κ = 0.480 + 1.574ω − 0.176ω 2

(6.7-8)

(6.7-9)

6.8 THE THIRD LAW OF THERMODYNAMICS In most treatises on thermodynamics, it is usual to refer to the laws of thermodynamics. The conservation of energy is referred to as the First Law of Thermodynamics, and this principle was discussed in detail in Chapter 3. The positive-definite nature of entropy generation used in Chapter 4, or any of the other statements such as those of Clausius or Kelvin and Planck, are referred to as the Second Law of Thermodynamics. The principle of conservation of mass precedes the development of thermodynamics, and therefore is not considered to be a law of thermodynamics. There is a Third Law of Thermodynamics, though it is less generally useful than the first two. One version of the third law is

Third law of thermodynamics

The entropy of all substances in the perfect crystalline state (for solids) or the perfect liquid state (for example, for helium) is zero at the absolute zero of temperature (0 K). Before we can use this statement, the perfect state must be defined. Here by “perfect” we mean without any disturbance in the arrangement of the atoms. That is, the substance must be without any vacancies, dislocations, or defects in the structure of the solid (or liquid) and not contain any impurities. The statement of the third law here is somewhat too constraining. A more correct statement is that all substances in the perfect state mentioned above should have the same value of entropy at 0 K, not necessarily a value of zero. It is mostly for convenience in the preparation of thermodynamic tables that a value of entropy of zero at 0 K is chosen. There are several implications of the above statement. The first obvious one is that there will be no entropy change on a chemical reaction at 0 K if each of the reacting substances is in a perfect state, to produce one or more products in perfect states. In fact, it was this observation that led to the formulation of the third law. A second implication, which is less obvious and is sometimes used as an alternative statement of the third law, is It is impossible to obtain a temperature of absolute zero. This statement is proved as follows. From Eq. 6.2-20, we have

CP dS = dT − T



∂V ∂T

 dP = P

CP dT T

(6.2-20)

for a change at constant pressure. To continue we divide by dT and take the limits as T → 0 to obtain

268 Chapter 6: The Thermodynamic Properties of Real Substances

 CP ∂S lim = lim (6.8-1) T →0 ∂T P T →0 T There is experimental evidence showing that the constant-pressure heat capacity is finite and positive in value at all temperatures, and zero at absolute zero. Since CP is positive and T is zero, we have   ∂S lim =∞ (6.8-2) T →0 ∂T P 

There are two possible conclusions from this equation. One is that as T decreases to absolute zero (so dT is negative), the entropy of any substance will become negative infinity. However, the experimental evidence is that the Gibbs energy of a substance, G = H − T S , converges to its enthalpy as absolute zero is approached, which means that the entropy must be finite. Therefore, the second, alternative conclusion is that it is not possible to reach a temperature of 0 K. This interpretation is the correct one, and in fact 0 K has not been attained in the laboratory. (However, with considerable effort temperatures of the order of 20 to 100 nK have been obtained.)

6.9 ESTIMATION METHODS FOR CRITICAL AND OTHER PROPERTIES To use either the generalized equations of state (such as the Peng-Robinson and Soave–Redlich-Kwong equations) or the method of corresponding states, one needs information on the critical and other properties of the fluids of interest. The discussion so far has been concerned with molecules for which such data are available. However, an issue that arises is what to do when one does not have such data, either because the data are not available, (e.g., the compound one wishes to study has not yet been made in the laboratory) or perhaps because one is interested in a preliminary identification of which compounds or which class of compounds might have certain desired properties. This is especially the case in “product engineering,” where one is interested in creating a compound or mixture of compounds with certain desired properties. This might be done by using a fast computational method to narrow the search of compounds with the desired properties, and then going to the library, searching the Web, or doing measurements in the laboratory to determine if the compounds so identified actually do have the desired properties. This task of identifying compounds with specific properties to make a new product is different from the usual job of a chemical engineer, which is “process engineering,” that is, designing a process to make a desired product. Also, the thermodynamic properties of most pharmaceuticals and naturally occurring biologically-produced chemicals are unknown, and this is another case where being able to make some estimates, even very approximate ones, can be useful in developing purification methods. The most common way to make properties estimates in the absence of experimental data is to use various group contribution methods. The basis of the method is that a molecule is thought of as a collection of functional groups, each of which makes an additive, though not necessarily linear, contribution to the properties of the molecule. Then as a result of summing up the contributions of each of the functional groups, the properties of the molecule are obtained. The underlying idea is that all molecules can be assembled from a limited number of functional groups (much in the same way that all of English literature can be created from only the 26 letters or functional groups in the alphabet). We will consider only one simple group contribution method for estimating

6.9 Estimation Methods for Critical and Other Properties 269 pure component properties here, that of Joback17 , though a number of other methods exist. However, before we proceed, a word of caution. Any group contribution method is inherently approximate and has some shortcomings. For example, it is assumed that a functional group makes the same contribution to the properties of a molecule independent of the molecule, and also which other functional groups it is bound to. This is a serious assumption, and one that is not generally true. For example, a methylene group, −CH2 −, makes a different contribution to the properties of a molecule if its binding partners are other methylene groups, halogens, or alcohols. Also, since simple group contribution methods are based on merely counting the number of each type of functional group, and not on their location within the molecule, they do not distinguish between isomers. In principle, group contribution methods can be improved by accounting for the first nearest neighbors, or the first and second nearest neighbors of each functional group. However, this makes the method much more difficult to apply, and would require extensive high-accuracy data and complicated data regression methods to obtain the contributions of each group. Instead, generally only simple group contributions methods are used, with the understanding that the results will be of uncertain accuracy and only of use for preliminary analysis, not for engineering design. The Joback group contribution method uses the following equations:

Tc (K) =

0.584 + 0.965 ·

Tb (K)  2 i νi Tc,i − ( i νi Tc,i )



1 Pc (bar) =  2  0.113 + 0.0032 · n − νi · Pc,i  Vc

3

cm mol

i

 = 17.5 +

Tb (K) = 198 +





νi · Vc,i

i

νi · Tb,i

i

Tf (K) = 122 +



νi · Tf,i

i

In these equations, the subscripts c, b, and f indicate the critical point, boiling point, and freezing point, respectively; the terms are the contributions of the group to each of the specified properties given in the following table; and νi is the number of functional groups of type i in the molecule. To use the generalized form of, for example, the Peng-Robinson or Soave–RedlichKwong equations of state, one also needs the acentric factor. If the vapor pressure of the substance is known as a function of temperature, and the critical properties are known, the acentric factor can be computed from its definition,

ω = −log

17 See,

P vap (T = 0.7Tc ) −1 Pc

for example, Chapter 2 of B. E. Poling, J. M. Prausnitz, and J. P. O’Connell, The Properties of Gases and Liquids, 5th ed., McGraw-Hill, New York, 2001

270 Chapter 6: The Thermodynamic Properties of Real Substances

Tc

P c

0.0141 0.0189 0.0100 0.0164

−0.0012 0.0000 0.0025 0.0020

CH− ring

0.0122

C nonring

− −

Table 6.9-1 Joback Group Contributions to Pure Component Properties Group

−

−

−CH3 nonring CH2 nonring −CH2 − ring CH− nonring

−

− −

− − −

−

−

−

−

−

C ring =CH2 nonring =CH− nonring =CH− ring =C nonring

−

− −

=C ring =C= nonring ≡CH nonring ≡C− nonring −F all −Cl all −Br all −I all −OH alcohol −OH phenol −O− nonring −O− ring C=O nonring

−

− − −

C=O ring O=CH− aldehyde −COOH acid −COO− nonring =O other −NH2 all NH nonring

−

− − −

NH ring

−

−

N− nonring −N= nonring −N= ring −CN all −NO2 all −SH all −S− nonring −S− ring

V c

Tb

Tf

65 56 48 41

23.58 22.88 27.15 21.74

−5.1 11.27 7.75 12.64

0.0004

38

21.78

19.88

0.0067

0.0043

27

18.25

46.43

0.0042 0.0113 0.0129 0.0082 0.0117

0.0061 −0.0028 −0.0006 0.0011 0.0011

27 56 46 41 38

21.32 18.18 24.96 26.73 24.14

60.15 −4.32 8.73 8.13 11.14

0.0143 0.0026 0.0027 0.0020 0.0111 0.0105 0.0133 0.0068 0.0741 0.0240 0.0168 0.0098 0.0380

0.0008 0.0028 −0.0008 0.0016 −0.0057 −0.0049 0.0057 −0.0034 0.0112 0.0184 0.0015 0.0048 0.0031

32 36 46 37 27 58 71 97 28 −25 18 13 62

31.01 26.15 9.2 27.38 −0.03 38.13 66.86 93.84 92.88 76.34 22.42 31.22 76.75

37.02 17.78 −11.18 64.32 −15.78 13.55 43.43 41.69 44.45 82.83 22.23 23.05 61.2

0.0284 0.0379 0.0791 0.0481 0.0143 0.0243 0.0295

0.0028 0.003 0.0077 0.0005 0.0101 0.0109 0.0077

55 82 89 82 36 38 35

94.97 72.24 169.09 81.1 −10.5 −10.5 50.17

75.97 36.9 155.5 53.6 2.08 2.08 52.66

0.0130

0.0114

29

52.82

101.51

0.0169 0.0255 0.0085 0.0496 0.0437 0.0031 0.0119 0.0019

0.0074 −0.0099 0.0076 −0.0101 0.0064 0.0084 0.0049 0.0051

9 0 34 91 91 63 54 38

11.74 74.6 57.55 125.66 152.54 63.56 68.78 52.1

48.84 0 68.4 59.89 127.24 20.09 34.4 79.93

6.9 Estimation Methods for Critical and Other Properties 271 When such information is not available, the following approximate equation can be used:

ω=

3 Tbr log Pc − 1 7 1 − Tbr

where Tbr = Tb /Tc . A very crude approximation, which was not suggested by Joback, is to use an estimated rather than a measured value for the normal boiling-point temperature in this equation should experimental data not be available. Illustration 6.9-1 Group Contribution Estimate of the Properties of a Pure Fluid Use the methods described above to estimate the properties of n-octane that has a boiling point of 398.8 K and ethylene glycol (1,2-ethanediol) that has a boiling point of 470.5 K. Also compare the estimates of using and not using the measured boiling points.

Solution The results for n-octane are as follows: Experiment

Using Tb

Not Using Tb

398.8 216.4 568.8 24.9 492 0.392

398.8 179.4 569.2 25.35 483.5 0.402

382.4 179.4 545.9 25.35 483.5 0.402

Experiment

Using Tb

Not Using Tb

470.5 260.2 645.0 77.0

470.5 233.4 645.5 66.5 185.5 1.094

429.5 233.4 589.3 66.5 185.5 1.094

Tb (K) Tf (K) Tc (K) Pc (bar) Vc (cc/mol) ω

The results for ethylene glycol are

Tb (K) Tf (K) Tc (K) Pc (bar) Vc (cc/mol) ω

? ?

We see that while none of the results are perfect, the estimates for n-octane are reasonably good, while those for ethylene glycol are less accurate. However, all the predictions are good enough to provide at least a qualitative estimate of the properties of these fluids. That is especially important when experimental data are not available, as is the case here for the critical volume and acentric factor of ethylene glycol. Note also that if a measured value of the normal boiling temperature is available and used, it results in a considerably more accurate value of the critical temperature than is the case if the information is not available.

Comment Another situation in which approximate group contribution methods are especially useful is in product design where an engineer can ask (on a computer) what changes in properties would

272 Chapter 6: The Thermodynamic Properties of Real Substances result if one or more functional groups were added, removed or replaced in a molecule. An example of a product design will be given later in this book However, it is important to remember that all group contribution methods are very approximate. Therefore, while the methods discussed here can be used to obtain a preliminary estimate of the properties of a molecule, they should be verified against measurements before being used in an engineering design.

6.10 SONIC VELOCITY When a pressure pulse is introduced into a gas (or any compressible fluid), for example, something so simple as a spoken word or as complex as an explosion, this pressure wave travels at a fixed velocity, referred to as the sonic velocity or simply the speed of sound, that depends on the properties of the fluid. It is the sonic velocity that we wish to compute here. So consider the system shown in Fig. 6.10-1 in which the small movement of a piston in a cylindrical tube initiates a pressure wave that travels at velocity c, the speed of sound. This pressure wave will have a very sharp wave front. We can consider the fluid in the tube to consist of the region immediately on the left through which the pressure wave has already passed, state 1, and the region on the right, state 2, that has not yet experienced the pressure wave. Note that the pressure wave and the fluids in both regions adjacent to the pressure wave are in motion because of the difference in pressure. The fluid adjacent to the pressure wave in region 2 has the velocity v2 = c and that in region 1 has a velocity v1 = c − δv ; the fluid far ahead of the pressure wave is quiescent.

State 1

State 2

n1

n2 = c

H1(P1, ρ1)

H2(P2, ρ2)

Pressure wave

Figure 6.10-1 Pressure wave in a gas

We start by writing the mass balance for a region of infinitesimal thickness that includes a small pressure wave and its immediate surrounding gas of molecular weight mw on both sides of the wave. The downstream enthalpy will be denoted by H 2 (P, ρ) and as the pressure difference between the regions upstream and downstream of the shock wave is small, the properties in the upstream region are at the slightly higher pressure P + δP and density ρ + δρ, and the enthalpy will be denoted as H 1 (P1 , ρ1 ) = H(P + δP, ρ + δρ) = H + δH . With this notation, the mass balance on the control volume is

N˙ 1 = ρ1 Av1 = N˙ 2 = ρ2 Av2 ;

(ρ + δρ)(c − δv) = ρc or δρ δρ − ρδv + cδρ − (δv)(δρ) = 0 so that δv = c ≈c ρ + δρ ρ Also

(6.10-1)

assuming that the pressure pulse is sufficiently small so that the density variation is small with respect to the total density

6.10 Sonic Velocity 273 The energy balance for the control volume is ! 1 ! 1 N˙ 1 H 1 (P1 , ρ1 ) + mw v12 = N˙ 1 H + δH + mw (c − v)2 2 2 !  1 1 = N˙ 2 H 2 (P2 , ρ2 ) + mw v22 = N˙ 2 H + mw c2 ; and since N˙ 1 = N˙ 2 2 2 1 1 2 1 2 δH + mw (c − v) = mw c or δH + mw (−cv + v 2 ) = 0 2 2 2 Now since the sonic veloity c is much larger than the fluid velocity v , neglecting the v 2 term with respect to the cv term, and using Eqn. 6.10-1 gives  δρ  ; or δH − mwcv = 0 and δH = mwcv = mwc c ρ  ρ δH ρ δH 2 c = and c = mw δρ mw δρ To complete the analysis, we have to determine the conditions under which the derivative δH/δρ is to be evaluated. Since the passage of the shock wave occurs very rapidly, the process is adiabatic. Less obvious is that the process can be considered to be reversible since each element of fluid differes in properties (temperature, pressure, density) only infinitesimally from its neighboring fluid elements. So that it is essentially in equilibrium with its neighbors, so that process considered to be reversible. can be ρ  ∂H  . Therefore, the speed of sound c is given by c = mw ∂ρ S Starting from  ∂P   ∂H   ∂P  dP dH = T dS +V dP = T dS + so that ρ = = −V 2 ρ ∂ρ S ∂ρ S ∂V S Now using the triple product rule gives

  ∂S  ∂P   ∂V   ∂S   ∂P  ∂V P = −1 or = −  ∂S ∂V S ∂S P ∂P V ∂V S ∂P

V

CP CP  ∂T  Using dS = dP gives = dT − T ∂T P ∂V P T ∂V P    ∂S  CV CV  ∂T  ∂P Also, starting from dS = dV gives = dT + T ∂T V ∂P V T ∂P V Therefore,     ∂T CP ∂T  ∂P  C P ∂V T ∂V P = −   = −  P and ∂T CV ∂T ∂V S CV ∂P T ∂P  ∂V 

 ∂S 

V

ρ

 ∂H  ∂ρ

= S

 ∂P  ∂ρ

S

= −V 2

V

 ∂P  ∂V

=V2 S

CP T CV T

 

∂T ∂V ∂T ∂P

 P V

274 Chapter 6: The Thermodynamic Properties of Real Substances

"   # CP ∂T # # 1 T ∂V P 2 c=# $ mw V CV  ∂T 

Also then

T

∂P

V

For the ideal gas

P V = RT,

so that

 ∂T  ∂V

P

=

P R

and

 ∂T  ∂P

= V

V R

Therefore, there speed of sound for an ideal gas is "   # %  CP ∂T # # 1 T ∂V 1 2 CP PR 1 CP P 2 #   = c=$ V V = RT V ∂T C mw mw CV R mw CV V T

∂P

V

So for the ideal gas, the speed of sound depends on the square root of the ratio of the constant pressure to constant volume heat capacities, on the square root of the temperature, and is inversely proportional to the square root of the molecular weight. Illustration 6.10-1 Compression of air For the purposes here, air can be considered to be a single-component gas (rather than a 21 mol % oxygen-79 mol % nitrogen mixture with a molecular weight of 29 and for which CP = 7R/2. Compute the velocity of sound in air at room temperature (298 K) and on a very cold morning in Minnesota when the temperature is −40◦ C.

Solution a. Since CP = 7R/2, then CV = CP − R = 5R/2 for the ideal gas. Therefore, at 298 K  1 CP RT = mw CV

c= =

√

119607



1 g 29 mol

J 7R/2 kg m2 g × 8.314 × 2 × × 298 K 5R/2 mol K s J kg

m m = 345.8 s s

b. At −40◦ C = 233.15 K 

 J 1 CP 1 7R/2 kg m2 g c= RT = × 8.314 × 2 × × 233.15 K g mw CV 29 mol 5R/2 mol K s J kg √ m m = 93578.4 = 305.9 s s

The reason that the speed of sound (or sonic velocity) is of interest to chemical engineers is that it the highest velocity possible that a gas can achieve. This is, for example, the highest gas velocity that could be obtained in the throat of a valve or in a converging nozzle, that is a nozzle whose cross-sectional area decreases in the direction of flow. Generally, for flow through a valve or converging nozzle, the mass flow rate will

6.11 More About Thermodynamic Partial Derivatives (Optional) 275 increase as the upstream pressure is increased or the downstream pressure is reduced. However, this will not occur if the velocity of the gas in the valve or nozzle was already at the sonic velocity. The flow then said to be choked, and the mass flow rate for the given geometry cannot be increased by higher upstream pressure or lower downstream pressure. The design of nozzles and valves to avoid choking flow is a problem that we leave to mechanical engineers and aerodynamicists. One final term to introduce, largely because it appears in both aerodynamics and the nontechnical literature, is the Mach number M that is defined as the ratio of the actual gas velocity v to the sonic velocity c at the same temperature and pressure. That is,

v c The Mach number is an important parameter in the flow of gases. However, it will not be considered further. M=

6.11 MORE ABOUT THERMODYNAMIC PARTIAL DERIVATIVES (OPTIONAL) This section appears on the website for this book.

PROBLEMS For some of these problems, it will be helpful to use the computer programs and/or MATHCAD worksheets on the website for this book and described in Appendix B in solving part of all of the problem. 6.1 For steam at 500◦ C and 10 MPa, using the Mollier diagram, a. Compute the Joule-Thomson coefficient μ = (∂T /∂P )H . b. Compute the coefficient κS = (∂T /∂P )S . c. Relate the ratio (∂H/∂S)T /(∂H/∂S)P to μ and κS , and compute its value for steam at the same conditions. 6.2 Derive Eqs. 6.4-29 and 6.4-30. 6.3 Evaluate the difference 

∂U ∂T





− P

∂U ∂T

 V

for the ideal and van der Waals gases, and for a gas that obeys the virial equation of state. 6.4 One of the beauties of thermodynamics is that it provides interrelationships between various state variables and their derivatives so that information from one set of experiments can be used to predict the results of a completely different experiment. This is illustrated here. a. Show that CP =

T2 μ



∂(V /T ) ∂T

 P

Thus, if the Joule-Thomson coefficient μ and the volumetric equation of state (in analytic or tabular form) are known for a fluid, CP can be computed. Alternatively, if CP and μ are known, (∂(V /T )/∂T )P can be calculated, or if CP and (∂(V /T )/∂T )P are known, μ can be calculated. b. Show that

V (P, T2 ) =

T2 V (P, T1 ) + T2 T1



P,T2 P,T1

μCP dT T2

so that if μ and CP are known functions of temperature at pressure P , and V is known at P and T1 , the specific volume at P and T2 can be computed. 6.5 Derive Eqs. 6.6-2 and 6.6-3, and show that Zc |van der Waals = 3/8. 6.6 Derive Eq. 6.6-11. 6.7 One hundred cubic meters of carbon dioxide initially at 150◦ C and 50 bar is to be isothermally compressed in a frictionless piston-and-cylinder device to a final pressure of 300 bar. Calculate i. The volume of the compressed gas ii. The work done to compress the gas iii. The heat flow on compression assuming carbon dioxide a. Is an ideal gas b. Obeys the principle of corresponding states of Sec. 6.6 c. Obeys the Peng-Robinson equation of state

276 Chapter 6: The Thermodynamic Properties of Real Substances 6.8 By measuring the temperature change and the specific volume change accompanying a small pressure change in a reversible adiabatic process, one can evaluate the derivative   ∂T ∂P

S

and the adiabatic compressibility κS = −

1 V



∂V ∂P

 S

Develop an expression for (∂T /∂P )S in terms of T , V , CP , α, and κT , and show that CV κS = κT CP

6.9 Prove that the following statements are true. a. (∂H/∂V )T is equal to zero if (∂H/∂P )T is equal to zero. b. The derivative (∂S/∂V )P for a fluid has the same sign as its coefficient of thermal expansion α and is inversely proportional to it. 6.10 By measuring the temperature change accompanying a differential volume change in a free expansion across a valve and separately in a reversible adiabatic expansion, the two derivatives (∂T /∂V )H and (∂T /∂V )S can be experimentally evaluated. a. Develop expressions for these derivatives in terms of the more fundamental quantities. b. Evaluate these derivatives for a van der Waals fluid. 6.11 a. Show for the Peng-Robinson equation of state (Eq. 6.4-2) that

and b = 0.07780RTc /Pc

b. Determine the critical compressibility of the PengRobinson equation of state. 6.12 Ethylene at 30 bar and 100◦ C passes through a heaterexpander and emerges at 20 bar and 150◦ C. There is no flow of work into or out of the heater-expander, but heat is supplied. Assuming that ethylene obeys the Peng-Robinson equation of state, compute the flow of heat into the heater-expander per mole of ethylene. 20 bar 150°C

Q

PV =1 RT (V →∞)   PV −1 =B lim V P →0 RT (V →∞)   PV B =C −1− lim V 2 P →0 RT V (V →∞) lim

P →0

a. Using these formulas, show that the van der Waals equation leads to the following expressions for the virial coefficients. B =b−

a RT

C = b2

b. The temperature at which 

lim

P →0 (or V →∞)

V

 PV −1 =B =0 RT

is called the Boyle temperature. Show that for the van der Waals fluid TBoyle = 3.375Tc

a(Tc ) = 0.45724R2 Tc2 /Pc

30 bar 100°C

6.13 A natural gas stream (essentially pure methane) is available at 310 K and 14 bar. The gas is to be compressed to 345 bar before transmission by underground pipeline. If the compression is carried out adiabatically and reversibly, determine the compressor outlet temperature and the work of compression per mole of methane. You may assume that methane obeys the Peng-Robinson equation of state. See Appendix A.II for heat capacity data. 6.14 Values of the virial coefficients B and C at a fixed temperature can be obtained from experimental P V T data by noting that

where Tc is the critical temperature of the van der Waals fluid given by Eqs. 6.6-3. (For many real gases TBoyle is approximately 2.5Tc !) 6.15 From experimental data it is known that at moderate pressures the volumetric equation of state may be written as P V = RT + BP

where the virial coefficient B is a function of temperature only. Data for nitrogen are given in the table. a. Identify the Boyle temperature (the temperature at which B = 0) and the inversion temperature [the temperature at which μ = (∂T /∂P )H = 0] for gaseous nitrogen. b. Show, from the data in the table, that at temperatures above the inversion temperature the gas temperature increases in a Joule-Thomson

Problems 277 T (K)

B (cm3 /mol)

75 100 125 150 200 250 300 400 500 600 700

−274 −160 −104 −71.5 −35.2 −16.2 −4.2 +9.0 +16.9 +21.3 +24.0

6.19

Source: J. H. Dymond and E. B. Smith, The Virial Coefficients of Gases, Clarendon Press, Oxford, 1969, p. 188.

expansion, whereas it decreases if the initial temperature is below the inversion temperature. c. Describe how you would find the inversion temperature as a function of pressure for nitrogen using Fig. 3.3-3 and for methane using Fig. 3.3-2. 6.16 Eighteen kilograms of the refrigerant HFC-134a at 150◦ C is contained in a 0.03-m3 tank. Compare the prediction you can make for the pressure in the tank with that obtained using Fig. 3.3-4. For data, see Table 6.6-1. 6.17 Calculate the molar volume, enthalpy, and entropy of carbon tetrachloride at 300◦ C and 35 bar using the Peng-Robinson equation of state and the principle of corresponding states of Sec. 6.6. The following data are available:

6.20

H(T = 16◦ C, ideal gas at P = 0.1 bar) = 0 S(T = 16◦ C, ideal gas at P = 0.1 bar) = 0 Tc = 283.2◦ C

6.21

Pc = 45.6 bar Zc = 0.272

ω = 0.194

∗

CP : see Appendix A.II

6.18 The Clausius equation of state is P (V − b) = RT

a. Show that for this volumetric equation of state CP (P, T ) = CV (P, T ) + R CP (P, T ) = CP∗ (T )

and

∗ CV (V , T ) = CV (T )

b. For a certain process the pressure of a gas must be reduced from an initial pressure P1 to the final pressure P2 . The gas obeys the Clausius equation of

6.22

state, and the pressure reduction is to be accomplished either by passing the gas through a flow constriction, such as a pressure-reducing valve, or by passing it through a small gas turbine (which we can assume to be both reversible and adiabatic). Obtain expressions for the final gas temperature in each of these cases in terms of the initial state and the properties of the gas. A tank is divided into two equal chambers by an internal diaphragm. One chamber contains methane at a pressure of 500 bar and a temperature of 20◦ C, and the other chamber is evacuated. Suddenly, the diaphragm bursts. Compute the final temperature and pressure of the gas in the tank after sufficient time has passed for equilibrium to be attained. Assume that there is no heat transfer between the tank and the gas and that methane: a. is an ideal gas; b. obeys the principle of corresponding states of Sec. 6.6; c. obeys the van der Waals equation of state; d. obeys the Peng-Robinson equation of state; Data: For simplicity you may assume CP∗ = 35.56 J/(mol K). The divided tank of the preceding problem is replaced with two interconnected tanks of equal volume; one tank is initially evacuated, and the other contains methane at 500 bar and 20◦ C. A valve connecting the two tanks is opened only long enough to allow the pressures in the tanks to equilibrate. If there is no heat transfer between the gas and the tanks, what are the temperature and pressure of the gas in each tank after the valve has been shut? Assume that methane a. Is an ideal gas b. Obeys the principle of corresponding states of Sec. 6.6 c. Obeys the Peng-Robinson equation of state Ammonia is to be isothermally compressed in a specially designed flow turbine from 1 bar and 100◦ C to 50 bar. If the compression is done reversibly, compute the heat and work flows needed per mole of ammonia if a. Ammonia obeys the principle of corresponding states of Sec. 6.6. b. Ammonia satisfies the Clausius equation of state P (V − b) = RT with b = 3.730 × 10−2 m3 /kmol. c. Ammonia obeys the Peng-Robinson equation of state. A tank containing carbon dioxide at 400 K and 50 bar is vented until the temperature in the tank falls to 300 K. Assuming there is no heat transfer between the gas and the tank, find the pressure in the tank at the end of the venting process and the fraction of the initial mass of gas remaining in the tank for each of the following cases.

278 Chapter 6: The Thermodynamic Properties of Real Substances a. The equation of state of carbon dioxide is P (V − b) = RT

b = 0.0441 m3 /kmol

with

b. Carbon dioxide obeys the law of corresponding states of Sec. 6.6. c. Carbon dioxide obeys the Peng-Robinson equation of state. The low-pressure (ideal gas) heat capacity of CO2 is given in Appendix A.II. 6.23 Derive the equations necessary to expand Illustration 6.4-1 to include the thermodynamic state variables internal energy, Gibbs energy, and Helmholtz energy. 6.24 Draw lines of constant Gibbs and Helmholtz energies on the diagrams of Illustration 6.4-1. 6.25 The speed of propagation of a small pressure pulse or sound wave in a fluid, vS , can be shown to be equal to  vS =

∂P ∂ρ

 S

where ρ is the molar density. a. Show that an alternative expression for the sonic velocity is 

vS =



γV 2

∂T ∂V

  P

∂P ∂T

 V

where γ = CP /CV . b. Show that γ = 1 + R/CV for both the ideal gas and a gas that obeys the Clausius equation of state P (V − b) = RT

and that γ is independent of specific volume for both gases. c. Develop expressions for vS for both the ideal and the Clausius gases that do not contain any derivatives other than CV and CP . 6.26 The force required to maintain a polymeric fiber at a length L when its unstretched length is L0 has been observed to be related to its temperature by F = γT (L − L0 )

where γ is a positive constant. The heat capacity of the fiber measured at the constant length L0 is given by CL = α + βT

where α and β are parameters that depend on the fiber length. a. Develop an equation that relates the change in entropy of the fiber to changes in its temperature and length, and evaluate the derivatives (∂S/∂L)T and (∂S/∂T )L that appear in this equation.

b. Develop an equation that relates the change in internal energy of the fiber to changes in its temperature and length. c. Develop an equation that relates the entropy of the fiber at a temperature T0 and an extension L0 to its entropy at any other temperature T and extension L. d. If the fiber at T = Ti and L = Li is stretched slowly and adiabatically until it attains a length Lf , what is the fiber temperature at Tf ? e. In polymer science it is common to attribute the force necessary to stretch a fiber to energetic and entropic effects. The energetic force (i.e., that part of the force that, on an isothermal extension of the fiber, increases its internal energy) is FU = (∂U/∂L)T , and the entropic force is FS = −T (∂S/∂L)T . Evaluate FU and FS for the fiber being considered here. 6.27 Derive the following Maxwell relations for open systems. a. Starting from Eq. 6.2-5a, 

 

∂T ∂V ∂T ∂N ∂P ∂N





 ∂P ∂S V,N   ∂G = ∂S V,N   ∂G =− ∂V S,N =−

S,N



S,V



S,V

b. Starting from Eq. 6.2-6a, 

 

∂T ∂P ∂T ∂N ∂V ∂N





= S,N





= S,P





= S,P

∂V ∂S ∂G ∂S ∂G ∂P

 

P,N

P,N



S,N

c. Starting from Eq. 6.2-7a, 

 

∂S ∂V ∂S ∂N ∂P ∂N



 ∂P ∂T V,N   ∂G =− ∂T V,N   ∂G =− ∂V T,N 

= T,N



T,V



T,V

d. Starting from Eq. 6.2-8a, 

 

∂S ∂P ∂S ∂N ∂V ∂N





=− T,N



T,P



T,P



∂V ∂T

 P,N

 ∂G =− ∂T P,N   ∂G = ∂P T,N

Problems 279 a. The gas is an ideal gas. b. The gas obeys the virial equation of state

6.28 For real gases the Joule-Thomson coefficient is greater than zero at low temperatures and less than zero at high temperatures. The temperature at which μ is equal to zero at a given pressure is called the inversion temperature. a. Show that the van der Waals equation of state exhibits this behavior, and develop an equation for the inversion temperature of this fluid as a function of its specific volume. b. Show that the van der Waals prediction for the inversion temperature can be written in the corresponding-states form

Trinv

C B PV + 2 =1+ RT V V

with B = −10.3 × 10−6 m3 /mol and C = 1517 × 10−12 m6 /mol2 . c. The gas is described by the van der Waals equation P =

or equivalently V PV a = − RT V −b V RT

3(3Vr − 1)2 = 4Vr2

c. The following graph shows the inversion temperature of nitrogen as a function of pressure.18 Plot on this graph the van der Waals prediction for the inversion curve for nitrogen.

6.30

350 300 250

6.31

Temperature, °C

200 150 100 50 0

6.32

–50 –100

6.33

–150 –200 0

40

80 120 160 200 240 280 320 360 400 Pressure, bar

6.29 Nitrogen is to be isothermally compressed at 0◦ C from 1 bar to 100 bar. Compute the work required for this compression; the change in internal energy, enthalpy; Helmholtz and Gibbs energies of the gas; and the heat that must be removed to keep the gas at constant temperature if 18 From

RT a − 2 V −b V

with a = 0.1368 Pa m6 /mol2 and b = 3.864 × 10−5 m3 /mol. d. The gas is described by the Peng-Robinson equation of state. For an isothermal process involving a fluid described by the Redlich-Kwong equation of state, develop expressions for the changes in a. Internal energy b. Enthalpy c. Entropy in terms of the initial temperature and the initial and final volumes. Steam is continuously expanded from a pressure of 25 bar and 300◦ C to 1 bar through a Joule-Thomson expansion valve. Calculate the final temperature and the entropy generated per kilogram of steam using a. The ideal gas law b. The van der Waals equation of state c. The Peng-Robinson equation of state d. The steam tables Repeat the calculations of Problem 6.13 if the mechanical efficiency of the adiabatic turbine is only 85 percent. In statistical mechanics one tries to find an equation for the partition function of a substance. The canonical partition function, Q(N, V, T ), is used for a closed system at constant temperature, volume, and number of particles N . This partition function can be written as a product of terms as follows: Q(N, V, T ) =

f (T )N · V N · Z(N/V, T ) N!

where f (T ) is the part that depends only on the properties of a single molecule, and Z(N/V, T ) is a normalized configuration integral that is unity for an ideal

B. F. Dodge, Chemical Engineering Thermodynamics, McGraw-Hill. Used with permission of McGrawHill Book Company.

280 Chapter 6: The Thermodynamic Properties of Real Substances gas and depends on the interaction energies among the molecules for a real gas. Also, the Helmholtz energy is related to the canonical partition function as follows: A(N, V, T ) = −kT ln Q(N/V , T )

where k is Boltzmann’s constant (the gas constant divided by Avogadro’s number). Write expressions for all the thermodynamic properties of a fluid in terms of its canonical partition function and its derivatives. 6.34 Any residual property θ is defined to be θres (T, P ) = θ(T, P ) − θIG (T, P )

where IG denotes the same property in the ideal gas state. Such a quantity is also referred to as a departure function. a. Develop general expressions for dH res , dU res , dS res , and dGres with temperature and pressure as the independent variables. b. Recognizing that as P → 0 all fluids become ideal gases, so that their residual properties are zero, develop explicit expressions for the residual properties H res , U res , S res , and Gres as functions of temperature and pressure for the van der Waals equation of state. c. Repeat part (a) for the Peng-Robinson equation of state. d. Repeat part (a) for the Redlich-Kwong equation of state. Problems Involving the Redlich-Kwong Equation of State (6.35–6.48). 6.35 a. Show for the Soave–Redlich-Kwong equation of state (Eq. 6.4-1) that a(T ) = 0.427 48 b = 0.086 64

R2 Tc2 α(T ) Pc

RTc Pc

b. Show that the critical compressibility of the Soave–Redlich-Kwong equation of state is 1/3. 6.36 Derive the expressions for the enthalpy and entropy departures from ideal gas behavior (that is, the analogues of Eqs. 6.4-29 and 6.4-30) for the Soave–Redlich-Kwong equation of state. 6.37 Repeat Illustration 6.4-1 using the Soave–RedlichKwong equation of state. 6.38 Repeat Illustration 6.7-1 using the Soave–RedlichKwong equation of state. 6.39 Redo Problem 6.7 with the Soave–Redlich-Kwong equation of state. 6.40 Redo Problem 6.12 with the Soave–Redlich-Kwong equation of state. 6.41 Redo Problem 6.22 with the Soave–Redlich-Kwong equation of state.

6.42 Using the Redlich-Kwong equation of state, compute the following quantities for nitrogen at 298.15 K. a. The difference CP − CV as a function of pressure from low pressures to very high pressures b. CP as a function of pressure from low pressures to very high pressures. [Hint: It is easier to first compute both CP − CV and CV for chosen values of volume, then compute the pressure that corresponds to those volumes, and finally calculate CP as the sum of (CP − CV ) + CV .] 6.43 The Boyle temperature is defined as the temperature at which the second virial coefficient B is equal to zero. a. Recognizing that any equation of state can be expanded in virial form, find the Boyle temperature for the Redlich-Kwong equation of state in terms of the parameters in that equation. b. The Redlich-Kwong parameters for carbon dioxide are a = 6.466 × 10−4 m6 MPa K0.5 mol−2 and b = 2.971 × 10−3 m3 mol−1 . Estimate the Boyle temperature for carbon dioxide, assuming that it obeys the Redlich-Kwong equation of state. 6.44 In the calculation of thermodynamic properties, it is convenient to have the following partial derivatives:  lim

P →0

∂Z ∂P



 and

lim

P →∞

T

∂Z ∂P

 T

where Z = (P V /RT ) is the compressibility factor. Develop expressions for these two derivatives for the Redlich-Kwong equation of state in terms of temperature and the Redlich-Kwong parameters. 6.45 The second virial coefficient B can be obtained from experimental P V T data or from an equation of state from   B = lim V P →0

PV −1 RT

a. Show that for the Redlich-Kwong equation P =

RT a −√ V −b T V (V + b)

the second virial coefficient is B =b−

a RT 3/2

b. Compute the second virial coefficient of n-pentane as a function of temperature from the RedlichKwong equation of state. 6.46 The Joule-Thomson coefficient, μ, given by  μ=

∂T ∂P

 =− H

V [1 − T α] CP

is a function of temperature. The temperature at which μ = 0 is known as the inversion temperature. a. Use the van der Waals equation of state to determine the inversion temperature of H2 , O2 , N2 , CO,

Problems 281

6.47

6.48

6.49 6.50 6.51

and CH4 . The van der Waals parameters for these gases can be found in Table 6.4-1. b. Repeat part (a) using the Redlich-Kwong equation of state. Using the Redlich-Kwong equation of state, compute and plot (on separate graphs) the pressure of nitrogen as a function of specific volume at the two temperatures: a. 110 K b. 150 K Use the information in Illustration 6.4-1 and the Soave–Redlich-Kwong equation of state to compute the thermodynamic properties of oxygen along the following two isotherms: a. 155 K b. 200 K Repeat Problem 5.9 assuming that helium is described by the Peng-Robinson equation of state. Repeat Problem 5.10 assuming that nitrogen is described by the Peng-Robinson equation of state. In this chapter, from the third law of thermodynamics, it has been shown that the entropy of all substances approaches a common value at 0 K (which for convenience we have taken to be zero). This implies that the value of the entropy at 0 K is not a function of volume or pressure. Use this information to show the following: CV = 0

at T = 0 K and that the coefficient of thermal expansion   α=

1 V

∂V ∂T

P

is zero at 0 K. 6.52 A fluid is described by the Clausius equation of state P =

RT V −b

where b is a constant. Also, the ideal gas heat capacity is given by CP∗ = α + βT + γT 2

For this fluid, obtain explicit expressions for a. A line of constant enthalpy as a function of pressure and temperature b. A line of constant entropy as a function of temperature and pressure c. Does this fluid have a Joule-Thomson inversion temperature? 6.53 A piston-and-cylinder device contains 10 kmol of npentane at −35.5◦ C and 100 bar. Slowly the piston is moved until the vapor pressure of n-pentane is reached, and then further moved until 5 kmol of the n-pentane is evaporated. This complete process takes

place at the constant temperature of −35.5◦ C. Assume n-pentane can be described by the Peng-Robinson equation of state. a. What is the volume change for the process? b. How much heat must be supplied for the process to be isothermal? 6.54 Develop an expression for how the constant-volume heat capacity varies with temperature and specific volume for the Peng-Robinson fluid. 6.55 A manuscript recently submitted to a major journal for publication gave the following volumetric and thermal equations of state for a solid: V (T, P ) = a + bT − cP

and

U (T, P ) = dT + eP

where a, b, c, d, and e are constants. Are these two equations consistent with each other? 6.56 The following equation of state describes the behavior of a certain fluid: P (V − b) = RT +

aP 2 T

where the constants are a = 10−3 m3 K/(bar mol) = 102 (J K)/(bar2 mol) and b = 8 × 10−5 m3 /mol. Also, for this fluid the mean ideal gas constant-pressure heat capacity, CP , over the temperature range of 0 to 300◦ C at 1 bar is 33.5 J/(mol K). a. Estimate the mean value of CP over the temperature range at 12 bar. b. Calculate the enthalpy change of the fluid for a change from P = 4 bar, T = 300 K to P = 12 bar and T = 400 K. c. Calculate the entropy change of the fluid for the same change of conditions as in part (b). 6.57 The van der Waals equation of state with a = 0.1368 Pa m3 /mol2 and b = 3.864 × 10−5 m3 /mol can be used to describe nitrogen. a. Using the van der Waals equation of state, prepare a graph of pressure (P ) of nitogen as a function of log V , where V is molar volume at temperatures of 100 K, 125 K, 150 K, and 175 K. The range of the plot should be V = 1 × 10−4 to 25 m3 /mol. b. One mole of nitrogen is to be isothermally compressed from 100 kPa to 10 MPa at 300 K. Determine (1) the molar volume at the initial and final conditions and (2) the amount of work necessary to carry out the isothermal compression. c. Repeat the calculation of part (b) for an ideal gas. 6.58 A bicycle pump can be treated as a piston-and-cylinder system that is connected to the tire at the “closed” end of the cylinder. The connection is through a valve that is initially closed, while the cylinder is filled with air at atmospheric pressure, following which the pumping in a cycle occurs as the plunger (piston) is pushed

282 Chapter 6: The Thermodynamic Properties of Real Substances further into the cylinder. When the air pressure in the cylinder increases until it reaches the same pressure as that in the tire, the valve opens and further movement of the piston forces air from the pump into the tire. The whole process occurs quickly enough that there is no significant heat flow to or from the surroundings to the pump or tire during the pumping process. Thus, a single pumping cycle can be considered a three-step process. First, air is drawn into the pump cylinder at constant pressure. Second, as the pumping begins, the valve is closed so that the pressure increases without a flow of air from the pump. Third, the valve opens and the pumping action forces air into the tire. Here we are interested in calculations only for the second step if the tire pressure is initially 60 psig (5.15 bar absolute). For reference, the cylinder of a bicycle hand pump is about 50 cm long and about 3 cm in diameter.

6.60 Redo Problem 4.45 if ethylene is described by the truncated virial equation with B (T ) = 5.86 · 10−5 − 0.056/T m3 /mol and T in K. 6.61 Redo Problem 4.45 if ethylene is described by the Peng-Robinson equation of state. 6.62 a. Show that (∂V /∂T )S (∂V /∂T )P

=−

CV κT V T



∂T ∂P

2 V

b. Use the result of part (a) to show that for a stable system at equilibrium (∂V /∂T )S and (∂V /∂T )P must have opposite signs. c. Two separate measurements are to be performed on a gas enclosed in a piston-and-cylinder device. In the first measurement the device is well insulated so there is no flow of heat to or from the gas, and the piston is slowly moved inward, compressing the gas, and its temperature is found to increase. In the second measurement the piston is free to move and the external pressure is lve constant. A small amount of heat is added to the Va Valve Plunger (piston) gas in the cylinder, resulting in the expansion of Air in pump cylinder the gas. Will the temperature of the gas increase or decrease? 6.63 Gasoline vapor is to be recovered at a filling station rather than being released into the atmosphere. In the a. If the ambient temperature is 25◦ C and air is treated scheme we will analyze, the vapor is first condensed to as an ideal gas with CP∗ = 29.3 J/(mol K), how far a liquid and then pumped back to a storage tank. Dedown will the piston have moved before the valve termine the work necessary to adiabatically pump the opens? What is the temperature of the gas? liquid gasoline from 0.1 MPa and 25◦ C to 25 MPa, and b. Repeat the calculation of part (a) if the Pengthe final temperature of the liquid gasoline. Consider Robinson equation is used to describe air, assuming gasoline to be n-octane and CP∗ = 122.2 J/(mol K) that the critical constants and acentric factor of air independent of temperature. are the same as for nitrogen. 6.64 A 3-m3 tank is in the basement of your house to store 6.59 The Euken coefficient ξ is defined as ξ = (∂T /∂V )U . propane that will be used for home and water heating a. Show that the Euken factor is also equal to and cooking. Your basement, and therefore the contents of the tank, remain at 20◦ C all year. Initially, ex  T (∂P/∂T )V − P cept for a small vapor space the tank is completely ξ=− full of liquid propane, but by January 60 percent of CV the propane has been used. Compute a. The fraction of the remaining propane that is b. What is the value of the Euken coefficient for an present as vapor ideal gas? b. The total flow of heat into the tank over the period c. Develop an explicit expression for the Euken coefin which the propane has been used ficient for a gas that is described by the truncated 6.65 Pipelines are used to transport natural gas over thouvirial equation sands of miles, with compressors (pumping stations) at regular intervals along the pipeline to compensate B(T ) PV =Z =1+ for the pressure drop due to flow. Because of the long RT V distances involved and the good heat transfer, the gas remains at an ambient temperature of 20◦ C. At each d. Develop an expression for the Euken coefficient for a gas that is described by the Peng-Robinson equapumping station the gas in compressed with a singletion of state. stage compressor to 4 MPa, and then isobarically

Problems 283 cooled back to 20◦ C. Because of safety and equipment constraints, the gas temperature during compression should not exceed 120◦ C. Assuming natural gas is methane that is described by the Peng-Robinson equation of state, how low a pressure is allowed before the gas should be recompressed? 6.66 In a continuous manufacturing process, chlorodifluoromethane (CHClF2 ), initially at 10 bar and 420 K, passes through an adiabatic pressure-reducing valve so that its pressure drops to 0.1 bar (this last pressure low enough that CHClF2 can be considered to be an ideal gas). At these operating conditions, the gas can be represented by a one-term virial equation of state:

a. Develop the energy and entropy balances for this one-dimensional material, assuming that there is no additional work term other than the stretching work, and that the volume of the material does not change on stretching (that is, the length change is compensated for by a change in cross-sectional area). b. An elastic material has the following properties at 733 K and a stress of 103 MPa. ασ =

1 L



∂L ∂T



= 16.8 × 10−6 K−1

σ

and B (T ) PV =1+ RT V

 Cσ = T

The following data are available for this gas: B (T ) = 297.6 −

CP∗ = 27.93 + 0.093T

J mol K

What is the temperature of the chlorodifluoromethane exiting the valve? 6.67 How much entropy is generated per mole of chlorodifluoromethane that passes through the pressurereducing valve of the previous problem? 6.68 One kilogram of saturated liquid methane at 160 K is placed in an adiabatic piston-and-cylinder device, and the piston will be moved slowly and reversibly until 25 percent of the liquid has vaporized. Compute the maximum work that can be obtained, assuming that methane is described by the Peng-Robinson equation of state. Compare your results with the solution to Problem 4.39. 6.69 The thermoelastic effect is the temperature change that results from stretching an elastic material or fiber. The work done on the material is given by ˙ = Force × Rate of change of distance = A · L · σ · dε W dt

where σ = Stress =

and

Force

Cross-sectional area

=

 = 38.1 σ

J (mol K)

Estimate the change in temperature per unit of strain at these conditions. 6.70 The following equation of state has been proposed for a fluid

256100 cm3 T mol

and

∂S ∂T

Force A

1 dL dε = Rate of strain = dt L dt

In the situation being considered here, the thermodynamic properties such as the heat capacity depend on the stress (just as for a gas the heat capacity depends on the pressure).

C B PV + 2 =1+ RT V V

where B and C are constants. a. Does this fluid exhibit a critcal point? Prove it. b. If you believe the answer to part a. is yes, derive expressions for B and C in terms of the critical temperature and pressure for this fluid. 6.71 Nitrogen at 15 bar and 100 K is to be adiabatically flashed to 1 bar. Determine the exiting temperature of nitrogen and the fractions that are vapor and liquid using the Peng-Robinson equation of state. 6.72 A residual thermodynamic property is defined as the difference between the property of the real fluid and that of an ideal gas at the same temperature and pressure, that is θres (N, P, T ) = θ (N, P, T ) − θIG (N, P, T )

Assuming that a fluid can be described by the viral equation of state with only the second virial coefficient, develop the expressions for the residual contributions to the volume, internal energy and Helmholtz energy at a fixed temperature and pressure. 6.73 A piston-and-cylinder device contains 10 kmols of n-pentane at −35.5◦ C and 100 bar. Slowly the piston is moved until the vapor pressure of n-pentane is reached, and then further moved until 5 kmols of the n-pentane is evaporated. This complete process takes place at the constant temperature of −35.5◦ C. Assume n-pentane can be described by the van der Waals equation of state. The following data are available from the vdW equation state at −35.5◦ C

284 Chapter 6: The Thermodynamic Properties of Real Substances P vap = 1.013 bar P = 100 bar

Z(liquid) = 0.009077 Z(liquid) = 0.69075 Z(vapor) = 0.94016

a. What is the volume change for the process? b. How much heat must be supplied for the process to be isothermal? 6.74 In a continuous manusfacturing process chlorodifluoromethane (CHClF2 ) initially at 10 bar and 420◦ C passes through and adiabatic pressure reducing valve so that its pressure is reduced to 0.1 bar (this last pressure is low enough that CHClF2 can be considered an ideal gas. At these operating conditions, this gas can be described by a one-term virial equation of state: B PV =1+ RT V

and the ideal heat capacity is CP∗ = 29.93 + J What is the temperature of the chlorodi0.093T molK fluoromethane exiting the valve? How much entropy is generated in the process per mole of CHClF2 flowing through the valve? 6.75 A gas is continuously passed through an adiabatic turbine at the rate of 2 mol/s. Its initial temperature is 600 K, its initial pressure is 5 bar and its exiting pressure is 1 bar. Determine the maximum rate at which work can be obtained in this process. The gas is described by an augmented Clausius equation of state P (V − b) = RT

with b = b0 +

b1 m3 , b0 = 3×10−5 T mol

and

b1 = 3×10−3

m3 · K mol

CP∗ (T ) = C0 + C1 ∗ T

with C0 = 25

J J and C2 = 1.8 × 10−2 mol · K mol · K2

R 6.76 Use the Estimation tool in Aspen Plus to estimate the physical properties of methyl vinyl ketone (MKK) after entering structure using the Molecular Structure tool. 6.77 Methane at 1 atm and 25◦ C is to be compressed to 5 bar in an adiabatic, isentropic compressor. What is the temperature of the methane stream exiting the compressor? If the flowrate of methane is 60 kmol/hr, how much work needs to be supplied by the compressor? 6.78 Nitrogen at 1 atm and 25◦ C is to be compressed to 5 bar in an adiabatic, isentropic compressor. What is the temperature of the nitrogen stream exiting the compressor? If the flowrate of nitrogen is 100 kmol/hr, how much work needs to be supplied by the compressor? R 6.79 Redo Problem 6.7 using Aspen Plus . R . 6.80 Redo Problem 6.12 using Aspen Plus R 6.81 Redo Problem 6.13 using Aspen Plus . R 6.82 Redo Problem 6.29 using Aspen Plus . R . 6.83 Redo Problem 6.31 using Aspen Plus R 6.84 Redo Problem 6.32 using Aspen Plus . R 6.85 Redo Problem 6.65 using Aspen Plus .

Chapter

7 Equilibrium and Stability in One-Component Systems Now that the basic principles of thermodynamics have been developed and some computational details considered, we can study one of the fundamental problems of thermodynamics: the prediction of the equilibrium state of a system. In this chapter we examine the conditions for the existence of a stable equilibrium state in a singlecomponent system, with particular reference to the problem of phase equilibrium. Equilibrium in multicomponent systems will be considered in following chapters.

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 7 The goals of this chapter are for the student to: • Identify the criterion for equilibrium in systems subject to different constraints

(Sec. 7.1)

• Use the concept of stability to identify when phase splitting into vapor and liquid

phases will occur (Sec. 7.2)

• Identify the conditions of phase equilibrium (Sec. 7.3) • Be able to identify the critical point of a fluid (Sec. 7.3) • Calculate the fugacity of a pure substance that is a gas or a liquid when a volumetric

equation of state is available (Sec. 7.4)

• Calculate the fugacity of a pure liquid or solid when a volumetric equation of state

is not available (Sec. 7.4)

• Use fugacity in the calculation of vapor-liquid equilibrium (Sec. 7.5) • Be able to determine the number degrees of freedom for a pure fluid (Sec. 7.6)

NOTATION INTRODUCED IN THIS CHAPTER f F φ I fsat (T ) HL HS HV

Fugacity (kPa) Number of degrees of freedom Fugacity coefficient = f /P Fugacity of phase I at a phase change at temperature T (kPa) Molar enthalpy of a liquid (kJ/mol) Molar enthalpy of a solid (kJ/mol) Molar enthalpy of a vapor (kJ/mol) 285

286 Chapter 7: Equilibrium and Stability in One-Component Systems vap H fus H sub H Pt P sub (T ) P vap (T ) Tt

Enthalpy change (or heat) of vaporization = H V − H L (kJ/mol) Enthalpy change (or heat) of melting = H L − H S (kJ/mol) Enthalpy change (or heat) of sublimation = H V − H S (kJ/mol) Triple-point pressure (kPa) Sublimation pressure at temperature T (kPa) Vapor pressure at temperature T (kPa) Triple-point temperature (K)

7.1 THE CRITERIA FOR EQUILIBRIUM In Chapter 4 we established that the entropy function provides a means of mathematically identifying the state of equilibrium in a closed, isolated system (i.e., a system in which M , U , and V are constant). The aim in this section is to develop a means of identifying the equilibrium state for closed thermodynamic systems subject to other constraints, especially those of constant temperature and volume, and constant temperature and pressure. This will be done by first reconsidering the equilibrium analysis for the closed, isolated system used in Sec. 4.1, and then extending this analysis to the study of more general systems. The starting points for the analysis are the energy and entropy balances for a closed system: Balance equations for a closed system

dU dV = Q˙ − P dt dt

(7.1-1)

Q˙ dS = + S˙ gen dt T

(7.1-2)

and

with Second law of thermodynamics

S˙ gen ≥ 0 (the equality holding at equilibrium or for reversible processes) (7.1-3) Here we have chosen the system in such a way that the only work term is that of the deformation of the system boundary. For a constant-volume system exchanging no heat with its surroundings, Eqs. 7.1-1 and 7.1-2 reduce to

dU =0 dt so that

U = constant (or, equivalently, U = constant, since the total number of moles, or mass, is fixed in a closed, one-component system) and

7.1 The Criteria for Equilibrium 287

dS = S˙ gen ≥ 0 dt

(7.1-4)

Since the entropy function can only increase in value during the approach to equilibrium (because of the sign of S˙ gen ), the entropy must be a maximum at equilibrium. Thus, the equilibrium criterion for a closed, isolated system is  S = maximum at equilibrium in a closed system at constant U and V or S = maximum (7.1-5) At this point you might ask how S can achieve a maximum value if U and V are fixed, since the specification of any two intensive variables completely fixes the values of all others. The answer to this question was given in Chapter 4, where it was pointed out that the specification of two intensive variables fixes the values of all other state variables in the uniform equilibrium state of a single-component, single-phase system. Thus, the equilibrium criterion of Eq. 7.1-5 can be used for identifying the final equilibrium state in a closed, isolated system that is initially nonuniform, or in which several phases or components are present. To illustrate the use of this equilibrium criterion, consider the very simple, initially nonuniform system shown in Fig. 7.1-1. There a single-phase, single-component fluid in an adiabatic, constant-volume container has been divided into two subsystems by an imaginary boundary. Each of these subsystems is assumed to contain the same chemical species of uniform thermodynamic properties. However, these subsystems are open to the flow of heat and mass across the imaginary internal boundary, and their temperature and pressure need not be the same. For the composite system consisting of the two subsystems, the total mass (though, in fact, we will use number of moles), internal energy, volume, and entropy, all of which are extensive variables, are the sums of these respective quantities for the two subsystems, that is,

N = N I + N II U = U I + U II V = V I + V II

(7.1-6)

and

S = S I + S II Now considering the entropy to be a function of internal energy, volume, and mole number, we can compute the change in the entropy of system I due to changes in N I , U I , and V I from Eq. 6.2-5c.

N I, S I, U I, V I

N II, S II, U II, V II

Figure 7.1-1 An isolated nonequilibrium system.

288 Chapter 7: Equilibrium and Stability in One-Component Systems

 I

dS = =

∂S I ∂U I



 I

dU + V I ,N I

∂S I ∂V I



 I

dV + U I ,N I

∂S I ∂N I

 dN I U I ,V I

1 P I I GI I dU + dV − I dN I TI TI T

In a similar fashion dS II can be computed. The entropy change of the composite system is

1 1 P I I P II II GI GII I II I dU + dU + dV + dV − dN − dN II TI T II TI T II TI T II However, the total number of moles, total internal energy, and total volume are constant by the constraints on the system, so that dS = dS I + dS II =

dN = dN I + dN II = 0 or dN I = −dN II dU = dU I + dU II = 0 or dU I = −dU II dV = dV I + dV II = 0 or dV I = −dV II Consequently,   I    I  1 P II GII 1 P G I I dS = − − − dU + dV − dN I T I T II T I T II T I T II

(7.1-7)

(7.1-8)

Now since S = maximum or dS = 0 for all system variations at constant N , U , and V (here all variations of the independent variables dU I , dV I , and dN I at constant total number of moles, total internal energy, and total volume), we conclude that   1 1 ∂S = 0 so that = II or T I = T II (7.1-9a) I I ∂U V I ,N I T T   ∂S P II PI = 0 so that = or P I = P II (7.1-9b) ∂V I U I ,N I TI T II (since it has already been shown that T I = T II )   GII ∂S GI = 0 so that = ∂N I U I ,V I TI T II

or

GI = GII

(7.1-9c)

Therefore, the equilibrium condition for the system illustrated in Fig. 7.1-1 is satisfied if both subsystems have the same temperature, the same pressure, and the same molar Gibbs energy. For a single-component, single-phase system, this implies that the composite system should be uniform. This is an obvious result, and it is reassuring that it arises so naturally from our development. The foregoing discussion illustrates how the condition dS = 0 may be used to identify a possible equilibrium state of the closed, isolated system, that is, a state for which S is a maximum. From calculus we know that dS = 0 is a necessary but not sufficient condition for S to achieve a maximum value. In particular, dS = 0 at a minimum value or an inflection point of S , as well as when S is a maximum. The condition d2 S < 0, when dS = 0, assures us that a maximum value of the entropy, and hence a true equilibrium state, has been identified rather than a metastable state (an inflection point) or

7.1 The Criteria for Equilibrium 289 an unstable state (minimum value of S ). Thus, the sign of d2 S determines the stability of the state found from the condition that dS = 0. The implications of this stability condition are considered in the next section. It is also possible to develop the equilibrium and stability conditions for systems subject to other constraints. For a closed system at constant temperature and volume, the energy and entropy balances are

dU = Q˙ dt and

Q˙ dS = + S˙ gen dt T

Eliminating Q˙ between these two equations, and using the fact that T dS = d(T S), since T is constant, gives

dA d(U − T S) = = −T S˙ gen ≤ 0 dt dt Here we have also used the facts that T ≥ 0 and S˙ gen ≥ 0, so that (−T S˙ gen ) ≤ 0. Using the same argument that led from Eq. 7.1-4 to Eq. 7.1-5 here yields  A = minimum for equilibrium in a closed system at constant T and V or A = minimum (7.1-10) ˙ It should be noted that this result is also a consequence of Sgen ≥ 0. If we were to use Eq. 7.1-10 to identify the equilibrium state of an initially nonuniform, single-component system, such as that in Fig. 7.1-1, but now maintained at a constant temperature and volume, we would find, following the previous analysis, that at equilibrium the pressures of the two subsystems are equal, as are the molar Gibbs energies (cf. Eqs. 7.1-9a, b, and c and Problem 7.4); since temperature is being maintained constant, it would, of course, be the same in the two subsystems. For a closed system maintained at constant temperature and pressure, we have

dV dU d = Q˙ − P = Q˙ − (P V ) dt dt dt and

dS Q˙ = + S˙ gen dt T

Again eliminating Q˙ between these two equations gives

dG d d dU d + (P V ) − (T S) = (U + P V − T S) = = −T S˙ gen ≤ 0 (7.1-11) dt dt dt dt dt so that the equilibrium criterion here is

290 Chapter 7: Equilibrium and Stability in One-Component Systems

G = minimum Most important of the equilibrium criteria

 for equilibrium in a closed system at constant T and P

or

G = minimum (7.1-12) This equation also leads to the equality of the molar Gibbs energies (Eq. 7.1-9c) as a condition for equilibrium. Of course, since both temperature and pressure are held constant, the system is also at uniform temperature and pressure at equilibrium. Finally, the equilibrium criterion for a system consisting of an element of fluid moving with the velocity of the fluid around it is also of interest, as such a choice of system arises in the study of continuous processing equipment used in the chemical industry. The tubular chemical reactor discussed in Chapter 14 is perhaps the most common example. Since each fluid element is moving with the velocity of the fluid surrounding it, there is no convected flow of mass into or out of this system. Therefore, each such element of mass in a pure fluid is a system closed to the flow of mass, and consequently is subject to precisely the same equilibrium criteria as the closed systems discussed above (i.e., Eq. 7.1-5, 7.1-10, or 7.1-12, depending on the constraints on the system). The equilibrium and stability criteria for systems and constraints that are of interest in this book are collected in Table 7.1-1. As we will see in Chapter 8, these equilibrium and stability criteria are also valid for multicomponent systems. Thus the entries in this table form the basis for the analysis of phase and chemical equilibrium problems to be considered during much of the remainder of this book. Using the method of analysis indicated here, it is also possible to derive the equilibrium criteria for systems subject to other constraints. However, this task is left to you (Problem 7.2). As a simple example of the use of the equilibrium conditions, we consider a problem to which we intuitively know the solution, but want to show that the answer arises naturally from the analysis of this section. Illustration 7.1-1 Using the Equilibrium Condition to Solve a Simple Problem Two identical metal blocks of mass M with initial temperatures T1,i and T2,i , respectively, are in contact with each other in a well-insulated (adiabatic), constant-volume enclosure. Find the Table 7.1-1 Equilibrium and Stability Criteria System Isolated, adiabatic fixed-boundary system Isothermal closed system with fixed boundaries Isothermal, isobaric closed system Isothermal, isobaric open system moving with the fluid velocity

Constraint

Equilibrium Criterion

Stability Criterion

U = constant V = constant

S = maximum dS = 0

d2 S < 0

T = constant V = constant

A = minimum dA = 0

d2 A > 0

T = constant P = constant

G = minimum dG = 0

d2 G > 0

T = constant P = constant M = constant

G = minimum dG = 0

d2 G > 0

7.1 The Criteria for Equilibrium 291 final equilibrium temperatures of the metal blocks. (Of course, intuituively, we know the solution, T1, f = T2, f . However, we want to show that this arises naturally, though with some work, from the equilibrium condition, as we will encounter many problems, especially when we deal with mixtures, where our intuition will not help us identify the equilibrium state.)

Adiabatic enclosure T2, i

T1, i

Solution We will choose the two metal blocks as the system for writing the balance equations. There is no exchange of mass between the blocks, so the mass balance does not provide any useful information. The energy balance for the system is ˆ1, f + M U ˆ2, f − M U ˆ1, i − M U ˆ2, i = 0 MU

or ˆV (T2, f − T2, ref )−M C ˆV (T1, f − T1, ref )+M C ˆV (T1, i − T1, ref )−M C ˆV (T2, i − T2, ref ) = 0 MC

which reduces to T1, f − T1, i + T2, f − T2, i = 0

(a)

The entropy balance cannot be used directly to provide useful information since the entropy generation due to heat transfer cannot be evaluated at this stage in the calculation. However, we can use the fact that for a system in which M, U , and V are constant, at equilibrium the entropy is a maximum with respect to the independent variations within the system. The final entropy of the system is  ˆV ln S = M Sˆ1, f + M Sˆ2, f = M C

T1, f T1, ref



 ˆV ln + MC

T2, f T2, ref



The only possible variations are in the two final temperatures, T1, f and T2, f . However, as the energy balance connects these two temperatures, only one of them can be considered to be independent, say T1, f . Therefore, to identify the equilibrium state we set the differential of the entropy with respect to T1, f equal to zero:      d dS ˆV ln T1, f + M C ˆV ln T2,f MC =0= dT1, f dT1, f T1,ref T2,ref =

ˆV ˆV dT2, f MC MC + T1, f T1, f dT1, f

or dT2, f T2, f =− dT1, f T1, f

(b)

292 Chapter 7: Equilibrium and Stability in One-Component Systems Now taking the derivative of the energy balance with respect to T1, f , we obtain  d  ˆ ˆV (T2, f − T2,i ) = 0 = M C ˆV dT2, f ˆV + M C M CV (T1, f − T1,i ) + M C dT1, f dT1, f

or

dT2, f = −1 dT1, f

(c)

Comparing Eqs. b and c, we have dT2, f T1, f = −1 = − dT1, f T2, f

Clearly, the only way this combined equation can be satisfied is if T1, f = T2, f = Tf ; that is, the final temperature of the two blocks must be equal, which is the intuitively obvious solution. To calculate this final equilibrium temperature, we use the energy balance, Eq. a, Tf − T1,i + Tf − T2,i = 2Tf − T1,i − T2,i = 0

which has the solution Tf =

T1,i + T2,i 2

Knowing the final state of the system, it is of interest to calculate the amount of entropy generated in the process of achieving equilibrium. We do this using the difference form of the entropy balance, Sf − Si = Sgen Sgen = Sf − Si

          Tf Tf T1,i T2,i ˆV ln = MC − M CˆV ln + ln + ln T1,ref T2,ref T1,ref T2,ref      2 ˆV ln T1,i + T2,i = M C ˆV ln (T1,i + T2,i ) ˆV ln T1,i + T2,i + M C = MC 2T1,i 2T2,i 4T1,i T2,i

Illustration 7.1-2 Proving the Equality of Gibbs Energies for Vapor-Liquid Equilibrium Use the information in the steam tables of Appendix A.III to show that Eq. 7.1-9c is satisfied at 100◦ C and 0.101 35 MPa.

Solution From the saturated steam temperature table, we have at 100◦ C and 0.101 35 MPa ˆ L = 419.04 kJ/kg H SˆL = 1.3069 kJ/(kg K)

ˆ V = 2676.1 kJ/kg H SˆV = 7.3549 kJ/(kg K)

ˆ=H ˆ − T Sˆ, we have further that Since G ˆ L = 419.04 − 373.15 × 1.3069 = −68.6 kJ/kg G

7.2 Stability of Thermodynamic Systems 293 and

ˆ V = 2676.1 − 373.15 × 7.3549 = −68.4 kJ/kg G

ˆL = G ˆ V (or, equivalently, that which, to the accuracy of the calculations here, confirms that G GL = GV ) at this vapor-liquid phase equilibrium state.

7.2 STABILITY OF THERMODYNAMIC SYSTEMS In the previous section we used the result dS = 0 to identify the equilibrium state of an initially nonuniform system constrained to remain at constant mass, internal energy, and volume. In this section we explore the information content of the stability criterion d2 S < 0 at constant M, U, and V [The stability analysis for closed systems subject to other constraints (i.e., constant T and V or constant T and P ) is similar to, and, in fact, somewhat simpler than the analysis here, and so it is left to you (Problem 7.3).] By studying the sign of the second differential of the entropy, we are really considering the following question: Suppose that a small fluctuation in a fluid property, say temperature or pressure, occurs in some region of a fluid that was initially at equilibrium; is the character of the equilibrium state such that d2 S < 0, and the fluctuation will dissipate, or such that d2 S > 0, in which case the fluctuation grows until the system evolves to a new equilibrium state of higher entropy? In fact, since we know that fluids exist in thermodynamically stable states (experimental observation 7 of Sec. 1.7), we will take as an empirical fact that d2 S < 0 for all real fluids at equilibrium, and instead establish the restrictions placed on the equations of state of fluids by this stability condition. We first study the problem of the intrinsic stability of the equilibrium state in a pure single-phase fluid, and then the mutual stability of two interacting systems or phases. We begin the discussion of intrinsic stability by considering further the example of Figure 7.1-1 of the last section, equilibrium in a pure fluid at constant mass (actually, we will use number of moles), internal energy, and volume. Using the (imaginary) subdivision of the system into two subsystems, and writing the extensive properties N , U , V , and S as sums of these properties for each subsystem, we were able to show in Sec. 7.1 that the condition dS = dS I + dS II = 0 (7.2-1) for all system variations consistent with the constraints (i.e., all variations in dN I , dV I , and dU I ) led to the requirements that at equilibrium

T I = T II P I = P II and

GI = GII Continuing, we write an expression for the stability requirement d2 S < 0 for this system and obtain

294 Chapter 7: Equilibrium and Stability in One-Component Systems I I I d2 S = SUU (dU I )2 + 2SUV (dU I )(dV I ) + SVV (dV I )2 I I I + 2SUN (dU I )(dN I ) + 2SVN (dV I )(dN I ) + SNN (dN I )2 II II II + SUU (dU II )2 + 2SUV (dU II )(dV II ) + SVV (dV II )2 II II II + 2SUN (dU II )(dN II ) + 2SVN (dV II )(dN II ) + SNN (dN II )2 < 0

In this equation we have used the abbreviated notation  I   2 I ∂  ∂S ∂ S I I SUU = S = UV ∂U 2 V,N ∂U V,N ∂V U,N

(7.2-2a)

etc.

Since the total number of moles, total internal energy, and total volume of the composite system are fixed, we have, as in Eqs. 7.1-7,

dN I = −dN II

dU I = −dU II

dV I = −dV II

and I II I II d2 S = (SUU + SUU )(dU I )2 + 2(SUV + SUV )(dU I )(dV I ) I II I II + SVV )(dV I )2 + 2(SUN + SUN )(dU I )(dN I ) + (SVV I II I II + SVN )(dV I )(dN I ) + (SNN + SNN )(dN I )2 + 2(SVN

(7.2-2b)

Furthermore, since the same fluid in the same state of aggregation is present in regions I and II, and since we have already established that the temperature, pressure, and molar Gibbs energy each have the same value in the two regions, the value of any state property must be the same in the two subsystems. It follows that any thermodynamic derivative that can be reduced to combinations of intensive variables must have the same value in the two regions of the fluid. The second derivatives (Eq. 7.2-2b), as we will see shortly, are combinations of intensive and extensive variables. However, the quantities N Sxy , where x and y denote U , V , or N , are intensive variables. Therefore, it follows that I II N I Sxy = N II Sxy

(7.2-3)

Using Eqs. 7.1-7 and 7.2-3 in Eq. 7.2-2 yields   I N + N II I I [N I SUU d2 S = (dU I )2 + 2N I SUV (dU I )(dV I ) N I N II I I (dV I )2 + 2N I SUN (dU I )(dN I ) + N I SVV I I (dV I )(dN I ) + N I SNN (dN I )2 ] < 0 + 2N I SVN

(7.2-4a)

The term (N I + N II )/N I N II must be greater than zero since mole numbers can only be positive. Also, we can eliminate the superscripts from the products N Sxy , as they are equal in the two regions. Therefore, the inequality Eq. 7.2-4a can be rewritten as

N SUU (dU I )2 + 2N SUV (dU I )(dV I ) + N SVV (dV I )2 + 2N SUN (dU I )(dN I ) + 2N SVN (dV I )(dN I ) + N SNN (dN I )2 < 0

(7.2-4b)

7.2 Stability of Thermodynamic Systems 295 Equations 7.2-3 and 4 must be satisfied for all variations in N I , U I , and V I if the fluid is to be stable. In particular, since Eq. 7.2-4b must be satisfied for all variations in U I (dU I = 0) at fixed values of N I and V I (i.e., dN I = 0 and dV I = 0), stable fluids must have the property that N SUU < 0 (7.2-5a) Similarly, by considering variations in volume at fixed internal energy and mole number, and variations in mole number at fixed volume and internal energy, we obtain

N SVV < 0

(7.2-5b)

N SNN < 0

(7.2-5c)

and as additional conditions for fluid stability. More severe restrictions on the equation of state result from demanding that Eq. 7.2-4b be satisfied for all possible and simultaneous variations in internal energy, volume, and mole number and not merely for variations in one property with the others held fixed. Unfortunately, the present form of Eq. 7.2-4b is not well suited for studying this more general situation since the cross-terms (i.e., dU I dV I , dU I dN I , and dV I dN I ) may be positive or negative depending on the sign of the variations dU I , dV I , and dN I , so that little can be said about the coefficients of these terms. By much algebraic manipulation (Problem 7.32), Eq. 7.2-4b can be written as

θ1 (dX1 )2 + θ2 (dX2 )2 + θ3 (dX3 )2 < 0

(7.2-6)

where

θ1 = N SUU 2 θ2 = (N SUU N SVV − N 2 SUV )/N SUU 2 2 (N SUU N SNN − N SUN ) (N SUU N SVN − N SUV N SUN )2 θ3 = − 2 N SUU N SUU (N SUU N SVV − N 2 SUV ) S S UV UN dX1 = dU I + dV I + dN I SUU SUU (SUU SVN − SUV SUN ) dX2 = dV I + dN I 2 (SUU SVV − SUV ) and

dX3 = dN I The important feature of Eq. 7.2-6 is that it contains only square terms in the system variations. Thus, (dX1 )2 , (dX2 )2 , and (dX3 )2 are greater than or equal to zero regardless of whether dU I , dV I , and dN I individually are positive or negative. Consequently, if

θ1 = N SUU ≤ 0 2 N SUU N SVV − N 2 SUV θ2 = ≤0 N SUU θ3 ≤ 0

(7.2-7a) (7.2-7b) (7.2-7c)

296 Chapter 7: Equilibrium and Stability in One-Component Systems Eq. 7.2-6, and hence Eq. 7.2-4b, will be satisfied for all possible system variations. Equations 7.2-7 provide more restrictive conditions for fluid stability than Eqs. 7.2-5a, b, and c. It now remains to evaluate the various entropy derivatives, so that the stability restrictions of Eqs. 7.2-7a, b, and c can be put into a more usable form. Starting from        N ∂T ∂  ∂(1/T ) ∂S N SUU = N =N =− 2 ∂U N,V ∂U N,V ∂U T ∂U N,V N,V and using that for the open system that  

dU = N CV dT + T 

leads to

∂U ∂T

  ∂P − P dV + G dN ∂T V

(7.2-8)

 = N CV V,N

1 0

First or thermal stability criterion

(7.2-9)

That is, the constant-volume heat capacity must be positive, so that internal energy increases as the fluid temperature increases. Next, we note that       ∂  ∂  ∂S ∂S N SUV = N =N ∂U V,N ∂V N,U ∂V U,N ∂U V,N      ∂  N ∂T 1 =− 2 =N ∂V U,N T T ∂V U,N and from Eq. 7.2-8,



∂T ∂V

 =− U,N

to obtain

N SUV =

1 Note,

    ∂P T −P ∂T V N CV

    ∂P T −P ∂T V CV T 2

(7.2-10a)1

from these equations, that N SUU , N SUV , and N SVV are intensive variables, as was suggested earlier, whereas SUU , SUV , and SVV are proportional to N −1 .

7.2 Stability of Thermodynamic Systems 297 Similarly, but with a great deal more algebra, we can show that 2      1 1 ∂P ∂P N SVV = − −P T T ∂V T CV T 2 ∂T V and

2 N SUU N SVV − N 2 SUV 1 θ2 = = N SUU T



∂P ∂V

(7.2-10b)1

 T

Thus, a further stability restriction on the equation of state of a substance, since T is positive, is that

 Second or mechanical stability criterion

∂P ∂V

or

1 κT = − V

 0

(7.2-11)

T

where κT is the isothermal compressibility of the fluid, introduced in Sec. 6.2. This result indicates that if a fluid is to be stable, its volumetric equation of state must be such that the fluid volume decreases as the pressure increases at constant temperature. As we will see shortly, this restriction has important implications in the interpretation of phase behavior from the equation of state of a fluid. Finally, and with a great deal more algebra (Problem 7.32), one can show that θ3 is identically equal to zero and thus does not provide any further restrictions on the equation of state. The main conclusion from this exercise is that if a fluid is to exist in a stable equilibrium state, that is, an equilibrium state in which all small internal fluctuations will dissipate rather than grow, the fluid must be such that

and



∂P ∂V

CV > 0

(7.2-12)

< 0 or κT > 0

(7.2-13)

 T

Alternatively, since all real fluids exist in thermodynamically stable states, Eqs. 7.2-12 and 7.2-13 must be satisfied for real fluids. In fact, no real fluid state for which either (∂P/∂V )T > 0 or CV < 0 has yet been found. Equations 7.2-12 and 7.2-13 may be thought of as part of the philosophical content of thermodynamics. In particular, thermodynamics alone does not give information on the heat capacity or the equation of state of any substance; such information can be gotten only from statistical mechanics or experiment. However, thermodynamics does provide restrictions or consistency relations that must be satisfied by such data; Eqs. 7.2-12 and 7.2-13 are examples of this. (Consistency relations for mixtures are discussed in later chapters.)

298 Chapter 7: Equilibrium and Stability in One-Component Systems Illustration 7.2-1 Using the Steam Tables to Show That the Stability Conditions Are Satisfied for Steam Show that Eqs. 7.2-12 and 7.2-13 are satisfied by superheated steam.

Solution It is easiest to use Eq. 7.2-13 in the form (∂P/∂V )T < 0, which requires that the volume decrease as the pressure increases at constant temperature. This is seen to be true by using the superheated steam table and observing that Vˆ decreases as P increases at fixed temperature. For example, at 1000◦ C P (MPa) Vˆ (m3 /kg)

0.50 1.1747

0.8 0.7340

1.0 0.5871

1.4 0.4192

1.8 0.3260

2.5 0.2346

Proving that CV > 0 or CˆV > 0 is a bit more difficult since  ˆV = C

ˆ ∂U ∂T

V

and the internal energy is not given at constant volume. Therefore, interpolation methods must be used. As an example of how the calculation is done, we start with the following data from the superheated vapor portion of the steam tables. P = 1.80 MPa

P = 2.00 MPa

◦

T ( C)

Vˆ (m /kg)

ˆ (kJ/kg) U

Vˆ (m3 /kg)

ˆ (kJ/kg) U

800 900 1000

0.2742 0.3001 0.3260

3657.6 3849.9 4048.5

0.2467 0.2700 0.2933

3657.0 3849.3 4048.0

3

To proceed, we need values of the internal energy at two different temperatures and the same specific volume. We will use P = 2.00 MPa and T = 1000◦ C as one point; at these conditions Vˆ = 0.2933 m3/kg and Uˆ = 4048.0 kJ/kg. We now need to find the temperature at which Vˆ = 0.2933 m3/kg at P = 1.80 MPa. We use linear interpolation for this: 0.2933 − 0.2742 Vˆ (T, 1.80 MPa) − Vˆ (800◦ C, 1.80 MPa) T − 800 = = 900 − 800 0.3001 − 0.2742 Vˆ (900◦ C, 1.80 MPa) − Vˆ (800◦ C, 1.80 MPa)

so that T = 873.75◦ C. Next we need the internal energy Uˆ at T = 873.75◦ C and P = 1.80 MPa (since at these conditions Vˆ = 0.2933 m3/kg). Again using linear interpolation, ˆ (800◦ C, 1.80 MPa) ˆ (873.75◦ C, 1.80 MPa) − U U 873.75 − 800 = ◦ ˆ ˆ 900 − 800 U (900 C, 1.80 MPa) − U (800◦ C, 1.80 MPa) =

ˆ (873.75◦ C, 1.80 MPa) − 3657.6 U 3849.9 − 3657.6

we find that ˆ (873.75◦ C, 1.80 MPa) = U ˆ (873.75◦ C, 0.2933 m3/kg) = 3799.4 kJ/kg U

7.2 Stability of Thermodynamic Systems 299 Finally, replacing the derivative by a finite difference, and for the average temperature [i.e.,

T = (1000 + 873.75)/2 = 936.9◦ C], we have

ˆV (T = 936.9◦ C, Vˆ = 0.2933 m3/kg) C ˆ (1000◦ C, 0.2933 m3/kg) − U ˆ (873.75◦ C, 0.2933 m3/kg) U

≈

1000◦ C − 873.75◦ C 4048.0 − 3799.4 kJ = = 1.969 >0 1000 − 873.75 kg K

Similarly, we would find that CˆV > 0 at all other conditions.

Next we consider the problems of identifying the equilibrium state for two interacting phases of the same molecular species but in different states of aggregation, and of determining the requirements for the stability of this state. An example of this is a vapor and a liquid in equilibrium. To be general, we again consider a composite system isolated from its environment, except that here the boundary between the two subsystems is the real interface between the phases. For this system, we have

S N V U

= S I + S II = N I + N II = constant = V I + V II = constant = U I + U II = constant

(7.2-14)

Since N , U , and V are fixed, the equilibrium condition is that the entropy should attain a maximum value. Now, however, we allow for the fact that the states of aggregation in regions I and II are different, so that the fluids in these regions may follow different equations of state (or the different vapor and liquid branches of the same equation of state). Using the analysis of Eqs. 7.1-6, 7.1-7, and 7.1-8, we find that at equilibrium (i.e., when dS = 0), Important criteria for phase equilibria under all constraints

T I = T II P I = P II GI = GII

(7.2-15a) (7.2-15b) (7.2-15c)

Here Eqs. 7.2-15a and b provide the obvious conditions for equilibrium, and since two different phases are present, Eq. 7.2-15c provides a less obvious condition for equilibrium. Next, from the stability condition d2 S < 0, we obtain (following the analysis that led to Eq. 7.2-2b) I II I II d2 S = {SVV + SVV }(dV I )2 + 2{SUV + SUV }(dU I )(dV I ) I II I II + {SUU + SUU }(dU I )2 + 2{SVN + SVN }(dV I )(dN I ) I II I II + SNN }(dN I )2 + 2{SUN + SUN }(dU I )(dN I ) + {SNN

(7.2-16)

Here, however, the two partial derivatives in each of the bracketed terms need not be equal, since the two phases are in different states of aggregation and thus obey different equations of state, or different roots of the same equation of state. It is clear from a comparison with Eq. 7.2-2b that a sufficient condition for Eq. 7.2-16 to be satisfied is

300 Chapter 7: Equilibrium and Stability in One-Component Systems that each phase to be intrinsically stable; that is, Eq. 7.2-16 is satisfied if, for each of the coexisting phases, the equations

 Must be satisfied in each stable phase

CV > 0

and

∂P ∂V

 T4 > T3 > T2 > T1 . The isotherm T3 has a single point, C , for which (∂P/∂V )T = 0; at all other points on this isotherm (∂P/∂V )T < 0. On the isotherms T4 and T5 , (∂P/∂V )T < 0 everywhere, whereas on the isotherms T1 and T2 , (∂P/∂V )T < 0 in some regions and (∂P/∂V )T > 0 in other regions (i.e., between the points A and B on isotherm T1 , and the points A and B  on T2 ). The criterion for fluid stability requires that (∂P/∂V )T < 0, which is satisfied for the isotherms T4 and T5 , but not in the aforementioned regions of the T1 and T2 isotherms. Thus we conclude that the regions A to B and A to B  of the isotherms T1 and T2 , respectively, are not physically realizable; that is, they will not be observed in any experiment since a fluid in these states is not stable, and instead would go to an appropriate stable state. This observation raises some question about the interpretation to be given to the T1 and T2 isotherms. We cannot simply attribute these oddities to a peculiarity of the

T5

P

T4

C

T3 T2

B' A'

B

T1

A V _

Figure 7.3-1 Isotherms of the van der Waals equation in the pressure-volume plane.

7.3 Phase Equilibria: Application of the Equilibrium and Stability Criteria to the Equation of State 301

P

T < Tc

b

Pb

II

Pα I

Pa

a V _α

V" _α V _

V' _α

Figure 7.3-2 A low-temperature isotherm of the van der Waals equation.

van der Waals equation because many other, more accurate equations of state show essentially the same behavior. Some insight into the physical meaning of isotherms such as T1 can be gained from Fig. 7.3-2, which shows this isotherm separately. If we look at any isobar (constant-pressure line) between PA and PB in this figure, such as Pα , we see that it intersects the equation of state three times, corresponding to the fluid volumes V α , V α , and V α . One of these, V α , is on the part of the isotherm that is unattainable by the stability criterion. However, the other two intersections, at V α and V α , are physically possible. This suggests that at a given pressure and temperature the system can have two different volumes, a conclusion that apparently contradicts the experimental observation of Chapter 1 that two state variables completely determine the state of a single-component, single-phase system. However, this can occur if equilibrium can exist between two phases of the same species that are in different states of aggregation (and hence density). The equilibrium between liquid water and steam at 100◦ C and 101.325 kPa (1 atm) is one such example. One experimental observation in phase equilibrium is that the two coexisting equilibrium phases must have the same temperature and pressure. Clearly, the arguments given in Secs. 7.1 and 7.2 establish this. Another experimental observation is that as the pressure is lowered along an isotherm on which a liquid-vapor phase transition occurs, the actual volume-pressure behavior is as shown in Fig. 7.3-3, and not as in Fig. 7.3-2.

T < Tc P

V _L

V _

V _V

Figure 7.3-3 A low-temperature isotherm of a real fluid.

302 Chapter 7: Equilibrium and Stability in One-Component Systems That is, there is a portion of the isotherm where the specific volume varies continuously at fixed temperature and pressure; this is the two-phase coexistence region, here the vapor-liquid coexistence region. The variation of the overall (or two-phase) specific volume in this region arises from the fact that although the specific volumes of the vapor and liquid phases are fixed (since the temperature and pressure are fixed in each of the one-phase equilibrium subsystems), the fraction of the mixture that is vapor, ω V , can vary continuously from 0 to 1. In the two-phase region the specific volume of the two-phase mixture is given by

V = ω V V V + ω L V L = ω V V V + (1 − ω V )V L

(7.3-1a)

where ω V and ω L are fractions of vapor and liquid, respectively, on a molar basis. (Equation 7.3-1a could also be written using volumes per unit mass and mass fractions.) These fractions can vary between 0 and 1 subject to the condition that ω V + ω L = 1. Solving for ω V yields

ωV =

Maxwell or lever rule

V −VL VV−VL

(7.3-1b)

and

ωV V −VL = 1 − ωV VV−V

(7.3-1c)

Equations analogous to those here also hold for the H , U , G, S , and A. Equations of the form of Eq. 7.3-1c are the Maxwell’s rules or lever rules first discussed in Sec. 3.3.

Illustration 7.3-1 Computing the Properties of a Two-Phase Mixture Compute the total volume, total enthalpy, and total entropy of 1 kg of water at 100◦ C, half by weight of which is steam and the remainder liquid water.

Solution From the saturated steam temperature table at 100◦ C, the equilibrium pressure is 0.101 35 MPa and Vˆ L = 0.001 004 m3/kg ˆ L = 419.04 kJ/kg H SˆL = 1.3069 kJ/(kg K)

Vˆ V = 1.6729 m3/kg ˆ V = 2676.1 kJ/kg H SˆV = 7.3549 kJ/(kg K)

Using Eq. 7.3-1a on a mass basis gives Vˆ = 0.5 × 0.001 004 + 0.5 × 1.6729 = 0.836 45 m3/kg

The analogous equation for enthalpy is ˆ = ωL H ˆ L + ωV H ˆ V = 0.5 × 419.04 + 0.5 × 2676.1 = 1547.6 kJ H kg

7.3 Phase Equilibria: Application of the Equilibrium and Stability Criteria to the Equation of State 303 and that for entropy is kJ Sˆ = ω L SˆL + ω V SˆV = 0.5 × 1.3069 + 0.5 × 7.3549 = 4.3309 kg K

Consequently, the continuous variation of specific volume of the vapor-liquid mixture at fixed temperature and pressure is a result of the continuous change in the fraction of the mixture that is vapor. The conclusion, then, is that an isotherm such as that shown in Fig. 7.3-2 is an approximate representation of the real phase behavior (shown in Fig. 7.3-3) by a relatively simple analytic equation of state. In fact, it is impossible to represent the discontinuities in the derivative (∂P/∂V )T that occur at V L and V V with any analytic equation of state. By its sigmoidal behavior in the two-phase region, the van der Waals equation of state is somewhat qualitatively and crudely exhibiting the essential features of vapor-liquid phase equilibrium; historically, it was the first equation of state to do so. We can improve the representation of the two-phase region when using the van der Waals or other analytic equations of state by recognizing that all van der Waals loops, such as those in Fig. 7.3-2, should be replaced by horizontal lines (isobars), as shown in Fig. 7.3-3. This construction ensures that the equilibrium phases will have the same temperature and pressure (see Eqs. 7.1-9a and b). The question that remains is at which pressure should the isobar be drawn, since any pressure such that

Pa < P < Pb will yield an isotherm like that in Fig. 7.3-3. The answer is that the pressure chosen must satisfy the last condition for equilibrium, GI = GII . To identify the equilibrium pressure, we start from Eq. 4.2-8b,

dG = V dP − S dT and recognize that for the integration between any two points along an isotherm of the equation of state, we have  P2

ΔG =

V dP P1

Vapor-liquid coexistence pressure or vapor pressure

Thus, for a given equation of state we can identify the equilibrium pressure for each temperature by arbitrarily choosing pressures Pα along the van der Waals loop, until we find one for which  Pa  Pb  Pα V L G −G =0= V dP + V dP + V dP (7.3-2) Pα

Pa

Pb

Here the specific volume in each of the integrations is to be computed from the equation of state for the appropriate part of the van der Waals loop. Alternatively, we can find the equilibrium pressure graphically by noting that Eq. 7.3-2 requires that areas I and II between the van der Waals loop and the constant pressure line in Fig. 7.3-2 be equal at the pressure at which the vapor and liquid exist in equilibrium. This vapor-liquid coexistence pressure, which is a function of temperature, is called the vapor pressure of the liquid and will be denoted by P vap (T ). We can continue in the manner described here to determine the phase behavior of the fluid for all temperatures and pressures. For the van der Waals fluid, this result is

304 Chapter 7: Equilibrium and Stability in One-Component Systems 1.4 1.3 1.2 1.1

T =3 Tc

Critical point

1.0

T = 2.5 Tc

0.9 0.8 P Pc

0.7

T = 2.0 Tc

Liquid vapor coexistence region

T = 1.5 Tc

0.6 0.5

T = 1.0 Tc

0.4 0.3 0.2

T = 0.9 Tc

0.1 0

1

2

3

4

5

6

7

T = 0.8 Tc 8

T = 0.7 Tc 9

10

V _ _Vc

Figure 7.3-4 The van der Waals fluid with the vapor-liquid coexistence region identified.

shown in Fig. 7.3-4. An important feature of this figure is the dome-shaped, two-phase coexistence region. The inflection point C of Fig. 7.3-1 is the peak of this dome, and therefore is the highest temperature at which the condensed phase (the liquid) can exist; this point is called the critical point of the liquid. It is worthwhile retracing the steps followed in identifying the existence and location of the two-phase region in the P -V plane: 1. The stability condition (∂P/∂V )T < 0 was used to identify the unstable region of an isotherm and thereby establish the existence of a two-phase region. 2. The conditions T I = T II and P I = P II were then used to establish the shape (but not the location) of the horizontal coexistence line in the P -V plane. 3. Finally, the equilibrium condition GI = GII was used to locate the position of the coexistence line. A more detailed representation of phase equilibrium in a pure fluid, including the presence of a single solid phase,2 is given in the three-dimensional P V T phase diagram of Fig. 7.3-5. Such complete phase diagrams are rarely available, although data may be available in the form of Fig. 7.3-4, which is a projection of the more complete diagram onto the P -V plane, and Fig. 7.3-6, which is the projection onto the P -T plane. The concepts of phase equilibrium and the critical point can also be considered from a somewhat different point of view. Presume it were possible to compute the Gibbs energy as a function of temperature and pressure for any phase, either from an equation of state, experimental data, or statistical mechanics. Then, at fixed pressure, one could

2 If several solid phases occur corresponding to different crystal structures, as is frequently the case, the solid region

is partitioned into several regions.

7.3 Phase Equilibria: Application of the Equilibrium and Stability Criteria to the Equation of State 305

Critical point uid

lid

p Va

T4 > Tc

liquid

Pressure, P

So

Solid a nd

T3 = Tc T2 < Tc

Liq

ui va d an po d r

Tr

ipl

So

lid

Sp

T5

or

Liq

an

fic v

olu

me

oin

t li

dv

eci

T1

ep

ne

ap

,T

ure

or

t era

mp

∧

V

Te

Figure 7.3-5 The P V T phase diagram for a substance with a single solid c phase. [Adapted from J. Kestin, A Course in Thermodynamics, vol. 1. 1966 by Blaisdell Publishing Co. (John Wiley & Sons, Inc.) Used with permission of John Wiley & Sons, Inc.]

Melting (fusion) curve Critical point (Tc , Pc)

P

Liquid

Solid

Vapor

Vapor-pressure curve Triple point (Tt , Pt) Sublimation pressure curve T

Figure 7.3-6 Phase diagram in the P-T plane.

306 Chapter 7: Equilibrium and Stability in One-Component Systems

Liquid

G _

θ Vapor P = constant

TP

T

Figure 7.3-7 The molar Gibbs energy as a function of temperature for the vapor and liquid phases of the same substance.

plot G as a function of T for each phase, as shown in Fig. 7.3-7 for the vapor and liquid phases. From the equilibrium condition that G be a minimum, one can conclude that the liquid is the equilibrium phase at temperatures below TP , that the vapor is the equilibrium phase above TP , and that both phases are present at the phase transition temperature TP . If such calculations are repeated for a wide range of temperatures and pressures, it is observed that the angle of intersection θ between the liquid and vapor Gibbs energy curves decreases as the pressure (and temperature) at which the intersection increases (provided P ≤ Pc ). At the critical pressure, the two Gibbs energy curves intersect, with θ = 0; that is, the two curves are collinear for some range of T around the critical temperature Tc . Thus, at the critical point,  L  V ∂G ∂G = ∂T P ∂T P Further, since (∂G/∂T )P = −S , we have that at the critical point

S L (Tc , Pc ) = S V (Tc , Pc ) Also, for the coexisting phases at equilibrium, we have

GL (Tc , Pc ) = GV (Tc , Pc ) by Eq. 7.2-15c. Since the molar Gibbs energy, molar entropy, temperature, and pressure each have the same value in the vapor and the liquid phases, the values of all other state variables must be identical in the two equilibrium phases at the critical point. Consequently, the vapor and liquid phases become indistinguishable at the critical point. This is exactly what is experimentally observed. At all temperatures higher than the critical temperature, regardless of the pressure, only the vapor phase exists. This is the reason for the abrupt terminus of the vapor-liquid coexistence line in the pressure-temperature plane (Fig. 7.3-6). [Thus, we have two ways of recognizing the fluid critical point: first, as the peak in the vapor-liquid coexistence curve in the P -V plane (Fig. 7.3-4), and second, as the terminus of the vapor-liquid coexistence curve in the P-T plane.] Also interesting is the fluid triple point, which is the intersection of the solid-liquid, liquid-vapor, and solid-vapor coexistence curves. It is the only point on the phase diagram where the solid, liquid, and vapor coexist at equilibrium. Since the solid-liquid coexistence curve generally has a steep slope (see Fig. 7.3-6), the triple-point temperature for most fluids is close to the normal melting temperature, that is, the melting temperature at atmospheric pressure (see Problem 7.10).

7.4 The Molar Gibbs Energy and Fugacity of a Pure Component 307 Although, in general, we are not interested in equilibrium states that are unstable to large perturbations (the metastable states of Chapter 1), superheated liquids and subcooled vapors do occur and are sufficiently familiar that we will briefly relate these states to the equilibrium and stability discussions of this chapter. For convenience, the van der Waals equation of state and Fig. 7.3-2 are the basis for this discussion, though the concepts involved are by no means restricted to this equation. We start by noticing that although the liquid phase is thermodynamically stable along the isotherm shown in Fig. 7.3-2 down to a pressure Pa , the phase equilibrium analysis indicates that the vapor, and not the liquid, is the equilibrium phase at pressures below the vapor pressure Pα = P vap (T1 ). If care is taken to avoid vapor-phase nucleation sites, such as by having only clean, smooth surfaces in contact with the liquid, it is possible to maintain a liquid at fixed temperature below its vapor pressure (but above its limit of stability, Pa ), or at fixed pressure at a temperature higher than its boiling temperature, without having the liquid boil. Such a liquid is said to be superheated. The metastability of this state is illustrated by the fact that a superheated liquid, if slightly perturbed, may vaporize with explosive violence. (To prevent this occurrence, “boiling stones” are used in chemistry laboratory experiments.) It is also possible, if no condensation nucleation sites, such as dust particles, are present, to prepare a vapor at a pressure higher than the liquid-vapor coexistence pressure or vapor pressure at the given temperature, but below its limit of stability [i.e., between Pα = P vap (T1 ) and Pb in Fig. 7.3-2] or, at a temperature lower than the liquid boiling temperature. Such a vapor is termed subcooled and is also metastable. (See Problem 7.8.) Why superheating and subcooling occur is discussed in Sec. 7.8. At sufficiently low temperatures, the van der Waals equation predicts that the limit of stability of the liquid phase occurs at negative values of pressure, that is, that a liquid could support a tensile force. In fact, such behavior has been observed with water in capillary tubes and is thought to be important in the vascular system of plants.3

7.4 THE MOLAR GIBBS ENERGY AND FUGACITY OF A PURE COMPONENT In this section we consider how one uses an equation of state to identify the states of vapor-liquid equilibrium in a pure fluid. The starting point is the equality of molar Gibbs energies in the coexisting phases,

GL (T, P ) = GV (T, P )

(7.1-9c)

To proceed, we note that from Eq. 6.2-8b,

dG = −S dT + V dP so that

and



∂G ∂T





∂G ∂P

= −S

(7.4-1)

=V

(7.4-2)

P

 T

3 For a review of water under tension, especially in biological systems, see P. F. Scholander, Am. Sci. 60, 584 (1972).

308 Chapter 7: Equilibrium and Stability in One-Component Systems Since our presumption here is that we have an equation of state from which we can compute V as a function of T and P , only Eq. 7.4-2 will be considered further. Integration of Eq. 7.4-2 between any two pressures P1 and P2 (at constant temperature) yields  P2

G(T1 , P2 ) − G(T1 , P1 ) =

V dP

(7.4-3)

P1

If the fluid under consideration were an ideal gas, then V IG = RT /P , so that  P2 RT IG IG G (T1 , P2 ) − G (T1 , P1 ) = (7.4-4) dP P P1 Subtracting Eq. 7.4-4 from Eq. 7.4-3 gives

 [G(T1 , P2 ) − GIG (T1 , P2 )] − [G(T1 , P1 ) − GIG (T1 , P1 )] =

P2 P1

  RT V − dP P (7.4-5a)

Further, (1) setting P1 equal to zero, (2) recognizing that at P = 0 all fluids are ideal gases so that G(T1 , P = 0) = GIG (T1 , P = 0), and (3) omitting all subscripts yields   P RT IG G(T, P ) − G (T, P ) = (7.4-5b) V − dP P 0 For convenience, we define a new thermodynamic function, the fugacity, denoted by the symbol f , as



Definition of fugacity

f = P exp

G(T, P ) − GIG (T, P ) RT



= P exp

1 RT



P

0



 RT dP V − P (7.4-6a)

and the related fugacity coefficient φ by Definition of the fugacity coefficient

f φ= = exp P



G(T, P ) − GIG (T, P ) RT



= exp

1 RT

 0

P



 RT dP V − P (7.4-6b)

From this definition it is clear that the fugacity has units of pressure, and that f → P as P → 0; that is, the fugacity becomes equal to the pressure at pressures low enough that the fluid approaches the ideal gas state.4 Similarly, the fugacity coefficient φ = f /P → 1 as P → 0. Both the fugacity and the fugacity coefficient will be used extensively throughout this book.

4 It

is tempting to view the fugacity as a sort of corrected pressure; it is, however, a well-defined function related to the exponential of the difference between the real and ideal gas Gibbs energies.

7.4 The Molar Gibbs Energy and Fugacity of a Pure Component 309 The fugacity function has been introduced because its relation to the Gibbs energy makes it useful in phase equilibrium calculations. The present criterion for equilibrium between two phases (Eq. 7.1-9c) is GI = GII , with the restriction that the temperature and pressure be constant and equal in the two phases. Using this equality and the definition of the fugacity (Eq. 7.4-6) gives

GIG (T, P ) + RT ln

f I (T, P ) f II (T, P ) = GIG (T, P ) + RT ln P P

Now recognizing that the ideal gas molar Gibbs energy at fixed temperature and pressure has the same value regardless of which phase is considered yields as the condition for phase equilibrium f II (T, P ) f I (T, P ) = ln ln P P or in terms of the fugacity,

f I (T, P ) = f II (T, P )

Important forms of equilibrium criterion

(7.4-7a)

and in terms of the fugacity coefficient,

φI (T, P ) = φII (T, P )

(7.4-7b)

Since these equations follow directly from the equality of the molar Gibbs energy in each phase at phase equilibrium, Eqs. 7.4-7a and b can be used as criteria for equilibrium. They will be used for this purpose in this book. Since the fugacity is, by Eq. 7.4-6, related to the equation of state, the equality of fugacities provides a direct way of doing phase equilibrium calculations using the equations of state. In practice, however, Eq. 7.4-6 is somewhat difficult to use because although the molar volume V is needed as a function of T and P , it is difficult to solve the equations of state considered in Sec. 6.4 explicitly for volume. In fact, all these equations of state are in a form in which pressure is an explicit function of volume and temperature. Therefore, it is useful to have an equation relating the fugacity to an integral over volume (rather than pressure). We obtain such an equation by starting with Eq. 7.4-6b and using Eq. 6.4-25 at constant temperature in the form

dP =

1 P P P d(P V ) − dV = dZ − dV V V Z V

to change the variable of integration to obtain (Problem 7.14)  V  RT f (T, P ) 1 ln − P dV − ln Z + (Z − 1) = ln φ = P RT V =∞ V

(7.4-8)

where Z = P V /RT is the compressibility factor defined earlier. [Alternatively, Eq. 7.4-8 can be obtained from Eq. 7.4-6 using

G(T, P ) − GIG (T, P ) = [H(T, P ) − H IG (T, P )] − T [S(T, P ) − S IG (T, P )] and Eqs. 6.4-27 and 6.4-28.]

310 Chapter 7: Equilibrium and Stability in One-Component Systems For later reference we note that from Eqs. 7.4-2 and 7.4-6 we have     ∂ ln f ∂G RT =V = ∂P T ∂P T

(7.4-9a)

Also, the temperature dependence of the fugacity is usually given as the temperature dependence of the logarithm of the fugacity coefficient ( f /P ), which is computed as   

f ∂  ∂ G(T, P ) − GIG (T, P ) ln = ∂T P P ∂T P RT

1 ∂ G(T, P ) − GIG (T, P ) IG = {G(T, P ) − G (T, P )} − RT ∂T RT 2

1 G(T, P ) − GIG (T, P ) IG {S(T, P ) − S (T, P )} − =− RT RT 2

1 {[G(T, P ) − T S(T, P )] − [GIG (T, P ) − T S IG (T, P )]} RT 2

H(T, P ) − H IG (T, P ) =− RT 2 (7.4-9b)

=−

In deriving this result we have used the relations (∂G/∂T )P = −S and G = H − T S . Since the fugacity function is of central importance in phase equilibrium calculations, we consider here the computation of the fugacity for pure gases, liquids, and solids.

a. Fugacity of a Pure Gaseous Species To compute the fugacity of a pure gaseous species we will always use a volumetric equation of state and Eq. 7.4-6b or General fugacity coefficient equation

ln

1 f V (T, P ) = P RT



V =Z V RT /P V =∞



RT −P V

 dV − ln Z V + (Z V − 1)

(7.4-8) where the superscript V is used to designate the fugacity and compressibility of the vapor phase. Thus, given a volumetric equation of state of a gas applicable up to the pressure of interest, the fugacity of a pure gas can be computed by integration of Eq. 7.4-8. At very low pressures, where a gas can be described by the ideal gas equation of state P V = RT or ZV = 1 we have

f V (T, P ) (7.4-10) =0 or f V (T, P ) = P P Thus, for a low-pressure gas, the fugacity of a species is just equal to the total pressure. The calculation of the fugacity when detailed thermodynamic data, such as the steam tables, are available is demonstrated in the following illstrations. ln

7.4 The Molar Gibbs Energy and Fugacity of a Pure Component 311 Illustration 7.4-1 Computing Fugacity from Volumetric Data Use the volumetric information in the steam tables of Appendix A.III to compute the fugacity of superheated steam at 300◦ C and 8 MPa.

Solution With tabulated volumetric data, as in the steam tables, it is most convenient to use Eq. 7.4-6a:  f = P exp

1 RT



P 0

   RT V − dP P

From the superheated vapor steam tables at 300◦ C, we have P (MPa)

Vˆ (m3 /kg)

V (m3 /mol)

[V − RT /P ] (m3 /mol) × 104

0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.80 1.0 1.2 1.4 1.6 1.8 2.0 2.5 3.0 3.5 4.0 4.5 5. 0 6. 0 7. 0 8. 0

26.445 5.284 2.639 1.316 2 0.875 3 0.654 8 0.522 6 0.433 4 0.324 1 0.257 9 0.213 8 0.182 28 0.158 62 0.140 21 0.125 47 0.098 90 0.081 14 0.068 42 0.058 84 0.051 35 0.045 32 0.036 16 0.029 47 0.024 26

0.476 41 0.095 191 0.047 542 0.023 711 0.015 769 0.011 796 0.009 414 6 0.007 807 7 0.005 838 7 0.004 646 1 0.003 851 6 0.003 283 8 0.002 857 5 0.002 525 9 0.002 260 3 0.001 781 7 0.001 461 7 0.001 232 6 0.001 060 0 0.000 925 07 0.000 816 44 0.000 651 42 0.000 530 90 0.000 437 04

−1.02 −1.121 −1.101 −1.145 −1.154 −1.167 −1.157 −1.342 −1.178 −1.191 −1.194 −1.199 −1.207 −1.214 −1.222 −1.244 −1.267 −1.289 −1.313 −1.339 −1.366 −1.428 −1.498 −1.586

Numerically evaluating the integral using the data above, we find

8 MPa 0

  RT m3 MPa V − dP ≈ −1.093 × 10−3 P mol

and ⎤ m3 MPa ⎥ mol ⎥ 3 ⎦ = 8 exp(−0.2367) MPa −6 MPa m 573.15 K × 8.314 × 10 mol K = 8 × 0.7996 MPa = 6.397 MPa ⎡

⎢ f = 8 MPa exp ⎢ ⎣

−1.093 × 10−3

312 Chapter 7: Equilibrium and Stability in One-Component Systems Also, the fugacity coefficient, φ, in this case is φ=

f = 0.7996 P

Comment Had the same calculation been done at a much higher temperature, the steam would be closer to an ideal vapor, and the fugacity coefficient would be closer to unity in value. For example, the result of a similar calculation at 1000◦ C and 10 MPa yields f = 9.926 MPa and φ = f /P = 0.9926.

Illustration 7.4-2 Alternative Fugacity Calculation Use other data in the superheated vapor steam tables to calculate the fugacity of steam at 300◦ C and 8 MPa, and check the answer obtained in the previous illustration.

Solution ˆ = 3076.5 kJ/kg and Sˆ = 9.2813 At 300◦ C and 0.01 MPa we have from the steam tables H kJ/(kg K). Therefore, ◦ ˆ ˆ − T Sˆ G(300 C, 0.01 MPa) = H

= 3076.5 − 573.15 × 9.2813 = −2243.1 kJ/kg

and G(300◦ C, 0.01 MPa) = −2243.1 J/g × 18.015 g/mol = −40 409 J/mol. Since the pressure is so low (0.01 MPa) and well away from the saturation pressure of steam at 300◦ C (which is 8.581 MPa), we assume steam is an ideal gas at these conditions. Then using Eq. 7.4-3 for an ideal gas, we have GIG (300◦ C, 8 MPa) = GIG (300◦ C, 0.01 MPa) +

= −40 409 +



8MPa

V IG dP 0.01MPa

8MPa

0.01MPa

RT dP P

8 0.01 = −40 409 + 8.134 × 573.15 ln 800 J = −8555.7 mol = −40 409 + RT ln

ˆ = 2785.0 kJ/kg and For real steam at 300◦ C and 8 MPa, we have, from the steam tables, H Sˆ = 5.7906 kJ/(kg K), so that kJ ◦ ˆ G(300 C, 8 MPa) = 2785.0 − 573.15 × 5.7906 = −533.88 kg

and G(300◦ C, 8 MPa) = −9617.9

J mol

Now using Eq. 7.4-6a in the form  f (T, P ) = P exp

G(T, P ) − GIG (T, P ) RT



7.4 The Molar Gibbs Energy and Fugacity of a Pure Component 313 results in f (300◦ C, 8 MPa) = 8 MPa exp



−9617.9 − (−8555.7) 8.314 × 573.15



= 8 MPa exp[−0.2229] = 6.402 MPa

which is in excellent agreement with the results obtained in the previous illustration.

Illustration 7.4-3 Calculation of the Fugacity of Saturated Steam Compute the fugacity of saturated steam at 300◦ C.

Solution The saturation pressure of steam at 300◦ C is 8.581 MPa, and from Illustration 7.3-1 we have

ˆ V = −520.5 kJ/kg and GV = −9376.8 J/mol. Following the previous illustration, we have G GIG (300◦ C, 8.581 MPa) = GIG (300◦ C, 0.01MPa) +



8.581 MPa 0.01 MPa

RT dP P

= −40 409 + 8.314 × 573.15 ln 858.1 J = −8221.6 mol

Therefore, f V (300◦ C, 8.581 MPa) = 8.581 MPa exp



−9376.8 − (−8221.6) 8.314 × 573.15



= 8.581 MPa × 0.7847 = 6.7337 MPa

Comment Note that since, at equilibrium, f V = f L , it is also true that f L (300◦ C, 8.581 MPa) = 6.7337 MPa

At low to moderate pressures, the virial equation of state truncated after the second virial coefficient, PV B(T ) (7.4-11) =Z =1+ RT V

Fugacity coefficient: virial equation of state

may be used, if data for B(T ) are available. Using Eq. 7.4-11 in Eq. 7.4-8, we obtain (Problem 7.14) f V (T, P ) 2P B(T ) 2B(T ) ln − ln Z = − ln Z (7.4-12) = P V ZRT where

 

4B(T )P B(T ) 1 Z =1+ = 1+ 1+ V 2 RT

314 Chapter 7: Equilibrium and Stability in One-Component Systems Illustration 7.4-4 Calculation of the Fugacity of a Gas Using the Virial Equation Compute the fugacities of pure ethane and pure butane at 373.15 K and 1, 10, and 15 bar, assuming the virial equation of state can describe these gases at these conditions. Data: BET (373.15 K) = −1.15 × 10−4 m3/mol BBU (373.15 K) = −4.22 × 10−4 m3/mol

[Source: E. M. Dontzler, C. M. Knobler, and M. L. Windsor, J. Phys. Chem. 72, 676 (1968).]

Solution Using Eq. 7.4-12, we find Ethane

Butane

P (bar)

Z

f (bar)

Z

f (bar)

1 10 15

0.996 0.961 0.941

0.996 9.629 14.16

0.986 0.838 0.714

0.986 8.628 11.86

Since the pressures are not very high, these results should be reasonably accurate. However, the virial equation with only the second virial coefficient will be less accurate as the pressure increases. In fact, at slightly above 15 bar and 373.15 K, n-butane condenses to form a liquid. In this case the virial equation description is inappropriate, as it does not show a phase change or describe liquids.

Fugacity coefficient: van der Waals equation of state

Fugacity coefficient: Peng-Robinson equation of state

At higher pressure, a more complicated equation of state (or higher terms in the virial expansion) must be used. By using the (not very accurate) van der Waals equation, one obtains     fV V a PV PV ln − + − 1 − ln = ln P V − b RT V RT RT or A = (Z V − 1) − ln(Z V − B) − V (7.4-13) Z For hydrocarbons and simple gases, the Peng-Robinson equation (Eq. 6.4-2) provides a more accurate description. In this case we have √   V  fV bP a Z + (1 + 2)bP/RT V V √ ln ln − √ = (Z − 1) − ln Z − P RT 2 2bRT Z V + (1 − 2)bP/RT √  V A Z + (1 + 2)B V V √ √ ln = (Z − 1) − ln(Z − B) − 2 2B Z V + (1 − 2)B (7.4-14a) 2

where in Eqs. 7.4-13 and 7.4-14a, A = aP/(RT ) and B = P b/RT . Of course, other equations of state could be used for the fugacity calculations starting from Eq. 7.4-8, though we do not consider such calculations here.

7.4 The Molar Gibbs Energy and Fugacity of a Pure Component 315 To use either the virial, van der Waals, Peng-Robinson, or other equation of state to calculate the fugacity of a gaseous species, the following procedure is used: (1) For a given value of T and P , use the chosen equation of state to calculate the molar volume V or, equivalently, the compressibility factor Z . When using cubic or more complicated equations of state, it is the low-density (large V or Z ) solution that is used. (2) This value of V or Z is then used in Eq. 7.4-12, 7.4-13, or 7.4-14, as appropriate, to calculate the species fugacity coefficient, f /P , and thus the fugacity. The Peng-Robinson equationof-state programs, the MATHCAD worksheets described in Appendix B on the website R can be used for this calculation. for this book or in Aspen Plus Illustration 7.4-5 Calculation of the Fugacity of a Gas Using the Peng-Robinson Equation of State Compute the fugacities of pure ethane and pure butane at 373.15 K and 1, 10, and 15 bar, assuming the Peng-Robinson equation describes these gases at these conditions.

Solution Using one of the omputer programs described in Appendix B.I on the website for this book (or R in Aspen Plus ), or the data in Table 4.6-1, we obtain the following (note that the results may differ slightly depending on the program used): Ethane

Butane

P (bar)

Z

f (bar)

Z

f (bar)

1 10 15

0.996 0.957 0.935

1.00 9.58 14.06

0.985 0.837 0.733

0.99 8.57 11.81

Comment We see that these results are only slightly different from those computed with the virial equation of state. The differences would become larger as the pressure increases or the temperature decreases.

For hand calculations it is simpler, but less accurate, to compute the fugacity of a species using a specially prepared corresponding-states fugacity chart. To do this, we note that since, for simple gases and hydrocarbons, the compressibility factor Z obeys a corresponding-states relation (see Sec. 6.6), the fugacity coefficient f /P given by Eq. 7.4-6 can also be written in corresponding-states form as follows:

 P  

 P fV PV dP 1 IG (V − V ) dP = exp −1 = exp P RT 0 RT P 0

 Pr [Z(Tr , Pr , ω) − 1] d ln Pr = exp 0

(7.4-15a) or Fugacity coefficient corresponding states

fV ln = P

 0

Pr

[Z(Tr , Pr , ω) − 1] d ln Pr

(7.4-15b)

316 Chapter 7: Equilibrium and Stability in One-Component Systems 1.5 1.4

1.40

Generalized fugacity coefficients of pure gases and liquids (Zc = 0.27)

1.3

1.30

1.2 1.25

1.1 Tr

1.0

1.50

ƒ P

0.80

Fugacity coefficient,

15.00

0.9

1.2

0.9

0

Sat

0.9

4

ura

0.8

0

tio

10.00

2.00

1.15

1.40

2.00

1.90 1.10

1.80

1.0

1.70

0

nl

1.20

6.00

1.05

ine

1.60 1.50

0.7

1.45

0.90

1.00 1.40

0.6

0.85

0.94

1.35

1.00

0.94

1.30

0.5

1.25

0.80

0.90 1.20

0.4

1.15

0.75

0.85

0.3

0.70

0.2

0.65

0.1 0

1.08

1.10

1.04

1.06

1.00

1.02

0.96 0.92

0.98 0.94 0.90

0.75 0.70

0.85

0.60

0.75 0.60

0.1

0.80

0.2

0.3 0.4 0.5

1.0

2.0

3.0 4.0 5.0

0.80

0.60

0.70

10

20

30

Reduced pressure, Pr

Figure 7.4-1 (Reprinted with permission from O. A. Hougen, K. M. Watson, and R. A. Ragatz, Chemical Process Principles Charts, 2nd ed., John Wiley & Sons, New York, 1960.)

Consequently, the fugacity coefficient can be tabulated in the corresponding-states manner. The corresponding-states correlation for the fugacity coefficient of nonpolar gases and liquids given in Fig. 7.4-1 was obtained using Eq. 7.4-15b and the compressibility correlation (Fig. 6.6-3).

b. The Fugacity of a Pure Liquid The fugacity of a pure liquid species can be computed in a number of ways, depending on the data available. If the equation of state for the liquid is known, we can again start from Eq. 7.4-8, now written as

f L (T, P ) 1 ln = P RT



V =Z L RT /P

V =∞



RT −P V

 dV − ln Z L + (Z L − 1) (7.4-18)

where the superscript L is used to indicate that the liquid-phase compressibility (high density, small V and Z ) solution of the equation of state is to be used, and it is the

7.4 The Molar Gibbs Energy and Fugacity of a Pure Component 317

Fugacity coefficient Peng-Robinson equation of state

liquid-phase fugacity that is being calculated. Using, for example, the Peng-Robinson equation of state in Eq. 7.4-8 yields √   L  fL bP a Z + (1 + 2)bP/RT L L √ = (Z − 1) − ln Z − ln ln − √ P RT 2 2bRT Z L + (1 − 2)bP/RT √  L A Z + (1 + 2)B √ ln = (Z L − 1) − ln(Z L − B) − √ 2 2B Z L + (1 − 2)B (7.4-14b) To use this equation, we first, at the specified value of T and P , solve the Peng-Robinson equation of state for the liquid compressibility, Z L , and use this value to compute f L (T, P ). Of course, other equations of state could be used in Eq. 7.4-8. The PengRobinson equation-of-state programs described in Appendix B on the website for this R can be used for this calculation for the Peng-Robinson equabook or in Aspen Plus tion of state. Illustration 7.4-6 Calculation of the Fugacity of a Liquid Using the Peng-Robinson Equation of State Compute the fugacity of pure liquid n-pentane and pure liquid benzene at 373.15 K and 50 bar using the Peng-Robinson equation of state.

Solution Using one of the Peng-Robinson equation-of-state programs in Appendix B on the website for this book with the liquid (high-density) root, we obtain for n-pentane ZPE (373.15 K, P = 50 bar) = 0.2058 fPE (373.15 K, P = 50 bar) = 6.22 bar

and for benzene ZBZ (373.15 K, P = 50 bar) = 0.1522 fBZ (373.15 K, P = 50 bar) = 1.98 bar

If one has some liquid volume data, but not an equation of state, it is more convenient to start with Eq. 7.4-6, which can be rearranged to  P fL RT dP V − RT ln (7.4-16a) = P P 0 and perform the integration. However, one has to recognize that a phase change from a vapor to a liquid occurs within the integration range as the pressure is increased from zero pressure to the vapor pressure, and that the molar volume of a fluid is discontinuous at this phase transition. Thus, the result of the integration is  L f = G(T, P ) − GIG (T, P ) RT ln P     P vap (T )  f RT dP + RT Δ ln V − = P P phase P =0 change    P RT V − + dP P (7.4-16b) P vap (T )

318 Chapter 7: Equilibrium and Stability in One-Component Systems The first term on the right side of this equation is the difference between the real and ideal gas Gibbs energy changes of compressing the vapor from zero pressure to the vapor pressure of the substance at temperature T . The second term allows for the Gibbs energy change at the phase transition. The last term is the difference between the liquid and ideal gas Gibbs energy changes on compression of the liquid from its vapor pressure to the pressure of interest. From Eq. 7.4-7 and the fact that the pressure is continuous and the fugacities are equal at a phase change, we have   f Δ ln =0 P phase change

and from Eq. 7.4-6, we further have that     P vap (T )  f RT dP = RT ln V − P P sat,T 0

(7.4-17)

where (f /P )sat,T is the fugacity coefficient of the saturated fluid (either vapor or liquid at the phase transition pressure, since these fugacities are equal) at temperature T . Finally, the last term in Eq. 7.4-16 can be partially integrated as follows:

   P  P RT 1 V − V dP − RT dP = dP P P vap (T ) P vap (T ) P vap (T ) P  P P V dP − RT ln vap = P (T ) vap P (T )



Poynting correction

P

Combining these terms yields the following expressions for the fugacity of a pure liquid:     P f 1 L vap f (T, P ) = P (T ) exp V dP P sat,T RT P vap (T ) (7.4-18)   P 1 V dP = fsat (T ) exp RT P vap (T ) The exponential term in this equation, known as the Poynting pressure correction, accounts for the increase in fugacity due to the fact that the system pressure is higher than the vapor pressure of the liquid. Since the molar volume of a liquid is generally much less than that of a gas (so that P V L /RT 1), the Poynting term is only important at high pressures. (An exception to this is for cryogenic systems, where T is very low.) Equation 7.4-18 lends itself to several levels of approximation. The simplest approximation is to neglect the Poynting and (f /P )sat terms and set the fugacity of the liquid equal to its vapor pressure at the same temperature, that is,

Simplest approximation for f L

f L (T, P ) = P vap (T )

(7.4-19)

The result is valid only when the vapor pressure is low and the liquid is at a low total pressure. Although this equation applies to most fluids at low pressure, it is not correct for fluids that associate, that is, form dimers in the vapor phase, such as acetic acid.

7.4 The Molar Gibbs Energy and Fugacity of a Pure Component 319 A more accurate approximation is Better approximation for f L

L

f (T, P ) =

L fsat (T )

 V

= f (T, P

vap

)=P

vap

(T )

f P

 (7.4-20) sat,T

which states that the fugacity of a liquid at a given temperature and any pressure is equal to its fugacity at its vapor pressure (as a vapor or liquid). This approximation is valid provided the system pressure is not greatly different from the species vapor pressure at the temperature of interest (so that the Poynting term is negligible). To evaluate f V (T, P vap ), any of the methods considered in Sec. 7.4a may be used. For example, if second virial coefficient data are available, Eq. 7.4-13 can be used, or if an equation of state is available for the vapor, but not for the liquid (as may be the case for nonhydrocarbon fluids), Eq. 7.4-8 (or its integrated form for the equations of state considered in Sec. 7.4a) can be used to estimate f V (T, P vap ). Alternatively, but less accurately, (f /P )sat,T can be gotten from the corresponding-states diagram of Fig. 7.4-1 using the saturation line and the reduced temperature of interest. Another approximation that can be made in Eq. 7.4-18 is to take into account the Poynting pressure correction, but to assume that the liquid is incompressible (Fig. 7.3-4 for the van der Waals equation of state, for example, does indicate that at constant temperature the liquid volume is only slightly pressure dependent). In this case

 Best approximation for f L

L

f (T, P ) = P

vap

(T )

f P



 sat,T

V (P − P vap ) exp RT

(7.4-21)

where the term on the right, before the exponential, is the same as in Eq. 7.4-20. Alternatively, one can use   f L f (T, P ) = P (7.4-22) P T,P where the fugacity coefficient is evaluated using the principle of corresponding states and Fig. 7.4-1. However, if an equation of state for the liquid is available, one should use Eq. 7.4-8. Illustration 7.4-7 Calculation of the Fugacity of Liquid Water from Density Data Compute the fugacity of liquid water at 300◦ C and 25 MPa.

Solution From Eq. 7.4-18, we have  f L (T, P ) = f sat (T ) exp

1 RT





P

V L dP P vap (T )

Since liquids are not very compressible, we can assume (away from the critical point of steam) that V L ∼ V Lsat , so that  f L (T, P ) = f sat (T ) exp

vap VL (T )) sat (P − P RT



320 Chapter 7: Equilibrium and Stability in One-Component Systems From Illustration 7.4-3 we have f sat (300◦ C) = 6.7337 MPa and from the steam tables, at 300◦ C, P vap = 8.581 MPa and Vˆ L = 0.001 404 m3/kg, so that g m3 1 kg × 3 × 18.015 kg 10 g mol m3 = 2.5293 × 10−5 mol

V sat = 1.404 × 10−3

Consequently,

⎤ 3 −5 m × 10 × (25 − 8.581) MPa 2.5293 ⎥ ⎢ mol ⎥ f L (300◦ C, 25 MPa) = 6.7337 MPa exp ⎢ 3 ⎦ ⎣ MPa m −6 8.314 × 10 × 573.15 K mol K = 6.7337 exp[0.08715] MPa ⎡

= 6.7337 × 1.0911 MPa = 7.347 MPa

c.

Fugacity of a Pure Solid Phase

The extension of the previous discussion to solids is relatively simple. In fact, if we recognize that a solid phase may undergo several phase transitions, and let V J be the molar volume of the Jth phase and P J be the pressure above which this phase is stable at the temperature T , we have      J+1 1  P f J S sat f (T, P ) = P (T ) exp V dP (7.4-23) P sat,T RT PJ J=1

Here P is generally equal to the sublimation pressure of the solid.5 Since the sublimation (or vapor) pressure of a solid is generally small, so that the fugacity coefficient can be taken equal to unity, it is usually satisfactory to approximate Eq. 7.4-23 at low total pressures by f S (T, P ) = P sat (T ) (7.4-24a) or, for a solid subject to moderate or high total pressures, by sat



Fugacity of a solid

S

f (T, P ) = P

sat

V S (P − P sat (T )) (T ) exp RT

(7.4-24b)

Illustration 7.4-8 Calculation of the Fugacity of Ice from Density Data Compute the fugacity of ice at −5◦ C and pressures of 0.1 MPa, 10 MPa, and 100 MPa.

5 Below

the triple-point temperature, as the pressure is reduced at constant temperature, a solid will sublimate directly to a vapor. However, near the triple-point temperature some solids first melt to form liquids and then vaporize as the ambient pressure is reduced at constant temperature (see Fig. 7.3-6). In such cases P sat (T ) in Eq. 7.4-23 is taken to be the vapor pressure of the liquid.

7.4 The Molar Gibbs Energy and Fugacity of a Pure Component 321 Data: At −5◦ C, the sublimation pressure of ice is 0.4 kPa and its specific gravity is 0.915.

Solution We assume that ice is incompressible over the range from 0.4 kPa to 100 MPa, so that its molar volume over this pressure range is V =

18.015 g/mol = 19.69 cc/mol = 1.969 × 10−5 m3/mol 0.915 g/cc

From Eq. 7.4-24b, we then have ⎤ 3 −5 m × 10 × (0.1 − 0.0004) MPa 1.969 ⎥ ⎢ mol ⎥ fice (−5◦ C, 0.1 MPa) = 0.4 kPa exp ⎢ 3 ⎦ ⎣ MPa m −6 268.15 K × 8.314 × 10 mol K = 0.4 exp[8.797 × 10−4 ] kPa ⎡

= 0.4 × 1.000 88 kPa = 0.4004 kPa

Similarly, ⎤ m3 ⎢ 1.969 × 10 mol × (10 − 0.0004) MPa ⎥ ⎥ fice (−5◦ C, 10 MPa) = 0.4 kPa exp ⎢ ⎣ MPa m3 ⎦ 268.15 K × 8.314 × 10−6 mol K = 0.4369 kPa ⎡

−5

and ⎤ m3 ⎢ 1.969 × 10 mol × (100 − 0.0004) MPa ⎥ ⎥ fice (−5◦ C, 100 MPa) = 0.4 kPa exp ⎢ ⎦ ⎣ MPa m3 268.15 K × 8.314 × 10−6 mol K = 0.9674 kPa ⎡

−5

Comment Generally, the change in fugacity of a condensed species (liquid or solid) with small pressure changes is small. Here we find that the fugacity of ice increases by 10 percent for an increase in pressure from the sublimation pressure to 100 bar, and by a factor of 2.5 as the pressure on ice increases to 1000 bar.

Illustration 7.4-9 Calculation of the Fugacity of Naphthalene from Density Data The saturation pressure of solid naphthalene can be represented by log10 P sat (bar) = 8.722 −

3783 T

(T in K)

322 Chapter 7: Equilibrium and Stability in One-Component Systems The density of the solid is 1.025 g/cm3 , and the molecular weight of naphthalene is 128.19. Estimate the fugacity of solid naphthalene at 35◦ C at its vapor pressure, and also at 1, 20, 30, 40, 50, and 60 bar. (This information is used in Sec. 12.1 in a study of supercritical phase behavior.)

Solution We start with Eq. 7.4-23 for a single solid phase with the assumption that the solid is incompressible:  f S (T, P ) = P sat (T )

f P



 exp sat,T

V (P − P sat (T )) RT



Using the data in the problem statement, P sat (35◦ C) = 10[8.722−(3783/(273.15+35))] = 2.789 × 10−4 bar

Since the sublimation pressure is so low, we can use (see Fig. 7.4-1) 

Therefore,

f P

 =1 sat

f S (T = 35◦ C, P sat ) = 2.789 × 10−4 bar

At any higher pressure, we have f S (T = 35◦ C, P ) = 2.789 × 10−4 (bar) ⎤ ⎡ 3 128.19 cm3 −4 −6 m × (P − 2.879 × 10 ) bar × 10 ⎢ 1.025 mol cm3 ⎥ ⎥ × exp ⎢ 3 ⎦ ⎣ bar m −5 8.314 × 10 × 308.15 K mol K

The result is P (bar)

f S (35◦ C, P ) (bar)

2.789 × 10−4

2.789 × 10−4 2.803 × 10−4 2.933 × 10−4 3.083 × 10−4 3.241 × 10−4 3.408 × 10−4 3.582 × 10−4 3.766 × 10−4

1 10 20 30 40 50 60

7.5 THE CALCULATION OF PURE FLUID-PHASE EQUILIBRIUM: THE COMPUTATION OF VAPOR PRESSURE FROM AN EQUATION OF STATE Now that the fugacity (or equivalently, the molar Gibbs energy) of a pure fluid can be calculated, it is instructive to consider how one can compute the vapor-liquid equilibrium pressure of a pure fluid, that is, the vapor pressure, as a function of temperature, using a volumetric equation of state. Such calculations are straightforward in principle, but, because of the iterative nature of the computation involved, are best done on a

7.5 The Calculation of Pure Fluid-Phase Equilibrium 323 computer. One starts by choosing a temperature between the melting point and critical point of the fluid of interest and guessing the vapor pressure. Next, for these values of T and P, the equation of state is solved to find the liquid (smaller V and Z ) and the vapor (larger V and Z ) roots, which we denote by Z L and Z V , respectively. These values are then used in the fugacity coefficient expression appropriate to the equation of state to obtain f L and f V . Finally, these values are compared. If f L is, within a specified tolerance, equal to f V , the guessed pressure is the correct vapor pressure for the temperature of interest. If, however, the liquid-phase fugacity is greater than the vapor-phase fugacity (i.e., f L > f V ), the guessed pressure is too low; it is too high if f V > f L . In either case, a new guess must be made for the pressure, and the calculation repeated. Figure 7.5-1 is a flow diagram for the calculation of the vapor pressure using the Peng-Robinson equation-of-state (the Peng-Robinson equation-of-state programs in R Appendix B on the website for this book or in Aspen Plus can be used for this calculation); clearly, other equations of state could have been used. Also, the algorithm could be modified slightly so that pressure is chosen, and the boiling temperature at this pressure found (in this case remember that, from Eq. 6.7-1, the a parameter in the equation of state is temperature dependent). Enter Tc , Pc , ω

Enter T and guessed value of P

Compute a and b using Eqs. 6.7-1, 6.7-2, and 6.7-3 Compute A and B, where A = aP/R 2T 2 and B = Pb/RT

Solve Eq. 6.7-5 for Z L and ZV

Compute f V by substituting ZV into Eq. 7.4-14a Compute f L by substituting ZL into Eq. 7.4-14b

Is No

P=P

fL fV

| ff

L V

|

– 1 < 0.0001? Yes Print the equilibrium (vapor) pressure

Exit or repeat calculation for another temperature

Figure 7.5-1 Flow sheet of a computer program for the calculation of the vapor pressure of a fluid using the Peng-Robinson equation of state.

324 Chapter 7: Equilibrium and Stability in One-Component Systems Figure 7.5-2 contains experimental vapor pressure versus temperature data for n-butane, together with vapor pressure predictions from (1) the van der Waals equation; (2) the Peng-Robinson equation, but with a = 0.457 24R2 Tc2 /Pc rather than the correct expression of Eq. 6.7-1; and (3) the complete Peng-Robinson equation of state, that is, with the a parameter being the function of temperature given in Eq. 6.7-1. From this figure we see that the vapor pressure predictions of the van der Waals equation are not very good, nor are the predictions of the simplified Peng-Robinson equation with the a parameter independent of temperature. However, the predictions with the complete Peng-Robinson equation are excellent. Indeed, the specific form of the temperature dependence of the α(T ) term in the a parameter was chosen so that good vapor pressure predictions would be obtained at all temperatures, and so that α(Tc ) = 1 to ensure that the critical-point conditions are met. Finally, it should be pointed out that in the calculation scheme suggested here, the initial guess for the vapor pressure at the chosen temperature (or temperature at fixed pressure) must be made with some care. In particular, the pressure (or temperature) must be within the range of the van der Waals loop of the equation of state, so that separate solutions for the vapor and liquid densities (or compressibilities) are obtained. If the guessed pressure is either too high so that only the high-density root exists, or too low so that only the low-density root exists, the presumed vapor and liquid phases will be identical, and the algorithm of Fig. 7.5-1 will accept the guessed pressure as the vapor pressure, even though it is an incorrect one-phase solution, not the correct twophase solution to the problem. The incorrect “solution” so obtained is referred to as the trivial solution (in which the two phases are identical) rather than the actual solution T (K) 500 400 100

300

250

200

10

1 P vap (bar)

a b

0.1

0.01 c

0.001 2

3

4 1000/T (K)

5

6

Figure 7.5-2 The vapor pressure of n-butane as a function of temperature. The points   are the experimental data. Line a is the prediction of the van der Waals equation, line b is the prediction of the Peng-Robinson equation with α = 1, and line c is the prediction of the complete Peng-Robinson equation [i.e., α = α(t)]. The reason for plotting ln P vap versus 1/T rather than P vap versus T is discussed in Sec. 7.7.

7.5 The Calculation of Pure Fluid-Phase Equilibrium 325 to the problem, in which vapor and liquid phases exist. The best method of avoiding this difficulty is to make a very good initial guess, though this becomes increasingly harder to do as one approaches the critical point of the fluid, and the van der Waals loop becomes very small (remember, it vanishes at the critical point). Of course, using an equation of state, not only can the vapor pressure of a fluid be calculated, but so can other thermodynamic properties along the vapor-liquid phase boundary. This is demonstrated in the following illustration, which is a continuation of Illustration 6.4-1, dealing with the thermodynamic properties of oxygen. Illustration 7.5-1 (Illustration 6.4-1 continued) Using an Equation of State to Calculate the Vapor Pressure of a Compound Using the data in Illustration 6.4-1, and the same reference state, compute the vapor pressure of oxygen over the temperature range of −200◦ C to the critical temperature, and also compute the specific volume, enthalpy, and entropy along the vapor-liquid equilibrium phase envelope. Add these results to Figs. 6.4-3, 6.4-4, and 6.4-5.

Solution Using the Peng-Robinson equation-of-state programs described in Appendix B on the website for this book, we obtain the results in Table 7.5-1. The vapor pressure as a function of temperature is plotted in Fig. 7.5-3. The specific volumes and molar enthalpies and entropies of the coexisting phases have been added as the two-phase envelopes in Figs. 6.4-3, 6.4-4, and 6.4-5.

Illustration 7.5-2 (Illustration 6.4-1 concluded) Completing the Construction of a Thermodynamic Properties Chart Using an Equation of State Complete the calculated thermodynamic properties chart for oxygen by considering temperatures between −100◦ C and −200◦ C.

Solution The calculation here is much like that of Illustration 6.4-1 except that for some pressures at temperatures below the critical temperature, three solutions for the compressibility or specific T (K) 4

160

140

120

100

80 40

3

20 10

2 1

2

0

1

Pvap (bar)

In Pvap (bar)

5

0.5

–1

0.2 –2 0.1 –3 0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

1/T (K) X 10 2

Figure 7.5-3 The vapor pressure of oxygen calculated using the Peng-Robinson equation of state.

326 Chapter 7: Equilibrium and Stability in One-Component Systems Table 7.5-1 Thermodynamic Properties of Oxygen Along the Vapor-Liquid Phase Boundary Calculated Using the Peng-Robinson Equation of State T (◦ C)

Vapor

Liquid

−200

P = 0.11 bar Z V H S

0.9945 53.3256 −6311.60 −20.87

0.0004 0.0232 −13 501.04 −119.14

−190

P = 0.47 bar Z V H S

0.9832 14.5751 −6064.97 −29.34

0.0016 0.0241 −13 013.02 −112.90

−180

P = 1.39 bar Z V H S

0.9611 5.3617 −5842.71 −35.65

0.0045 0.0252 −12 517.98 −107.30

−175

P = 2.19 bar Z V H S

0.9452 3.5151 −5744.89 −38.25

0.0070 0.0259 −12 261.51 −104.64

−170

P = 3.31 bar Z V H S

0.9256 2.3964 −5658.53 −40.60

0.0103 0.0266 −11 997.44 −102.05

−160

P = 6.76 bar Z V H S

0.8746 1.2175 −5529.07 −44.75

0.0204 0.0284 −11 440.52 −96.99

T (◦ C)

Vapor

Liquid

−150

P = 12.30 bar Z V H S

0.8062 0.6713 −5475.94 −48.49

0.0371 0.0309 −10 829.37 −91.95

−140

P = 20.54 bar Z V H S

0.7170 0.3864 −5534.19 −52.20

0.0641 0.0346 −10 131.95 −86.72

−130

P = 32.15 bar Z V H S

0.5983 0.2215 −5780.50 −56.43

0.1107 0.0410 −9273.89 −80.84

−125

P = 39.44 bar Z V H S

0.5191 0.1621 −6043.66 −59.19

0.1501 0.0469 −8716.50 −77.23

−120

P = 47.85 bar Z V H S

0.4000 0.1065 −6599.85 −63.62

0.2253 0.0600 −7885.19 −72.01

V [=] m3/kmol; H [=] J/mol = kJ/kmol; S [=] J/mol K = kJ/(kmol K).

volume are obtained. To choose the correct solution in such cases, the vapor pressure calculated in the previous illustration is needed. If three solutions are obtained and the system pressure is above the vapor pressure, the liquid is the stable phase and the smallest compressibility (or specific volume) is the correct solution, and it should be used in the calculation of all other thermodynamic properties. Conversely, if the system pressure is below the calculated vapor pressure, the vapor is the stable phase and the largest compressibility or specific volume root is to be used. Using this calculational procedure, which is incorporated into the programs on the website for this book, the entries in Table 7.5-2 were obtained. These values also appear in Figs. 6.4-3, 6.4-4, and 6.4-5.

7.5 The Calculation of Pure Fluid-Phase Equilibrium 327 Table 7.5-2 The Thermodynamic Properties of Oxygen in the Low-Temperature Range Calculated Using the Peng-Robinson Equation of State T (◦ C)

−125

−150

P = 1 bar Z V H S

0.9915 12.2138 −4302.26 −19.97

0.9863 10.0988 −4995.07 −25.09

0.9756 7.9621 −5684.10 −31.35

0.0038 0.0232 −13 491.36 −119.04

P = 2 bar Z V H S

0.9830 6.0543 −4330.39 −25.85

0.9723 4.9777 −5031.67 −31.04

0.9502 3.8775 −5734.76 −37.42

0.0076 0.0232 −13 489.39 −119.04

P = 5 bar Z V H S

0.9569 2.3574 −4416.92 −33.84

0.9285 1.9014 −5146.94 −39.24

0.0158 0.0258 −12 257.99 −104.68

0.0191 0.0232 −13 484.22 −119.07

P = 10 bar Z V H S

0.9116 1.1229 −4569.30 −40.27

0.8478 0.8681 −5362.75 −46.14

0.0316 0.0258 −12 251.78 −104.75

0.0382 0.0232 −13 475.52 −119.11

P = 20 bar Z V H S

0.8119 0.5000 −4915.26 −47.61

0.0598 0.03063 −10 834.90 −92.19

0.0630 0.0257 −12 239.17 −104.88

0.0762 0.0232 −13 458.10 −119.19

P = 30 bar Z V H S

0.6912 0.2838 −5356.44 −53.14

0.0889 0.0304 −10 840.08 −92.48

0.0943 0.0256 −12 226.32 −105.01

0.1142 0.0232 −13 440.62 −119.26

P = 40 bar Z V H S

0.1512 0.0466 −8734.00 −77.37

0.1176 0.0301 −10 843.32 −92.75

0.1254 0.0256 −12 213.23 −105.14

0.1521 0.0231 −13 423.11 −119.34

P = 50 bar Z V H S

0.1740 0.0429 −8937.76 −79.04

0.1458 0.0299 −10 844.89 −93.01

0.1563 0.0255 −12 199.92 −105.27

0.1899 0.0231 −13 405.54 −119.42

(continued)

−175

−200

328 Chapter 7: Equilibrium and Stability in One-Component Systems Table 7.5-2 (Continued) T (◦ C)

−125

−150

−175

−200

P = 60 bar Z V H S

0.1987 0.0408 −9054.31 −80.11

0.1737 0.0296 −10 845.02 −93.25

0.1871 0.0256 −12 186.40 −105.39

0.2277 0.0231 −13 387.93 −119.49

P = 70 bar Z V H S

0.2235 0.0393 −9134.97 −80.92

0.2013 0.0294 −10 843.88 −93.48

0.2178 0.0254 −12 172.67 −105.51

0.2654 0.0231 −13 370.28 −119.57

P = 80 bar Z V H S

0.2481 0.0382 −9195.27 −81.59

0.2285 0.0293 −10 841.61 −93.70

0.2483 0.0253 −12 158.77 −105.62

0.3030 0.0230 −13 352.59 −119.64

P = 90 bar Z V H S

0.2724 0.0373 −9242.27 −82.16

0.2555 0.0291 −10 838.35 −93.91

0.2787 0.0253 −12 144.68 −105.74

0.3405 0.0230 −13 334.86 −119.71

P = 100 bar Z V H S

0.2964 0.0365 −9279.82 −82.67

0.2823 0.0289 −10 834.20 −94.11

0.3089 0.0252 −12 130.43 −105.85

0.3780 0.0230 −13 317.09 −119.78

V [=] m3/mol; H [=] J/mol = kJ/kmol; S [=] J/mol K = kJ/(kmol K).

Note This completes the calculation of the thermodynamic properties of oxygen. You should remember, however, that these results were obtained using only a simple, generalized three-parameter (Tc , Pc , ω ) cubic equation of state. Therefore, although they are good estimates, the results are not of as high an accuracy as would be obtained using a considerably more complicated equation of state. For example, an equation with 59 constants specific to water was used to compute the very accurate properties of steam in the steam tables (Appendix A.III). Clearly, a very large amount of carefully obtained experimental data, and sophisticated numerical analysis, must be used to obtain the 59 equation-of-state constants. This was done for water because in various industrial applications, especially in the design and evaluation of steam boilers and turbines, the properties of steam are needed to high accuracy.

It must be emphasized that the generalization of the Peng-Robinson equation-ofstate parameters given by Eqs. 6.7-2, 6.7-3, and 6.7-4 is useful only for hydrocarbons and inorganic gases (O2 , N2 , CO2 , etc.). For polar fluids (water, organic acids, alcohols, etc.), this simple generalization is not accurate, especially at low temperatures

7.5 The Calculation of Pure Fluid-Phase Equilibrium 329 and pressures. A number of alternative procedures have been suggested in this case. One is to add additional species properties in the generalization for the equation-of-state parameters, such as the dipole moment, polarizability, and/or other properties. A simpler and generally more accurate procedure is to make the equation-of-state parameters specific for each fluid. One such procedure, and the only one that we shall introduce here, is due to Stryjek and Vera,6 which is to replace Eq. 6.7-4 with

κ = κ0 + κ1 (1 + Tr0.5 )(0.7 − Tr )

(7.5-1)

where

κ0 = 0.378893 + 1.4897153ω + 0.17131848ω 2 + 0.0196554ω 3

(7.5-2)

Here κ1 is a parameter specific to each pure compound that is optimized to accurately fit low-temperature vapor pressure information. We refer to this modification of the Peng-Robinson equation as the PRSV equation. We also note (for later reference) that in this case √   R 2 Tc 1 − Tr da(T ) × = 0.457 24 dT Pc Tr      

0.7 − Tr − (1 + Tr ) Tr κ1 (1 − Tr ) − κ 2 (7.5-3) Illustration 7.5-3 Vapor Pressure Calculations for Water with the Peng-Robinson Equation of State a. Compare the predictions for the vapor pressure of water from the Peng-Robinson equation of state with generalized coefficients with data in the saturated steam tables. b. Use the PRSV equation of state with Eqs. 7.5-1 and 7.5-2 with κ1 = −0.0665 to calculate the vapor pressure of water, and compare the results with data in the steam tables.

Solution In the table below are the vapor pressure data from the steam tables and as calculated from the Peng-Robinson (PR) and PRSV equations of state.

Comment Clearly, the PRSV equation of state, with the fluid-specific parameter κ1 , leads to more accurate water-vapor pressure predictions than the Peng-Robinson equation of state with generalized parameters. This is not surprising since the PRSV equation has an extra parameter that can be fit to data for each fluid. In general, the difference in accuracy between the Peng-Robinson and PRSV equations will be larger the more different the fluid is from a simple hydrocarbon, with the PRSV equation being more accurate. For comparison, values of the vapor pressure calculated from the Peng-Robinson equation of state with α equal to a constant value of unity are also given. These values are in poor agreement with the experimental data, and demonstrate why α has been made a function of temperature. Finally, since the critical properties were used in determining the parameters in the Peng-Robinson (with α a constant or a function of temperature)

6 R.

Stryjek and J. H. Vera, Can. J. Chem. Eng. Sci. 64, 334, 820 (1986).

330 Chapter 7: Equilibrium and Stability in One-Component Systems Vapor Pressure (kPa) T (K)

Steam Tables

PR (α(T ))

PR (α = 1)

PRSV (κ1 = −0.0665)

273.16 283.15 293.15 303.15 323.15 343.15 373.15 393.15 423.15 448.15 474.15 523.15 573.15 623.15 643.15

0.6113 1.2276 2.339 4.246 12.35 31.19 101.4 198.5 475.8 892.0 1 554 3 973 8 581 16 513 21 030

0.4827 0.9969 1.947 3.617 10.94 28.50 95.98 191.3 467.4 885.9 1 555 4 012 8 690 16 646 21 030

121.7 164.8 218.6 284.8 460.7 705.9 1 233 1 712 2 626 3 668 4 976 8 228 12 750 18 654 21 440

0.6094 1.2233 2.330 4.229 12.30 31.08 101.0 198.0 474.4 889.5 1 550 3 970 8 602 16 579 21 030

and PRSV equations, both equations must give the same critical point, so that the difference between these equations is greatest far away from the critical point, at low temperatures and pressures.

In Chapter 10 we consider vapor-liquid equilibria in mixtures. For such calculations it is important to have the correct pure component vapor pressures if the mixture behavior is to be predicted correctly. Therefore, for equation-of-state calculations involving polar fluids, the PRSV equation will be used.

7.6 SPECIFICATION OF THE EQUILIBRIUM THERMODYNAMIC STATE OF A SYSTEM OF SEVERAL PHASES: THE GIBBS PHASE RULE FOR A ONE-COMPONENT SYSTEM As we have already indicated, to completely fix the equilibrium thermodynamic state of a one-component, single-phase system, we must specify the values of two state variables. For example, to fix the thermodynamic state in either the vapor, liquid, or solid region of Fig. 7.3-6, both the temperature and pressure are needed. Thus, we say that a one-component, single-phase system has two degrees of freedom. In addition, to fix the total size or extent of the system we must also specify its mass or one of its extensive properties, such as total volume or total energy, from which the mass can be calculated. In this section we are interested in determining the amount of information, and its type, that must be specified to completely fix the thermodynamic state of an equilibrium single-component, multiphase system. That is, we are interested in obtaining answers to the following questions: 1. How many state variables must be specified to completely fix the thermodynamic state of each phase when several phases are in equilibrium (i.e., how many degrees of freedom are there in a single-component, multiphase system)? 2. How many additional variables need be specified, and what type of variable should they be, to fix the distribution of mass (or number of moles) between the phases, and thereby fix the overall molar properties of the composite, multiphase system? 3. What additional information is needed to fix the total size of the multiphase system?

7.6 Specification of the Equilibrium Thermodynamic State of a System of Several Phases 331 To specify the thermodynamic state of any one phase of a single-component, multiphase system, two thermodynamic state variables of that phase must be specified; that is, each phase has two degrees of freedom. Thus, it might appear that if P phases are present, the system would have 2P degrees of freedom. The actual number of degrees of freedom is considerably fewer, however, since the requirement that the phases be in equilibrium puts constraints on the values of certain state variables in each phase. For example, from the analysis of Secs. 7.1 and 7.2 it is clear that at equilibrium the temperature in each phase must be the same. Thus, there are P − 1 relations of the form

T I = T II T I = T III .. . that must be satisfied. Similarly, at equilibrium the pressure in each phase must be the same, so that there are an additional P − 1 restrictions on the state variables of the form

P I = P II P I = P III .. . Finally, at equilibrium, the molar Gibbs energies must be the same in each phase, so that

GI (T, P ) = GII (T, P ) GI (T, P ) = GIII (T, P ) .. . which provides an additional P − 1 restrictions on the phase variables. Since there are a total of 3(P − 1) restrictions on the 2P state variables needed to fix the thermodynamic state of each of the P phases, the number of degrees of freedom for the single-component, multiphase system is Gibbs phase rule for a single-component system

⎞ Restrictions on Number of state ⎜ these state variables ⎟ ⎟ Number of degrees ⎜ variables needed to ⎟ ⎜ F= =⎝ − ⎜as a result of each of the⎟ ⎠ fix the state of each of freedom ⎝ ⎠ phases being in of the P phases equilibrium ⎛

⎞

⎛

= 2P − 3(P − 1) =3−P (7.6-1) Thus, the specification of 3 − P state variables of the individual phases is all that is needed, in principle, to completely fix the thermodynamic state of each of the phases

332 Chapter 7: Equilibrium and Stability in One-Component Systems in a one-component, multiphase system. Of course, to fix the thermodynamic states of the phases, we would, in fact, need appropriate equation-of-state information. It is easy to demonstrate that Eq. 7.6-1 is in agreement with our experience and with the phase diagram of Fig. 7.3-6. For example, we have already indicated that a singlephase (P = 1) system has two degrees of freedom; this follows immediately from Eq. 7.6-1. To specify the thermodynamic state of each phase in a two-phase system (i.e., vapor-liquid, vapor-solid, or solid-liquid coexistence regions), it is clear from Fig. 7.3-6 that we need specify only the temperature or the pressure of the system; the value of the other variable can then be obtained from the appropriate coexistence curve. Setting P equal to 2 in Eq. 7.6-1 confirms that a two-phase system has only a single degree of freedom. Finally, since the solid, liquid, and vapor phases coexist at only a single point, the triple point, a single-component, three-phase system has no degrees of freedom. This also follows from Eq. 7.6-1 with P equal to 3. The character of the variable to be specified as a degree of freedom is not completely arbitrary. To see this, consider Eq. 7.3-1a, which gives the molar volume of a twophase mixture as a function of ω V and the two single-phase molar volumes. Clearly, a specification of either V V or V L is sufficient to fix the thermodynamic state of both phases because the two-phase system has only one degree of freedom. However, the specification of the two-phase molar volume V can be satisfied by a range of choices of temperatures along the coexistence curve by suitably adjusting ω V , so that V or any other molar property of the two phases combined is not suitable for a degree-of-freedom specification. Consequently, to fix the thermodynamic state of each of the P phases in equilibrium, we must specify 3 − P properties of the individual phases. Next we want to consider how many variables, and of what type, must be specified to also fix the distribution of mass or number of moles between the phases, so that the molar thermodynamic properties of the system consisting of several phases in equilibrium can be determined. If there are P phases, there are P values of ω i , the mass fraction in phase i, which must be determined. Since the mass fractions must sum to unity, we have

ω + ω + ··· = I

II

P 

ωi = 1

(7.6-2)

i=1

as one of the relations between the mass fractions. Thus, P − 1 additional equations of the form P 

ω i Vˆ i = Vˆ

i=1

P 

ˆ ˆi = H ωi H

(7.6-3a)

i=1

or generally, P 

ω i θˆi = θˆ

(7.6-3b)

i=1

are needed. In these equations, θˆi is the property per unit mass in phase i, and θˆ is the property per unit mass of the multiphase mixture.

7.6 Specification of the Equilibrium Thermodynamic State of a System of Several Phases 333 From these equations, it is evident that to determine the mass distribution between the phases, we need to specify a sufficient number of variables of the individual phases to fix the thermodynamic state of each phase (i.e., the degrees of freedom F) and P − 1 thermodynamic properties of the multiphase system in the form of Eq. 7.6-3. For example, if we know that steam and water are in equilibrium at some temperature T (which fixes the single-degree freedom of this two-phase system), the equation of state or the steam tables can be used to obtain the equilibrium pressure, specific enthalpy, entropy, and volume of each of the phases, but not the mass distribution between the phases. If, in addition, the volume (or enthalpy or entropy, etc.) per unit mass of the two-phase mixture were known, this would be sufficient to determine the distribution of mass between the two phases, and then all the other overall thermodynamic properties. Once the thermodynamic properties of all the phases are fixed (by specification of the F = 3 − P degrees of freedom) and the distribution of mass determined (by the specification of an additional P − 1 specific properties of the multiphase system), the value of any one extensive variable (total volume, total enthalpy, etc.) of the multiphase system is sufficient to determine the total mass and all other extensive variables of the multiphase system. Thus, to determine the thermodynamic properties per unit mass of a single-component, two-phase mixture, we need to specify the equivalent of one single-phase state variable (the one degree of freedom) and one variable that provides information on the mass distribution. The additional specification of one extensive property is needed to determine the total mass or size of the system. Similarly, to fix the thermodynamic properties of a single-component, three-phase mixture, we need not specify any single state variable (since the triple point is unique), but two variables that provide information on the distribution of mass between the vapor, liquid, and solid phases and one extensive variable to determine the total mass of the three-phase system. Illustration 7.6-1 Use of the Gibbs Phase Rule a. Show, using the steam tables, that fixing the equilibrium pressure of a steam-water mixture at 1.0135 bar is sufficient to completely fix the thermodynamic states of each phase. (This is an experimental verification of the fact that a one-component, two-phase system has only one degree of freedom.) b. Show that additionally fixing the specific volume of the two-phase system Vˆ at 1 m3/kg is sufficient to determine the distribution of mass between the two phases. c. What is the total enthalpy of 3.2 m3 of this steam-water mixture?

Solution a. Using the saturation steam tables of Appendix A.III, we see that fixing the pressure at 1.0135 bar is sufficient to determine the temperature and the thermodynamic properties of each phase: T = 100◦ C Vˆ L = 0.001 044 m3/kg ˆ L = 419.04 kJ/kg H SˆL = 1.3069 kJ/(kg K)

Vˆ V = 1.6729 m3/kg ˆ V = 2676.1 kJ/kg H SˆV = 7.3549 kJ/(kg K)

Alternatively, specifying only the temperature, the specific volume, the specific enthalpy, or in fact any one other intensive variable of one of the phases would be sufficient to fix the thermodynamic properties of both phases. However, a specification of only the system

334 Chapter 7: Equilibrium and Stability in One-Component Systems pressure, temperature, or any one-phase state variable is not sufficient to determine the relative amounts of the vapor and liquid phases. b. To determine the relative amounts of each of the phases, we need information from which ω V and ω L can be determined. Here the specific volume of the two-phase mixture is given, so we can use Eq. 7.3-1, Vˆ = ω V Vˆ V + (1 − ω V )Vˆ L

and the relation ωV + ωL = 1

to find the distribution of mass between the two phases. For the situation here, 1 m3/kg = ω V (1.6729 m3/kg) + (1 − ω V )(0.001 044 m3/kg)

so that ω V = 0.5975

and ω L = 0.4025

c. Using the data in the problem statement, we know that the total mass of the steam-water mixture is M =

3.2 m3 V = = 3.2 kg ˆ 1.0 m3/kg V

From the results in parts (a) and (b) we can compute the enthalpy per unit mass of the two-phase mixture: ˆ = ωL H ˆ L + ωV H ˆV H = 0.4025(419.04) + 0.5975(2676.1) = 1767.7 kJ/kg

Therefore, ˆ = 3.2 × 1767.7 = 5303 kJ H = MH

7.7 THERMODYNAMIC PROPERTIES OF PHASE TRANSITIONS In this section several general properties of phase transitions are considered, as well as a phase transition classification system. The discussion and results of this section are applicable to all phase transitions (liquid-solid, solid-solid, vapor-solid, vapor-liquid, etc.), although special attention is given to vapor-liquid equilibrium. One phase transition property important to chemists and engineers is the slope of the coexistence curves in the P -T plane; the slope of the vapor-liquid equilibrium line gives the rate of change of the vapor pressure of the liquid with temperature, the slope of the vapor-solid coexistence line is equal to the change with temperature of the vapor pressure of the solid (called the sublimation pressure), and the inverse of the slope of the liquid-solid coexistence line gives the change of the melting temperature of the solid with pressure. The slope of any of these phase equilibrium coexistence curves can be found by starting with Eq. 7.2-15c,

GI (T, P ) = GII (T, P )

(7.7-1)

7.7 Thermodynamic Properties of Phase Transitions 335 where the superscripts label the phases. For a small change in the equilibrium temperature dT , the corresponding change of the coexistence pressure dP (i.e., the change in pressure following the coexistence curve) can be computed from the requirement that since Eq. 7.7-1 must be satisfied all along the coexistence curve, the changes in Gibbs energies in the two phases corresponding to the temperature and pressure changes must be equal, that is, dGI = dGII (7.7-2) Using Eq. 6.2-8b in Eq. 7.7-2 gives

V I dP − S I dT = V II dP − S II dT or, since P and T are the same in both phases for this equilibrium system (see Sec. 7.2)  I    S − S II ∂P ΔS = (7.7-3) = I II ∂T GI =GII ΔV V −V From Eq. 7.7-1, we also have

GI = H I − T S I = GII = H II − T S II or

S I − S II =

H I − H II T

so that Eq. 7.7-3 can be rewritten as

 Clapeyron equation

∂P sat ∂T

 = GI =GII

ΔS ΔH = ΔV T ΔV

(7.7-4)

where Δθ = θI − θII , and we have used P sat to denote the equilibrium coexistence pressure.7 Equation 7.7-4 is the Clapeyron equation; it relates the slope of the coexistence curve to the enthalpy and volume changes at a phase transition. Figure 7.3-6 is, in many ways, a typical phase diagram. From this figure and Eqs. 7.7-3 and 7.7-4 one can make several observations about property changes at phase transitions. First, since none of the coexistence lines has zero slope, neither the entropy change nor the enthalpy change is equal to zero for solid-liquid, liquid-vapor, or solidvapor phase transitions. Also, since the coexistence lines do not have infinite slope, ΔV is not generally equal to zero. (In general, both the heat of fusion, Δfus H = H L − H S , and the volume change on melting, Δfus V = V L − V S , are greater than zero for the liquid-solid transition, so that the liquid-solid coexistence line is as shown in the figure. One exception to this is water, for which Δfus H > 0 but Δfus V < 0, so the ice-water coexistence line has a negative slope.) From Sec. 7.3 we know that at the fluid critical point the coexisting phases are indistinguishable. Therefore, we can conclude that ΔH , ΔV , and ΔS are all nonzero away from the fluid critical point and approach zero as the critical point is approached.

7 P vap

is used to denote the vapor pressure of the liquid, P sub to denote the vapor pressure of the solid, and P sat to designate a general equilibrium coexistence pressure; equations containing P sat are applicable to both the vapor pressures and sublimation pressures.

336 Chapter 7: Equilibrium and Stability in One-Component Systems Of particular interest is the application of Eq. 7.7-4 to the vapor-liquid coexistence curve because this gives the change in vapor pressure with temperature. At temperatures for which the vapor pressure is not very high, it is found that V V  V L and ΔV ≈ V V . If, in addition, the vapor phase is ideal, we have Δvap V = V V = RT /P , so that

P vap Δvap H dP vap = dT RT 2

Clausius-Clapeyron equation

and

ln

Δvap H d ln P vap = dT RT 2

or

P vap (T2 ) = P vap (T1 )



T2 T1

(7.7-5a)

Δvap H dT RT 2

(7.7-5b)

which relates the fluid vapor pressures at two different temperatures to the heat of vaporization, Δvap H = H V − H L . Equation 7.7-5a is referred to as the Clausius-Clapeyron equation. The heat of vaporization is a function of temperature; however, if it is assumed to be independent of temperature, Eq. 7.7-5 can be integrated to give Approximate integrated Clausius-Clapeyron equation

Δvap H P vap (T2 ) ln vap =− P (T1 ) R



1 1 − T2 T1

 (7.7-6)

a result that is also valid over small temperature ranges even when Δvap H is temperature dependent. Equation 7.7-6 has been found to be fairly accurate for correlating the temperature dependence of the vapor pressure of liquids over limited temperature ranges. (Note that Eq. 7.7-6 indicates that ln P vap should be a linear function of 1/T , where T is the absolute temperature. It is for this reason that Figs. 7.5-2 and 7.5-3 were plotted as ln P vap versus 1/T .) Illustration 7.7-1 Use of the Clausius-Clapeyron Equation The vapor pressure of liquid 2,2,4-trimethyl pentane at various temperatures is given below. Estimate the heat of vaporization of this compound at 25◦ C. Vapor pressure (kPa) Temperature (◦ C)

0.667

1.333

2.666

5.333

8.000

13.33

26.66

53.33

101.32

−15.0

−4.3

7.5

20.7

29.1

40.7

58.1

78.0

99.2

Solution Over a relatively small range of temperature (say from 20.7 to 29.1◦ C), Δvap H may be taken to be constant. Using Eq. 7.7-6, we obtain Δvap H − ln[P vap (T2 )/P vap (T1 )] − ln(8.000/5.333) = = = 4287.8 K 1 1 1 1 R − − T2 T1 302.25 293.85

so that Δvap H = 35.649 kJ/mol

7.7 Thermodynamic Properties of Phase Transitions 337 T (K) 375

350

325

300

275

100 50

Pvap (kPa)

20 10 5

2 1 0.5 2.6

3.0

3.5

3.9

1 × 10 3 (K –1) T

Figure 7.7-1 The vapor pressure of 2,2,4-trimethyl pentane as a function of temperature.

One can obtain an estimate of the temperature variation of the heat of vaporization by noting that the integration of Eq. 7.7-5 can be carried out as an indefinite rather than definite integral. In this case we obtain Δvap H ln P vap (T ) = − +C RT

where C is a constant. Therefore, if we were to plot ln P vap versus 1/T , we should get a straight line with a slope equal to −Δvap H/R if the heat of vaporization is independent of temperature, and a curve if Δvap H varies with temperature. Figure 7.7-1 is a vapor pressure-temperature plot for the 2,2,4-trimethyl pentane system. As is evident from the linearity of the plot, Δvap H is virtually constant over the whole temperature range. This illustration is a nice example of the utility of thermodynamics in providing interrelationships between properties. In this case we see how data on the temperature dependence of the vapor pressure of a fluid can be used to determine its heat of vaporization. The equation developed in the illustration can be rewritten as ln P vap (T ) = A −

B T

(7.7-7)

with B = Δvap H/R, and it is reasonably accurate for estimating the temperature dependence of the vapor pressure over small temperature ranges. More commonly, the Antoine equation Antoine equation

ln P vap (T ) = A −

B T +C

(7.7-8)

is used to correlate vapor pressures accurately over the range from 1 to 200 kPa. Antoine constants for many substances are given by Poling, Prausnitz, and O’Connell.8 Other commonly used vapor pressure correlations include the Riedel equation ln P vap (T ) = A +

B + C ln T + DT 6 T

(7.7-9)

8 B. E. Poling, J. M. Prausnitz, and J. P. O’Connell, The Properties of Gases and Liquids, 5th ed., McGraw-Hill, New York (2001).

338 Chapter 7: Equilibrium and Stability in One-Component Systems and the Harlecher-Braun equation, ln P vap (T ) = A +

DP vap (T ) B + C ln(T ) + T T2

(7.7-10)

which must be solved iteratively for the vapor pressure, but is reasonably accurate from low vapor pressure up to the critical pressure.

Illustration 7.7-2 Interrelating the Thermodynamic Properties of Phase Changes The following vapor pressure data are available T (◦ C)

P vap (mm Hg)

Ice

−4 −2

Water

+2 +4

3.280 3.880 5.294 6.101

Estimate each of the following: a. b. c. d.

Heat of sublimation of ice Heat of vaporization of water Heat of fusion of ice The triple point of water

Solution a. Here we use Eq. 7.7-6 in the form 

3.880 3.280



ln Δsub H ln[P sub (T2 )/P sub (T1 )] =− =− = 6130 K 1 1 1 1 R − − T2 T1 271.15 K 269.15 K

so that Δsub H = 6130 K × 8.314

J kJ J = 50 965 = 50.97 mol K mol mol

b. Similarly, here we have  ln

6.101 5.294



Δvap H ln[P vap (T2 )/P vap (T1 )] = 5410 K =− =− 1 1 1 1 R − − T2 T1 277.15 K 275.15 K

or Δvap H = 5410 K × 8.314

J J kJ = 44 979 = 44.98 mol K mol mol

7.7 Thermodynamic Properties of Phase Transitions 339 c. Since

Δsub H = H(vapor) − H(solid)

and

Δvap H = H(vapor) − H(liquid)

it then follows that Δfus H = H(liquid) − H(solid) = Δsub H − Δvap H = 50.97 − 44.98 = 5.99

kJ mol

d. At the triple-point temperature Tt the sublimation pressure of the solid and the vapor pressure of the liquid are equal; we denote this triple-point pressure as Pt . Using Eq. 7.7-6 for both the solid and liquid phases gives  Δsub H = 6130 = − R

ln

and

Pt P sub (T ) 1 1 − Tt T

 Δvap H = 5410 = − R

ln

Pt vap P (T ) 1 1 − Tt T





 Pt 3.880 =− 1 1 − Tt 271.15 ln





 Pt 5.294 =− 1 1 − Tt 275.15 ln

The solution to this pair of equations is Tt = 273.279 K, and Pt = 4.627 mm Hg. The reported triple point is 273.16 K and 4.579 mm Hg, so our estimate is quite good.

Comment This example illustrates the value of thermodynamics in interrelating properties in that from two sublimation pressure and two vapor pressure data points, we were able to estimate the heat of sublimation, the heat of vaporization, the heat of fusion, and the triple point. Further, we can now use the information we have obtained and write the equations  ln

and

 ln

P sub 3.880

P vap 5.294



 = −6130



 = −5410

1 1 − T 271.15

1 1 − T 275.15





Consequently, we can also calculate the sublimation pressure and vapor pressure of ice and water, respectively, at other temperatures. Using these equations, we find P sub (−10◦ C) = 1.951 mm Hg

which compares favorably with the measured value of 1.950 mm Hg. Also, P vap (+10◦ C) = 9.227 mm Hg

compared with the measured value of 9.209 mm Hg.

340 Chapter 7: Equilibrium and Stability in One-Component Systems An important characteristic of the class of phase transitions we have been considering so far is that, except at the critical point, certain thermodynamic properties are discontinuous across the coexistence line; that is, these properties have different values in the two coexisting phases. For example, for the phase transitions indicated in Fig. 7.3-6 there is an enthalpy change, an entropy change, and a volume change on crossing the coexistence line. Also, the constant-pressure heat capacity CP = (∂H/∂T )P becomes infinite at a phase transition for a pure component because temperature is constant and the enthalpy increases across the coexistence line. In contrast, the Gibbs energy is, by Eq. 7.2-15c, continuous at a phase transition. These observations may be summarized by noting that for the phase transitions considered here, the Gibbs energy is continuous across the coexistence curve, but its first derivatives     ∂G ∂G = −S and =V ∂T P ∂P T are discontinuous, as are all higher derivatives. This class of phase transition is called a first-order phase transition. The concept of higher-order phase transitions follows naturally. A second-order phase transition (at constant T and P ) is one in which G and its first derivatives are continuous, but derivatives of second order and higher are discontinuous. Third-order phase transition is defined in a similar manner. One example of a second-order phase change is the structural rearrangement of quartz, where G, S , and V are continuous across the coexistence line, but the constant-pressure heat capacity, which is related to the second temperature derivative of the Gibbs energy,   2   CP ∂ G ∂S, =− =− 2 ∂T P ∂T P T is not only discontinuous but has a singularity at the phase change (see Fig. 7.7-2). Another example of a second-order phase transition is the change from the ferromagnetic to paramagnetic states in some materials. No phase transitions higher than second order have been observed in the laboratory.

CP (J/g K)

1.51

1.34

1.17

1.0 200

400 T (°C)

600

Figure 7.7-2 The specific heat of quartz near a second-order phase transition. (Reprinted with permission from H. B. Callen, Thermodynamics, John Wiley & Sons, New York, 1960.)

7.8 Thermodynamic Properties of Small Systems, or Why Subcooling and Superheating Occur 341

7.8 THERMODYNAMIC PROPERTIES OF SMALL SYSTEMS, OR WHY SUBCOOLING AND SUPERHEATING OCCUR A drop of liquid in a vapor consists of molecules in the bulk of the liquid interacting with many other molecules, and molecules at the surface that on the liquid side of the interface interact with many molecules and on the vapor side interact with only a few molecules. Consequently, the energy of interaction (a part of the internal energy) of the molecules at the interface between the two phases is different from that of molecules in the bulk. Other cases of molecules at the surface having different energies than those in the bulk occur at liquid-liquid, liquid-solid, and solid-vapor interfaces. Except for very small drops that have large surface areas relative to their volume, the number of molecules at the surface is very much smaller than the number of molecules in the bulk, so that the effect of the difference in energies between molecules in the bulk and surface molecules can be neglected. However, for very small drops, this surface effect is important. The contribution to the energy from the surface is usually written as σA, where σ is the surface tension for the liquid-vapor interface (or the interfacial tension for a liquidliquid interface) and A is the surface area. The σA contribution is the two-dimensional analogue of the P V term for bulk fluids. Including the effect of changing surface area in the energy balance, just as we have included the effect of changing volume, gives for a closed system without shaft work

dV dA dU −σ = Q˙ − P dt dt dt

(7.8-1)

where P is the external pressure on the system. The negative sign on the σ(dA/dt) term is due to the fact that the system must do work to increase its surface area, just as the system must do work against the surroundings to expand (increase its volume). Our interest here will be in spherical drops, as droplets that occur in nature are generally spherical or almost so. In this case V = 43 πr3 and A = 4πr2 , where r is the drop radius. Therefore, the energy balance is     2σ dr 2σ dV dr dU 2 dr 2 ˙ ˙ ˙ = Q−4πr P −8πrσ = Q−4πr P + = Q− P + dt dt dt r dt r dt or simply

dU dV = Q˙ − Pint dt dt

(7.8-2)

where Pint = (P + 2σ/r) is the internal pressure in the droplet as a result of the external pressure and the surface tension acting on the surface of the droplet. (The same relation between the internal pressure and the external pressure can be obtained from a force balance.) If the drop is very large (i.e., r → ∞), the external pressure and the internal pressure are equal. However, for a very small drop (that is, as r → 0), the internal pressure is significantly larger than the external pressure. (The expression for the internal pressure indicates that it becomes infinite in the limit of a drop of the size of a single molecule. However, when only one or a few molecules are involved, one must use a statistical mechanical description, not the macroscopic thermodynamic description in this book. Therefore, we will not consider the case of r being of molecular dimensions.)

342 Chapter 7: Equilibrium and Stability in One-Component Systems Table 7.8-1 Surface Tension at a Liquid-Vapor Interface Liquid

Temperature (◦ C)

σ (dyne/cm)∗

Water

20 25 20 20 20 20 20 20 20 20 20 20 20 139 1073

72.9 72.1 22.7 22.1 27.6 28.9 43.4 64.0 11.9 20.4 21.6 26.7 487 198 115

Methanol Ethanol 1-Octanol Benzene Aniline Glycerol Perfluorohexane n-Heptane n-Octane Propionic acid Mercury Sodium Sodium chloride ∗

Divide by 1000 for J/m2 .

Table 7.8-2 Interfacial Tension at a Liquid-Liquid Interface Liquid Water/n-butyl alcohol Water/mercury Water/benzaldehyde Water/diethylene glycol Mercury/n-hexane ∗

Temperature (◦ C) 20 20 20 25 20

σ (dyne/cm)∗ 1.8 415 15.5 57 378

Divide by 1000 for J/m2 .

Some values of the surface tension (for vapor-liquid interfaces) and interfacial tensions (for liquid-liquid interfaces) are given in Table 7.8-1. We now consider two cases. The first is a bulk liquid at a temperature T subject to an external pressure of P . The fugacity of this liquid, f L (T, P ), can be computed by any of the methods described earlier in this chapter. The second case is that of a small droplet of the same liquid at the same temperature and external pressure. The fugacity of this droplet is     2σ 2σ L L L L L = f (T, P ) exp fdrop (T, P ) = f (T, Pint ) = f T, P + V r rRT (7.8-3) where V L is the molar volume of the liquid, and we have introduced the Poynting correction (so that the fugacity of the liquid drop can be expressed as the product of the fugacity of the bulk liquid at the same temperature and same external pressure) and a correction factor for the surface effect. Clearly, the contribution from the surface term is always greater than unity, so that the fugacity of the drop is always greater than the fugacity of the bulk liquid at the same temperature and external pressure. Further, this difference will be very large if the drop is very small, as shown in Fig. 7.8-1. There are several implications of this result. The first is that since at a given temperature the fugacity of a drop is higher than that of a bulk liquid, it will attain a value of

7.8 Thermodynamic Properties of Small Systems, or Why Subcooling and Superheating Occur 343 104

Gdrop − G = RT ln[fdrop(T, P)/f(T, P)] (J/mol)

103 100 10 1 0.1 0.01 10–3 10–4 10–5 10–6 10–7 10–9

10–8

10–7

10–6

10–5

10–4

10–3

0.01

0.1

1

10

r (m)

Figure 7.8-1 Difference in Gibbs energy between a bubble of radius r and the bulk fluid for water at 20◦ C.

1.013 bar at a lower temperature than that of the bulk liquid; that is, the boiling temperature of small drops at atmospheric pressure will be lower than that of the bulk liquid (Problem 7.72). Or, more generally, small drops will vaporize more easily than the bulk fluid because of their higher fugacity (a thermodynamic effect) and their greater surface area per unit mass (promoting mass transfer). Also, as vaporization of the drop occurs and the drop radius becomes smaller, both the fugacity of the liquid in the drop and the surface-to-volume ratio increase, so that the vaporization process accelerates. It is for this reason that to vaporize a liquid quickly in consumer products—for example, in an air freshener—an atomization process is used that produces small droplets. From the relationship between fugacity and molar Gibbs energy, we know that the molar Gibbs energy of a droplet is higher than that of the bulk fluid at the same temperature and external pressure. This is shown in Fig. 7.8-1. In particular, as the droplet radius goes to zero, this difference in molar Gibbs energy becomes infinite. Instead of vaporization, which is the disappearance of a phase, consider the opposite process of condensation, in which a liquid phase is created. The liquid will form when the fugacity (or, equivalently, molar Gibbs energy) of the liquid is less than that of the vapor. However, we see from Fig. 7.8-1 that the first liquid droplet that forms, of infinitesimal size, will have a very large Gibbs energy. Therefore, a vapor can be cooled below its normal condensation temperature (as long as there is no dust or other nucleation sites present) but may not condense, because while the Gibbs energy of the bulk liquid will be less than that of the vapor, the Gibbs energy of the very small droplets that will form first is greater than that of the vapor. This is the phenomenon of subcooling. As the vapor continues to be cooled, the difference between the vapor and bulk liquid Gibbs energies will become large enough to overcome the Gibbs energy penalty required for the formation of the first droplet of liquid. Once this first droplet is formed, since the Gibbs energy of a droplet decreases by the droplet growing in size, droplet growth (condensation) then occurs rapidly.

344 Chapter 7: Equilibrium and Stability in One-Component Systems Similarly, to initiate boiling in a liquid, an infintesimal bubble of vapor must first be created. However, here too there is a very large Gibbs energy barrier to the formation of the first, very small bubble, and unless boiling chips or other devices are used to induce the nucleation of bubbles, on heating the normal boiling temperature will be exceeded before boiling starts. This is superheating. Note that the cause of both subcooling and superheating is the cost in Gibbs energy of producing a new phase (the liquid droplet in subcooling and a vapor bubble in superheating). Similar phenomena occur in other phase transitions, such as solidification or crystallization. That is, on cooling a liquid, initially a temperature below its normal freezing point will be reached without freezing (liquid-solid subcooling), and then suddenly a significant amount of freezing or crystallization will occur. There is a related process that occurs after nucleation of a new phase and is especially evident in crystallization. Initially, crystals of many different sizes form when crystallization (or freezing) first occurs. If the system is adiabatic (so that no further heating or cooling is supplied the small crystals will decrease in size and eventually disappear, while the large crystals will grow, until there may be only a few (or even just one) large crystals. This phenomenon is referred to as Ostwald ripening. We can easily understand why this occurs by examining Fig. 7.8-1, where we see that crystals of large size (radius) have a lower Gibbs energy than crystals of small size. Therefore, in the evolution toward equilibrium, the state of lowest Gibbs energy, the Gibbs energy of the system is lowered by replacing many small crystals with a few larger ones. PROBLEMS 7.1 By doing some simple calculations and plotting several graphs, one can verify some of the statements made in this chapter concerning phase equilibrium and phase transitions. All the calculations should be done using the steam tables. a. Establish, by direct calculation, that GL = GV

for steam at 2.5 MPa and T = 224◦ C. b. Calculate GV at P = 2.5 MPa for a collection of temperatures between 225 and 400◦ C and extrapolate this curve below 224◦ C. c. Find GL at 160, 170, 180, 190, 200, and 210◦ C. Plot this result on the same graph as used for part (b) and extrapolate above 224◦ C. (Hint: For a liquid H and S can be taken to be independent of pressure. Therefore, the values of H and S for the liquid at any pressure can be gotten from the data for the saturated liquid at the same temperature.) How does this graph compare with Fig. 7.3-7? d. Plot V versus T at P = 2.5 MPa for the temperature range of 150 to 400◦ C, and show that V is discontinuous. e. Plot CP versus T at P = 2.5 MPa over the temperature range of 150 to 400◦ C, and thereby establish that CP is discontinuous. f. Using the data in the steam tables, show that the Gibbs energies per unit mass of steam and liquid water in equilibrium at 300◦ C are equal.

7.2 a. Show that the condition for equilibrium in a closed system at constant entropy and volume is that the internal energy U achieve a minimum value subject to the constraints. b. Show that the condition for equilibrium in a closed system at constant entropy and pressure is that the enthalpy H achieve a minimum value subject to the constraints. 7.3 a. Show that the intrinsic stability analysis for fluid equilibrium at constant temperature and volume leads to the single condition that 

∂P ∂V

 b exhibits a van der Waals loop. However, there is also interesting behavior of the equation in the ranges V < b and V < 0, though these regions do not have any physical meaning. At supercritical temperatures the van der Waals loop disappears, but much of the structure in the V < 0 region remains. Establish this by drawing a P -V plot for n-butane at temperatures of 0.6 Tc , 0.8 Tc , 1.0 Tc , 1.5 Tc , and 5 Tc for all values of V . 7.40 The van’t Hoff corollary to the third law of thermodynamics is that whenever two solid forms of a substance are known, the one with the greater specific heat will be the more stable one at higher temperatures. Explain why this is so. 7.41 Consider the truncated virial equation of state B(T ) PV =1+ RT V

where B(T ) is the second virial coefficient. Obtain the constraint on B(T ) if the fluid is to be thermodynamically stable. 7.42 Show that the requirement for stability of a closed system at constant entropy and pressure leads to the condition that CP > 0 for all stable fluids. (Hint: You do not need to make a change of variable to show this.) 7.43 A fluid obeys the Clausius equation of state P =

RT V − b(T )

where b(T ) = b0 + b1 T . The fluid undergoes a JouleThomson expansion from T1 = 120.0◦ C and P1 = 5.0 MPa to a final pressure P2 = 1.0 MPa. Given that CP = 20.97J/(mol K), b0 = 4.28 × 10−5 m3 /mol, and b1 = 1.35 × 10−7 m3 /(mol K), determine the final temperature of the fluid. 7.44 Does a fluid obeying the Clausius equation of state have a vapor-liquid transition? 7.45 Can a fluid obeying the virial equation of state have a vapor-liquid transition? 7.46 Derive the expression for the fugacity coefficient of a species described by the Soave–Redlich-Kwong equation of state (the analogue of Eq. 7.4-14b).

7.47 Using the Redlich-Kwong equation of state, compute and plot (on separate graphs) the pressure and fugacity of nitrogen as a function of specific volume at the two temperatures a. 110 K b. 150 K 7.48 The vapor pressure of liquid ethanol at 126◦ C is 505 kPa, and its second virial coefficient at this temperature is −523 cm3 /mol. a. Calculate the fugacity of ethanol at saturation at 126◦ C assuming ethanol is an ideal gas. b. Calculate the fugacity of ethanol at saturation at 126◦ C assuming ethanol is described by the virial equation truncated after the second virial coefficient. (Note: Since this calculation is being done at saturation conditions, the value computed is for both the vapor and liquid, even though the calculation has been done using only the vapor pressure and vapor-phase properties.) 7.49 Liquid ethanol, for which κT = 1.09 × 10−6 kPa−1 , has a relatively high isothermal compressibility, which can have a significant effect on its fugacity at high pressures. Using the data in the previous problem, calculate the fugacity of liquid ethanol at 25 MPa and 126◦ C a. Assuming liquid ethanol is incompressible b. Using the value for the isothermal compressibility of liquid ethanol given above 7.50 Redo Illustration 7.4-6 using the Soave–RedlichKwong equation of state. 7.51 Redo Illustration 7.5-1 using the Soave–RedlichKwong equation of state. 7.52 Redo Illustration 7.5-2 using the Soave–RedlichKwong equation of state. 7.53 Adapt the program PR1, or one of the other PengRobinson programs, or develop a program of your own to use the Soave–Redlich-Kwong equation of state rather than the Peng-Robinson equation to calculate the thermodynamic properties of a fluid. 7.54 Adapt the program PR1, or one of the other PengRobinson programs, or develop a program of your own using the Peng-Robinson equation of state to do the calculations for a Joule-Thomson expansion of a liquid under pressure to produce a vapor-liquid mixture at ambient pressure. The output results should include the outlet temperature and the fractions of the outlet stream that are liquid and vapor, respectively. 7.55 Adapt the program PR1, or one of the other PengRobinson programs, or develop a program of your own using the Peng-Robinson equation of state to do the calculations for an isentropic expansion of a liquid under pressure to produce a vapor-liquid mixture at ambient pressure. The output results should include

Problems 351

7.56 7.57 7.58

7.59 7.60

the outlet temperature and the fractions of the outlet stream that are liquid and vapor, respectively. Redo Problem 7.54 with the Soave–Redlich-Kwong equation of state. Redo Problem 7.55 with the Soave–Redlich-Kwong equation of state. Redo Illustration 7.5-3 with the Soave–RedlichKwong equation of state. Compare the results obtained with those for the Peng-Robinson equation in the illustration. Also, repeat the calculations for the Soave– Redlich-Kwong equation but with the simplifying assumption that the α function of Eqs. 6.7-6 and 6.7-8 is a constant and equal to unity, rather than a function of temperature. Redo Illustration 6.7-1 using the Soave–RedlichKwong equation of state. A fluid obeys the equation of state

7.63

C B PV + 2 =1+ RT V V

where B and C are constants. Also,

7.64

∗ CV = a + bT

a. For what values of the constants B and C will this fluid undergo a vapor-liquid phase transition? b. What is the molar internal energy change if this fluid is heated at a constant pressure P from T1 to T2 , and how does this compare with the molar internal energy change for an ideal gas with the same ideal gas heat capacity undergoing the same change? c. Develop an expression relating the differential change in temperature accompanying a differential change in volume for a reversible adiabatic expansion of this fluid. What would be the analogous expression if the fluid were an ideal gas? 7.61 It has been suggested that a simple cubic equation of state can also be used to describe a solid. However, since in a solid certain molecular contacts are preferred, compared with a fluid (vapor or liquid) in which molecules interact in random orientations, it has also been suggested that different equation-of-state parameters be used for the fluid and solid phases. Assume both the fluid and solid phases obey the equation of state RT a P = − 2 V

V

with the parameters af and as , respectively. Develop expressions for the fugacity coefficients of the gas, liquid, and solid. Discuss how you would use such an equation to calculate the vapor-liquid, liquid-solid, vapor-solid, and vapor-liquid-solid equilibria. 7.62 Proteins can exist in one of two states, the active, folded state and the inactive, unfolded state. Protein

7.65

7.66

7.67

folding is sometimes thought of as a first-order phase transition from folded to unfolded (denaturation) with increasing temperature. (We will revisit this description in Chapter 15.) Denaturation is accompanied by an increase in molecular volume and a significant positive latent heat (i.e., F→D V > 0 and F→D H > 0). a. Sketch the Gibbs energy function for both D (denatured) and F (folded) proteins, and identify T ∗ , the phase transition temperature. Which state has the higher entropy? b. Given the observed volume and enthalpy changes upon denaturation, how would increasing the pressure affect T ∗ ? Should biopharmaceuticals (i.e., folded proteins) be stored under high pressure? Calculate the vapor pressure of methane as a function of temperature using the Peng-Robinson equation of state. Compare your results with (a) literature values and (b) predictions using the Peng-Robinson equation of state in which the temperature-dependent parameter α (T ) is set equal to unity at all temperatures. Calculate the vapor pressure of n-butane as a function of temperature using the Peng-Robinson equation of state. Compare your results with (a) literature values and (b) predictions using the Peng-Robinson equation of state in which the temperature-dependent parameter α (T ) is set equal to unity at all temperatures. Calculate the vapor pressure of n-decane as a function of temperature using the Peng-Robinson equation of state. Compare your results with (a) literature values and (b) predictions using the Peng-Robinson equation of state in which the temperature-dependent parameter α (T ) is set equal to unity at all temperatures. The critical temperature of benzene is 289◦ C and its critical pressure is 4.89 MPa. At 220◦ C its vapor pressure is 1.91 MPa. a. Calculate the fugacity of liquid benzene in equilibrium with its pure vapor at 220◦ C. b. Repeat the calculation for the case in which liquid benzene is under an atmosphere of hydrogen such that the total pressure is 13.61 MPa. The average density of liquid benzene at these conditions is 0.63 g/cc, and you can assume that hydrogen is insoluble in benzene at this temperature. The molecular weight of benzene is 78.02. The sublimation pressure of carbon dioxide as a function of temperature is

T (K) P (kPa)

130 0.032

155 1.674

185 44.02

194.5 101.3

205 227

and the molar volume of CO2 is 2.8 × 10−5 m3 /mol. a. Determine the heat of sublimation of CO2 at 190 K. b. Estimate the fugacity of solid CO2 at 190 K and 200 bar.

352 Chapter 7: Equilibrium and Stability in One-Component Systems 7.68 The figure below shows a pressure-enthalpy diagram submitted by Joe Udel as part of a homework assignment. On the diagram isotherms (T1 < T2 < T3 ), isochores (V 1 < V 2 < V 3 ), and isentropes (S 1 < S 2 < S 3 ) are shown. Is this figure consistent with requirements for a stable equilibrium system? V3

V2 V1 S1 S2

P

the external pressure is constant. A small amount of heat is added to the gas in the cylinder, resulting in the expansion of the gas. Will the temperature of the gas increase or decrease? 7.71 Using a Knudsen effusion cell, Svec and Clyde11 measured the sublimation pressures of several amino acids as a function of temperature. a. The results they reported for glycine are T (K) P (torr)

S3 T1

T2

453

457

466

471

5.87 × 10−5

4.57 × 10−5

1.59 × 10−4

× 10−4

2.43

Use these data to estimate the heat of sublimation of glycine over this temperature range. b. The results they reported for l-alanine are

T3

H

7.69 A relatively simple cubic equation of state is P =

RT a −b− √ V TV 2

where a is a constant. a. Determine the relationship between the a and b parameters in this equation of state and the critical temperature and pressure of the fluid. b. Obtain an expression for the constant-volume heat capacity as a function of temperature, volume, the critical temperature and pressure, and the ideal heat capacity. c. The constant-volume heat capacities of real fluids diverge as the critical point is approached. Does the constant-volume heat capacity obtained from this equation of state diverge as the critical point is approached? 7.70 a. Show that

T (K) P (torr)

b. Use this result to show that, for a stable system at equilibrium, (∂V /∂T )S and (∂V /∂T )P must have opposite signs. c. Two separate measurements are to be performed on a gas enclosed in a piston-and-cylinder device. In the first measurement the device is well insulated so there is no flow of heat to or from the gas, and the piston is slowly moved inward, compressing the gas, and its temperature is found to increase. In the second measurement the piston is free to move and

11 H.

460

5.59

× 10−5

1.22

× 10−4

465 2.03

× 10−4

469 2.58

× 10−4

Use these data to estimate the heat of sublimation of l-alanine over this temperature range. [1 torr = 133.32 Pa] 7.72 Determine the boiling temperature of a water droplet at 1.013 bar as a function of the droplet radius. 7.73 The following equation of state has been proposed for a fluid B C PV =1+ + 2 RT V V

7.74



 ∂V   ∂T S ∂T 2 CV   =− ∂V T V κT ∂P V ∂T P

453

7.75 7.76 7.77 7.78

7.79

where B and C are constants. a. Does this fluid exhibit a critical point? Prove it. b. If you believe the answer to part (a) is yes, derive expressions for B and C in terms of the critical temperature and pressure for this fluid. Estimate the vapor pressures of the refrigerants 1,1,1,2-tetrafluoroethane, dichlorodifluoromethane and 2-chloro-1,1,1,2-tetrafluoroethane over the temperature range of −40 to +85◦ C. R Redo Problem 7.1 using Aspen Plus . R Redo Problem 7.33 using Aspen Plus . R . Redo Problem 7.34 using Aspen Plus Explain the nonmonotonic behavior of the fugacity coefficent along the T = 1.50 isotherm in Fig. 7.4-1 using the van der Waals equation of state. R Use the Analysis>Pure tool in Aspen Plus with the Peng-Robinson equation of state to compute the vapor pressures of the refrigerants R12 (dichlorodifluoromethane, or Freon 12), R124 (1-chloro-1,2,2,2,-tetrafluoroethane), and R134a (1,1,1,2-tetrafluoroethane) over the temperature range of 0 to 100◦ C.

J. Svec and D. C. Clyde, J. Chem. Eng. Data 10, 151 (1965).

Chapter

8 The Thermodynamics of Multicomponent Mixtures In this chapter the study of thermodynamics, which was restricted to pure fluids in the first part of this book, is extended to mixtures. First, the problem of relating the thermodynamic properties of a mixture, such as the volume, enthalpy, Gibbs energy, and entropy, to its temperature, pressure, and composition is discussed; this is the problem of the thermodynamic description of mixtures. Next, the equations of change for mixtures are developed. Using the equations of change and the positive definite nature of the entropy generation term, the general equilibrium criteria for mixtures are considered and shown to be formally identical to the pure fluid equilibrium criteria. These criteria are then used to establish the conditions of phase and chemical equilibrium in mixtures, results that are of central importance in the remainder of this book.

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 8 The goals of this chapter are for the student to: • Understand the difference between partial molar properties and pure component

properties (Sec. 8.1)

• Be able to use the Gibbs-Duhem equation to simplify equations (Sec. 8.2) • Be able to identify a set of independent reactions • Be able to use the mass, energy, and entropy balance equations for mixtures

(Secs. 8.4 and 8.5)

• Be able to compute partial molar properties from experimental data (Sec. 8.6) • Be able to derive the criteria for phase and chemical equilibria in multicomponent

systems (Secs. 8.7 and 8.8)

NOTATION INTRODUCED IN THIS CHAPTER A¯i Partial molar Helmholtz energy of species i (J/mol) C C¯P,i f¯i ¯i G

Number of components Partial molar heat capacity of species i (J/mol K) Fugacity of species i in a mixture (kPa) Partial molar Gibbs energy of species i (J/mol) 353

354 Chapter 8: The Thermodynamics of Multicomponent Mixtures

Δf Goi Δmix G Δrxn G Δrxn G◦ ¯i H Δmix H Δf H oi Δrxn H Δrxn H ◦ M P S¯i V¯i Δmix V Xj wi wI φ¯i νi,j

Standard state molar Gibbs energy of formation of species i (kJ) Gibbs energy change on mixing (kJ) Gibbs energy change of reaction (kJ) Standard state Gibbs energy change on reaction (kJ) Partial molar enthalpy of species i (J/mol) Enthalpy energy change on mixing (kJ) Standard State molar enthalpy of formation of species i (kJ) Enthalpy change (or heat) of reaction (kJ) Standard state enthalpy change on reaction (kJ) Number of independent chemical reactions Number of phases Partial molar entropy of species i [J/(mol K)] Partial molar volume of species i (m3 /mol) Volume change on mixing (m3 ) Molar extent of reaction of the jth independent chemical reaction (mol) Mass fraction of species i in a phase Fraction of total system mass in phase I Fugacity coefficient of species i in a mixture (= f¯i /P ) Stoichiometric coefficient of species i in independent chemical reaction j

8.1 THE THERMODYNAMIC DESCRIPTION OF MIXTURES In Chapter 1 we pointed out that the thermodynamic state of a pure, single-phase system is completely specified by fixing the values of two of its intensive state variables (e.g., T and P or T and V ), and that its size and thermodynamic state is fixed by a specification of its mass or number of moles and two of its state variables (e.g., N , T , and P or N , T , and V ). For a single-phase mixture of C components, one intuitively expects that the thermodynamic state will be fixed by specifying the values of two of the intensive variables of the system and all but one of the species mole fractions (e.g., T , P , x1 , x2 , . . . , xC−1 ), the remaining mole fraction being calculable from the fact that the mole fractions must sum to 1. (Though we shall largely use mole fractions in the discussion, other composition scales, such as mass fraction, volume fraction, and molality, could be used. Indeed, there are applications in which each is used.) Similarly, one anticipates that the size and state of the system will also be fixed by specifying the values of two state variables and the mole numbers of all species present. That is, for a C-component mixture we expect equations of state of the form

U = U (T, P, N1 , N2 , . . . , NC ) or U = U (T, P, x1 , x2 , . . . , xC−1 ) V = V (T, P, N1 , N2 , . . . , NC ) or V = V (T, P, x1 , x2 , . . . , xC−1 ) and so on. Here xi is the mole fraction of species i—that is, xi = Ni /N , where Ni is the number of moles of species i and N is the total number of moles—and U and V are the internal energy and volume per mole of the mixture.1 Our main interest here will be in determining how the thermodynamic properties of the mixture depend on species concentrations. That is, we would like to know the concentration dependence of the mixture equations of state. 1 In

writing these equations we have chosen T and P as the independent state variables; other choices could have been made. However, since temperature and pressure are easily measured and controlled, they are the most practical choice of independent variables for processes of interest to chemists and engineers.

8.1 The Thermodynamic Description of Mixtures 355 A naive expectation is that each thermodynamic mixture property is a sum of the analogous property for the pure components at the same temperature and pressure weighted with their fractional compositions. That is, if U i is the internal energy per mole of pure species i at temperature T and pressure P , then it would be convenient if, for a C-component mixture,

U (T, P, x1 , . . . , xC−1 ) =

C 

xi U i (T, P )

(8.1-1)

i=1

ˆi is the internal energy per unit mass of pure species i, then the Alternatively, if U development of a mixture equation of state would be simple if ˆ (T, P, w1 , . . . , wC−1 ) = U

C 

ˆi (T, P ) wi U

(8.1-2)

i=1

where wi is the mass fraction of species i. Unfortunately, relations as simple as Eqs. 8.1-1 and 8.1-2 are generally not valid. This is evident from Fig. 8.1-1, which shows the enthalpy of a sulfuric acid–water mixture at various temperatures. If equations of the form of Eq. 8.1-2 were valid, then

ˆ = w1 H ˆ 1 + w2 H ˆ 2 = w1 H ˆ 1 + (1 − w1 )H ˆ2 H

Definition of the volume change on mixing

and the enthalpy of the mixture would be a linear combination of the two purecomponent enthalpies; this is indicated by the dashed line connecting the purecomponent enthalpies at 0◦ C. The actual thermodynamic behavior of the mixture, given by the solid lines in the figure, is quite different. In Fig. 8.1-2 are plotted the volume change on mixing, defined to be2

Δmix V = V (T, P, x) − x1 V 1 (T, P ) − x2 V 2 (T, P ) and the enthalpy change on mixing,

Definition of the enthalpy change on mixing

Δmix H = H(T, P, x) − x1 H 1 (T, P ) − x2 H 2 (T, P ) for several mixtures. (The origin of such data is discussed in Sec. 8.6.) If equations like Eq. 8.1-1 were valid for these mixtures, then both Δmix V and Δmix H would be equal to zero at all compositions; the data in the figures clearly show that this is not the case. There is a simple reason why such elementary mixture equations as Eqs. 8.1-1 and 8.1-2 fail for real mixtures. We have already pointed out that one contribution to the internal energy (and other thermodynamic properties) of a fluid is the interactions between its molecules. For a pure fluid this contribution to the internal energy is the result of molecules interacting only with other molecules of the same species. However, in a binary fluid mixture of species 1 and 2, the total interaction energy is a result of 1-1, 2-2, and 1-2 interactions. Although the 1-1 and 2-2 molecular interactions that occur in the mixture are approximately taken into account in Eqs. 8.1-1 and 8.1-2 through the pure-component properties, the effects of the 1-2 interactions are not. Therefore, it is

2 In

these equations, and most of those that follow, the abbreviated notation θ(T, P, x1 , x2 , . . . , xC−1 ) = θ(T, P, x) will be used.

356 Chapter 8: The Thermodynamics of Multicomponent Mixtures 450

ng po int at 0.1 M 204 Pa .4° C

0.1

250

Liquid + Vapor

a MP

B oili

200

232.2°C

°C 176. 7°C

150 100

14

8.9

50 0 93

–50

.3

°C 1.1

°C

12

C

C .8° 37 1°C .

.6°

65

–100 21

Enthalpy, (kJ/kg) of mixture including vapor

287.8°C 260°C 232.2°C

176.7°C

t at

oin

300

287.8°C 260°C

gp

ilin

148.9°C 121.1°C 115.6°C

110°C

104.4°C

Bo

350

204.4°C

100°C 400

–150

C

0°

–200

Liquid

–250 –300 –350 0

20

40

60

80

100

Wt. percentage H2SO4

Figure 8.1-1 Enthalpy-concentration diagram for aqueous sulfuric acid at 0.1 MPa. The sulfuric acid percentage is by weight (which, in the two-phase region, includes the vapor). Reference states: The enthalpies of the pure liquids at 0◦ C and their vapor pressures are zero. (Based on Fig. 81, p. 325 in O. A. Hougen, K. M. Watson, and R. A. Ragatz, Chemical Process Principles: I. Material and Energy Balances, 2nd ed. Reprinted with permission from John Wiley & Sons, New York, 1954.) This figure appears as an Adobe PDF file on the website for this book, and may be enlarged and printed for easier reading and for use in solving problems.

not surprising that these equations are not good approximations for the properties of most real fluid mixtures. What we must do, instead of making simple guesses such as Eqs. 8.1-1 and 8.1-2, is develop a general framework for relating the thermodynamic properties of a mixture to the composition variables. Since the interaction energy of a mixture is largely determined by the number of molecular interactions of each type, the mole fractions, or, equivalently, the mole numbers, are the most logical composition variables for the analysis. We will use the symbol θ to represent any molar property (molar volume, molar enthalpy, etc.) of a mixture consisting of N1 moles of species 1, N2 moles of C species 2, and so forth, N = i Ni being the total number of moles in the system. For the present our main interest is in the composition variation of θ at fixed T and P , so

8.1 The Thermodynamic Description of Mixtures 357 200 H 0

0.20

–200

ΔH _ mix (J/mol)

ΔV _ mix (cm3/mol)

0.15 0.10 0.05

–400

F

–600

Cl Br

–800 0 –1000

–0.05 –0.10

I

–1200 0

0.2

0.4

0.6

0.8

1

0

0.2

xMF

0.4 xC

0.6

0.8

1

6F5Y

(a)

(b)

Figure 8.1-2 (a) Volume change on mixing at 298.15 K: ◦ methyl formate + methanol, • methyl formate + ethanol. [From J. Polak and B. C.-Y. Lu, J. Chem. Thermodyn., 4, 469 (1972). Reprinted with permission from Academic Press.] (b) Enthalpy change on mixing at 298.15 K for mixtures of benzene (C6 H6 ) and aromatic fluorocarbons (C6 F5 Y), with Y = H, F, Cl, Br, and I. [From D. V. Fenby and S. Ruenkrairergsa, J. Chem. Thermodyn., 5, 227 (1973). Reprinted with permission from Academic Press.]

we write

N θ = φ(N1 , N2 , . . . , NC ) at constant T and P (8.1-3) 3 to indicate that N θ is a function of all the mole numbers. If the number of moles of each species in the mixture under consideration were doubled, so that N1 → 2N1 , N2 → 2N2 , . . . , and N → 2N (thereby keeping the relative amounts of each species, and the relative numbers of each type of molecular interaction in the mixture, unchanged), the value of θ per mole of mixture, θ, would remain unchanged; that is, θ(T, P, N1 , N2 , . . .) = θ(T, P, 2N1 , 2N2 , . . .). Rewriting Eq. 8.1-3 for this situation gives 2N θ = φ(2N1 , 2N2 , . . . , 2NC )

(8.1-4)

while multiplying Eq. 8.1-3 by the factor 2 yields

2N θ = 2φ(N1 , N2 , . . . , NC ) so that

φ(2N1 , 2N2 , . . . , 2NC ) = 2φ(N1 , N2 , . . . , NC )

3 The

total property θ = N θ is a function of T , P , and the mole numbers; that is, N θ = N θ(T, P, x1 , x2 , . . .) = θ(T, P, N1 , N2 , . . .)

(8.1-5)

358 Chapter 8: The Thermodynamics of Multicomponent Mixtures This relation indicates that as long as the relative amounts of each species are kept constant, φ is a linear function of the total number of moles. This suggests φ should really be considered to be a function of the mole ratios. Thus, a more general way of indicating the concentration dependence of θ is as follows:   NC N2 N3 N θ = N1 φ1 , ,..., (8.1-6) N1 N1 N1 where we have introduced a new function, φ1 , of only C − 1 mole ratios in a Ccomponent mixture (at constant T and P ). In this equation the multiplicative factor N1 establishes the extent or total mass of the mixture, and the function φ1 is dependent only on the relative amounts of the various species. (Of course, any species other than species 1 could have been singled out in writing Eq. 8.1-6.) To determine how the property N θ varies with the number of moles of each species present, we look at the derivatives of N θ with respect to mole number. The change in the value of N θ as the number of moles of species 1 changes (all other mole numbers, T , and P being held constant) is   Ni   ∂ C   N1 ∂ ∂φ N2   1 (N θ) = φ1 , . . . + N1 (8.1-7) ∂N1 N1 ∂N1 Ni T,P,N2 ,N3 ,... i=2 ∂ N1 where the last term arises from the chain rule of partial differentiation and the fact that φ1 is a function of the mole ratios. Now multiplying by N1 , and using the fact that



Ni ∂ N1 ∂N1

   

i=1

=−

Ni N12

gives

⎫ ⎧   C  ⎨  ∂ 1 ∂φ1 ⎬ N2  N1 (N θ) = N1 φ1 ,... − Ni   ∂N1 N1 N1 ⎩ Ni ⎭ T,P,Nj=1 i=2 ∂ N1 (8.1-8a) C  ∂φ1 Ni   = Nθ − Ni i=2 ∂ N1 or

N θ = N1

C   ∂ ∂φ1  (N θ) + Ni   ∂N1 Ni T,P,Nj=1 i=2 ∂ N1

(8.1-8b)

Here we have introduced the notation Nj=1 in the partial derivative to indicate that all mole numbers except N1 are being held constant. Similarly, for the derivative with

8.1 The Thermodynamic Description of Mixtures 359 respect to Ni , i = 1, we have

∂(N θ)   ∂Ni T,P,Nj=i



Nj ∂ C  N1 ∂φ  1 = N1 ∂N Nj i j=2 ∂ N1

 =

∂φ  1 Ni ∂ N1

(8.1-9)

since (∂Nj /∂Ni ) is equal to 1 for i = j, and 0 for i = j. Using Eq. 8.1-9 in Eq. 8.1-8b yields C   ∂ ∂(N θ)   N θ = N1 (N θ) + Ni  ∂N1 ∂Ni T,P,Nj=i T,P,Nj=1 i=2

or

Nθ = Dividing by N =



C 

Ni

i=1

(8.1-10)

Ni gives θ=

C 

xi

i=1

Definition of a partial molar property

∂(N θ)   ∂Ni T,P,Nj=i

∂(N θ)   ∂Ni T,P,Nj=i

(8.1-11)

Common thermodynamic notation is to define a partial molar thermodynamic property θi as4

θi = θi (T, P, x) =

∂(N θ)   ∂Ni T,P,Nj=i

(8.1-12)

so that Eq. 8.1-11 can be written as Mixture property in terms of partial molar properties

θ=

C 

xi θi (T, P, x)

(8.1-13)

i=1

Although this equation is similar in form to the naive assumption of Eq. 8.1-1, there is the very important difference that θi , which appears here, is a true mixture property to be evaluated experimentally for each mixture (see Sec. 8.6), whereas θi , which appears in Eq. 8.1-1, is a pure component property. It should be emphasized that generally θi = θi ; that is, the partial molar and pure component thermodynamic properties are not equal! Some common partial molar thermodynamic properties are listed in Table 8.1-1. An important problem in the thermodynamics of mixtures is the measurement or estimation of these partial molar properties; this will be the focus of part of this chapter and all of Chapter 9.

4 Note

that the temperature and pressure (as well as certain mole numbers) are being held constant in the partial molar derivative. Later in this chapter we will be concerned with similar derivatives in which T and/or P are not held constant; such derivatives are not partial molar quantities!

360 Chapter 8: The Thermodynamics of Multicomponent Mixtures Table 8.1-1 Partial Molar Property 

Partial molar U i = internal

Molar Propert of the Mixture

=

energy Partial molar Vi = = volume Hi =

Si =

Gi =

Partial molar = enthalpy Partial molar = entropy Partial molar = Gibbs energy

energy









Partial molar Ai = Helmholtz



=



∂(N U ) ∂Ni

U =  V =

∂(N A) ∂Ni



xi V i

T,P,Nj=i

∂(N H) ∂Ni

∂(N G) ∂Ni

xi U i

T,P,Nj=i

∂(N V ) ∂Ni

∂(N S) ∂Ni



 H=



xi H i

T,P,Nj=i

 S=



xi S i

T,P,Nj=i

 G=



xi Gi

T,P,Nj=i

 A=



xi Ai

T,P,Nj=i

The fact that the partial molar properties of each species in a mixture are different from the pure-component molar properties gives rise to changes in thermodynamic properties during the process of forming a mixture from the pure components. For example, consider the formation of a mixture from N1 moles of species 1, N2 moles of species 2, and so on. The total volume and enthalpy of the unmixed pure components are

V =

C 

Ni V i (T, P )

i

and

H=

C 

Ni H i (T, P )

i

whereas the volume and enthalpy of the mixture at the same temperature and pressure are, from Eq. 8.1-13,

V (T, P, N1 , N2 , . . .) = N V (T, P, x) =

C 

Ni V i (T, P, x)

i

and

H(T, P, N1 , N2 , . . .) = N H(T, P, x) =

C  i

Ni H i (T, P, x)

8.1 The Thermodynamic Description of Mixtures 361 Therefore, the isothermal volume change on mixing, Δmix V , and the isothermal enthalpy change on mixing or the heat of mixing, Δmix H , are Volume change on mixing in terms of partial molar volumes

Δmix V (T, P, N1 , N2 , . . .) = V (T, P, N1 , N2 , . . .) − =

C 

C 

Ni V i (T, P )

i=1

(8.1-14)

Ni [V i (T, P, x) − V i (T, P )]

i=1

and Heat of mixing in terms of partial molar enthalpies

Δmix H(T, P, N1 , N2 , . . .) = H(T, P, N1 , N2 , . . .) − =

C 

C 

Ni H i (T, P )

i=1

Ni [H i (T, P, x) − H i (T, P )]

i=1

(8.1-15) Other thermodynamic property changes on mixing can be defined similarly. In Chapter 4 the Helmholtz and Gibbs energies of pure components were introduced by the relations A = U − TS (8.1-16a) G = H − TS (8.1-16b) These definitions are also valid for mixtures, provided the values of U , H , and S used are those for the mixture. That is, the Helmholtz and Gibbs energies of a mixture bear the same relation to the mixture internal energy, enthalpy, and entropy as the pure component Gibbs energies do to the pure component internal energy, enthalpy, and entropy. Equations like Eqs. 8.1-16a and b are also satisfied by the partial molar properties. To see this, we multiply Eq. 8.1-16a by the total number of moles N and take the derivative with respect to Ni at constant T , P , and Nj=i to obtain    ∂ ∂ ∂    (N A) = (N U ) −T (N S) ∂Ni ∂Ni ∂Ni T,P,Nj=i T,P,Nj=i T,P,Nj=i or

Ai = U i − T S i Similarly, starting with Eq. 8.1-16b, one can easily show that Gi = H i − T S i . The constant-pressure heat capacity of a mixture is given by   ∂H CP = ∂T P,Nj where H is the mixture enthalpy, and the subscript Nj indicates that the derivative is to be taken at a constant number of moles of all species present; the partial molar heat capacity for species i in a mixture is defined to be  ∂  C P,i = (N CP ) ∂Ni T,P,Nj=i

362 Chapter 8: The Thermodynamics of Multicomponent Mixtures Now

N CP =

∂(N H)   ∂T P,Nj

and

C P,i

 ∂ ∂(N CP )  ∂   = = (N H)   ∂Ni T,P,Nj=i ∂Ni T,P,Nj=i ∂T P,Nj   ∂  ∂(N H)  ∂H i = =   ∂T P,Ni ∂Ni T,P,Nj=i ∂T P,Nj

so that C P,i = (∂H i /∂T )P,Nj for species i in a mixture, just as CP,i = (∂H i /∂T )P,N for the pure species i. In a similar fashion a large collection of relations among the partial molar quantities can be developed. For example, since (∂Gi /∂T )P,N = −S i for a pure fluid, one can easily show that (∂Gi /∂T )P,Ni = −S i for a mixture. In fact, by extending this argument to other mixture properties, one finds that for each relationship among the thermodynamic variables in a pure fluid, there exists an identical relationship for the partial molar thermodynamic properties in a mixture!

Illustration 8.1-1 Calculating the Energy Release of an Exothermic Mixing Process Three moles of water and one mole of sulfuric acid are mixed isothermally at 0◦ C. How much heat must be absorbed or released to keep the mixture at 0◦ C?

Solution Water has a molecular weight of 18.015, and that of sulfuric acid is 98.078. Therefore, the mixture will contain 3 × 18.015 + 1 × 98.078 = 152.12 g, and will have a composition of 98.078 g × 100% = 64.5 wt % sulfuric acid 152.12 g

From Fig. 8.1-1 the enthalpy of the mixture is about −315 kJ/kg. Therefore, when 3 mol water and 1 mol sulfuric acid are mixed isothermally, ˆ =H ˆ mix − w1 H ˆ 1 − w2 H ˆ 2 = −315 kJ Δmix H kg ˆ 1 (T = 0◦ C) = 0 and H ˆ 2 (T = 0◦ C) = 0, so that a total of −315 kJ/kg × 0.152 kg = since H −47.9 kJ of energy must be removed to keep the mixture at a constant temperature of 0◦ C.

Comment Sulfuric acid and water are said to mix exothermically since energy must be released to the environment to mix these two components at constant temperature. The temperature rise that occurs when these two components are mixed adiabatically is considered in Illustration 8.4-1. Note also that to solve this problem we have, in effect, used an energy balance without explicitly writing a detailed balance equation. We will consider the balance equations (mass, energy, and entropy) for mixtures in Sec. 8.4.

8.2 The Partial Molar Gibbs Energy and the Generalized Gibbs-Duhem Equation 363 The meaning of a partial molar property may be understood as follows. For a property θ(T, P, N1 , N2 ) in a binary mixture, we have

θ(T, P, N1 , N2 ) = N1 θ1 (T, P, x) + N2 θ2 (T, P, x)

(8.1-17a)

Now suppose we add a small amount of species 1, ΔN1 , which is so small compared with the total number of moles of species 1 and 2 that x1 and x2 are essentially unchanged. In this case we have

θ(T, P, N1 + ΔN1 , N2 ) = (N1 + ΔN1 )θ1 (T, P, x) + N2 θ2 (T, P, x) (8.1-17b) Subtracting Eq. 8.1-17a from Eq. 8.1-17b, we find that the change in the property θ is

Δθ = θ(T, P, N1 + ΔN1 , N2 ) − θ(T, P, N1 , N2 ) = ΔN1 θ1 (T, P, x) Therefore, the amount by which a small addition of a species to a mixture changes the mixture property is equal to the product of the amount added and its partial molar property, that is, how the species behaves in a mixture, and not its pure component property.

8.2 THE PARTIAL MOLAR GIBBS ENERGY AND THE GENERALIZED GIBBS-DUHEM EQUATION Since the Gibbs energy of a multicomponent mixture is a function of temperature, pressure, and each species mole number, the total differential of the Gibbs energy function can be written as      C   ∂G ∂G ∂G dG = dT + dP + dNi ∂T P,Ni ∂P T,Ni ∂Ni T,P,N i=1 j=i (8.2-1) C  = −S dT + V dP + Gi dNi i=1

Chemical potential

Here the first two derivatives follow from Eqs. 6.2-12a, b and c for the pure fluid, and the last from the definition of the partial molar Gibbs energy. Historically, the partial molar Gibbs energy has been called the chemical potential and designated by the symbol μi . Since the enthalpy can be written as a function of entropy and pressure (see Eq. 6.2-6), we have      C   ∂H ∂H ∂H dH = dP + dS + dNi ∂P S,Ni ∂S P,Ni ∂Ni P,S,N i=1 j=i (8.2-2)  C   ∂H = V dP + T dS + dNi ∂Ni P,S,N i=1

j=i

In this equation it should be noted that (∂H/∂Ni )P,S,Nj=i is not equal to the partial molar enthalpy, which is H i = (∂H/∂Ni )T,P,Nj=i (see Problem 8.1).

364 Chapter 8: The Thermodynamics of Multicomponent Mixtures However, from H = G + T S and Eq. 8.2-1, we have5  dH = dG + T dS + S dT = −S dT + V dP + Gi dNi + T dS + S dT

= V dP + T dS +



i

(8.2-3)

Gi dNi

i

Comparing Eqs. 8.2-2 and 8.2-3 establishes that     ∂H ∂G = Gi = ∂Ni P,S,N ∂Ni T,P,N j=i

j=i

so that the derivative (∂H/∂Ni )P,S,Nj=i is equal to the partial molar Gibbs energy. Using the procedure established here, it is also easily shown (Problem 8.1) that

dU = T dS − P dV +

C 

Gi dNi

(8.2-4)

i

dA = −P dV − S dT +

C 

Gi dNi

(8.2-5)

i

with

 Gi =

∂U ∂Ni



 =

S,V,Nj=i

∂A ∂Ni

 T,V,Nj=i

From these equations we see that the partial molar Gibbs energy assumes special importance in mixtures, as the molar Gibbs energy does in pure fluids. Based on the discussion of partial molar properties in the previous section, any thermodynamic function θ can be written as  Nθ = Ni θi so that

d(N θ) =



Ni dθi +



θi dNi

(8.2-6)

by the product rule of differentiation. However, N θ can also be considered to be a function of T , P , and each of the mole numbers, in which case we have, following Eq. 8.2-1, C  ∂(N θ)  ∂(N θ)  ∂(N θ)  dT + dP + dNi    ∂T P,Ni ∂P T,Ni ∂Ni T,P,Nj=i i=1 (8.2-7)     C  ∂θ ∂θ =N dT + N dP + θi dNi ∂T P,Ni ∂P T,Ni

d(N θ) =

i=1

5 You

should compare Eqs. 8.2-3, 8.2-4, and 8.2-5 with Eqs. 6.2-6a, 6.2-5a, and 6.2-7a, respectively.

8.2 The Partial Molar Gibbs Energy and the Generalized Gibbs-Duhem Equation 365 Subtracting Eq. 8.2-7 from Eq. 8.2-6 gives the relation     C  ∂θ ∂θ −N dT − N dP + Ni dθi = 0 ∂T P,Ni ∂P T,Ni

(8.2-8a)

i=1

and dividing by the total number of moles N yields

 −

Generalized Gibbs-Duhem equation

∂θ ∂T



 dT − P,Ni

∂θ ∂P

 dP +

C 

T,Ni

xi dθi = 0

(8.2-8b)

i=1

These results are forms of the generalized Gibbs-Duhem equation. For changes at constant temperature and pressure, the Gibbs-Duhem equation reduces to C 

 Ni dθi T,P = 0

(8.2-9a)

xi dθi |T,P = 0

(8.2-9b)

i=1

and C 

Gibbs-Duhem equation at constant T and P

i=1

Finally, for a change in any property Y (except mole fraction6 ) at constant temperature and pressure, Eq. 8.2-9a can be rewritten as   C  ∂θi Ni =0 (8.2-10) ∂Y T,P i

whereas for a change in the number of moles of species j at constant temperature, pressure, and all other mole numbers, we have   C  ∂θi Ni =0 (8.2-11) ∂Nj T,P,N i

k=j

Since in much of the remainder of this book we are concerned with equilibrium at constant temperature and pressure, the Gibbs energy will be of central interest. The Gibbs-Duhem equations for the Gibbs energy, obtained by setting θ = G in Eqs. 8.2-8a and b, are C  0 = S dT − V dP + Ni dGi (8.2-12a) i=1

0 = S dT − V dP +

C 

xi dGi

(8.2-12b)

i=1 6 Since

mole fractions can be varied in several ways—for example, by varying the number of moles of only one species while holding all other mole numbers fixed, by varying the mole numbers of only two species, and so on, a more careful derivation than that given here must be used to obtain the mole fraction analogue of Eq. 8.2-10. The result of such an analysis is Eq. 8.2-18.

366 Chapter 8: The Thermodynamics of Multicomponent Mixtures and the relations analogous to Eqs. 8.2-9, 8.2-10, and 8.2-11 are C 

  Ni dGi 

i=1

C 

  xi dGi 

i=1

C 

 Ni

i=1

∂Gi ∂Y

=0

(8.2-13a)

=0

(8.2-13b)

T,P

T,P

 =0

(8.2-14)

T,P

and C 

 Ni

i=1

∂Gi ∂Nj

 =0

(8.2-15)

T,P,Nk=j

Each of these equations will be used later. The Gibbs-Duhem equation is a thermodynamic consistency relation that expresses the fact that among the set of C + 2 state variables, T , P , and C partial molar Gibbs energies in a C-component system, only C + 1 of these variables are independent. Thus, for example, although temperature, pressure, G1 , G2 , . . . , and GC−1 can be independently varied, GC cannot also be changed at will; instead, its change dGC is related to the changes dT, dP, dG1 , . . . , dGC−1 by Eq. 8.2-12a, so that

 C−1  1 dGC = −S dT + V dP − Ni dGi NC i=1

Thus, the interrelationships provided by Eqs. 8.2-8 through 8.2-15 are really restrictions on the mixture equation of state. As such, these equations are important in minimizing the amount of experimental data necessary in evaluating the thermodynamic properties of mixtures, in simplifying the description of multicomponent systems, and in testing the consistency of certain types of experimental data (see Chapter 10). Later in this chapter we show how the equations of change for mixtures and the Gibbs-Duhem equations provide a basis for the experimental determination of partial molar properties. Although Eqs. 8.2-8 through 8.2-11 are well suited for calculations in which temperature, pressure, and the partial molar properties are the independent variables, it is usually more convenient to have T , P , and the mole fractions xi as the independent variables. A change of variables is accomplished by realizing that for a C-component mixture there are only C − 1 independent mole fractions (since C i=1 xi = 1). Thus we write

 dθi =

∂θi ∂T



 dT +

P,x

∂θi ∂P

 dP + T,x

 C−1   ∂θi j=1

∂xj

T,P

dxj

(8.2-16)

8.3 A Notation for Chemical Reactions 367 where T, P, x1 , . . . , xC−1 have been chosen as the independent variables. Substituting this expansion in Eq. 8.2-8b gives     ∂θ ∂θ 0=− dT − dP ∂T P,Ni ∂P T,Ni       C C−1    ∂θi ∂θi ∂θi xi dT + dP + dxj + ∂T P,x ∂P T,x ∂xj T,P i=1 j=1 (8.2-17) Since C 

 xi

i=1

∂θi ∂T



∂  dT =  ∂T P,x P,x



C 

 xi θi

 dT =

i=1

∂θ ∂T

 dT P,x

and since holding all the mole numbers constant is equivalent to keeping all the mole fractions fixed, the first and third terms in Eq. 8.2-17 cancel. Similarly, the second and fourth terms cancel. Thus we are left with  C C−1    ∂θi xi dxj = 0 (8.2-18) ∂xj T,P i=1

j=1

which for a binary mixture reduces to 2 

 xi

i=1

or, equivalently

 x1

Gibbs-Duhem equations for binary mixture at constant T and P

∂θ1 ∂x1

∂θi ∂x1

 dx1 = 0

(8.2-19a)

T,P



 + x2

T,P

∂θ2 ∂x1

 =0

(8.2-19b)

T,P

For the special case in which θ is equal to the Gibbs energy, we have

 x1

∂G1 ∂x1



 + x2

T,P

∂G2 ∂x1

 =0

(8.2-20)

T,P

Finally, note that several different forms of Eqs. 8.2-19 and 8.2-20 can be obtained by using x2 = 1 − x1 and dx2 = −dx1 .

8.3 A NOTATION FOR CHEMICAL REACTIONS Since our interest in this chapter is with the equations of change and the equilibrium criteria for general thermodynamic systems, we need a convenient notation for the description of chemical reactions. Here we will generalize the notation for a

368 Chapter 8: The Thermodynamics of Multicomponent Mixtures chemical reaction introduced in Sec. 2.3. There we wrote the reaction

αA + βB + · · · −→ ←− ρR + · · · where α, β, . . . are the molar stoichiometric coefficients, as

ρR + · · · − αA − βB − · · · = 0 

or

νi I = 0

(8.3-1)

i

In this notation, from Chapter 2, νi is the stoichiometric coefficient of species I, so defined that νi is positive for reaction products, negative for reactants, and equal to zero for inert species. (You should remember that in this notation the reaction H2 + 12 O2 = H2 O is written as H2 O − H2 − 12 O2 = 0, so that νH2 O = +1, νH2 = −1, and νO2 = − 12 .) Using Ni to represent the number of moles of species i in a closed system at any time t , and Ni,0 for the initial number of moles of species i, then Ni and Ni,0 are related through the reaction variable X , the molar extent of reaction, and the stoichiometric coefficient νi by Ni = Ni,0 + νi X (8.3-2a) and Ni − Ni,0 X= (8.3-2b) νi The reason for introducing the reaction variable X is that it has the same value for each molecular species involved in a reaction, as was shown in Illustration 2.3-1. Thus, given the initial mole numbers of all species and X (or the number of moles of one species from which X can be calculated) at time t, one can easily compute all other mole numbers in the system. In this way the complete progress of a chemical reaction (i.e., the change in mole numbers of all the species involved in the reaction) is given by the value of the single variable X . Also, the rate of change of the number of moles of species i resulting from the chemical reaction is   dNi = νi X˙ (8.3-3) dt rxn where the subscript rxn indicates that this is the rate of change of species i attributable to chemical reaction alone, and X˙ is the rate of change of the molar extent of reaction.7 The total number of moles in a closed system at any time (i.e., for any extent of reaction) is

N=

C  i=1

C C C    Ni = (Ni,0 + νi X) = Ni,0 + X νi i=1

i=1

(8.3-4)

i=1

However, our interest can also extend to situations in which there are multiple chemical reactions, and the notation introduced in Sec. 2.3 and reviewed here needs to be extended to such cases. Before we do this, it is necessary to introduce the concept of independent chemical reactions. The term independent reactions is used here to 7 For

an open system the number of moles of species i may also change due to mass flows into and out of the system.

8.3 A Notation for Chemical Reactions 369 designate the smallest collection of reactions that, on forming various linear combinations, includes all possible chemical reactions among the species present. Since we have used the adjective smallest in defining a set of independent reactions, it follows that no reaction in the set can itself be a linear combination of the others. For example, one can write the following three reactions between carbon and oxygen:

C + O2 = CO2 2C + O2 = 2CO 2CO + O2 = 2CO2 If we add the second and third of these equations, we get twice the first, so these three reactions are not independent. In this case, any two of the three reactions form an independent set. As we will see in Chapter 13, it is not necessary to consider all the chemical reactions that can occur between the reactant species in order to describe a chemically reacting system, only the independent reactions. Furthermore, the molar extent of reaction for any chemical reaction among the species can be computed from an appropriate linear combination of the known extents of reaction for the set of independent chemical reactions (see Sec. 13.3). Thus, in the carbon-oxygen reaction system just considered, only two reaction variables (and the initial mole numbers) need be specified to completely fix the mole numbers of each species; the specification of a third reaction variable would be redundant and may result in confusion. To avoid unnecessary complications in the analysis of multiple reactions, we restrict the following discussion to sets of independent chemical reactions. Consequently, we need a method of identifying a set of independent reactions from a larger collection of reactions. When only a few reactions are involved, as in the foregoing example, this can be done by inspection. If many reactions occur, the methods of matrix algebra8 can be used to determine a set of independent reactions, though we will employ a simpler procedure developed by Denbigh.9 In the Denbigh procedure one first writes the stoichiometric equations for the formation, from its constituent atoms, of each of the molecular species present in the chemical reaction system. One of the equations that contains an atomic species not actually present in the atomic state at the reaction conditions is then used, by addition and/or subtraction, to eliminate that atomic species from the remaining equations. In this way the number of stoichiometric equations is reduced by one. The procedure is repeated until all atomic species not present have been eliminated. The equations that remain form a set of independent chemical reactions; the molar extents of reaction for these reactions are the variables to be used for the description of the multiple reaction system and to follow the composition changes in the mixture.10 As an example of this method, consider again the oxidation of carbon. We start by writing the following equations for the formation of each of the compounds:

2O = O2 C + O = CO C + 2O = CO2 8 N. R. Amundson, Mathematical Methods in Chemical Engineering, Prentice Hall, Englewood Cliffs, N.J. (1966), p. 50. 9 K. Denbigh, Principles of Chemical Equilibrium, 4th ed., Cambridge University Press, Cambridge (1981), pp. 169–172. 10 When considering chemical reactions involving isomers—for example, ortho-, meta-, and para-xylene—one proceeds as described here, treating each isomer as a separate chemical species.

370 Chapter 8: The Thermodynamics of Multicomponent Mixtures Since a free oxygen atom does not occur, the first equation is used to eliminate O from the other two equations to obtain

2C + O2 = 2CO C + O2 = CO2 All the remaining atomic species in these equations (here only carbon) are present in the reaction system, so no further reduction of the equations is possible, and these two equations form a set of independent reactions. Note that two other, different sets of independent reactions are obtained by using instead the second or third equation to eliminate the free oxygen atom. Though different sets of independent reactions are obtained, each would contain only two reactions, and either set could be used as the set of independent reactions. For the multiple-reaction case, Xj is defined to be the molar extent of reaction (or simply the reaction variable) for the jth independent reaction, and νij the stoichiometric coefficient for species i in the jth reaction. The number of moles of species i present at any time (in a closed system) is

Ni = Ni,0 +

M 

νij Xj

(8.3-5)

j=1

where the summation is over the set of M independent chemical reactions. The instantaneous rate of change of the number of moles of species i due only to chemical reaction is   M  dNi = νij X˙ j (8.3-6) dt rxn j=1

ˆ j = Xj /V defined to be the molar extent of reaction per unit volume, Finally, with X Eq. 8.3-6 can be written as 

dNi dt

 = rxn

M  j=1

νij

M M   d ˆ j ) = dV ˆj + V νij X νij rj (V X dt dt j=1

(8.3-7)

j=1

ˆ j /dt, the rate of reaction per unit volume, is the reaction variable most where rj = dX frequently used by chemists and chemical engineers in chemical reactor analysis.

8.4 THE EQUATIONS OF CHANGE FOR A MULTICOMPONENT SYSTEM The next step in the development of the thermodynamics of multicomponent systems is the formulation of the equations of change. These equations can, in completely general form, be considerably more complicated than the analogous pure component equations since (1) the mass or number of moles of each species may not be conserved due to chemical reactions, and (2) the diffusion of one species relative to the others may occur if concentration gradients are present. Furthermore, there is the computational difficulty that each thermodynamic property depends, in a complicated fashion, on the temperature, pressure, and composition of the mixture. To simplify the development of the equations, we will neglect all diffusional processes, since diffusion has very little effect on the thermodynamic state of the system. (This assumption is equivalent to setting the average velocity of each species equal to the mass average velocity of the fluid.) In Chapter 2 the kinetic and potential energy

8.4 The Equations of Change for a Multicomponent System 371 terms were found to make a small contribution to the pure component energy balance; the relative importance of these terms to the energy balance for a reacting mixture is even less, due to the large energy changes that accompany chemical transformations. Therefore, we will neglect the potential and kinetic energy terms in the formulation of the mixture energy balance. This omission will cause serious errors only in the analysis of rocket engines and similar high-speed devices involving chemical reaction. With these assumptions, the formulation of the equations of change for a multicomponent reacting mixture is not nearly so formidable a task as it might first appear. In fact, merely by making the proper identifications, the equations of change for a mixture can be written as simple generalizations of the equations of change for a single-component system. The starting point is Eq. 2.1-4, which is rewritten as ⎛ ⎞   Rate at which θ enters dθ Rate of change of = ⎝ the system across ⎠ = θ in the system dt system boundaries ⎛ ⎞ Rate at which θ leaves − ⎝ the system across ⎠ (2.1-4) system boundaries ⎞ ⎛ Rate at which θ is + ⎝ generated within ⎠ the system The balance equation for species i is obtained by setting θ equal to Ni , the number of moles of species i; letting (N˙ i )k equal the molar flow rate of species i into the system at the k th port; and recognizing that species i may be generated within the system by chemical reaction;

  dNi  ˙ Rate at which species i is being = (Ni )k + produced by chemical reaction dt k=1   K  dNi ˙ (Ni )k + = dt rxn K

Species mass balance for a reacting system

=

k=1 K 

(N˙ i )k +

M 

(8.4-1a)

νij X˙ j

j=1

k=1

We can obtain a balance equation on the total number  of moles in the system by summing Eq. 8.4-1a over all species i, recognizing that Ci=1 Ni = N is the total number of moles, and that C  K 

where N˙ k = Total mass balance for a reacting system

C

(N˙ i )k =

i=1 k=1

˙

i=1 (Ni )k

K  C 

(N˙ i )k =

k=1 i=1

K 

N˙ k

k=1

is the total molar flow rate at the k th entry port, so that C

M

  dN νij X˙ j = N˙ k + dt K

k=1

i=1 j=1

(8.4-1b)

372 Chapter 8: The Thermodynamics of Multicomponent Mixtures Since neither the number of moles of species i nor the total number of moles is a conserved quantity, we need information on the rates at which all chemical reactions occur to use Eqs. 8.4-1a and b. That is, we need detailed information on the reaction processes internal to the black-box system (unless, of course, no reactions occur, in which case X˙ j = 0 for all j). Although we will be interested in the equations of change mainly on a molar basis, for completeness and for several illustrations that follow, some of the equations of change will also be given on a mass basis. To obtain a balance equation for the mass of species i, we need only multiply Eq. 8.4-1a by the molecular weight of species i, mi , and use ˙ i )k = mi (N˙ i )k to get the notation Mi = mi Ni and (M K

M

k=1

j=1

 dMi  ˙ = (Mi )k + mi νij X˙ j dt

(8.4-2a)

Also, summing the equation over species i, we get the overall mass balance equation

 dM = M˙ k dt K

(8.4-2b)

k=1

˙ k = C (M˙ i )k is the total mass flow at the k th entry port. In Eq. 8.4-2b the where M i=1 chemical reaction term vanishes since total mass is a conserved quantity (Problem 8.5). If θ in Eq. 2.1-4 is now taken to be the total energy of the system (really only the internal energy, since we are neglecting the kinetic and potential energy terms), and the same energy transfer mechanisms identified in Sec. 3.1 are used here, we obtain  dU ˙ s − P dV (N˙ H)k + Q˙ + W = dt dt K

(molar basis)

(8.4-3)

k=1

and Energy balance for a reacting system

 dU ˆ k + Q˙ + W ˙ s − P dV (M˙ H) = dt dt K

(mass basis)

k=1

Here (N˙ H)k is the product of the molar flow rate and the molar enthalpy of the fluid, ˙ H) ˆ k the product of mass flow rate and enthalpy per unit mass, both at the k th and (M entry port. Since some or all of the flow streams may be mixtures, to evaluate a term such as (N˙ H)k the relation

(N˙ H)k =

C 

(N˙ i H i )k

(8.4-4)

i=1

must be used, where (N˙ i )k is the molar flow rate of species i in the k th flow stream, and (H i )k is its partial molar enthalpy. Similarly, the internal energy of the system must be obtained from C  U= Ni U i i=1

where Ni is the number of moles of species i and U i is its partial molar internal energy. At this point you might be concerned that the heat of reaction (i.e., the energy released or absorbed on reaction since the total energy of the chemical bonds in the reactant

8.4 The Equations of Change for a Multicomponent System 373 and product molecules are not equal) does not explicitly appear in the energy balance equation. Since the only assumptions made in deriving this equation were to neglect the potential and kinetic energy terms and diffusion, neither of which involves chemical reaction, the heat of reaction is in fact contained in the energy balance, though it appears implicitly rather than explicitly. Although energy balances on chemical reactors are studied in detail in Chapter 14, to demonstrate that the heat of reaction is contained in Eq. 8.4-3, we will consider its application to the well-stirred, steady-state chemical reactor in Fig. 8.4-1. Here by steady state we mean that the rate at which each species i leaves the reactor just balances the rate at which species i enters the reactor and is produced within it, so that Ni does not vary with time, that is,

dNi (8.4-5a) = 0 = (N˙ i )in + (N˙ i )out + νi X˙ dt and, further, that the rates of flow of energy into and out of the reactor just balance, so that the internal energy of the contents of the reactor does not change with time C

C

i=1

i=1

  dU (N˙ i H i )in + (N˙ i H i )out + Q˙ =0= dt

Energy balance for a continuous-flow stirred-tank reactor

(8.4-5b)

For the very simple isothermal case in which the inlet and outlet streams and the reactor contents are all at temperature T , and with the assumption that the partial molar enthalpy of each species is just equal to its pure-component enthalpy, we obtain

Q˙ = −

C 

[(N˙ i )out + (N˙ i )in ]H i

i=1

Now if there were no chemical reaction (N˙ i )out = −(N˙ i )in and the heat flow rate Q˙ should be equal to zero to maintain the constant temperature T . However, when a chemical reaction occurs, (N˙ i )out and (N˙ i )in are not equal in magnitude, and the steady heat flow Q˙ required to keep the reactor at constant temperature is

Q˙ = −

C C   [(N˙ i )out + (N˙ i )in ]H i = X˙ νi H i i=1

= X˙

C 

i=1

νi H i = Δrxn H X˙

i=1

Reactor inlet stream

Stirrer

Heating and cooling coil

Reactor outlet stream

Figure 8.4-1 A simple stirred-tank reactor.

374 Chapter 8: The Thermodynamics of Multicomponent Mixtures

 so that the heat flow rate is equal to the product of Δrxn H = νi H i , the isothermal ˙ heat of reaction, and X , the reaction rate. Thus, we see that the heat of reaction is implicitly contained in the energy balance equation through the difference in species mole numbers in the inlet and outlet flow streams. For nonideal mixtures (i.e., mixtures for which H i = H i ) and nonisothermal reactors, the expression for the heat of reaction is buried somewhat deeper in the energy balance equation; this is discussed in Chapter 14. To derive the macroscopic entropy balance equation for mixtures we now set θ = S in Eq. 2.1-4 and make the same identification for the entropy flow terms as was made in Sec. 4.1, to obtain K dS  ˙ Q˙ (N S)k + + S˙ gen = dt T

(8.4-6)

k=1

where

(N˙ S)k =

C 

(N˙ i S i )k

i=1

Thus the entropy balance, like the energy balance equation, is of the same form for the pure fluid and for mixtures. As the final step in the development of the entropy balance, an expression for the entropy generation term should be obtained here, as was done for the pure fluid in Sec. 4.6. The derivation of such an expression is tedious and would require us to develop detailed microscopic equations of change for a mixture. Rather than doing this, we will merely write down the final result, referring the reader to the book by de Groot and Mazur for the details.11 Their result, with the slight modification of writing the chemical reaction term on a molar rather than a mass basis, is  ˙ Sgen = σ˙ s dV where

M C λ μ 2 1  2 σ˙ s = 2 (∇T ) + φ − νij Gi X˙ j T T T

(8.4-7)

j=1 i=1

(see Secs. 3.6 and 4.6 for the definitions of λ, φ, and the gradient operator ∇). The last term, which is new, represents the contribution of chemical reactions to the entropy generation rate. One can establish that this term makes a positive contribution to the entropy generation term. It is of interest to note that had diffusional processes also been considered, there would be an additional contribution to S˙ gen due to diffusion. That term would be in the form of a diffusional flux times a driving force for diffusion and would also be greater than or equal to zero. In Chapter 4 we defined a reversible process to be a process for which S˙ gen = 0. This led to the conclusion that in a reversible process in a one-component system only infinitesimal temperature and velocity gradients could be present. Clearly, on the basis of Eq. 8.4-7, a reversible process in a multicomponent system is one in which only

11 S. R. de Groot and P. Mazur, Non-equilibrium Thermodynamics, North-Holland, Amsterdam (1962), Chapter 3.

8.4 The Equations of Change for a Multicomponent System 375 infinitesimal gradients in temperature, velocity, and concentration may be present, and in which all chemical reactions proceed at only infinitesimal rates. The differential form of the multicomponent mass, energy, and entropy balances can be integrated over time to get difference forms of the balance equations, as was done in Chapters 2, 3, and 4 for the pure-component equations. The differential and difference forms of the balance equations are listed in Tables 8.4-1 and 8.4-2, respectively. It is left to the reader to work out the various simplifications of these equations that arise for special cases of closed systems, adiabatic processes, and so forth. With the equations of change for mixtures, and given mixture thermodynamic data, such as the enthalpy data for sulfuric acid–water mixtures in Fig. 8.1-1, it is possible to solve many thermodynamic energy flow problems for mixtures. One example is given in the following illustration. Table 8.4-1 The Differential Form of the Equations of Change for a Multicomponent System Species balance M

  dNi (N˙ i )k + νij X˙ j = dt j=1 k=1 K

(molar basis)

M

  dMi (M˙ i )k + mi νij X˙ j = dt j=1 k=1 K

Overall mass balance

C

(mass basis)

M

  dN N˙ k + = νij X˙ j dt i=1 j=1 k=1 K

(molar basis)

 dM M˙ k = dt k=1 K

(mass basis)

Energy balance  dU ˙ s − P dV (N˙ H)k + Q˙ + W = dt dt k=1

(molar basis)

 dU ˆ k + Q˙ + W ˙ s − P dV (M˙ H) = dt dt k=1

(mass basis)

K

K

Entropy balance K  Q˙ dS = + S˙ gen (N˙ S)k + dt T k=1

(molar basis)

K ˙  dS ˆ k + Q + S˙ gen (M˙ S) = dt T k=1

(mass basis)

where N˙ k =

C  (N˙ i )k i=1

(N˙ H)k =

C  (N˙ i H i )k i=1

and M˙ k =

C  (M˙ i )k i=1

(N˙ S)k =

C  (N˙ i S i )k i=1

376 Chapter 8: The Thermodynamics of Multicomponent Mixtures Table 8.4-2 The Difference Form of the Equations of Change for a Multicomponent Mixture Species balance Ni (t2 ) − Ni (t1 ) =

K   k=1

Mi (t2 ) − Mi (t1 ) =

K   k=1

Overall mass balance N (t2 ) − N (t1 ) =

K   k=1

M (t2 ) − M (t1 ) =

K   k=1

Energy balance U (t2 ) − U (t1 ) =

K   k=1

U (t2 ) − U (t1 ) =

K   k=1

Entropy balance S(t2 ) − S(t1 ) =

K   k=1

S(t2 ) − S(t1 ) =

K   k=1

t2

(N˙ i )k dt +

t1 t2

M 

˙ i )k dt + (M

M 

t1 t2

νij mi Xj

j=1

N˙ k dt +

t1 t2

νij Xj

j=1

C  M 

νij Xj

i=1 j=1

˙ k dt M

t1 t2

 (N˙ H)k dt + Q + Ws −

P dV

t1 t2

 ˙ H) ˆ k dt + Q + Ws − (M

P dV

t1 t2



t1 t2

t2

(N˙ S)k dt +

t1

 ˙ S) ˆ k dt + (M

t1

t2 t1

Q˙ dt + Sgen T Q˙ dt + Sgen T

where (N˙ )k , (N˙ H)k , (N˙ S)k , and (M˙ )k are defined in Table 8.4-1

Illustration 8.4-1 Temperature Change on Adiabatic Mixing of an Acid and Water Three moles of water and one mole of sulfuric acid, each at 0◦ C, are mixed adiabatically. Use the data in Fig. 8.1-1 and the information in Illustration 8.1-1 to estimate the final temperature of the mixture.

Solution From the closed-system mass balance, we have M f = MH2 O + MH2 SO4 = 3 × 18.015 + 1 × 98.078 = 152.12 g

and from the energy balance, we have ˆ mix = MH O H ˆ H O + MH SO H ˆ H SO = 3 × 0 + 1 × 0 = 0 kJ U f = Hf = Hi = M f H 2 2 2 4 2 4

Thus finally we have a mixture of 64.5 wt % sulfuric acid that has an enthalpy of 0 kJ/kg. (Note that we have used the fact that for liquids and solids at low pressure, the internal energy and enthalpy are essentially equal.) From Fig. 8.1-1 we see that a mixture containing 64.5 wt % sulfuric acid has an enthalpy of 0 kJ/kg at about 150◦ C. Therefore, if water and sulfuric acid are adiabatically mixed in the ratio of 3:1, the mixture will achieve a temperature of 150◦ C, which

8.4 The Equations of Change for a Multicomponent System 377 is just below the boiling point of the mixture. This large temperature rise, and the fact that the mixture is just below its boiling point, makes mixing sulfuric acid and water an operation that must be done with extreme care.

Comment If instead of starting with refrigerated sulfuric acid and water (at 0◦ C), one started with these components at 21.2◦ C and mixed them adiabatically, the resulting 3:1 mixture would be in the liquid + vapor region; that is, the mixture would boil (and splatter). Also note that because of the shape of the curves on the enthalpy-concentration diagram, adding sulfuric acid to water adiabatically (i.e., moving to the right from the pure water edge of the diagram) results in a more gradual temperature rise than adding water to sulfuric acid (i.e., moving to the left from the pure– sulfuric acid edge). Therefore, whenever possible, sulfuric acid should be added to water, and not vice versa.

Illustration 8.4-2 Mass and Energy Balances on a Nonreacting System A continuous-flow steam-heated mixing kettle will be used to produce a 20 wt % sulfuric acid solution at 65.6◦ C from a solution of 90 wt % sulfuric acid at 0◦ C and pure water at 21.1◦ C. Estimate 90 wt % sulfuric acid, T = 0°C

Water T = 21.1°C

Heating coil

20 wt % sulfuric acid, T = 65.6°C

a. The kilograms of pure water needed per kilogram of initial sulfuric acid solution to produce a mixture of the desired concentration b. The amount of heat needed per kilogram of initial sulfuric acid solution to heat the mixture to the desired temperature c. The temperature of the kettle effluent if the mixing process is carried out adiabatically

Solution We choose the contents of the mixing kettle as the system. The difference form of the equations of change will be used for a time interval in which 1 kg of concentrated sulfuric acid enters the kettle. a. Since there is no chemical reaction, and the mixing tank operates continuously, the total and sulfuric acid mass balances reduce to 0=

3  k=1

˙k M

and

3  (MH2 SO4 )k = 0 k=1

Denoting the 90 wt % acid stream by the subscript 1 and its mass flow by M˙ 1 , the water stream by the subscript 2, and the dilute acid stream by subscript 3, we have, from the total mass balance, ˙1 +M ˙ 2 + M˙ 3 = M˙ 1 + Z M ˙1 +M ˙3 0=M

378 Chapter 8: The Thermodynamics of Multicomponent Mixtures or ˙ 3 = −(1 + Z)M ˙1 M

where Z is equal to the number of kilograms of water used per kilogram of the 90 wt % acid. Also, from the mass balance on sulfuric acid, we have ˙ 2 + 0.20M˙ 3 = 0.90M˙ 1 − 0.20(1 + Z)M ˙1 0 = 0.90M˙ 1 + 0M

Therefore, 1+Z =

0.90 = 4.5 0.20

and Z = 3.5

so that 3.5 kg of water must be added to each 1 kg of 90 wt % acid solution to produce a 20 wt % solution. b. The steady-state energy balance is 0=

 ˙ H) ˆ k + Q˙ (M k



since Ws = 0 and P dV = 0. From the mass balance of part (a), ˙ 2 = 3.5M ˙1 M ˙ M3 = −4.5M˙ 1

From the enthalpy-concentration chart, Fig. 8.1-1, we have ˆ 1 = H(90 ˆ H wt % H2 SO4 , T = 0◦ C) = −183 kJ/kg ˆ 2 = H(pure ˆ H H2 O, T = 21.1◦ C) = 91 kJ/kg ˆ 3 = H(20 ˆ H wt % H2 SO4 , T = 65.56◦ C) = 87 kJ/kg

so that Q = (4.5 × 87 − 3.5 × 91 − 1 × (−183)) = (391.5 − 318.5 + 183) = 256 kJ/kg of initial acid solution

c. For adiabatic operation, the energy balance is 0=

 ˆ k (M˙ H) k

or ˆ 3 − 3.5(91) − 1(−183) 0 = 4.5H ˆ 3 = 135.5 4.5H

ˆ 3 = 30.1 kJ/kg and H

Referring to the enthalpy-concentration diagram, we find that T ∼ 50◦ C.

8.5 THE HEAT OF REACTION AND A CONVENTION FOR THE THERMODYNAMIC PROPERTIES OF REACTING MIXTURES In the first part of this book we were interested in the change in the internal energy, enthalpy, and entropy accompanying a change in the thermodynamic state of a pure substance. For convenience in such calculations, one state of each substance was chosen as the reference state, with zero values for both enthalpy and entropy, and the values of the thermodynamic properties of the substance in other states were given relative to

8.5 The Heat of Reaction and a Convention for the Thermodynamic Properties of Reacting Mixtures 379 this reference state. Thus in the steam tables in Appendix A.III the internal energy and entropy are zero for liquid water at the triple point, whereas the zero values of these thermodynamic properties for methane (Fig. 2.4-2) and nitrogen (Fig. 2.4-3) correspond to different conditions of temperature and pressure. The reference states for sulfuric acid and water in the enthalpy-concentration diagram of Fig. 8.1-1 were also chosen on the basis of convenience. Such arbitrary choices for the reference state were satisfactory because our interest was only with the changes in thermodynamic properties in going from one thermodynamic state to another in a nonreacting system. However, when chemical reactions occur, the reference state for the thermodynamic properties of each species must be chosen with great care. In particular, the thermodynamic properties for each species must be such that the differences between the reactant and product species in any chemical transformation will be equal to those that would be measured in the appropriate experiment. For example, in the ideal gas-phase reaction

H2 + 12 O2 → H2 O it is observed that 241.82 kJ are liberated for each mole of water vapor produced when this reaction is run in an isothermal, constant-pressure calorimeter at 25◦ C and 1 bar with all species in the vapor phase. Clearly, then, the enthalpies of the reacting molecules must be related as follows:

Δrxn H(T = 25◦ C, P = 1 bar) = H H2 O (vapor, T = 25◦ C, P = 1 bar) − H H2 (gas, T = 25◦ C, P = 1 bar) − 12 H O2 (gas, T = 25◦ C, P = 1 bar)

= −241.82

kJ mol H2 O produced

so that we are not free to choose the values of the enthalpy of hydrogen, oxygen, and water vapor all arbitrarily. A similar argument applies to the other thermodynamic properties such as the entropy and Gibbs free energy. By considering a large collection of chemical reactions, one finds that, to be consistent with the heat-of-reaction data for all possible reactions, the zero value of internal energy or enthalpy can be set only once for each element. Thus, one could set the enthalpy of each element to be zero at 25◦ C and 1 bar (or some other reference state) and then determine the enthalpy of every compound by measuring the heat liberated or absorbed during its production, by isothermal chemical reaction, from its elements. Although such a procedure is a conceptually pleasing method of devising a thermodynamic enthalpy scale, it is experimentally difficult. For example, to determine the enthalpy of water vapor, one would have to measure the heat liberated on its formation from hydrogen and oxygen atoms. However, since hydrogen and oxygen molecules, and not their atoms, are the thermodynamically stable species at 25◦ C, one would first have to dissociate the oxygen and hydrogen molecules into their constituent atoms and then accurately measure the heat evolved when the atoms combine to form a water molecule—a very difficult task. Instead, the thermodynamic energy scale that is most frequently used is based on choosing as the zero state of both enthalpy and Gibbs energy12 for each atomic species its simplest thermodynamically stable state at 25◦ C 12 For chemical equilibrium the Gibbs energy, rather than the entropy, is of central importance. Therefore, generally

the enthalpy and the Gibbs energy are set equal to zero in the reference state.

380 Chapter 8: The Thermodynamics of Multicomponent Mixtures and 1 bar. Thus, the reference state for argon and helium is the atomic gases; for oxygen, nitrogen, and hydrogen, it is the molecular gases; for mercury and bromine, it is the atomic and molecular liquids, respectively; whereas for iron and carbon, it is as the α-crystalline and graphite solids, respectively, all at 25◦ C and 1 bar. Starting from this basis, and using the measured data from a large number of heats of reaction, heats of mixing, and chemical equilibrium measurements, the enthalpy and Gibbs energy content of all other molecular species relative to their constituent atoms in their reference states can be determined; these quantities are called the enthalpy of formation, Δf H ◦ and the Gibbs energy of formation, Δf G◦ . Thus, by definition,

Δf H ◦H2 O (vapor, 25◦ C, 1 bar) = H H2 O (vapor, 25◦ C, 1 bar) − H H2 (gas, 25◦ C, 1 bar) − 12 H O2 (gas, 25◦ C, 1 bar) Appendix A.IV contains a listing of Δf H ◦ and Δf G◦ for a large collection of substances in their normal states of aggregation at 25◦ C and 1 bar. Isothermal heats (enthalpies) and Gibbs energies of formation of species may be summed to compute the enthalpy change and Gibbs free energy change that would occur if the molecular species at 25◦ C, 1 bar, and the state of aggregation listed in Appendix A.IV reacted to form products at 25◦ C, 1 bar, and their listed state of aggregation. We will denote these changes by Δrxn H ◦ (25◦ C, 1 bar) and Δrxn G◦ (25◦ C, 1 bar), respectively. For example, for the gas-phase reaction

3NO2 + H2 O = 2HNO3 + NO we have

Δrxn H ◦ (25◦ C, 1 bar) = 2H HNO3 (25◦ C, 1 bar) + H NO (25◦ C, 1 bar) − 3H NO2 (25◦ C, 1 bar) − H H2 O (25◦ C, 1bar)

= 2[Δf H ◦HNO3 + 32 H O2 + 12 H H2 + 12 H N2 ]25◦ C,1bar + [Δf H ◦NO + 12 H O2 + 12 H N2 ]25◦ C,1bar − 3[Δf H ◦NO2 + H O2 + 12 H N2 ]25◦ C,1bar − [Δf H ◦H2 O + H H2 + 12 H O2 ]25◦ C,1bar

= [2Δf H ◦HNO3 + Δf H ◦NO − 3Δf H ◦NO2 − Δf H ◦H2 O ]25◦ C,1bar  = νi Δf H ◦i (25◦ C, 1 bar) Note that the enthalpies of the reference state atomic species cancel; by the conservation of atomic species, this will always occur. Thus, we have, as general results, Standard state heat of reaction

Δrxn H ◦ (25◦ C, 1 bar) =

 i

νi Δf H ◦i (25◦ C, 1 bar)

(8.5-1)

8.5 The Heat of Reaction and a Convention for the Thermodynamic Properties of Reacting Mixtures 381 and

Δrxn G◦ (25◦ C, 1 bar) =



νi Δf G◦i (25◦ C, 1 bar)

(8.5-2)

i

Standard state Gibbs energy change on reaction

We use the term standard state of a substance to indicate the state of aggregation given in Appendix A.IV (or, in certain cases, the reference states discussed in Chapter 13), at the temperature T and pressure of 1 bar, with the following qualifications. For a gas here (as in Chapter 6) we use the ideal gas at 1 bar, not the real gas. As was mentioned in Chapter 6, the P = 0 state cannot be used since the entropy would then be positive infinity and the Gibbs energy would be negative infinity. For substances that are typically solutes—sodium chloride, for example—properties in two or more standard states are given. One standard state is as a solid (in the crystalline state) and the other is in solution. The latter case is a bit complicated in that the enthalpy of formation may be given at a specified concentration whereas the Gibbs energy formation is that in a (hypothetical) ideal 1-molal solution (see Sec. 9.8). The pure liquid cannot be used as the standard state since the solute does not exist as a liquid at these conditions, and the very high (infinite) dilution state cannot be used since the Gibbs energy of the solute goes to negative infinity at infinite dilution. What is used, instead, is a hypothetical ideal state of finite concentration, here 1 molal, with properties obtained from the extrapolation of data obtained for the solute highly diluted in aqueous solution, as will be discussed in Chapter 9. The standard heat of reaction at any temperature T , Δrxn H ◦ (T ), is defined as the change in enthalpy that results when a stoichiometric amount of the reactants, each in their standard states at the temperature T , chemically react to form the reaction products in their standard states at the temperature T . Thus, in analogy with Eqs. 8.5-1 and 8.5-2,  Δrxn H ◦ (T, P = 1 bar) = νi Δf H ◦i (T, P = 1 bar) (8.5-3) i

and

Δrxn G◦ (T, P = 1 bar) =



νi Δf G◦i (T, P = 1 bar)

(8.5-4)

i

where the subscript i indicates the species and the superscript ◦ denotes the standard state. From  T ◦ ◦ ◦ H i (T, P = 1 bar) = H i (T0 , P = 1 bar) + CP,i (T, P = 1 bar) dT T0

we have ◦

Δrxn H (T, 1 bar) =



◦

◦

νi Δf H i (25 C, 1 bar) +

i

i

◦

◦

= Δrxn H (25 C, 1 bar) +

 i





νi

νi



T

T =25◦ C

T

T =25◦ C

◦ is the heat capacity of species i in its standard state. where CP,i

◦ CP,i dT

(8.5-5) ◦

CP,i dT

382 Chapter 8: The Thermodynamics of Multicomponent Mixtures In certain instances, such as in the study of large organic compounds that require a complicated synthesis procedure or of biochemical molecules, it is not possible to measure the heat of formation directly. A substitute procedure in these cases is to measure the energy change for some other reaction, usually the heat of combustion, that is, the energy liberated when the compound is completely oxidized (all the carbon is oxidized to carbon dioxide, all the hydrogen to water, etc.). The standard heat of combustion Δc H ◦ (T ) is defined to be the heat of combustion with both the reactants and the products in their standard states, at the temperature T . The standard heat of combustion at 25◦ C and 1 bar is listed in Appendix A.V for a number of compounds. (Note that there are two entries in this table, one corresponding to the liquid phase and the other to the vapor phase, being the standard state for water.) Given the standard heat of combustion, the standard heat of reaction is computed as indicated in Illustration 8.5-1. Illustration 8.5-1 Calculation of the Standard Heat of Reaction at 25◦ C Compute the standard heat of reaction for the hydrogenation of benzene to cyclohexane, C6 H6 + 3H2 → C6 H12

from the standard-heat-of-combustion data.

Solution The standard heat of reaction can, in principle, be computed from Eq. 8.5-3; however, for illustration, we will use the heat of combustion for cyclohexane. From the standard heat-of-combustion data in Appendix A.V, we have Δc H ◦ = −3 919 906 J/mol of cyclohexane for the following reaction: C6 H12 (l) + 9O2 → 6CO2 + 6H2 O(l)

Thus Δc H ◦C6 H12 = 6H CO2 + 6H H2 O − H C6 H12 − 9H O2 = −3 919 906 J/mol

or H C6 H12 = −Δc H ◦C6 H12 − 9H O2 + 6H CO2 + 6H H2 O

Similarly, H C6 H6 = −Δc H ◦C6 H6 − 7 12 H O2 + 6H CO2 + 3H H2 O

and 3H H2 = −3Δc H ◦H2 − 1 12 H O2 + 3H H2 O

Therefore, Δrxn H ◦ = −Δc H ◦C6 H12 − 9H O2 + 6H CO2 + 6H H2 O + Δc H ◦C6 H6

+ 7 12 H O2 − 6H CO2 − 3H H2 O + 3Δc H ◦H2 + 1 21 H O2 − 3H H2 O  = −Δc H ◦C6 H12 + Δc H ◦C6 H6 + 3Δc H ◦H2 = − νi Δc H ◦i i

= 3 919 906 − 3 267 620 − 3 × 285 840 = −205 234 = −205.23

kJ mol benzene

J mol benzene

8.5 The Heat of Reaction and a Convention for the Thermodynamic Properties of Reacting Mixtures 383 Comment



Note that in the final equation, Δrxn H ◦ = − νi Δc H ◦i , the enthalpies of the reference-state atomic species cancel, as they must due to conservation of atomic species on chemical reaction. R Aspen Plus can be used to solve this illustration as shown in Wiley website for this book in the folder Aspen Illustrations>Chapter 8>8.5-1. The calculation is done there using the RStoic reactor block in the Simulation mode with the ideal gas model. Using 1 kmol/hr flow rate of benzene with an isothermal 25◦ C reactor. Looking the Results Summary>Model gives in a heat duty of −57272.2 J S 1 hr kmol 1 kJ × 3600 × ×1 3 × 3 S hr 1 kmol benzene 10 mol 10 mol kJ = −206.18 mol benzene

Watts = −57272.2

This is in agreement with the result above.

The equation developed in this illustration,  Δrxn H ◦ (T = 25◦ C, 1 bar) = − νi Δc H ◦i (T = 25◦ C, P = 1 bar) i

is always valid and provides a way of computing the standard heat of reaction from standard-heat-of-combustion data. Illustration 8.5-2 Calculation of the Standard Heat of Reaction as a Function of Temperature Compute the standard-state heat of reaction for the gas-phase reaction N2 O4 = 2NO2 over the temperature range of 200 to 600 K.13 Data: See Appendices A.II and A.IV.

Solution The heat of reaction at a temperature T can be computed from Δrxn H ◦ (T ) =



νi Δf H ◦i (T )

i

◦

At T = 25 C we find, from the data in Appendix A.IV, that for each mole of N2 O4 reacted, Δrxn H ◦ (T = 25◦ C) = [2 × 33.18 − 9.16] kJ/mol = 57.20 kJ/mol

To compute the heat of reaction at any temperature T, we start from Eq. 8.5-5 and note that since ◦ ∗ = CP,i . Therefore, the standard state for each species is a low-pressure gas, CP,i Δrxn H ◦ (T ) = Δrxn H ◦ (T = 25◦ C) +



T T =298.2K



∗ νi CP,i dT

i

13 Since, as we will see, H = H for an ideal gas mixture, the standard state heat of reaction and the actual heat i i of reaction are identical in this case.

384 Chapter 8: The Thermodynamics of Multicomponent Mixtures For the case here we have, from Appendix A.II, 

∗ ∗ ∗ = 2CP,NO − CP,N νi CP,i 2 2 O4

= 12.804 − 7.239 × 10−2 T + 4.301 × 10−5 T 2 + 1.5732 × 10−8 T 3

Thus Δrxn H ◦ (T ) = 57 200 +



T

J mol K

(12.804 − 7.239 × 10−2 T

298.2 −5

T 2 + 1.5732 × 10−8 T 3 ) dT 7.239 × 10−2 (T 2 − 298.152 ) = 57 200 + 12.804(T − 298.15) − 2 4.301 1.5732 + × 10−5 (T 3 − 298.153 ) + × 10−8 (T 4 − 298.154 ) 3 4 = 56 189 + 12.804T − 3.619 × 10−2 T 2 + 1.4337 × 10−5 T 3 + 4.301 × 10

+ 3.933 × 10−9 T 4 J/mol N2 O4

Values of Δrxn H ◦ for various values of T are given in the following table. T (K) ◦

Δrxn H (kJ/mol N2 O4 )

200

300

400

500

600

57.423

57.192

56.538

55.580

54.448



R can be used to compute the standard heat of reaction using the folder Aspen Aspen Plus Illustrations>Chapter 8>8.5-2 on Wiley website for this book following the procedure used in Illustration 8.5-1. One difference is that calculation is repeated for each of the temperatures with an isothermal reactor (reactor feed and exit at the same temperatures). The results are

T (K)

200 300 400 500 600

Watts 15134.7 15920.5 16169.9 16087.3 15828.9

kJ/mol N2 O4 54.48 57.31 58.21 57.91 56.98

These results are in only approximate agreement with those in the illustration above. The differ R ences are the result of differences in the databases in the appendices and in Aspen Plus .

Comment The heat of reaction can be, and in fact usually is, a much stronger function of temperature than is the case here.

Third law reference state

In some databases—for example, in the very extensive NIST Chemistry WebBook14 —the data reported for each substance are the the standard state heat of formation Δf H ◦ and the absolute entropy S o , both at 25◦ C. Here by absolute entropy is meant entropy based on the third law of thermodynamics as defined in Sec. 6.8. The reason for reporting these two quantities is that they are determined directly by thermal or calorimetric measurements, unlike the Gibbs energy of formation, which is obtained by measuring chemical equilibrium constants. 14 http://webbook.nist.gov/chemistry/

8.6 The Experimental Determination of the Partial Molar Volume and Enthalpy 385 If the standard state heat of formation and the absolute entropy of each substance are known, the Gibbs energy of reaction can be computed as follows. First, the heat of reaction is computed using

Δrxn H ◦ =

C 

νi Δf H ◦i

i=1

and then the entropy change for the reaction is computed from ◦

Δrxn S =

C 

νi S ◦

i=1

The standard-state Gibbs energy change on reaction at T = 25o C can then be computed from ◦

◦

◦

Δrxn G = Δrxn H − 298.15 · Δrxn S =

C 

νi [Δf H i ◦ − 298.15 · S ◦i ]

i=1

Then using the heat capacities reported in the NIST Chemistry WebBook and Eq. 8.5-5, the Gibbs free energy of reaction at any other temperature can be obtained. As an example of the use of data in this form, we return to the gas-phase reaction of hydrogen and oxygen to form water considered at the beginning of this section. Using the NIST Chemistry WebBook, we obtain the following data.

kJ mol

Species

Δf H o

H2 O2 H2 O

0 0 −241.826

J mol K 130.680 205.150 188.835

So

This results in Δrxn H = −241.826 kJ/mol, Δrxn S = 44.43 J/(mol K), and Δrxn G = −228.579 kJ/mol. This last value agrees with the Gibbs energy of formation for water as a vapor in Appendix A.IV.

8.6 THE EXPERIMENTAL DETERMINATION OF THE PARTIAL MOLAR VOLUME AND ENTHALPY Experimental values for some of the partial molar quantities can be obtained from laboratory measurements on mixtures. In particular, mixture density measurements can be used to obtain partial molar volumes, and heat-of-mixing data yield information on partial molar enthalpies. Both of these measurements are considered here. In Chapter 10 phase equilibrium measurements that provide information on the partial molar Gibbs energy of a component in a mixture are discussed. Once the partial molar enthalpy and partial molar Gibbs energy are known at the same temperature, the partial molar entropy can be computed from the relation S i = (Gi − H i )/T . Table 8.6-1 contains data on the mixture density, at constant temperature and pressure, for the water(1)–methanol(2) system. Column 4 of this table contains calculated values for the molar volume change on mixing (i.e., Δmix V = V − x1 V 1 − x2 V 2 ) at various compositions; these data have also been plotted in Fig. 8.6-1. The slope of the

386 Chapter 8: The Thermodynamics of Multicomponent Mixtures Table 8.6-1 Density Data for the Water(1)–Methanol(2) System at T = 298.15 K x ρ (kg/m3 ) V (m3 /mol) × 106 * Δ V (m3 /mol) × 106 1

mix

0.0 0.1162 0.2221 0.2841 0.3729 0.4186 0.5266 0.6119 0.7220 0.8509 0.9489 1.0

786.846 806.655 825.959 837.504 855.031 864.245 887.222 905.376 929.537 957.522 981.906 997.047

40.7221 37.7015 35.0219 33.5007 31.3572 30.2812 27.7895 25.9108 23.5759 20.9986 19.0772 18.0686

0

−0.3883 −0.6688 −0.7855 −0.9174 −0.9581 −1.0032 −0.9496 −0.7904 −0.4476 −0.1490 0

*Note that the notation V (m3 /mol) × 106 means that the entries in the table have been multiplied by the factor 106 . Therefore, for example, the volume of pure methanol V (x1 = 0) is 40.7221 × 10−6 m3 /mol = 40.7221 cm3 /mol.

0

–0.2

3 ΔmixV _ × 106 m

) mol )

–0.4

–0.6

–0.8 A

∂(ΔmixV) _ –x1 ∂x1

ΔmixV _ _ ∂(ΔmixV) _ = V2 – V _2 ∂x1 _ ∂(ΔmixV) _ B = ΔmixV = V1 – _V1 _ – x2 ∂x1

0

0.2

0.4

0.6

∂(ΔmixV) _ ∂x1

B

A = ΔmixV _ – x1

–1.2

–1.4

x2

0.8

1.0

Mole fraction water

Figure 8.6-1 The isothermal volume change on mixing for the water(1)–methanol(2) system.

8.6 The Experimental Determination of the Partial Molar Volume and Enthalpy 387

Δmix V versus mole fraction curve at any mole fraction is related to the partial molar volumes. To obtain this relationship we start from Δmix V = (x1 V 1 + x2 V 2 ) − x1 V 1 − x2 V 2 = x1 (V 1 − V 1 ) + x2 (V 2 − V 2 ) (8.6-1) so that

    ∂V 1 ∂V 2 ∂(Δmix V )  = (V 1 − V 1 ) + x1 − (V 2 − V 2 ) + x2  ∂x1 ∂x1 T,P ∂x1 T,P T,P (8.6-2) Here we have used the facts that the pure-component molar volumes are independent of mixture composition (i.e., (∂V i /∂x1 )T,P = 0), and that for a binary mixture x2 = 1 − x1 , so (∂x2 /∂x1 ) = −1. From the Gibbs-Duhem equation, Eq. 8.2-19b with θi = V i , we have     ∂V 1 ∂V 2 x1 + x2 =0 ∂x1 T,P ∂x1 T,P so that Eq. 8.6-2 becomes

∂(Δmix V )  = (V 1 − V 1 ) − (V 2 − V 2 )  ∂x1 T,P

(8.6-3)

Now multiplying this equation by x1 and subtracting the result from Eq. 8.6-1 gives Equation for the calculation of the partial molar volume

Δmix V − x1

∂(Δmix V )  = (V 2 − V 2 )  ∂x1 T,P

(8.6-4a)

Similarly,

Δmix V − x2

∂(Δmix V )  ∂(Δmix V )  = ΔV + x   2 mix ∂x2 ∂x1 T,P T,P = (V 1 − V 1 )

(8.6-4b)

Therefore, given data for the volume change on mixing as a function of concentration, so that Δmix V and the derivative ∂(Δmix V )/∂x1 can be evaluated at x1 , we can immediately compute (V 1 − V 1 ) and (V 2 − V 2 ) at this composition. Knowledge of the pure-component molar volumes, then, is all that is necessary to compute V 1 and V 2 at the specified composition x1 . By repeating the calculation at other values of the mole fraction, the complete partial molar volume versus composition curve can be obtained. The results of this computation are given in Table 8.6-2. It is also possible to evaluate (V 1 − V 1 ) and (V 2 − V 2 ) in a more direct, graphical manner. At a given composition, say x∗1 , a tangent line to the Δmix V curve is drawn; the intersections of this tangent line with the ordinates at x1 = 0 and x1 = 1 are designated by the symbols A and B in Fig. 8.6-1. The slope of the this tangent line is

∂(Δmix V )   ∗ ∂x1 x1

388 Chapter 8: The Thermodynamics of Multicomponent Mixtures Table 8.6-2 The Partial Molar Volumes of the Water(1)–Methanol(2) System at T = 298.15 K x1

0 0.1162 0.2221 0.2841 0.3729 0.4186 0.5266 0.6119 0.7220 0.8509 0.9489 1.0

V1−V1 (m3 /mol) × 106

V1 (m3 /mol) × 106

V2−V2 (m3 /mol) × 106

−3.8893 −2.9741 −2.3833 −2.0751 −1.6452 −1.4260 −0.9260 −0.5752 −0.2294 −0.0254 −0.0026

14.180* 15.095 15.686 15.994 16.424 16.643 17.143 17.494 17.840 18.044 18.072 18.069†

−0.0530 −0.1727 −0.2773 −0.4884 −0.6321 −1.0822 −1.5464 −2.2363 −2.9631 −3.1689 −3.0348

0

0

V2 (m3 /mol) × 106 40.722† 40.669 40.549 40.445 40.234 40.090 39.640 39.176 38.486 37.759 37.553 37.687*

*Value of partial molar volume at infinite dilution. †Value of pure-component molar volume.

so that −x∗1

∂(Δmix V )   ∗ ∂x1 x1

is the distance indicated in Fig. 8.6-1. Referring to the figure, it is evident that the numerical value on the ordinate at point A is equal to the left side of Eq. 8.6-4a and, therefore, equal to the value of (V 2 −V 2 ) at x∗1 . Similarly, the intersection of the tangent line with the ordinate at x1 = 1 (point B ) gives the value of (V 1 − V 1 ) at x∗1 . Thus, both (V 1 − V 1 ) and (V 2 − V 2 ) are obtained by a simple graphical construction. For more accurate calculations of the partial molar volume (or any other partial molar property), an analytical, rather than graphical, procedure is used. First, one fits the volume change on mixing, Δmix V , with a polynomial in mole fraction, and then the necessary derivative is found analytically. Since Δmix V must equal zero at x1 = 0 and x1 = 1 (x2 = 0), it is usually fit with a polynomial of the Redlich-Kister form: Redlich-Kister expansion (can be used for the change on mixing of any molar property)

Δmix V = x1 x2

n 

ai (x1 − x2 )i

(8.6-5a)

i=0

(Similar expansions are also used for Δmix H, Δmix U , and the other excess properties to be defined in Chapter 9.) Then, rewriting Eq. 8.6-5a, we have

Δmix V = x1 (1 − x1 )

n 

ai (2x1 − 1)i

i=0 n n   ∂Δmix V  =− ai (2x1 − 1)i+1 + 2x1 (1 − x1 ) ai i(2x1 − 1)i−1  ∂x1 T,P i=0

i=0

(8.6-5b)

8.6 The Experimental Determination of the Partial Molar Volume and Enthalpy 389 and

Δmix V − x1

∂(Δmix V )  =V2−V2  ∂x1 T,P  = x21 ai [(x1 − x2 )i − 2ix2 (x1 − x2 )i−1 ] i

(8.6-6a) Also

∂(Δmix V )  V 1 − V 1 = Δmix V + x2  ∂x1 T,P  2 i = x2 ai [(x1 − x2 ) + 2ix1 (x1 − x2 )i−1 ]

(8.6-6b)

i

An accurate representation of the water-methanol data has been obtained using Eq. 8.6-5 with (in units of m3 /mol)

a0 a1 a2 a3

= −4.0034 × 10−6 = −0.177 56 × 10−6 = 0.541 39 × 10−6 = 0.604 81 × 10−6

and the partial molar volumes in Table 8.6-2 have been computed using these constants and Eqs. 8.6-6a and b. Finally, we note that for the water-methanol system the volume change on mixing was negative, as were (V 1 − V 1 ) and (V 2 − V 2 ). This is not a general characteristic in that, depending on the system, these three quantities can be positive, negative, or even positive over part of the composition range and negative over the rest. The partial molar enthalpy of a species in a binary mixture can be obtained by a similar analysis, but using enthalpy change on mixing (or heat of mixing) data. Such measurements are frequently made using the steady-state flow calorimeter schematically indicated in Fig. 8.6-2. Two streams, one of pure fluid 1 and the second of pure fluid 2, both at a temperature T and a pressure P , enter this steady-state mixing device, and a single mixed stream, also at T and P , leaves. Heat is added or removed to

Pure fluid 1 at temperature T

Pure fluid 2 at temperature T

Stirrer Heating or cooling coil to maintain calorimeter at temperature T

Fluid mixture at temperature T

Figure 8.6-2 An isothermal flow calorimeter.

390 Chapter 8: The Thermodynamics of Multicomponent Mixtures maintain the temperature of the outlet stream. Taking the contents of the calorimeter to be the system, the mass and energy balances (Eqs. 8.4-1 and 8.4-3) are

0 = N˙ 1 + N˙ 2 + N˙ 3

(8.6-7)

0 = N˙ 1 H 1 + N˙ 2 H 2 + N˙ 3 H mix + Q˙

(8.6-8)

and Thus

Q˙ = [N˙ 1 + N˙ 2 ]H mix − N˙ 1 H 1 − N˙ 2 H 2 = [N˙ 1 + N˙ 2 ]Δmix H and

˙ N˙ 1 + N˙ 2 ] Δmix H = Q/[ where Δmix H = H mix − x1 H 1 − x2 H 2 . Therefore, by monitoring N˙ 1 and N˙ 2 as well as the heat flow rate Q˙ necessary to maintain constant temperature, the heat of mixing ˙ N˙ 1 + N˙ 2 ] can be determined at the composition x1 = N˙ 1 /[N˙ 1 + N˙ 2 ]. Δmix H = Q/[ Measurements at a collection of values of the ratio N˙ 1 /N˙ 2 give the complete heat of mixing versus composition curve at fixed T and P . Once the composition dependence of the heat of mixing is known, H 1 and H 2 may be computed in a manner completely analogous to the procedure used for the partial molar volumes. In particular, it is easily established that ∂(Δmix H)  Δmix H − x1 = (H 2 − H 2 ) (8.6-9a)  ∂x1 T,P and ∂(Δmix H)  ∂(Δmix H)  Δmix H − x2 = Δ H + x   mix 2 (8.6-9b) ∂x2 ∂x1 T,P T,P = (H 1 − H 1 ) so that either the computational or graphical technique may also be used to calculate the partial molar enthalpy. Table 8.6-3 and Fig. 8.6-3 contain the heat of mixing data for the water–methanol system. These data have been used to compute the partial molar enthalpies given in Table 8.6-4. Note that one feature of the heat of mixing data of Fig. 8.6-3 is that it is skewed, with the largest absolute value at x1 = 0.73 (and not x1 = 0.5). The entries in Tables 8.6-2 and 8.6-4 are interesting in that they show that the partial molar volume and partial molar enthalpy of a species in a mixture are very similar to the pure component molar quantities when the mole fraction of that species is near unity and are most different from the pure component values at infinite dilution, that is, as the species mole fraction goes to zero. (The infinite dilution values in Tables 8.6-2 and 8.6-4 were obtained by extrapolating both the V and H versus mole fraction data for each species to zero mole fraction.) This behavior is reasonable because in a strongly nonequimolar mixture the molecules of the concentrated species are interacting most often with like molecules, so that their environment and thus their molar properties are very similar to those of the pure fluid. The dilute species, on the other hand, is interacting mostly with molecules of the concentrated species, so that its molecular environment, and consequently its partial molar properties, will be unlike those of its pure component state. Since the environment around a molecule in a mixture is most dissimilar from its pure component state at infinite dilution, the greatest difference between the pure component molar and partial molar properties usually occurs in this limit.

8.6 The Experimental Determination of the Partial Molar Volume and Enthalpy 391 Table 8.6-3 Heat of Mixing Data for the Water(1)–Methanol(2) System at T = 19.69◦ C x1

Q+ (kJ/mol MeOH)

Q (kJ/mol) = Δmix H

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95

−0.134 −0.272 −0.419 −0.569 −0.716 −0.862 −1.017 −1.197 −1.398 −1.632 −1.896 −2.218 −2.591 −3.055 −3.666 −4.357 −5.114 −5.989 −6.838

−0.127 −0.245 −0.356 −0.455 −0.537 −0.603 −0.661 −0.718 −0.769 −0.816 −0.853 −0.887 −0.907 −0.917 −0.917 −0.871 −0.767 −0.599 −0.342

Source: International Critical Tables, Vol. 5, McGraw-Hill, New York, 1929, p. 159. Q = (1 − x1 )Q+ .

–1.0 –0.9 –0.8

ΔmixH _ (kJ/mol)

–0.7 –0.6 –0.5 –0.4 –0.3 –0.2 –0.1 0

0.1

0.2

0.3

0.4

0.5 x1

General equation relating the partial molar property to the pure component property and the property change on mixing

0.6

0.7

0.8

0.9

1.0

Figure 8.6-3 Heat of mixing data for the water(1)–methanol(2) system at T = 19.69◦ C.

Finally, the analyses used here to obtain expressions relating V i and H i to Δmix V and Δmix H , respectively, are easily generalized, yielding the following for the partial molar property of any extensive function θ:

θ1 (T, P, x) − θ1 (T, P ) = Δmix θ(T, P, x) − x2

∂(Δmix θ)   ∂x2 T,P

(8.6-10a)

392 Chapter 8: The Thermodynamics of Multicomponent Mixtures Table 8.6-4 The Difference between the Partial Molar and Pure Component Enthalpies for the Water(1)–Methanol(2) System at T = 19.69◦ C x1

0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

H 1 − H 1 (kJ/mol)

H2 − H 2 (kJ/mol)

−2.703* −2.482 −2.251 −2.032 −1.838 −1.678 −1.551 −1.456 −1.383 −1.325 −1.270 −1.209 −1.131 −1.028 −0.898 −0.740 −0.560 −0.371 −0.193 −0.056

0

−0.006 −0.025 −0.056 −0.097 −0.143 −0.191 −0.237 −0.280 −0.323 −0.373 −0.441 −0.548 −0.719 −0.992 −1.412 −2.036 −2.935 −4.192 −5.905 −8.188*

0

*Indicates value at infinite dilution.

and

θ2 (T, P, x) − θ2 (T, P ) = Δmix θ(T, P, x) − x1

∂(Δmix θ)   ∂x1 T,P

(8.6-10b)

One can also show that if θmix is any molar property of the mixture (not the change on mixing, which is Δmix θ), we have

θ1 (T, P, x) = θmix (T, P, x) − x2

∂(θmix )   ∂x2 T,P

(8.6-11a)

θ2 (T, P, x) = θmix (T, P, x) − x1

∂(θmix )   ∂x1 T,P

(8.6-11b)

and

where, for a binary mixture, x is used to represent one pair of mole fractions x1 and x2 . Illustration 8.6-1 Calculations of Partial Molar Enthalpies from Experimental Data Using the data in Fig. 8.1-1, determine the partial molar enthalpy of sulfuric acid and water at 50 mol % sulfuric acid and 65.6◦ C.

8.6 The Experimental Determination of the Partial Molar Volume and Enthalpy 393 Solution First we must obtain values of enthalpy versus concentration at 65.6◦ C. The values read from this figure and converted to a molar basis are given below. 

wt % H2 SO4

ˆ H

0 20 40 60 80 90 100

kJ kg



 ˆ Δmix H

+278 +85 −78 −175 −153 −60 92

kJ kg



0

−155.8 −281.6 −341.4 −282.2 −170.6 0



mol kg

xH2 SO4 0 0.0439 0.1091 0.2160 0.4235 0.6231 1.000

Δmix H 0

−3.354 −7.533 −12.055 −14.652 −11.582

46.45 37.38 28.32 19.26 14.73

0

These data are fit reasonably well with the simple expression  Δmix H

kJ mol

 = xH2 SO4 xH2 O (−82.795 + 56.683xH2 SO4 ) = xH2 SO4 (1 − xH2 SO4 )(−82.795 + 56.683xH2 SO4 ) = −82.795xH2 SO4 + 139.478x2H2 SO4 − 56.683x3H2 SO4

and dΔmix H = −82.795 + 278.965xH2 SO4 − 170.049x2H2 SO4 dxH2 SO4

Therefore,

and

dΔmix H

dxH2 SO4 xH

dΔmix H

dxH2 O xH

2 SO4 =0.5

= 14.17 2 SO4 =0.5

dΔmix H

=− dxH2 SO4 xH

kJ mol

= −14.17 2 SO4 =0.5

kJ mol

Also, Δmix H(xH2 SO4 = 0.5) = −13.61

kJ mol

H H2 SO4 = 92

kJ kJ 1 kg 98.708 g × × = 9.02 kg 1000 g mol mol

H H2 O = 278

kJ kJ 1 kg 18.015 g = 5.01 × × kg 1000 g mol mol

and

Finally, from Eq. 8.6-9b, we have H H2 SO4 (xH2 SO4 = 0.5, T = 65.6◦ C) = H H2 SO4 (T = 65.6◦ C) + Δmix H(xH2 SO4 = 0.5) + xH2 O kJ = 9.02 − 13.61 + 0.5(−14.17) = −11.68 mol

kJ mol

∂Δmix H

∂xH2 SO4 xH

2 SO4 =0.5



394 Chapter 8: The Thermodynamics of Multicomponent Mixtures and H H2 O (xH2 SO4 = 0.5, T = 65.6◦ C) = H H2 O (T = 65.6◦ C) + Δmix H(xH2 SO4 = 0.5) + xH2 SO4

∂Δmix H

∂xH2 O xH

kJ = 5.01 − 13.61 + 0.5(14.17) = −1.52 mol

2 SO4 =0.5

Note that in this case the pure component and partial molar enthalpies differ considerably. Consequently, we say that this solution is quite nonideal, where, as we shall see in Chapter 9, an ideal solution is one in which some partial molar properties (in particular the enthalpy, internal energy, and volume) are equal to the pure component values. Further, here the solution is so nonideal that at the temperature chosen the pure component and partial molar enthalpies are even of different signs for both water and sulfuric acid. For later reference we note that, at xH2 SO4 = 0.5, we have H H2 SO4 − H H2 SO4 = −20.7

kJ mol

and

H H2 O − H H2 O = −6.5

kJ mol

Generally, any partial molar property differs most from the pure component property in the limit of the component being in high dilution, or at infinite dilution. Therefore, except for components that associate to form dimers, for example, the largest differences between the partial molar and pure component molar properties are   ∂(Δmix θ)  ∂(Δmix θ)  θ1 (T, P, x1 → 0) − θ1 (T, P ) = − =+ ∂x2 x2 =1 ∂x1 x1 =0 (8.6-12a) and

  ∂(Δmix θ)  ∂(Δmix θ)  θ2 (T, P, x2 → 0) − θ2 (T, P ) = − =+ ∂x1 x1 =1 ∂x2 x2 =0 (8.6-12b) Illustration 8.6-2 Calculation of Infinite Dilution Partial Molar Enthalpies from Experimental Data Compute the difference between the infinite dilution partial molar enthalpy and the pure component molar enthalpy for sulfuric acid and water at 65.6◦ C using the information in the previous illustration.

Solution From the previous illustration 

∂Δmix H ∂xH2 SO4

 = −82.795 + 278.965xH2 SO4 − 170.049x2H2 SO4 T,P

Therefore, 

∂Δmix H ∂xH2 SO4

 = +26.11 xH2 SO4 =1

kJ mol

 and

∂Δmix H ∂xH2 SO4

 = −82.80 xH2 SO4 =0

kJ mol

8.6 The Experimental Determination of the Partial Molar Volume and Enthalpy 395 so that H H2 SO4 (T = 65.6◦ C, xH2 SO4 = 0) − H H2 SO4 (T = 65.6◦ C) = −82.80

kJ mol

in which case H H2 SO4 (T = 65.6◦ C, xH2 SO4 = 0) = 9.02 − 82.80 = −73.80

kJ mol

H H2 O (T = 65.6◦ C, xH2 O = 0) − H H2 O (T = 65.6◦ C) = −26.11

and H H2 O (T = 65.6◦ C, xH2 O = 0) = 5.01 − 26.11 = −21.1

kJ mol

kJ mol

Note that for the sulfuric acid + water system at T = 65.6◦ C the differences between the pure component and partial molar properties at infinite dilution are considerably greater than at the mole fraction of 0.5 in the previous illustration.

There is a simple physical interpretation for partial molar properties at infinite dilution. In general, we have from Eq. 8.1-13 for any total property θ(T, P, x) in a binary mixture that θ(T, P, x) = N1 θ1 (T, P, x) + N2 θ2 (T, P, x) (8.6-13) Now consider the case when N2 = 1 and N1 N2 so that x1 ∼ 1, in which case ∼ θ (T, P ), since species 1 is essentially at the pure component θ1 (T, P, x1 ∼ 1) = 1 limit. Also, θ2 (T, P, x1 ∼ 1) is the partial molar property of species 2 at infinite dilution, so that at in this limit

θ(T, P, x1 ∼ 1) = N1 θ(T, P ) + θ2 (T, P, x2 ∼ 0)

(8.6-14)

From this equation we see that the infinite dilution partial molar property θ2 (T, P, x2 ∼ 0) is the amount by which the total property θ changes as a result of the addition of one mole of species 2 to an infinitely large amount of species 1 (so that x2 remains about zero). Note that if the solution were ideal, the total property θ would change by an amount equal to the pure component molar property θ2 ; however, since most solutions are nonideal, the change is instead equal to θ2 . Illustration 8.6-3 Calculation of the Isothermal Enthalpy Change of Mixing One mole of sulfuric acid at 65.6◦ C is added to 1000 moles of water at the same temperature. If the mixing is done isothermally, estimate the change in enthalpy of the mixture.

Solution From Eq. 8.6-14, H(T = 65.6◦ C, NH2 O = 1000, NH2 SO4 = 1) − H(T = 65.6◦ C, NH2 O = 1000, NH2 SO4 = 0) = H H2 SO4 (T = 65.6◦ C, xH2 SO4 = 0.001 ∼ 0) = −73.80

kJ mol

= Δmix H(1000 mol H2 O + 1 mol H2 SO4 )

[The numerical value for H H2 SO4 (T = 65.6◦ C, xH2 SO4 ∼ 0) was obtained from the previous illustration.]

396 Chapter 8: The Thermodynamics of Multicomponent Mixtures

8.7 CRITERIA FOR PHASE EQUILIBRIUM IN MULTICOMPONENT SYSTEMS An important observation in this chapter is that the equations of change for a multicomponent mixture are identical, in form, to those for a pure fluid. The difference between the two is that the pure fluid equations contain thermodynamic properties (U , H , S , etc.) that can be computed from pure fluid equations of state and heat capacity data. Whereas in the multicomponent case these thermodynamic properties can be computed only if the appropriate mixture equation of state and heat capacity data or enthalpy-concentration and entropy-concentration data are given, or if we otherwise have enough information to evaluate the necessary concentration-dependent partial molar quantities at all temperatures, pressures, and compositions of interest. Although this represents an important computational difference between the pure fluid and mixture equations, it has no effect on their fundamental structure. Consequently, for a closed system, we have for both the pure component and multicomponent cases

dU ˙ S − P dV = Q˙ + W dt dt and

where

dS Q˙ = + S˙ gen dt T  U = N U (T, P ) for the pure component system (molar basis) S = N S(T, P )

(8.7-1)

(8.7-2)

and

U=

C 

Ni U i (T, P, x)

i=1

S=

C 

 for a multicomponent system (molar basis)

(8.7-3)

Ni S i (T, P, x)

i=1

Since the form of the balance equations is unchanged, we can use, without modification, the analysis of the last chapter to establish that the equilibrium criteria for a closed multicomponent mixture are (Problem 8.23)

S = maximum for equilibrium at constant M, U, and V A = minimum for equilibrium at constant M, T, and V G = minimum for equilibrium at constant M, T, and P

(8.7-4)

Thus, although it may be computationally more difficult to identify the equilibrium state in a multicomponent mixture than is the case for a pure fluid, the basic criteria used in this identification are the same. As the first application of these criteria, consider the problem of identifying the state of equilibrium in a closed, nonreacting multicomponent system at constant internal energy and volume. To be specific, suppose N1 moles of species 1, N2 moles of species 2, and so on are put into an adiabatic container that will be maintained at constant volume, and that these species are only partially soluble in one another, but do not chemically react. What we would like to be able to do is to predict the composition of each of the phases present at equilibrium. (A more difficult but solvable problem is to also predict the number of phases that will be present. This problem is briefly considered in

8.7 Criteria for Phase Equilibrium in Multicomponent Systems 397 Chapter 11.) In the analysis that follows, we develop the equation that will be used in Chapters 10, 11, and 12 to compute the equilibrium compositions. The starting point for solving this problem are the general equilibrium criteria of Eq. 8.7-4. In particular, the equilibrium criterion for a closed, adiabatic, constant-volume system is

S = maximum subject to the constraints of constant U , V , and total number of moles of each species Ni . For the two-phase system, each extensive property (e.g., Ni , S , U , V ) is the sum of the properties for the individual phases, for example,

Ni = NiI + NiII where the superscripts I and II refer to the phase. In general, the problem of finding the extreme value of a function subject to constraints is not a straightforward task, as will become evident later. However, here this can be done easily. We start by setting the differential of the entropy for the two-phase system equal to zero,

dS = dS I + dS II = 0

(8.7-5)

and then use Eq. 8.2-4, rearranged as

dS =

C 1 P 1 Gi dNi dU + dV − T T T i=1

for each phase. Now recognizing that since the total internal energy, total volume, and the number of moles of each species are fixed, we have

dU II = −dU I dV II = −dV I dNiII = −dNiI which can be used in Eq. 8.7-5 to obtain      I C  I  1 P II Gi GII 1 P i I I dU + dV − dNiI = 0 dS = − − − T I T II T I T II T I T II i=1 (8.7-6) The condition for equilibrium is that the differential of the entropy be zero with respect to all variations of the independent and unconstrained variables, here dU I , dV I , and each of the dNiI . In order for Eq. 8.7-6 to be satisfied, we must have (1) that   ∂S =0 ∂U I V I ,N I i

which implies First criterion for phase equilibrium

1 1 = II I T T (2) that



or simply

∂S ∂V I

T I = T II

 =0 U I ,NiI

(8.7-7a)

398 Chapter 8: The Thermodynamics of Multicomponent Mixtures which implies

PI P II = TI T II or, in view of Eq. 8.7-7a, Second criterion for phase equilibrium

P I = P II and (3) that (∂S/∂NiI )U I ,V I ,N I

j=i

GIi = GII i

Third criterion for phase equilibrium

(8.7-7b)

= 0 for each species i, which implies

or μIi = μII i

for each species i

(8.7-7c)

since T I = T II . In Eq. 8.7-7c, μi = Gi is the chemical potential of species i. Thus, for phase equilibrium to exist in a closed, nonreacting multicomponent system at constant energy and volume, the pressure must be the same in both phases (so that mechanical equilibrium exists), the temperature must be the same in both phases (so that thermal equilibrium exists), and the partial molar Gibbs energy of each species must be the same in each phase (so that equilibrium with respect to species diffusion exists).15 Note that with the replacement of the partial molar Gibbs free energy by the pure component Gibbs energy, Eqs. 8.7-7a, b, and c become identical to the conditions for phase equilibrium in a one-component system derived in Sec. 7.1 (see Eqs. 7.1-9a, b, and c). In principle, we could now continue to follow the development of Chapter 7 and derive the conditions for stability of the equilibrium state. However, this task is algebraically complicated and will not be considered here.16 To derive the conditions for phase equilibrium in a closed system at constant (and, of course, uniform) temperature and pressure, we start from the equilibrium criterion that G be a minimum and set the differential of G for the two-phase system equal to zero, that is,

dG = dGI + dGII = 0

(8.7-8)

Now recognizing that at constant T and P (from Eq. 8.2-1)

dG|T,P =

C 

Gi dNi

i=1

and that the total number of moles of each species is fixed, so that Ni = NiI + NiII or dNiII = −dNiI , we obtain

dG =

C  i=1

15 Clearly,

GIi dNiI +

C  i=1

II GII i dNi =

C 

I (GIi − GII i ) dNi = 0

i=1

from these results, it is a species partial molar Gibbs energy difference between phases, rather than a concentration difference, that is the driving force for interphase mass transfer in the approach to equilibrium. 16 See, for example, of J. W. Tester and M. Modell, Thermodynamics and Its Applications, 3rd ed., Prentice Hall, Englewood Cliffs, N.J. (1997) Chapter 7.

8.8 Criteria for Chemical Equilibrium, and Combined Chemical and Phase Equilibrium 399 Setting the derivative of the Gibbs energy with respect to each of its independent variables (here the mole numbers NiI ) equal to zero yields   ∂G = 0 = GIi − GII or GIi = GII i i ∂NiI N I j=i

so that here, as in Chapter 7, we find that the equality of Gibbs free energies is a necessary condition for the existence of phase equilibrium for systems subject to a variety of constraints (see Problem 8.3). Although we will not do so here, it is easy to show that these analyses for two-phase equilibrium are easily generalized to multiphase equilibrium and yield III GIi = GII i = Gi = · · ·

(8.7-9)

8.8 CRITERIA FOR CHEMICAL EQUILIBRIUM, AND COMBINED CHEMICAL AND PHASE EQUILIBRIUM Equations 8.7-4 also provide a means of identifying the equilibrium state when chemical reactions occur. To see this, consider first the case of a single chemical reaction occurring in a single phase (both of these restrictions will be removed shortly) in a closed system at constant temperature and pressure.17 The total Gibbs energy for this system, using the reaction variable notation introduced in Sec. 8.3, is

G=

C 

Ni Gi =

i=1

C 

(Ni,0 + νi X)Gi

i=1

Since the only variation possible in a one-phase, closed system at constant temperature and pressure is in the extent of reaction X , the equilibrium criterion is   ∂G =0 ∂X T,P which yields Criterion for chemical equilibrium of a single reaction

0=

C 

νi Gi

(8.8-1)

i=1

(Note that

C  Ni (∂Gi /∂X)T,P is equal to zero by the Gibbs-Duhem equation, i

Eq. 8.2-14 with Y = X .) It is possible to show that the criterion for chemical equilibrium developed here is also applicable to systems subject to constraints other than constant temperature and pressure (Problem 8.4). In fact, Eq. 8.8-1, like the phase equilibrium criterion of Eq. 8.7-9, is of general applicability. Of course, the difficulty that arises in using either of these equations is translating their simple form into a useful prescription for equilibrium 17 Since

chemists and chemical engineers are usually interested in chemical and phase equilibria at constant temperature and pressure, the discussions that follow largely concern equilibrium under these constraints.

400 Chapter 8: The Thermodynamics of Multicomponent Mixtures calculations by relating the partial molar Gibbs energies to quantities of more direct interest, such as temperature, pressure, and mole fractions. This problem will be the focus of much of the rest of this book. The first generalization of the analysis given here is to the case of multiple chemical reactions in a closed, single-phase, constant-temperature and constant-pressure system. Using the notation of Sec. 8.3, the number of moles of species i present at any time is

Ni = Ni,0 +

M 

νij Xj

(8.35)

j=1

where the summation is over the independent reactions. The total Gibbs energy of the system is  C C  M    Ni,0 + G= Ni Gi = νij Xj Gi i=1

=

C 

i=1

Ni,0 Gi +

i=1

j=1

C  M 

(8.8-2)

νij Xj Gi

i=1 j=1

The condition for chemical equilibrium in this multireaction system is G = minimum or dG = 0 for all variations consistent with the stoichiometry at constant temperature, pressure, and total mass. For the present case this implies   ∂G = 0 j = 1, 2, . . . , M (8.8-3) ∂Xj T,P,X i=j

so that



∂G ∂Xj

 =0= T,P,Xi=j

C  i=1

νij Gi +

C  i=1

 Ni

∂Gi ∂Xj

 T,P,Xk=j

for all independent reactions j = 1, 2, . . . , M

 Since the sum Ci=1 Ni (∂Gi /∂Xj )T,P,Xk=j vanishes by the Gibbs-Duhem equation, the equilibrium criterion is Chemical equilibrium criteria for multiple reactions

C 

νij Gi = 0 j = 1, 2, . . . , M

(8.8-4)

i=1

This equation is analogous to Eq. 8.8-1 for the single-reaction case. The interpretation of Eq. 8.8-4 is clear: In a system in which several chemical reactions occur, chemical equilibrium is achieved only when each reaction is itself in equilibrium. The final case to be considered is that of combined phase and chemical equilibrium in a closed system at constant temperature and pressure. At this point you can probably guess the final result: If both phase changes and chemical transformations are possible, equilibrium occurs only when each possible transformation is itself in equilibrium. Thus, Eqs. 8.7-7 and 8.8-4 must be simultaneously satisfied for all species and all reactions in all phases. To prove this assertion, it is first useful to consider the mathematical technique of Lagrange multipliers, a method used to find the extreme (maximum or minimum) value

8.8 Criteria for Chemical Equilibrium, and Combined Chemical and Phase Equilibrium 401 of a function subject to constraints. Rather than develop the method in complete generality, we merely introduce it by application to the problem just considered: equilibrium in a single-phase, multiple–chemical reaction system. We identified the equilibrium state for several chemical reactions occurring in a single-phase system at constant temperature and pressure by finding the state for which C  G= Ni Gi was equal to a minimum subject to the stoichiometric constraints i=1

Ni = Ni,0 +

M 

νij Xj

for all species i

j=1

The procedure used was to incorporate the constraints directly into the Gibbs energy function and then find the minimum value of the resulting unconstrained equation (Eq. 8.8-2) by setting each of the M derivatives (∂G/∂Xj )T,P,Xi=j equal to zero. However, this direct substitution technique can be very cumbersome when the constraints are complicated, as is the case in the problem of combined chemical and phase equilibrium. An alternative method of obtaining a solution to the multiple-reaction, single-phase equilibrium problem is to use the method of Lagrange multipliers.18 Here one first rewrites the constraints as M  Ni − Ni,0 − νij Xj = 0 i = 1, 2, . . . , C (8.8-5) j=1

and then creates a new function G by adding the constraints, each with a multiplying parameter αi , a Lagrange multiplier, to the original Gibbs function:   C C M    G= Ni Gi + αi Ni − Ni,0 − νij Xj (8.8-6) i=1

i=1

j=1

The independent variables of this new function are N1 , N2 , . . . , NC ; X1 , X2 , . . . , XM ; and α1 , α2 , . . . , αC . To determine the state for which the Gibbs energy is a minimum subject to the stoichiometric constraints of Eq. 8.8-5, the partial derivatives of this new unconstrained function G with respect to each of its independent variables are set equal to zero. From this procedure we obtain the following sequence of simultaneous equations to be solved:     C  ∂G ∂Gi = 0 = Gk + αk + Ni ∂Nk T,P,Xj ,N ∂Nk T,P,Xj ,N j=k

i=1

j=k

where the last term is zero by the Gibbs-Duhem equation, so that

αk = −Gk

k = 1, 2, . . . , C

(8.8-7)

Also,



18 A

∂G ∂Xk

 =0=− T,P,Ni ,Xj=k

C 

νik αi

k = 1, 2, . . . , M

(8.8-8)

i=1

more complete discussion of Lagrange multipliers may be found in M. H. Protter and C. B. Morrey, College Calculus with Analytic Geometry, Addison-Wesley, Reading, Mass. (1964), pp. 708–715; and V. G. Jenson and G. V. Jeffreys, Mathematical Methods in Chemical Engineering, Academic Press, New York (1963), pp. 482–483.

402 Chapter 8: The Thermodynamics of Multicomponent Mixtures or, using Eq. 8.8-7, C 

νik Gi = 0 k = 1, 2, . . . , M

(8.8-9)

i=1

and finally,



∂G ∂αi

 = 0 = Ni − Ni,0 − T,P,Xj ,Nk

M 

νij Xj

i = 1, 2, . . .

(8.8-10)

j=1

Clearly, Eq. 8.8-9 gives the same equilibrium requirement as before (see Eq. 8.8-4), whereas Eq. 8.8-10 ensures that the stoichiometric constraints are satisfied in solving the problem. Thus the Lagrange multiplier method yields the same results as the direct substitution or brute-force approach. Although the Lagrange multiplier method appears awkward when applied to the very simple problem here, its real utility is for complicated problems in which the number of constraints is large or the constraints are nonlinear in the independent variables, so that direct substitution is very difficult or impossible. To derive the criteria for combined chemical and phase equilibrium, the following notation will be used: Nik and Gki are, respectively, the number of moles and partial molar Gibbs energy of species i in the kth phase; Ni,0 is the initial number of moles of species i in the closed system; and Xj is the overall molar extent of reaction (reaction variable) for the jth independent reaction, regardless of which phase or in how many different phases the reaction occurs. Thus ⎛ ⎞ P M Total number of   k ⎝moles of species⎠ = Ni = Ni,0 + νij Xj (8.8-11) i in all P phases j=1 k=1 and



Total Gibbs free energy of system

 =G=

P  C 

Nik Gki

(8.8-12)

k=1 i=1

where P is the number of phases, C is the number of components, and M is the number of independent reactions. The equilibrium state for a closed system at constant temperature and pressure is that state for which the Gibbs energy G achieves a minimum value from among all the states consistent with the reaction stoichiometry. The identification of the equilibrium state is then a problem of minimizing the Gibbs energy subject to the stoichiometric constraints of Eq. 8.8-11. Since the easiest way of solving this problem is to use the method of Lagrange multipliers, we define a set of Lagrange multipliers, α1 , α2 , . . . , αC and construct the augmented function

P  P  C C M     k k k G= Ni Gi + αi Ni − Ni,0 − νij Xj (8.8-13) k=1 i=1

i=1

k=1

j=1

whose minimum we wish to find for all variations of the independent variables X1 , X2 , . . . , XM , N1I , N1II , . . . , NCP , and α1 , α2 , . . . , αC .

8.8 Criteria for Chemical Equilibrium, and Combined Chemical and Phase Equilibrium 403 Setting each of the partial derivatives of G with respect to N1I , N1II , . . . equal to zero, remembering that the N ’s, X ’s, and α’s are now to be treated as independent variables, yields19    k P  C  ∂G ∂Gi I k = G1 + + α1 = GI1 + α1 = 0 Ni ∂N1I ∂N1I k=1 i=1



∂G ∂N1II

 =

GII 1

+

P  C 

 Nik

k=1 i=1

∂Gki ∂N1II

 + α1 = GII 1 + α1 = 0

(8.8-14)

.. . In each case the double-summation term vanishes by application of the Gibbs-Duhem equation (Eq. 8.2-15) to each phase. The net information content of Eqs. 8.8-14 is P GI1 = GII 1 = · · · = G1 = −α1

and by generalization, P GIi = GII i = · · · = Gi = −αi

i = 1, 2, . . . , C

(8.8-15)

These equations establish that one of the equilibrium conditions in a multiple-reaction, multiphase system is that phase equilibrium be established for each of the species among the phases in which the species is present. Another set of equilibrium criteria is obtained by minimizing G with respect to each of the reaction variables Xj ( j = 1, . . . , M). Thus20



∂G ∂X1

 =0=

P  C 

 Nik

k=1 i=1

∂Gki ∂X1

 −

C 

αi νi1 = 0

(8.8-16)

i=1

The first term on the right side of this equation vanishes by the Gibbs-Duhem equation, k and from Eq. 8.8-15, we can set GIi = GII i = · · · = Gi = · · · = −αi and so obtain



∂G ∂X1

 =0=

C 

νi1 Gki

(8.8-17)

i=1

Similarly, from ∂ G /∂X2 = 0, ∂ G /∂X3 = 0, . . . , we obtain C  i=1

νij Gki = 0

for all phases k = I, II, . . . ,P and all reactions j = 1, 2, . . . , M

(8.8-18)

which establishes that a further condition for equilibrium in a multiphase, multireaction system is that each reaction be in equilibrium in every phase. In fact, since at 19 Though we have not listed the variables being held constant, you should recognize that all variables in the set T ,

P , Nik (i = 1, . . . , C; k = 1, . . . , P), Xj (j = 1, . . . , M), and αi (i = 1, . . . , C), except the one being varied in the derivative, have been held constant. 20 See footnote 18.

404 Chapter 8: The Thermodynamics of Multicomponent Mixtures equilibrium the partial molar Gibbs energy of each species is the same in every phase (see Eq. 8.8-15), if Eq. 8.8-18 is satisfied in any phase, it is satisfied in all phases. Finally, setting the partial derivatives of G with respect to each of the Lagrange multipliers αi equal to zero yields the stoichiometric constraints of Eq. 8.8-11. The equilibrium state is that state for which Eqs. 8.8-11, 8.8-15, and 8.8-18 are simultaneously satisfied. Thus, we have proved the assertion that in the case of combined chemical and phase equilibria the conditions of phase equilibrium must be satisfied for all species in each of the phases and, furthermore, that chemical equilibrium must exist for each reaction in each phase. (The fact that each reaction must be in chemical equilibrium in each phase does not imply that each mole fraction will be the same in each phase. This point is demonstrated in Chapter 13.)

8.9 SPECIFICATION OF THE EQUILIBRIUM THERMODYNAMIC STATE OF A MULTICOMPONENT, MULTIPHASE SYSTEM; THE GIBBS PHASE RULE As has been mentioned several times, the equilibrium state of a single-phase, onecomponent system is completely fixed by the specification of two independent, intensive variables. From this observation we were able, in Sec. 7.6, to establish a simple relation for determining the number of degrees of freedom for a single-component, multiphase system. Here an analogous equation is developed for determining the number of degrees of freedom in a reacting multicomponent, multiphase system; this relationship is called the Gibbs phase rule. The starting point for the present analysis is the observation in Sec. 8.1 that the equilibrium thermodynamic state of a single-phase C-component system can be fixed by specifying the values of two intensive variables and C − 1 mole fractions. Alternatively, the specification of any C + 1 independent state variables could be used to fix the state of this system.21 Thus, we can say that a C-component, single-phase system has C + 1 degrees of freedom, that is, we are free to adjust C + 1 independent intensive thermodynamic properties of this system; however, once this is done, all the other intensive thermodynamic properties are fixed. This is equivalent to saying that if T , P , x1 , x2 , . . . , xC−1 are taken as the independent variables, there exist equations of state in nature of the form

V = V (T, P, x1 , . . . , xC−1 ) S = S(T, P, x1 , . . . , xC−1 ) G = G(T, P, x1 , . . . , xC−1 ) although we may not have been clever enough in our experiments to have determined the functional relationship between the variables. Our interest here is in determining the number of degrees of freedom in a general multicomponent, multiphase chemically reacting system consisting of C components distributed among P phases and in which M independent chemical reactions occur. Since C + 1 variables are required to completely specify the state of each phase, and 21 By C+1 independent state variables we mean C+1 nonredundant pieces of information about the thermodynamic

state of the system. For example, temperature, pressure, and C − 1 mole fractions form a set of C + 1 independent  variables; temperature and C mole fractions are not independent, however, since Ci xi = 1, so that only C − 1 mole fractions are independent. Similarly, for a gas mixture composed of ideal gases, the enthalpy or internal energy, temperature, and C − 1 mole fractions do not form an independent set of variables, since H and U are calculable from the mole fractions and the temperature. However, H, P , and C − 1 mole fractions are independent.

8.9 Specification of the Equilibrium Thermodynamic State of a Multicomponent, Multiphase System 405 there are P phases present, it would appear that a total of P (C + 1) variables must be specified to fix the state of each of the phases. Actually, the number of variables that must be specified is considerably fewer than this, since the requirement that equilibrium exists provides a number of interrelationships between the state variables in each of the phases. In particular, the fact that the temperature must be the same in all phases,

T I = T II = T III = · · · = T P

(8.9-1)

results in (P − 1) restrictions on the values of the state variables of the phases. Similarly, the requirement that the pressure be the same in all phases,

P I = P II = P III = · · · = P P

(8.9-2)

provides an additional (P − 1) restrictions. Since each of the partial molar Gibbs energies is in principle calculable from equations of state of the form j Gji = Gji (T, P, x1j , x2j , . . . , xC− 1)

(8.9-3)

the condition for phase equilibrium, P GIi = GII i = · · · = Gi

i = 1, 2, . . . , C

(8.9-4)

provides C (P − 1) additional relationships among the variables without introducing any new unknowns. Finally, if M independent chemical reactions occur, there are M additional relations of the form C  νij Gi = 0 j = 1, 2, . . . , M (8.9-5) i=1

(where we have omitted the superscript indicating the phase since, by Eq. 8.9-4, the partial molar Gibbs energy for each species is the same in all phases). Now designating the number of degrees of freedom by the symbol F, we have Gibbs phase rule

⎛ ⎞   Number of unknown Number of independent relations F = ⎝ thermodynamic ⎠ − among the unknown parameters parameters = P (C + 1) =C−M−P +2

(8.9-6)

− [2(P − 1) + C (P − 1) + M]

Therefore, in a C-component, P-phase system in which M independent chemical reactions occur, the specification of C − M − P + 2 state variables of the individual phases completely fixes the thermodynamic state of each of the phases. This result is known as the Gibbs phase rule. In practice, temperature, pressure, and phase composition are most commonly used to fix the thermodynamic state of multicomponent, multiphase systems, though any other information about the thermodynamic state of the individual phases could be used as well. However, thermodynamic information about the composite multiphase system is not useful in fixing the state of the system. That is, we could use the specific volume of any one of the phases as one of the C − M − P + 2 degrees of freedom, but not the molar volume of the multiphase system. Finally, we note that for a pure fluid C = 1

406 Chapter 8: The Thermodynamics of Multicomponent Mixtures and M = 0, so that Eq. 8.9-6 reduces to F =3−P

(8.9-7)

the result found in Sec. 7.6. Illustration 8.9-1 Application of the Gibbs Phase Rule In Chapter 13 we will consider the reaction equilibrium when styrene is hydrogenated to form ethylbenzene. Depending on the temperature and pressure of the system, this reaction may take place in the vapor phase or in a vapor-liquid mixture. Show that the system has three degrees of freedom if a single phase exists, but only two degrees of freedom if the reactants and products form a two-phase mixture.

Solution The styrene-hydrogen-ethylbenzene system is a three-component (C = 3), single-reaction (M = 1) system. Thus F =C−M−P +2=4−P Clearly, if only the vapor phase exists (P = 1), there are three degrees of freedom; if, however, both the vapor and liquid are present (P = 2), the system has only two degrees of freedom.

Illustration 8.9-2 Another Application of the Gibbs Phase Rule Determine the number of degrees of freedom for each of the following mixtures. a. A one-component vapor-liquid mixture b. A nonreacting two-component vapor-liquid mixture c. A vapor-liquid mixture of ortho-, meta-, and para-xylenes and ethylbenzene at temperatures high enough that the xylenes can undergo isomerization

Solution a. This system has one component (C = 1) and two phases (P = 2), and there are no chemical reactions (M = 0). Therefore, F =C−M−P +2=1−0−2+2=1 Consequently, this system has one degree of freedom. If the temperature is set, the pressure is fixed; or if the pressure is set, the temperature is fixed. We see this when boiling water in a pot open to the atmosphere. At 101.3 kPa (1 atm), the temperature of the boiling water will be 100◦ C and will remain at this temperature no matter how much of the water boils away, provided the water is pure. In order to change the temperature of this vapor-liquid mixture, the pressure must change. b. This system has two components (C = 2) and two phases (P = 2), and there are no chemical reactions occurring (M = 0). The number of degrees of freedom in this system is F =C−M−P +2=2−0−2+2=2 Therefore, the values of two state parameters—for example, temperature and pressure, temperature and the mole fraction of one of the species, or pressure and the mole fraction of one of the species—must be set to fix the state of this mixture. This suggests that at a fixed pressure the boiling temperature of the mixture will be a function of its composition. To see the implication of this, consider the experiment of preparing a mixture

8.9 Specification of the Equilibrium Thermodynamic State of a Multicomponent, Multiphase System 407 of two species (composition initially known) at one atmosphere and heating this mixture to its boiling point, and removing the vapor as the boiling continues. For most mixtures (an azeotropic mixture, to be discussed in Sec. 10.2, is the exception) as boiling occurs the composition of the vapor will be different from that of the prepared mixture, so that (by a mass balance) the composition of the remaining liquid will change during the boiling process (unless the vapor is continually condensed and replaced). As a result, at fixed pressure, the boiling temperature of this mixture will continually change as the process of boiling continues. This behavior is different from the boiling of a one-component mixture considered above, in which the temperature remains constant as the boiling process continues at fixed pressure. Also, by changing the composition of this mixture, a range of equilibrium temperatures can be obtained at the same pressure, or a range of equilibrium pressures can occur at a fixed temperature. c. There are three independent reactions for this system. One set of such independent reactions is m-xylene ↔ o-xylene m-xylene ↔ p-xylene m-xylene ↔ ethylbenzene This system has four components (C = 4) and two phases (P = 2), and there are three independent chemical reactions occurring (M = 3). The number of degrees of freedom in this system is F =C−M−P +2=4−3−2+2=1 Therefore, specification of the value of only one state variable—temperature, pressure, or the mole fraction of one of the species in one of the phases—completely fixes the twophase state of this mixture. For example, consider the experiment of preparing a mixture of these species, and heating the mixture under pressure to a temperature that is high enough that vaporization and chemical reaction occur. In this mixture, once such a temperature is fixed, the pressure, liquid composition, and vapor composition are all fixed. If an additional amount of one of the components is added to this mixture, the total number of moles of vapor and liquid will change, but the pressure, the vapor mole fractions, and the liquid mole fractions will not change. Also for this mixture, as with a pure component, the vaporliquid equilibrium temperature will change when the pressure is changed, but will remain constant as boiling occurs and the vapor is removed.

It is also of interest to determine the amount and type of additional information needed to fix the relative amounts of each of the phases in equilibrium, once their thermodynamic states are known. We can obtain this from an analysis that equates the number of variables to the number of restrictions on these variables. It is convenient for this discussion to write the specific thermodynamic properties of the multiphase system in terms of the distribution of mass between the phases. The argument could be based on a distribution of numbers of moles; however, it is somewhat more straightforward on a mass basis because total mass, and not total moles, is a conserved quantity. Thus, we will use wi to represent the mass fraction of the ith phase.22 Clearly the wi must satisfy the equation 1 = wI + wII + wIII + · · · + wP (8.9-8) The total volume per unit mass, Vˆ , the total entropy per unit mass, Sˆ, and so on, are related to the analogous quantities in each of the phases by the equations

Vˆ = wI Vˆ I + wII Vˆ II + · · · + wP Vˆ P Sˆ = wI SˆI + wII SˆII + · · · + wP SˆP

(8.9-9)

hope the notation is not too confusing. Here wI is the fraction of the total mass of the system in phase I, while earlier in this chapter wi was used to represent the mass fraction of species i in a phase.

22 I

408 Chapter 8: The Thermodynamics of Multicomponent Mixtures In writing these equations we are presuming that the specific volumes, entropies, and so forth for each phase (denoted by the superscript) are known from the equations of state or experimental data for the individual phases and a previous specification of the C − M − P + 2 degrees of freedom. Since there are P unknown mass distribution variables, each of the wi , it is evident that we need P equations to determine the relative amounts of each of the phases. Therefore, Eq. 8.9-8, together with the specification of P − 1 intensive thermodynamic variables for the multiphase system (which can be written in the form of Eq. 8.9-9), are needed. This is in addition to the C − M − P + 2 intensive variables of the individual phases that must be specified to completely fix the thermodynamic state of all of the phases. (You should convince yourself that this conclusion is in agreement with Illustration 7.6-1.) Thus far we have not considered the fact that the initial composition of a chemical or phase equilibrium system may be known. Such information can be used in the formulation of species mass balances and the energy balance, which lead to additional equations relating the phase variables. Depending on the extent of initial information available and the number of phases present, the initial state information may or may not reduce the number of degrees of freedom of the system. This point is most easily demonstrated by reference to specific examples, so the effect of initial state information will be considered in the illustrations of the following chapters, not here. We should point out that the Gibbs phase rule is of use in deciding whether or not an equilibrium problem is “well posed,” that is, whether enough information has been given for the problem to be solvable, but it is not of use in actually solving for the equilibrium state. This too is demonstrated by examples later in this book. The Gibbs phase rule, being general in its scope and application, is regarded as another part of the philosophical content of thermodynamics.

8.10 A CONCLUDING REMARK The discussion in this chapter essentially concludes our development of thermodynamic theory. The remainder of this book is largely concerned with how this theory is used to solve problems of interest to the chemical process industry. Since the partial molar Gibbs free energy has emerged as the central function in equilibrium computations, Chapter 9 is concerned with the techniques used for estimating this quantity in gaseous, liquid, and solid mixtures. Chapters 10 to 15 are devoted to the use of thermodynamics in explaining and predicting the great diversity of physical, chemical, and biochemical equilibria that occur in mixtures.

PROBLEMS 8.1 Prove that 



∂H = H i − T S i = Gi ∂Ni P,S,Nj=i     ∂U ∂A b. Gi = = ∂Ni S,V,Nj=i ∂Ni V,T,Nj=i

a.

8.2 Derive the analogues of the Gibbs-Duhem equations (Eqs. 8.2-8 and 8.2-9) for the constraints of

a. Constant temperature and volume b. Constant internal energy and volume c. Constant entropy and volume 8.3 In Sec. 8.7 we established that the condition for equilibrium between two phases is GIi = GII i

(for all species present in both phases)

for closed systems either at constant temperature and pressure or at constant internal energy and volume.

Problems 409 Show that this equilibrium condition must also be satisfied for closed systems at a. Constant temperature and volume b. Constant entropy and volume 8.4 Show that the criterion for chemical equilibrium developed in the text, C 

νi Gi = 0

i

for a closed system at constant temperature and pressure, is also the equilibrium condition to be satisfied for closed systems subject to the following constraints: a. Constant temperature and volume b. Constant internal energy and volume 8.5 Prove that since total mass is conserved during a chemical reaction, C 

νi mi = 0

8.9 a. In vapor-liquid equilibrium in a binary mixture, both components are generally present in both phases. How many degrees of freedom are there for such a system? b. The reaction between nitrogen and hydrogen to form ammonia occurs in the gas phase. How many degrees of freedom are there for this system? c. Steam and coal react at high temperatures to form hydrogen, carbon monoxide, carbon dioxide, and methane. The following reactions have been suggested as being involved in the chemical transformation: C + 2H2 O = CO2 + 2H2 C + H2 O = CO + H2 C + CO2 = 2CO C + 2H2 = CH4 CO + H2 O = CO2 + H2 CO + 3H2 = CH4 + H2 O

for a single-reaction system

i=1

and C  i=1

νij mi = 0

j = 1, 2, . . . , M

for a multiple-reaction system

where mi is equal to the molecular weight of species i. Also show, by direct substitution, that the first of these equations is satisfied for the reaction

8.10 a.

H2 O = H2 + 12 O2

8.6 Show that the partial molar volumes computed from Eqs. 8.6-4a and b and the partial molar enthalpies computed from Eqs. 8.6-9a and b must satisfy the Gibbs-Duhem equation. 8.7 Compute the partial molar volumes of methyl formate in methanol–methyl formate and ethanol–methyl formate mixtures at 298.15 K for various compositions using the experimental data in Fig. 8.1-2a and the following pure-component data:

b.

c.

V MF = 0.062 78 m3 /kmol V M = 0.040 73 V E = 0.058 68

8.8 Compute the difference between the pure-component and partial molar enthalpies for both components at 298.15 K and various compositions in each of the following mixtures using the data in Fig. 8.1-2b. a. benzene–C6 F5 H b. benzene–C6 F6 c. benzene–C6 F5 Cl d. benzene–C6 F5 Br e. benzene–C6 F5 I

8.11 a.

b.

How many degrees of freedom are there for this system? [Hint: (1) How many independent chemical reactions are there in this sequence? (2) How many phase equilibrium equations are there?] In vapor-liquid equilibrium, mixtures sometimes occur in which the compositions of the coexisting vapor and liquid phases are the same. Such mixtures are called azeotropes. Show that a binary azeotropic mixture has only one degree of freedom. In osmotic equilibrium, two mixtures at different pressures and separated by a rigid membrane permeable to only one of the species present attain a state of equilibrium in which the two phases have different compositions. How many degrees of freedom are there for osmotic equilibrium in a binary mixture? The phase equilibrium behavior of furfural (C5 H4 O2 )–water mixtures is complicated because furfural and water are only partially soluble in the liquid phase. (i) How many degrees of freedom are there for the vapor-liquid mixture if only a single liquid phase is present? (ii) How many degrees of freedom are there for the vapor-liquid mixture if two liquid phases are present? What is the maximum number of phases that can coexist for a mixture of two nonreacting components? How would the answer in part (a) change if the two components could react to form a third component?

410 Chapter 8: The Thermodynamics of Multicomponent Mixtures 8.12 Consider a reaction that occurs in a vessel containing a semipermeable membrane that allows only one of the components to pass through it (for example, a small molecule such as hydrogen) but will not allow the passage of large molecules. With such a membrane, the chemical potential of the permeable component can be kept constant in the reaction vessel. a. Derive the equilibrium criterion for a one-phase reaction system described above in which the temperature, pressure, and chemical potential of one component are held constant. b. Derive the equilibrium criterion for a one-phase reaction system described above in which the temperature, volume, and chemical potential of one component are held constant. 8.13 The following set of reactions is thought to occur between nitrogen and oxygen at high temperatures N2 + 12 O2 = N2 O N2 + O2 = 2NO N2 + 2O2 = N2 O4 N2 O4 = 2NO2 2NO + 32 O2 = N2 O5 N2 O + NO2 = 3NO

a. Find an independent set of reactions for the nitrogen-oxygen system. b. How many degrees of freedom are there for this system? c. If the starting oxygen-to-nitrogen ratio is fixed (as in air), how many degrees of freedom are there? 8.14 The temperature achieved when two fluid streams of differing temperature and/or composition are adiabatically mixed is termed the adiabatic mixing temperature. Compute the adiabatic mixing temperature for the following two cases: a. Equal weights of aqueous solutions containing 10 wt % sulfuric acid at 20◦ C and 90 wt % sulfuric acid at 70◦ C are mixed. b. Equal weights of aqueous solutions containing 10 wt % sulfuric acid at 20◦ C and 60 wt % sulfuric acid at 0◦ C are mixed. Explain why the adiabatic mixing temperature is greater than that of either of the initial solutions in one of these cases, and intermediate between those of the initial solutions in the other case. 8.15 The molar integral heat of solution Δs H is defined as the change in enthalpy that results when 1 mole of solute (component 1) is isothermally mixed with N2 moles of solvent (component 2) and is given by 23 International

Δs H = (1 + N2 )H mix − H 1 − N2 H 2 = H 1 + N2 H 2 − H 1 − N 2 H 2 Δs H is easily measured in an isothermal calorimeter by monitoring the heat evolved or absorbed on successive additions of solvent to a given amount of solute. The table below gives the integral heat-ofsolution data for 1 mol of sulfuric acid in water at 25◦ C (the negative sign indicates that heat is evolved in the dilution process). N2 (moles of water)

0.25 8242

1.0 28 200

1.5 34 980

−Δs H (J)

4.0 54 440

5.44 58 370

9.0 62 800

N2 (moles of water) −Δs H (J)

10.1 64 850

19.0 70 710

20.0 71 970

−Δs H (J) N2 (moles of water)

2.33 44 690

a. Calculate the heat evolved when 100 g of pure sulfuric acid is added isothermally to 100 g of water. b. Calculate the heat evolved when the solution prepared in part (a) is diluted with an additional 100 g of water. c. Calculate the heat evolved when 100 g of a 60 wt % solution of sulfuric acid is mixed with 75 g of a 25 wt % sulfuric acid solution. d. Relate (H 1 − H 1 ) and (H 2 − H 2 ) to only N1 , N2 , ΔH s , and the derivatives of ΔH s with respect to the ratio N2 /N1 . e. Compute the numerical values of (H 1 − H 1 ) and (H 2 − H 2 ) in a 50 wt % sulfuric acid solution. 8.16 The following data have been reported for the constant-pressure heat capacity of a benzene–carbon tetrachloride mixture at 20◦ C.23

wt % CCl4

CP (J/g ◦ C)

wt % CCl4

CP (J/g ◦ C)

0 10 20 30 40 50

1.7655 1.630 1.493 1.358 1.222 1.100

60 70 80 90 100

1.004 0.927 0.858 0.816 0.807

Critical Tables, Vol. 5, McGraw-Hill, New York (1929).

Problems 411 On a single graph plot the constant-pressure partial molar heat capacity for both benzene and carbon tetrachloride as a function of composition. 8.17 A 20 wt % solution of sulfuric acid in water is to be enriched to a 60 wt % sulfuric acid solution by adding pure sulfuric acid. a. How much pure sulfuric acid should be added? b. If the 20 wt % solution is available at 5◦ C, and the pure sulfuric acid at 50◦ C, how much heat will have to be removed to produce the 60 wt % solution at 70◦ C? How much heat will have to be added or removed to produce the 60 wt % solution at its boiling point? 8.18 Develop a procedure for determining the partial molar properties for each constituent in a three-component (ternary) mixture. In particular, what data would you want, and what would you do with the data? Based on your analysis, do you suppose there is much partial molar property data available for ternary and quaternary mixtures? 8.19 The partial molar enthalpies of species in simple binary mixtures can sometimes be approximated by the following expressions: H 1 = a1 + b1 x22

and H 2 = a2 + b2 x21

a. For these expressions show that b1 must equal b2 . b. Making use of the fact that lim θi = θi

xi →1

for any thermodynamic property θ, show that

Alcohol wt % 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

Density at 20◦ C mol % (kg m−3 ) × 10−3 Water 0.9982 0.9894 0.9819 0.9751 0.9686 0.9617 0.9538 0.9449 0.9352 0.9247 0.9138 0.9026 0.8911 0.8795 0.8677 0.8556 0.8434 0.8310 0.8180 0.8042 0.7893

Heat Evolved on Mixing at 17.33◦ C (kJ/mol of Ethanol) 0.042 0.092 0.167 0.251 0.335 0.423 0.519 0.636 0.757 0.946 1.201 1.507 1.925 2.478 3.218 4.269 5.821 7.801 9.818

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95

8.21 Using the information in Problems 7.13 and 8.20, estimate the heat of vaporization for the first bit of ethanol from ethanol-water solutions containing 25, 50, and 75 mol % ethanol and from a solution infinitely dilute in ethanol. How do these heats of vaporization compare with that for pure ethanol computed in Problem 7.13? Why is there a difference between the various heats of vaporization? 8.22 The volume of a binary mixture has been reported in the following polynomial form: V (T, P, x1 , x2 ) = x1 b1 + x2 b2 + x1 x2

n 

ai (x1 − x2 )i

i=0

a1 = H 1

a2 = H 2

and

ΔH mix = b1 x1 x2

8.20 A partial molar property of a component in a mixture may be either greater than or less than the corresponding pure-component molar property. Furthermore, the partial molar property may vary with composition in a complicated way. Show this to be the case by computing (a) the partial molar volumes and (b) the partial molar enthalpies of ethanol and water in an ethanol-water mixture. (The data that follow are from Volumes 3 and 5 of the International Critical Tables, McGraw-Hill, New York, 1929.)

a. What values should be used for b1 and b2 ? b. Derive, from the equation here, expressions for V 1 , ex V 2 , V ex 1 = V 1 − V 1 and V 2 = V 2 − V 2 . c. Derive, from the equation here, expressions for the partial molar excess volumes of each species at infinite dilution, that is, V ex 1 (T, P, x1 → 0) and V ex (T, P, x → 0) . 2 2 8.23 Prove the validity of Eqs. 8.7-4. 8.24 The definition of a partial molar property is  Mi =

∂(N M ) ∂Ni

 T,P,Nj=i

412 Chapter 8: The Thermodynamics of Multicomponent Mixtures It is tempting, but incorrect, to assume that this equation can be written as  Mi =

∂M ∂xi

 T,P

8.27 When water and n-propanol are isothermally mixed, heat may be either absorbed (Q > 0) or evolved (Q < 0), depending on the final composition of the mixture. Volume 5 of the International Critical Tables (McGraw-Hill, New York, 1929) gives the following data:

Prove that the correct result is  Mi = M +

∂M ∂xi



−



 xj

T,P,xk=i

∂M ∂xj



mol % Water

Q, kJ/mol of n-Propanol

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95

+0.042 +0.084 +0.121 +0.159 +0.197 +0.230 +0.243 +0.243 +0.209 +0.167 +0.084

T,P,xk=j

8.25 In some cases if pure liquid A and pure liquid B are mixed at constant temperature and pressure, two liquid phases are formed at equilibrium, one rich in species A and the other in species B . We have proved that the equilibrium state at constant T and P is a state of minimum Gibbs energy, and the Gibbs energy of a twophase mixture is the sum of the number of moles times the molar Gibbs energy for each phase. What would the molar Gibbs free energy versus mole fraction curve look like for this system if we could prevent phase separation from occurring? Identify the equilibrium compositions of the two phases on this diagram. The limit of stability of a single phase at constant temperature and pressure can be found from d2 G = 0 or 

∂2G ∂x21

 =0 T,P

Identify the limits of single-phase stability on the Gibbs energy versus mole fraction curve. 8.26 Mattingley and Fenby [J. Chem. Thermodyn. 7, 307 (1975)] have reported that the enthalpies of triethylamine-benzene solutions at 298.15 K are given by H mix − [xB H B + (1 − xB )H EA ] = xB (1 − xB ){1418 − 482.4(1 − 2xB ) + 187.4(1 − 2xB )3 }

where xB is the mole fraction of benzene and H mix , H B , and H EA are the molar enthalpies of the mixture, pure benzene, and pure triethylamine, respectively, with units of J/mol. a. Develop expressions for (H B − H B ) and (H EA − H EA ). b. Compute values for (H B − H B ) and (H EA − H EA ) at xB = 0.5. c. One mole of a 25 mol % benzene mixture is to be mixed with one mole of a 75 mol % benzene mixture at 298.15 K. How much heat must be added or removed for the process to be isothermal? [Note: H mix − xB H B − (1 − xB )H EA is the enthalpy change on mixing defined in Sec. 8.1.]

−0.038 −0.201 −0.431 −0.778 −1.335 −2.264 −4.110 −7.985

Plot (H W − H W ) and (H NP − H NP ) over the whole composition range. 8.28 The heat-of-mixing data of Featherstone and Dickinson [J. Chem. Thermodyn., 9, 75 (1977)] for the n-octanol + n-decane liquid mixture at atmospheric pressure is approximately fit by Δmix H = x1 x2 (A + B(x1 − x2 ))

J/mol

where A = −12 974 + 51.505T

and

B = +8782.8 − 34.129T

with T in K and x1 being the n-octanol mole fraction. a. Compute the difference between the partial molar and pure-component enthalpies of n-octanol and of n-decane at x1 = 0.5 and T = 300 K. b. Compute the difference between the partial molar and pure-component heat capacities of n-octanol and of n-decane at x1 = 0.5 and T = 300 K. c. An x1 = 0.2 solution and an x1 = 0.9 solution are to flow continuously into an isothermal

Problems 413 mixer in the mole ratio 2:1 at 300 K. Will heat have to be added or removed to keep the temperature of the solution leaving the mixer at 300 K? What will be the heat flow per mole of solution leaving the mixer? 8.29 Two streams containing pyridine and acetic acid at 25◦ C are mixed and fed into a heat exchanger. Due to the heat-of-mixing effect, it is desired to reduce the temperature after mixing to 25◦ C using a stream of chilled ethylene glycol as indicated in the diagram. Calculate the mass flow rate of ethylene glycol needed. The heat capacity of ethylene glycol at these conditions is approximately 2.8 kJ/(kg K), and the enthalpy change of mixing (Δmix H) is given below.

Pyridine 1 kmol/min T = 25°C

Ethylene glycol T = 25 °C

Acetic acid 1 kmol/min T = 25°C

Pyridine + acetic acid 2 kmol/min T = 25°C

Ethylene glycol T = 5°C

Data: Heat of mixing for pyridine (C5 H5 N) and acetic acid at 25◦ C [H. Kehlen, F. Herold and H.J. Rademacher, Z. Phys. Chem. (Leipzig), 261, 809 (1980)].

Pyridine Mole Fraction 0.0371 0.0716 0.1032 0.1340 0.1625 0.1896 0.2190 0.2494 0.2760 0.3006 0.3234 0.3461 0.3671 0.3874 0.3991

Δmix H

(J/mol) −1006 −1851 −2516 −3035 −3427 −3765 −4043 −4271 −4440 −4571 −4676 −4760 −4819 −4863 −4882

Pyridine Mole Fraction 0.4076 0.4235 0.4500 0.4786 0.5029 0.5307 0.5621 0.5968 0.6372 0.6747 0.7138 0.7578 0.8083 0.8654 0.9297

8.31 For the study of the oxidation of methane, an engineer devises the following set of possible reactions: CH4 + 2O2 → CO2 + 2H2 O CH4 + 32 O2 → CO + 2H2 O CO + H2 O → CO2 + H2 C + O2 → CO2

CO + 12 O2 → CO2 CH4 → C + 2H2 C + 12 O2 → CO H2 + 12 O2 → H2 O

How many independent chemical reactions are there in this system? 8.32 Calculate the standard heats and Gibbs energies of reaction at 25◦ C for the following reactions: a. N2 (g) + 3H2 (g) = 2NH3 (g) b. C3 H8 (g) = C2 H4 (g) + CH4 (g) c. CaCO3 (s) = CaO(s) + CO2 (g) d. 4CO(g) + 8H2 (g) = 3CH4 (g) + CO2 (g) + 2 H2 O (g)

8.33 Steele et al. [J. Phys. Chem. Soc., 96, 4731 (1992)] used bomb calorimetry to compute the standard enthalpy of combustion of solid buckminsterfullerene (C60 ) at 298.15 K to be 26 033 kJ/mol. Calculate the standard state ΔH of transition from graphite to buckminsterfullerene and its standard enthalpy of formation. 8.34 The following data are available for the isothermal heat of mixing of trichloromethane (1) and ethanol (2) at 30◦ C [reference: J. P. Shatas, M. M. Abbott, and H. C. Van Ness, J. Chem. Eng. Data, 20, 406 (1975)].

Δmix H

(J/mol)

x1 , Mole

Δmix H

x1 , Mole

Δmix H

−4880 −4857 −4855 −4833 −4765 −4669 −4496 −4253 −3950 −3547 −3160 −2702 −2152 −1524 −806

Fraction

(J/mol)

Fraction

(J/mol)

0.0120 0.0183 0.0340 0.0482 0.0736 0.1075 0.1709 0.1919 0.2301 0.2636 0.2681 0.2721 0.3073 0.3221 0.3486 0.3720 0.3983 0.4604 0.4854

−68.8 −101.3 −179.1 −244.4 −344.6 −451.1 −565.3 −581.0 −585.0 −566.1 −561.9 −557.8 −519.6 −508.0 −468.5 −424.4 −369.1 −197.1 −135.4

0.5137 0.5391 0.5858 0.6172 0.6547 0.7041 0.7519 0.7772 0.7995 0.8239 0.8520 0.8784 0.8963 0.9279 0.9532 0.9778 0.9860 0.9971

−66.1 −1.9 117.1 186.5 266.9 360.3 436.6 470.5 495.9 510.0 515.8 505.3 486.0 420.5 329.2 184.7 123.3 25.1

8.30 Use the data in problem 8.29 to compute the partial molar enthalpies of pyridine and acetic acid in their mixtures at 25◦ C over the whole composition range.

414 Chapter 8: The Thermodynamics of Multicomponent Mixtures Compute the partial molar enthalpies of trichloromethane and ethanol in their mixtures at 30◦ C over the whole composition range. 8.35 An equimolar mixture of nitrogen and acetylene enters a steady-flow reactor at 25◦ C and 1 bar of pressure. The only reaction occurring is N2 (g) + C2 H2 (g) = 2 HCN(g)

The product leaves the reactor at 600◦ C and contains 24.2 percent mole fraction of HCN. How much heat is supplied to the reactor per mole of HCN? 8.36 Using the data below, calculate the partial molar enthalpies of 1-propanol and water as a function of composition at both 25◦ C and 50◦ C.

25◦ C

Mole Fraction Propanol

(J/mol)

0.027 0.034 0.054 0.094 0.153 0.262 0.295 0.349 0.533 0.602 0.739

−223.16 −290.15 −329.50 −384.35 −275.07 −103.41 −81.22 −11.35 133.98 168.31 177.94

H ex

50◦ C

Mole Fraction Propanol

(J/mol)

0.031 0.043 0.082 0.098 0.206 0.369 0.466 0.587 0.707 0.872

−76.20 −121.84 −97.55 −52.75 125.60 370.53 435.43 473.11 460.55 238.23

H ex

data: V. P. Belousov, Vent. Leningrad Univ. Fiz., Khim, 16(1), 144 (1961). 8.37 Following are the slightly smoothed heat-of-mixing data of R. P. Rastogi, J. Nath, and J. Misra [J. Chem. Thermodyn., 3, 307 (1971)] for the system trichloromethane (component 1) and 1,2,4-trimethyl benzene at 35◦ C.

Δmix H x1

(kJ/mol)

0.2108 0.2834 0.3023 0.4285 0.4498 0.5504

−738 −900 −933 −1083 −1097 −1095

Δmix H x1

(kJ/mol)

0.5562 0.6001 0.6739 0.7725 0.8309

−1096 −1061 −976 −780 −622

a. From the information in this table, calculate the quantity (Δmix H)/(x1 x2 ) at each of the reported compositions. b. Compute the difference between each of the partial molar and pure-component molar enthalpies for this system at each composition. 8.38 We want to make a simplified estimate of the maximum amount of work that can be obtained from gasoline, which we will assume to be adequately represented by n-octane (C8 H18 ). The processes that occur in the cylinder of an automobile engine are that first the gasoline reacts to form a high-temperature, high-pressure combustion gas consisting of carbon dioxide, water, and the nitrogen initially present in the air (as well as other by-products that we will neglect), and then work is extracted from this combustion gas as its pressure and temperature are reduced. a. Assuming that n-octane (vaporized in the fuel injector) and a stoichiometric amount of air (21 vol. % oxygen, 79 vol. % nitrogen) initially at 1 bar and 25◦ C react to completion at constant volume, calculate the final temperature and pressure of the combustion gas, assuming further that there is no heat loss to the pistons and cylinders of the automobile engine. b. Calculate the final temperature of the combustion gas and the work that can be obtained from the gas if it is adiabatically expanded from the pressure found in part (a) to 1 bar. c. Calculate the maximum amount of additional work that can be obtained from the combustion gas as its temperature is isobarically lowered from the temperature found in part (b) to an exhaust temperature of 150◦ C. Data: Consider the gas to be ideal, in which case the partial molar enthalpies of the components are equal to their pure component molar enthalpies at the same temperature and pressure.

C8 H18 CO2 H2 O N2 O2

Δf H ◦ (kJ/mol)

CP J/(mol K)

−208.4 −393.5 −241.8 0.0 0.0

51.25 39.75 32.43 33.45

Note that for the calculations here the molar heat capacity of the mixture is just the mole fraction– weighted sum of the heat capacities of the individual components, CP,mixture =

 i

xi CP,i

Problems 415 Below is a diagram of the three steps in the process. Constant volume

Adiabatic expansion

1

2

Isobaric cooling 3

4

The hexafluoride gas produced from naturally occurring uranium ores contains 0.72 mole percent U235 F6 the remainder being U238 F6 . It is desired to recover 25% of the U235 F6 in a gas stream enriched to 2 mole percent. What is the minimum heat and work flows necessary to accomplish this enrichment for a 1000 mole gaseous stream at ambient conditions? 8.42 The following is a modified van der Waals equation with an improved temperature dependence P =

8.39 “Duhem’s theorem” states that for any number of components, phases, and chemical reactions in a closed system, if the initial amounts of all the species are specified, the further specification of two independent state variables completely fixes the state of the system. Prove that this theorem is either valid or invalid. 8.40 Calculate the minimum work required to separate air (79 mole % nitrogen) into pure oxygen and nitrogen assuming an isothermal, steady flow process at 300 K. The inlet air pressure is 10 bar and each stream is to exit at 10 bar and 300 K. 8.41 To obtain uranium 235 (U235 ) for use in nuclear power plants, uranium containing ore containing U238 and U235 is crushed, subjected to a solvent extraction process, and then reacted with fluorine to produce uranium hexafluoride, which is a gas at ambient conditions the U235 is then separated from the U238 by various multi-stage methods that take advantage of the small mass difference between U235 F6 and U238 F6 , such as gas centrifuges or gaseous diffusion.

RT a − 2 √ V mix V mix T

The usual van der Waals one-fluid mixing rules are used with this equation of state. Develop an expression for the mixture constant volume heat capacity of a mixure at elevated pressures in terms of the mixture volume V mix , the mole fractions, temperature, the equation of state parameters, and each of the pure component ideal gas heat capacities. 8.43 The volume change on mixing in cm3 /mol for ethanol(1) + methyl butyl ether(2) mixtures at 25◦ C is given by the following equation Δmix V (x) = x1 x2 [−1.026 + 0.220(x1 − x2 )]

Given that V 1 = 58.63 cm3 /mol and V 2 = 118.46 cm3 /mol, calculate the following when 750 cm3 of pure ethanol is mixed with 1500 cm3 of methyl butyl ether at 25◦ C: a. The volume of the solution if an ideal mixture was formed. b. The actual volume of the solution. c. The partial molar volumes of both components in the mixture above.

Chapter

9 Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture The most important ingredient in the thermodynamic analysis of mixtures is information about the partial molar properties of each species in the mixture. The partial molar Gibbs energy is of special interest since it is needed in the study of phase and chemical equilibria, which are considered in great detail in the following chapters. For many mixtures the partial molar property information needed for equilibrium calculations is not available. Consequently, in this chapter we consider methods for estimating the partial molar Gibbs energy and its equivalent, the fugacity. Before proceeding with this detailed study, we will consider two very simple cases, a mixture of ideal gases (Sec. 9.1) and the ideal mixture (Sec. 9.3), for which the partial molar properties are simply related to the pure component properties.

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 9 The goals of this chapter are for the student to: • Be able to distinguish between ideal mixtures and nonideal mixtures (Sec. 9.3) • Understand the concepts of excess properties and activity coefficients (Secs. 9.3

and 5)

• Be able to calculate the fugacity of a component in a vapor mixture and in a liquid

mixture if an equation of state is available (Sec. 9.4)

• Be able to calculate the fugacity of a component in a vapor mixture and in a liquid

mixture if an equation of state is not available (Sec. 9.4)

• Be able to use correlative activity coefficient models with experimental data

(Sec. 9.5)

• Be able to use predictive activity coefficient models when there are no experimen-

tal data (Sec. 9.6)

• Be able to compute the fugacity of a species in a mixture when, as a pure compo-

nent it would be a supercritical gas or a solid (Sec. 9.7)

• Be able to use different standard states in thermodynamic calculations

(Secs. 9.7–9.9)

416

• Be able to do calculations involving electrolyte solutions (Sec. 9.10)

9.1 The Ideal Gas Mixture 417

NOTATION INTRODUCED IN THIS CHAPTER θIGM Pi Δmix θIGM θIGM θIG θIG IGM θi θIM θIM θIM i Δmix θIM θmix θex γi (T, P, x) fi φi θi δi φi ψi Xm Φm Ψm Hi Mi Hi I γi∗ γi γ± M± z + , z− ν+ , ν−

Property θ of an ideal gas mixture Partial pressure of species i in a mixture = yi P (kPa) Change in molar property θ on forming an ideal gas mixture Molar property θ of an ideal gas mixture Pure-component ideal gas property θ Pure-component molar property θ in an ideal gas Partial molar property θ of species i in an ideal gas mixture Property θ of an ideal mixture Molar property θ of an ideal mixture Partial molar property θ of species i in an ideal mixture Change in molar property θ on forming an ideal mixture Molar property θ of a mixture Excess molar property = θmix (T, P, x) − θIM mix (T, P, x) Activity coefficient of species i at T , P and x Fugacity of species i in a mixture (kPa) Fugacity coefficient of species i in a mixture Partial molar property of species i Solubility parameter of species i Volume fraction of species i in a mixture Surface area fraction of species i in a mixture Mole fraction of functional group m in a mixture Volume fraction of functional group m in a mixture Surface area fraction of functional group m in a mixture Henry’s law constant of species i in a mixture (kPa) Molality of species i (moles of species i per 1000 g of solvent) Molality-based Henry’s law constant of species i (kPa/M) Ionic strength (M) Activity coefficient of species i in Henry’s law based on mole fraction Activity coefficient of species i in Henry’s law based on molality Mean ionic activity coefficient in an electrolyte solution Mean ionic activity molality in an electrolyte solution Charge on cation and anion, respectively, in an electrolyte solution Numbers of cations and anions, respectively, from the ionization of an electrolyte

9.1 THE IDEAL GAS MIXTURE As defined in Chapter 3, the ideal gas is a gas whose volumetric equation of state at all temperatures, pressures, and densities is

P V = NRT

or P V = RT

(9.1-1)

and whose internal energy is a function of temperature only. By the methods of statistical mechanics, one can show that such behavior occurs when a gas is sufficiently dilute that interactions between the molecules make a negligible contribution to the

418 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture total energy of the system. That is, a gas is ideal when each molecule in the gas is (energetically) unaware of the presence of other molecules. An ideal gas mixture is a gas mixture with a density so low that its molecules do not appreciably interact. In this case the volumetric equation of state of the gas mixture will also be of the form of Eq. 9.1-1, and its internal energy will merely be the sum of the internal energies of each of the constituent ideal gases, and thus a function of temperature and mole number only. That is, Hypothetical ideal gas mixture

P V IGM = (N1 + N2 + · · · )RT =

 C 

 Nj RT = N RT

(9.1-2)

j=1

and

U

IGM

(T, N ) =

C 

Nj U IG j (T )

(9.1-3)

j=1

Here we have used the superscripts IG and IGM to indicate properties of the ideal gas and the ideal gas mixture, respectively, and taken pressure and temperature to be the independent variables. From Eq. 8.1-12 it then follows that for the ideal gas mixture Partial molar properties for the ideal gas mixtures

U IGM (T, x) i

 ∂U IGM (T, N )  =  ∂Ni T,P,N

j=i

 ∂  = ∂Ni T,P,N

C 

IG Nj U IG j (T ) = U i (T )

j=i j=1

(9.1-4) and

V

IGM (T, P, x) i

 ∂V IGM (T, P, N )  =  ∂Ni T,P,N =

j=i

 ∂  = ∂Ni T,P,N

C  j=i

j

Nj

RT P

RT = V IG i (T, P ) P (9.1-5)

Here and throughout the rest of this book, we use the notation x to represent all the mole fractions. That is, any property written as θ(T, P, x) is meant to indicate that θ is a function of temperature, pressure, and all the mole fractions (x1 , x2 , x3 , . . . , etc.). Equation 9.1-4 indicates that the partial molar internal energy of species i in an ideal gas mixture at a given temperature is equal to the pure component molar internal energy of that component as an ideal gas at the same temperature. Similarly, Eq. 9.1-5 establishes that the partial molar volume of species i in an ideal gas mixture at a given temperature and pressure is identical to the molar volume of the pure component as an ideal gas at that temperature and pressure. Consider now the process of forming an ideal gas mixture at temperature T and pressure P from a collection of pure ideal gases, all at that temperature and pressure. From the discussion here it is clear that for each species V i (T, P, x) = V i (T, P ) and U i (T, x) = U i (T ). It then follows immediately from equations such as Eqs. 8.1-14

9.1 The Ideal Gas Mixture 419 and 8.1-15 that Δmix V = 0 and Δmix U = 0 for this process. Also, Δmix H ≡ Δmix U + P Δmix V = 0. The partial pressure of species i in a gas mixture, denoted by Pi , is defined for both ideal and nonideal gas mixtures to be the product of the mole fraction of species i and total pressure P , that is,

Pi = xi P

(9.1-6)

For the ideal gas mixture,

PiIGM (N, V, T, x)

Ni Ni = C P = C   Nj Nj j=i

=



C  j=1

RT Nj V



j=i

Ni RT = P IG (Ni , V, T ) V

Thus, for the ideal gas mixture, the partial pressure of species i is equal to the pressure that would be exerted if the same number of moles of that species, Ni , alone were contained in the same volume V and maintained at the same temperature T as the mixture. Since there is no energy of interaction in an ideal gas mixture, the effect on each species of forming an ideal gas mixture at constant temperature and total pressure is equivalent to reducing the pressure from P to its partial pressure in the mixture Pi . Alternatively, the effect is equivalent to expanding each gas from its initial volume Vi = Ni RT /P to the volume of the mixture V = i Ni RT /P . Thus, from Eqs. 6.4-2 and 6.4-3, we have

S IGM (T, P, x) − S IG i i (T, P ) = −R ln or

S IGM (T, V, x) i

−

S IG i (T, Vi )

Pi = −R ln xi P

(9.1-7)

 V Nj RT /P = R ln = R ln = −R ln xi Vi Ni RT /P

Consequently,

Δmix S

IGM

=

C 

Ni [S IGM (T, P, x) i

−

S IG i (T, P )]

= −R

i=1

C 

Ni ln xi

i=1

and

C

Δmix S

IGM

 Δmix S IGM = −R = xi ln xi N

(9.1-8)

i=1

The statistical mechanical interpretation of Eq. 9.1-8 is that an ideal gas mixture is a completely mixed or random mixture. This is discussed in Appendix A9.1. Using the energy, volume, and entropy changes on mixing given here, one can easily compute the other thermodynamic properties of an ideal gas mixture (Problem 9.1). The results are given in Table 9.1-1. Of particular interest are the expressions for

420 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Properties of an Ideal Gas Mixture (Mixing at Constant T and P ) U IGM (T, x) = U IG i i (T )

Δmix U IGM = 0

H IGM (T, x) = H IG i i (T )

Δmix H IGM = 0

Internal energy Enthalpy Volume

V

Entropy

S IGM (T, P, x) = S IG i i (T, P ) − R ln xi

IGM (T, x) i

=V

IG i (T )

Δmix V IGM = 0 Δmix S IGM = −R

C 

xi ln xi

i=1

GIGM (T, P, x) = GIG i i (T, P ) + RT ln xi

Gibbs energy

Δmix GIGM = RT

C 

xi ln xi

i=1

Helmholtz energy

AIGM (T, P, x) = AiIG (T, P ) + RT ln xi i

Δmix AIGM = RT

C 

xi ln xi

i=1

U IGM (T, x) =

Internal energy Enthalpy Volume Entropy Gibbs energy Helmholtz energy



xi U IG i (T )  H IGM (T, P, x) = xi H IG i (T, P )  V IGM (T, P, x) = xi V IG i (T, P )   IGM IG S (T, P, x) = xi S i (T, P ) − R xi ln xi   GIGM (T, P, x) = xi GIG xi ln xi i (T, P ) + RT   IGM IG A (T, P, x) = xi Ai (T, P ) + RT xi ln xi

GIGM (T, P, x) and Δmix GIGM : i GIGM (T, P, x) = H IGM (T, P, x) − T S IGM (T, P, x) i i i IG = H IG i (T, P ) − T (S i (T, P ) − R ln xi )

(9.1-9)

= GIG i (T, P ) + RT ln xi Δmix GIGM =

C  i=1

= RT

xi {GIGM (T, P, x) − GIG i i (T, P )} 

(9.1-10)

xi ln xi

From Eqs. 9.1-1 to 9.1-10, we then have for an ideal gas mixture

U IGM (T, x) = V IGM (T, P, x) = H IGM (T, P, x) = S IGM (T, P, x) =

   

xi U IG i (T )

(9.1-11)

xi V IG i (T, P )

(9.1-12)

xi H IG i (T, P ) xi S IG i (T, P ) − R



xi ln xi  GIGM (T, P, x) = xi GIG xi ln xi i (T, P ) + RT   AIGM (T, P, x) = xi AIG xi ln xi i (T, P ) + RT 

(9.1-13) (9.1-14) (9.1-15) (9.1-16)

9.2 The Partial Molar Gibbs Energy and Fugacity 421 Illustration 9.1-1 Examination of Whether a Gas Mixture Is an Ideal Gas Mixture The following data are available for the compressibility factor Z = P V /RT of nitrogen-butane mixtures at 444.3 K. Mole Fraction Butane

P = 13.79 MPa Zmix

P = 68.95 MPa Zmix

0 0.10 0.30 0.50 0.70 0.90 1.0

1.06 1.04 0.96 0.81 0.64 0.52 0.51

1.47 1.51 1.61 1.72 1.83 1.95 1.98

Is the nitrogen-butane mixture an ideal gas mixture at these conditions?

Solution From Eq. 9.1-2 we have, for a gas mixture to be an ideal gas mixture, that IGM P V IGM P Vmix mix = = Zmix = 1 N RT RT

at all temperatures, pressures, and compositions. Clearly, the nitrogen-butane mixture is not an ideal gas mixture at the temperature and two pressures in this illustration.

9.2 THE PARTIAL MOLAR GIBBS ENERGY AND FUGACITY Unfortunately, very few mixtures are ideal gas mixtures, so general methods must be developed for estimating the thermodynamic properties of real mixtures. In the discussion of phase equilibrium in a pure fluid of Sec. 7.4, the fugacity function was especially useful; the same is true for mixtures. Therefore, in an analogous fashion to the derivation in Sec. 7.4, we start from

dG = −S dT + V dP +

C 

Gi dNi

(8.2-1)

i=1

and, using the commutative property of second derivatives of the thermodynamic functions (cf. Eq. 8.1-3),

 ∂  ∂Ni T,P,N

j=i

∂G ∂T

P,Nj



∂G ∂  = ∂T P,Nj ∂Ni T,P,N

T,Nj



∂G ∂  = ∂P T,Nj ∂Ni T,P,N



j=i

and

 ∂  ∂Ni T,P,N

j=i

∂G ∂P



j=i

422 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture to obtain the two equations

Si = −

∂Gi ∂T

(9.2-1) P,Nj

and

Vi =

∂Gi ∂P

(9.2-2) T,Nj

As in the pure component case, the second of these equations is more useful than the first, and leads to the relation  P2 Gi (T1 , P2 , x) − Gi (T1 , P1 , x) = V i dP P1

In analogy with Eq. 7.4-6, the fugacity of species i in a mixture, denoted by f i , is defined with reference to the ideal gas mixture as follows: Fugacity of a species in a mixture

Fugacity coefficient of a species in a mixture



Gi (T, P, x) − GIGM (T, P, x) i f i (T, P, x) = xi P exp RT

 P 1 IG = xi P exp (V i − V i ) dP RT 0

Gi (T, P, x) − GIG i (T, P ) = P exp RT

(9.2-3a)

so that f i → xi P ≡ Pi as P → 0. Here Pi is the partial pressure of species i, and the superscript IGM indicates an ideal gas mixture property. The fact that as the pressure goes to zero all mixtures become ideal gas mixtures (just as all pure fluids become ideal gases) is embedded in this definition. Also, the fugacity coefficient for a component in a mixture, φi , is defined as

fi Gi (T, P, x) − GIGM (T, P, x) i φi = = exp xi P RT (9.2-3b)

 P 1 IGM (V i − V i ) dP = exp RT 0 The multicomponent analogue of Eq. 7.4-9a, obtained by differentiating ln f i with respect to pressure at a constant temperature and composition, is



∂ ln f i ∂Gi RT = =Vi (9.2-4) ∂P ∂P T,x T,x To relate the fugacity of pure component i to the fugacity of component i in a mixture, we first subtract Eq. 7.4-9a from Eq. 9.2-4 and then integrate between P = 0 and the pressure of interest P , to obtain

9.2 The Partial Molar Gibbs Energy and Fugacity 423

RT ln

f i (T, P, x) f i (T, P → 0, x)

− RT ln

fi (T, P ) fi (T, P → 0)



P

= P →0

(V i − V i ) dP (9.2-5)

We now use the fact that as P → 0, f i → xi P and fi → P , to obtain

 P f i (T, P, x) = RT ln (V i − V i ) dP xi fi (T, P ) 0

(9.2-6)

Therefore, for a mixture in which the pure component and partial molar volumes are identical [i.e., V i (T, P, x) = xi Vi (T, P ) at all conditions], the fugacity of each species in the mixture is equal to its mole fraction times its pure-component fugacity evaluated at the same temperature and pressure as the mixture f i (T, P, x) = xi fi (T, P ). However, if, as is generally the case, V i = V i , then f i and fi are related through the integral over all pressures of the difference between the species partial molar and pure-component molar volumes. The temperature dependence of the fugacity f i (actually, the fugacity coefficient φi = f i /xi P ) can be gotten by differentiating Eq. 9.2-3 with respect to temperature at constant pressure and composition,



∂ ln(f i /xi P ) ∂T



(Gi − GIG 1 i ) =− + 2 RT RT P, x



) ∂(Gi − GIGM i ∂T

 (9.2-7) P, x

and then using dG = V dP − S dT and G = H − T S , to obtain



   ∂ ln f i xi P ∂T



P, x

∂ ln φi = ∂T





H i − H IGM i =− 2 RT P, x

 (9.2-8)

It is useful to have an expression for the change in partial molar Gibbs energy of a species between two states of the same temperature and pressure, but of differing composition. To derive such an equation, we start by writing

ΔGi = Gi (T, P, xII ) − Gi (T, P, xI ) where the superscripts I and II denote phases of different composition. Now substituting the logarithm of Eq. 9.2-3,

Gi (T, P, x) =

GIGM (T, P, x) i

+ RT ln

f i (T, P, x) xi P

= GIGM (T, P, x) + RT ln φi (T, P, x) i and using Eq. 9.1-9,

GIGM (T, P, x) = GIG i i (T, P ) + RT ln xi

(9.2-9)

424 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture yields

f i (T, P, xII ) Gi (T, P, x ) − Gi (T, P, x ) = + RT ln xII i P

f i (T, P, xI ) IGM I − Gi (T, P, x ) − RT ln xIi P

f i (T, P, xII ) IG II = Gi (T, P ) + RT ln xi + RT ln xII i P

f i (T, P, xI ) I − GIG (T, P ) − RT ln x − RT ln i i xIi P

II

I

GIGM (T, P, xII ) i

or, finally,

Gi (T, P, xII )−Gi (T, P, xI ) = RT ln

f i (T, P, xII ) f i (T, P, xI )

= RT ln

II xII i φi (T, P, x ) xIi φi (T, P, xI ) (9.2-10a)

A special case of this equation is when state I is a pure component:

Gi (T, P, x) − Gi (T, P ) = RT ln

f i (T, P, x) fi (T, P )

= RT ln

xi φi (T, P, x) φi (T, P )

(9.2-10b) The fugacity function has been introduced because its relation to the Gibbs energy makes it useful in phase equilibrium calculations. The present criterion for equilibrium between two phases is that GIi = GII i for all species i, with the restriction that the temperature and pressure be constant and equal in both phases. Using Eqs. 9.2-10a and b and the equality of partial molar Gibbs free energies yields Criterion for phase equilibrium

f Ii = f i (T, P, xI ) = f i (T, P, xII ) = f II i

(9.2-11)

Therefore, at equilibrium, the fugacity of each species must be the same in the two phases. Since this result follows directly from Eq. 8.7-10, it may be substituted for it. Furthermore, since we can make estimates for the fugacity of a species in a mixture in a more direct fashion than for partial molar Gibbs energies, it is more convenient to use Eq. 9.2-11 as the basis for phase equilibrium calculations. In analogy with Eqs. 7.4-6 and 9.2-3, the overall fugacity of a mixture f is defined by the relation

G(T, P, x) − GIGM (T, P, x) f = P exp (9.2-12) RT

9.3 Ideal Mixture and Excess Mixture Properties 425 Note that the fugacity of species i in a mixture, f i , as defined by Eq. 9.2-3 is not a partial molar fugacity, that is,

∂(N f ) f i = ∂Ni T,P,N j=i

Finally, note that Eq. 7.4-6, for computing the fugacity of a pure component, and Eq. 9.2-3, for computing the fugacity of a species in a mixture, both require that the volumetric equation of state be solved explicitly for the volume in terms of the temperature and pressure. However, as was discussed in Sec. 7.4, equations of state are usually pressure explicit (that is, easily solved for pressure as a function of temperature and volume, and not vice versa; see Eqs. 6.4-1 though 6.4-3), so that calculations based on Eqs. 7.4-6 and 9.2-3 can be difficult. Starting from Eq. 9.2-3, using Eq. 6.4-25 in the form 1 P P P dP = d(P V ) − dV = dZ − dV V V Z V and the triple-product rule (Eq. 8.1-6a),





∂V ∂P ∂Ni ∂Ni T,P,N ∂V T,Nj ∂P T,V,N j=i

in the form



∂V ∂Ni

= −1

j=i



dP = −

T,P,Nj=i

∂P ∂Ni

dV T,V,Nj=i

we obtain for the fugacity (actually the fugacity coefficient) of a species in a mixture Fugacity coefficient for a species in a mixture from an equation of state

1 f i (T, P, x) ln φi = ln = xi P RT



V =ZRT /P

V =∞



RT −N V



∂P ∂Ni



 dV − ln Z

T,V,Nj=i

(9.2-13) (see Problem 9.24), which is to be compared with the expression we previously obtained for a pure fluid,

 V =ZRT /P f (T, P ) RT 1 ln φ = ln − P dV − ln Z + (Z − 1) (7.4-8) = P RT V =∞ V for the fugacity of a pure component. Equation 9.2-13 is especially useful for computing the fugacity of a species in a mixture from a pressure-explicit equation of state, as we will see in Sec. 9.4.

9.3 IDEAL MIXTURE AND EXCESS MIXTURE PROPERTIES The estimation of the thermodynamic properties of a real fluid or fluid mixture in the absence of direct experimental data is a very complicated problem involving detailed spectroscopic, structural, and interaction potential data and the use of

426 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture statistical mechanics. Such a calculation is beyond the scope of this book and would not be of interest to most engineers. Instead, we use procedures for estimating mixture thermodynamic properties that are far simpler than starting from “first principles.” In particular, we will either use equations of state or, as in this section, choose a state for each system in which the thermodynamic properties are reasonably well known and then try to estimate how the departure of the real system from the chosen reference state affects the system properties. This last procedure is philosophically similar to the method used in Chapter 6, where the properties of real fluids were computed as the sum of an ideal gas contribution plus the departure from ideal gas behavior. Clearly, the accuracy of a property estimation technique based on such a procedure increases as the difference between the reference state and actual state of the system diminishes. Therefore, choosing the reference state to be an ideal gas mixture at the same temperature and composition as the mixture under consideration is not very satisfactory because the reference state and actual state, particularly for liquids, may be too dissimilar. It is in this context that we introduce the concept of an ideal mixture. An ideal mixture, which may be either a gaseous or liquid mixture, is defined to be a mixture in which

Ideal mixture

H IM i (T, P, x) = H i (T, P )

(9.3-1a)

V IM i (T, P, x) = V i (T, P )

(9.3-1b)

and

for all temperatures, pressures, and compositions (the superscript IM indicates an ideal mixture property). Note that unlike the ideal gas mixture, here neither H i nor V i is an ideal gas property. From Eqs. 8.1-14 and 8.1-15 it is evident that for such a mixture

Δmix V

IM

(T, P, x) =

C 

  Ni V IM i (T, P, x) − V i (T, P ) = 0

(9.3-2a)

  Ni H IM i (T, P, x) − H i (T, P ) = 0

(9.3-2b)

i=1

and

Δmix H

IM

(T, P, x) =

C  i=1

so that there are no volume or enthalpy changes on the formation of an ideal mixture =Vi from its pure components at the same temperature and pressure. Also, since V IM i at all temperatures, pressures, and compositions, we have, from Eq. 9.2-6,

f IM i (T, P, x) = xi fi (T, P )

(9.3-3)

Next, using Eqs. 9.3-1, it is easily established that

U IM i (T, P, x) = U i (T, P ) and from Eq. 9.2-10b,

(9.3-4a)

9.3 Ideal Mixture and Excess Mixture Properties 427

f IM i (T, P, x) − Gi (T, P ) = RT ln fi (T, P )

xi fi (T, P ) = RT ln = RT ln xi fi (T, P )

GIM i (T, P, x)

Consequently, we obtain

GIM i (T, P, x) = Gi (T, P ) + RT ln xi AIM i (T, P, x) = Ai (T, P ) + RT ln xi

(9.3-4b)

S IM i (T, P, x) = S i (T, P ) − R ln xi It then follows from Eqs. 9.3-1 to 9.3-4 that for an ideal mixture  U IM (T, P, x) = xi U i (T, P )  V IM (T, P, x) = xi V i (T, P )  H IM (T, P, x) = xi H i (T, P )   S IM (T, P, x) = xi S i (T, P ) − R xi ln xi   GIM (T, P, x) = xi Gi (T, P ) + RT xi ln xi   AIM (T, P, x) = xi Ai (T, P ) + RT xi ln xi

(9.3-5a) (9.3-5b) (9.3-5c) (9.3-5d) (9.3-5e) (9.3-5f)

Illustration 9.3-1 Determining Whether a Mixture Is an Ideal Mixture Even Though It Is Not an Ideal Gas Mixture Determine whether the nitrogen-butane mixture of Illustration 9.1-1 is an ideal mixture.

Solution For the nitrogen-butane mixture to be an ideal mixture, from Eq. 9.3-5b at all temperatures, pressures, and compositions, V IM (T, P, x) =



xi V i (T, P )

or, alternatively, after multiplying by P/RT and recognizing that Z = P V /RT , IM Zmix (T, P, x) =



(*)

xi Zi (T, P )

IM Below we tabulate Zmix and Zmix at the conditions in Illustration 9.1-1.

Mole Fraction Butane 0 0.1 0.3 0.5 0.7 0.9 1.0

P = 13.79 MPa

P = 68.95 MPa

Zmix

IM Zmix

Zmix

IM Zmix

1.06 1.04 0.96 0.81 0.64 0.52 0.51

1.06 1.01 0.90 0.79 0.68 0.57 0.51

1.47 1.51 1.61 1.72 1.83 1.95 1.98

1.47 1.52 1.62 1.73 1.83 1.93 1.98

428 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Clearly, Eq. (*) is much closer to being satisfied with the higher-pressure data than with the lower-pressure data. However, for an ideal mixture, this equation must be satisfied at all pressures, temperatures, and compositions. Therefore, we conclude that the nitrogen-butane mixture is not quite an ideal mixture at the conditions considered here, and it certainly is not an ideal gas mixture based on Illustration 9.1-1.

Equations 9.3-3 to 9.3-5 resemble those obtained in Sec. 9.1 for the ideal gas mixture. There is an important difference, however. In the present case we are considering an ideal mixture of fluids that are not ideal gases, so each of the pure-component properties here will not be an ideal gas property, but rather a real fluid property that must either be measured or computed using the techniques described in Chapter 6. Thus, the molar volume V i is not equal to RT /P , and the fugacity of each species is not equal to the pressure. Also, from Eq. 9.3-3,

φIM i (T, P, x) = φi (T, P ) and since all derivatives at constant composition must be equal,     ∂ ln(f IM ∂ ln(fi /P ) i /xi P ) = ∂T ∂T P, x P

(9.3-6)

Note that an ideal mixture identically satisfies the Gibbs-Duhem equation (Eq. 8.2-12b):

S IM dT − V IM dP +



xi dGIM i    = xi S IM xi d(Gi + RT ln xi ) xi V IM i dT − i dP +     xi ln xi dT − xi dGi xi V i dP + = xi S i dT − R   +R xi ln xi dT + RT xi d ln xi   = xi (S i dT − V i dP + dGi ) + RT dxi ≡0



(9.3-7)  xi = 1, so that d xi =

since dG = V dP − S dT for a pure component, and  dxi = d(1) = 0. Any property change on mixing Δmix θ of a real mixture can be written in terms of the analogous property of an ideal mixture plus an additional term as follows:

Δmix θ(T, P, x) = Δmix θIM (T, P, x) + [Δmix θ(T, P, x) − Δmix θIM (T, P, x)] = Δmix θIM + θex where Excess mixing property

θex = Δmix θ(T, P, x)−Δmix θIM (T, P, x) =

 i

  xi θi − xi θIM = xi (θi −θIM i i ) i

i

(9.3-8)

9.3 Ideal Mixture and Excess Mixture Properties 429 is the excess mixing property, that is, the change in θ that occurs on mixing at constant temperature and pressure in addition to that which would occur if an ideal mixture were formed. Excess properties θex are generally complicated, nonlinear functions of the composition, temperature, and pressure, and usually must be obtained from experiment. The hope is, however, that the excess properties will be small (particularly when θ is the Gibbs energy or entropy) compared with Δmix θIM so that even an approximate theory for θex may be sufficient to compute Δmix θ with reasonable accuracy. Table 9.3-1 also contains a list of excess thermodynamic properties that will be of interest in this book. Finally, in analogy with Eq. 6.1-12, we define a partial molar excess quantity by the relation    ∂(N θex )  ∂  ex θi = = Nk (θk − θIM k )   ∂Ni T,P,N ∂Ni T,P,N j=i

j=i

k

which reduces to IM θex i = θi − θi

(9.3-9)

since

 k

Nk

 ∂  (θk − θIM =0 k ) T,P,N j=i ∂Ni

by the Gibbs-Duhem equation (Eq. 8.2-11). For the case in which θi is equal to the partial molar Gibbs energy, we have IM IGM Gex ) + (GIGM − GIM i = (Gi − Gi ) = (Gi − Gi i i )

= (Gi − GIGM ) + (GIG i i + RT ln xi − Gi − RT ln xi ) ) + (GIG = (Gi − GIGM i i − Gi ) (9.3-10) fi fi − RT ln xi P P

 P fi φi = RT ln [V i − V i ] dP = RT ln = xi fi φi 0 If equations of state (or specific volume data) are available for both the pure fluid and the mixture, the integral can be evaluated. However, for liquid mixtures not describable by an equation of state, common thermodynamic notation is to define an activity coefficient γi (T, P, x), which is a function of temperature, pressure, and composition, by the equation

= RT ln

Definition of the activity coefficient

f Li (T, P, x) = xi γi (T, P, x)fiL (T, P )

(9.3-11)

where the superscript L has been used to denote the liquid phase. Alternatively,

RT ln γi (T, P, x) =

Gex i

=

∂N Gex ∂Ni

(9.3-12) T,P,Nj=i

430 Helmholtz energy

Gibbs energy

i=1

i=1

xi Gi + RT

i=1

i=1

C 

xi ln xi

AIM = i=1

C 

xi Ai + RT

i=1

i=1

C 

xi ln xi

AIM = Ai + RT ln xi ; i C  xi ln xi ; Δmix AIM = RT

GIM =

C 

GIM = Gi + RT ln xi ; i C  xi ln xi ; Δmix GIM = RT

i=1

xi S i − R xi ln xi

S IM = S i − R ln xi ; i C  xi ln xi ; Δmix S IM = −R

Entropy

S IM =

H IM = H i; i

Enthalpy

C 

Δmix H IM = 0;

U IM = U i; i

Internal energy

i=1

Δmix U IM = 0;

V IM = V i; i

Volume

C 

Δmix V IM = 0;

Ideal Mixtures

Property

H IM =

U IM =

V IM =

i=1

H ex = Δmix H;

xi ln xi + S ex

A=

i=1

C 

i=1

xi Gi + RT

xi Ai + RT

i=1

C 

xi ln xi + Gex

xi ln xi + Aex

i=1

i=1

C 

i=1

i=1

C 

Aex i = Ai − Ai − RT ln xi ; C  Aex = Δmix A − RT xi ln xi ;

G=

C 

i=1

xi S i − R

i=1

Gex i = Gi − Gi − RT ln xi ; C  Gex = Δmix G − RT xi ln xi ;

S=

C 

S ex i = S i − S i + R ln xi ; C  S ex = Δmix S + R xi ln xi ;

H ex i = H i − H i;

H=

i=1

C 

xi H i

C 

U =

i=1

U ex = Δmix U ;

C 

i=1

U ex i = U i − U i;

V = C 

xi U i

V ex = Δmix V ;

C 

V ex i = V i − V i; i=1

xi V i

Real Mixtures

i=1

C 

Table 9.3-1 Mixture Thermodynamic Properties at Constant T and P

xi H i + Δmix H

xi U i + Δmix U

xi V i + Δmix V

9.3 Ideal Mixture and Excess Mixture Properties 431 [Note that the fugacity of the pure liquid, fiL (T, P ), in Eq. 9.3-11 can be found from the methods of Sec. 7.4b.] As will be seen in Chapters 10 to 12, the calculation of the activity coefficient for each species in a mixture is an important step in many phase equilibrium calculations. Therefore, much of this chapter deals with models (equations) for Gex and activity coefficients. Illustration 9.3-2 Analyzing the Gibbs Energy of a Mixture to Determine Whether It Is an Ideal Mixture Experimentally (as will be described in Sec. 10.2) it has been found that the Gibbs energy for a certain binary mixture has the form Gmix (T, P, x) =

2 

2 

xi Gi (T, P ) + RT

i=1

xi ln xi + ax1 x2

(**)

i=1

where a is a constant. Determine whether this mixture is an ideal mixture.

Solution For an ideal mixture Eqs. 9.3-1a and b must be satisfied at all conditions. To see if this is so, we start with the relation between the Gibbs energy and volume and use the equation above to obtain  V mix =

∂Gmix ∂P

 = T,N

2  i=1

 xi

∂Gi ∂P

 =



xi V i (T, P )

T,N

Therefore,

V 1 (T, P ) =

 ∂(N V mix )  ∂ (N1 V 1 (T, P ) + N2 V 2 (T, P ))T,N2 = V 1 (T, P ) =  ∂N1 ∂N1 T,N2

Similarly, V 2 (T, P ) = V 2 (T, P ), so Eq. 9.3-1b is satisfied. We now move on to Eq. 9.3-1a by starting with      G 1 1 ∂G ∂  G (H − T S) H − 2 = (−S) − =− 2 = ∂T P T T ∂T P T T T2 T

or H = −T

2

   G ∂  ∂T P T

Therefore,     a ∂   Gi ln x + x x + R x x i i i 1 2 ∂T P T T      H a xi − 2i − 2 x1 x2 = xi H i + ax1 x2 = −T 2 T T

H mix = −T 2

432 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Then     ∂  ∂  aN1 N2 N N H = H + N H + 1 1 2 2 mix ∂N1 T,P,N2 ∂N1 T,P,N2 N 1 + N2   N2 N1 N2 − = H 1 (T, P ) + a N 1 + N2 (N1 + N2 )2

H 1 (T, P, x) =

= H 1 (T, P ) + a(x2 − x1 x2 ) = H 1 (T, P ) + ax2 (1 − x1 ) = H 1 (T, P ) + ax22

Similarly, H 2 (T, P, x) = H 2 (T, P ) + ax21 . Therefore, Eq. 9.3-1a is not satisfied, and the mixture described by Eq. ** above is not an ideal mixture.

Comment Clearly, for this mixture Gex = ax1 x2 and Aex = ax1 x2 . Since the excess Gibbs energies of mixing are not zero, this also shows that the mixture cannot be ideal.

Illustration 9.3-3 Developing Expressions for Activity Coefficients and Species Fugacities from the Gibbs Energy Develop an expression for the activity coefficients and species fugacities using the Gibbs energy function of the previous illustration.

Solution We start from Gex = Gmix − GIM mix  2  2 2 2     = xi Gi + RT xi ln xi + ax1 x2 − xi Gi + RT xi ln xi = ax1 x2 i=1

i=1

i=1

i=1

Then, by definition,    ∂N Gex ∂  ∂  aN1 N2 = N ax x = 1 2   ∂N1 ∂N ∂N (N 1 1 1 + N2 ) T,P,N2 T,P,N2 T,P,N2   N1 N2 N2 = a(x2 − x1 x2 ) − =a N 1 + N2 (N1 + N2 )2 

Gex 1 =

= ax2 (1 − x1 ) = ax22 = RT ln γ1 (x1 )

or 

ax22 γ1 (x1 ) = exp RT





a(1 − x1 )2 = exp RT



Similarly,  2 2 Gex 2 = ax1 = a(1 − x2 )

and

γ2 (x2 ) = exp

ax21 RT



 = exp

a(1 − x2 )2 RT



9.3 Ideal Mixture and Excess Mixture Properties 433 Finally,  f 1 (T, P, x1 ) = x1 exp

   ax22 a(1 − x1 )2 f1 (T, P ) = x1 exp f1 (T, P ) RT RT

and  f 2 (T, P, x2 ) = x2 exp

 a(1 − x2 )2 f2 (T, P ) RT

Comment The results above for the activity coefficients and fugacities are not general, but are specific to the very simple excess Gibbs energy function we have used, Gex (T, P, x) = ax1 x2

As we shall see, in order to have excess Gibbs energy functions that more accurately fit experimental data, more complicated activity coefficient and fugacity expressions are needed. However, the expressions here do show that the activity coefficients are strong functions of composition or mole fraction. Indeed, in this example the activity coefficients are exponential functions of the square of the mole fraction. There is a very important point in this illustration that is easily overlooked. To obtain the activity coefficient from the expression for the excess Gibbs energy, we used the definition  Gex 1 = RT ln γ1 (x1 ) =

Correct!

∂N Gex ∂N1

 T,P,N2

We did not use  Gex 1 =

Incorrect!

dGex dx1

 T,P

which is an incorrect equation and gives the wrong result (try it and see).

From Eq. 9.3-3, the activity coefficient is equal to unity for species in ideal mixtures, and for nonideal mixtures ex

 P   1 Gi = exp γi (T, P, x) = exp V i (T, P, x) − V i (T, P ) dP RT RT 0 (9.3-13) Since both real and ideal mixtures satisfy the Gibbs-Duhem equation of Sec. 8.2,

0 = S dT − V dP +

C 

xi dGi

(8.2-12b)

i=1

and

0 = S IM dT − V IM dP +

C  i=1

xi dGIM i

(9.3-7)

434 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture we can subtract these two equations to obtain a form of the Gibbs-Duhem equation applicable to excess thermodynamic properties: Gibbs-Duhem equation for excess properties

0 = S dT − V ex

ex

dP +

C 

xi dGex i

(9.3-14)

i=1

Now using Eq. 9.3-12, in the form of Gex i = RT ln γi , and  ex Gex = xi Gex − T S ex i = H we obtain C

 H ex xi d ln γi dT − V ex dP + RT T

0=

(9.3-15)

i=1

Thus, for changes in composition at constant temperature and pressure, we have

0=

C 

 xi d ln γi T,P

(9.3-16)

i=1

Gibbs-Duhem equation for activity coefficients

which is a special case of Eqs. 8.2-9b and 8.2-13b. For a binary mixture, we have, from Eq. 8.2-20,



∂ ln γ1 ∂ ln γ2 x1 + x2 =0 (9.3-17) ∂x1 T,P ∂x1 T,P Illustration 9.3-4 Testing Whether an Activity Coefficient Model Satisfies the Gibbs-Duhem Equation Determine whether the activity coefficient expressions derived in the previous illustration satisfy the Gibbs-Duhem equation, Eq. 9.3-17.

Solution From the previous illustration, we have ln γ1 (x1 ) =

a 2 a x = (1 − x1 )2 RT 2 RT

ln γ2 (x2 ) =

a 2 a x = (1 − x2 )2 RT 1 RT

and

Consequently, 

∂ ln γ1 ∂x1

so that

 x1



−2a

=

RT

T,P

∂ ln γ1 ∂x1

as required by Eq. 9.3-17.

(1 − x1 ) =



 + x2

T,P

−2ax2

∂ ln γ2 ∂x1

RT

 and

 =− T,P

∂ ln γ2 ∂x1

 = T,P

2ax1 RT

2ax1 x2 2ax1 x2 + =0 RT RT

9.3 Ideal Mixture and Excess Mixture Properties 435 Comment All activity coefficients derived from an excess Gibbs energy expression that satisfies the boundary conditions of being zero at x1 = 0 and 1 will satisfy the Gibbs-Duhem equation. Can you prove this?

To determine the dependence of the activity coefficient γi on temperature and pressure, we rewrite Eq. 9.3-13 as  P   Gex (T, P, x) 1 ln γi (T, P, x) = i V i (T, P, x) − V i (T, P ) dP (9.3-18) = RT RT 0 Taking the derivative with respect to pressure at constant temperature and composition, we obtain



∂ ln γi (T, P, x) ∂P

T,x

∂ = ∂P =



1 RT



  V i (T, P, x) − V i (T, P ) dP

P

0

T,x

V i (T, P, x) − V i (T, P ) V ex (T, P, x) = i RT RT

(9.3-19)

Therefore,



P2

V ex i (T, P, x) γi (T, P2 , x) = γi (T, P1 , x) exp dP RT P1   ex V i (T, x)(P2 − P1 ) γi (T, P1 , x) exp RT

 (9.3-20)

In obtaining the last expression in Eq. 9.3-20 we have assumed that the excess partial molar volume is independent of pressure. (Note that although Eqs. 9.3-18 and 9.3-19 are correct, they are difficult to use in practice since the activity coefficient description is applied to fluid mixtures not well described by an equation of state.) Next, taking the temperature derivative of Eq. 9.3-18 at constant pressure and composition, we obtain

ex

∂ Gi H ex (T, P, x) ∂ ln γi (T, P, x) = =− i (9.3-21) ∂T ∂T RT P,x RT 2 P,x so that Temperature variation of activity coefficients

  γi (T2 , P, x) = γi (T1 , P, x) exp −

T2 T1

H ex i (T, P, x) dT RT 2

 (9.3-22)

For small temperature changes, or if the excess partial molar enthalpy is independent of temperature, we have

  ex H i (x) 1 1 γi (T2 , P, x) = γi (T1 , P, x) exp − (9.3-23) R T2 T1

436 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture These equations, especially Eqs. 9.3-20, 9.3-22, and 9.3-23, are useful when one wants to correct the numerical values of activity coefficients obtained at one temperature and pressure for use at other temperatures and pressures. For example, either Eq. 9.3-22 or 9.3-23 would be needed for the prediction of low-temperature liquidliquid phase equilibrium (Sec. 11.1) if the only activity coefficient data available were those obtained from higher-temperature vapor-liquid equilibrium measurements (Sec. 10.2). Alternatively, if activity coefficients have been measured at several pressures, Eq. 9.3-20 can be used to calculate partial molar excess volumes, or if activity coefficients have been measured at several temperatures, Eq. 9.2-23 can be used to calculate partial molar excess enthalpies (and heats of mixing).

9.4 FUGACITY OF SPECIES IN GASEOUS, LIQUID, AND SOLID MIXTURES The fugacity function is central to the calculation of phase equilibrium. This should be apparent from the earlier discussion of this chapter and from the calculations of Sec. 7.5, which established that once we had the pure fluid fugacity, phase behavior in a pure fluid could be predicted. Consequently, for the remainder of this chapter we will be concerned with estimating the fugacity of species in gaseous, liquid, and solid mixtures.

9.4-1 Gaseous Mixtures Several methods are commonly used for estimating the fugacity of a species in a gaseous mixture.1 The most approximate method is based on the observation that some gaseous mixtures follow Amagat’s law, that is,

Vmix (T, P, N1 , N2 , . . .) =

C 

Ni V i (T, P )

(9.4-1)

i=1

which, on multiplying by P/N RT , can be written as

Zmix (T, P, y) =

C 

yi Zi (T, P )

(9.4-2)

i=1

Here Z = P V /N RT is the compressibility of the mixture, Zi is the compressibility of the pure component i, and yi = Ni /N is the mole fraction of species i. (Hereafter yi will be used to indicate a gas-phase mole fraction and xi to denote a mole fraction in the liquid phase or in equations that are applicable to both gases and liquids.) The data in Fig. 9.4-1 for the nitrogen-butane system show that the mixture compressibility is nearly a linear function of mole fraction at both low and high pressures, so that Eq. 9.4-2 is approximately satisfied at these conditions, but fails in the intermediate pressure range. If we nonetheless accept Eq. 9.4-2 as being a reasonable approximation over the whole pressure range, we then have, from Eq. 9.4-1 and the definition of the partial molar volume, V i (T, P, y) = V i (T, P ) and, from Eq. 9.2-6, V fV i (T, P, y) = yi fi (T, P )

Lewis-Randall rule

1 See

also Secs. 9.7 and 9.9.

(9.4-3)

9.4 Fugacity of Species in Gaseous, Liquid, and Solid Mixtures 437 2.0 1.9 1.8 1.7 a

Z = P/NRT

1.6

95

68.

MP

6

1.5

62.0

1.4

55.16

1.3

48.27

41.37

1.2

34.48 27.5 8 2 0 13.79 .69

1.1 1.0

0

1.38 0.9 4.1

4

0.8

52

5.

0.7

2.76

0.6 0.5 0.4 0.3

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Mole fraction butane

Figure 9.4-1 Compressibility factors for nitrogen-butane mixtures at 444.3 K. [Data of R. B. Evans and C. M. Watson, Chem. Eng. Data Ser., 1, 67 (1956). Based on figure in J. M. Prausnitz, Molecular Thermodynamics of Fluid-Phase Equilibria, 1969. Reprinted with permission from Prentice Hall, Englewood Cliffs, N.J.]

This result, which relates the fugacity of a species in a gaseous mixture only to its mole fraction and the fugacity of the pure gaseous component at the same temperature and pressure, is known as the Lewis-Randall rule. Illustration 9.4-1 Approximate Species Fugacity Calculation Using the Lewis-Randall Rule Compute the fugacities of ethane and n-butane in an equimolar mixture at 323.15 K at 1, 10, and 15 bar total pressure assuming that the Lewis-Randall rule is correct.

Solution From Illustration 7.4-4, we have fET (373.15 K, 1 bar) = 0.996 bar

fBU (373.15 K, 1 bar) = 0.986 bar

fET (373.15 K, 10 bar) = 9.629 bar

fBU (373.15 K, 10 bar) = 8.628 bar

fET (373.15 K, 15 bar) = 14.16 bar

fBU (373.15 K, 15 bar) = 11.86 bar

438 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Then, by the Lewis-Randall rule, f i (T, P, y) = yi fi (T, P ), we have f ET (373.15 K, 1 bar, yET = 0.5) = 0.5 × fET (373.15, 1 bar) = 0.5 × 0.996 bar = 0.498 bar f ET (373.15 K, 10 bar, yET = 0.5) = 4.815 bar f ET (373.15 K, 15 bar, yET = 0.5) = 7.08 bar

and f BU (373.15 K, 1 bar, yBU = 0.5) = 0.493 bar f BU (373.15 K, 10 bar, yBU = 0.5) = 4.314 bar f BU (373.15 K, 15 bar, yBU = 0.5) = 5.930 bar

A more accurate way to estimate the fugacity of a species in a gaseous mixture is to start with Eq. 9.2-13,

fV i (T, P, y) = ln φV i (T, P, y) yi P 

 V =Z V RT /P  1 RT ∂P = −N dV − ln Z V RT V =∞ V ∂Ni T,V,N j=i (9.2-13) and use an appropriate equation of state. At low pressures, the truncated virial equation of state ln

Bmix (T, y) PV = Zmix =1+ RT V

(9.4-4)

can be used if data for the mixture virial coefficient as a function of composition are available.2 From statistical mechanics it is known that  Bmix (T, y) = yi yj Bij (T ) (9.4-5) i

j

where each Bij (T ) is a function only of temperature and Bij = Bji . Using Eqs. 9.4-4 and 9.4-5 in Eq. 9.2-13 yields (see Problem 9.6)

Fugacity coefficient from virial equation of state

ln

fV 2  i (T, P, y) yj Bij (T ) − ln Zmix = yi P V j

2P  = yj Bij (T ) − ln Zmix Zmix RT

(9.4-6)

j

2 The book The Virial Coefficients of Pure Gases and Mixtures: A Critical Compilation, by J. H. Dymond and E. B. Smith, Clarendon Press, Oxford, 1980, contains second-virial-coefficient data for more than one thousand mixtures.

9.4 Fugacity of Species in Gaseous, Liquid, and Solid Mixtures 439 where

1 = 2

Zmix





1+

4Bmix P 1+ RT

 (9.4-7)

Illustration 9.4-2 Species Fugacity Calculation Using the Virial Equation of State Compute the fugacities of ethane and n -butane in an equimolar mixture at 373.15 K and 1, 10, and 15 bar using the virial equation of state. Data: BET−ET = −1.15 × 10−4 m3 /mol −4

BET−BU = −2.15 × 10

BBU−BU = −4.22 × 10−4 m3 /mol

3

m /mol

Solution We start by noting from Eq. 9.4-5 that Bmix =



2 2 yi yj Bij = yET BET−ET + 2yET yBU BET−BU + yBU BBU−BU

= −2.417 × 10−4 m3 /mol

Then by using Eqs. 9.4-6 and 9.4-7 we find that

P

Z

f ET (373.15, P, yET = 0.5)

f BU (373.15, P, yBU = 0.5)

1 bar 10 bar 15 bar

0.992 0.915 0.865

0.499 bar 4.866 bar 7.211 bar

0.494 bar 4.367 bar 6.074 bar

Comment These results are slightly higher than those computed from the Lewis-Randall rule, and are more accurate.

At higher pressures, Eq. 9.2-13 can still be used to compute the fugacity of a species in a gaseous mixture, but more accurate equations of state must be used. One can, for example, use the Peng-Robinson equation of state

P =

a(T ) RT − V − b V (V + b) + b(V − b)

(6.4-2)

where the parameters a and b are now those for the mixture, amix and bmix . To obtain these mixture parameters one starts with the a and b parameters for the pure components obtained from either fitting pure component data or the generalized correlations of Sec. 6.7 and then uses the mixing rules

440 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Equation-of-state mixing rule

amix =

C  C 

yi yj aij

i=1 j=1

bmix =

C 

(9.4-8)

yi bi

i=1

where aii and bi are the parameters for pure component i, and the combining rule

aij =

Equation-of-state combining rule

√ aii ajj (1 − kij ) = aji

(9.4-9)

Here a new parameter kij , known as the binary interaction parameter, has been introduced to result in more accurate mixture equation-of-state calculations. This parameter is found by fitting the equation of state to mixture data (usually vapor-liquid equilibrium data, as discussed in Chapter 10). Values of the binary interaction parameter kij that have been reported for a number of binary mixtures appear in Table 9.4-1. Equations 9.4-8 and 9.4-9 are referred to as the van der Waals one-fluid mixing rules. The term one-fluid derives from the fact that the mixture is being described by the same equation of state as the pure fluids, but with concentration-dependent parameters. As a result of the mole fraction dependence of the equation-of-state parameters, the pressure is a function of mole fraction or, alternatively, of the number of moles of each species present. Evaluating the derivative (∂P/∂Ni )T,V,Nj=i , which appears in Eq. 9.2-13, using the Peng-Robinson equation of state yields Fugacity coefficient from Peng-Robinson equation of state

ln

fV i (T, P, y) yi P = ln φV i (T, P, y)

bmix P V = − 1) − ln Zmix − bmix RT ⎡  ⎤ ⎡  √  bmix P ⎤ 2 yj aij V + 1 + 2 Z ⎢ j mix amix bi ⎥ RT ⎥ ⎢ ⎥ ln ⎢ − √ − ⎣ √ bmix P ⎦ bmix ⎦ 2 2bmix RT ⎣ amix V Z + (1 − 2) RT Bi V (Z V − 1) − ln(Zmix − Bmix ) = Bmix mix  ⎡ ⎤ 2 yj Aij √   V Amix ⎢ Bi ⎥ Zmix + (1 + 2)Bmix j ⎢ ⎥ √ − √ − ln V Bmix ⎦ 2 2Bmix ⎣ Amix Zmix + (1 − 2)Bmix bi

V (Zmix

(9.4-10)

441

0.016 0.001 0.007

0.033 0.089

0.022

C3 H8

C3 H6

C2 H6

−0.003 0.010

C2 H4

−0.007 −0.014 −0.007

0.026

i-C4 H10

0.003 0 .0

0.019 0.092 0.010

n-C4 H10

0.011

−0.006

i-C5 H12

−0.006

0.017 0.06

0.001

−0.04

0.040

n-C6 H14

0.027

0.008

0.026

n-C5 H12

0.018 0.010

0.023

0.055 0.031 0.042

C6 H6

0.004 −0.004 0.013

0.018

0.039

c-C6 H12

0.007 −0.008 0.001

0.003

0.006

0.035 0.014 0.007

n-C7 H16

0 .0

0.003

0 .0

0.007

0 .0

0.019

0.050

n-C8 H18

0 .1

0.008

0 .0

0.049 0.025 0.014

n-C10 H22

N2

0.14 0 .1 0 .1 0.11

0.092 0.10 0.15 0.164

0.078 0.10 0.087

0.030 0.086 0.044 0.09

CO

0.012

0.10 0.04 0.04 0.04

0.04 0.04 0.04 0.11

0.03 0.04 0.04

0.030 −0.022 0.026 0.026

−0.02 0.03

0.105 0.10 0.12 0.114

0.121 0.125 0.11 0.077

0.12 0.13 0.135

0.09 0.056 0.130 0.093

CO2

SO2

0.136

0.08

0.015

0.136

H2 S

0.17 0.054 0.097

0.06 0.06 0.033

0.06 0.063 0.06

0.08 0.047 0.07

0.086 0.08

0.08

*Obtained from data in “Vapor-Liquid Equilibria for Mixtures of Low-Boiling Substances,” by H. Knapp, R. D¨oring, L. Oellrich, U. Pl¨ocker, and J. M. Prausnitz, DECHEMA Chemistry Data Series, Vol. VI, Frankfurt/Main, 1982, and other sources. Blanks indicate no data are available from which the k12 could be evaluated. In such case use estimates from mixtures of similar compounds.

H2 S

N2 CO CO2 SO2

c-C6 H12 n-C7 H16 n-C8 H18 n-C10 H22

i-C5 H12 n-C5 H12 n-C6 H14 C6 H6

C3 H8 i-C4 H10 n-C4 H10

CH4 C2 H4 C2 H6 C3 H6

Table 9.4-1 Binary Interaction Parameters k12 for the Peng-Robinson Equation of State*

442 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture where, again, A = aP/(RT )2 , B = bP/RT , and the superscript V is used to remind us that the vapor-phase compressibility factor is to be used. Also, amix and bmix (and therefore Amix and Bmix ) are computed using the vapor-phase mole fractions. To calculate the fugacity of each species in a gaseous mixture using Eq. 9.4-10 at specified values of T , P , and mole fractions of all components y1 , y2 , . . . , yC , the following procedure is used: 1. Obtain the parameters aii and bi for each pure component of the mixture either from the generalized correlations given in Sec. 6.7 or, if necessary, from pure component data. 2. Compute the a and b parameters for the mixture using the mixing rules of Eq. 9.4-8. (If the value for any binary interaction parameter, kij , is not available, either use a value for similar mixtures or assume it is zero.) 3. Solve the cubic equation of state for the vapor compressibility Z V . 4. Use this value of Z V to compute the vapor fugacity for each species using Eq. 9.4-10 repeatedly for i = 1, 2, . . . , C, where C is the number of components. Computer programs and a MATHCAD worksheet to do this calculation are discussed R can also be used. in Appendix B. Aspen Plus Illustration 9.4-3 Species Fugacity Calculation Using the Peng-Robinson Equation of State Compute the fugacities of ethane and n -butane in an equimolar mixture at 373.15 K and 1, 10, and 15 bar using the Peng-Robinson equation of state.

Solution R Using the computer programs in Appendix B on the website for this book or Aspen Plus and from Table 9.4-1, with kET−BU = 0.010, the results below are obtained.

P

Z

f ET (373.15, P, yET = 0.5)

f BU (373.15, P, yBU = 0.5)

1 bar 10 bar 15 bar

0.991 0.910 0.861

0.498 bar 4.836 bar 7.143 bar

0.493 bar 4.333 bar 6.024 bar

Comment These results are similar to those obtained using either the Lewis-Randall rule or the virial equation of state. However, greater differences between the methods occur for gases as the pressure is increased. At higher pressures the results from the Peng-Robinson equation of state are expected to be the most accurate.

Illustration 9.4-4 Calculation of the Fugacity of a Species in a Mixture by Two Methods Compute the fugacity of both carbon dioxide and methane in an equimolar mixture at 500 K and 500 bar using (a) the Lewis-Randall rule and (b) the Peng-Robinson equation of state.

9.4 Fugacity of Species in Gaseous, Liquid, and Solid Mixtures 443 Solution a. The Lewis-Randall rule is  f i (T, P, y) = yi fi (T, P ) = yi

f P

 P i

From Fig. 7.4-1, we have

CO2 CH4

Tr

Pr

f /P

1.644 2.623

6.78 10.87

∼ 0.77 ∼ 1.01*

*Since it is very difficult to read the fugacity coefficient curve in the range of the reduced temperature and pressure for methane in this problem, this entry was obtained from the tables in O. A. Hougen, K. M. Watson, and R. A. Ragatz, Chemical Process Principles, Part II, Thermodynamics, 2nd ed., John Wiley & Sons, New York (1959).

Thus f CO2 = 0.5 × 0.77 × 500 bar = 192.5 bar

and f CH4 = 0.5 × 1.01 × 500 bar = 252.5 bar

b. Using the Peng-Robinson equation of state (Eqs. 6.4-2 and 9.4-8 through 9.4-10) and AsR pen Plus or the computer programs discussed in Appendix B, we find, for k12 = 0.0, f CO2 = 208.71 bar

and

f CH4 = 264.72 bar

and using k12 = 0.09 (from Table 9.4-1), f CO2 = 212.81 bar

and

f CH4 = 269.35 bar

These last values should be the most accurate.

9.4-2 Liquid Mixtures The estimation of species fugacities in liquid mixtures is done in two different ways, depending on the data available and the components in the mixture. For liquid mixtures involving only hydrocarbons and dissolved gases, such as nitrogen, hydrogen sulfide, and carbon dioxide, simple equations of state may be used to describe liquid-state behavior. For example, if the approximate Peng-Robinson equation of state and the van der Waals mixing rules are used, the fugacity of each species in the mixture is, following the same development as for gaseous mixtures, given by

444 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture

ln

f Li (T, P, x) xi P = ln φLi (T, P, x)



bmix P L − 1) − ln Zmix − = bmix RT ⎡  ⎤ ⎡  √  bmix P ⎤ 2 xj aij L + 1 + 2 Z ⎢ j mix amix bi ⎥ RT ⎥ ⎢ ⎥ ln ⎢ − √ − ⎣ √ ⎣ ⎦ bmix P ⎦ amix bmix 2 2bmix RT L Zmix + (1 − 2) RT bi

=

L (Zmix

Bi L (Z L − 1) − ln(Zmix − Bmix ) Bmix mix ⎡  ⎤ xj Aij 2 √   L Amix ⎢ Bi ⎥ Zmix + (1 + 2)Bmix j ⎢ ⎥ √ − √ − ln L Bmix ⎦ 2 2Bmix ⎣ Amix Zmix + (1 − 2)Bmix (9.4-11)

where, again, A = aP/(RT )2 , B = bP/RT , and Z L is the liquid (high density or small Z) root for the compressibility factor. Here, however, amix and bmix (and therefore Amix and Bmix ) are computed using the liquid-phase mole fractions. Consequently, the calculational scheme for species fugacities in liquid mixtures that can be described by equations of state is similar to that for gaseous mixtures, except that it is the liquidphase, rather than the vapor-phase, compositions and compressibility factor that R are used in the calculations. The Appendix B computer programs or Aspen Plus can be used for this calculation as well. Illustration 9.4-5 Species Fugacity Calculation for a Hydrocarbon Mixture Compute the fugacities of n -pentane and benzene in their equimolar mixture at 373.15 K and 50 bar.

Solution The simplest estimate is obtained using the Lewis-Randall rule and the results of Illustration 7.4-6. In this case we obtain f PE (373.15 K, 50 bar, xPE = 0.5) = 0.5 × fPE (373.15 K, 50 bar) = 0.5 × 6.22 bar = 3.11 bar

and f BZ (373.15 K, 50 bar, xBZ = 0.5) = 0.5 × 1.98 bar = 0.99 bar

A better estimate is obtained using the Peng-Robinson equation of state with a value kPE−BZ = 0.018 from Table 9.4-1. The results are f PE (373.15 K, 50 bar, xPE = 0.5) = 3.447 bar f BZ (373.15 K, 50 bar, xBZ = 0.5) = 1.129 bar

9.4 Fugacity of Species in Gaseous, Liquid, and Solid Mixtures 445 For liquid mixtures in which one or more of the components cannot be described by an equation of state in the liquid phase—for example, mixtures containing alcohols, organic or inorganic acids and bases, and electrolytes—another procedure for estimating species fugacities must be used. The most common starting point is Eq. 9.3-11,

f i (T, P, x) = xi γi (T, P, x)fi (T, P )

(9.3-11)

where γi is the activity coefficient of species i, and fi (T, P ) is the fugacity of pure species i as a liquid at the temperature and pressure of the mixture.3 However, since volumetric equation-of-state data are not available, the activity coefficient is obtained not from integration of Eq. 9.3-13, but rather from

∂N Gex ex RT ln γi (T, P, x) = Gi = (9.4-12) ∂Ni T,P,N j=i

ex

and approximate models for the excess Gibbs energy G of the liquid mixture. Several such liquid solution models are considered in Sec. 9.5.

9.4-3 Solid Mixtures The atoms in a crystalline solid are arranged in an ordered lattice structure consisting of repetitions of a unit cell. To form a solid solution, it is necessary to replace some molecules in the lattice with molecules of another component. If the two species are different in size, shape, or in the nature of their intermolecular forces, a solid solution can be obtained only by greatly distorting the crystalline structure. Such a distortion requires a great deal of energy, and, therefore, is energetically unfavorable. That is, the total energy (and in this case the Gibbs energy) of the distorted crystalline solid solution is greater than the energy of the two pure solids (each in its undistorted crystalline state). Consequently, with the exception of metals and metal alloys, the energetically preferred state for many solid mixtures is as heterogeneous mixtures consisting of regions of pure single species in their usual crystalline state. If these regions are large enough that macroscopic thermodynamics can be applied without making corrections for the thermodynamic state of the molecules at the interface between two solid regions, the solid mixture can be treated as an agglomeration of pure species, each having its own pure component solid fugacity. Thus

f Si (T, P, x) = fiS (T, P )

Fugacity of a species in many mixed solids

(9.4-13)

The point that the discrete pure component regions in a solid mixture should not be too small is worth dwelling on. A collection of molecules can be thought of as being composed of molecules at the surface of the region and molecules in the interior, as was discussed in Sec. 7.8. The interior molecules interact only with similar molecules and therefore are in the same energy state as molecules in a macroscopically large mass of the pure substance. The surface or interfacial molecules, however, interact with both like and unlike molecules, and hence their properties are representative of molecules in a mixture. The assumption being made in Eq. 9.4-13 is that the number of interfacial 3 If

the pure species does not exist as a liquid at the temperature and pressure of interest, as, for example, in the case of a gas or solid dissolved in a liquid, other, more appropriate starting points are used, as discussed in Sec. 9.7.

446 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture molecules is small compared with the number of interior molecules, so that even if the properties of the interfacial molecules are poorly represented, the effect on the total thermodynamic properties of the system is small. While Eq. 9.4-13 will generally be used to calculate the fugacity of a species in a solid mixture, especially involving organic compounds, there are cases of true solid mixtures, especially involving metals and inorganic compounds. While we defer considerations of such mixtures until Sec. 12.4, it is useful to note that the activity coefficient for a species in a true solid mixture, following Eq. 9.3-11, can be written as

Fugacity of a species in a true solid mixture

f Si (T, P, xS ) = xSi γiS (T, P, xS )fiS (T, P )

(9.4-14)

9.5 SEVERAL CORRELATIVE LIQUID MIXTURE ACTIVITY COEFFICIENT MODELS In this section we are interested in mixtures for which equation-of-state models are inapplicable. In such cases attempts are made to estimate Gex or Δmix G directly, either empirically or semitheoretically. In making such estimates it is useful to distinguish between simple and nonsimple liquid mixtures. We define a simple liquid mixture to be one that is formed by mixing pure fluids, each of which is a liquid at the temperature and pressure of the mixture. A nonsimple liquid mixture, on the other hand, is formed by mixing pure species, at least one of which is not a liquid at the temperature and pressure of the mixture. Examples of nonsimple liquid mixtures are solutions of dissolved solids in liquids (for example, sugar in water) and dissolved gases in liquids (for example, carbon dioxide in soda water). In this section we are concerned with mixture models that are applicable to simple liquid mixtures. Nonsimple liquid mixtures are considered in Sec. 9.7 and elsewhere in this book. Also, for simplicity, most of the excess Gibbs energy models we will consider here are for two-component (binary) mixtures. Multicomponent-mixture versions of these models are listed in Appendix A9.2. From Eq. 9.3-8, we have that the Gibbs energy change on forming a simple liquid mixture is

Δmix G = Δmix GIM + Gex = RT

C 

xi ln xi + Gex (x)

i=1

The excess Gibbs energy of mixing for the benzene–2,2,4-trimethyl pentane system (a simple liquid mixture) is shown in Fig. 9.5-1 (the origin of these data is discussed in Sec. 10.2); data for the excess Gibbs energy for several other mixtures are shown in Fig. 9.5-2. Each of the curves in these figures is of simple form. Thus, one approach to estimating the excess Gibbs energy of mixing has been merely to try to fit results, such as those given in Figs. 9.5-1 and 9.5-2, with polynomials in the composition. The hope is that given a limited amount of experimental data, one can determine the parameters in the appropriate polynomial expansion, and then predict the excess Gibbs energy and liquid-phase activity coefficients over the whole composition range. Of course, any expression chosen for the excess Gibbs energy must satisfy the Gibbs-Duhem equation (i.e., Eqs. 9.3-14 through 9.3-17) and, like the data in the figures, go to zero as x1 → 0 and x1 → 1.

9.5 Several Correlative Liquid Mixture Activity Coefficient Models 447 400

G _ ex, J/mol

300

200

100

0

0

0.2

0.4

0.6

0.8

1.0

xB

Figure 9.5-1 The excess Gibbs energy for the benzene–2,2,4trimethyl pentane system at 55◦ C.

1

600

400

2 3

G _ ex, J/mol

200

0

–200

–400

–600 4 –800

0

0.2

0.4

0.6

0.8

1.0

x1

Figure 9.5-2 The excess Gibbs energy for several mixtures. Curve 1: Trimethyl methane (1) and benzene at 0◦ C. [Data from V. Mathot and A. Desmyter, J. Chem. Phys. 21, 782 (1953).] Curve 2: Trimethyl methane (1) and carbon tetrachloride at 0◦ C. [Data from ibid.] Curve 3: Methane (1) and propane at 100 K. [Data from A. J. B. Cutler and J. A. Morrison, Trans. Farad. Soc., 61, 429 (1965).] Curve 4: Water (1) and hydrogen peroxide at 75◦ C. [From the smoothed data of G. Scatchard, G. M. Kavanagh, and L. B. Ticknor, J. Am. Chem. Soc., 74, 3715 (1952).]

448 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture The simplest polynomial representation of Gex satisfying these criteria is One-constant Margules G ex

Gex = Ax1 x2 for which

Gex 1

(9.5-1)



∂(N Gex )  ∂ AN1 N2 = = ∂N1 T,P,N2 ∂N1 N1 + N2

N1 N2 N2 − = Ax22 =A N1 + N2 (N1 + N2 )2

(9.5-2)

so that

γ1 (x) = exp

Gex 1 RT

and similarly one can show that

γ2 (x) = exp

= exp



Ax21 RT

Ax22 RT



= exp

= exp

A(1 − x1 )22 RT

A(1 − x2 )22 RT



(9.5-3)

Consequently, for this solution model we have

f Li (T, P, x) = xi γi fiL (T, P )

(9.5-4)

RT ln γ1 = Ax22 RT ln γ2 = Ax21

(9.5-5)

where One-constant Margules activity coefficients

These relations, known as the one-constant Margules equations, are plotted in Fig. 9.5-3. Two interesting features of the one-constant Margules equations are apparent from this figure. First, the two species activity coefficients are mirror images of each other as a function of the composition. This is not a general result, but follows from the choice of a symmetric function in the compositions for Gex . Second, γi → 1 as xi → 1, 3

γ2

γ1 γ

2

1

0

0.2

0.4

0.6 x1

0.8

1.0

Figure 9.5-3 The activity coefficients for the one-constant Margules equation with (A/RT ) = 1.0.

9.5 Several Correlative Liquid Mixture Activity Coefficient Models 449 so that f i → fi in this limit. This makes good physical sense since one expects the fugacity of a component in a mixture to tend toward that of the pure liquid as the mixture becomes concentrated in that component. Conversely, γi exhibits a greater departure from unity (and greater departure from ideal solution behavior) the greater the dilution of species i, as would be expected from the discussion in Chapter 8. Note that the parameter A can be either positive or negative, so that Eq. 9.5-5 can lead to activity coefficients that are greater than 1 (A > 0) or less than 1 (A < 0). It is interesting to note that in the ideal solution model (i.e., f i = xi fi ), the only role of components other than the one of interest is as diluents, so that they affect f i only through xi . No account is taken of the fact that the energy of a species 1–species 2 interaction could be different from that of species 1–species 1 or species 2–species 2 interactions. For the nonideal solution of Eq. 9.5-1 the parameter A is dependent on both species, or more precisely, on the differences in species interaction energies involved. This results in the behavior of species 1 being influenced by both the nature and composition of species 2, and vice versa. The value of the parameter A depends on the macroscopic and molecular properties of both species in the mixture and is difficult to estimate a priori; its value may be either positive or negative and generally is a function of temperature. Over small temperature ranges A may be assumed to be a constant, so that its value found from experiments at one temperature can be used at neighboring temperatures. The one-constant Margules equation provides a satisfactory representation for activity coefficient behavior only for liquid mixtures containing constituents of similar size, shape, and chemical nature. For more complicated systems, particularly mixtures of dissimilar molecules, simple relations such as Eq. 9.5-1 or 9.5-5 are not valid. In particular, the excess Gibbs energy of a general mixture will not be a symmetric function of the mole fraction, and the activity coefficients of the two species in a mixture should not be expected to be mirror images. One possible generalization of Eq. 9.5-1 to such cases is to set

Gex = x1 x2 {A + B(x1 − x2 ) + C(x1 − x2 )2 + · · · }

Redlich-Kister expansion

(9.5-6)

where A, B, C, . . . are temperature-dependent parameters. This expression for Gex is another example of the Redlich-Kister expansion used for the representation of excess thermodynamic properties, which was discussed in Sec. 8.6. The number of terms retained in this expansion depends on the shape of the Gex curve as a function of composition, the accuracy of the experimental data, and the goodness of the fit desired. When A = B = C = · · · = 0, the ideal solution model is recovered; for A = 0, B = C = · · · = 0, the one-constant Margules equation is obtained. For the case in which A = 0, B = 0, but C = D = · · · = 0, one obtains (see Problem 9.8)

RT ln γ1 = α1 x22 + β1 x32

Two-constant Margules expansion

RT ln γ2 = α2 x21 + β2 x31 where

αi = A + 3(−1)i+1 B and

βi = 4(−1)i B

(9.5-7)

450 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture In this equation i denotes the species and has values of 1 and 2. These results are known as the two-constant Margules equations. In this case the excess Gibbs energy is not symmetric in the mole fractions and the two activity coefficients are not mirror images of each other as a function of concentration. The expansion of Eq. 9.5-6 is certainly not unique; other types of expansions for the excess Gibbs energy could also be used. Another expansion is that of Wohl,4

Gex = 2a12 z1 z2 + 3a112 z12 z2 + 3a122 z1 z22 RT (x1 q1 + x2 q2 ) + 4a1112 z13 z2 + 4a1222 z1 z23 + 6a1122 z12 z22 + ···

(9.5-8)

where qi is some measure of the volume of molecule i (e.g., its liquid molar volume or van der Waals b parameter) and the a ’s are parameters resulting from the unlike molecule (i.e., species 1–species 2) interactions. The zi in Eq. 9.5-8 are, essentially, volume fractions defined by xi qi zi = x1 q1 + x2 q2 Equation 9.5-8 is modeled after the virial expansion for gaseous mixtures, and, in fact, the constants in the expansion (2, 3, 3, 4, 4, 6, etc.) are those that arise in that equation. The liquid-phase activity coefficients for the Wohl expansion can be obtained from Eq. 9.5-8 by taking the appropriate derivatives:

∂ Gex N Gex ln γi = i = RT ∂Ni RT T,P,N j=i

In particular, for the case in which we assume a12 = 0 and a112 = a122 = · · · = 0, we have

2a12 x1 q1 x2 q2 Gex = (x1 q1 + x2 q2 )2a12 z1 z2 = RT x1 q1 + x2 q2 and (see Problem 9.8)

α ln γ1 =  2 α x1 1+ β x2

van Laar equations

and

β ln γ2 =  2 β x2 1+ α x1

(9.5-9)

where α = 2q1 a12 and β = 2q2 a12 . Equations 9.5-9 are known as the van Laar equations;5 they are frequently used to correlate activity coefficient data. Other, more complicated, activity coefficient equations can be derived from Eq. 9.5-8 by retaining additional terms in the expansion, though this is not done here. The values of the parameters in the activity coefficient equations are usually found by fitting these equations to experimental activity coefficient data (see Problem 9.22) 4 K. 5 J.

Wohl, Trans. AIChE, 42, 215 (1946). J. van Laar, Z. Physik. Chem., 72, 723 (1910); 83, 599 (1913).

9.5 Several Correlative Liquid Mixture Activity Coefficient Models 451 over the whole composition range. Alternatively, if only limited data are available, the van Laar equations, which can be written as

2 x2 ln γ2 α= 1+ ln γ1 x1 ln γ1

2 x1 ln γ1 β = 1+ ln γ2 x2 ln γ2 (9.5-10) so that data for γ1 and γ2 at only a single mole fraction can be used to evaluate the two van Laar constants, and hence to compute the activity coefficients at all other compositions. This procedure is used in Chapter 10. However, if activity coefficient data are available at several compositions, a regression procedure can be used to obtain the “best” fit values for α and β . Table 9.5-1 contains values that have been reported for these parameters for a number of binary mixtures. Table 9.5-1 The van Laar Constants for Some Binary Mixtures Component 1–Component 2 Acetaldehyde–water Acetone–benzene Acetone–methanol Acetone–water Benzene–isopropanol Carbon disulfide–acetone Carbon disulfide–Carbon tetrachloride Carbon tetrachloride–benzene Ethanol–benzene Ethanol–cyclohexane Ethanol–toluene Ethanol–water Ethyl acetate–benzene Ethyl acetate–ethanol Ethyl acetate–toluene Ethyl ether–acetone Ethyl ether–ethanol n-Hexane–ethanol Isobutane–furfural Isopropanol–water Methanol–benzene Methanol–ethyl acetate Methanol–water Methyl acetate–methanol Methyl acetate–water n-Propanol–water Water–phenol

Temperature Range (◦ C) 19.8–100 56.1–80.1 56.1–64.6

25 56.1–100 71.9–82.3 39.5–56.1 46.3–76.7 76.4–80.2 67.0–80.1 66.3–80.8 76.4–110.7 25 71.1–80.2 71.7–78.3 77.2–110.7 34.6–56.1 34.6–78.3 59.3–78.3

37.8 51.7 82.3–100 55.5–64.6 62.1–77.1

25 64.6–100 53.7–64.6 57.0–100 88.0–100 100–181

α

β

1.59 0.405 0.58 1.89 2.05 1.36 1.28 0.23 0.12 1.946 2.102 1.757 1.54 1.15 0.896 0.09 0.741 0.97 1.57 2.62 2.51 2.40 0.56 1.16 0.58 0.83 1.06 2.99 2.53 0.83

1.80 0.405 0.56 1.66 1.50 1.95 1.79 0.16 0.11 1.610 1.729 1.757 0.97 0.92 0.896 0.58 0.741 1.27 2.58 3.02 2.83 1.13 0.56 1.16 0.46 0.51 1.06 1.89 1.13 3.22

Source: This table is an adaptation of one given in J. H. Perry, ed., Chemical Engineers’ Handbook, 4th ed., McGraw-Hill, New York (1963), p. 13–7. α Note: When α = β , the van Laar equation ln γ1 = reduces to the Margules form 2 [1 + (αx1 /βx2 )]

ln γ1 = αx22 .

452 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture 0.8 A/RT = 0.676 A/RT = 0.397

0.6

γ2

γ1

ln γ 0.4

0.2

0

0

0.2

0.4

0.6

0.8

1.0

x1

Figure 9.5-4 Experimental activity coefficient data for the benzene–2,2,4-trimethyl pentane mixture and the correlation of these data obtained using the one-constant Margules equation.

Illustration 9.5-1 Use of Activity Coefficient Models to Correlate Data The points in Figs. 9.5-4 and 9.5-5 represent smoothed values of the activity coefficients for both species in a benzene–2,2,4-trimethyl pentane mixture at 55◦ C taken from the vapor-liquid equilibrium measurements of Weissman and Wood (see Illustration 10.2-4). Test the accuracy of the one-constant Margules equation and the van Laar equations in correlating these data.

Solution a. The one-constant Margules equation. From the data presented in Fig. 9.5-4 it is clear that the activity coefficient for benzene is not the mirror image of that for trimethyl pentane. Therefore, the one-constant Margules equation cannot be made to fit both sets of activity coefficients simultaneously. (It is interesting to note that the Margules form, RT ln γi = Ai x2j , will fit these data well if A1 and A2 are separately chosen. However, this suggestion does not satisfy the GibbsDuhem equation! Can you prove this?) 0.8

α = 0.415 β = 0.706

0.6 γ1

γ2

ln γ 0.4

0.2

0

0

0.2

0.4

0.6

0.8

1.0

x1

Figure 9.5-5 Experimental activity coefficient data for the benzene–2,2,4-trimethyl pentane mixture and the correlation of these data obtained using the van Laar equation.

9.5 Several Correlative Liquid Mixture Activity Coefficient Models 453 b. The van Laar equation. One can use Eqs. 9.5-10 and a single activity coefficient–composition data point (or a least-squares analysis of all the data points) to find values for the van Laar parameters. Using the data at x1 = 0.6, we find that α = 0.415 and β = 0.706. The activity coefficient predictions based on these values of the van Laar parameters are shown in Fig. 9.5-5. The agreement between the correlation and the experimental data is excellent. Note that these parameters were found using the MATHCAD worksheet ACTCOEFF on the website for this book, and discussed in Appendix B.III.

The molecular-level assumption underlying the Redlich-Kister expansion is that completely random mixtures are formed, that is, that the ratio of species 1 to species 2 molecules in the vicinity of any molecule is, on the average, the same as the ratio of their mole fractions. A different class of excess Gibbs energy models can be formulated by assuming that the ratio of species 1 to species 2 molecules surrounding any molecule also depends on the differences in size and energies of interaction of the chosen molecule with species 1 and species 2. Thus, around each molecule there is a local composition that is different from the bulk composition. From this picture, the several binary mixture models have been developed. The first model we consider of this type is the two-parameter (Λ12 , Λ21 ) Wilson equation6

Gex = −x1 ln(x1 + x2 Λ12 ) − x2 ln(x2 + x1 Λ21 ) RT

(9.5-11)

for which



 Λ21 Λ12 ln γ1 = − ln(x1 + x2 Λ12 ) + x2 − x1 + x2 Λ12 x1 Λ21 + x2   Λ21 Λ12 ln γ2 = − ln(x2 + x1 Λ21 ) − x1 − x1 + x2 Λ12 x1 Λ21 + x2

Wilson equation

(9.5-12a)

The two infinite-dilution activity coefficients in this model are

ln γ1∞ = − ln Λ12 + 1 − Λ12 and ln γ2∞ = − ln Λ21 + 1 − Λ21

(9.5-12b)

The second model is the three-parameter (α, τ12 , τ21 ) nonrandom two-liquid (NRTL) equation7

τ12 G12 Gex τ21 G21 + (9.5-13) = x1 x2 RT x1 + x2 G21 x2 + x1 G12 with ln G12 = −ατ12 and ln G21 = −ατ21 , for which

 NRTL model



ln γ1 = x22 τ21  ln γ2 = x21 τ12

6 G. 7 H.



G21 x1 + x2 G21 G12 x2 + x1 G12

2

2

M. Wilson, J. Am. Chem. Soc., 86, 127 (1964). Renon and J. M. Prausnitz, AIChE J., 14, 135 (1968).

τ12 G12 + (x2 + x1 G12 )2 τ21 G21 + (x1 + x2 G21 )2

 

(9.5-14a)

454 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture and

ln γ1∞ = τ21 + τ12 G12 = τ21 + τ12 exp (−ατ12 ) ln γ2∞ = τ12 + τ21 G21 = τ12 + τ21 exp (−ατ21 )

(9.5-14b)

Note that in these models there are different weightings of the mole fractions of the species due to the parameters (Λij and τij ), the values of which depend on differences in size and interaction energies of the molecules in the mixture. (The multicomponent forms of the Wilson and NRTL models are given in Appendix A9.2.) Another model, the Flory and Huggins model, is meant to apply to mixtures of molecules of very different size, including solutions of polymers. This solution model contains two parts. The first is an expression for the entropy of mixing per mole:

Δmix S = −R(x1 ln φ1 + x2 ln φ2 )

(9.5-15a)

S ex = Δmix S − Δmix S IM

φ1 φ2 = −R x1 ln + x2 ln x1 x2

(9.5-15b)

or

Here

φ1 =

x1 v1 x1 = x1 v1 + x2 v2 x1 + mx2

and φ2 =

mx2 x1 + mx2

are the volume fractions, with vi being some measure of the volume of species i molecules, and m = v2 /v1 . The assumption in Eqs. 9.5-15a and b is that for molecules of different size it is the volume fractions, rather than the mole fractions, that determine the entropy of mixing. The second part of the model is that the enthalpy of mixing, or excess enthalpy, can be expressed by the simple one-constant term in volume fractions (rather than mole fractions, as in the case of the one-constant Margules equation)

Δmix H = H ex = χRT (x1 + mx2 )φ1 φ2

(9.5-16)

where χ is an adjustable parameter referred to as the Flory interaction parameter or simply the Flory parameter (and sometimes as the chi parameter). Combining Eqs. 9.5-15 and 9.5-16 gives

Flory-Huggins equations

H ex − T S ex Gex = RT   RT (9.5-17) φ1 φ2 + χ(x1 + mx2 )φ1 φ2 + x2 ln = x1 ln x1 x2 which is the Flory-Huggins model. The first term on the right side of this equation is the entropic contribution to the excess Gibbs energy, and the second term is the enthalpic contribution. These two terms are also referred to as the combinatorial and residual terms, respectively. The activity coefficient expressions for this model (Problem 9.29) are

1 φ1 φ2 + χφ22 ln γ1 = ln + 1− x1 m (9.5-18) φ2 2 ln γ2 = ln + (m − 1)φ1 + mχφ1 x2

9.5 Several Correlative Liquid Mixture Activity Coefficient Models 455 We will consider only one additional activity coefficient equation here, the UNIQUAC (universal quasichemical) model of Abrams and Prausnitz.8 This model, based on statistical mechanical theory, allows local compositions to result from both the size and energy differences between the molecules in the mixture. The result is the expression Gex (combinatorial) Gex (residual) Gex (9.5-19) = + RT RT RT where the first term accounts for molecular size and shape differences, and the second term accounts largely for energy differences. These terms, in multicomponent form, are given by

φi z  θi Gex (combinatorial)  xi ln + xi qi ln = RT xi 2 φi

UNIQUAC equation

i

and

(9.5-20)

i

    Gex (residual) =− qi xi ln θj τji RT i

(9.5-21)

j

where

ri = volume parameter for species i qi = surface area parameter for species i  θi = area fraction of species i = xi qi xj qj j

φi = segment or volume fraction of species i = xi ri



xj rj

j

(uij − ujj ) RT with uij being the average interaction energy for a species i–species j interaction and z being the average coordination number, that is, the number of molecules around a central molecule, usually taken to be 10. Combining Eqs. 9.5-19, 9.5-20, and 9.5-21 gives ln τij = −

ln γi = ln γi (combinatorial) + ln γi (residual) (9.5-22)  φi z φi φi ln γi (combinatorial) = ln − qi ln + li − xj lj (9.5-23a) xi 2 θi xi j ⎡ ⎤     θj τij ⎥ ⎢ ⎥ (9.5-23b) ln γi (residual) = qi ⎢ θj τji −  ⎣1 − ln ⎦ θ τ k kj j j

UNIQUAC expression for activity coefficients

k

where li = (ri − qi )z/2 − (ri − 1). Since the size and surface area parameters ri and qi can be evaluated from molecular structure information, as will be discussed next, the UNIQUAC equation contains only 8 D.

S. Abrams and J. M. Prausnitz, AIChE J., 21, 116 (1975).

456 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture two adjustable parameters, τ12 and τ21 (or, equivalently, u12 –u22 and u21 –u11 ) for each binary pair. Thus, like the van Laar or Wilson equations, it is a two-parameter activity coefficient model. It does have a better theoretical basis than these models, though it is somewhat more complicated. Instead of listing the volume (r) and surface area (q ) parameters for each molecular species for use in the UNIQUAC model, these parameters are evaluated by a group contribution method. The underlying idea is that a molecule can be considered to be a collection of functional groups, and that volume Ri and surface area Qi of functional group i will be approximately the same in any molecule in which that group occurs. For example, we expect the contribution to the total volume and surface area of a molecule from a methyl (CH3 –) group to be the same independent of whether the methyl group is at the end of an ethane, propane, or dodecane molecule. Thus, the volume and surface area parameters r and q of a molecule are obtained from a sum over its functional groups of the R and Q parameters. The advantage of this group contribution approach is that from a relatively small number of functional groups, the properties of the millions upon millions of different molecules can be obtained. Table 9.5-2 contains the R and Q parameters for 106 functional groups referred to as subgroups in the table (the main group vs. subgroup terminology will be explained in the following section). All the values that appear in the table have been normalized to the properties of a methylene group in polymethylene, and therefore are unitless. There are several things to note about the entries in this table. First, several molecules, such as water and furfural, have such unique properties that they have been treated as functional groups. Such molecule functional groups appear in bold in the table. Second, similar groups, such as –CH3 and –C–, may have different surface area or Q parameters. This is because a –CH3 group, being at the end of a molecule, increases its surface area, whereas the –C– group, which is at the interior of a molecule, makes no contribution to its surface area. Finally, as the functional group method continues to evolve, new groups are added and group parameters are subject to change. Illustration 9.5-2 Computation of Volume and Surface Area Fractions for Use in the UNIQUAC Model One mole each of benzene and 2,2,4-trimethyl pentane are mixed together. Using the data in Table 9.5-2, compute the volume fraction and surface area fractions of benzene and 2,2,4-trimethyl pentane in this mixture.

Solution Benzene consists of six aromatic CH (ACH) groups. Therefore, rB = 6 × 0.5313 = 3.1878 qB = 6 × 0.4000 = 2.4000

The structure of 2,2,4-trimethyl pentane is CH3

—

— —

CH3

CH3 — C — CH2 — CH — CH3 CH3

This molecule consists of five CH3 groups, one CH2 group, one CH group, and one C group. Thus rTMP = 5 × 0.9011 + 0.6744 + 0.4469 + 0.2195 = 5.8463

9.5 Several Correlative Liquid Mixture Activity Coefficient Models 457 Table 9.5-2 The Group Volume and Surface Area Parameters, R and Q, for Use with the UNIQUAC and UNIFAC Models* Main Group CH2

C=C

ACH ACCH2

OH

CH3 OH Water ACOH CH2 CO CHO CCOO HCOO CH2 O

CNH2

CNH

(C)3 N ACNH2 Pyridines

CCN COOH HCOOH

Subgroup

R

Q

CH3 CH2 CH C CH2 =CH CH=CH CH2 =C CH=C C=C ACH AC ACCH3 ACCH2 ACCH OH(p) OH(s) OH(t) CH3 OH H2 O ACOH CH3 CO CH2 CO CHO CH3 COO CH2 COO HCOO CH3 O CH2 O CHO CH3 NH2 CH2 NH2 CHNH2 CNH2 CH3 NH CH2 NH CHNH CH3 N CH2 N ACNH2 AC2 H2 N AC2 HN AC2 N CH3 CN CH2 CN COOH HCOOH

0.6325 0.6325 0.6325 0.6325 1.2832 1.2832 1.2832 1.2832 1.2832 0.3763 0.3763 0.9100 0.9100 0.9100 1.2302 1.0630 0.6895 0.8585 1.7334 1.0800 1.7048 1.7048 0.7173 1.2700 1.2700 1.9000 1.1434 1.1434 1.1434 1.6607 1.6607 1.6607 1.6607 1.3680 1.3680 1.3680 1.0746 1.0746 1.1849 1.4578 1.2393 1.0731 1.5575 1.5575 0.8000 0.8000

1.0608 0.7081 0.3554 0.0000 1.6016 1.2489 1.2489 0.8962 0.4582 0.4321 0.2113 0.9490 0.7962 0.3769 0.8927 0.8663 0.8345 0.9938 2.4561 0.9750 1.6700 1.5542 0.7710 1.6286 1.4228 1.8000 1.6022 1.2495 0.8968 1.6904 1.3377 0.9850 0.9850 1.4332 1.0805 0.7278 1.1760 0.8240 0.8067 0.9022 0.6330 0.3539 1.5193 1.1666 0.9215 1.2742

Example Assignments n-Hexane: 4 CH2 , 2 CH3 n-Heptane: 5 CH2 , 2 CH3

2-Methylpropane: 1 CH, 3 CH3 Neopentane: 1 C, 4 CH3 1-Hexene: 1 CH2 =CH, 3 CH2 , 1 CH3 2-Hexene: 1 CH=CH, 2 CH3 , 2 CH2 2-Methyl-1-butene: 1 CH2 =C, 1 CH2 , 2 CH3 2-Methyl-2-butene: 1 CH=C, 2 CH3 12,3-Dimethyl-2-butene: 1 C=C, 4 CH3 Benzene: 6 ACH Styrene: 1 CH2 =C, 5 ACH2 , 1 ACH Toluene: 5 ACH, 1 ACCH3 Ethylbenzene: 5 ACH, 1 ACCH2 , 1 CH3 Isopropylbenzene: 5 ACH, 1 ACCH, 2 CH3 1-Propanol: 1 OH(p), 1 CH3 , 2 CH2 2-Propanol: 1 OH(s), 2 CH3 , 1 CH tert-Butanol: 1 OH(t), 3 CH3 , 1 C Methanol Water Phenol: 1 ACOH, 5 ACH 2-Butanone: 1 CH3 CO, 1 CH3 , 1 CH2 2-Pentanone: 1 CH2 CO, 2 CH3 , 1 CH2 Propionic aldehyde: 1 CHO, 1 CH3 , 1 CH2 Butyl acetate: 1 CH3 COO, 1 CH3 , 3 CH2 Methyl propionate: 1 CH2 COO, 2 CH3 Ethyl formate: 1 HCOO, 1 CH3 , 1 CH2 Dimethyl ether: 1 CH3 CO, 1 CH3 Diethyl ether: 1 CH2 O, 2 CH3 , 1 CH2 Diisopropyl ether: 1 CHO, 4 CH3 , 1 CH Methylamine: CH3 NH2 Ethylamine: 1 CH2 NH2 , 1 CH3 Isopropylamine: 1 CHNH2 , 2 CH3 tert-Butylamine: 1 CNH2 , 3 CH3 Dimethylamine: CH3 NH, 1 CH3 Diethylamine: 1 CH2 NH, 2 CH3 , 1 CH2 Diisopropylamine: 1 CHNH, 4 CH3 , 1 CH Trimethylamine: 1 CH3 N, 2 CH3 Triethylamine: 1 CH2 N, 2 CH2 , 3 CH3 Aniline: 1 ACNH2 , 5 ACH Pyridine: 1 AC2 H2 N, 3ACH 2-Methylpyridine: 1 AC2 HN, 3 ACH, 1 CH3 2,5-Methylpyridine: 1 AC2 N, 3 ACH, 2 CH3 Acetonitrile Propionitrile: 1 CH2 CN, 1 CH3 Acetic acid: 1 COOH, 1 CH3 Formic acid

*The parameters for the UNIQUAC and UNIFAC models have been supplied by Prof. J. Gmehling of the University of Oldenburg, Germany, supported by the UNIFAC Consortium.

(continued)

458 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Table 9.5-2 (Continued) Main Group CCl

CCl2

CCl3 CCl4 ACCl CNO2

ACNO2 CS2 CH3 SH Furfural Diol I Br C≡C DMSO ACRY Cl(C=C) ACF DMF CF2

COO cy-CH2

cy-CH2 O

cy-CON-C

ACS

Subgroup

R

Q

CH2 Cl CHCl CCl CH2 Cl2 CHCl2 CCl2 CHCl3 CCl3 CCl4 ACCl CH3 NO2 CH2 NO2 CHNO2 ACNO2 CS2 CH3 SH CH2 SH furfural (CH2 OH)2 I Br CH≡C C≡C DMSO ACRY Cl(C=C) ACF DMF HCON(CH2 )2 CF3 CF2 CF COO cy-CH2 cy-CH cy-C cy-CH2 OCH2 cy-CH2 O(CH2 )1/2 cy-(CH2 )1/2 O(CH2 )1/2 cy-CON-CH3 cy-CON-CH2 cy-CON-CH cy-CON-C AC2 H2 S AC2 HS AC2 S

0.9919 0.9919 0.9919 1.8000 1.8000 1.8000 2.4500 2.6500 2.6180 0.5365 2.6440 2.5000 2.8870 0.4656 1.2400 1.2890 1.5350 1.2990 2.0880 1.0760 1.2090 0.9214 1.3030 3.6000 1.0000 0.5229 0.8814 2.0000 2.3810 1.2840 1.2840 0.8215 1.6000 0.7136 0.3479 0.3470 1.7023 1.4046 1.0413 3.9819 3.7543 3.5268 3.2994 1.7943 1.6282 1.4621

1.3654 1.0127 0.6600 2.5000 2.1473 1.7946 2.8912 2.3778 3.1836 0.3177 2.5000 2.3040 2.2410 0.3589 1.0680 1.7620 1.3160 1.2890 2.4000 0.9169 1.4000 1.3000 1.1320 2.6920 0.9200 0.7391 0.7269 2.0930 1.5220 1.2660 1.0980 0.5135 0.9000 0.8635 0.1071 0.0000 1.8784 1.4000 1.0116 3.2000 2.8920 2.5800 2.3520 1.3400 1.0600 0.7800

Example Assignments 1-Chlorobutane: 1 CH2 Cl, 1 CH3 , 2 CH2 2-Chloropropane: 1 CHCl, 2 CH3 tert-Butyl chloride: 1 CHCl, 3 CH3 Dichloromethane: 1 CH2 Cl2 1,1-Dichloroethane: 1 CHCl2 , 1 CH3 2,2-Dichloropropane: 1 CCl2 , 2 CH3 Chloroform 1,1,1-Trichloroethane: 1 CCl3 , 1 CH3 Tetrachloromethane Chlorobenzene: 1 ACCl, 5 ACH Nitromethane 1-Nitropropane: 1 CH2 NO2 , 1 CH3 , 1 CH2 2-Nitropropane: 1 CHNO2 , 2 CH3 Nitrobenzene: 1 ACNO2 , 5 ACH Carbon disulfide Methanethiol Ethanethiol: 1 CH2 SH, 1 CH3 Furfural 1,2-Ethanediol (ethylene glycol) Ethyl iodide: 1 I, 1 CH3 , 1 CH2 Ethyl bromide: 1 Br, 1 CH3 , 1 CH2 1-Hexyne: 1 CH≡C, 1 CH3 , 3 CH2 2-Hexyne: 1 C≡C, 2 CH3 , 2 CH2 Dimethyl sulfoxide Acrylonitrile Trichloroethylene: 3 Cl(C=C), 1 CH=C Hexafluorobenzene: 6 ACF N, N -Dimethylformamide N, N -Diethylformamide: 1 HCON(CH2 )2 , 2 CH3 1,1,1-Trifluoroethane: 1 CF3 , 1 CH3 Perfluorohexane: 4 CF2 , 2 CF3 Perfluoromethylcyclohexane: 1 CF, 5 CF2 , 1 CF3 Methyl acrylate: 1 COO, 1 CH3 , 1 CH2 =CH Cyclohexane: 6 cy-CH2 Methylcyclohexane: 1 cy-CH, 5 cy-CH2 , 1 CH3 1,1-Dimethylcyclohexane: 1 cy-C, 5 cy-CH2 , 2 CH3 Tetrahydrofuran: 1 cy-CH2 OCH2 , 2 cy-CH2 1,3-Dioxane: 2 cy-CH2 O(CH2 )1/2 , 1 cy-CH2 1,3,5-Trioxane: 3 cy-(CH2 )1/2 O(CH2 )1/2 N -Methylpyrrolidone: 1 cy-CON-CH3 , 3 cy-CH2 N -Ethylpyrrolidone: 1 cy-CON-CH2 , 3 cy-CH2 , 1 CH3 N -Isopropylpyrrolidone: 1 cy-CON-CH, 3 cy-CH2 , 2 CH3 N -tert-Butylpyrrolidone: 1 cy-CON-C, 3 cy-CH2 , 3 CH3 Thiophene: 1 AC2 H2 S, 2 ACH 2-Methylthiophene: 1 AC2 HS, 2 ACH, 1 CH3 2,5-Dimethylthiophene: 1 AC2 S, 2 ACH, 2 CH3

Note: A (as in ACH) denotes a group in an aromatic ring, cy- denotes a group in a cyclic structure, and functional groups in bold without any example assignments, such as water, formic acid, etc. are specific to that molecule.

9.5 Several Correlative Liquid Mixture Activity Coefficient Models 459 and

qTMP = 5 × 0.8480 + 0.5400 + 0.2280 + 0.0 = 5.0080

Consequently, θB =

0.5 × 2.4000 = 0.324 0.5 × 2.4000 + 0.5 × 5.0080

θTMP = 0.676

φB =

0.5 × 3.1878 = 0.353 0.5 × 3.1878 + 0.5 × 5.8463

φTMP = 0.647

Finally, it is useful to note that the Gibbs-Duhem equation can be used to get information about the activity coefficient of one component in a binary mixture if the concentration dependence of the activity coefficient of the other species is known. This is demonstrated in the next illustration. Illustration 9.5-3 Interrelating the Two Activity Coefficients in a Binary Mixture The activity coefficient for species 1 in a binary mixture can be represented by ln γ1 = ax22 + bx32 + cx42

where a, b, and c are concentration-independent parameters. What is the expression for ln γ2 in terms of these same parameters?

Solution For a binary mixture at constant temperature and pressure we have, from Eq. 8.2-20 or 9.3-17, x1

∂ ln γ1 ∂ ln γ2 + x2 =0 ∂x2 ∂x2

Since x1 = 1 − x2 and ln γ1 = ax22 + bx32 + cx42 , we have ∂ ln γ2 x1 ∂ ln γ1 =− ∂x2 x2 ∂x2 (1 − x2 ) (2ax2 + 3bx22 + 4cx32 ) x2 = −2a + (2a − 3b)x2 + (3b − 4c)x22 + 4cx32 =−

Now, by definition, γ2 (x2 = 1) = 1, and ln γ2 (x2 = 1) = 0. Therefore, 

x2 x2 =1

∂ ln γ2 dx2 = ln γ2 (x2 ) − ln γ2 (x2 = 1) = ln γ2 (x2 ) ∂x2  x2 = [−2a + (2a − 3b)x2 + (3b − 4c)x22 + 4cx32 ] dx2 x2 =1

= −2a(x2 − 1) + +

4c 4 (x − 1) 4 2

(2a − 3b) 2 (3b − 4c) 3 (x2 − 1) + (x2 − 1) 2 3

460 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Again using x1 + x2 = 1, or x2 = 1 − x1 , yields ln γ2 = −2a(1 − x1 − 1) + 12 (2a − 3b)(1 − 2x1 + x21 − 1) + 13 (3b − 4c)(1 − 3x1 + 3x21 − x31 − 1) + c(1 − 4x1 + 6x21 − 4x31 + x41 − 1)   3b ln γ2 = a + + 2c x21 − (b + 83 c)x31 + cx41 2

Finally, we should point out that any of the liquid-state activity coefficient models discussed here can also be used for mixtures in the solid state. However, in practice, simpler models, such as the one introduced in Sec. 12.4, are generally used.

9.6 TWO PREDICTIVE ACTIVITY COEFFICIENT MODELS Because the variety of organic compounds of interest in chemical processing is very large, and the number of possible binary, ternary, and other mixtures is essentially uncountable, situations frequently arise in which engineers need to make activity coefficient predictions for systems or at conditions for which experimental data are not available. Although the models discussed in the previous section are useful for correlating experimental data, or for making predictions given a limited amount of experimental information, they are of little value in making predictions when no experimental data are available. This is because in these models there is no theory that relates the values of the parameters to molecular properties. Most of the recent theories of liquid solution behavior have been based on welldefined thermodynamic or statistical mechanical assumptions, so that the parameters that appear can be related to the molecular properties of the species in the mixture, and the resulting models have some predictive ability. Although a detailed study of the more fundamental approaches to liquid solution theory is beyond the scope of this book, we consider two examples here: the theory of van Laar,9 which leads to regular solution theory; and the UNIFAC group contribution model, which is based on the UNIQUAC model introduced in the previous section. Both regular solution theory and the UNIFAC model are useful for estimating solution behavior in the absence of experimental data. However, neither one is considered sufficiently accurate for the design of a chemical process. The theory of van Laar (i.e., the argument that originally led to Eqs. 9.5-9) is based on the assumptions that (1) the binary mixture is composed of two species of similar size and energies of interaction, and (2) the van der Waals equation of state applies to both the pure fluids and the binary mixture.10 The implication of assumption 1 is that the molecules of each species will be uniformly distributed throughout the mixture (see Appendix A9.1) and the intermolecular spacing will be similar to that in the pure fluids. Consequently, it is not unreasonable to expect in this case that at a given temperature and pressure

Δmix V = 0 or V ex = 0 9 See

Footnote 5. Laar was a student of van der Waals.

10 van

9.6 Two Predictive Activity Coefficient Models 461 and

Δmix S = −R

2 

xi ln xi

or S ex = 0

(9.6-1)

i=1

so that

Gex = U ex + P V ex − T S ex = U ex for such liquid mixtures. Thus to obtain the excess Gibbs energy change, and thereby the activity coefficients for this liquid mixture, we need only compute the excess internal energy change on mixing. Since the internal energy, and therefore the excess internal energy, is a state function, U ex may be computed along any convenient path leading from the pure components to the mixture. The following path is used. I Start with x1 moles of pure liquid 1 and x2 moles of pure liquid 2 (where x1 + x2 = 1) at the temperature and pressure of the mixture, and, at constant temperature, lower the pressure so that each of the pure liquids vaporizes to an ideal gas. II At constant temperature and (very low) pressure, mix the ideal gases to form an ideal gas mixture. III Now compress the gas mixture, at constant temperature, to a liquid mixture at the final pressure P . The total internal energy change for this process (which is just the excess internal energy change since Δmix U IM = 0) is the sum of the internal energy changes for each of the steps:

Gex = U ex = ΔU I + ΔU II + ΔU III Noting that



∂U ∂V



=T T

∂P ∂T

(9.6-2)

−P V

and using the facts that V → ∞ as P → 0 and that ΔU II = Δmix U IGM = 0, one obtains    

∞ ∂P Gex = ΔU = x1 − P dV T ∂T V V1 pure fluid 1    

∞ ∂P + x2 − P dV T ∂T V V2 pure fluid 2    

V mix ∂P + − P dV (9.6-3) T ∂T V ∞ mixture

Each of the bracketed terms represents the internal energy change on going from a liquid to an ideal gas and is equal to the internal energy change on vaporization.

462 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Next we use assumption 2, that the van der Waals equation of state is applicable to both pure fluids and the mixture. From

a RT − (V − b) V 2

P = it is easily established that

∂U ∂V



=T T

∂P ∂T

(9.6-4)



a V2

−P = V

so that Eq. 9.6-3 becomes



∞

a1 2 dV + x2 V V1 a1 a2 amix = x1 + x2 − V1 V 2 V mix

Gex = ΔU = x1



∞ V2

a2 dV − V2



∞ V mix

amix dV V2

(9.6-5)

where the molar volumes appearing in this equation are those of the liquid. Since the molecules of a liquid are closely packed, liquids are relatively incompressible; that is, enormous pressures are required to produce relatively small changes in the molar volume. Such behavior is predicted by Eq. 9.6-4 if V ≈ b. Making this substitution in Eq. 9.6-5 gives a1 a2 amix Gex = ΔU = x1 + x2 − b1 b2 bmix Now using the mixing rules of Eq. 9.4-8 (with kij = 0), we obtain the following expression for the excess internal energy of a van Laar mixture: √ √ 2 a1 a2 x1 x2 b1 b2 ex ex U = G = ΔU = − x1 b1 + x2 b2 b1 b2 Finally, differentiating this expression with respect to composition yields Gex i and, from Eq. 9.3-12, one obtains the van Laar equations for the activity coefficients,

α ln γ1 =  2 αx1 1+ βx2

and

β ln γ2 =  2 βx2 1+ αx1

with

b1 α= RT

√

√ 2 a1 a2 − b1 b2

b2 and β = RT

√

√ 2 a1 a2 − b1 b2

(9.6-6)

This development provides both a justification for the van Laar equations and a method of estimating the van Laar parameters for liquid-phase activity coefficients from the parameters in the van der Waals equation of state. Since we know the van der Waals equation is not very accurate, it is not surprising that if α and β are treated as adjustable parameters, the correlative value of the van Laar equations is greater than when α and β are determined from Eqs. 9.6-6 (Problem 9.9).

9.6 Two Predictive Activity Coefficient Models 463 Scatchard,11 based on the observations of Hildebrand,12 concluded that although V ex and S ex were approximately equal to zero for some solutions (such mixtures are called regular solutions), few obeyed the van der Waals equation of state. Therefore, he suggested that instead of using an equation of state to predict the internal energy change on vaporization, as in the van Laar theory, the experimental internal energy change on vaporization (usually at 25◦ C) be used. It was also suggested that the internal energy change of vaporization for the mixture be estimated from the approximate mixing rule ! 2  ! V 1 Δvap U 1 V 2 Δvap U 2 (Δvap U )mix = x1 + x2 (9.6-7) V mix V mix since experimental data on the heat of vaporization of mixtures are rarely available. In this equation all molar volumes are those of the liquids, and V mix = x1 V 1 + x2 V 2 , since Δmix V = 0 by the first van Laar assumption. Defining the volume fraction Φi and solubility parameter δi of species i by

Φi ≡ and

δi ≡

xi V i V mix

Δvap U i Vi

(9.6-8)

1/2 (9.6-9)

one obtains (see Eq. 9.6-3)

Gex = U ex = x1 Δvap U 1 + x2 ΔU vap − (Δvap U )mix 2 2 = (x1 V 1 + x2 V 2 )Φ1 Φ2 [δ1 − δ2 ] which, on differentiation, yields

RT ln γ1 = V 1 Φ22 [δ1 − δ2 ]2 RT ln γ2 = V 2 Φ21 [δ1 − δ2 ]2

Regular solution model activity coefficients

(9.6-10)

Thus, we have a recipe for estimating the activity coefficients of each species in a binary liquid mixture from a knowledge of the pure-component molar volumes, the mole (or volume) fractions, and the solubility parameters (or internal energy changes on vaporization) of each species. Table 9.6-1 gives a list of the solubility parameters and molar volumes for a number of nonpolar liquids. Note, however, that the assumptions contained in regular solution theory (i.e., V ex = S ex = 0 and Eq. 9.6-7) are not generally applicable to polar substances. Therefore, regular solution theory should be used only for the components listed in Table 9.6-1 and similar compounds. Regular solution theory is functionally equivalent to the van Laar theory since, with the substitutions

α= 11 G.

V1 (δ1 − δ2 )2 RT

and β =

V2 (δ1 − δ2 )2 RT

Scatchard, Chem Rev., 8, 321 (1931).

12 J. H. Hildebrand, J. Am. Chem. Soc., 41, 1067 (1919). This is discussed in J. H. Hildebrand, J. M. Prausnitz, and

R. L. Scott, Regular and Related Solutions, Van Nostrand-Reinhold, Princeton, N.J. (1970).

464 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Table 9.6-1 Molar Liquid Volumes and Solubility Parameters of Some Nonpolar Liquids V L (cc/mol)

δ (cal/cc)1/2

38.1 37.1 29.0 28.0 35.3 46.0 45.7

5.3 5.7 6.8 7.2 7.4 8.3 9.5

Liquefied gases at 90 K Nitrogen Carbon monoxide Argon Oxygen Methane Carbon tetrafluoride Ethane Liquid solvents at 25◦ C Perfluoro-n-heptane Neopentane Isopentane n-Pentane n-Hexane 1-Hexene n-Octane n-Hexadecane Cyclohexane Carbon tetrachloride Ethyl benzene Toluene Benzene Styrene Tetrachloroethylene Carbon disulfide Bromine

226 122 117 116 132 126 164 294 109 97 123 107 89 116 103 61 51

6.0 6.2 6.8 7.1 7.3 7.3 7.5 8.0 8.2 8.6 8.8 8.9 9.2 9.3 9.3 10.0 11.5

Source: J. M. Prausnitz, Molecular Thermodynamics of Fluid-Phase Equilibria. 1969. Reprinted with permission from Prentice-Hall, Englewood Cliffs, N.J. Note: In regular solution theory the solubility parameter has traditionally been given in the units shown. For this reason the traditional units, rather than SI units, appear in this table.

Eqs. 9.5-9 and 9.6-10 are identical. The important advantage of regular solution theory, however, is that we can calculate its parameters without resorting to activity coefficient measurements. Unfortunately, the parameters derived in this way are not as accurate as those fitted to experimental data. Illustration 9.6-1 Test of the Regular Solution Model Compare the regular solution theory predictions for the activity coefficients of the benzene– 2,2,4-trimethyl pentane mixture with the experimental data given in Illustration 9.5-1.

Solution From Table 9.6-1 we have V L = 89 cc/mol and δ = 9.2 (cal/cc)1/2 for benzene. The parameters for 2,2,4-trimethyl pentane are not given. However, the molar volume of this compound is approximately 165 cc/mol, and the solubility parameter can be estimated from

9.6 Two Predictive Activity Coefficient Models 465  δ=

Δvap U VL

1/2

 =

Δvap H − RT VL

1/2

where in the numerator we have neglected the liquid molar volume with respect to that of the vapor, and further assumed ideal vapor-phase behavior. Using the value of Δvap H found in Illustration 7.7-1, we obtain δ = 6.93 (cal/cc)1/2 . Our predictions for the activity coefficients, together with the experimental data, are plotted in Fig. 9.6-1. It is evident that although the regular solution theory prediction leads to activity coefficient behavior that is qualitatively correct, the quantitative agreement in this case is, in fact, poor. This example should serve as a warning that theoretical predictions may not always be accurate and that experimental data are to be preferred.

The extension of regular solution theory to multicomponent mixtures is an algebraically messy task; the final result is

RT ln γi = V i (δi − δ)2

(9.6-11)

where

⎛

⎞ Volume fraction  δ = ⎝averaged solubility⎠ = Φj δj parameter j

1.4

1.2

1.0

γMP

0.8 ln γ γB 0.6

0.4

0.2

0

0

0.2

0.4

0.6

0.8

1.0

xB

Figure 9.6-1 The regular solution theory predictions for the activity coefficients of the benzene– 2,2,4-trimethyl pentane mixture. The points indicated by ◦ are the experimental data.

466 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture and

xj V j Φj =  xk V k k

where each of the sums is over all the species in the mixture. Several other characteristics of regular solution theory are worth noting. The first is that the theory leads only to positive deviations from ideal solution behavior in the sense that γi ≥ 1. This result can be traced back to the assumption of Eq. 9.6-7, which requires that (Δvap U )mix always be intermediate to Δvap U of the two pure components. Also, the solubility parameters are clearly functions of temperature since δ → 0 as T → Tc for each species; however, (δ1 − δ2 ) is often nearly temperature independent, at least over limited temperature ranges. Therefore, the solubility parameters listed in Table 9.6-1 may be used at temperatures other than the one at which they were obtained (see Eq. 9.3-22, however). Also, referring to Eq. 9.6-10, it is evident that liquids with very different solubility parameters, such as neopentane and carbon disulfide, can be expected to exhibit highly nonideal solution behavior (i.e., γi > 1), whereas adjacent liquids in Table 9.6-1 will form nearly ideal solutions. This is useful information. The most successful recent activity coefficient prediction methods are based on the idea of group contributions, in which each molecule is considered to be a collection of basic building blocks, the functional groups discussed in the last section. Thus, a mixture of molecules is considered to be a mixture of functional groups, and its properties result from functional group interactions. Because the number of different types of functional groups is very much smaller than the number of different molecular species, it is possible, by the regression of experimental data, to obtain a fairly compact table of parameters for the interaction of each group with all others. From such a table, the activity coefficients in a mixture for which no experimental data are available can be estimated from a knowledge of the functional groups present. The two most developed group contribution methods are the ASOG (Analytical Solution Of Groups)13 and UNIFAC (UNIquac Functional-group Activity Coefficient)14 models, both of which are the subjects of books. We will consider only the UNIFAC model here. UNIFAC is based on the UNIQUAC model of Sec. 9.5. This model, you will remember, has a combinatorial term that depends on the volume and surface area of each molecule, and a residual term that is a result of the energies of interaction between the molecules. In UNIQUAC the combinatorial term was evaluated using group contributions to compute the size parameters, whereas the residual term had two adjustable parameters for each binary system that were to be fit to experimental data. In the UNIFAC model, both the combinatorial and residual terms are obtained using group contribution methods. When using the UNIFAC model, one first identifies the functional subgroups present in each molecule using the list in Table 9.5-2. Next the activity coefficient for each species is written as

ln γi = ln γi (combinatorial) + ln γi (residual)

13 K.

Kojima and T. Tochigi, Prediction of Vapor-Liquid Equilibrium by the ASOG Method, Elsevier, Amsterdam (1979). 14 A. Fredenslund, J. Gmehling, and P. Rasmussen, Vapor-Liquid Equilibrium Using UNIFAC, Elsevier, Amsterdam (1977).

9.6 Two Predictive Activity Coefficient Models 467 The combinatorial term is evaluated from a modified form of the comparable term in the UNIQUAC equation (Eq. 9.5-23a):

φi φi φi  φi  z ln γi (combinatorial) = ln +1− − qi 1 + ln − (9.6-12a) xi xi 2 θi θi  3/4 3/4 xj rj , with the r and q parameters evaluated using the group where φi = xi ri j

contribution discussed in the previous section. Here, however, the residual term is also evaluated by a group contribution method, so that the mixture is envisioned as being a mixture of functional groups rather than of molecules. The residual contribution to the logarithm of the activity coefficient of group k in the mixture, ln Γk , is computed from the group contribution analogue of Eq. 9.5-23b, which is written as     Θm Ψkm ' & ln Γk = Qk 1 − ln Θm Ψmk − (9.6-12b)  Θn Ψnm m m n

where

⎛

⎞ Surface area Xm Qm Θm = ⎝ fraction of ⎠ =  Xn Qn group m n

Xm = mole fraction of group m in mixture and

(9.6-13a)



Ψmn

 & −a ' −(umn − unn ) mn = exp = exp kT T

(9.6-13b)

where umn is a measure of the interaction energy between groups m and n, and the sums are over all groups in the mixture. Finally, the residual part of the activity coefficient of species i is computed from

ln γi (residual) =

The UNIFAC model



vk(i) [ln Γk − ln Γ(i) k ]

(9.6-14)

k (i)

(i)

Here vk is the number of k groups present in species i, and Γk is the residual contribution to the activity coefficient of group k in a pure fluid of species i molecules. This last term is included to ensure that in the limit of pure species i, which is still a mixture of groups (unless species i molecules consist of only a single functional group), ln γi (residual) is zero or γi (residual) = 1. The combination of Eqs. 9.6-12 through 9.6-14 is the UNIFAC model. Since the volume (Ri ) and surface (Qi ) parameters are known (Table 9.5-2), the only unknowns are binary parameters, anm and amn , for each pair of functional groups. Continuing with the group contribution idea, it is next assumed that any pair of functional groups m and n will interact in the same manner, that is, have the same value of amn and anm , independent of the mixtures in which these two groups occur. Thus, for example, it is assumed that the interaction between an alcohol (—OH) group and a methyl (—CH3 ) group will be the same, regardless of whether these groups occur in ethanol–n-pentane, isopropanol-decane, or 2-octanol–2,2-4-trimethyl pentane mixtures.

468 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Consequently, by a regression analysis of very large quantities of activity coefficient (or, as we will see in Sec. 10.2, actually vapor-liquid equilibrium) data, the binary parameters anm and amn for many group-group interactions can be determined. These parameters can then be used to predict the activity coefficients in mixtures (binary or multicomponent) for which no experimental data are available. In the course of such an analysis it was found that (1) experimental data existed to determine many of, but not all, the binary group parameters anm and amn and (2) that some very similar groups, such as the —CH3 , —CH2 , —CH, and —C groups, interact with other groups in approximately the same way and, therefore, have the same interaction parameters with other groups. Such very similar subgroups are considered to belong to the same main group. Consequently, in Table 9.5-2 there are 46 main groups among the 106 subgroups. All the subgroups within a main group (i.e., the CH3 , CH2 , CH, and C subgroups within the CH2 main group) have the same values of the binary parameters for interactions with other main groups and zero values for the interactions with other subgroups in their own main group. Appendix B describes several programs that use the UNIFAC model for activity coefficient predictions that are the on the website for this book. The Visual Basic and MATLAB programs use the recent UNIFAC model described in this chapter, whereas the DOS Basic version uses an earlier model (the storage limitations in DOS Basic did not allow updating this program to accommodate the many new parameters that had been added). Therefore, the Visual Basic or MATLAB versions of the program can be used; the DOS-based version has been retained should the reader wish to examine the difference in the two methods. HowR contains the UNIFAC model with the latest parameters and has an ever, Aspen Plus easier to use user interface therefore is preferred. [See the folder Aspen Illustrations> Chapter 9>9.6-5 on Wiley website for this book.] Illustration 9.6-2 Test of the UNIFAC Model Compare the UNIFAC predictions for the activity coefficients of the benzene–2,2,4-trimethyl pentane mixture with the experimental data given in Illustration 9.5-1.

Solution Benzene consists of six aromatic CH groups (i.e., subgroup 10 of Table 9.5-2), whereas 2,2,4trimethyl pentane contains five CH3 groups (subgroup 1), one CH2 group (subgroup 2), one CH R group (subgroup 3), and one C group (subgroup 4). Using the UNIFAC program of Aspen Plus with the folder Aspen Illustrations>Chapter 9>9.6-2 on the Wiley website for this book, the results plotted in Fig. 9.6-2 are obtained. It is clear from this figure that, for this simple system, the UNIFAC predictions are good— much better than the regular solution predictions of Fig. 9.6-1. Although the UNIFAC predictions for all systems are not always as good as for the benzene–2,2,4-trimethyl pentane system, UNIFAC with its recent improvements is the best activity coefficient prediction method currently available. The

Excel file Illustration 9.6-2 in Appendix C on the website for this book gives the results.

9.7 FUGACITY OF SPECIES IN NONSIMPLE MIXTURES To estimate the fugacity of a species in a gaseous mixture using the Lewis-Randall rule, V fV i (T, P, y) = yi fi (T, P )

(9.7-1)

9.7 Fugacity of Species in Nonsimple Mixtures 469 0.8

0.6

ln γ 0.4 γTMP γB

0.2

0

0

0.2

0.4

0.6

0.8

1.0

xB

Figure 9.6-2 UNIFAC predictions for the activity coefficients of the benzene–2,2,4-trimethyl pentane system. The points are activity coefficients derived from experimental data.

we need the fugacity of pure species i as a vapor at the temperature and pressure of the mixture. In a nonsimple gaseous mixture at least one of the pure components does not exist as a vapor at the mixture temperature and pressure, so fiV (T, P ) for that species cannot be computed without some sort of approximation or assumption. To estimate the fugacity of a species in a liquid mixture from an activity coefficient model, one uses

f Li (T, P, x) = xi γi (T, P, x)fiL (T, P )

(9.7-2)

and one would need the fugacity of pure species i as a liquid at the temperature and pressure of the mixture. In a nonsimple liquid mixture, at least one of the components, if pure, would be a solid or a vapor at the temperature and pressure of the mixture, so it is not evident how to compute fiL (T, P ) in this case. The study of vapor-liquid equilibria (Sec 10.1) of the solubility of gases in liquids (Sec. 11.1), and of the solubility of solids in liquids (Sec. 12.1), all involve nonsimple mixtures. To see why this occurs, consider the criterion for vapor-liquid equilibrium:

f Li (T, P, x) = f V i (T, P, y) Using Eqs. 9.7-1 and 9.7-2, we have

xi γi (T, P, x)fiL (T, P ) = yi fiV (T, P ) To use this equation we must estimate the pure component fugacity of each species as both a liquid and a vapor at the temperature and pressure of the mixture. However, at this temperature and pressure either the liquid or the vapor will be the stable phase for each species, generally not both.15 Consequently, at least one of the phases will be a nonsimple mixture. In most vaporliquid problems both phases will be nonsimple mixtures, in that species with pure component vapor pressures greater than the system pressure appear in the liquid phase, and those with vapor pressures less than the system pressure appear in the vapor phase. For those situations one can proceed in several ways. The simplest, and most accurate when it is applicable, is to use equations of state to compute species fugacities in 15 Unless,

fortuitously, the system pressure is equal to the vapor pressure of that species.

470 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture mixtures, thereby avoiding the use of Eqs. 9.7-1 and 9.7-2 and the necessity of computing a pure component fugacity in a thermodynamic state that does not occur. Thus, when an equation of state (virial, cubic, etc.) can be used for the vapor mixture, a nonsimple gaseous mixture can be treated using the methods of Sec. 9.4. Similarly, if a nonsimple liquid mixture can be described by an equation of state, which is likely to be the case only for hydrocarbons and perhaps hydrocarbons with dissolved gases, it can also be treated by the methods of Sec 9.4. If an equation of state cannot be used, one can, in principle, proceed in either of two ways. The first procedure is to write the Gibbs energy of mixing as

Δmix G = ΔG + Δmix GIM + Gex where Δmix GIM and Gex have their usual meanings, and ΔG is the Gibbs energy change of converting to the same phase as the mixture those species that exist in other phases as pure substances. For example, ΔG might be the Gibbs energy change of producing pure liquids from either gases or solids before forming a liquid mixture. The partial molar Gibbs energy and fugacity of each species in the mixture would then be computed directly from Δmix G. In practice, however, the calculation of ΔG can be difficult, since it may involve the estimation of the Gibbs energy of substances in hypothetical states (i.e., a liquid above its critical point). A second, more straightforward procedure is to use Eqs. 9.7-1 and 9.7-2 and the models for γi considered in Secs. 9.5 and 9.6, but with estimates for the pure component fugacities of the hypothetical gases and liquids obtained by simple extrapolation procedures. Of course, such extrapolation schemes have no real physical or chemical basis; they are, rather, calculational methods that have been found empirically to lead to satisfactory predictions, provided the extrapolation is not too great. The first extrapolation scheme to be considered is for the liquid-phase fugacity of a species that would be a vapor as a pure component at the temperature and pressure of the mixture. The starting point here is the observation that the fugacity of a pure liquid is equal to its vapor pressure if the vapor pressure is not too high (or the product of P vap and the fugacity coefficient at higher vapor pressures). Consequently, one takes the fugacity of the hypothetical liquid at the temperature T and the pressure P to be equal to the vapor pressure of the real liquid at the same temperature [even though P vap (T ) > P ]. Thus, in Illustration 10.1-2, where we consider vapor-liquid equilibrium for an n-pentane, n-hexane, and n-heptane mixture at 69◦ C and 1 bar, the fugacity of “liquid” n-pentane, for use in Eq. 9.7-2, will be taken to be equal to its vapor pressure at 69◦ C, 2.721 bar. To calculate the fugacity of a pure vapor from corresponding states that, at the conditions of the mixture, exists only as a liquid, we will use Eq. 7.8-1 with fiV (T, P ) equal to the total pressure, if the pressure is low enough, or

f V fi (T, P ) = P P i at higher pressures. Here, however, the fugacity coefficient is obtained not from Fig. 7.4-1, but rather from Fig. 9.7-1, which is a corresponding states correlation in which the fugacity coefficient for gases has been extrapolated into the liquid region. (You should compare Figs. 7.4-1 and 9.7-1.) Extrapolation schemes may also be used in some circumstances where the desired phase does not exist at any pressure for the temperature of interest—for example, to estimate the fugacity of a “liquid” not too far above its critical temperature, or not much

9.7 Fugacity of Species in Nonsimple Mixtures 471 3.5 3

1.0

1.6 1.2 1.0

0.9

2.5

ƒ P

0.95

0.8

0.80

0.85

0.7

0.70

0.75

0.55 0.60

0.65

Low pressure range

0.6

0

Tr = 0.50

0.1

0.2

Pr

0.3

1.0 0.9 0.8 0.7

0.4

0.5

6 8 0 1 2.0

2.0

1.5

1.5

1.3 1.1

1.3

1.0

0.6 0.5

0.8 Re d

0.4 0.35

uc

ed

0.3

0.6

te

1.1

0.9

0.7

1. 0

Fugacity coefficient

1.1

0.90

2 1.5

1.4

m

pe

8

0.9

ra t

0.25

ur eT

r

0.2

=

0.5

0.96

0.9

0.15

0.9

4

0.1 0.2

0.94

0.3

0.4

0.5

0.7

1.0

1.5

2

2.5

3

4

5

6

7 8 9 10

15

20

25

30

40

Reduced pressure Pr = P/Pc

Figure 9.7-1 Fugacity coefficients of gases and vapors. (Reprinted with permission from O. A. Hougen and K. M. Watson, Chemical Process Principles Charts, John Wiley & Sons, New York, 1946.) In this figure Zc = 0.27.

below its triple-point temperature. As an example of the methods used, we consider the estimation of the liquid-phase fugacity at temperatures below its triple point (so that the solid is the stable phase) and also at temperatures above its critical temperature (where the gas is the stable phase) for the substance whose pure component phase diagram is given in Fig. 9.7-2a. In either case the first step in the procedure is to extend the vapor pressure curve, either analytically (using the Clausius-Clapeyron equation) or graphically as indicated in Fig. 9.7-2b, to obtain the vapor pressure of the hypothetical liquid.16 In the case of the subcooled liquid, which involves an extrapolation into the solid region, the vapor pressure is usually so low that the fugacity coefficient is close to unity, and the fugacity of this hypothetical liquid is equal to the extrapolated vapor pressure. For the supercritical liquid, however, the extrapolation is above the critical temperature of the liquid and yields very high vapor pressures, so that the fugacity of this hypothetical liquid is equal to the product of the extrapolated vapor pressure and the fugacity coefficient (which is taken from the corresponding-states plot of Fig. 9.7-1). Another way to estimate the “subcooled” liquid fugacity f1L below the melting point is to use heat (enthalpy) of fusion data and, if available, the heat capacity data for both 16 For

accurate extrapolations ln P vap should be plotted versus 1/T as in Sec. 7.5.

472 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture Vapor pressure of supercritical liquid

Critical point Solid

Liquid

Solid

P

P Triple point Triple point Vapor

Critical point

Liquid

Solid Critical point

Liquid

P Triple point

Vapor pressure of supercooled liquid Vapor

Vapor

T

T

T

(a)

(b)

(c)

Sublimation pressure of superheated solid

Figure 9.7-2 (a) P -T phase diagram for a typical substance. (b) P -T phase diagram with dashed lines indicating extrapolation of the liquid-phase vapor pressure into the solid and supercritical regions. (c) P -T phase diagram with a dashed line indicating extrapolation of the sublimation pressure of the solid into the liquid region.

the solid and liquid to compute the Gibbs energy of fusion Δfus G(T ), which is related to the fugacity ratio as follows:

Δfus G(T, P ) GL (T, P ) − GS1 (T, P ) f L (T, P ) = 1 = ln 1S RT RT f1 (T, P )

(9.7-3)

The value of Δfus G(T ) is computed by separately calculating Δfus H(T ) and Δfus S(T ), and then using the relation Δfus G(T ) = Δfus H(T ) − T Δfus S(T ). To compute the enthalpy and entropy changes of fusion, we suppose that the melting of a solid (below its normal melting point) to form a liquid is carried out in the following three-step constant-pressure process: 1. The solid is heated at fixed pressure from the temperature T to its normal melting temperature Tm . 2. The solid is then melted to form a liquid. 3. The liquid is cooled without solidification from Tm back to the temperature of the mixture. The enthalpy and entropy changes for this process are





Tm

CPS

Δfus H(T ) =

T

CPL dT

dT + Δfus H(Tm ) +

T



Tm

(9.7-4)

T

= Δfus H(Tm ) +

ΔCP dT Tm



Tm

CPS Δfus S(T ) = dT + Δfus S(Tm ) + T T  T ΔCP dT = Δfus S(Tm ) + T Tm where ΔCP = CPL − CPS .



T

Tm

CPL dT T (9.7-5)

9.7 Fugacity of Species in Nonsimple Mixtures 473 Note that Eqs. 9.7-4 and 9.7-5 relate the enthalpy and entropy changes of fusion at any temperature T to those changes at the melting point at the same pressure. Now since G = H − T S , and Δfus G(T = Tm ) = 0, Eq. 9.7-5 can be rewritten as  T Δfus H(Tm ) ΔCP dT Δfus S(T ) = + (9.7-6) Tm T Tm and, therefore,

Δfus G(T ) = Δfus H(T ) − T Δfus S(T )    T  T T ΔCP dT ΔCP dT − T = Δfus H(Tm ) 1 − + Tm T (9.7-7) Tm Tm   L f (T, P ) ≡ RT ln 1S f1 (T, P ) Thus

f1L (T, P ) =

f1S



 T   T T CP 1 + fus H (T ) 1 − CP dT − T dT (T, P ) exp RT Tm T Tm Tm 

(9.7-8a) As heat capacity data may not be available, this equation is usually simplified to



f1L

(T, P ) =

f1S

fus H (T ) (T, P ) exp RT



 T 1− Tm

(9.7-8b)

Therefore, if the sublimation pressure of the solid, which is equal to the solid fugacity f1S and the heat of fustion at the melting point, are known, the fugacity of the hypothetical liquid, or liquid below its melting temperature, can be computed. The fugacity of a hypothetical superheated solid can be estimated by extrapolating the sublimation pressure line into the liquid region of the phase diagram. This is indicated in Fig. 9.7-2c. The fugacity coefficient is usually equal to unity in this case. For species whose thermodynamic properties are needed in hypothetical states far removed from their stable states, such as a liquid well above its critical point, the extrapolation procedures discussed here are usually inaccurate. In some cases special correlations or prescriptions are used; one such correlation is discussed in Chapter 11. In other cases different procedures, such as those discussed next, are used. The fugacity of a very dilute species in a liquid mixture (e.g., a dissolved gas or solid of limited solubility) is experimentally found to be linearly proportional to its mole fraction at low mole fractions, that is,

f Li (T, P, x) = xi Hi (T, P ) as xi → 0

(9.7-9)

The value of the “constant of proportionality,” called the Henry’s law constant, is dependent on the solute-solvent pair, temperature, and pressure. At higher concentrations the linear relationship between f Li (T, P, x) and mole fraction fails; a form of

474 Chapter 9: Estimation of the Gibbs Energy and Fugacity of a Component in a Mixture γi* >1

_

ƒiL (T, P, xi = 1) = ƒi* (T, P)

_

ƒi

γi* 1

ƒi

_

ƒi (T, P, Mi = 1) γi 10 mL

since the first 0.2 M × 10 mL = 2 mmol of hydroxyl ions neutralize the hydrogen ions resulting from the dissociation of the weak acid. Therefore, MH+ =

10−14 (10 + X) · 10−14 = M MOH− 0.2 · X − 2

and



(10 − X ) · 10−14 pH = −log 2·X −2

X > 10 mL

The results are shown in the following figure. 15

10 pH 5

0 0

5

10 X

15

20

It is of interest to compare the titration curves for the weak acid with a strong base with the titration curve for a strong acid with a strong base. In Illustration 13.5-2 and here we have used equal amounts of acidic solutions of equal concentrations, and titrated both with the same sodium hydroxide solution. However, we see that the initial parts of the titration curve look somewhat different while the parts beyond neutrality are identical. In particular, the titration curve for the strong acid starts at a much lower pH than the titration curve for the weak acid. Also, the initial shape of the titration curve has more curvature for the weak acid than is the case here for the strong acid.

Buffer

A buffer is a compound that stabilizes the pH of a solution in spite of the addition of an acid or a base. To be specific, sodium acetate, a salt that dissociates completely, can be used to buffer an acidic solution such as acetic acid. For example, in an aqueous solution of sodium acetate (which we will represent as NaAc), which has a pH of 7, and acetic acid (HAc), the following reactions occur:

NaAc → Na+ + Ac− KHA

HAc  H+ + Ac− and KW

H2 O  H+ + OH− In addition, we have the mass balances MNa+ = (MAc− )salt ≡ b

(13.5-17)

816 Chapter 13: Chemical Equilibrium

Total acetate = MHAc + (MAc− )from acid + (MAc− )from salt = MHAc + MAc− ≡ a + b Note that with these definitions, a is the initial (before dissociation) molal concentration of acetic acid MHAc,o , and b is the molal concentration of sodium acetate. Finally, from the condition that the solution must be electrically neutral, we have

MH+ + MNa+ = MOH− + MAc− so that

or MNa+ = b = MOH− + MAc− − MH+

MAc− = b + MH+ − MOH−

and

a = MHAc + MAc− − b = MHAc + MH+ − MOH−

or

MHAc = a − MH+ + MOH− These expressions are used in the equilibrium relations as follows:

KHA =

MH+ · MAc− M + · (b + MH+ − MOH− ) = H MHAc (a + MOH− − MH+ )

Also,

KW = MH+ · MOH−

or MOH− =

KW MH+

and

KHA

) * $ & MH+ · b + MH+ − MKW+ MH+ · bMH+ + MH2 + − KW H ) * $ & = = aMH+ + KW − MH2 + a + MKW+ − MH+ H

Therefore,

 b + 10−pH − 10(pH−pKW ) = pH − log a + 10(pH−pKW ) − 10−pH   MNaAc + 10−pH − 10(pH−pKW ) = pH − log MHAc,o + 10(pH−pKW ) − 10−pH 

pKHA

or



MNaAc + 10−pH − 10(pH−pKW ) pH = pKHA + log MHAc,o + 10(pH−pKW ) − 10−pH

 (13.5-18)

Since pKW is of the order of 14, for values of pH in the approximate range of 4 ≤ pH ≤ 10, this equation reduces to

b pKHA = pH − log a

13.6 Ionization of Biochemicals 817 or HendersonHasselbalch equation





b MNaAc = pKHA + log pH = pKHA + log a MHAc,o

(13.5-19)

which is referred to as the Henderson-Hasselbalch equation. The result of Eq. 13.5-18 for the pH of a solution in the presence of a buffer is analytically and numerically very different from Eq. 13.5-12b, obtained earlier when no buffering salt is present: ( ! 4 · MHA,o pH = pKHA − log 1+ − 1 + log 2 (13.5-12b) KHA The calculated results for the pH of acetic acid solutions with various amounts of sodium acetate are given in the table below. There we see that the presence of sodium acetate keeps the pH of acetic acid solutions closer to neutrality (pH = 7), though greater amounts of the salt are needed as the acid concentration is increased. MHAc,o

MNaAc

pH

1.0 1.0 1.0 1.0 1.0 5.0 5.0 5.0 5.0 5.0

0.0 0.5 1.0 2.0 5.0 0.0 0.5 1.0 2.0 5.0

2.375 4.449 4.750 5.051 5.449 2.026 3.750 4.051 4.352 4.750

(no buffer)

(no buffer)

(At these concentrations, we should include the effect of electrolyte solution nonideality. That would make the analysis is more difficult, and consequently this is left to the reader as an exercise.)

13.6 IONIZATION OF BIOCHEMICALS There are chemicals, especially biochemicals, that have more than a single ionization state. Indeed, proteins, composed of many amino acids, have numerous ionization sites and therefore multiple ionization states with different net charges. As a simple introduction to species with multiple ionization sites, consider phthalic acid, a so-called dibasic acid that ionizes as follows: COO−

COOH K1

COOH

+ H+ COOH

818 Chapter 13: Chemical Equilibrium and COO−

COO− K2

+ H+ COO−

COOH

Ionization of a dibasic acid

with pK1 = 2.95 and pK2 = 5.41. To be general, we will write the reactions for an acidic species with two ionization sites as K1

H2 A  H+ + HA−

K2

and HA−  H+ + A2−

(13.6-1)

The reaction equilibrium relations using the apparent equilibrium constants are

K1 =

MH+ · MHA− MH2 A

and K2 =

MH+ · MA2− MHA−

(13.6-2a)

Combining these two equations,

K1 K2 =

(MH+ )2 · MA2− MH2 A

(13.6-2b)

Our interest is in determining the fraction of the initial amount of the dibasic acid in each of the states (H2 A, HA− , and A2− ) at each pH of the solution, which we assume is being adjusted by adding a strong acid or base. That is, we will take the pH to be known, and we want to compute the fraction of acid in each ionization state. The total concentration of acidic species (that is, of all forms of the dibasic acid) Macid is

Macid = MH2 A + MHA− + MA2 − = MH2 A +   K1 · K2 K1 = MH2 A 1 + + MH+ (MH+ )2

MH2 A · K1 MH2 A · K1 K2 + MH+ (MH+ )2 (13.6-3)

Therefore, the fraction of undissociated dibasic acid is

fH2 A =

=

MH2 A 1 (MH+ )2 =" # = Macid K1 K1 · K2 (MH+ )2 + K1 MH+ + K1 · K2 1+ + MH+ (MH+ )2 10−2pH 10−2pH

+

10−(pK1 +pH)

+ 10−(pK1 +pK2 )

(13.6-4a)

13.6 Ionization of Biochemicals 819 Similarly,

fHA− =

MHA− K1 · MH2 A = · Macid MH+

1   · K K K 1 1 2 MH2 A · 1 + + MH+ (MH+ )2

=

K1 · MH+ (MH+ ) + K1 · MH+ + K1 · K2

=

10−(pK1 +pH) 10−2pH + 10−(pK1 +pH) + 10−(pK1 +pK2 )

2

(13.6-4b)

and

fA2− = =

K1 · K2 (MH+ ) + K1 · MH+ + K1 · K2 2

10−(pK1 +pK2 ) 10−2pH + 10−(pK1 +pH) + 10−(pK1 +pK2 )

(13.6-4c)

Illustration 13.6-1 The Ionization of a Dibasic Acid Compute the fraction of phthalic acid in each of its ionization states as a function of pH. Also, compute the average charge on the phthalic acid species as a function of pH.

Solution Using the pK1 and pK2 values given earlier and the equations above, we obtain the results in Fig. 13.6-1 for the fraction of each species present as a function of pH. Though each molecule of phthalic acid can have a charge of only 0, −1, or −2, the average charge on the phthalic acid components (which need not be an integer) is computed as the sum of the fractions of molecules in each of these charge states times the integer charge in that state, that is, Charge = 0 · fH2 A − 1 · fHA− − 2 · fA2−

The results are shown in Fig. 13.6-2. There we see that at low pH, where there is no ionization of phthalic acid, it has no net charge. However, as the pH is increased, there is an increasingly negative charge on the acid until a pH of 7 is reached, at which point the acid is fully dissociated with a charge of −2. Note that the fraction singly ionized is fHA− =

10−2pH

+

10−(pK1 +pH) + 10−(pK1 +pK2 )

10−(pK1 +pH)

1 10−pK1 10−pK2 10−2pH + 1 + 10−pK1 10−pH 10−pK1 10−pH 1 = −pH pK 10 10 1 + 1 + 10pH 10−pK2 =

820 Chapter 13: Chemical Equilibrium 1 H2A

HA–

A2–

f 0.5

0 1

2

3

4

5

6

7

8

pH

Figure 13.6-1 The fractions of undissociated fH2 A (dashed line), singly ionized fHA− (dotted line), and completely dissociated fA2− (solid line) phthalic acid as a function of pH. 0

Charge

– 0.5

–1

–1.5

–2 1

2

3

4

5

6

7

8

pH

Figure 13.6-2 The net charge on phthalic acid as a function of pH. For fHA¯ to the a maximum, denominator must be a minimum with respect to pH  d  −pH −pK1 d(denom) = 10 10 + 1 + 10pH 10−pK2 = 0 dpH dpH

Since

d(au ) d(a−u ) = au loge a and = −a−u loge a du du d(denom) = −10−pH (loge 10) 10pK1 + 0 + 10pH (loge 10) 10−pK2 dpH

⇒ 10−pH 10pK1 = 10pH 10−pK2 ⇒ 102pH = 10pK1 +pK2 pH for maximum fHA¯ ⇒ 2pH = pK1 + pK2 or pH =

pK1 + pK2 2.95 + 5.41 = = 4.18 2 2

Amino acids consist of acidic and basic groups on the same molecule, and at neutral pH can be considered to be neutral or to have a negatively charged anionic part and a positively charged cationic part. This is shown below

13.6 Ionization of Biochemicals 821 — —

R — C — COOH

or

— —

NH+3

NH2

R — C — COO−

H

H

Figure 13.6-3 Amino acid as a zwitterion.

where different amino acids have different R groups as side chains. For example, if R = H the amino acid is glycine, if R = CH3 the amino acid is alanine, and in lysine the side group is R = (CH2 )NH2 . Compounds such as amino acids with both anionic and cationic parts are referred to as zwitterions (coming from the German word zwitter meaning “hybrid”), and their charge changes with pH as a result of the reactions shown below for glycine

H — C — COOH

K1

H

NH+3 — —

— —

NH+3

H — C — COO− +H+ H

(A)

(B)

−

H — C — COO H

K1

NH2 — —

— —

NH+3

H — C — COO− +H+ H

(B)

(C)

Figure 13.6-4 Ionization reactions of glycine (ignoring the zwitterion effect).

in which the sequence shown starts at low pH (high hydrogen ion concentration), where the amino acid has a positive charge. As the pH is raised (the hydrogen ion concentration is lowered) the first reaction occurs, releasing a hydrogen ion from the amine group and resulting in no net charge on the amino acid. As the pH increases further, the second reaction occurs releasing a second hydrogen ion, now from the carboxylic acid, and resulting in the negatively charged form of the amino acid. The coupled set of equations that result from the coupled reactions, using an apparent equilibrium constant, with the species denoted as A, B, and C as indicated in Fig. 13.6-4, are MB MH+ K1 = (13.6-5a) MA and MC MH+ K2 = (13.6-6a) MB Using MB,o to represent the initial molar concentration of the uncharged amino acid in neutral solution (pH = 7), we have

MB,o = MA + MB + MC

822 Chapter 13: Chemical Equilibrium A simplification that we can use in solving these equations is that since the equilibrium constants are generally so different in value for amino acids that one reaction will have gone to completion before the other reaction has occurred to an appreciable extent. For example, for glycine we have pK1 = 2.34 or K1 = 4.57 × 10−3 and pK2 = 9.6 or K2 = 2.51 × 10−10 , so that the positively charged (A) form of the acid is only presentto an appreciable extent only below pH = 7, and the negatively charged (C) form is present to an appreciable extent only above pH = 7. MB = MB,o − MA , so that to an appreciable extent only below pH = 7, and the negatively charged (C) form is present to an appreciable extent only above pH = 7. MB = MB,o − MA , so that

K1 =

(MB,o − MA ) MH+ MA

(13.6-5b)

which has the solution

MH+ /K1 10(pK1 −pH) MA = = MB,o MH+ /K1 + 1 1 + 10(pK1 −pH)

(13.6-5c)

The equilibrium relation for the second reaction is

K2 =

MC MH+ MB

(13.6-6a)

Now using MB = MB,o − MC , we obtain

MC 1 1 = = (pK MB,o 1 + MH+ /K2 1 + 10 2 −pH)

(13.6-6b)

The average charge f on an amino acid, then, is

10(pK1 −pH) MA MC 1 + (−1) × = − (pK (pK − pH) MB,o MB,o 1 + 10 1 1 + 10 2 −pH) (13.6-7a) Also, it is easy to show (and left to the reader) that for the case here of an amino acid with a single carboxylic acid group and a single amine group, the isoelectric point, pI, that is, the pH at which the the amino acid has zero net charge, is f (pH) = (+1) ×

pH =

pK1 + pK2 2

(13.6-7b)

[Here the isoelectric point has been taken to be the pH at which the substance has no net charge. The experimental definition is the pH at which the substance has zero mobility in an electrophoresis cell. In fact, these two values of pH differ so slightly, they are frequently used interchangeably, as is done here.] Illustration 13.6-2 The Charge on an Amino Acid as a Function of pH The ionization equilibrium constants for glycine are pK1 = 2.34 and pK2 = 9.6. Determine the fraction of glycine molecules in the positively charged, negatively charged, and neutral states as a function of pH. Also, determine the average net charge on glycine as a function of pH, and the isoelectric point of glycine.

13.6 Ionization of Biochemicals 823 Solution Using Eqs. 13.6-5c and 13.6-6b, and the pK values given above, we can easily calculate the fraction of glycine in each of the ionization states, and by difference, the fraction that is uncharged. The results are shown in Figs. 13.6-5 and 6.

1 B (uncharged form) 0.8

C (– charge)

Fraction

0.6

0.4

A (+ charge)

0.2

0 0

5

10

15

pH

Figure 13.6-5 Fractions of neutral and charged forms of glycine as a function of pH.

1

Charge

0.5

0

–0.5

–1 0

5

10

15

pH

Figure 13.6-6 Average net charge on glycine as a function of pH.

Note that, as assumed, the positively charged (A) species have essentially disappeared at pH = 5, and the negatively charged (C) species begin to appear only above about pH = 8. So our assumption that the first reaction goes to completion before the second reaction occurs is justified. Because the charge on glycine is near zero over the pH range between about 3 and 9.5, it is very difficult to determine the isoelectric point from Fig. 13.6-6. However, from Eq. 13.6-7b, the isoelectric point is calculated to be pH = 5.97.

824 Chapter 13: Chemical Equilibrium A more detailed description of the ionization states of the amino acid glycine (H2 NCH2 COOH) is shown below, and is a bit more complicated than discussed above since the neutral form should be considered to be a zwitterion that exists in two different states. (GH2+)

+ H NCH COOH 3 2 K1 (GH)

K2

H2NCH2COOH K3

+ H NCH COO− 3 2

K4

H2NCH2COO−

(GH )

(G)

(13.6-8) In this diagram the letters in parentheses indicate the notation that will be used to represent each of these species. The equilibrium relations for the glycine reactions, neglecting solution nonidealities, are

K1 =

MGH · MH+ MGH+

K2 =

2

MGH · MH+ MGH+

K3 =

2

and

K4 =

MG− · MH+ MGH

MG− · MH+ MGH

Consequently,

K1 K3 =

MGH · MH+ MG− · MH+ M − · (MH+ )2 · = G MGH+ MGH MGH+

(13.6-9a)

MGH · MH+ MG− · MH+ M − · (MH+ )2 · = G MGH+ MGH MGH+

(13.6-9b)

2

and

K2 K4 =

2

2

2

so that K1 K3 = K2 K4 . In laboratory measurements we cannot distinguish between the GH and GH states, as they contain the same number of protons, and by hydrogen transfer, the molecule will jump between these two states. Therefore, we define new equilibrium constants that can be measured as follows: K1 =

MGH · MH+ M · MH+ (MGH + MGH ) · MH+ = + GH = K1 + K2 MGH+ MGH+ MGH+ 2 2 2 (13.6-10a)

and K2 =

MG− · MH+ (MGH + MGH )

13.6 Ionization of Biochemicals 825 so that

1 MGH MGH 1 1 K3 + K4 (MGH + MGH ) = = + = + = K2 MG− · MH+ MG− · MH+ MG− · MH+ K3 K4 K3 K4 or

K3 K4 K3 + K4 Thus the glycine reaction system can be represented as K2 =

K1

+ GH+ 2  GH + H

(13.6-10b)

K2

and GH  G− + H+

with GH now representing the sum of the two neutral forms of glycine, GH and GH . It is this description that is generally used in describing amino acids, as was done in Illustration 13.6-2. Proteins are long-chain molecules formed by polymerization reactions between amino acids. For example, the first step in such a reaction to form a peptide is shown below. −

=

H O

=

−

H − C − C − N − C − C − OH −

−→

−

R2

NH2 O −

NH2 − C − C − OH −

+

=

−

−

R1

H O

=

−

NH2 O

H − C − C − OH

R1

H

R2

+

H2O

The carbon-nitrogen bond is referred to as the peptide bond. A polypeptide chain is the result of a number of such reactions, and the characteristic of all peptides, like the amino acids from which they have been formed, is that one end of the molecule is an amino group and the other end is a carboxylic acid. Different peptides are characterized by their lengths and their side chains (R groups). In fact, only twenty types of side chains, with different sizes, shapes, charges, chemical reactivity, and hydrogen-bonding capacity, have so far been found in the proteins on all species on earth, from bacteria to man. Since proteins are large polypeptides molecules and are made up of a number of acidic (usually carboxylic acid) and basic groups (usually simple amino groups, but also histidine, lysine, arginine, and guanidine), including on their side chains, proteins have many ionization states. Therefore, the charge on a protein varies greatly with pH; however, like the amino acids from which they are formed, they will have a positive charge at low pH and a negative charge at high pH. As an example, the charge on the protein hen egg white lysozyme as a function of pH is shown in Fig. 13.6-7. [Lysozyme is a small protein that protects mammals (including us) from bacterial infection by attacking the protective cell walls of bacteria.] It has 32 ionizable groups, some anionic and others cationic, and as a result, its charge changes from +19 in very acidic solutions to −13 in very basic solutions. Notice that at a pH of 11.2, there is zero net charge on a lysozyme molecule, so this is the isoelectric point of this protein. A rule of thumb is that an amino acid or protein with multiple ionization sites will always include a primary carboxylic acid (−COOH) group that ionizes at low pH (usually 2 to 3), a primary amino (−NH2 ) group that ionizes in the pH range between 8 and 10, and other ionizable groups on side chains.

826 Chapter 13: Chemical Equilibrium 20 15

Charge of lysozyme

10 5 0 –5 –10 –15

0

1

2

3

4

5

6

7

8

9

10 11 12 13 14

pH

Figure 13.6-7 Average net charge on hen egg white lysozyme as a function of pH.

The isoelectric point, the pI, depends on the proportion of acidic and basic ionizable side chains. If the number of basic (largely ammonium) side chains is greater than the number of acidic (largely carboxyl) side chains, the pI of the protein will be high and the net charge on the protein will be positive over most of the pH range. It is for this reason that the pI of lysozyme, which has 19 basic groups and 13 acid groups, is 11.2. However, if the number of basic side chains is fewer than the number of acidic groups, the pI of the protein will be low, and its net charge will negative over most of the pH range as occurs in the albumins, which have pI values in the range from 4.6 to 4.9. One way that the charge on a protein changes with pH is used in the laboratory is in a separation method called isoelectric focusing, a form of electrophoresis. The protein or mixture of proteins is put on a conducting gel plate that has a fixed pH gradient and is also subject to a electrostatic potential gradient in the same direction. A protein that is in a region of the plate of pH below its pI will be positively charged and will move towards the cathode; a protein that is a region above its pI will migrate towards the anode. Once the protein reaches a region of the gel with a pH that is equal to its pI it will have no net charge and stop migrating. In this way, a mixture proteins can be separated, with each protein located in the pH region of the gel corresponding to its pI value. As a result of their highly polar, ionizable groups, amino acids are very soluble in polar solvent such as water, and only slightly soluble in nonpolar organic liquids. Also, by releasing hydrogen ions at high pH and accepting hydrogen ions at low pH, zwitterions act as good buffers by keeping a solution closer to neutrality upon the addition of an acid or a base. Illustration 13.6-3 The pH of a Solution Containing an Amino Acid as a Buffer Twenty-five milliliters of a 0.1-M aqueous solution of glycine is to be titrated with a 0.1-M sodium hydroxide solution. Determine the pH of the solution, the fraction of glycine in the − GH+ 2 , GH, and G ionization states, and the charge on glycine as a function of the amount of sodium hydroxide added.

13.6 Ionization of Biochemicals 827 Solution Initially there are 25 mL of a 0.1-M solution, so the titration starts with 2.5 mmol of glycine. Let α be the number of mmoles of GH and β the number of mmoles of G− present at any time. Also, as the 0.1-M NaOH solution is added, the OH− ions resulting from the sodium hydroxide ionization react with the H+ ions from the ionization of glycine to produce water. Therefore, the amount of each species present after the addition of X mL of NaOH is Species

Initial (mmol)

Final (mmol)

2.5 0 0 0

2.5 − α α−β β α + β − 0.1X

GH+ 2 GH G− H+

Also, the volume of the solution is 25+X , where X is the milliliters of the 0.1-M NaOH solution added. Therefore, the equilibrium relations are K1 =

(α − β) · (α + β − 0.1X) MGH · MH+ = (2.5 − α) · (25 + X) MGH+ 2

and K2 =

MG− · MH+ β · (α + β − 0.1X) = (α − β) · (25 + X) MGH

In principle, these equations need to be solved simultaneously. As already discussed, the equilibrium constants are so different in value (pK1 = 2.34 and pK2 = 9.6) that the first reaction goes to completion (i.e., α = 2.5) before the second reaction occurs to a significant extent. Therefore, one needs solve only the simpler equations K1 = 10−2.34 = and K2 = 10−9.6 = Also, MH+ =

α · (α − 0.1X) (2.5 − α) · (25 + X)

β · (2.5 + β − 0.1X) (2.5 − β) · (25 + X)

α + β − 0.1X M 25 + X

for X ≤ 25 mL

for 25 < X ≤ 50 mL 

pH = −log

and

α + β − 0.1X 25 + X



When X > 50 mL, all the hydrogen ions resulting from the ionization of glycine have been neutralized, and the only hydrogen ions remaining are from the ionization of water, so that MOH− =

0.1X 25 + X

and

 pH = −log MH+ = −log

and

MH+ =

10−14 (25 + X ) 0.1X

10−14 10−14 (25 + X) = MOH− 0.1X



 = 14 + log

0.1X 25 + X

 for X > 50 mL

− Finally, the fractions f of glycine molecules in the GH+ 2 , GH, and G states can be computed from 2.5 − α α−β β

fGH+ = 2

2.5

fGH =

2.5

and

fGH =

2.5

The results are shown in Figs. 1 and 2. For comparison, note that in water without any added sodium hydroxide, 81 percent of the glycine is in the GH+ 2 state, 19 percent is in the GH or GH state, and the pH of the solution is 1.72.

828 Chapter 13: Chemical Equilibrium 14 12

Without glycine

10 8 pH

With glycine 6 4 2 0 0

10 20 30 40 50 ml of added 0.1 M NaOH solution

60

Figure 13.6-8 The pH of a glycine solution as a function of added sodium hydroxide (solid line). The dashed line is the pH of water upon NaOH addition without glycine, showing the effectiveness of the zwitterion glycine as a buffer for pH values near 2 and near 8. 1 0.8

GH12

G–

GH

0.6 f 0.4

0.2 0 0

10

20

30 X

40

50

60

− Figure 13.6-9 Fractions f of glycine in the GH+ 2 , GH, and G states as a function of the milliliters of 0.1-M sodium hydroxide solution added.

If the glycine were not present, so that the starting solution was only 25 mL of pure water, the molality of hydrogen ions (neglecting solution nonideality) would be given by MH+ =

KW 10−14 = MOH− MNaOH

so that

pH = 14 + log MNaOH

since the strong base, sodium hydroxide, is fully ionized. The molality of sodium hydroxide when X mL of a 0.1-M NaOH solution is added is MNaOH =

0.1X 25 + X 

and pH = 14 + log

0.1X 25 + X



13.6 Ionization of Biochemicals 829 The pH of the solution without glycine is shown in Fig. 1. Note that without glycine the solution is initially at pH 7, and that the pH initially rises rapidly upon the addition of sodium hydroxide and then more gradually. However, with glycine present the initial pH is low and, while gradually increasing, remains low until 25 mL of the NaOH solution has been added, as the hydrogen ion released by carboxylic acid portion of the glycine reacts with the hydroxide ion of NaOH to form water. Further addition of NaOH results in a sharp increase in pH to slightly above pH = 8. Then continued additions of the strong base lead to the release of hydrogen ions from the amino groups that react with the sodium hydroxide, resulting in only a slight increase in pH until 50 mL of NaOH solution has been added. At that point, there are no additional hydrogen ions that can be released, and the pH of the solution increases rapidly. From Fig. 1, we see that glycine, being a zwitterion, is an effective buffer in the pH range near 2 and also in the pH range slightly above 8.

Consider next the reaction between benzoyl tyrosine and glycinamide when they are in the same solution

benzoyl tyrosine (BT) + glycinamide (GA) ↔ benzoyltyrosyl glycylamide (BTGA) The equilibrium constant for this reaction based on standard states of ideal 1-molal solutions is

K=

[BTGA] = 0.49 M or pK= 0.31 [BTCOOH] [GANH2 ]

As BT and GA are present in both the neutral and deprotonated forms, what has is actually measured is an apparent equilibrium constant κ

[BTGA] " " #& $ #& κ= $ [BTCOOH] + BTCOO− [GANH2 ] + GANH+ 3 since analytically it may not be easy to distinguish between the neutral and ionic species in solution. Now using the ionization equilibrium relations we have

κ=

[BTCOOH] 1 +

K [BTGA]



=

+ H [H+ ] [ ] KA KA 1 + KB 1 + H+ [NH2 GA] 1 + KB [H+ ] [ ]

or

κ=

K 

 1+

=

KA H+

[

]

[H+ ]



=

1+ K B

1+

KA 10−pH

K 

−pH

1+ 10K

B



=

0.49  M

−3.7

1+ 10−pH 10

−pH 1+ 10−7.93 10



0.49 M (1+10pH−3.7 )(1+107.93−pH )

Figure 13.6-10 is a plot of the apparent equilibrium constant κ as a function of pH. We see that in the pH range of about 5.8 to 7.7, κ is equal to the true equilibrium constant K. However at very low pH and very high pH, the apparent equilibrium constant κ is essentially zero. That is, there is an appreciable extent of reaction around the neutral pH of 7, but the reaction essentially does not occur very far away from neutral pH.

830 Chapter 13: Chemical Equilibrium 0.6

0.4 κ 0.2

0

0

5

10

15

pH

Figure 13.6-10 Apparent equilibrium constant for the reaction between benzoyltyrosine and glycinaminde as a function of pH.

The discussion so far has been for processes at constant temperature; here we consider how the change in a protein charge changes with temperature. Acid and base ionizations are chemical reactions, and have temperature-dependent equilibrium constants. For a small temperature change,



Δrxn H K(T2 ) Δrxn H 1 1 1 1 or pK (T2 ) = pK (T1 )+ ln − − =− K(T1 ) R T2 T1 2.303R T2 T1 As an example, the heat of ionization of the ammonium group in glycine is about -44.77 kJ/mol, while that of the carboxylic group is approximately 0 kJ/mol. Therefore, while the pK for ammonium group in glycine is 9.60 at 25o C, using the equation above, it is expected to be 9.90 at body temperature of 37o C and 8.88 at 0o C. However, the pK for the carboxylic acid group is essentially constant over that same temperature range. This results in small changes in the charge versus pH diagram as shown in Fig. 13.6-9 for the three temperatures. Note also (though not obvious in that figure) that the isoelectric point shifts slightly from pH = 5.97 at 25o C, to 5.61 at 37o C and 6.12 at 0o C. 1 Glycine 0.5 f25 f0

0

f37 –0.5

–1

0

5

10 pH

15

Figure 13.6-11 Charge on glycine as a function of pH at 25◦ C (solid line), 0◦ C (dotted line) and 37◦ C (dashed line).

The variation of the solubility of a protein with temperature provides information on its heat of dissolution. In particular, using the solubility product K sp , asolution Ksp = = asolution ≈ c asolid where asolution is the activity of biomolecule in solution which, at low concentrations, can be taken to be its concentration, and its activity as a pure solid, asolid , is unity. Consequently, knowing the solubility of a biomolecule at two temperatures, we can use

Ksp (T2 ) c(T2 ) Δsol H 1 1 ln − = ln =− Ksp (T1 ) c(T1 ) R T2 T1

13.7 Partitioning of Amino Acids and Proteins Between Two Liquids 831 to calculate its heat of solution. For example, the solubility of lactoglobulin in water is 0.35 g/liter at 5◦ C and 0.58 g/liter at 25◦ C, from which using Eq. 13.6-11 we estimate its heat of solution is 17.4 kJ/mol, and can then compute that its solubility at body temperature of 37◦ C to be 0.76 g/liter. This calculation of the heat of solution contains the assumption that the protein has not ionized in the dissolution process, that is, it is near its pI. If the protein ionizes on dissolution, the apparent heat of solution determined using Eq. 13.6-11 will include the heat of ionization.

13.7 PARTITIONING OF AMINO ACIDS AND PROTEINS BETWEEN TWO LIQUIDS Next consider the partitioning of an acid between an organic and an aqueous liquid, as occurs in liquid-liquid extraction. The measured partition coefficient K P is the ratio of the total concentration of the acid in the organic (upper) phase (O) to that in the aqueous phase (W)

KP =

[RCOOH]O # " [RCOOH]W + RCOO− W

(13.7-1)

where we have recognized that the weak acid may be partially (or completely) ionized, but only in the aqueous phase denoted by the subscript W. The equilibrium constant for the ionization of an acid at high dilution so that all activity coefficients can be considered to be unity, is " # RCOO− W [H+ ]W KA = (13.7-2) [RCOOH]W Combining these two equations gives

KP =

[RCOOH]O [RCOOH]O = + [RCOOH]W (1 + KA / [H ]) [RCOOH]W (1 + 10pH−pKA )

(13.7-3)

It then follows that there can be a significant variation of the partition coefficient K P with pH. Now consider the case in which the concentration of the weak acid is at saturation in both phases. The ratio R of partition coefficients for the saturated acid concentration in two separated and different aqueous solutions of differing pH but in equilibrium with the same organic second phase (assuming the acid does not deprotonate in the organic phase), is given by

R=

KP (pH2 ) 1 + 10pH1 −pKA = KP (pH1 ) 1 + 10pH2 −pKA

(13.7-4)

Since the saturated acid concentration in the organic phase is independent of the saturated acid concentration in the aqueous phase, and the saturation concentration of undissociated acid in the aqueous phase is independent of pH. In the discussion above, we have neglected the effect of solution nonidealities. To correct for this, we should have written " # RCOO− W [H+ ]W γ±2 KA = (13.7-5) [RCOOH]W γA

832 Chapter 13: Chemical Equilibrium where γA is the activity coefficient of the unionized acid. Using that pH is more properly defined as pH = √ − log ([H+ ] γ±√) and for simplicity using the Debye-H¨uckel limiting law γ± = e−1.176 I = 10−0.5116 I gives √

1 + γA 10pH1 −pKA +0.5116 I1 KP (pH2 , I2 ) √ = R= (13.7-6) KP (pH1 , I1 ) 1 + γA 10pH2 −pKA +0.5116 I2 This equation shows the effects of pH and ionic strength on the partitioning of a weak acid between organic and aqueous phases. The analogous result for the ratio of the partition functions at saturation for an amino acid at different pHs and ionic strengths is √

√

KP (pH2 , I2 ) 10pK1 −pH1 ++0.5116 I1 + 1 + 10pH1 −pK2 +0.5116 I1 √ R= = pK −pH ++0.5116√I 2 + 1 + 10pH2 −pK2 +0.5116 I2 2 KP (pH1 , I1 ) 10 1 (13.7-7) In Fig. 13.7-1 is shown the predicted partition function ratio R for glycine between a upper organic phase and lower aqueous phase as a function of pH of the aqueous phase compared to a neutral (pH = 7) aqueous phase. The solid line is the result for a solution of low ionic strength and the dashed line is for I = 1 molal (using only the simple Debye-H¨uckel limiting law). Though glycine will partition very differently between the aqueous and organic phases, we see in this figure that for a range of pH from about 3 to 9, the partitioning of glycine is essentially independent of the pH of the aqueous solution and is the same as when the aqueous phase is pH 7 water. However, for pH9, where glycine is ionized there will be more glycine in the aqueous phase (i.e., RChapter 14>14.3-2 on the Wiley website for this book that uses both the REquil and RGibbs reactor blocks, which led to the same answers, the following results were obtained: T = 640.8 K yB = yE = 0.0949 yEB = 0.8102



These results are not in good agreement with the results above for this illustration.

In Illustration 14.3-2 the reaction was exothermic (Δrxn H < 0). This implies that (1) energy is released on reaction, which must appear as an increase in temperature of the reactor effluent since the reactor is adiabatic, and (2) the equilibrium conversion of reactants to products decreases with increasing temperature. Consequently, the equilibrium extent of reaction achieved in an adiabatic reactor is less than that which would be obtained had the reactor been operating isothermally at the reactor inlet conditions. For an endothermic reaction (Δrxn H > 0), the equilibrium conversion decreases with decreasing temperature and, since energy is absorbed on reaction, the reactor effluent is at a lower temperature than the feed stream. Thus, the equilibrium and energy balance curves are mirror images of those in Illustration 14.3-2 (see Problem 14.13), and again the equilibrium extent of reaction is less than would be obtained with an isothermal reactor operating at the feed temperature. Another interesting application of the multicomponent balance equations for reacting systems is in the estimation of the maximum work that can be obtained from an isothermal chemical reaction that occurs in a steady-state fuel cell or other workproducing device. The starting point here is the steady-state isothermal energy and entropy balances of Table 8.4-1: C C   ˙s 0= (N˙ i H i )in − (N˙ i H i )out + Q˙ + W (14.3-12) i=1

and

0=

C 

i=1

(N˙ i S i )in −

C 

i=1

i=1

Q˙ (N˙ i S i )out + + S˙ gen T

(14.3-13)

Solving the second of these equations for the heat flow Q˙ ,

Q˙ = T

C  i=1

(N˙ i S i )out − T

C  i=1

(N˙ i S i )in − T S˙ gen

(14.3-14)

14.3 Overall Reactor Balance Equations and the Adiabatic Reaction Temperature 867 and then eliminating the heat flow from Eq. 14.3-12 yields

˙s= W

C 

[N˙ i (H i − T S i )]out −

i=1

=

C 

C 

[N˙ i (H i − T S i )]in + T S˙ gen

i=1

[N˙ i Gi ]out −

i=1

C 

(14.3-15)

[N˙ i Gi ]in + T S˙ gen

i=1

˙ s will be negative. The maximum Since a fuel cell is a work-producing device, W ˙ work (i.e., the largest negative value of Ws for given inlet conditions) is obtained when  the terms Ci=1 [N˙ i Gi ]out and T S˙ gen are as small as possible. Since S˙ gen ≥ 0, with the equality holding only for reversible processes, one condition for obtaining the maximum work is that the chemical reaction and mechanical energy production process be carried out reversibly, so that S˙ gen is equal to zero. The first term on the right side  of Eq. 14.3-15, Ci=1 [N˙ i Gi ]out , is the flow of Gibbs energy accompanying the mass flow out of the system. The minimum value of this term for a given mass flow at fixed temperature occurs when the exit stream is in chemical equilibrium (so that the Gibbs energy per unit mass is a minimum). Thus, another condition for obtaining maximum work from an isothermal flow reactor is that the exit stream be in chemical equilibrium. When both these conditions are met, ˙ max = W

C 

(N˙ i Gi )out −

i=1

C 

(N˙ i Gi )in

(14.3-16a)

(Ni Gi )in

(14.3-16b)

i=1

and Maximum work from a fuel cell at constant T and P

Wmax =

C 

(Ni Gi )out −

i=1

C  i=1

It is also possible to compute the maximum work attainable from an isothermal, constant-volume batch reactor. It is left to you to show that

Wmax =

C 

(Ni Ai )final −

i=1

C 

(Ni Ai )initial

(14.3-17)

i=1

Illustration 14.3-3 Calculation of Energy Produced in a Fuel Cell 4

The November 1972 issue of Fortune magazine discussed the possibility of the future energy economy of the United States being based on hydrogen. In particular, the use of hydrogen fuel in automobiles was considered. Assuming that pure hydrogen gas and twice the stoichiometric amount of dry air, each at 400 K and 1 bar, are fed into a catalytic fuel cell, and that the reaction products leave at the same temperature and pressure, compute the maximum amount of work that can be obtained from each mole of hydrogen.

4 Fuels

cells are discussed in some detail in Sec. 14.7

868 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry Air Fuel cell

O2, N2, H2, H2O

H2

Ws

Data: Air may be considered to be 21 mol % oxygen and 79 mol % nitrogen; the heats and Gibbs energies of formation are given in Appendix A.IV, and the heat capacities in Appendix A.II.

Solution From Eq. 14.3-16 we have Wmax =

C 

{(Ni Gi )out − (Ni Gi )in }

i=1

where the outlet compositions are related by the fact that chemical equilibrium exists. Here, since both the reactants and products are gases (presumably ideal gases at the reaction conditions), we have Gi (T, P, y) = Gi (T, P ) + RT ln yi

Since hydrogen enters the fuel cell as a separate, pure stream, we have Species

(yi )in

(Ni )in (mol)

(Ni )out (mol)

(yi )out

H2 O2

1 0.21

1−X 1 − 12 X

(1 − X)/Σ (1 − 12 2X)/Σ

N2

0.79

3.762

3.762/Σ

H2 O

0

1 1 0.79 × 1 = 3.762 0.21

X

X/Σ

Σ = 5.762 −

Total

1 X 2

Now, from the equilibrium relation, we have Ka = =

aH2 O 1/2

aH2 aO2

yH2 O 1/2

yH2 yO2 (P/1 bar)1/2

=

X(5.762 − 12 X)1/2

(1 − X)(1 − 12 X)1/2

Using the data in the problem statement and Eq. 13.1-16, we find that Ka = 1.754 × 10+29 ; a huge number. Therefore, X ∼ 1. Also, using Eq. 13.1-23b and the heat capacity data, the values of the pure-component Gibbs energies at 400 K and 1 bar, relative to the reference states in Appendix A.IV, can be computed; the results are Species H2 O2 N2 H2 O

Gi (T = 400 K, P = 1 bar)

−453.1 J/mol −466.9 −457.5

−224 600

(Ni )out

(yi )out

0 0.5 3.762 1

0 0.095 0.715 0.190

14.4 Thermodynamics of Chemical Explosions 869 Therefore, Wmax (T = 400 K) =

  (Ni Gi )out − (Ni Gi )in i

i

  = [Ni (Gi + RT ln yi )]out − [Ni (Gi + RT ln yi )]in i

i

= {0 + 0.5[−466.9 + 8.314 × 400 ln(0.095)] + 3.762[−457.5 + 8.314 × 400 ln(0.715)] + 1[−224, 600 + 8.314 × 400 ln(0.190)]}

− {1[−453.1 + 8.314 × 400 ln(1)] + 1[−466.9 + 8.314 × 400 ln(0.21)] + 3.762[−457.5 + 8.314 × 400 ln(0.79)] + 0} = (0 − 4147.5 − 5918.2 − 230 122.9) − (−453.1 − 5657.0 − 4670.1) J/mol = −229 408.4 J/mol = −229.4 kJ/mol

Comments 1. In this illustration the Gibbs energy of mixing terms make a relatively small contribution to Wmax . This is only because for the H2 + 12 O2 = H2 O reaction, the dominant term in the calculation is Δrxn G. For many reactions Δrxn G is not so large, and the Gibbs energy of mixing terms can be important. 2. Although the simple black-box thermodynamic analysis here permits us to calculate the maximum work obtainable from the fuel cell, it does not provide any indication as to how to design the fuel cell (i.e., as an internal combustion engine, electrolytic cell, etc.) to obtain this work. The design of efficient fuel cells is currently an important unsolved engineering problem, much like the design of heat engines at the time of Sadi Carnot.

14.4 THERMODYNAMICS OF CHEMICAL EXPLOSIONS In Sec. 5.3 we considered the thermodynamics of explosions that did not involve chemical reaction. Here we extend this discussion to explosions with chemical reaction. Since the energy released on an exothermic reaction may be very large, chemical explosions are generally more devastating than purely mechanical explosions. In a chemical explosion there are extremely rapid temperature and pressure rises as some or all of the reactants are oxidized, followed by an expanding shock wave resulting in a decrease in the temperature and pressure of the combustion products. It is the rapid overpressurization at the expanding shock-wave front that causes most of the damage in an explosion. In order for a chemical explosion to occur, the heat of reaction must be reasonably large and negative (Δrxn H < 0), that is, the reaction must be exothermic, and the equilibrium constant for the reaction should be large (Keq 0, or Δrxn G < 0). In this case, the reaction will proceed to an appreciable extent and a significant amount of energy released. The equations describing a chemical explosion are the same as those for a mechanical explosion (see Sec. 5.3); that is, to compute the maximum energy released in an explosion, we assume the process within the region bounded by the shock wave is reversible and adiabatic so that

U f − U i = U (T f , P f , N f ) − U (T i , P i , N i ) = W S f − S i = S(T f , P f , N f ) − S(T i , P i , N i ) = Sgen = 0

(14.4-1) (14.4-2)

870 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry Here the notation N has been used to indicate all the mole numbers N = (N1 , N2 , . . . , NC ), and we have written these equations in terms of total thermodynamic properties rather than molar or specific properties since, as a result of the chemical reaction, there is a change in the number of moles of each species. Also, as before, we assume that the entropy generation term is zero, thereby obtaining an upper bound on the energy released in an explosion. Multiplying Eq. 14.4-2 by the initial temperature T i and subtracting it from Eq. 14.4-1 gives

U (T f , P f , N f ) − T i S(T f , P f , N f ) − [U (T i , P i , N i ) − T i S(T i , P i , N i )] = U (T f , P f , N f ) − T i S(T f , P f , N f ) − A(T i , P i , N i ) = W (14.4-3) i i i i i i i i i i where A(T , P , N ) = U (T , P , N ) − T S(T , P , N ) is the Helmholtz energy of the initial state. [Why is U (T f , P f , N f ) − T i S(T f , P f , N f ) not the Helmholtz energy of the final state?] As in Sec. 5.3 the final pressure can be taken to be 1.013 bar when the shock wave has dissipated, and all the damage has been done. What are unknown are the final temperature and the composition of the reacting mixture. The composition is especially difficult to estimate because the rates of chemical reaction can be slow compared with the rates of change of temperature and pressure in an explosion. Consequently, it is not certain that chemical equilibrium will be achieved in a chemical explosion. Further, because of the rapid drop in temperature in the expanding shock wave, and as the rate of reaction decreases with decreasing temperature, even if chemical equilibrium were achieved, it is not clear to which temperature this equilibrium would correspond. That is, chemical equilibrium could be achieved at some high or intermediate temperature, and then the reaction may be quenched by the rapid temperature drop at a composition corresponding to that unknown temperature. Another question that arises is the amount of oxidant available for the reaction. In a vapor cloud explosion one can generally assume that there is sufficient oxygen present for the complete oxidation of all carbon to carbon dioxide and all hydrogen to water. In this case the final composition is known. The situation for explosions involving liquids and solids in which there is a limited amount of oxidizing agent available is more complicated and will be considered shortly. The final temperature of the explosive mixture, after the shock wave expands so that the internal pressure is 1.013 bar, will generally be higher than ambient. Since the quantity U f − T i S f increases with increasing temperature, to obtain an upper bound on the energy released in a chemical explosion, we will assume that the final temperature and pressure are ambient, as are the initial temperature and pressure. In this case we obtain from Eq. 14.4-3 U (25◦ C, 1.013 bar, N f ) − (298.15 K) × S(25◦ C, 1.013 bar, N f ) − U (25◦ C, 1.013 bar, N i ) − (298.15 K) × S(25◦ C, 1.013 bar, N i ) = A(25◦ C, 1.013 bar, N f ) − A(25◦ C, 1.013 bar, N i ) = W (14.4-4) Since Gibbs energy data are more readily available than Helmholtz energies, we note that G = A + P V ≈ A for solids and liquids and G = A + P V = A + N V RT for vapors

14.4 Thermodynamics of Chemical Explosions 871 where in the last equation we have used the ideal gas law (as the pressure is low) and the notation N V to indicate the total number of moles in the gas phase. With this substitution we have

W = G(25◦ C, 1.013 bar, N f ) − G(25◦ C, 1.013 bar, N i ) − [N V,f − N V,i ]RT (14.4-5) This is the equation that will be used to compute the energy released in an explosion. In using this equation, we will make one further simplification. The total Gibbs energies in this equation should be computed from  G(T, P, N ) = Ni Gi (T, P, x) (14.4-6) For simplicity, we assume here that vapors and liquids form ideal solutions, since the contribution of the solution nonideality to the energy is small compared with the chemical reaction term (you should verify this), so that

Gi = Δf G◦i + RT ln xi

(14.4-7)

and that solid phases are pure so that

Gi = Δf G◦i

(14.4-8)

In a vapor-phase explosion it is generally assumed that sufficient oxygen is present for the substance to be completely oxidized. This makes estimating the explosive energy release simpler, as the reaction stoichiometry is known. Also, because of the large amount of nitrogen present before and after the explosion (air contains about 79 percent nitrogen), the entropy change on mixing is small and frequently can be neglected, as is shown in the following illustration. Illustration 14.4-1 Estimating the Energy Released in a Vapor-Phase Explosion Estimate the energy released in a vapor-phase explosion of 1 kg of ethylene with a stoichiometric amount of air.

Solution The reaction stoichiometry is C2 H4 + 3O2 + 3 ×

0.79 0.79 N2 = 2CO2 + 2H2 O + 3 × N2 0.21 0.21

where we have recognized that with 1 mole of oxygen in air there is 0.79/0.21 moles of nitrogen. To proceed further we note that N V,i = 1 + 3 + 3 × (0.79/0.21) = 15.29 moles of gas present initially, and N V,f = 2 + 2 + 3 × (0.79/0.21) = 15.29 moles of gas present finally, so that there is no contribution to the total energy release as a result of a change in the number of moles present in the gas phase. Next, ◦ ◦ G(N i ) = Δf GC + 3 × Δf G◦O2 + 11.29 × Δf GN 2 H4 2

    1 3 + RT 1 × ln + 3 × ln 1 + 3 + 11.29 1 + 3 + 11.29   11.29 +11.29 × ln 1 + 3 + 11.29

= 68.5 + 3 × 0 + 11.29 × 0 + 8.314 × 10−3 × 298.15 × (−2.7 − 4.9 − 3.4) = 68.5 − 27.3 = 41.2 kJ/mol

872 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry and ◦ ◦ ◦ G(N f ) = 2 × Δf GCO + 2 × Δ f GH + 11.29 × Δf GN 2 2O

  2   2 2 + RT 2 × ln + 2 × ln 2 + 2 + 11.29 2 + 2 + 11.29   11.29 + 11.29 × ln 2 + 2 + 11.29

= 2 × (−394.4) + 2 × (−228.6) + 11.29 × 0 + 8.314 × 10−3 × 298.15 × (−4.1 − 4.1 − 3.4) = −488.8 − 457.2 − 28.8 = −1274.8 kJ/mol

Therefore, W = −1274.8 − 41.2 = −1316.0

kJ mol

and the total energy released by the explosion of 1 kg of ethylene is W = −1316.0

kJ kJ 1 mol 1000 g × × = −46 910 mol 28.054 g 1 kg kg

Note that the energy released on the explosion of ethylene, 46 910 kJ/kg, is more than 10 times that for TNT, which is 4570 kJ/kg. It is of interest to note that in this example of the total energy release of −1316 kJ/mol, the amount −1314.5 [= 2 × (−394.4) + 2 × (−228.6) − 68.5] kJ/mol results from the standard-state Gibbs energy change on reaction, which is the dominant contribution, and only −28.8 − (−27.3) = −1.5 kJ/mol is from the entropy-of-mixing term.

As indicated earlier, one of the difficulties in chemical explosions is ascertaining the final composition of the exploding mixture. In explosions of a solid or a liquid, because of mass transfer limitations usually only the oxygen present in the original compound is involved in the reaction. In this case there may be an infinite number of partial oxidation reaction stoichiometries possible. Illustration 14.4-2 Examining Different Reaction Stoichiometries List some of the possible reactions for TNT, C7 H5 N3 O6 .

Solution Among the many possible reactions are C7 H5 N3 O6 → 1.0C + 2.5H2 + 1.5N2 + 6.0CO

→ 5.25C + 1.75CO2 + 1.5N2 + 2.5H2 O → 0.96C + 6.0CO + 1.48N2 + 0.04HCN + 2.48H2 → 3.7C + 0.4CO + 2.2CO2 + 1.5N2 + 1.1H2 O + 0.7CH4 However, there is some experimental evidence that the reaction that occurs is actually C7 H5 N3 O6 → 3.65C + 1.98CO + 1.25CO2 + 1.32N2 + 0.16NH3 + 0.02HCN + 0.46H2 + 1.60H2 O + 0.10CH4 + 0.004C2 H6

which has an explosive energy release of about 4570 kJ/kg.

14.4 Thermodynamics of Chemical Explosions 873 Generally what is done in dealing with reactions involving liquids or solids to make a conservative (which in this context means worst-case) estimate of the explosive energy release of a chemical is to examine the different reaction stoichiometries, and then choose the reaction with the largest negative Gibbs energy change (or equivalently, the largest equilibrium constant). However, this assumes that the reaction is a rapid one. Illustration 14.4-3 Estimating the Energy Released on the Explosion of a Liquid Nitromethane (a liquid) undergoes the following reaction on explosion: CH3 NO2 → 0.2CO2 + 0.8CO + 0.8H2 O + 0.7H2 + 0.5N2

Estimate the energy released on the explosion of nitromethane. Data:

◦ Δf GNM = −13.0 kJ/mol Δf H ◦NM = −111.7 kJ/mol

Solution There are a number of terms that contribute to the total energy released. First we note that N V,f = 0.2 + 0.8 + 0.8 + 0.7 + 0.5 = 3 moles of gas present finally, and N V,i = 0 moles of gas initially. Next, ◦ G(N i ) = Δf GNM = −13.0 kJ/mol ◦ ◦ + 0.5Δf GN G(N f ) = 0.2Δf G◦CO2 + 0.8Δf G◦CO + 0.8Δf G◦H2 O + 0.7Δf GH 2 2

 0.2 0.8 0.8 0.7 0.5 + RT 0.2 ln + 0.8 ln + 0.8 ln + 0.7 ln + 0.5 ln 3 3 3 3 3 kJ J 1 kJ = −371.5 × 298.15 K × (−4.571) × + 8.314 mol mol K 1000 J kJ = −371.5 − 11.3 = −382.8 mol

Therefore, W = −382.8 − (−13.0) − 3 ×

kJ 8.314 × 298.15 = −377.2 1000 mol

Comments It is again interesting to assess the contributions of the various terms to the total energy release on explosion. By far the largest contribution is the difference in Gibbs energies of the components, that is ◦ ◦ ◦ ◦ ◦ ΔG = 0.2Δf GCO + 0.8Δf GCO + 0.8Δf GH + 0.7Δf G◦H2 + 0.5Δf GN − Δf GNM 2 2O 2

= −371.5 − (−13.0) = −358.5 kJ/mol

The Gibbs energy change due to the entropy of mixing is −11.1 kJ/mol, and the energy released due to the vapor-phase mole number change is −7.4 kJ/mol. Therefore, the majority of the explosive energy release comes from the Gibbs energies of formation of the species involved. A reasonable approximation in many cases, accurate to ±10 percent, is to neglect the gas-formation and entropy-of-mixing terms and merely use W = Δrxn G.

874 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry A second condition for an explosion is that the heat of reaction be negative (i.e., the reaction should be exothermic). For the reaction here, we have Δrxn H = 0.2Δf H ◦CO2 + 0.8Δf H ◦CO + 0.8Δf H ◦H2 O + 0.7Δf H ◦H2 + 0.5Δf H ◦N2 − Δf H ◦NM = 0.2 × (−393.5) + 0.8 × (−110.5) + 0.8 × (−241.8) + 0.7 × (0) + 0.5 × (0) − (−111.7) = −78.7 − 88.4 − 193.4 + 111.7 = −248.8

kJ mol

which is large and negative. Therefore, nitromethane is a likely candidate for a chemical explosion.

Illustration 14.4-4 Estimating the Energy Released on the Explosion of a Solid One possible reaction for 2,4,6-trinitrotoluene (TNT) is as follows: C7 H5 O6 N3 → C + 6CO + 2.5H2 + 1.5N2

Estimate the energy released on an explosion of TNT. Data: Δf G◦TNT = 273 kJ/mol

Solution Since TNT is a solid, N V,i = 0; also N V,f = 10. Using the standard-state Gibbs energy of formation tables in Appendix A.IV and Eqs. 14.4-7 and 14.4-8, we have at 25◦ C GTNT = 273

kJ , mol

GC = 0

J × 298.15 K 6 mol K ln = −110.5 + J 10 1000 kJ kJ = −110.5 − 1.3 = −111.8 mol J × 298.15 K 2.5 8.314 kJ mol K ln =0+ = −3.4 J 10 mol 1000 kJ 8.314

GCO

GH2

and 8.314 GN2 = 0 +

J × 298.15 K 1.5 kJ mol K ln = −4.7 J 10 mol 1000 kJ

Therefore, W = [1 × 0 + 6 × (−111.8) + 2.5 × (−3.4) + 1.5 × (−4.7)] 8.314 × 298.15 − [273] − (10 − 0) × 1000 kJ = −670.8 − 8.5 − 7.1 − 273 − 2.48 = −961.9 mol TNT kJ kJ kJ 1 mol TNT × = −4.059 = −4059 = −961.9 mol TNT 237 g g TNT kg TNT

14.5 Maximum Useful Work and Availability in Chemically Reacting Systems 875 The experimentally measured energy release on a TNT explosion is about 4570 kJ/kg. The reason for the discrepancy is that, as discussed in Illustration 14.4-2, the actual reaction stoichiometry is different from that assumed here. It is of interest to note that in this example of the total energy release of −961.9 kJ, −6 × 110.5 − 273 = −936 kJ arises from the Gibbs energy change on reaction, which is the dominant contribution; −2.48 kJ is a result of the formation of 10 moles of gas per mole of TNT; and RT × [6 × ln(0.6) + 2.5 × ln(0.25) + 1.5 × ln(0.15)] = −23.24 kJ results from the entropy-ofmixing term.

14.5 MAXIMUM USEFUL WORK AND AVAILABILITY IN CHEMICALLY REACTING SYSTEMS Here we are interested in the maximum useful work that can be obtained from a chemically reacting system. This section is an extension of Section 4.6 to mixtures and chemical reactions. As in that section, to obtain the maximum useful shaft work we require that the temperature of the exiting stream be T amb , that any heat transfer from the system occur at the ambient temperature, that the exiting pressure should be Pamb, that the velocity of the exit stream should be zero, that the height of the exit stream should be a ground level, i.e., h2 = 0, and that the process should operate reversibly, that is S˙ gen = 0. The energy and entropy balances for the steady-state flow system of constant volume in Fig. 4.6-1, now written on a molar (rather than mass) basis are:   ˙ S,max = 0 ¯ i (T1 , P1 , x1 ) − ¯ (Tamb , Pamb , x2 ) + Q˙ + W N˙ 1,i H N˙ 2,i H species i



species i

N˙ 1,i S¯ (T1 , P1 , x1 ) −

species i



species i

Q˙ N˙ 2,i S¯ (Tamb , Pamb , x2 ) + =0 Tamb (14.5-1)

where stream 2 is the exit stream. Now using the entropy balance to solve for the heat flow, and the mass balance to eliminate the exit stream molar flows in terms of the inlet flows results in    ¯ i (T1 , P1 , x1 ) − Tamb S¯i (T1 , P1 , x1 ) N˙ 1,i H species i

−



  ˙ S,max = 0 ¯ i (Tamb , Pamb , x2 ) − Tamb S¯i (Tamb , Pamb , x1 ) + W N˙ 2,i H

species i

(14.5-2) Defining the partial molar steady-flow availability of species i to be

¯ i (T, P, x) − Tamb S i (T, P, x) B¯i (T, P, x) = H

(14.5-3)

leads to the following   ˙ S,max = 0 N2,i B¯i (Tamb , Pamb , x2 ) + W N˙ 1,i B¯i (T1 , P1 , x1 ) − species i

species i

or   ˙ S,max = W N˙ 2,i B¯i (Tamb , Pamb , x2 ) − N˙ 1,i B¯i (T1 , P1 , x1 ) species i

species i

(14.5-4)

876 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry Integrating this steady-state equation over any time interval gives for the maximum shaft work   WS,max = N2,i B¯i (Tamb , Pamb , x2 ) − N1,i B¯i (T1 , P1 , x1 ) (14.5-5) species i

species i

¯ i (Tamb , Pamb , x) − Tamb S i (Tamb , Pamb , x) = Now since B¯i (Tamb , Pamb , x) = H ¯ i (Tamb , Pamb , x) G WS,max =



¯ i (Tamb , Pamb , x2 ) − N2,i G

species i



N1,i B¯i (T1 , P1 , x1 )

(14.5-6)

species i

Since the partial molar properties depend on composition, one has to solve the chemical equilibrium problem to determine the exit stream compositions. Note that

¯ iex (T, P, x) and that ¯ i (T, P, x) = H i (T, P ) + H H ex S¯i (T, P, x) = S i (T, P ) + S¯i (T, P, x) + R ln xi So that

¯ i (T, P, x) − Tamb S¯i (T, P, x) Bi (T, P, ) = G = Gi (T, P ) − Tamb (S i (T, P ) − R ln xi ) = Bi (T, P ) + RTamb ln xi Now generally if there is a chemical reaction the energy and Gibbs energy changes are of the order of 100s kJ/mol or more, while the excess enthalpy and entropy and the RTln x are of much lower magnitude (and may cancel between the inlet and outlet streams). Therefore, a reasonable simplification is to replace the partial molar properties with the pure component properties in the availability function to a first approximation so that

¯ i (T, P, x) B¯i (T, P, x) = H ∼ H i (T, P ) − Tamb S i (T, P ) = B i (T, P ) ¯ −Tamb Si (T, P, x) = so that

WS,max =

 species i

N2,i Gi (Tamb , Pamb ) −



N1,i Bi (T1 , P1 )

(14.5-7)

species i

Note that the temperature and pressure dependences of the pure component availability B are given in Eq. 4.6-6 and 8. Also, that to compute the pure component Gibbs energy we use Gi (25o C, 1 bar) = Δf Goi given in Appendix A.IV. Two examples of the application of the availability concept of the simplified and exact availability equations to a combustion process are given in the following illustrations. Illustration 14.5-1 Determine the maximum useful shaft work that can be obtained from natural gas that we will model as methane, which is available at 25o C and 1 bar. The methane is to be burned in air that contains nitrogen and oxygen in the mole ratio of 3.76 to 1. [a] Use the approximate availability equation, and [b] Use the exact availability equation.

14.5 Maximum Useful Work and Availability in Chemically Reacting Systems 877 Solution The stoichiometry of the reaction is CH4 + 2O2 = CO2 +2H2 O [a] Since methane is available at ambient conditions Eq. (14.5-7) above neglecting solution nonidealities and the RTlnx term reduces to WS,max

=



N2,i Gi (25o C, 1 bar) −

species i

=





N1,i Gi (25o C, 1 bar)

species i



N2,i ΔGf,i (25 C, 1 bar) − o

species i

N1,i ΔGf,i (25o C, 1 bar)

species i

= ΔGf,CO2 (25o C, 1 bar) + 2ΔGf,H2 O (25o C, 1 bar) − ΔGf,CH4 (25o C, 1 bar) −2ΔGf,O2 (25o C, 1 bar) = −394.4 − 2 × 228.6 − (−50.5 + 0) = −801.1 kJ/mol methane

Thus, the maximum amount of shaft work that can be produced per mol of methane is 801.1 kJ. [b] Now the stoichiometry of the reaction is CH4 + 2O2 + 2x3.76N2 = CO2 +2H2 O + 2x3.76N2 The stoichiometric table for the inlet conditions is Species CH4 O2 N2 H2 CO2 Total

Initial Number of Moles 1.0 2.0 7.56 0 0

Mole fraction 0.0947 0.1894 0.7159 0 0

Ni ln yi

−2.3570 −3.3278 −2.5267 0 0

−8.2115

10.56

The stoichiometric table for the outlet conditions is Species CH4 O2 N2 H2 CO2 Total

Final Number of Moles 0 0 7.56 2 1.0 10.56

Mole fraction 0 0 0.7159 0.1894 0.0947

Ni ln y i 0 0

−2.5267 −3.3278 −2.3570 −8.2115

Comments Note that for combustion reactions such as the one above, the reaction generally goes to completion, that is all the fuel burns, so we do not have to solve the chemical equilibrium problem. Also, in computation the Gibbs energy of formation at 25o C has been used for the molar Gibbs energy. At other than 25o C, these values would have to be corrected for temperature. Note also by the fortuitous of this reaction and that a stoichiometric amount of air has been used, the sum of the Ni lnyi terms cancel out, and the approximate and exact equations give the same result, This would not be the case if the stoichiometry were different or excess air was used to insure complete combustion (Problems 14.23 to 27) was used to insure complete combustion.

878 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry A measure of the efficiency of an electricity-generating power plant is Efficiency =

electrical power generated −WS,max

(14.5-8)

By this measure the U. S. Department of Energy (http://energy.gov/fe/how-gas-turbinepower-plants-work) estimates that the average efficiency of gas-fired power plants is only about 35%. However, this can be improved to about 60% in co-generation, in which the waste heat from the power plant, that is heat contained in the high-temperature exit stream, is used elsewhere in commercial processes or for residential heating. (See Problem 14.28). Illustration 14.5-2 Determine the maximum useful shaft work that can be obtained from n-octane, a model for gasoline, which is available at 25o C and 1 bar. The octane is to be burned in air that contains nitrogen and oxygen in the mole ratio of 3.76 to 1, and that n-octane liquid at ambient conditions [a] Use the approximate availability equation, and [b] Use the exact availability equation.

Solution The stoichiometry of the reaction is C8 H18 + 12.5xO2 +12.5x3.76N2 = 8CO2 +9H2 O + 12.5x3.76N2 [a] Since n-octane is available at ambient conditions Eq. (14.5-7) at which it is a liquid, neglecting solution nonidealities and the RTlny term reduces to WS,max

=



N2,i Gi (25o C, 1 bar) −

species i

=





N1,i Gi (25o C, 1 bar)

species i

N2,i ΔGf,i (25 C, 1 bar) − o

species i



N1,i ΔGf,i (25o C, 1 bar)

species i

= 8ΔGf,CO2 (25 C, 1 bar) + 9ΔGf,H2 O (25o C, 1 bar) − ΔGf,C8H18 (25o C, 1 bar) −2ΔGf,O2 (25o C, 1 bar) o

= −8 × 394.4 − 9 × 228.6 − (−16.3) = −4565.3 kJ/mol octane

Thus, the maximum amount of shaft work that can be produced per mol of methane is 4565.3 kJ/mol octane. [b] The stoichiometric table for the inlet conditions is, noting that n-octane is a liquid at ambient conditions, Species C8 H18 O2 N2 H2 CO2 H2 O Total

Initial Number of Moles (1.0) 12.5 47 0 0 0 59.5

Mole fraction 0 0.2101 0.7899 0 0 0

Ni ln yi 0

−19.503 −11.0843 0 0 0

−30.587

14.5 Maximum Useful Work and Availability in Chemically Reacting Systems 879 The stoichiometric table for the outlet conditions is Species

Final Number of Moles

C8 H18 O2 N2 H2 CO2 H2 O

0 0 47 9 1.0 9

Total

Mole fraction

Ni ln y i

0 0 0.6144 0.1177 0.0947 0.1177

0 0

−22.896 −3.3278 −19.603 −19.261 −83.608

76.5

Comments So the maximum shaft work available is WS,max = −4565.26 + RTamb (83.608 − 30.587)/1000 = = −4565.26 + 8.314 × 298.15 × 53.021/1000 = −4565.26 + 131.42J/mol = −4433.8kJ/mol which is not insignificant, and the error in using the approximate equation is about 2.9% in this case.

The maximum amount of useful work that can be obtained between inlet state 1 and any exit state 2 is

WS,max =





N2,i B¯i (T2 , P2 , x2 ) −

species i

N1,i B¯i (T1 , P1 , x1 )

(14.5-9)

species i

The mass, energy, and entropy balances for the non-flow system integrated over time, neglecting the potential and kinetic energy terms that are generally negligible if a reaction occurs), are

Ni,2 = Ni,1 +



νij Xj

reactions j

U2 (T2 , P2 ) − U1 (T1 , P1 ) =



¯i (T2 , P2 , x2 ) − Ni,2 U

species i

= WS + Q −

¯i (T1 , P1 , x1 ) Ni,1 U

species i V2

P dV V1

and S2 (T2 , P2 ) − S1 (T1 , P1 )





=



Ni,2 S¯i (T2 , P2 , x2 ) −

species i

=

Q + Sgen T



Ni,1 S¯i (T1 , P1 , x1 )

species i

(14.5-10)

To obtain the maximum shaft work available from the system the final state should be at ambient temperature and pressure, the heat flow should occur at the ambient

880 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry temperature and the expansion work should be at ambient pressure. Therefore, the balance equations are:  Ni,2 = Ni,1 + νij Xj reactions j

U2 (Tamb , Pamb , x2 ) − U1 (T1 , P1 , x1 ) = WS,max + Q − Pamb (V2 − V1 )   ¯i (Tamb , Pamb , x2 ) − ¯ 1 , P1 , x1 ) Ni,2 U Ni,1 U(T = species i

species i

and S2 (Tamb , Pamb , x2 ) − S1 (T1 , P1 , x1 ) = 

=

Ni,2 S¯i (Tamb , Pamb , x2 ) −

species i

Q Tamb 

= Ni,1 S¯i (T1 , P1 , x1 )

species i

(14.5-11) Now eliminating the heat flow Q between these equations gives    ¯i (Tamb , Pamb , x2 ) − Tamb S¯i (Tamb , Pamb , x2 ) Ni,2 U species i

−



  ¯i (T1 , P1 , x1 ) − Tamb S¯i (T1 , P1 , x1 ) Ni,1 U

species i

= WS,max − Pamb (V2 − V1 )   ¯i (Tamb , Pamb , x2 ) + Pamb V¯i (Tamb , Pamb , x2 ) Ni,2 U  ¯i (Tamb , Pamb , x2 ) S − T amb    ¯i (T1 , P1 , x1 ) + Pamb V¯i (T1 , P1 , x1 ) − Tamb S¯i (T1 , P1 , x1 ) − Ni,1 U

species i

species i

=

WS,max

or 

Ni,2 A¯i (Tamb , Pamb , x2 ) −

species i

and

WS,max =



Ni,1 A¯i (T1 , P1 , x1 ) = WS,max

species i



Ni,1 A¯i (Tamb , Pamb , x1 ) −

species i



Ni,2 A¯i (T2 , P2 , x2 ) (14.5-12)

species i

where

¯i (T, P, x) + Pamb V¯i (T, P, x) − Tamb S¯i (T, P, x) A¯i (T, P, x) = U

(14.5-13)

[Generally, solution nonidealities are very small compared to the enthalpies of reaction. Therefore, if these are neglected, the equation for computing the maximum useful work that can be extracted from a non-flow system simplifies to   WS,max = Ni,2 Ai (Tamb , Pamb ) − Ni,1 Ai (T1 , P1 ) (14.5-14) species i

species i

14.5 Maximum Useful Work and Availability in Chemically Reacting Systems 881 Note that the chemical equilibrium problem must still be solved to compute the values of Ni,2 .] Finally, the maximum useful shaft work that can be obtained from the change from any initial state 1 to any final state 2 (neglecting solution nonidealities) is   WS,max = Ni,2 A¯i (T2 , P2 , x2 ) − Ni,1 A¯i (T1 , P1 , x1 ) (14.5-15) species i

species i

Illustration 14.5-3 The common sugar glucose C6 H12 O6 is oxidized essentially continuously in the human body to produce the energy needed for various metabolic processes. The reaction is C6 H12 O6 +6O2 → 6CO2 +6H2 O

This reaction is done biochemically in the body and is referred to as glycolysis. The specific path is referred to as the Krebs cycle, and part of the energy released in the reaction is stored by producing two adenosine triphosphate (ATP) molecules from 2 adenosine diphosphate (ADT) molecules. The dissociation of ATP to ADT is then used as an energy source elsewhere in the body to drive biochemical processes. The following data are available at 25◦ C: a. Complete the table above. b. Calculate the availability of each component and the maximum flow shaft work that can be obtained from this reaction with respect to ambient conditions of 25◦ C. c. Calculate the maximum flow shaft work from this reaction with respect to body temperature of 37◦ C.

Solution a. The enthalpy, Gibbs free energy, and entropy of formation are at 25◦ C. Therefore the missing items in the table are computed using Δf G = Δf H − T Δf S = Δf H − 298.15 · Δf S . Also, here at 25◦ C the availability is equal to the Gibbs energy of formation. The results are: Δf H (kJ/mol) Δf G (kJ/mol) Δf S (J/mol · K) CP (J/mol · K) B (kJ/mol)

Glucose O2 CO2 H2 O

−1271 0 −393.5 −285.8

−1333.4 0 −394.4 −237.1

209.2 0 3.0 −163.3

218.6 29.4 37.0 73.4

−1333.4 0 −394.4 −237.1

b. At 25◦ C the flow availability of each compound is equal to its Gibbs free energy of formation. Therefore, the maximum shaft work available from this reactio; is Ws,max

= 6B CO2 + 6B H2 O − B glucose − 6B O2 = 6 × (−394.4) + 6 × (−237.1) − (−1333.4) − 6 × 0 = −2455.6 kJ/mol glucose

c. As the first step in the calculation, we have to correct the enthalpy and the entropy to 37◦ C. This is accomplished using the following equations  310.15 H(37o C) = H(25o C)+ 298.15 CP dT = H(25o C) + 12 × CP    310.15 S(37o C) = S(25o C)+ 298.15 CTP dT = S(25o C)+CP × ln 310.15 298.15 = S(25o C) + 0.03946 × CP Also, here the flow availability is B (37o C) =H(37o C) − 310.15 × S(37o C).

882 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry The results at 37◦ C are given in the following table. H (kJ/mol) S (J/mol · K) B (kJ/mol)

Glucose O2 CO2 H2 O

−1268.4 −0.350 −393.1 −284.8

217.9 1.12 4.48 −160.44

−1336.0 0.01 −394.5 −235.2

and WS,max = −2441.6 kJ/mol glucose

This maximum work at 310.15 K is only slightly less than at 298.15 K in part b.

Comments In humans approximately 32% of the energy released on the glycolysis reaction is used in the ADT to ATP reaction, the rest is released as heat. Therefore, the glycolysis reaction is only 32% efficient.

14.6 INTRODUCTION TO ELECTROCHEMICAL PROCESSES Electrical work can be obtained from carefully controlled chemical reactions, but not if the reaction is allowed to proceed spontaneously. For example, in the standard state of pure gases of hydrogen and chlorine at 1 bar and 298 K, the reaction to form hydrogen chloride in aqueous solution 1 H (g) 2 2

+ 12 Cl2 (g) = HCl(aq)

has a Gibbs energy change of

Δrxn G◦ = −13.12 − [0 + 0] = −13.12

kJ mol HCl

Therefore, there is a possibility of the reaction occurring spontaneously; however, this will produce no useful work. An alternative is to run the reaction in an electrolytic cell in which hydrogen and chlorine gases are metered into separate electrodes and the electromotive force (EMF) or voltage produced is almost balanced by the application of an external voltage. In this way the reaction will occur at a very slow rate and electrical work can be obtained from the cell. In this process chemical energy is directly converted to electrical energy. Electrochemical processes occur in batteries, fuel cells, electrolysis, electrolytic plating, and corrosion (generally an undesirable process). Electrochemical processes can be used to produce electricity, to recover metals from solution, and for the measurement of the thermodynamic properties of electrolyte solutions. The device used to study electrochemical reactions is an electrochemical cell, which consists of two electrodes (metallic conductors) in electrolytes that are usually liquids containing salts, but may be solids, as in solid-state batteries. The two electrodes may be in the same electrolyte, as shown in Fig. 14.6-1a, or each electrode may be in a separate compartment with its own electrolyte, as in Fig. 14.6-1b. In this case the two compartments are connected by a salt bridge, an electrolyte that completes the electrical circuit. A third alternative, not shown, is for the two compartments to be in direct contact through a porous membrane.

14.6 Introduction to Electrochemical Processes 883 Electrodes

Electrodes Salt bridge

Electrolyte

(a)

First electrolyte

Second electrolyte

Electrode compartments (b)

Figure 14.6-1 Two types of electrochemical cells. (a) A cell with two electrodes and a shared electrolyte. One example of such a cell contains a copper electrode, a zinc electrode, and a zinc sulfate and copper sulfate electrolyte solution. The overall cell reaction is Cu2+ (aq) + Zn(s) → Cu(s) + Zn2+ (aq). (b) A cell with two separate compartments connected by a salt bridge. If the same electrodes as in the previous case were used, one compartment would contain a copper electrode and a CuSO4 solution, the other would have a zinc electrode and a ZnSO4 solution as the electrolyte, and the two compartments would be connected by a bridge containing, for example, a sodium chloride solution.

When the two electrodes are connected through a potentiometer or electrical resistance, and electricity is produced by the chemical reaction that occurs spontaneously, the electrochemical cell is referred to as a galvanic cell; it is considered to be a fuel cell if the reagents are continually supplied to the cell. Batteries are galvanic cells. The term electrolytic cell is used to indicate an electrochemical cell operated in the reverse manner to that just described, in that an external voltage is used to cause a nonspontaneous reaction to occur, as in the electrolysis of water. An automobile battery and other storage batteries can be considered galvanic cells when they are supplying electricity, and electrolytic cells when they are being recharged. There are several ways that thermodynamics is used to analyze electrochemical cells. One is to compute the work, or equivalently the voltage, that can be produced by a galvanic cell. Alternatively, measured cell voltages can be used to determine the equilibrium constant of the reaction taking place within the cell. A third use of electrochemical cells is to measure the thermodynamic activity or activity coefficients of the ions in electrolyte solutions. The main processes occurring in electrochemical cells are simultaneous oxidation and reduction reactions,5 or redox reactions. At one electrode, the anode, a reduced species is oxidized here meaning to release electrons, while at the other electrode, the cathode, an oxidized species absorbs electrons and is reduced. It is common to think of an electrochemical cell as consisting of two half-cells (one containing the anode and the second containing the cathode) and to describe the processes in terms of halfcell reactions. For example, one common cell consists of a copper cathode in a copper sulfate solution, and a zinc anode in a zinc sulfate solution. The overall reaction is

Cu2+ (aq) + Zn(s) → Cu(s) + Zn2+ (aq) (redox reaction) 5 Here the term oxidation is being used as in general chemical terminology to indicate a release of electrons. In this

sense it is not necessary for oxygen to be involved in an oxidation reaction, as unusual as this may seem.

884 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry which is the sum of the two half-cell reactions

Cu2+ (aq) + 2e− → Cu(s) (reduction, absorption of electrons) and

Zn(s) → Zn2+ (aq) + 2e−

(oxidation, release of electrons)

The following abbreviated notation is used to describe the complete cell

Zn(s) | ZnSO4 (aq)  CuSO4 (aq) | Cu(s) The electrochemical processes occurring in this cell are the oxidation of zinc and the production of zinc sulfate and electrons at the anode, the absorption of electrons and the reduction and deposition of copper at the cathode, the flow of electrons through an external electrical circuit (resulting in electrical work), and a balancing flow of sulfate ions through the salt bridge. The metallic electrode described above is the simplest of the electrode types. Another type of electrode is the insoluble-salt electrode, in which a metal is covered with one of its insoluble salts; silver chloride deposited on silver is one such example. The oxidation reaction in this case is

AgCl(s) + e− → Ag(s) + Cl− (aq) and the half-cell description is

Ag | AgCl | Cl− A third type of electrode is the gas electrode, in which a gas is in equilibrium with a solution of its ions in a half-cell that contains an inert metal conductor. The hydrogen electrode, in which one bubbles hydrogen through a solution and across a platinum electrode, is perhaps the best known of this type. The half-cell chemical reaction is

H2 (g) → 2H+ (aq) + 2e− and the description used is

Pt | H2 | H+ (aq) Since hydrogen must be continually supplied from outside the cell, an electrochemical device using a hydrogen gas electrode would be considered a fuel cell. From Sec. 4.3 we have that for any process occurring at constant temperature and pressure, the manner in which electrochemical cells are operated, the maximum work that can be obtained is equal to the change in Gibbs energy of the process, that is,

W max = ΔG

(14.6-1)

This maximum work is obtained if the process is sufficiently slow that there are no irreversibilities, for example, no resistive heating as a result of the current flow. This implies that the rate of reaction is very slow, and that the electrical potential produced is just balanced by an external potential so that the current flow is infinitesimal. This electrical potential produced by the cell (or of the balancing external potential) will be referred to as the zero-current cell potential and designated by E. The work done by the electrical cell Welec in moving n moles of electrons across a potential difference of E is

Welec = −nF E

(14.6-2)

14.6 Introduction to Electrochemical Processes 885 where F = 96 485 C/mol is the Faraday constant, and the negative sign indicates that work is done by the cell on the surroundings if the cell potential is positive. [A coulomb (abbreviated C) is a unit of electrical charge; moving one coulomb of charge through a potential difference of one volt requires one joule of energy. Also, for later reference we note that at 25◦ C, the quantity RT /F is equal to 25.7 mV.] Consequently, we have

ΔG = W max = Welec = −nF E or simply

ΔG = −nF E

(14.6-3)

We write a generic electrochemical reaction as z+

z+

ν1+ M1+1 (aq) + ν2 M2 (s) + ν1 M1 (s) + ν2+ M2+2 (aq) = 0 or, more generally (using our standard notation for chemical reactions),  νi Mi = 0 i

For example, the reaction

Cu2+ (aq) + Zn(s) → Cu(s) + Zn2+ (aq) will be written as Zn2+ (aq) + Cu(s) − Zn(s) − Cu2+ (aq) = 0, so that νZn2+ = 1, νCu = 1, νZn = −1, and νCu2+ = −1. Also, the Gibbs energy of any species can be written as

Gi (T, P, x) = G◦i (T, P ◦ , x◦i ) + RT ln ◦

◦

f i (T, P, x) f ◦i (T, P ◦ , x◦i )

(14.6-4)

◦

= Gi (T, P , xi ) + RT ln ai (T, P, x) where P ◦ and x◦i are the standard-state pressure and composition (in units appropriate to the standard state chosen), f ◦i (T, P ◦ , x◦i ) and Gi (T, P ◦ , x◦i ) are the standard-state fugacity and Gibbs energy of species i, and ai is its activity. Combining these last two equations, we have

Δrxn G = Δrxn G◦ + RT ln



aνi i = −nF E = −nF E ◦ + RT ln

i



aνi i (14.6-5)

i

where E ◦ would be the zero-current cell potential if the ions were in their standard states, while E is the actual (measurable) zero-current cell potential with the ions at the concentration of the cell. The last relation is known as the Nernst equation. Remembering from Eq. 13.1-18 that −

Δrxn G◦ = ln Ka RT

(13.1-8)

where Ka is the chemical equilibrium constant, we have

ln Ka =

nF E ◦ RT

(14.6-6)

886 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry Table 14.6-1 Standard Half-Cell Potentials at 25◦ C Oxidizing Agent or Oxidant Au+ Cl2 Br2 Ag+ Hg22+ Fe3+ Cu2+ AgCl 2H+ Fe3+ Pb2+ Zn2+ Al3+ Mg2+ Na+ Li+

Reducing Agent or Reductant

E ◦ (V)

Δrxn G

Au 2Cl− 2Br− Ag 2Hg Fe2+ Cu Ag + Cl− H2 Fe Pb Zn Al Mg Na Li

+1.69 +1.36 +1.09 +0.80 +0.79 +0.77 +0.34 +0.22 0.0 −0.04 −0.13 −0.76 −1.66 −2.36 −2.71 −3.05

−69.79 −105.88 −84.86 −33.14 −61.50 −29.97 −26.47 −8.56

+e− +2e− +2e− +e− +2e− +e− +2e− +e− +2e− +3e− +2e− +2e− +3e− +2e− +e− +e−

RT

0 +4.67 +10.12 +59.17 +193.86 +183.74 +105.19 +118.73

Δrxn G(kJ)

−173.0 −262.5 −210.4 −77.2 −154.4 −74.3 −65.6 −21.2 0 (by definition) +11.6 +25.09 +146.7 +480.8 +455.5 +261.5 +294.3

This equation allows one to compute the chemical equilibrium constant from measured standard-state electrochemical cell potentials (usually referred to as standard cell potentials). Some standard half-cell potentials are given in Table 14.6-1. The standard potential of an electrochemical cell is obtained by combining the two relevant half-cell potentials. Since two connected half-cells are needed to measure a voltage, any one half-cell potential cannot be measured directly. Therefore, by convention, half-cell potentials are referenced to the standard hydrogen electrode (referred to as the SHE) consisting of a platinum electrode in an aqueous solution and into which pure hydrogen gas is bubbled through the electrode. When the hydrogen gas is pure at 1 bar pressure so that the hydrogen ion activity in solution is unity the standard hydrogen potential is, by convention, taken to be zero at all temperatures, and all other half-cell potentials are referenced to it. Conceptually this is analogous to constructing a fuel cell or battery in which one half-cell is the standard hydrogen electrode and then attributing all of the measured voltage to the other half-cell. Table 14.6-1 was constructed in this way. One point to note when using this table is that

Δrxn G◦ = −nF E ◦

or E ◦ = −

Δrxn G◦ nF

where n is the number of electrons transferred. Consequently, for example, for the reaction Cl2 + 2e− → 2Cl− , E ◦ is 1.36 V, and Δrxn G◦ = −2 × 1.36F = −2.72F . However, if instead we wrote the reaction as 12 Cl2 + e− → Cl− , then Δrxn G◦ would be one-half its previous value, or −1.36F . However, now n equals unity, so that again

E◦ = −

Δrxn G◦ (−1.36F ) =− = 1.36 V nF 1·F

Consequently, if we multiply any of the half-cell reactions in Table 14.6-1 by an integer or fractional constant, the standard-state Gibbs energy change will change by that same factor, but the standard-state half-cell potential will be unchanged.

14.6 Introduction to Electrochemical Processes 887 Illustration 14.6-1 Computation of the Standard Cell Potential Compute the standard cell potential and the equilibrium constant for the reaction Cu2+ (aq) + Zn(s) → Cu(s) + Zn2+ (aq)

Solution The reaction above is the sum of the two half-cell reactions Cu2+ (aq) + 2e− → Cu(s) Zn(s) → Zn2+ (aq) + 2e−

half-cell standard potential = +0.34 V half-cell standard potential = −(−0.76 V) = +0.76 V

where the negative of the reported half-cell standard potential has been used for the second reaction, since its direction is opposite to that given in the table. Therefore, the standard cell potential for the overall reaction is E ◦ = +0.34 + 0.76 = +1.10 V

The equilibrium constant for this reaction is then nF E o ln Ka = = RT

C × 1.10 V mol = 85.6 J CV 8.314 × 298.15 K × 1 mol K mol 2 × 9.6485 × 104

or

Ka = 1.5 × 1037

or equivalently, 1000 mV 2 × 1.10 V × nF E o nE o V  = ln Ka = =  = 85.6 RT RT 25.7 mV F

or

Ka = 1.5 × 1037

Illustration 14.6-2 Calculation of the Equilibrium Constant from Standard Half-Cell Potentials Determine the equilibrium constant for the dissolution and dissociation of silver chloride in water, and the silver chloride solubility in water.

Solution The reaction is AgCl → Ag+ + Cl−

which in terms of half-cell reactions we write as AgCl + e− → Ag + Cl− → Ag+ + e− Ag

half-cell standard potential = +0.22 V half-cell standard potential = −(+0.80 V)

Thus E ◦ = +0.22 − (+0.80) = −0.58 V, and nF E ◦ ln Ka = = RT

C × (−0.58)V mol = −22.568 J CV 8.314 × 298.15 K × 1 mol K mol 9.6485 × 104

888 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry or Ka = 1.58 × 10−10

which compares well with the value of 1.607 × 10−10 computed in Illustration 13.3-2. The equilibrium relation is then Ka = 1.58 × 10−10 =

 2 aAg+ aCl− = MAg+ MCl− = MAg+ aAgCl

so that MAg+ = MAgCl = 1.257 × 10−5 M = 1.257 × 10−5

mol kg water

In writing these last equations we have recognized that the activity of the pure silver chloride solid is unity, assumed that the ion concentrations will be so low that the activity coefficients would be unity, and used the fact that by stoichiometry the concentrations of the silver and chloride ions must be equal.

Following upon the illustration above, quite generally we can write  ν ΔG = Δrxn G◦ + RT ln ai i = −nF E ◦

= Δrxn G + RT

i ν ν ν1 ln aM1 (s) aM2+ a 1+ aν2 2+ (aq) M1+ (aq) M2 (s) ν

ν

= −RT ln Ka + RT ln aνM11 (s) aM2+ a 1+ aν2 2+ (aq) M1+ (aq) M2 (s) ◦

= Δrxn G + RT

(14.6-7)

ν ν ln aM2+ a 1+ 2+ (aq) M1+ (aq) ν

ν

= −RT ln Ka + RT ln aM2+ a 1+ 2+ (aq) M1+ (aq) = −nF E ◦ + RT ln aM2+ a 1+ 2+ (aq) M1+ (aq) ν

ν

In writing these equations we have recognized that the activities of the pure metal electrodes are unity. Note that if the electrochemical cell is in chemical equilibrium, that is, if ν

ν

Ka = aM2+ a 1+ 2+ (aq) M1+ (aq)

(14.6-8)

then E is equal to zero (there is no voltage produced by the cell even though E ◦ is nonzero), and ΔG = 0, as must be the case for a process at equilibrium. We have shown earlier that it is possible to produce work if there is a temperature difference between two subsystems (for example, by connecting them through a Carnot or other cycle). Similarly, if there is a pressure difference between two subsystems, this can be used to drive a fluid through a turbine, again producing work. An electrochemical cell is one way of producing work if there is a concentration difference between two subsystems. To see how this could be done, suppose we had two beakers containing a salt, say copper sulfate, at different concentrations. We could put a metallic copper electrode connected to a potentiometer or other electrical circuit in each beaker and then connect the electrolytes in the beakers by a salt bridge. In the beaker containing the dilute copper sulfate solution, which we designate as beaker 1, the reaction would be

Cu → Cu2+ + 2e−

14.6 Introduction to Electrochemical Processes 889 while in the beaker containing the concentrated solution, which we will refer to as beaker 2, the reaction would be

Cu2+ + 2e− → Cu In this case the two half-cell potentials cancel, but as a result of the concentration difference, we have

    aCu2+ aSO2− MCu2+ MSO2− γCu2+ γSO2− 4 4 4 1 = RT ln  1 ΔG = nF E = RT ln  aCu2+ aSO2− MCu2+ MSO2− γCu2+ γSO2− 4 4 4 2 2   MCu2+ MSO2− γ±2 4 1 = RT ln  2 MCu2+ MSO2− γ± 4 2 (14.6-9) To proceed, we use that by stoichiometry and from the fact that copper sulfate is fully dissociated into ions, MCu2+ = MSO2− = MCuSO4 4

Also, for the purposes of illustration, we will use the simple Debye-H¨uckel limiting law of Eq. 9.10-15 for the mean ionic activity coefficient, which here becomes √   ln γ± = −α|z+ z− | I = −α4 4MCuSO4 = −α8 MCuSO4 since 2 I = 12 (M+ z+ + M− z−2 ) = 12 MCuSO4 ((2)2 + (−2)2 ) = 4MCuSO2

Putting this all together gives

 2    MCuSO4 γ±2 1 MCu2+ MSO4 2− γ±2 1  = RT ln  2  ΔG = −nF E = −2F E = RT ln  MCu2+ MSO4 2− γ±2 2 MCuSO4 γ±2 2 [MCuSO4 ]1 [γ± ]1 + 2RT ln [MCuSO4 ]2 [γ± ]2   [MCuSO4 ]1 = 2RT ln − RT α8 [I]1 − [I]2 [MCuSO4 ]2    [MCuSO4 ]1 − RT α16 = 2RT ln [MCuSO4 ]1 − [MCuSO4 ]2 [MCuSO4 ]2 = 2RT ln

(14.6-10)

Illustration 14.6-3 Computing the Cell Voltage That Is Produced as a Result of the Concentration Difference One beaker contains copper sulfate at a concentration of 0.0001 M and another contains a 0.01-M solution of the same salt. Compute the maximum voltage that could be obtained at 25◦ C with an electrochemical cell that used these two solutions as electrolytes.

890 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry Solution The starting point is the preceding equation written as follows:

   [MCuSO4 ]1 RT RT [MCuSO4 ]1 − [MCuSO4 ]2 ln − ·α·8· [MCuSO4 ]2 F F

 √ √ 0.0001 M − 1.178 · 8 · 0.0001 − 0.01 = 25.7 mV ln 0.01 M = 25.7 mV [ln 0.01 − 1.178 · 8 · (0.01 − 0.1)]

−E =

= 25.7 mV[−4.605 + 0.848] = −96.5 mV

so that E = 96.5 mV.

When a half-cell reaction involves hydrogen ions, the cell potential will depend upon the hydrogen ion concentration of the solution, or the pH, where the pH scale is defined as follows:   MH+ + + pH = − log a(H ) = − log a(H3 O ) ≈ − log (14.6-11) 1 molal In writing this equation we have recognized that in solution the hydrogen ion is actually present as a hydronium ion, and the last expression is valid only if the hydrogen-ion concentration is so low that its activity coefficient is unity. It is interesting to compute how the cell potential varies with changes in pH. To do this we consider an electrochemical cell that contains a hydrogen electrode, and leave the other half-cell reaction unspecified. The overall cell reaction is written as

H2 (g) + 2M+ = 2H+ + 2M(s) for which the cell potential will be   2 2   2 aH+ aM aH+ RT RT ◦ ◦ E=E + =E + ln ln 2F aH2 a2M+ 2F aH2 a2M+ RT RT = E◦ + ln [aH+ ] − ln[aH2 a2M+ ] F 2F RT RT pH − ln[aH2 a2M+ ] = E ◦ − 2.303 F 2F At 25◦ C we have

E = E ◦ − 59.2 · pH − 12.86 · ln[aH2 a2M+ ] mV

(14.6-12)

(14.6-13)

We see from this equation that the actual potential produced by an electrochemical cell involving a hydrogen (or hydronium) ion depends linearly on the pH of the solution. The total cell potential also depends on the activity of the metal ion in the other half-cell, which usually would be approximately constant, and the activity of molecular hydrogen, which can be controlled by its partial pressure. Consequently, the primary variation of the cell potential is with pH. This result suggests that an electrochemical cell can be used to measure the pH of a solution. This is actually done in the laboratory, but by using specially chosen liquid electrodes rather than a hydrogen gas electrode, which is not convenient to use. By the appropriate choice of an electrochemical cell, it is possible to measure the mean ionic activity coefficient of an electrolyte. As an example, consider a cell consisting of a hydrogen electrode and a silver–silver chloride electrode, both in the same

14.7 Fuel Cells and Batteries 891 solution with hydrochloric acid as the electrolyte. It is the mean ionic activity coefficient of hydrochloric acid that can be measured. The overall cell reaction is 1 H (g) 2 2

+ AgCl(s) = HCl(aq) + Ag(s)

The cell potential for this reaction, since n = 1, is     a a a RT RT HCl(aq) Ag(s) HCl(aq) = E◦ − E = E◦ − ln 0.5 ln 2F aH2 (g) aAgCl(s) F a0.5 H2 (g)

(14.6-14)

since the solids silver and silver chloride are at unity activity. Further, the activity of hydrogen is regulated by its partial pressure, so that we can consider it to be independently fixed and known. Therefore,

E = E◦ +

RT RT ln [aH2 (aq) ] − ln [aHCl(aq) ] 2F F

But

aH2 (g) =

(14.6-15)

PH2 1 bar

and

MH+ MCl− γH+ γCl− (M = 1 molal)2 2 2   M± MHCl 2 = γ± = γ±2 M = 1 molal M = 1 molal

aHCl = (aH+ ) (aCl− ) =

since the hydrochloric acid, being a strong electrolyte, is fully ionized. Consequently,

E = E◦ +

    PH2 MHCl 2RT 2RT RT ln ln ln γ± (14.6-16) − − 2F 1 bar F M = 1 molal F

or

     2RT F RT PH2 MHCl E◦ − E + − ln ln 2RT 2F 1 bar F M = 1 molal     F 1 MHCl PH2 ◦ = − ln (E − E) + ln 2RT 4 1 bar M = 1 molal (14.6-17) So that by fixing the value of the partial pressure of hydrogen and measuring the cell potential E , the value of the mean ionic activity coefficient of hydrogen chloride can be determined. ln γ± =

14.7 FUEL CELLS AND BATTERIES In this section we consider electrochemical cells, fuel cells, and batteries. All electrochemical cells consist of two electrodes, an anode at which an oxidation reaction occurs and a cathode at which a reduction reaction occurs, and both of which are immersed in one or more electrolyte solutions or conductors as described below. The voltage

892 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry produced in a fuel cell or battery, or required when operated in reverse as in an electroplating process or charging a battery, is a result of the difference in the free energies of the anode and cathode reactions. This can be expressed as free energies, or more commonly in electrochemistry as voltages as in Table 14.6-1. So for example, an electrochemical cell at which the oxidizing, electron-releasing reaction at the anode is

H2 → 2H+ (aq) + 2e− or equivalently

1 H2 (g) → 2H+ (aq) + e− 2 and the reduction electron-absorbing reaction at the cathode is 1 O2 (g) + 2H+ (aq) + 2e− → H2 O 2 with all species in their standard states, is found to have a voltage of 1.229 V. So using the standard convention in Table 14.6-1, E o = 0 for the first reaction and E ◦ = 1.229 V for the second half-cell reaction. However, it should be remembered that these standard potentials are for all the species being at unit activity. We consider later the correction when the species are not at unit activity. Fuel Cells In Chapter 5 we considered the conversion of thermal energy from a high temperature heat source to mechanical energy using a number of cyclic processes. Generally, though not explicitly stated, the high temperature was produced by the combustion of a fuel. The efficiency of any such cyclic processes is limited by the Carnot efficiency (Eq. 4.3-9). An alternative is to use a fuel cell, which converts the chemical energy of the fuel directly into electrical energy and is not limited to the Carnot efficiency. A fuel cell, as shown in Fig. 14.7-1, has two electrodes that may or may not be chemically similar and have catalytic surfaces in a conducting electrolyte, for example, an aqueous solution containing a salt, a conducting polymer proton exchange membrane (PEM, as in Fig. 14.7-1), or other electrolytes as will be mentioned later. The electrodes are connected with wires to the load. Load

oxygen

hydrogen +

H proton

Pt

Pt

PEM water +

−

H2 → 2H + 2e

Figure 14.7-1 Proton Exchange Membrane Fuel Cell

½ O2 + 2e + 2H+ → H2O −

14.7 Fuel Cells and Batteries 893 A hydrogen fuel cell is conceptually the simplest. In this case in an acidic solution hydrogen is supplied to the anode where it is oxidized and oxygen is supplied to the cathode where it is reduced. The reactions that occur are as follows:

Anode reaction :

H2 → 2H+ + 2e−

Cathode reaction : 2H+ + 12 O2 + 2e− → H2 O Overall reaction :

H2 + 12 O2 → H2 O

E o (25o C) = 1.229 V

The charge carrier is the hydrogen ion, H+ . The result is the production of electrical energy that is then consumed in the external circuit with its load. The electrical work obtained per mole of hydrogen, We , is

We = −2F ΔE = Δrxn G [Note that the factor of 2 arises since as written, the reaction involves the transfer of 2 electrons.] The standard cell voltage is ΔE = 1.229 V if the hydrogen and oxygen gases and the water are at unit activity. Note that the second law, because of system irreversibilities, requires that We ≤ Δrxn G In other fuel cells different electrolytes and electrode types are used for reasons including the temperature range of interest and the voltage produced. An alkaline fuel cell using, for example, an aqueous solution of potassium hydroxide in a porous matrix, has the following reactions

Anode reaction :

H2 + 2OH− → 2H2 O + 2e−

Cathode reaction :

1 O 2 2

Overall reaction :

H2 + 12 O2 → H2 O

+ H2 O + 2e− → 2OH− E o (25o C) = 1.229 V

So while the overall reaction is the same as in fuel cell above, here charge carrier is OH− ion, and has a useful wide operating temperature range of 60 to 250◦ C. A potential disadvantage of the KOH fuel cell is that if carbon dioxide is present in the fuel or oxidant, it will react with the KOH producing K2 CO3 and poisoning the cell. Alkaline fuel cells have been used by the NASA since the mid-1960s, starting with the Apollo series of flights and continuing to the Space Shuttle. The electrolyte in phosphoric acid fuel cell is a highly concentrated H3 PO4 saturating a silicon carbide matrix, and the electrodes are carbon paper coated with a finely dispersed platinum catalyst. The electrode reactions are the same in the hydrogen fuel cell, and the hydrogen ion is the charge carrier. The useful operating range is 150 to 200◦ C and has an electrical generating efficiency of about 40%. However, an advantage of the phosphoric acid fuel cell is that the steam produced can be used for heating air or water. In this way, as much as 70% of the energy content of the fuel (hydrogen in the discussion here, but could be methane, other hydrocarbons, and even gasoline) can be used to produce the combination of electricity and heating. Phosphoric acid fuel cells were the first to be commercialized, and have been used for stationary power generators producing 100 to 400 kW, and more recently in buses and other large vehicles. Solid oxide fuel cells use a solid oxide ceramic electrolyte at high temperatures (750 to 1000◦ C) and generally use a mixture of hydrogen and carbon monoxide that is

894 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry produced internally in the fuel cell by a reforming reaction between a hydrocarbon fuel and steam. The reforming reaction, in a different region of the fuel cell and with a different catalyst, is (starting with methane as an example)

CH4 + H2 O → CO + 3H2 The electrochemical reactions are

Anode reactions :

H2 + O2− → 2H2 O + 2e− CO + O2− → CO2 + 2e−

Cathode reaction : O2 + 4e− → 2O2− Overall reaction :

H2 + CO + O2 → H2 O + CO2

For this overall reaction the free energy change is

Δrxn G (25o C) =



νi Δf G = −228.6 − 394.4 + 137.2 = −485.8

i

kJ mol

and

485.8 Δrxn G (25o C) = V = 1.259 V nF 4 × 96.485 and the O2− ion is the charge carrier. The molten carbonate fuel cell uses a molten alkali metal (Li/K or Li/Na) in a porous LiAlO2 matrix, and the electrode catalysts are based on nickel. Such fuel cells operate at temperatures of 650o C and above, and have an electrical generating efficiency of almost 60%, and when the waste heat is used productively, have an overall thermal efficiency as high as 85%. The electrochemical reactions are: E o (25o C) = −

Anode reaction :

H2 + CO3−2 → H2 O + CO2 + 2e−

Cathode reaction :

1 CO2 2

Overall reaction :

H2 + 12 O2 → H2 O

+ 2e− → CO3−2 E o (25o C) = 1.229 V

In this case the charge carrier is the carbonate ion CO3−2 . The last type of fuel cell to be considered here is the polymer-membrane based Proton Exchange Membrane fuel cell or PEM fuel cell. In this case a solid, typically a perfluorosulfonic polymer is the electrolyte separating the electrodes that are coated with noble metal catalysts. The half-cell electrode and overall reactions are

Anode reaction :

H2 → 2H+ + 2e−

Cathode reaction : 2H+ + 12 O2 + 2e− → H2 O Overall reaction :

H2 + 12 O2 → H2 O

E o (25o C) = 1.229 V

It is interesting to note that all the fuel cells considered here produce standard state voltages between 1.23 and 1.26 V (largely because all but one of the reactions is between hydrogen and oxygen). In actual operation the voltage obtained will be lower. First because in most fuel cells the reacting species will not be in states of unit activity.

14.7 Fuel Cells and Batteries 895 Second, the voltages calculated here are the open circuit voltages, that is, those that would be measured using a voltameter connected to the two electrodes, but with no other connection or load between the electrodes. The actual closed circuit voltage, that is the voltage across a resistance or other load between the electrodes, would be lower due to irreversibilities in the fuel cell operation, such as diffusion of the charge carriers in the electrolytes or diffusion across the membrane, concentration gradients (polarization) that occur near the electrodes and electrical resistance in the circuit. The voltage produced by a fuel cell in actual operation, depending on the operating conditions and the states of the reactants, may be from 70 to 85% of the E ◦ value. The rate at which the charge carrier can move across the cell determines the rate at which electrons can be delivered to a load, that is the amperage produced by the cell. The deliverable power density, the product of the cell voltage and the amperage, is an important characterization factor of a fuel cell. The voltage can be increased by stacking a number of fuel cells in series, and the limiting amperage can be increased by increasing the cross-sectional area (membranes and electrodes) of each of the cells. The values for the half-cell potentials discussed so far are based on all the species involved being in states of unit activity. That is, pure gases at 1 bar and ionic species at unit activity in an ideal 1 molal solution. If this is not the case, then the voltage must be corrected using the Nernst equation, Eq. 14.6-5. When applied to the standard hydrogen electrode written in the form of 1 H+ (aq) + e− → H2 (g) 2 the voltage is

E = E◦ −

RT  vi RT a + ai = E ◦ + ln ln  H 1/2 2F F PH2 i 1 bar   RT PH2 RT ◦ =E + ln aH+ − ln F 2F 1 bar

or

2.303RT RT E=E − pH − ln F 2F ◦



PH2 1 bar



Similar corrections can be made for all the other half cell voltages. Batteries While a fuel cell produces electrical energy, a battery only stores electrical energy. Like a fuel cell, a battery has two electrodes, an anode and a cathode separated by an electrolyte. The most common battery is the lead-acid battery used, for example, in motor vehicles. The cathode, the positive electrode, is lead oxide, the anode, the negative electrode, is pure lead, and the electrolyte is a concentrated sulfuric acid. It is the oldest of the rechargeable batteries. Anode reaction: + − Pb (s) + HSO− 4 (aq) → PbSO4 (s) + H (aq) + 2e

Cathode reaction: + − PbO2 (s) + HSO− 4 (aq) + 3H (aq) + 2e → PbSO4 (s) + 2H2 O (l)

896 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry The overall reaction is

Pb (s) + PbO2 (s) + 2H2 SO4 (aq) → 2PbSO4 (s) + 2H2 O (l) The hydrogen ion is the charge carrier and the lead-acid battery has an open circuit (no load) voltage is 2.05 V. A typical lead-acid battery for an automobile delivers 12 volts by having six cells in series. The reaction is reversed on charging by applying an external voltage greater than the cell voltage. Generally, the charging-discharging process has an energy efficiency of about 75 to 80%. Lithium-ion batteries are currently being used as the main energy storage device in consumer electronics such as laptop computers, cell phones, and on a large scale in hybrid and all-electric motor vehicles. The anode is usually carbon and the cathode is frequently lithium cobalt oxide (LiCoO2 ), lithium iron phosphate or other lithiumbased materials. The electrolyte is typically a non-aqueous mixture of carbonates such as ethylene or diethyl carbonates with lithium salts, such as LiPF6 . The cathode half-cell reaction is LiCoO2 → Li1−x CoO2 + xLi+ + xe− The anode half cell reaction is xLi+ + xe− + xC6 → xLiC6 and the overall reaction is LiCoO2 + 6C → CoO2 + LiC6 The lithium ion is the charge carrier and the battery (or each cell of the battery) has a no-load voltage of 3.7 V. The energy efficiency of a lithium ion battery is approximately 85%, and its cost per kilowatt-hour is approximately twice that of a lead acid battery. Among the other batteries are the nickel-cobalt, nickel-metal hydride and sodium sulfur batteries. These will not be considered here. Charging or recharging a battery requires forcing all the reactions to go in the opposite direction to those indicated above by providing an external voltage that is greater than the voltage produced by the battery. Generally, but within limits, the greater the overvoltage, that is, the difference between the applied voltage and the battery voltage, the faster the rate of recharge. Electroplating The process of running an electrochemical cell with an overvoltage is used in the process of electroplating. For example, putting a coating of gold on a metal such as silver or copper. The electrode to be coated, the cathode, is connected to the external source negative voltage and the electrode that is supplying the coating material is the anode and is connected to the positive source of the external voltage. So, as an example, for the electroplating of gold

Au → Au+ + e−

E o (25o C) = −1.69 V

Cathode reaction : Au+ + e− → Au

E o (25o C) = +1.69 V

Anode reaction :

Overall reaction :

Au → Au

E o (25o C) = 0 V

Therefore, any applied voltage in the direction indicated above will result in gold dissolving from the anode and being deposited at the cathode. Of course, the larger the applied voltage the more rapidly the process will occur. However, a too rapid rate of

Problems 897 coating, that is too high an applied voltage, will result in imperfections in the coating, including the formation of whiskers. Capacitors Another type of electrical energy storage device is the capacitor. In its simplest form, it consists of two electrically conducting electrodes separated by an insulating medium, for example, air. When subjected to an external potential difference, equal amounts of positive and negative charges accumulate on the surface of the electrodes. The advantage of a capacitor is that the stored energy can be released extremely quickly since the discharge process is the rapid flow of electrons through wires. This is to be compared with the discharge rate of a battery that is determined by the rate at which the ion carriers can travel through the electrolyte. Also, the initial voltage of a capacitor is determined by the applied external potential unlike a battery whose voltage is determined by the two half-cell potentials. The disadvantage of a traditional capacitor is that since the charges are stored on the surface of the electrodes, its total energy capacity is very limited. Recently, supercapacitors have been developed using porous materials rather than simple plates as the electrodes. In these supercapacitors the total volume of the electrode, rather than just its surface, is used for energy storage resulting in energy storage densities of 10,000 times that of a simple capacitor, and a rate of discharge (power delivered) of ten or more times that of a lithium-ion battery. Supercapacitors have begun to appear in motor vehicles that require high power, such as race cars, buses, trucks, and other heavy vehicles that require high power when starting from a stop. PROBLEMS (Note: The Chemical Engineer’s Handbook, McGraw-Hill, New York, contains a comprehensive list of standard-state Gibbs energies and enthalpies of formation.) 14.1 The flame temperature attained in a torch or a burner can be computed using the adiabatic reaction temperature analysis of Sec. 14.3 if it is assumed that the radiant heat loss from the flame is negligible. Compute the flame temperature and exit composition in a hydrogen torch if a. A stoichiometric amount of pure oxygen is used as the oxidant. b. Oxygen is the oxidant, but a 100 percent excess is used. c. Twice the stoichiometric amount of air is used as the oxidant. In each case the hydrogen and oxidant entering the torch are at room temperature (298.15 K), and the torch pressure is 1.013 bar. 14.2 Compute the flame temperature of an oxyacetylene torch using pure acetylene and 50 percent more pure oxygen than is needed to convert all the acetylene to carbon dioxide and water. Both the oxygen and

acetylene are initially at room temperature and atmospheric pressure. The following reactions may occur: HCCH + 32 O2 = 2CO + H2 O HCCH + 52 O2 = 2CO2 + H2 O CO + 12 O2 = CO2 CO + H2 O = CO2 + H2 HCCH = 2C(s) + H2

14.3 One mole of ethylene and one mole of benzene are fed to a constant-volume batch reactor and heated to 600 K. On the addition of a catalyst, an equilibrium mixture of ethylbenzene, benzene, and ethylene is formed: C6 H6 (g) + C2 H4 (g) −→ ←− C6 H5 C2 H5 (g)

The pressure in the reactor, before addition of the catalyst (i.e., before any reaction has occurred), is 1.013 bar. Calculate the equilibrium conversion and the heat that must be removed to maintain the reactor temperature constant at 600 K.

898 Chapter 14: The Balance Equations for Chemical Reactors, Availability, and Electrochemistry 14.4 How would you determine the entropy change and the enthalpy change of an electrochemical cell reaction? 14.5 One of the purposes of the kidneys is to transfer useful chemicals from the urine to the blood, and toxins from the blood to the urine. In the transport of glucose from the urine to the blood, the kidneys are transporting glucose against a concentration gradient (that is, the direction of transport is from a low concentration to a high concentration). This can occur only because the transport is coupled with a chemical reaction, a process called active transport. If the concentration of glucose initially in the urine is 5 × 10−5 mol/kg and after leaving the kidney is 5 × 10−6 mol/kg, and the concentration of glucose in the blood (which has a much larger volume) is approximately constant at 5 × 10−3 mol/kg, compute the minimum work that must be done (or Gibbs energy supplied) per mole of glucose transported across the kidney. 14.6 There are two beakers, each one liter in total volume. One of the beakers contains copper sulfate at a concentration of 0.0001 M and the other contains a 0.01-M solution of the same salt. Compute the maximum total work that can be obtained at 25◦ C with an electrochemical cell that used these two solutions as electrolytes. 14.7 Estimate the maximum amount of work that can be obtained from the combustion of gasoline, which we will take to be represented by n -octane (C8 H18 ), in an automobile engine. For this calculation, assume that n -octane vapor and a stoichiometric amount of air (21 vol % oxygen, 79 vol % nitrogen) initially at 1 bar and 25◦ C react to completion, and that the exit gas is at 1 bar and 150◦ C. (You may want to compare this result with that of Problem 8.38.) 14.8 Methane is to be burned in air. Determine the adiabatic flame temperature as a function of the methane-to-air ratio at a pressure of 1 bar. 14.9 Estimate the maximum amount of work that can be obtained from the controlled combustion of methane in air at 1 bar as a function of the methane-to-air ratio. Assume that the process is as follows. First the methane is burned adiabatically so that the adiabatic flame temperature is obtained. Next a Carnot cycle is used to extract heat from the combustion products until they are cooled to 25◦ C. Note that the work cannot all be extracted from the combustion gases at the adiabatic flame temperature, but rather is extracted over a range of temperatures starting at the adiabatic flame temperature and ending at 25◦ C. 14.10 Compare your answer above with the maximum amount of work that could be obtained from methane if a nonthermal energy conversion route, such as a fuel cell, was used. 14.11 a. What is the solubility product for silver sulfate, Ag2 SO4 , in water?

14.12

14.13

14.14

14.15 14.16 14.17 14.18 14.19 14.20 14.21 14.22

14.23

14.24 14.25

14.26

14.27

b. An electrolytic silver-producing cell consists of a copper cathode, a silver anode, and a solution that initially contains Ag2 SO4 at its solubility limit and 0.5 M CuSO4 . Compute the minimum electrical potential that must initially be applied to electrolytically deposit silver in this cell. Redo Problem 14.3 with the change that the reactor is to operate at a constant pressure of 1.0 bar (instead of at constant volume) and that the initial pressure is 1.0 bar. Equal amounts of pure nitrogen and pure oxygen, each at 3000 K and 1 bar, are continuously fed into a chemical reactor, and the reactor effluent, consisting of the reaction product nitric oxide and unreacted nitrogen and oxygen, is continually withdrawn. Assuming that the reactor is adiabatic and that the reactor effluent is in chemical equilibrium, determine the temperature and composition of the effluent. An unsecured tank contains 20 kg of n -butane at its vapor pressure at 25◦ C. The cylinder falls over, breaking off the valve and releasing the entire contents of the cylinder, and the resulting vapor cloud comes in contact with an ignition source and explodes. Estimate the energy released. Derive Eq. 14.3-17. R Redo Problem 14.1 using Aspen Plus .  Redo Problem 14.2 using Aspen Plus R . R Redo Problem 14.3 using Aspen Plus .  Redo Problem 14.8 using Aspen Plus R . R . Redo Problem 14.12 using Aspen Plus  Redo Problem 14.13 using Aspen Plus R . Methanol and acetic acid can react in the liquid phase to form methyl acetate and water. Determine the equilibruim compositions of this mixture at 150◦ C and 50 bar using the NRTL model and either the RGibbs R . or REquil model in Aspen Plus Octane can be used as a model for gasoline. If octane is available at ambient conditions and used in an automobile engine, which can be considered to be a continuous flow reactor, what is the maximum shaft work that can be obtained per kilogram of octane if it is used with 20% more air than is needed for complete combustion.. (Hint: Here you need to include the RTlnyi terms to see the impact on the answer. Do these terms turn out to be important?) Repeat the calculation of the previous problem if 100% excess air is used If octane is burned in a closed, constant volume reactor with a stoichiometric amount of oxygen, what is the maximum amount of shaft work that can be obtained? If octane is burned in a closed, constant volume reactor with 20% excess air, what is the maximum amount of shaft work that can be produced? Repeat Illustration 14.5-1 if 50% excess air is used.

Problems 899 14.28 One suggestion that has been made to conserve energy is that all new electrical power generation plants should be cogeneration facilities. In a typical power plant the combustion of coal is used to produce steam that is run through a turbine and the only useful energy that results is electricity. In such cases the pressure at the downstream end of the turbine is kept as low as possible to produce the most work (electricity). This is done by having a condenser after the turbine cooled by (frequently river) water or air. Another alternative is a co-generation power plant in which the temperature of the exiting steam is kept higher so that the steam leaving the turbine can be used for heating purposes (as process steam in a chemical plant or for residential heating in a city). In a co-generation plant the useful energy obtained is the sum of the electrical energy and energy that can be used for heating. Calculate

a. b. -

-

-

the useful energy and the overall energy efficiency obtained from: a standard power generation plant, and a co-generation plant. The following data are available: The heat of combustion of coal is used to produce steam at 900◦ C and 25 bar from water at 1 bar and 25◦ C. This is accomplished with 80% efficiency (that is 20% of the heating value of the coal is lost in the process); The steam turbines are adiabatic and reversible; The condensed saturated steam leaving the standard power plant is at 45◦ C which sets the pressure at the exit of the turbine; and The steam exit pressure in the co-generation plant is 1 bar, and this steam is then used as a heat transfer fluid until it is condensed and its temperature is 40◦ C.

Chapter

15 Some Additional Biochemical Applications of Thermodynamics Though most of the applications of thermodynamics in this textbook have dealt with chemicals and petrochemicals, there have been a few examples dealing with biochemical processes. In this chapter we focus on the use of the principles of thermodynamics, that is, mass, energy, and entropy balances, and the concept of the equilibrium state, to some applications involving biochemical reactions, biochemical processing, and processes occurring in living cells. These cases are somewhat more complicated to deal with than those of traditional chemical processing because of the large number of chemical species that are involved, and because many of the species (cells, proteins, enzymes, etc.) may be incompletely specified. Also, aqueous solutions of electrolytes are generally involved, and pH and ionic strength have significant effects on the thermodynamics of such systems, including the solubility and biological activity of proteins, cells, and other biomaterials, and on the extent of biological reactions. Therefore, we start with a discussion of the acidity of solutions and pH, which you may have encountered in courses in general chemistry and physical chemistry. We then move on to a number of applications of thermodynamics to biological, physiological, and biochemical processes. There are several very good references on the subject of biothermodynamics.1,2

INSTRUCTIONAL OBJECTIVES FOR CHAPTER 15 The goals of this chapter are for the student to: • Be able to calculate the solubility of amino acids, proteins and pharmaceuticals as

a function of pH, ionic strength and temperature (Secs. 15.1 and 15.2)

• Be able to compute the equilibrium state in biochemical reactions including ligand

binding. (Secs. 15.3 and 15.4)

• Be able to compute the equilibrium denaturation (unfolding) of proteins (Sec. 15.5) • Understand coupled biochemical reactions such as the ADP-to-ATP reaction

(Sec. 15.6) 1 E.

J. Cohn and J. T. Edsall, Proteins, Amino Acids and Peptides, Reinhold, New York (1943). T. Edsall and H. Gutfreund, Biothermodynamics: The Study of Biochemical Processes at Equilibrium, John Wiley & Sons, New York (1983). 2 J.

900

15.1 Solubilities of Weak Acids, Weak Bases, and Amino Acids as a Function of pH 901 • Be able to develop a thermodynamic description of fermenters and other biochem-

ical reactors (Sec. 15.7)

• Be able to compute the equilibrium state, osmotic pressure, and membrane poten-

tials of proteins and other charged species (Gibbs-Donnan equilibrium) (Sec. 15.8) • Understand how proteins can be concentrated in an ultracentrifuge (Sec. 15.9)

NOTATION INTRODUCED IN THIS CHAPTER H O N C K K So ST

YB/S YP/S YN/S YO2 /S YW/S YC/S YQ/S Z θ ω ξ

Stoichiometric coefficient of hydrogen in a biochemical compound Stoichiometric coefficient of oxygen in a biochemical compound Stoichiometric coefficient of nitrogen in a biochemical compound Stoichiometric coefficient of carbon in a biochemical compound An apparent chemical equilibrium constant based on concentration ratios A product of equilibrium constants Saturation solubility of an electrically neutral substance (M) Total solubility of the neutral and ionized forms of a substance (M) C-moles of biomass produced per C-mole of substrate consumed C-moles of product per C-mole of substrate consumed Moles of nitrogen source consumed per C-mole of substrate consumed Moles of O2 consumed per C-mole of substrate consumed Moles of water consumed per C-mole of substrate consumed Moles of CO2 produced per C-mole of substrate consumed Heat flow per C-mole of substrate consumed Charge on a protein Extent of coverage in ligand binding Rotational speed in an ultracentrifuge (1/s) Generalized degree of reduction

15.1 SOLUBILITIES OF WEAK ACIDS, WEAK BASES, AND AMINO ACIDS AS A FUNCTION OF pH Here we consider some acids, bases, proteins, pharmaceuticals, or other compounds that ionize, and in which the undissociated species is only partially soluble in aqueous solution while the portion that ionizes is completely soluble. The precipitation of proteins and other pharmaceuticals from solution is important for several reasons. For example, the precipitation of a drug from a processing broth can be useful step during purification. Especially if a crystal can be obtained, the product will be of high purity. Conversely, the precipitation of a drug from an aqueous liquid formulation solution can indicate instability of the product, poor bioavailability, and difficulties in delivery. Also, it is of interest to know how far removed a drug product is from its solubility limit since stresses (thermal, shear, pressure, etc.) during processing may result in undesired precipitation. It is in these areas that classical thermodynamics, and particular the Gibbs energy of the phases, can provide some useful information. The solubility of proteins and amino acids in solution can be considered from a number of perspectives. One is the solubility of the various ionic forms of these compounds as a function pH, as a result of the ionic form at fixed ionic strength and temperature. A second is the solubility as a function ionic strength at fixed temperature and pH.

902 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Still a third is the solubility as a function of temperature at fixed ionic strength and pH. Each of these will be considered separately, though they also can be combined into a model of broader applicability. Still another issue is the character of the solid precipitate that forms when the solubility limit is exceeded, which might be a crystal (in one or more of its polymorphic forms), an amorphous solid (or floc), a glass, or a gel. The accurate prediction of the character of the solid that will form is beyond present capabilities because of the complexity of the molecules involved, that the different possible phases may have very small differences in Gibbs energy beyond the accuracy of present prediction methods, and that as crystallization is a slow process, the type of solid formed may be determined by kinetics and not thermodynamics. Nonetheless, a correlation between a measurable thermodynamic property, the osmotic second virial coefficient, and the conditions for crystallization is discussed here. The charge on an amino acid or protein is important in determining its solubility. Consider a compound in solution in equilibrium with its pure solid. Generally, the solubility in water of such compounds, because of their chemical composition and hydrophobic character, is limited when in the neutral form and higher when in any ionized form. It is this change in solubility that we want to model. We first consider some acids, bases, or other compouns that ionize, and in which the undissociated species is only partially soluble in aqueous solution while the portion that inoizes is completely soluble. We start with an analysis of the equations for the ionization and dissolution of an acid. We consider KHA

HA  H+ + A− for which the equilibrium constant is

KHA =

MH+ · MA− MHA

where, as usual, the apparent equilibrium constant has been used and solution nonidealities have been neglected. Let So be the solubility of the uncharged acidic species, which could be a pharmaceutical drug. This is the value of the solubility at such a low pH that ionization is suppressed. As long as there is sufficient undissolved acid for the solution to remain saturated, we have So = MHA (15.1-1) We will use ST to represent total solubility of the acid, that is, the concentration of both undissociated and ionized acid:

ST = MHA + MA− = So + MA− or

MA− = ST − So

Now

KHA =

MH+ · MA− M + · (ST − So ) = H MHA So

(15.1-2)

15.1 Solubilities of Weak Acids, Weak Bases, and Amino Acids as a Function of pH 903



or

pKHA

(ST − So ) = pH − log So



Then the total acid solubility at any pH, ST (pH), is   ST (pH) = So · 1 + 10−[pKHA −pH] or

ST (pH) = So + Clearly, if

MH+  KHA

while if

So · KHA MH+

(15.1-3)

then ST = So

KHA  MH+

there is a significant increase in solubility. In particular, at pH = 7 (pure water), we have   ST (pH = 7) = So · 1 + 10−[pKHA −7] (15.1-4a)

Solubility of a weak acid

so that the total solubility of a compound at any pH is related to its solubility in pure water (which is the solubility that is usually reported) by   1 + 10−[pKHA −pH] ST (pH) = ST (pH = 7) · (15.1-4b) (1 + 10−[pKHA −7] )

Illustration 15.1-1 The Solubility of a Weak Acid as a Function of pH Hexanoic acid (C6 H12 O2 ), in biology usually referred to as caproic acid, has a solubility in water of 9.67 g/kg at pH = 7 and 25◦ C, and its pKHA is 4.85. Estimate the solubility of hexanoic acid in the following body fluids: Blood Saliva Stomach contents Urine

pH = 7.4 pH = 6.4 pH = 1.0 to 3.0 pH = 5.8

Solution Using Eq. 15.1-4b, the solubility of hexanoic acid in blood is found to be 24.2 g/kg, in saliva it is 2.48 g/kg, in stomach contents the solubility is 0.068 g/kg, and in urine the solubility of hexanoic acid is 0.68 g/kg. The observation from this calculation is that there is very different solubility of an organic chemical (or pharmaceutical) in different bodily fluids. This can be especially important in the formulation of a drug, where one wants to ensure its solubility and bioavailability. The complete solubility versus pH curve is shown in the figure below. Note that the solubility of hexanoic acid begins to increase when the value of the pH is close to the pKHA value. This is a general result for acids, since, as we can see from Eqs. 15.1-4, there will be a significant

904 Chapter 15: Some Additional Biochemical Applications of Thermodynamics change in the slope of the solubility curve of a partially soluble acid at a pH that is close to its pKHA value. 100

ST, g/kg

10 1 0.1

0

2

4

6

8

10

pH

Solubility of hexanoic acid (g/kg) in aqueous solutions as a function of pH.

Next we consider the solubility of a partially soluble weak base. The extent of ionization of the weak base KBOH BOH  B+ + OH− can be computed from the equilibrium relation

KBOH =

MB+ · MOH− MBOH

Now, using the notation

So = MBOH

ST = MBOH + MB+ = So + MB+

we obtain

KBOH = and

Solubility of a weak base

and MOH− =

KW MH+ (15.1-5)

(ST − So ) · KW /MH+ So

    KBOH · MH+ = So · 1 + 10−[pKBOH +pH−pKW ] ST = So · 1 + KW

or, using pH = 7 (pure water) as the reference,   1 + 10−[pKBOH +pH−pKW ] ST (pH) = ST (pH = 7) · (1 + 10−[pKBOH +7−pKW ] )

(15.1-6)

(15.1-7)

In the biological literature, equilibrium constants are sometimes written for the reaction ∗ KBOH

BOH + H+  B+ + H2 O

which can be considered the combination of KBOH

BOH  B+ + OH−

(15.1-8)

15.1 Solubilities of Weak Acids, Weak Bases, and Amino Acids as a Function of pH 905 and 1/KW

H+ + OH−  H2 O Summing these last two reactions gives the reaction of Eq. 15.1-8. Also,

KBOH =

MB+ · MOH− MBOH

1 MH2 O = KW MH+ · MOH−

and

so that ∗ KBOH =

KBOH KW

∗ or pKBOH = pKBOH − pKW

Therefore,

  ∗ ST (pH = 7) = So · 1 + 10−[pKBOH +pH]

(15.1-9a)

so that an alternative expression is

  ∗ 1 + 10−[pKBOH +pH]  ST (pH) = ST (pH = 7) ·  ∗ 1 + 10−[pKBOH +7]

(15.1-9b)

Illustration 15.1-2 The Solubility of a Weak Base as a Function of pH The pyrimidine base thymine (C5 H6 N2 O2 , also known as 5-methyluracil) has a solubility in water of 4.0 g/kg (pH = 7) at 25◦ C and its pKBOH is 9.9 (there is a further dissociation at higher pH that we will neglect here). Estimate the solubility of thymine in aqueous solutions over the pH range of 1 to 13.

Solution The results in the following figure calculated with Eq. 15.1-7 show a significant decrease of thymine solubility with increasing pH until about pH = 4, beyond which the solubility remains approximately constant at 4.0 g/kg. 1.103

100

10

1 0

2

4

6

8

10

12

14

pH

Thymine solubility (g/kg) in aqueous solutions as a function of pH

906 Chapter 15: Some Additional Biochemical Applications of Thermodynamics In this figure there is a marked change in the slope of the solubility curve of a base when the pOH = (pKW − pH) = (14 − pH) is equal to the pKBOH value. (This is seen here in the change in slope at about pH = 4, corresponding to pOH = 10, which is close to the pKBOH value of 9.9.)

Note As an aside, it is of interest to note that deoxyribonucleic acid, or DNA, which has been called the molecule of heredity, consists of a repeated backbone with a collection of side chain groups, the sequence of which is different in different species (and in individuals within the same species) and contains the genetic information. The side chains on the DNA backbone consist of only four types of bases, the two purines adenine (A) and guanine (G), and the two pyrimidines thymine (T) and cytosine (C). What is commonly referred to as the genetic code is a long sequence of the four letters A, G, T, and C in the order in which these bases appear on the DNA chain.

Some slightly soluble amino acids and other weak acids and bases have more than a single ionization state. For example, at low pH the simple amino acid glycine is positively charged; it is uncharged around pH = 7 and has a negative charge at higher pH, as shown in Fig. 2 of Illustration 13.6-2. Using the A (positively charged), B (neutral), and C (negatively charged) species identifications of that illustration, we have

K1 =

MB · MH+ MA

K2 =

MC · MH+ MB

and

Letting So represent the equilibrium saturation solubility of the uncharged amino acid, that is, using So = MB , we have

K1 =

So · MH+ MA

and K2 =

MC · MH+ So

Therefore, the total solubility

ST = So + MA + MC  MH+ K2 K2 MH+ = So + So + So = So 1 + + K1 MH+ K1 MH+ is and

 ST = So · 1 + 10(pK1 −pH) + 10−(pH−pK2 ) ST = 1 + 10(pK1 −pH) + 10−(pH−pK2 ) So

(15.1-10a) (15.1-10b)

Note that while So is the solubility of the uncharged species, it is usually the solubility of the amino acid in water pH = 7 that is reported, and that may be somewhat different from the solubility of the unchanged amino acid, unless its isoelectric point is pH = 7. At other pH values the total solubility will be the result of the solubility of the uncharged and charged species. The total solubility at pH = 7 is   ST (pH = 7) = So 1 + 10(pK1 −7) + 10−(7−pK2 ) (15.1-11a)

15.1 Solubilities of Weak Acids, Weak Bases, and Amino Acids as a Function of pH 907 Now eliminating the unknown solubility of the uncharged amino acid gives the final result for the total solubility at any pH value in terms of the measured solubility in neutral water at pH = 7:

  1 + 10(pK1 −pH) + 10−(pH−pK2 ) ST (pH) = ST (pH = 7) · (1 + 10(pK1 −7) + 10−(7−pK2 ) )

(15.1-11b)

The discussion that led to this equation is easily extended to amino acids, proteins, and other biomolecules with more than two ionizable sites (Problems 15.1 and 15.4). Illustration 15.1-3 The Solubility of a Weak Amino Acid as a Function of pH Tyrosine (C9 H11 NO3 ), another amino acid in proteins, has two dominant ionizable groups with pKHA values of 2.24 and 9.04 at 25◦ C (and a third, less easily ionizable group with a pK value of 10.10, which we will neglect here). Its water solubility at this temperature is 0.46 g/kg of water. a. Estimate the total solubility of tyrosine in aqueous solutions over the pH range of 0.4 to 11. b. Estimate the solubility of the uncharged tyrosine.

Solution a. Using Eq. 15.1-11b, we obtain the results shown in the following figure. Note that the solubility of tyrosine varies little (only between 0.46 and 0.56 g/kg of water) over the pH range of 2.9 of 8.4, and then increases rapidly as the pH either decreases below 2.9 or increases above 8.9. The pH range of limited solubility is also where tyrosine has approximately zero average net charge (Problem 15.20). 100

S

10

1

0.1

5

10 pH

The solubility S of tyrosine (g/kg) in aqueous solutions as a function of pH. b. Using Eq. 15.1-11a, we have ST (pH = 7) = 0.46

g = So × 1.009 kg

and

So = 0.456

g kg

908 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Understanding the solubility of weak acids and bases as a function of pH is important in the formulation of pharmaceuticals for general use. From Illustration 15.1-1 we see that if we had to deliver a drug with low solubility in water that had a pK value similar to hexanoic acid, it would have to be delivered in a slightly basic solution to have appreciable solubility or concentration in the aqueous solution. For example, at pH = 7.4 (the pH of blood), its solubility would be 2.5 times greater than in water, while its solubility would be considerably less in the fluids in the stomach, and may indeed precipitate out of solution there. However, for a drug with a pK similar to that of thymine (9.9), the solubility decreases with increasing pH at low pH, and is essentially unchanged from its neutral water solubility by increasing the pH to that of blood. Understanding the solubility of proteins and other biochemicals is also important in developing methods for their purification. For example, an important method for the purification of antibiotics is by liquid-liquid extraction. The procedure is as follows. The pH of the solution is first adjusted to the isoelectric point at which the desired biochemical compound has no net charge and limited solubility, while the other biochemicals in the mixture are charged and are more soluble. This solution is then contacted with an organic solvent, which results in the extraction of the desired biochemical compound into the organic phase. Then by contacting this organic solution with an aqueous phase at a pH at which desired biochemical has a charge, it is extracted back into an aqueous phase. This method is used to separate components with different isoelectric points. Also, the differing solubilities of species as a function of pH can be used to recover one biochemical from a mixture by selective precipitation (or crystallization). As a final example in this section we consider the solubility in water of the amino acid dl-alanine HCOO-CNH2 H-CH3 it needs to be recognized that it exists in the following four forms in solution K1

+ − + NH+ 3 CH2 COOH  H + NH3 CH2 COO K2

− − + NH+ 3 CH2 COO  H + NH2 CH2 COO KD

NH2 CH2 COOH 

− NH+ 3 CH2 COO

pK1 = 2.348 pK2 = 9.867 pKD = −5.41

The equilibrium relations are



 − [H+ ] · NH+ [H+ ] · NH3 CH2 COO− 3 CH2 COO

+

+  ;  ; K2 = K1 = NH3 CH2 COOH NH3 CH2 COO−

+  NH3 CH2 COO− KD = [NH2 CH2 COOH]

(15.1-12)

Therefore

+  NH3 CH2 COO− = KD · [NH2 CH2 COOH] = [NH2 CH2 COOH] · 10−pKD −

+  [H+ ]·[NH+ [H+ ]·[NH2 CH2 COOH]·10−pKD 3 CH2 COO ] NH3 CH2 COOH = = K1 K1 = [NH2 CH2 COOH] · 10pK1 −pH−pKD −

 K2 ·[NH+ 3 CH2 COO ] = [NH2 CH2 COOH] · 10pH−pK2 −pKD NH2 CH2 COO− = + H [ ] (15.1-13)

15.1 Solubilities of Weak Acids, Weak Bases, and Amino Acids as a Function of pH 909 These relations can now be used to determine the fraction of the alanine in each of the possible protonation states as follows (assuming the activity coefficients are unity)

fNH2 CH2 COOH = =

[NH2 CH2 COOH] + − − [NH2 CH2 COOH]+[NH+ 3 CH2 COO ]+[NH3 CH2 COOH]+[NH2 CH2 COO ]

1 1+10−pKD +10pK1 −pH−pKD +10pH−pK2 −pKD

fNH+ CH2 COO− =

10−pKD 1+10−pKD +10pK1 −pH−pKD +10pH−pK2 −pKD

fNH+ CH2 COOH =

10pK1 −pH−pKD 1+10−pKD +10pK1 −pH−pKD +10pH−pK2 −pKD

3

3

fNH2 CH2 COO− =

;

10pH−pK2 −pKD 1+10−pKD +10pK1 −pH−pKD +10pH−pK2 −pKD

(15.1-14) Using the values of the equilibrium constants given above, we obtain the following distribution of dl-alanine among its possible states 1 Alanine

f 0.5

0 0

5

10

15

pH

Figure 15.1-1 Fraction of alanine in the zwiterion (solid line), cation (dotted line) and anion (dashed line) states as a funciton of pH.

We see from this Fig. 15.1-1 that the NH+ 3 CH2 COOH form is dominant in very acidic (low pH) solutions, the NH2 CH2 COO− form is dominant in very basic solu− tions, and in the mid pH range the neutral dipolar form NH+ 3 CH2 COO dominates with virtually none of the NH2 CH2 COOH form present. [Note that since the pK values for dl-alanine are similar to that for glycine (Sec. 13.6), a plot of the charge on dl-alanine as a function of pH would look very similar to Fig. 13.6-6.] At neutral pH, the solubility of dl-alanine has been reported to be 1.893 molal. This information and the pK values can be used to compute the solubility of dl-alanine at other pH values as follows. Based on the analysis above, using the assumption that the amount of NH2 CH2 COOH present can be neglected, we have −

+   [H+ ]·[NH+

− 3 CH2 COO ] NH3 CH2 COOH = = NH+ · 10pK1 −pH 3 CH2 COO K1

 NH2 CH2 COO− =

− K2 ·[NH+ 3 CH2 COO ]

H+

 − = NH+ · 10pH−pK2 3 CH2 COO (15.1-15)

910 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Therefore, the total solubility of dl-alanine, S, which is the sum of the solubilities in each of the protonation states, is

 



− S = NH+ + NH2 CH2 COO− + NH+ 3 CH2 COO 3 CH2 COOH

  − = NH+ (15.1-16) 1 + 10pH−pK2 + 10pK1 −pH 3 CH2 COO   pH−pK2 pK1 −pH + 10 = 1.893 1 + 10 The predictions based on this equation together with the experimental data of Pradhan and Vera3 are shown in the following figure. 3.5

3

Alanine solubility

S 2.5

2

1.5 2

4

6

8

pH

10

Figure 15.1-2 Alanine solubility as a function of pH.

While the agreement is not perfect, the trends and magnitudes are correctly predicted. One way to improve the predictions, though it will not be done here, is to include the neglected the activity coefficients, especially the mean ionic activity coefficients of the − ionized alanine species NH+ 3 CH2 COOH and NH2 CH2 COO . As Pradhan and Vera used appreciable concentrations of KOH and HNO3 to achieve the far-from-neutral pH values of their measurements, the solution nonideality effect on the equilibrium calculations should be included. Also, as a result of specific ion effects discussed next, slightly different solubilities were obtained when instead NaOH and HCl were used to adjust the pH of the solution. As a general conclusion, we have that the solubility of an amino acid, and by extension of a protein, is a minimum near its isoelectric point. Information on how the solubility of proteins and amino acids vary with pH can be used to design a selective precipitation separation processes. For example, albumin could be recovered as a precipitate from a reasonably concentrated albumin-lysozyme mixture by lowering the pH of the solution to its pI of about 4.8, while lysozyme would be recovered as a precipitate by raising the solution pH to its pI of 11.2.

3 A.

A. Pradhan and J. H. Vera, Fluid Phase Eq. 152, 121-132 (1988)

15.2 The Solubility of Amino Acids and Proteins as a funciton of Ionic Strength and Temperature 911

15.2 THE SOLUBILITY OF AMINO ACIDS AND PROTEINS AS A FUNCITON OF IONIC STRENGTH AND TEMPERATURE (a) Solubility as a Function of Ionic Strength and Salt Type (The Salting In and Salting out of Amino Acids and Proteins) The addition of a salt, amino acid, or other compounds can also change the solubility protein due to nonideal solution behavior, that is, the effect on the activity coefficients. To start, consider the equilibrium between the solid protein and the protein in a nonsalt containing solution, and separately the equilibrium between the same solid protein and the protein in a salt- containing solution. The equilibrium relations for these two cases are

μsolid = μoP (ideal 1 M solution) + RT ln (MP,ns γns ) and P solid μP = μoP (ideal 1 M solution) + RT ln (MP,s γs )

(15.2-1)

where the subscripts P, ns and s indicate the protein, the no-salt and the salt-containing solutions, respectively. From the equilibrium relation, we have     MP,s = ln γγnss and ln (MP,ns γns ) = ln (MP,s γs ) or ln MP,ns (15.2-2) γns MP,s = MP,ns γs There are a number of contributions to the activity coefficient of a protein in solution, a entropic contribution due to size and shape differences between the solute and the solvent, weak dispersion forces, and the strong electrostatic contributions as described, for example, by the Debye-Hückel or the extended Debye-Hückel models. Of these, the latter generally dominate except near the isoelectric point, so the former will be ignored here and then added later. In this case γ will be replaced by γ± . Further, since the ionic strength in the salt-free solution will be quite small, we can take γ±,ns ≈ 1, so that  MP,ns MP,s = − ln γ±,s MP,s = or ln (15.2-3) γ±,s MP,ns There are two factors that influence the value of the mean ionic activity coefficient of a protein upon the addition of a salt. One is the increase in ionic strength as usually accounted for by an extended Debye-Hückel models, which I will denote as γ± |DH , and is represented by √ B |z+ z− | I √ ln γ± |DH = − (15.2-4) 1+A I where A is a parameter dependent of the size of the protein, and the value of the parameter B depends on the solvent and its dielectric constant; its value is 1.178 for pure water at 25o C. The second effect is that the addition of a salt also changes the value of the dielectric constant D of the solution. The latter effect on the activity coefficient, which will be denoted by γ± |D , is obtained by starting with the relation4 4 “Proteins, Amino Acids and Peptides as Ions and Dipolar Ions”, by E. J. Cohn and J. T. Edsall, American Chemical

Society Monograph Series, Reinhold Pub. Co., New York, 1943.

912 Chapter 15: Some Additional Biochemical Applications of Thermodynamics

D = Do + δs Ms so that order  to first  δs Ms 1 1 1 = ≈ Do 1 − Do = D1o − D Do (1+ δsDMs ) o

δs Ms Do2

(15.2-5)

where δ s is the dielectric increment of the added salt or solute S and M is its molality (or some measure of concentration consistent with the value of δ s ). The dielectric increment, that is, the amount by which the dielectric constant of the solution changes per unit of concentration change of the added solute, may be positive (for example, for some amino acids) or negative (as is the case for many inorganic salts). The reason for writing the equation in this form is that the Born equation of electrostatics shows that chemical potential (or the logarithm of the activity coefficient) of a charged particle depends on the reciprocal of the dielectric constant of the surrounding medium. Without going through the details of that equation, at fixed ionic strength,  δs Ms 1 1 = −α 2 = −β  δs Ms ln γ±,s |D = α − (15.2-6) D Do Do

or equivalently

ln γ±,s |D = −βδs Is

(15.2-7)



where α, β and β are combinations of parameters from electrostatic theory. Note that if this were the only contribution to the activity coefficient we would have

ln

MP,s  = ln γ±,s |D = ks Ms = ks Is MP,ns

(15.2-8)

which provides the theoretical basis for the empirical Cohn equation5 with kS and  kS being empirical salt (or other solute) and protein-dependent parameters. [The inter relationship between ks and ks is a numerical constant that depends on the charge on the  cation and anion, and the concentration units used. Note that ks is generally referred to as the Setschenow constant. ] If ks (or δ s ) is positive, the saturation concentration of the protein increases with increasing solute concentration; the protein is said to salt in. If ks (or δ s ) is negative, the protein salts out, that is its saturation concentration decreases with increasing solute concentration. As an example, the value of δ in water5 is −1.4 Debye/mol for methanol, −3.2 for acetone, and −7.5 for sucrose, all of which will initially reduce the protein solubility upon their addition, while its value is +2.7 for urea and +22.6 for glycine, and their addition will increase protein solubility. An example of this is that the solubility of l-asparagine in water is 0.184 moles/liter, while it is 0.211 in an aqueous solution that also contains 1.0 moles/liter glycine, and 0.247 in a solution containing 2.80 moles/liter of glycine. The overall effect of the addition of a salt is the product of the change in dielectric constant, and the change the ionic strength with salt addition, that is

γ±,s = γ±,s |DH × γ±,s |D or lnγ±,s = ln γ±,s |DH + ln γ±,s |D √ 1.178 |z+ z− | I √ so that lnγ±,s = − − βδs Is 1+A I

(15.2-9)

where z+ and z− are the charges on the protein and its counter-ion. Therefore, the activity coefficients of the salt and of the protein or amino acid are the result of two contributions of differing dependence on ionic strength, and sometimes of different sign. In these cases the concentration range over which the simple Cohn equation is valid is significantly reduced. Indeed, it is not unusual to see the solubility of a biomolecule first increase and then decrease with increasing salt concentration (or more correctly,

15.2 The Solubility of Amino Acids and Proteins as a funciton of Ionic Strength and Temperature 913 ionic strength). This can be accounted for by combining the contributions to the activity coefficient from the Debye-Hückel term and the change in dielectric constant leading to √   1.178 |z+ z− | Is √ ln MP,s = ln MP,ns − ln γ±,s = ln MP,ns + + ks Is 1 + A Is (15.2-10) or √   1.178 |z+ z− | Is √ MP,s = MP,ns exp + ks Is (15.2-11) 1 + A Is where I s is the ionic strength of the salt. Empirically, it is known that different salts affect the salting out or salting in of proteins and amino acids differently. Also, anions of the salt have a greater effect than the cations, and the strength of the effect of different anions generally (but not always) follows the Hofmeister series5 − − F− ∼ SO24− > HPO24− > acetate > Cl− > NO− 3 > Br > ClO3 − − > I− > ClO4 > SCN and for cations + + 2+ + > Ca2+ > guanidium NH+ 4 > K > Na > Li > Mg Thus, ammonium sulfate can be expected to have a greater effect on protein solubility than sodium chloride, and this is what is generally observed in that on a molality basis ammonium sulfate is found to be a better precipitant than sodium chloride. The ordering by Hofmeister6 in 1888 was based of observations of the denaturation (unfolding) of proteins, a topic to be considered later. As an example of the applicability of Eq. 15.2-11 for use in correlating the solubility of proteins, including salting in and salting out effects, consider the data of Green7 solubility of carboxyhemoglobin (in units of g/1000 g water) at 25o C as a function of the concentration of three different salts at pH about 6.7 (remember, the charge on a protein can change significantly with pH as shown previously). The experimental solubility data of carboxyhemoglobin in solutions containing NaCl (diamonds), Na2 SO4 (squares) and MgSO4 (squares) and the respective correlations using Eq. 15.2-10 are shown in Fig. 15.2-1.

Chb

100

10

1 0

1

2 sqrt I

5 List

from Wikipedia.

6 Hofmeister 7 A.

A. Green, J. Biol. Chem. XCV, 47-66 (1932).

3

Figure 15.2-1 The solubility of carboxyhemoglobin (g/kg water) as a function of the ionic strength of NaCl (diamonds), Na2 SO4 (squares) and MgSO4 (squares).

914 Chapter 15: Some Additional Biochemical Applications of Thermodynamics There we see that sodium chloride increases the solubility (decreases the activity coefficient) of the carboxyhemoglobin at all concentrations studied, and therefore does not precipitate, that is salt out, this protein. Magnesium sulfate does salt out this protein as does sodium sulfate. Of the two, sodium sulfate is more effective in that it results in a lower carboxyhemoglobin solubility (greater salting out) at the same ionic strength. (Note however, that I s = 2.5 M in a Na2 SO4 solution and 4 M in a MgSO4 solution. So that I s = 4 M is achieved in a 1 M MgSO4 solution and 1.6 M Na2 SO4 solution.) We also see in Fig. 15.2-1 that the results for the solubility of carboxyhemoglobin are in accord with the Hofmeister series in that the sulfate anion is more effective than the chloride anion, indeed, there is no salting out with NaCl, but there is with both sulfate salts. Further, also in agreement with the Hofmeister series, the sodium cation is more effective than magnesium as seen by Na2 SO4 being a more effective salting out agent than MgSO4 . Based on verifications such as this, an initial choice of possible salting out agents is frequently made by Hofmeister series considerations. (b) Solubility as a Function of Temperature For proteins and other large molecules, the usual progression with increasing temperature is from a crystal to a glass (a less well ordered solid that can be considered an amorphous solid) and then to a liquid melt. Since the amorphous solid is the most likely pre-melting phase for proteins, we will only consider the amorphous-to-solution transition here. In particular, the conditions for the formation of an amorphous (glassy) phase of a protein or drug component from solution as a function of temperature and composition (at fixed pH and ionic strength) are considered. The starting point for this phase equilibrium calculation is again the equality of chemical potentials (or equivalently partial molar Gibbs energies or fugacities) of the precipitating species in the solid (denoted by s) and solution or liquid (denoted by l) phases, that is

μsi (T, P ) = μli (T, P, xi ) or Δs→l G (T, P, xi ) = μli (T, P, xi ) − μsi (T, P ) = 0 (15.2-12) Since we will not be concerned with pressure variations here, it will be eliminated from the equations. Also, we will only be interested in the Gibbs energy change of the precipitating component at incipient phase separation, so the compositions of the other components are not changed. Therefore, the Gibbs energy change of the solution at incipient phase separation is only that of the precipitating component. Thus Δs→l G (T, P, xi ) = μli (T, P, xi ) − μsi (T, P ) = μli (T, P ) + RT ln xi γi − μsi (T, P ) = Δs→l G (T, P ) + RT ln xi γi = Δs→l G (T, P ) + RT ln ai = 0 (15.2-13) based on the pure component standard state, and ai is the activity of precipitating species. Since the Gibbs energy change is from the pure solid to the pure molten liquid, this term will be written as Δs→l G (T, P ) in what follows. Using Δs→l H and Δs→l S to represent the enthalpy and entropy changes on protein melting, Δs→l CP to represent the difference in heat capacities between the solid and liquid states, and T m to be the normal melting temperature gives T Δs→l H (T ) = Δs→l H (Tm ) + Tm Δs→l CP (T ) dT ≈ Δs→l H (Tm ) +Δs→l CP (T − Tm )

15.2 The Solubility of Amino Acids and Proteins as a funciton of Ionic Strength and Temperature 915

and   ≈ Δs→l S (Tm ) + Δs→l CP ln TTm (15.2-14) This equation is derived in standard thermodynamics textbooks, and contains the assumption of constant the heat capacities of the solid and liquid above and below the melting point, respectively. At the T m the Gibbs energies of the solid and (molten) liquid states are equal at this temperature, so that Δs→l S (T ) = Δs→l S (Tm ) +

T

Δs→l CP (T ) dT T Tm

Δs→l G (Tm ) = Δs→l H (Tm ) − Tm Δs→l S (Tm ) = 0 and Δs→l S (Tm ) =

Δs→l H(Tm ) Tm

(15.2-15)

and

Δs→l G (T ) = Δs→l H (T ) − T Δs→l S (T )

T = Δs→l H (Tm ) + Δs→l CP (T − Tm ) − Δs→l H (Tm ) Tm  T −T Δs→l CP ln T   m  T T + Δs→l CP T − Tm − T ln = Δs→l H (Tm ) 1 − Tm Tm (15.2-16)

Note that all the parameters that appear in this equation, the enthalpy change, heat capacity change, and the melting temperature are readily measurable by calorimetry, and especially differential scanning calorimetry (DSC), as long as the substance does not decompose on melting. These parameters have been reported for a number of proteins. The second is that the equation is nonlinear in temperature. For proteins and other large molecules, the usual progression with increasing temperature is from a crystal to a glass (a less well ordered solid that can be considered an amorphous solid) and then to a liquid melt. Since the amorphous solid is the most likely phase pre-melting phase for proteins, we will only consider the amorphous to solution transition here. Fig. 15.2-2 below8 are the DSC scans for two drug products (identified only as Material A and Material B for intellectual property reasons). The authors of that reference used these scans to deduce the following information (I have eliminated the information their error estimates.) Table I Property Material A Tg (K) 393.4 416 Tm (K) Δs→l H (kJ mol−1 ) 16 CP,c ( kJ mol−1 K−1 ) crystal 0.6 at 373 K CP,a ( kJ mol−1 K−1 ) amorphous 1.0 at 423 K CP,l ( kJ mol−1 K−1 ) liquid 0.7 at 373 K 8 From

Material B 361 483.6 76 1.38 at 448 K 1.9 at 373 K 1.4 at 348 K

D. S. Hsieh, B. A. Sarsfield, M. Davidovich, L. M. Dimemmo, S.-Y. Chang and S. Kiang, J. Pharm. Sci. 99, 4096-4105 (2010).

916 Chapter 15: Some Additional Biochemical Applications of Thermodynamics 3.5

15 14 13

3.0

12 11

2.5 Cp/kJ*mol–1*K–1

Cp/kJ*mol–1*K–1

10 2.0

1.5

9 8 7 6 5

1.0

4 3

0.5

2 Tg

0.0 250

300

1

Tm 450

350 400 T (K)

Tg

0 500

300

Tm

350

500

400 450 T (K)

550

Figure 15.2-2 Heat capacities of Material A (left) and Material B (right) determined by differential scanning calorimetry with the glass transition Tg and melting Tm temperatures indicated.

Using these data, Fig. 15.2-3 for Material A shows the temperature of incipient precipitation to an amorphous solid, TA, as a function of its activity aA (which is equal to the mole fraction in an ideal solution). What is seen is the typical freezing point lowering upon dilution of a component in the liquid solution, which starts at the pure component melting point, and results in decreasing transition temperature as the substance is diluted. From this figure one can conclude that it may be difficult to process this substance in solution at mole fractions (activities) of less than 0.73 because the precipitation temperature drops sharply with further dilution. The situation is quite different for Material B, largely because of its much higher heat of fusion. From Fig. 15.2-3 (right, note scale changes), it is seen that this substance can be processed over a very large range of liquid compositions without precipitation. In fact, for mole fractions (more strictly, activities) above 0.1, precipitation is unlikely as long as the temperature is above 400 K. 420

500

400

450

TA (K)

TB (K) 380

360 0.7

400

0.8

0.9 aA

1

350 0

0.2

0.4

0.6

0.8

1

aB

Figure 15.2-3 Amorphous precipitation temperature as a function of concentration in solution for material A (left) and material B (right).

15.3 Binding of a Ligand to a Substrate 917

15.3 BINDING OF A LIGAND TO A SUBSTRATE Another type of chemical reaction that occurs, especially among biological molecules, is the reversible binding of a smaller molecule (referred to as a ligand) to a binding site on a protein or other large molecule (referred to as a substrate). We consider first the simplest case of a substrate that has only a single binding site. The reaction is ligand (L) plus protein (P) to form a ligated protein (PL): K

L + P  PL

(15.3-1)

for which the apparent (or measured) concentration-based equilibrium constant is

K=

MPL ML · MP

(15.3-2)

Using this apparent equilibrium constant, we have MPL = K · ML · MP . Further, data are frequently reported as the coverage θ, which is the ratio of concentration of the ligated protein MPL to the total concentration of protein in solution MP + MPL . That is,

θ=

Fractional ligand coverage

MPL K · ML · MP K · ML = = MP + MPL MP + K · ML · MP 1 + K · ML

(15.3-3)

With increasing ligand concentration, the coverage initially increases linearly (though the linear range may be small if the value of K is large), and at high ligand concentration θ saturates (for this case of a single binding site) to unity. This form of the coverage versus concentration relation was originally derived for molecules adsorbing on a solid surface, and is referred to as the Langmuir isotherm (isotherm designating a process occurring at constant temperature), after Irving Langmuir, who developed the equation to describe the adsorption of gases on metal surfaces. Consider next a protein molecule with two different binding sites. We will designate proteins with a single bound ligand as PL and LP, depending on which of the two sites binds the ligand, and use LPL to represent the protein with the two ligands. The binding reactions are K1

L + P  LP K2

L + P  PL K3

L + LP  LPL and

K4

L + PL  LPL

Reaction 1 Reaction 2 Reaction 3 Reaction 4

with

K1 =

MLP ML · MP

K2 =

MPL ML · MP

K3 =

MLPL ML · MLP

and K4 =

MLPL ML · MPL

The coverage θ, defined as the moles of bound ligand per mole of protein in any form, is

θ=

MLP + MPL + 2 · MLPL moles of bound ligand = moles of protein MP + MLP + MPL + MLPL

(15.3-4)

918 Chapter 15: Some Additional Biochemical Applications of Thermodynamics where the factor of 2 in the numerator arises because two ligands are attached in each LPL combination resulting from reactions 3 and 4. In this case the saturation coverage θ will be 2 when each protein binds two ligands. Now using the equilibrium constant relations, we obtain

θ=

K1 · ML · MP + K2 · ML · MP + K3 · ML · MLP + K4 · ML · MPL MP + K1 · ML · MP + K2 · ML · MP + 12 (K3 · ML · MLP + K4 · ML · MPL )

=

K1 · ML · MP + K2 · ML · MP + K3 · K1 · ML2 · MP + K4 · K1 · ML2 · MP MP + K1 · ML · MP + K2 · ML · MP + 12 (K3 · K1 · ML2 · MP + K4 · K1 · ML2 · MP )

=

K1 · ML + K2 · ML + K3 · K1 · ML2 + K4 · K1 · ML2 1 + K1 · ML + K2 · ML + 12 (K3 · K1 · ML2 + K4 · K1 · ML2 )

(15.3-5)

However, also note that from the equilibrium relations,

MLPL = K3 · ML · MLP = K3 · K1 · ML2 · MP = K4 · ML · MPL = K4 · K2 · ML2 · MP (15.3-6) so that K3 · K1 = K4 · K2 , and therefore θ=

(K1 + K2 ) · ML + 2 · K1 · K3 · ML2 1 + (K1 + K2 ) · ML + K1 · K3 · ML2

Of special interest is the case of a protein with four indistinguishable binding sites (i.e., a case where experimentally we cannot distinguish between the LP and PL states as above), but because of steric interference at binding sites or attractions between bound ligands on the formation of successive ligand bonds, the equilibrium constants for each successive ligand binding are different. The reason this is interesting is because this is exactly the case of oxygen binding to hemoglobin in our blood. The oxygen + hemoglobin (Hb) binding reactions are K1

Hb + O2  HbO2

Hemoglobin + oxygen binding

K2

HbO2 + O2  Hb(O2 )2 K3

Hb(O2 )2 + O2  Hb(O2 )3 K4

Hb(O2 )3 + O2  Hb(O2 )4 and the reported apparent equilibrium constants9 are

MHbO2 = 4.0 × 104 M−1 MHb · MO2 MHb(O2 )2 K2 = = 2.73 × 104 M−1 MHbO2 · MO2 MHb(O2 )3 K3 = = 5.94 × 104 M−1 MHb(O2 )2 · MO2 K1 =

9 G.

K. Ackers, Biophys. J. 32, 331 (1980).

15.3 Binding of a Ligand to a Substrate 919 and

K4 =

MHb(O2 )4 = 2.02 × 106 M−1 MHb(O2 )3 · MO2

Therefore

θ= =

Number of bound oxygen molecules Number of hemoglobin molecules

MHbO2 + 2 · MHb(O2 )2 + 3 · MHb(O2 )3 + 4 · MHb(O2 )4 MHb + MHbO2 + MHb(O2 )2 + MHb(O2 )3 + MHb(O2 )4

which, after some algebra, becomes

θ=

K1 · MO2 + 2 · K1 ·K2 ·M2O2 + 3 · K1 ·K2 ·K3 ·M3O2 + 4 · K1 ·K2 ·K3 · K4 · M4O2 1 + K1 · MO2 +K1 ·K2 ·M2O2 + K1 ·K2 ·K3 · M3O2 + K1 ·K2 ·K3 · K4 · M4O2 (15.3-7)

Clearly, at saturation θ = 4. The binding isotherm as a function of oxygen concentration is shown in Fig. 15.3-1. Note that as a result of the sigmoidal shape of the curve, there is a large change in the number of oxygen molecules bound to hemoglobin in the range between 0.5 × 10−5 M and 1.5 × 10−5 M oxygen. This is especially important in human and animal physiology since, as a result of the reverse reactions, a small decrease in oxygen concentration (or partial pressure) will lead to a significant release of oxygen from hemoglobin in blood. As the arterial blood passes through the body, it experiences a decrease in oxygen partial pressure due to the metabolic use of oxygen and also the small pressure drop due to flow in the capillaries, resulting in the release of a significant amount of oxygen. It is in this way that oxygen is delivered to cells throughout the body. 4

3

θ 2

1

0

0

–

–

–

–

–

–

5 .10 6 1.10 5 1.5 .10 5 2 .10 5 2.5 .10 5 3 .10 5 MO2

Figure 15.3-1 The number of oxygen molecules (θ) bound to a molecule of hemoglobin as a function of oxygen concentration.

The efficiency of this four-site adsorption/desorption process can be seen by noting that for a physiologically important change from 5 × 10−6 M oxygen to 5 × 10−5 M,

920 Chapter 15: Some Additional Biochemical Applications of Thermodynamics 3.51 molecules of oxygen are released for each molecule of hemoglobin. In contrast, if there were only a single binding site with K = 4.0 × 104 M (the value of the first equilibrium constant of the actual four sites), the shape of the isotherm would be of the form given by Eq. 15.3-3, with a release of only 0.50 oxygen molecules per hemoglobin molecule for the same change in oxygen concentration. Further, if we considered the case of single-site adsorption and used the largest value of the four equilibrium constants, K = 2.02 × 106 M, the isotherm would also have the shape given by Eq. 15.3-3, and the oxygen release would be only 0.080 molecules of oxygen per molecule of hemoglobin since the isotherm is near saturation over this range of oxygen concentrations, as shown in Figure 15.3-2. Thus, it is a result of the cooperative four-site oxygen binding to hemoglobin that large animals (such as humans) are possible. This sigmoidal shape of the ligand-binding isotherm, which occurs in hemoglobinoxygen binding, is a result of what is referred to as cooperative binding. That is, as a result of the increasing values of the apparent equilibrium constants (i.e., K3 > K1 and K2 , and K4 > K1 , K2 , and K3 ), the binding of oxygen molecules beyond the first becomes more favorable. The mechanistic reason that this occurs is that the hemoglobin molecule, which is actually a tetramer, undergoes structural rearrangements as each oxygen binds, resulting in the binding of the third and fourth oxygens being more favorable than the first. Note: Hemoglobin consists of four chains, each with a heme or iron group that can bind an oxygen molecule. Myoglobin is a simpler molecule consisting of a single chain that is of the same general structure as each of the chains in hemoglobin. The oxygencarrying and releasing capacity of myoglobin, which has only a single oxygen binding site, is much like that shown in Figure 15.3-2, and is much less than hemoglobin. Consequently, the association of myoglobin chains into hemoglobin was important to the development of life as we know it. 1

0.8

0.6 θ 0.4

0.2

0 0

–

5 .10 6

–

1.10 5

–

1.5 .10 5 Mi

–

2 .10 5

–

2.5 .10 5

–

3 .10 5

Figure 15.3-2 The number of oxygen molecules (θ) bound to hemoglobin if it had a single binding site with a high value of the binding constant.

15.3 Binding of a Ligand to a Substrate 921 Illustration 15.3-1 The Binding of Warfarin to Human Albumin Warfarin, also known as coumadin, is used to prevent blood clots in humans and also, at high body doses (referred to as a high body burden in the biomedical literature), as a rat poison by promoting internal bleeding. Warfarin binds with human plasma albumin in the reaction K

W + A  WA

The following measured or apparent (not standard-state) data at 25◦ C are available for this binding reaction, Δbnd G = −30.8 kJ mol−1 , Δbnd H = −13.1 kJ mol−1 , and Δbnd CP ∼ 0, where these apparent changes are for the following equilibrium constant: Δbnd G = RT ln K

where

K=

MWA MW · MA

where M is the concentration in molality. a. What is the entropy change for this reaction at 25◦ C? b. Starting with concentrations of warfarin and human plasma albumin of 0.0001 M, what is the fraction of unbound (unreacted) albumin over the temperature range of 0 to 50◦ C?

Solution a. Since ΔG = ΔH − T ΔS , at 25◦ C ΔH − ΔG −13 100 − (−30 800) kJ mol−1 ΔS =

=

T

298.15 K

= 59.37

J mol K

b. Since Δbnd CP = 0, Δbnd H is independent of temperature, and the equilibrium constant as a function of temperature is computed from K (T ) = ln K (T = 298.15 K)



T 298.15 K

Δbnd H Δbnd H dT = · RT 2 R



1 1 − T 298.15



The results are given in the table that follows. Also, letting α be the fraction of the initial warfarin (or albumin) that has reacted, we have MWA = α · MA,o = 0.0001α

so that K=

and

MW = MA = 0.0001 (1 − α)

0.0001α α = (0.1)2 · (1 − α)2 0.0001 · (1 − α)2

the solutions of which are also given in the table. T (◦ C)

0 10 20 30 40 50

K(T ) 4.039 × 10 3.295 × 105 2.725 × 105 2.282 × 105 1.933 × 105 1.654 × 105 5

α

1−α

0.855 0.840 0.826 0.811 0.797 0.783

0.145 0.160 0.174 0.189 0.203 0.217

Consequently, we see that warfarin is strongly bound to human plasma albumin at all temperatures.

922 Chapter 15: Some Additional Biochemical Applications of Thermodynamics

15.4 SOME OTHER EXAMPLES OF BIOCHEMICAL REACTIONS In this section we consider several other examples of physiologically important biochemical reactions. Illustration 15.4-1 The Trimerization of a Pancreatic Hormone Glucagon is a 29-amino peptide hormone of the pancreas. At pH = 10.6 it is known that glucagon can associate or trimerize, and the following standard-state (ideal 1-molal solution at 25◦ C) information is available: K

G  13 G3 Δrxn Go = −30.6

kJ mol

Δrxn H o = −131

kJ mol

and

Δrxn CPo = −1.8

kJ mol K

If the initial concentration of glucagon is 0.01 M (which is sufficiently low for the solution to be considered ideal), what is the fraction of glucagon trimerized over the temperature range from 0 to 50◦ C?

Solution The equilibrium relation is Ka = e −(Δrxn G

o /RT )

=

(aG3 )1/3 ∼ (MG3 )1/3 · (1 M)2/3 = aG MG

and the mass balances are MG = MG,o − α = 0.01 − α

so that

 α 1/3 Ka =

3

and

MG3 =

α 3

· (1 M)2/3

0.01 − α To calculate the equilibrium constant at any temperature, we use ln

Ka (T ) = Ka (T = 298.15 K)



Δrxn H ◦ (T ) dT RT 2

T 298.15 K

with



T

Δrxn H o (T ) = Δrxn H o (T = 298.15 K) +

rxn CPo dT

298.15 K

= Δrxn H o (T = 298.15 K) + rxn CPo · (T − 298.15 K)

so that ln

Ka (T ) = Ka (T = 298.15 K)



T

298.15 K



Δrxn H o (T = 298.15 K) dT RT 2

Δrxn CPo · (T − 298.15 K) dT RT 2 298.15 K   1 1 o − = −Δrxn H (T = 298.15 K) · T 298.15 K    o 1 T Δrxn CP 1 − ln + (298.15 K) · + R 298.15 K T 298.15 K +

T

15.4 Some Other Examples of Biochemical Reactions 923 We obtain the following values: T (◦ C)

0 25 50

Ka (T )

α(M)

Percent Trimerized

1.237 × 107 2.297 × 105 1.944 × 103

0.01000 0.009999 0.009923

100.0 99.99 99.23

Therefore, over the temperature range of 0 to 50◦ C glucagon is almost completely trimerized.

Polymers are formed by the polymerization of monomers, and proteins can aggregate (or associate) as in the illustration above. It is sometimes of use to know the relationship between the partial molar Gibbs energies of the polymer and the monomer or protein from which it is formed. To be completely general, suppose species A (whether it be a protein or polymer) polymerizes according to the reaction

βA  Aβ

(15.4-1)

The total Gibbs energy of the system is

G = NA GA + NAβ GAβ If the initial number of moles of species A is N, after chemical equilibrium is established we have α NA = N − α and NAβ = β where α is the number of moles of species A that has reacted. Therefore,

α GA β β   At equilibrium at constant temperature and pressure ∂G = 0, and using the Gibbs∂α T,P Duhem equation (Eq. 8.2-9a) gives G = (N − α) GA +

−GA +

1 GA = 0 or GAβ = βGA β β

(15.4-2)

That is, the partial molar Gibbs energy of a polymer containing β monomeric units is just β times the partial molar Gibbs energy of the monomer. Note that this is true whether or not the solution is ideal. We next consider the pH dependence of a biochemical reaction. To be specific we examine the oxidation reaction of ethanol by dehydrogenase enzyme and the oxidized form of nicotinamide adenine dinucleotide (NAD+ ) to produce acetaldehyde and the reduced form NADH: Ka

pH dependence of a biochemical reaction

NAD+ + CH3 CH2 OH  CH3 CHO + NADH + H+ The equilibrium relation for this reaction is

Δrxn Go = −RT ln Ka = −RT ln

aCH3 CHO · aNADH · aH+ aNAD+ · aCH3 CH2 OH

924 Chapter 15: Some Additional Biochemical Applications of Thermodynamics based on ideal, 1-molal standard states. This equilibrium relation can be rewritten as − RT ln Ka = −RT ln

aCH3 CHO · aNADH − RT ln aH+ aNAD+ · aCH3 CH2 OH

= −RT ln

aCH3 CHO · aNADH − RT ln(10) log(aH+ ) aNAD+ · aCH3 CH2 OH

= −RT ln

aCH3 CHO · aNADH + 2.303RT · pH aNAD+ · aCH3 CH2 OH

(15.4-3)

At reaction conditions, the pH will normally be controlled externally—for example, by a buffer solution—rather than being determined by the ionization of water or by hydrogen ion production from the reaction. Consequently, we can write this equation as aCH3 CHO · aNADH ln Ka + 2.303 · pH = ln aNAD+ · aCH3 CH2 OH So increasing the pH of the solution results in an increased equilibrium dehydrogenation of ethanol to acetaldehyde. (Note: The dehydrogenation of ethanol is an enzymatic reaction. In this reaction NAD+ is what is referred to as a cofactor, that is, a substance that is needed for the reaction and that will be regenerated by another reaction, but is not a catalyst for the reaction; the enzymes involved are the catalysts. Vitamins and minerals are common cofactors.) Illustration 15.4-2 The pH Dependence of a Biochemical Reaction At 25◦ C and pH = 7, Δrxn Go for the reaction above based on ideal 1-molal standard states is +23.8 kJ/mol. Assuming an initial concentration of both NAD+ and ethanol of 0.1 M, compute the extent of reaction as a function of the externally controlled pH.

Solution The mass balance, using α as the fraction of ethanol oxidized, is MNAD+ = MCH3 CH2 OH = 0.1(1 − α)

and

MNADH = MCH3 CHO = 0.1α

Now neglecting solution nonidealities (or equivalently, assuming they cancel in the numerator and denominator of the equilibrium relation), we have ln Ka + 2.303 · pH = ln

aCH3 CHO · aNADH 0.12 α2 α2 = ln 2 2 = ln aNAD+ · aCH3 CH2 OH (1 − α)2 0.1 (1 − α)

and α2 (1 − α)

2

= Ka · e 2.303·pH

or

α 1−α

=

Alternatively, this can be written as 2.303 α = e 2(pH−pKa ) 1−α

√

Ka · e(2.303·pH/2)

15.5 The Denaturation of Proteins 925 and α=

8.224 × 10−3 × e(2.303·pH/2) 1 + 8.224 × 10−3 × e(2.303·pH/2)

The figure below shows how the extent of this reaction, and the fraction of ethanol oxidized, change dramatically with the pH of the solution, with very little reaction at low pH (acid solution), a rapid increase with pH between pH = 2 and pH = 6, and the reaction being essentially complete above pH = 7. 1 0.8 0.6 α 0.4 0.2 0 0

2

4

6 pH

8

10

12

Extent of the reaction α as a function of pH Note that α = 0.5 when pH = pKa , also α < 0.5 when pH < pKa , and the extent of reaction α > 0.5 when pH > pKa . Thus in the case of reactions, as already mentioned in the discussion on solubility, large changes occur at pH values near the pKa value.

15.5 THE DENATURATION OF PROTEINS Protein unfolding (denaturation)

The unfolding of proteins can be considered to be a type of biochemical reaction. Proteins in the native state exist in a specific folded state that is necessary for their biological catalytic activity. With changes in temperature, pressure, or chemical environment (e.g., changes in pH or ionic strength, or the addition of denaturants such as urea), the protein will unfold or denature and lose its biological activity. This unfolding and denaturation, depending on the protein, may be reversible or irreversible. Though many intermediate steps are involved, protein unfolding is frequently treated as a twostate process (folded or unfolded). Also, since the transition occurs over a range of conditions (for example, over a temperature range rather than at a single temperature), the process is considered to be a unimolecular chemical reaction, not a first-order phase transition (such as the boiling of a pure liquid that occurs at a single temperature). In the biological literature the temperature at which half of the initial amount of protein has unfolded is referred to as the “melting point” of the protein, though that is a misnomer since the protein does not melt (which would be a first-order phase transition). Also, this melting point can be a function of state conditions such as pH, pressure, ionic strength, etc. We consider first protein denaturation as a result of changes in temperature. Eq. 15.2-13 discussed earlier can be used, but now with the interpretation of Δunf H and Δunf S to being the enthalpy and entropy changes on protein unfolding, Δunf CP being the difference in heat capacities between the unfolded and folded states, and T m

926 Chapter 15: Some Additional Biochemical Applications of Thermodynamics referred to in the biological literature as the melting temperature (though experimentally it is the temperature at which half the proteins in a sample are denatured) gives

Δunf G (T ) = Δunf H (T ) − T Δunf S (T ) = Δunf H (Tm ) + Δunf CP (T − Tm ) − −T Δunf CP ln



 = Δunf H (Tm ) 1 −

T Tm T Tm

T Δunf H Tm

(Tm )

 

   + Δunf CP T − Tm − T ln TTm (15.5-1)

Illustration 15.5-1 Unfolding of a Protein as a Function of Temperature G-actin is a globular dimer of actin, a cytoskeleton protein. In solutions at pH = 7.5 the following data are available for the unfolding of G-actin at 25◦ C: Δunf G = 27

kJ mol

Δunf H = 197

kJ mol

and

Δunf CP = 27

J mol K

Determine the temperature at which half of the G-actin is unfolded (that is, find its “melting temperature”).

Solution The apparent equilibrium constant for this unimolecular chemical reaction is K(T ) =

α(T ) · Mo · γunfolded (T ) α(T ) Munfolded (T ) = = (Mo − α(T ) · Mo ) · γfolded (T ) (1 − α(T )) Mfolded (T )

where α(T ) is the fraction of the initial G-actin that has been denatured at the temperature T, and since there is only a conformational change between the folded and unfolded protein, we have assumed that their activity coefficients are equal, and therefore cancel from the equation. Also, we are interested in the melting temperature, that is, the temperature Tm at which α = 1/2, so that K = 1 and ln K = 0. Therefore, the equation to be solved is ln K(Tm ) = 0

However,

unf G(25◦ C) unf H(25◦ C) ln K(Tm ) = − − R · 298.15 R



1 1 − Tm 298.15 +

unf CP R

 

 ln

Tm 298.15

 +

298.15 −1 Tm



or 0=−

27 000 197 000 P − 8.314 · 298.15 8.314



1 1 − Tm 298.15



+

    Tm 298.15 54 000 −1 ln + R 298.15 Tm

15.5 The Denaturation of Proteins 927 which has the solution of Tm = 330.0 K = 56.8◦ C, which is close to the measured value of 57.1◦ C. It is also of interest to determine the fraction of G-actin that is unfolded at any temperature. To do this we solve the equation K(T ) =

α 1−α

or

α=

K(T ) 1 + K(T )

where  K(T ) = exp −

27 000 197 000 − 8.314 · 298.15 8.314



1 1 − T 298.15 +



    T 298.15 54 000 −1 ln + R 298.15 T

The results are shown in the following figure. Note that the unfolding of G-actin occurs over a temperature interval of more than 20◦ C. 1

α 0.5

0 310

320

330 T (K)

340

350

Fraction α of the protein G-actin unfolded (“melted”) as a function of temperature.

Note that all the parameters that appear in Eq. 15.5-1, the enthalpy change, heat capacity change, and unfolding temperature can be obtained from calorimetry, and these parameters have been reported for a number of proteins. An interesting characteristic of this equation is that it is nonlinear in temperature, and so can have two values of T m for a value ofΔunf G. Privalov and Khechinashvili10 reported data for the denaturation of proteins; here we consider only one, metmyoglobin, which is the oxidized form of the oxygen-carrying protein myoglobin (which is brown in color, and is why meat turns brown as it ages and oxidizes). Reading from the graphical presentation of their data and retaining their original units, Δunf CP = 0.16 cal/gr · K and Δunf H (50o C) = 4.1 cal/gr in a buffered solution in the pH range of 9.5 to 12.5. Using these values, the calculated enthalpy and the product of temperature and entropy (the difference of which is the Gibbs energy) for unfolding as a function of temperature are shown in Fig. 15.5-1. 10 P.

L. Privalov and N. N. Khechinashvili, J. Mol. Biol. 86, 665 (1974)

928 Chapter 15: Some Additional Biochemical Applications of Thermodynamics 10

5

0

–5

–10 260

280

300

320

340

T (K)

Figure 15.5-1 The enthalpy (solid line) and the product of temperature times entropy (dashed line) for the unfolding of metmyoglobin as a function of temperature. Units of kcal/g.

In this figure we see that the changes on unfolding of the enthalpy and the product of the temperature and the entropy T Δunf S are very close in value, and that below about 274 K and above 323 K the entropic term is slightly greater than the enthalpic term, while between these temperatures the enthalpy term is slightly larger. Further, at these two temperatures, the two terms are equal, and therefore the Gibbs energy of unfolding is zero. These are the equilibrium (or melting) temperatures for unfolding. The Gibbs energy of unfolding Δunf G calculated above is shown on a greatly expanded scale in the Fig. 15.5-2. There are several things to note in this figure. First is that the magnitude of the Gibbs energy of unfolding is small, indeed one to two orders of magnitude less than the separate enthalpy and T Δunf S contributions. Therefore, protein unfolding is a result of a delicate balance between these two contributions. Second, is that the unfolded state is favorable when the Gibbs energy change on unfolding is negative, and the native or folded state is favored when the Gibbs energy change is positive. Consequently, we find that the unfolded native state of metmyoglobin will dominate between 274 K and 323 K, but the protein will unfold and denature below 274 K (the “cold denaturation temperature”) and above 323 K (the “hot denaturation temperature”).

0.1

0

ΔunfG –0.1

–0.2

–0.3 260

270

280

290

300 T (K)

310

320

330

340

Figure 15.5-2 The Gibbs energy of unfolding of metmyoglobin as a function of temperature. The two temperatures at which T Δunf G are zero the cold and hot denaturation temperatures. Note that T Δunf G is of the order of 0.1 kcal/g, while T Δunf H and T Δunf S are at least an order of magnitude greater over most of the temperature range.

15.5 The Denaturation of Proteins 929 There is a temperature between the cold and hot denaturation temperatures at which the protein is most stable to unfolding, that is, there is temperature at which Δunf G has a maximum value. For metmyoglobin is occurs at 298.5 K. An interesting result is that at the temperature of maximum stability  ∂Δunf G = 0 = −Δu S (15.5-2) ∂T P where this relationship between the temperature derivative of the Gibbs energy and the entropy comes from classical thermodynamics. So the conclusion is that at the temperature of maximum stability there is no entropy change on unfolding of the protein. That is, at this temperature the stabilization of the native protein structure is completely enthalpic in nature. The reversible unfolding of many proteins is being considered here as a two-state process,u  f and using αto represent the fraction of the initial amount of protein that has unfolded gives  Δunf G Fα α [unf] or Keq = = = = exp − F (1 − α) RT [f] (1 − α)  Δ G (15.5-3) unf exp − RT  Δunf G  α= 1 + exp − RT The fraction of metmyoglobin molecules unfolded, α,is shown as a function of temperature in Fig. 15.5-3. The Gibbs energy penalty for denaturation at the point of maximum stability is about 0.158 cal/g or 11.2 kJ/mol. 1

α

0.5

0 260

280

300

320

T (K) Figure Fraction of Proteins Unfolded

340

Figure 15.5-3 Fraction of metmyoglobin denatured α as a function of temperature.

Human prions (proteinaceous infectious particles) are proteins that cause encephalitis, Creutzfeldt-Jakob disease, and Mad Cow disease. Human prions have a value11 of Δunf G of 26.3 kJ/mol, Δunf H of 294 kJ/mol, and a denaturation (melting) temperature of approximately 70o C. Consequently prions are much more stable with respect to temperature (and also chemical agents) than most other proteins, and is the reason 11 W.

Swietnicki, R. B. Petersen, P. Gambetti, and W. K. Surewicz, J. Biol. Chem. 273, 31048 (1998).

930 Chapter 15: Some Additional Biochemical Applications of Thermodynamics that their deactivation and disposal is difficult. The fraction of human prions denatured as a function of temperature is shown Fig. 15.5-4. 1

α

0.5

0 320

340

360

380

T (K)

Figure 15.5-4 The fraction α of human prion denatured as a function of temperature.

Fraction of Prions Denatured

There are also proteins that have adapted to withstand high temperatures (hyperthermophiles), high salinity (halophiles), pH (acidophiles or alkaliphiles) and pressure (barophiles) or other extreme conditions; collectively these are called extremophiles. As an example, one such thermophile is the β -lactoglobulin12 that at pH = 2 has a reported Δunf H of 312 kJ/mol at its (hot) thermal denaturation temperature of 351 K and Δunf CP of 5.58 kJ/mol K. The calculated stability curve for this protein (in units of kJ/mol) from these data is a shown as the solid line Fig. 15.5-5. So in addition to the hot denaturation temperature of 351 K, this protein has a predicted cold denaturation temperature of 250 K. This 101 K stability range is to be compared with 49 K (274 to 323 K) for metmyoglobin. Denaturation temperatures are also a function of other components in the solution. This is illustrated in Fig. 15.5-5 in which the solid lines are for the protein in a ureafree solution, and the dashed lines are for the protein in a 2 M urea solution. The range of thermal stability is reduced by more than 30o C by the addition of urea as shown in Fig. 15.5-6. 40

20

0 ΔG ΔGu

–20

–40

–60

–80 200

250

300

350

T (K)

12 Yu.

V. Griko and P. L. Privalov, Biochem. 31, 8810 (1992).

400

Figure 15.5-5 Gibbs energy of unfolding of the thermophile protein β -lactoglobulin as a function of temperature; solid line protein In a urea-free solution; dashed line protein in 2 M urea solution.

15.5 The Denaturation of Proteins 931 1

0.8

0.6

α αu

0.4

0.2

0 220

240

260

280

320

300

340

Temperature (K)

360

380

Figure 15.5-6 Fraction of α of β -lactoglobulin unfolded as a function of temperature; solid line protein in buffer; dashed line protein in 2 M urea solution.

Proteins can also unfold or denature as a result of increasing external pressure, presumably because a pressure increase results in an increase in solution density that produces a structural rearrangement of the protein. Consequently, the melting temperature of a protein will also be a function of pressure. To determine how the melting temperature of a protein changes with pressure, we start from the observation that at the melting temperature at any pressure

unf G(Tm ) = −RT ln Ka (Tm ) = 0 = unf H(Tm ) − Tm · unf S(Tm ) (15.5-4a) since Ka (Tm ) = 1. Therefore,

unf S (Tm ) =

unf H (Tm ) Tm

(15.5-4b)

Since, along the melting curve

Gnat (T, P ) = Gunf (T, P ) where the subscripts nat and unf refer to the native (folded) and unfolded proteins, respectively. Therefore, for a small change in pressure following the melting curve, we have dGnat (T, P ) = dGunf (T, P )

V nat dP − S nat dT = V unf dP − S unf dT or



dP dT

= Gnat =Gunf

unf S

unf H S unf − S nat = = V unf − V nat

unf V Tm · unf V

(15.5-5)

(Compare this equation with the Clapeyron equation, Eq. 7.7-4.) Consequently, along the melting curve  dT

unf H Tm = · (15.5-6)

unf V dP Gnat =G unf

932 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Illustration 15.5-2 Unfolding of a Protein as a Function of Pressure It has been observed13 that at 1.013 bar the cell membrane compound dipalmitoylphosphatidylcholine (DPPC) melts at 313 K, and that the melting temperature increases by 0.04 K for each increase of 1 bar in pressure. Also, when this compound unfolds, there is a volume change of 26 cm3 /mol. Use these data to estimate a. The enthalpy change on the denaturation of DPPC b. The entropy change on this pressure-induced denaturation of DPPC

Solution a. To obtain the enthalpy change, we use 

unf H = Tm · unf V = 20 345

dP dT



= 313 K × 26 Gnat =Gunf

cm3 1 bar m3 105 J × × 10−6 3 × mol 0.04 K cm bar m3

J kJ = 20.345 mol mol

b. To compute the entropy change, we use

unf S (Tm ) =

unf H (Tm ) Tm

=

J 20 345 mol J = 65 313 K mol K

Proteins can also be unfolded or denatured as a result of chemical interaction or changes in solution conditions. Common chemical denaturants include urea and guanidine hydrochloride; we do not consider this type of denaturation here.

15.6 COUPLED BIOCHEMICAL REACTIONS: THE ATP-ADP ENERGY STORAGE AND DELIVERY MECHANISM The most important energy storage method in cells is as the chemical adenosine triphosphate (ATP). In the dissociation of ATP to adenosine diphosphate (ADP) and a phosphate ion, schematically shown as OH HO

P

OH O

O

P

OH O

O

P

O

H

O

Adenine

CH2 O

+ H2O

H

H

H OH OH

OH HO

P O

OH O

P O

O

Adenine

CH2 O H H

+ H2PO4

H H

OH OH

13 D.

B. Mountcastle, R. L. Biltonen, and M. J. Halsey, Proc. Nat’l. Acad. Sci. USA 75, 4906 (1978).

15.6 Coupled Biochemical Reactions: The ATP-ADP Energy Storage and Delivery Mechanism 933 which is usually written simply as

ATP ⇐⇒ ADP + phosphate

(15.6-1)

a large amount of Gibbs energy is released that, for example, muscles convert to mechanical work by contraction, and by coupled reactions is used elsewhere in the body to provide the Gibbs energy for the synthesis of physiologically important molecules. In fact, ATP, which is present in the cytoplasm and nucleoplasm, provides the energy for most processes in cells. For example, as mentioned earlier, the sodium ion concentration within a cell is usually about 10 mM, while the concentration in the external plasma outside the cell membrane is much higher, about 140 mM. This concentration difference is maintained by a set of biological reactions and transport processes, the sodium ion pump, that uses Gibbs energy from ATP hydrolysis to expel sodium ions from the cell. Also, ATP hydrolysis provides energy for the functioning of nerves and muscle. Consequently, ATP has been called the energy currency of biological systems; it can be used to store energy in the body, supply energy when necessary, and be regenerated by a series of chemical reactions coupled to the partial oxidation of carbohydrates. For this and other biological reactions the standard state chosen depends on temperature, the ionic strength, pH of the solution, and the presence of other ions (especially magnesium ions for this reaction). At very low ionic strength, pH = 7, and in the absence of magnesium ions, the values14 of the standard-state Gibbs energy of reaction, based on ideal 1-molal aqueous solutions, is 33.5 kJ/mol at 25◦ C and 34.0 kJ/mol at 37◦ C. However, physiological conditions are quite different, and the actual free energy change, neglecting solution nonidealities, is computed from

   MP     MADP · · a a ADP P M =1  MATP M  =1 = rxn G◦ + RT ln

rxn G = rxn G◦ + RT ln aATP M =1 (15.6-2) While the concentrations of ATP, ADP, and the phosphate ion are neither constant nor known exactly, it is possible to make an estimate of the Gibbs energy released on the ATP hydrolysis reaction in muscles at body temperature (37◦ C) as in the illustration below. Illustration 15.6-1 Energy Release by ATP Hydrolysis in the Body Assuming the following concentrations in human muscle, MATP = 1 × 10−3 M

MADP = 1 × 10−5

and

MP = 1 × 10−3 M

estimate the Gibbs energy released on the hydrolysis of 1 mol of ATP at body temperature (37◦ C).

Solution Using these values of the concentration, we have

rxn G = −34.0 + 8.314 × 10−3 × 310.15 × ln

14 As



1 × 10−5 × 1 × 10−3 1 × 10−3

 = −63.7

kJ mol

reported in Biothermodynamics: The Study of Biochemical Processes at Equilibrium, by J. T. Edsall and H. Gutfreund, John Wiley & Sons, New York (1983).

934 Chapter 15: Some Additional Biochemical Applications of Thermodynamics which is almost twice that of the standard-state Gibbs energy release. (It has been suggested15 that the Gibbs energy change for this reaction in vivo is about −50 kJ/mol, though the concentrations of the species were not reported, and undoubtedly are different from the estimates used here.) Nonetheless, we can conclude that 50 to 60 kJ of Gibbs energy becomes physiologically available per mole of ATP hydrolyzed in muscles or other parts of the body.

It should be pointed out that the mechanism by which the reaction occurs is much more complicated than Eq. 15.6-1 implies. Many intermediate compounds, and also enzymes acting as biological catalysts, are involved. The precise mechanism of this reaction is discussed in basic biology courses. However, as the sum of all the reactions is as shown in Eq. 15.6-1, from the point of view of thermodynamics, that is the only reaction we need to consider if our interest is solely in the net energy release (Sec. 13.1). Given that there is such a large Gibbs energy release on the hydrolysis reaction of ATP to form ADP, we can ask the question of how ATP can be formed by the reverse reaction so that, for example, a relaxed muscle can regenerate ATP in preparation for the next contraction. We consider this next, but first have to introduce the concept of coupled chemical reactions. From the second law of thermodynamics, any spontaneous process must occur with a net production of entropy. As shown in Chapter 5, a result of the second law is that the Gibbs energy of a system must decrease for a spontaneous process to occur at constant temperature and pressure. At first glance it might appear that this would mean that every process that occurs spontaneously must lower the Gibbs energy of the system. However, if there is a sequence of coupled processes, and they must be coupled, then as long as the total Gibbs energy of the system is lowered, the collection of coupled processes can occur even if some of the intermediate steps by themselves would increase the Gibbs energy of the system. Before considering how this may occur in chemical reactions, we consider an analogy with the simple mechanical system shown in Fig. 15.6-1a. There we see two weights; weight A is at a low level, and the somewhat heavier weight B is at a higher level. Raising weight A to the higher level would increase its potential energy (and had we included mechanical potential and kinetic energy in our thermodynamic description, would increase its Gibbs energy as well); we know that this process will not occur spontaneously. Moving weight B to the lower level will decrease its potential energy and could occur easily, for example, if the weight were given a slight sideways push off the ledge. Now suppose that we couple the two weights, for example, by connecting them with the rope-and-pulley system shown in Fig. 15.6-1b. Since weight B is somewhat heavier than weight A, in this arrangement it is possible to raise weight A to the higher level, lower weight B, and still decrease the total potential (or Gibbs) energy of the system. Thus by coupling the two processes, it is possible to make one process occur that has a positive Gibbs energy change by driving it with a coupled process that has a larger negative free energy change. A similar coupling of processes occurs frequently in nature. In fact, many chemical reactions consist of a number of molecular-level steps, the so-called elementary steps, one or more of which have a positive Gibbs energy change and would not occur spontaneously. However, the total reaction process, which consists of coupled elementary steps in which the products of one elementary step are the reactants for the next step, results in a net decrease in the Gibbs energy of the system. 15 A.

L. Lehninger, Bioenergetics, 2nd ed., Benjamin, London (1971).

15.6 Coupled Biochemical Reactions: The ATP-ADP Energy Storage and Delivery Mechanism 935

B

B

A

A (a)

(b)

Figure 15.6-1 (a) Two weights that are not coupled. (b) Two weights that are coupled.

As a simple example of coupled reactions, consider an electrochemical cell consisting of lead and magnesium electrodes in a lead sulfate electrolyte. The half-cell reactions are

Pb2+ + 2e− → Pb

E = −0.13 V

ΔG◦ = 0.13 F = 12.5 kJ

and

Mg → Mg2+ + 2e−

ΔG◦ = −2.36 F = −227.6 kJ (15.6-3) In this case the standard-state Gibbs energy change for the first half-cell reaction is positive, so that the reaction is not favorable, while the standard-state Gibbs energy change for the second half-cell reaction is strongly negative. For the overall process E = +2.36 V

Pb2+ + Mg → Pb + Mg2+

ΔG◦ = −215.1 kJ

(15.6-4)

The Gibbs energy change for this overall process is very negative, so that the reaction can (and will) occur, even though it is the result of one process with a positive Gibbs energy change coupled with a second process with a larger negative Gibbs energy change. It is only by looking at the details of the process, here the two half-cell reactions, that we see that a process with a positive Gibbs energy change is being driven by another process with a larger negative Gibbs energy change. This is not evident by looking only at the overall process of Eq. 15.6-4. Many naturally occurring processes are of the type discussed above, in that the overall process results in a negative Gibbs energy change, even though one or more of the elementary steps that are part of the overall process result in a positive Gibbs energy change. Some of the most important examples of this occur in biological processes. Of interest here is the phosphorylation of adenosine diphosphate (ADP) to regenerate adenosine triphosphate (ATP). The reaction is

ADP + phosphate → ATP

ΔG◦ = 33.5 kJ

(15.6-5)

Since the standard-state Gibbs energy change is positive, one would expect that this reaction would occur to only a very limited extent in biological systems. (Of course,

936 Chapter 15: Some Additional Biochemical Applications of Thermodynamics the reason it would occur to any extent is that the reacting components are not in their standard states.) However, through a complicated series of enzymatic reactions, the phosphorylation of ATP is coupled to a reaction in which glucose is oxidized with an extremely large negative Gibbs energy change,

C6 H12 O6 + 6O2 → 6CO2 + 6H2 O

ΔG◦ = −2870.2 kJ

(15.6-6)

Indeed, the Gibbs energy change is so large that in many biological systems the oxidation of one molecule of glucose can result in the production of 38 molecules of ATP, that is,

C6 H12 O6 + 6O2 + 38ADP + 38 phosphate → 6CO2 + 6H2 O + 38ATP (15.6-7) for which ΔG◦ = −1756.8 kJ. Written in this manner, we would conclude that this reaction is thermodynamically possible and will occur to an appreciable extent. One measure of the efficiency of this process is to note that 38 × 33.5 = 1273 kJ of Gibbs energy (in the standard state) is stored as ATP for each mole of glucose oxidized, which releases 2870.2 kJ, so that only about 44 percent of the Gibbs energy released on oxidation is stored, the rest going to heat. The actual “in vivo” efficiency will be somewhat different, as the reactants and products are not in their standard states. The important observation here is that it is from coupled reactions resulting from the oxidation of glucose that ATP is regenerated in cells. As we have seen by looking one level deeper, the reaction of Eq. 15.6-7 is composed of two reactions, one of which is thermodynamically favored and the second of which is not. If we examined the mechanism still further by considering all the coupled reactions, as biologists do, we would find that there is a very large collection of enzymatically catalyzed intermediate reactions that occur, and that some of them have negative Gibbs energy changes and others have positive Gibbs energy changes. Thermodynamics provides us with the information that the overall reaction is possible regardless of the internal mechanisms that might be necessary to drive this reaction. Nature has developed a coupled reaction mechanism in biological systems that is more clever than any developed by humans to make this reaction occur. However, even nature is constrained by the fact that the overall Gibbs energy change for any process at constant temperature and pressure must be negative! It is only within this limitation that interesting biological processes, such as active transport (the transport of species against a concentration gradient), occur. The discussion of ATP synthesis above is for the case in which sufficient oxygen is present for the reaction to occur as described. That is, the glucose oxidation reaction occurs aerobically. It is also possible that as a result of extreme exercise or circulatory impairment there is not sufficient oxygen for the oxidation of glucose to occur. Under anaerobic conditions glucose can be used to form ATP from ADP, but by a different biochemical pathway (referred to as anaerobic glycolysis). In this process only about 2 percent of the energy in a glucose molecule is stored in ATP. However, even this small amount of ATP generated by anaerobic glycolysis can be important to the survival of a cell (and of organisms) during a temporary disruption of its oxygen supply. While this book is not the appropriate place for a detailed discussion of the ADPATP cycle, perhaps a bit more mechanistic detail is useful. The aerobic metabolism of glucose to produce ATP from ADP can be considered to consist of three parts: the fermentation of glucose to form pyruvate (and a small amount of ATP from ADP), the conversion of the pyruvate to carbon dioxide in the citric acid cycle in which NAD+

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 937 and FAD (the oxidized form of flavin adenine dinucleotide) are converted to NADH and FADH2 , and their oxidation in the respiratory chain, resulting in the formation of a larger quantity of ATP from ADP. If oxygen is not available, so the reaction is anaerobic, only the first step, formation of pyruvate, occurs with a small amount of ATP production. In this discussion the concept of the reactions being coupled has been stressed, and this is an important point. If there were two reactions that were not coupled—for example, by the product of one reaction not being the reactant of the second—then the reactions would be independent, and the Gibbs energy of the system would be minimized when the Gibbs energy of each reaction is separately a minimum. That is, each reaction would achieve an equilibrium state independent of the other (except for the possibility of a weak coupling as a result of solution nonidealities that depend on the compositions of all species present). However, if the two reactions are coupled as in the examples here, the equilibrium state of minimum Gibbs energy is a function of both reactions and is in general different from the state in which each reaction separately goes to equilibrium.

15.7 THERMODYNAMIC ANALYSIS OF FERMENTERS AND OTHER BIOREACTORS A bioreactor or fermenter is a chemical reactor in which microbes (e.g., bacteria or yeast) act on an organic material (referred to as a substrate) to produce additional microbes and other desired or undesired products. A schematic diagram of a bioreactor is given in Fig. 15.7-1. Mass balances for a biochemical reactor or fermenter are slightly more complicated than those for a usual chemical reactor because one of the products is cells or biomass (in the case of a wastewater or sewage treatment plant, which is another form of bioreactor, is referred to as sludge, which is a complicated chemically-undefined mixture of biochemical species). Likewise, other products of a bioreactor may also not be well-defined compounds of known molecular weight. The usual procedure, then, is that an elemental analysis is done on such products, and reported as the number of atoms of hydrogen, oxygen, and nitrogen (perhaps also sulfur, phosphorus, and other trace atoms that we will neglect here) for each carbon atom. For example, Roels16 has suggested that in the analysis of bioreactors, one can use an “average biomass” of CH1.8 O0.5 N0.2 to represent cells and bacteria. An elemental analysis can also be used for the substrate, especially if it contains a mixture of compounds, as, for example, in the mixed organic wastes entering a sewage treatment facility from an industrial or farming process. In these cases we do not have well-defined chemical species, but cells, other products, or substrates identified only by their atom ratios (usually referred to carbon). Consequently, mass balances for bioreactors are usually done for each atomic element, rather than for molecules, as was the case elsewhere in this book. As a simple example, consider the fermentation of carbon monoxide in aqueous solution using the bacterium Methylotrophicum to produce acetic acid, butyric acid, and additional cells, which has the reported stoichiometry

CO + xO2 + yN2 + zH2 O → 0.52CO2 + 0.21CH3 COOH + 0.006CH3 CH2 CH2 COOH + 0.036 C-mole cells

16 J.

A. Roels, Energetics and Kinetics in Biotechnology, Elsevier Biomedical Press, Amsterdam (1983).

938 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Oxygen (NO )4 2

Nitrogen (NN )4 2

W

Carbondioxide (NC)4

Substrate (N S)1, CHH OO NN S

S

S

Substrate (NS)3

Nitrogen Source (NN)1, CCNHHNOONNNN

Nitrogen Source (NN)3 Water (NW)3

Water (NW)1, H2O

Biomass (NB)3, CHH OO NN B

B

Product (NP)3, CHH OO NN P

Oxygen (NO )2 2

P

B

P

Q

Nitrogen (NN )2 2

Figure 15.9-1 A schematic diagram of a fermenter (bioreactor) showing mass, heat, and work flows. The heat flow is generally to keep the reactor at constant temperature, and the work flow, which is usually small, is for stirring the reactor contents.

where we have used the abbreviated notation C-mole cells to indicate the number of moles of elemental carbon present in cells. In the biochemical literature it is common notation to write stoichiometric equations in terms of each species on a C-mole basis, even though familiar compounds then appear in a less familiar way. For example, the reaction above would be written as

CO + xO2 + yN2 + zH2 O → 0.52CO2 + 0.42CH2 O + 0.024CH2 O0.5 + 0.036 C-mole cells where in this second equation the stoichiometric coefficients have changed since acetic acid contains two carbon atoms (so its initial stoichiometric coefficient is multiplied by 2 in this abbreviated notation), and butyric acid has four carbon atoms. Illustration 15.7-1 Mass Balance on a Bioreactor Assuming that cells produced in this biochemical reaction can be represented by the Roels average biomass (that is, 1 C-mole cells = CH1.8 O0.5 N0.2 ), how many moles of oxygen, nitrogen, and water are consumed for each mole of carbon monoxide consumed?

Solution First, we test the consistency of the reported stoichiometry by checking that the number of moles of carbon atoms balance on each side of the equation. C-balance: 1 = 0.52 + 0.42 + 0.024 + 0.036 = 1 So the carbon-atom balance is correct. The other atom balances are as follows. O-balance:

1 + 2x + z = 0.52 · 2 + 0.42 + 0.024 · 0.5 + 0.036 · 0.5 = 1.49

N-balance: H-balance:

2 · y = 0.036 · 2 = 0.0072

or

y = 0.0036

2 · z = 0.42 · 2 + 0.024 · 2 + 0.036 · 1.8 = 0.9528

or

z = 0.4764

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 939 Using this last result in the oxygen-atom balance, we obtain 2 · x = 1.49 − 1 − z = 0.49 − 0.4764 = 0.0136

or

x = 0.0068

so that the complete stoichiometry for this biochemical reaction (based on the assumed composition of the average biomass) is CO + 0.0068 · O2 + 0.0036 · N2 + 0.4764 · H2 O → 0.52 · CO2 + 0.42 · CH2 O + 0.024 · CH2 O0.5 + 0.036CH1.8 O0.5 N0.2

(a)

or, returning to common chemical notation, CO + 0.0068 · O2 + 0.0036 · N2 + 0.4764 · H2 O → 0.52 · CO2 + 0.21 · CH3 COOH + 0.006 · CH3 CH2 CH2 COOH + 0.036CH1.8 O0.5 N0.2

In the analysis of biochemical reactors, it is common to report results in terms of yield factors that relate the production or consumption of one species relative to that of another. For example, the yield factor for biomass produced per mole of substrate consumed YB/S is defined to be

YB/S =

Number of C-moles of biomass produced Number of C-moles of substrate consumed

The substrate is the carbonaceous material consumed, carbon monoxide in the example above. Some common yield factors are listed in Table 15.7-1. Table 15.7-1. Yield Factors Based on Substrate Consumed YB/S =

Number of C-moles of biomass produced Number of C-moles of substrate consumed

Yi/S =

Number of C-moles of product i produced Number of C-moles of substrate consumed

YN/S =

Number of C-moles of nitrogen source consumed Number of C-moles of substrate consumed

YO2 /S =

Number of moles of O2 consumed Number of C-moles of substrate consumed

YW/S =

Number of moles of water consumed Number of C-moles of substrate consumed

YC/S =

Number of moles of CO2 produced Number of C-moles of substrate consumed

Illustration 15.7-2 Yield Factors for a Biochemical Reaction What are the yield factors for the reaction in the previous illustration?

Solution From Eq. (a) of the previous illustration, we have YB/S = 0.036, YCH2 O/S = 0.42, YCH2 O0.5 /S = 0.024, YN/S = 0.0036, YO2 /S = 0.0068, YC/S = 0.52, and YW/S = 0.4764.

940 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Note that for generality, the reaction so far considered in this section could be written as

CO + YO2 /S · O2 + YN2 /S · N2 + YW/S · H2 O → YC/S · CO2 + YCH2 O/S · CH2 O + YCH2 O0.5 /S · CH2 O0.5 + YB/S C-mol cells Consider next the fermentation of some nonnitrogenous, but otherwise unspecified, substrate (for example, a sugar such as glucose) in aqueous solution with dissolved ammonia as the nitrogen source to produce “average” biomass, carbon dioxide, and water. On a carbon mole basis, this reaction is Substrate + YO2 /S · O2 + YN/S · NH3 + YW/S · H2 O → YB/S · CH1.8 O0.5 N0.2 + YC/S · CO2 The carbon balance is 1 = YB/S + YC/S

or YB/S = 1 − YC/S

and the nitrogen balance is

YN2 /S = 0.2 · YB/S

or YB/S = 5 · YN2 /S

Since we do not know the oxygen or hydrogen contents of the substrate, we cannot write any other complete balance equations for this system or solve for the other yield factors. All that we can say with certainty is (1) the number of C-moles of biomass produced per C-mole of substrate consumed will be less than unity by the number of moles of carbon dioxide produced per mole of substrate consumed, and (2) five C-moles of biomass will be produced for each mole of ammonia substrate consumed. We now want to consider the completely general steady-state mass balance equations for the bioreactor of Fig. 15.7-1. The balances will be written for each species of atoms C, H, N, and O, and we will use the following notation: [The equations to follow equally well be written for one cycle of a batch reactor in which case (N˙ I )i is replaced with (NI )i ] Flows: (N˙ I )i = molar flow rate of species I in flow stream i Subscripts: S = substrate N = nitrogen source substance B = biomass produced P = other product C = carbon dioxide W = water N2 = nitrogen O2 = oxygen Atomic stoichiometric notation: CS = moles of carbon per C-mole of substrate = 1 CN = moles of carbon per C-mole of nitrogen source (e.g., 0 in ammonia) CB = moles of carbon per C-mole of biomass = 1 CP = moles of carbon per C-mole of product = 1 CC = moles of carbon per mole of carbon dioxide = 1 NS = moles of nitrogen atoms per C-mole of substrate NN = moles of nitrogen atoms per C-mole of nitrogen source (e.g., 1 in ammonia)

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 941 NB = moles of nitrogen atoms per C-mole of biomass (e.g., 0.2 in average biomass) NP = moles of nitrogen atoms per C-mole of product NN2 = moles of nitrogen atoms per mole of molecular nitrogen = 2 HS = moles of hydrogen atoms per C-mole of substrate HN = moles of hydrogen atoms per C-mole of nitrogen source (e.g., 3 in ammonia) HB = moles of hydrogen atoms per C-mole of biomass (e.g., 1.8 in average biomass) HP = moles of hydrogen atoms per C-mole of product HW = moles of hydrogen atoms per mole of water = 2 OS = moles of oxygen atoms per C-mole of substrate ON = moles of oxygen atoms per C-mole of nitrogen source (e.g., 0 in ammonia) OB = moles of oxygen atoms per C-mole of biomass (e.g., 1.8 in average biomass) OP = moles of oxygen atoms per C-mole of product OW = moles of oxygen atoms per mole of water = 1 OO2 = moles of oxygen atoms per mole of molecular oxygen = 2 With this notation, we have for the steady-state carbon balance

0 = (N˙ S )1 + (N˙ N )1 · CN − (N˙ S )3 − (N˙ N )3 · CN − (N˙ B )3 − (N˙ P )3 − (N˙ C )4 · CC or

    0 = (N˙ S )1 − (N˙ S )3 + (N˙ N )1 − (N˙ N )3 · CN − (N˙ B )3 − (N˙ P )3 − (N˙ C )4 · CC and

0=1+

(N˙ N )1 − (N˙ N )3 (N˙ B )3 · CN − (N˙ S )1 − (N˙ S )3 (N˙ S )1 − (N˙ S )3 −

(N˙ P )3 (N˙ C )4 − · CC (N˙ S )1 − (N˙ S )3 (N˙ S )1 − (N˙ S )3

Finally, in terms of the yield factors, we obtain the simpler equation

1 + YN/S · CS = YB/S + YP/S + YC/S

Carbon balance

(15.7-1)

since in the notation used here to describe the flows into and out of the reactor,

YN/S =

(N˙ N )1 − (N˙ N )3 (N˙ S )1 − (N˙ S )3 YC/S =

YB/S =

(N˙ C )4 (N˙ S )1 − (N˙ S )3

and

YO2 /S =

(N˙ B )3 (N˙ S )1 − (N˙ S )3 YW/S =

YP/S =

(N˙ P )3 (N˙ S )1 − (N˙ S )3

(N˙ W )1 − (N˙ W )3 − (N˙ W )4 (N˙ S )1 − (N˙ S )3

(N˙ O2 )2 − (N˙ O2 )4 (N˙ S )1 − (N˙ S )3

Note that defined in this way, the yield factors should be positive, with the exception of YW/S , which may be positive if water is consumed in the reaction or negative if water is produced in the reaction.

942 Chapter 15: Some Additional Biochemical Applications of Thermodynamics The hydrogen balance is as follows:

0 = (N˙ S )1 · HS + (N˙ N )1 · HN + (N˙ W )1 · HW − (N˙ S )3 · HS − (N˙ N )3 · HN − (N˙ B )3 · HB − (N˙ P )3 · HP − (N˙ W )3 · HW − (N˙ W )4 · HW or

0 = HS +

(N˙ N )1 − (N˙ N )3 (N˙ W )1 −(N˙ W )3 −(N˙ W )4 · HN + · HW (N˙ S )1 − (N˙ S )3 (N˙ S )1 − (N˙ S )3 −

(N˙ B )3 (N˙ P )3 · HB − · HP (N˙ S )1 − (N˙ S )3 (N˙ S )1 − (N˙ S )3

and HS + YN/S · HN + YW/S · HW = YB/S · HB + YP/S · HP or Hydrogen balance

HS + YN/S · HN + YW/S · 2 = YB/S · HB + YP/S · HP

(15.7-2)

The nitrogen balance is

0 = (N˙ S )1 · NS + (N˙ N )1 · NN + (N˙ N2 )2 · NN2 − (N˙ S )3 · NS − (N˙ N )3 · NN − (N˙ B )3 · NB − (N˙ P )3 · NP − (N˙ N2 )4 · NN2 which reduces to NS + YN/S · NN = YB/S · NB + YP/S · NP

Nitrogen balance

(15.7-3)

In the last form of this equation we have assumed that the number of moles of N2 entering the reactor (for example, in the air) equals that leaving, since molecular nitrogen is not usually produced in a biochemical reaction. Finally, the oxygen balance is

0 = (N˙ S )1 · OS + (N˙ N )1 · ON + (N˙ W )1 · OW + (N˙ O2 )2 · OO2 −(N˙ S )3 · OS − (N˙ N )3 · ON − (N˙ B )3 · OB − (N˙ P )3 · OP − (N˙ W )3 · OW − (N˙ O2 )4 · OO2 − (N˙ W )4 · OW which reduces to Oxygen balance

OS + YN/S · ON + 2 · YO2 /S + YW/S = 2 · YC/S + YB/S · OB + YP/S · OP (15.7-4) The advantage of using the yield factor notation can be seen by comparing the rather complicated initial forms of each of these balance equations for the atomic species in terms of the flows or flow rates to the much simpler final equations in terms of the yield factors. Note that while the steady-flow fermenter was considered here, precisely the same equations, Eqs. 15.7-1, 2, 3, and 4, would be obtained from the analysis of one cycle of a batch fermenter.

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 943 The following two illustrations are based on data in the book by Roels. Illustration 15.7-3 Mass Balance Using the Yield Factors The yeast Saccharomyces cerevisiae is used in beer making and other fermentation processes to produce ethanol from the common plant sugar glucose in an anaerobic (no added oxygen) process in which dissolved ammonia is used as the nitrogen source. The production of biomass also occurs to the extent of 0.14 C-moles per C-mole of glucose. The biomass has an elemental composition of CH1.8 O0.5 N0.2 . What is the fractional conversion of glucose to ethanol (on a C-mole basis), the amount of ammonia used, and the amount of carbon dioxide produced per C-mole of glucose consumed?

Solution The chemical formula for glucose is C6 H12 O6 , which on a C-mole basis is CH2 O. The formula for ethanol is C2 H5 OH or CH3 O0.5 . Therefore, the set of algebraic equations to be solved is C-balance: 1 = YB/S + YP/S + YC/S H-balance:

2 + 3 · YN/S + 2 · YW/S = 1.8 · YB/S + 3 · YP/S

N-balance: and O-balance:

YN/S = 0.2 · YB/S

1 + YW/S = 2 · YC/S + 0.5 · YB/S + 0.5 · YrmP/S

The solution to this set of algebraic equations is YP/S = 0.569; that is, 0.569 C-moles of the product ethanol are produced for each C-mole of glucose consumed, or 1.707 ( = 0.569 × 6/2) moles of ethanol are produced per mole of glucose consumed. YC/S = 0.291; that is, 0.291 C-moles of carbon dioxide are produced for each C-mole of glucose consumed, or 1.748 ( = 0.291 × 6) moles of carbon dioxide are produced per mole of glucose consumed. YW/S = −0.063; that is, 0.063 moles of water are produced for each C-mole of glucose consumed, or 0.378 ( = 0.063 × 6) moles of water are produced per mole of glucose consumed. YN/S = 0.028; that is, 0.028 moles of ammonia are consumed for each C-mole of glucose consumed, or 0.168 ( = 0.028 × 6) moles of ammonia are consumed per mole of glucose consumed.

Illustration 15.7-4 Production of Saccharomyces cerevisiae The yeast Saccharomyces cerevisiae of the previous illustration is grown in a continuous, aerobic (oxygen-consuming) fermentation using glucose as the substrate and dissolved ammonia as the nitrogen source. It has been observed that the consumption of 4.26 C-moles of glucose results in 1 C-mole of the yeast and 1.92 C-moles of ethanol. Determine how much carbon dioxide is evolved and how much oxygen, ammonia, and water are consumed per C-mole of yeast produced. The elemental composition of the yeast is CH1.8 O0.56 N0.17 .

Solution From the problem statement, we have YB/S =

1 = 0.235 4.26

and

YP/S =

1.92 = 0.451 4.26

944 Chapter 15: Some Additional Biochemical Applications of Thermodynamics To calculate the other information needed, we use the four balance equations. C-balance: 1 = YB/S + YP/S + YC/S H-balance: 2 + 3 · YN/S + 2 · YW/S = 1.8 · YB/S + 3 · YP/S N-balance: YN/S = 0.17 · YB/S

and O-balance: 1 + 2 · YW/S + YW/S = 2 · YC/S + 0.56 · YB/S + 0.5 · YP/S

The solution to this set of equations is YC/S = 0.3145; that is, 0.3145 C-moles of carbon dioxide are produced for each C-mole of glucose consumed, or 1.887 ( = 0.3145 × 6) moles of carbon dioxide are produced per mole of glucose consumed. YN/S = 0.0399; that is, 0.0399 moles of ammonia are consumed for each C-mole of glucose consumed, or 1.748 ( = 0.291 × 6) moles of ammonia are consumed per mole of glucose consumed. YW/S = −0.1725; that is, 0.1725 moles of water are produced for each C-mole of glucose consumed, or 1.035 ( = 0.1725 × 6) moles of water are produced per mole of glucose consumed. YO2 /S = 0.07923; that is, 0.00792 moles of molecular oxygen are consumed for each C-mole of glucose consumed, or 0.475 ( = 0.0792 × 6) moles of O2 are consumed per mole of glucose consumed.

Available experimental data indicate that about 0.35 moles of O2 are consumed for each mole of yeast produced. Since 4.26 C-moles of glucose are required to produce 1 C-mole of yeast, it follows that 0.0792 × 4.26 = 0.3375 moles of molecular oxygen are required to produce 1 C-mole of yeast, which is close to the experimental value.

The energy balance for a steady-state bioreactor is

˙ 0 = (N˙ H)1 + (N˙ H)2 + (N˙ H)3 + (N˙ H)4 + Q˙ + W

(15.7-5)

where we have recognized that at steady state the reactor volume is not changing. Proceeding further, we have

0 = (N˙ S H S )1 + (N˙ N H N )1 + (N˙ W H W )1 + (N˙ O2 H O2 )2 + (N˙ N2 H N2 )2 − (N˙ S H S )3 − (N˙ N H N )3 − (N˙ W H W )3 − (N˙ B H B )3 − (N˙ P H P )3 ˙ − (N˙ O2 H O2 )4 − (N˙ N2 H N2 )4 − (N˙ CO2 H CO2 )4 + Q˙ + W Because of the number of flow streams and components involved, here we are departing from the usual notation in this textbook by writing the energy balance so that each flow rate of each species N˙ is positive, and the sign in front of the term indicates whether it is a flow into the reactor (+) or a flow out of the reactor (−). Generally, in chemical reactions, the effect of solution nonidealities on the enthalpy is small compared with the very much larger heats of reaction. Therefore, as an approximation, the partial molar enthalpies can be replaced by the pure-component enthalpies. Also, for simplicity, we will assume that the temperatures of the streams entering and leaving the fermenter are the same. (This may not be exactly true, but this effect will also be small compared with the heat of reaction, and could easily be included, if needed.) Finally, unless the solution is very viscous, the energy input from stirring is small compared with the heat of reaction. Therefore, we will also neglect the ˙ . work term W

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 945 With these assumptions, the energy balance is       0 = (N˙ S )1 − (N˙ S )3 HS + (N˙ N )1 − (N˙ N )3 HN + (N˙ W )1 − (N˙ W )3 HW     + (N˙ O2 )2 − (N˙ O2 )4 H O2 + (N˙ N2 )2 − (N˙ N2 )4 H N2 − (N˙ B )3 HB − (N˙ P )3 HP − (N˙ CO2 )4 H CO2 + Q˙ Note that in this equation, the enthalpies of oxygen, nitrogen, carbon dioxide, and water are on a per-mole basis, while that of the substrate, the product, and the biomass are on a per–C-mole basis. The enthalpy of the nitrogen source is on a per-mole basis if the source does not contain carbon (for example, ammonia), and on a per–C-mole basis for a carbon-containing nitrogen source (for example, glutamic acid C5 H9 NO4 , which on a C-mole basis is CH1.8 N0.2 O0.8 ). Next we use the yield factors defined earlier and introduce one additional yield factor

Q˙ (15.7-6) (N˙ S )1 − (N˙ S )3 to obtain the energy balance written on a per C-mole of substrate consumed basis and in terms of yield factors, YQ/S =

Bioreactor energy balance I

HS + YN/S HN + YO2 /S H O2 + YW/S HW + YN2 /S H N2 + YQ/S = YB/S HB + YP/S HP + YCO2 /S H CO2

(15.7-7)

The first difficulty in using the energy balance is that enthalpy of formation data may not be available, especially for the biochemical species that are not completely defined. Also, the substrate may be a mixture of substances, and an incompletely characterized mixture of products may be produced in a fermenter. The information that is more likely to be available are the elemental analysis and the heat of combustion (it is relatively easy to put any substance in a calorimeter and measure the heat released on combustion). As a simple example, suppose we wanted to know the enthalpy of benzene at 25◦ C. We could put one mole of benzene in an isothermal calorimeter with 7.5 moles of oxygen and provide an ignition source to cause the reaction

C6 H6 + 7.5O2 −→ 6CO2 + 3H2 O The energy balance is

0 = H C6 H6 + 7.5H O2 − 6H CO2 − 3H H2 O + Q where Q is the heat flow as a result of the combustion process, which we know from experience will be a negative number since heat is released on combustion, that is, Q = − c H . Here a negative sign is used because while there is an energy release on combustion, the heat of combustion is reported as a positive number and applies to complete combustion—that is, for all carbon being converted to carbon dioxide, all hydrogen to water, and all nitrogen to N2 (and other elements, if present, to their oxides)17 :

0 = H C6 H6 + 7.5H O2 − 6H CO2 − 3H H2 O − c H C6 H6 17 In fact, two heats of combustion may be reported for a compound, one in which all the water formed is present as

liquid, and a smaller value in which water is a vapor. The two values will differ by the product of heat of vaporization of water times the number of moles of water produced. We will generally use the liquid water reference state here.

946 Chapter 15: Some Additional Biochemical Applications of Thermodynamics It is important to note that when using heats of combustion, the reference states (i.e., the states of zero enthalpy) are nitrogen as molecular nitrogen, hydrogen as water, carbon as carbon dioxide, oxygen as molecular oxygen, and other elements that are present in their oxidized states (e.g., sulfur as SO2 ). This is different from the reference states used for heats of formation, in which each element is in its simplest pure, stable state. Therefore, in the example being considered, if all species enter and leave at 25◦ C,

H O2 = 0 H CO2 = 0 and H H2 O = 0 so that H C6 H6 = c H C6 H6 Now using the heat of combustion in the general energy balance for a bioreactor, we have

c HS + YN/S c HN + YQ/S = YB/S c HB + YP/S c HP

(15.7-8a)

YQ/S = YB/S c HB + YP/S c HP − c HS − YN/S c HN

(15.7-8b)

or Bioreactor energy balance II

Generalized degree of reduction

The next difficulty in using the energy balance is that thermodynamic property data, such as the heat of combustion, may not be known for some identified molecular species compounds, and is not likely to been measured for incompletely defined species such as the “average biomass” CH1.8 O0.5 N0.2 . To estimate the properties in these cases, we will use the generalized degree of reduction,18 ξ , defined as follows:

ξi =

4 × Ci + Hi − 2 × Oi Ci

for carbon-containing compounds

and

ξi = Hi − 2 × Oi

for compounds without carbon but with H or O atoms (15.7-9)

where the numerical coefficients in Eq. 15.7-9 are the valences of the atoms. (Note that there are alternative definitions of the generalized degree of reduction. In the one used here, nitrogen does not appear, so that ξ = 0 for N2 and ξ = 3 for NH3 . Also, ξ = 0 for O2 .) To a reasonable approximation, based on the data in Table 15.7-2 for the organic compounds,

kJ kJ (15.7-10) and Δc H = 110.9 · ξ C-mole C-mole Equations 15.7-10 are referred to as the energy regularity principles in the biochemical literature. Δc G = 112 · ξ

Illustration 15.7-5 Properties of Average Biomass The “average biomass” used to represent many yeasts and bacteria has an elemental composition, on a C-mole basis, of CH1.8 O0.5 N0.2 . Compute its generalized degree of reduction, and estimate its Gibbs energy and heat of combustion.

Solution The generalized degree of reduction for the average biomass is ξ=

18 V.

4×C+H−2×O

C

=

4 × 1+1.8 − 2 × 0.5 = 4.8 1

K. Eroshin and I. J. Minkevich, Biotech. Bioeng., 24, 2263 (1982).

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 947 Table 15.7-2 The Degree of Reduction ξ , Gibbs Energy, and Heat of Combustion of Biochemicals and Hydrocarbons on a Carbon Mole Basis (kJ/C-mole)a,b Compound Acetic acid Propionic acid Butyric acid Valeric acid Palmitic acid Lactic acid Oxalic acid Succinic acid Fumaric acid Malic acid Citric acid Glucose Ethanol i-Propanol n-Butanol Ethylene glycol Glycerol Glucitol Acetone Acetaldehyde Alanine Arginine Asparagine Glutamic acid Aspartic acid Glutamine Glycine Leucine Isoleucine Phenylalanine Serine Threonine Trytophane Tyrosine Valine Guanine n-Hexane (l) n-Heptane (l) n-Octane (l) n-Decane (l) n-Dodecane (l) n-Hexadecane (l) n-Eicosane (l) Toluene (l) Cyclohexane (l) Ethyl acetate (l) Average

ξ

c G (kJ/C-mole)

c H (kJ/C-mole)

4.00 4.67 5.00 5.20 5.75 4.00 1.00 3.50 3.00 3.00 3.00 4.00 6.00 6.00 6.00 5.00 4.67 4.33 5.33 5.00 5.00 5.67 4.50 4.20 3.75 4.80 4.50 5.50 5.50 4.78 4.33 4.75 4.73 4.56 5.40 4.60 6.33 6.29 6.25 6.20 6.17 6.13 6.10 5.14 6.00 5.00

447.0 511.0 543.3 562.6 612.5 459.0 163.5 399.8 362.0 361.0 357.8 478.7 659.5 648.7 648.0 585.0 547.7 514.0 578.0 561.5 547.3 631.0 499.8 463.0 421.5 525.6 505.5 594.2 594.0 516.3 500.7 532.5 513.6 498.1 584.0 522.4 670.5 665.7 662.2 657.1 653.7 649.5 646.8 547.7 636.1 552.0

438.0 509.7 548.5 568.2 624.3 456.3 123.0 373.3 334.3 332.3 327.2 467.8 684.5 663.0 670.0 590.5 554.3 508.2 597.7 584.0 569.0 624.0 484.0 450.0 402.0 514.0 487.0 598.0 598.0 517.0 485.0 526.0 512.0 493.0 584.0 500.0 693.9 688.1 683.9 677.7 673.5 667.9 663.9 563.0 652.0 563.6

c G/ξ

c H/ξ

111.8 109.5 108.7 108.3 106.5 114.8 163.5 114.2 120.7 120.3 119.3 119.7 109.9 108.1 108.0 117.0 117.4 118.6 108.4 112.3 109.5 111.4 111.1 110.2 112.4 109.5 112.3 108.0 108.0 108.0 115.6 112.1 108.6 109.3 108.2 113.6 105.9 105.9 106.0 106.0 106.0 106.0 106.0 106.6 106.0 110.0 112.0

109.5 109.2 109.7 109.3 108.6 114.1 123.0 106.6 111.4 110.8 109.1 117.0 114.1 110.5 111.7 118.1 118.8 117.3 112.1 116.8 113.8 110.1 107.6 107.1 107.2 107.1 108.2 108.7 108.7 108.2 111.9 110.7 108.3 108.2 108.2 108.7 109.6 109.5 109.4 109.3 109.2 109.0 108.8 109.7 108.7 112.7 110.9

948 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Table 15.7-2 (Continued) The Degree of Reduction ξ , Gibbs Energy, and Heat of Combustion of Biochemicals and Hydrocarbons on a Carbon Mole Basis (kJ/C-mole)a,b Compound Other compounds Ammonia (l, 0.01 M) Hydrogen Carbon monoxide Nitric acid Hydrazine Hydrogen sulfide Sulfuric acid

ξ

c G (kJ/C-mole)

c H (kJ/C-mole)

3.00 2.00 2.00 −5.00 4.00 2.00 −6.00

391.9 238.0 257.0 7.3 602.4 323.0 −507.4

348.1 286.0 283.0 −30.0 622.0 247.0 −602.0

c G/ξ

c H/ξ

130.6 119.0 128.5 −1.5 150.6 161.5 84.6

116.0 143.0 141.5 6.0 155.6 123.5 100.3

a The heat and Gibbs energy of combustion are based on liquid water and gaseous carbon dioxide and nitrogen as combustion products. Also, the Gibbs energy is based on liquid phase reactants and products at unit molality. b Based on a table in Roels.

Therefore, from the energy regularity approximation Δc G = 112 · ξ = 112 · 4.8 = 537.6

kJ

C-mole

and Δc H = 110.9 · ξ = 110.9 · 4.8 = 532.3

kJ

C-mole

While measured enthalpies or heats of combustion should be used whenever possible, the energy regularity estimate for the heat of combustion can be used in the energy balance, Eq. 15.9-8, for those species for which such data are not available. The most approximate form for the energy balance obtained when using the energy regularity estimate for all species is Approximate bioreactor energy balance

YQ/S = 110.9 (YB/S ξB + YP/S ξP − ξS − YN/S ξN ) kJ

(15.7-11)

for each C-mole of substrate consumed. Illustration 15.7-6 Energy Balance on an Isothermal Fermenter In Illustration 15.7-4 we considered the production of ethanol from glucose using the yeast Saccharomyces cerevisiae. In that illustration 0.451 C-moles of ethanol and 0.235 C-moles of biomass of elemental composition CH1.8 O0.56 N0.17 were produced per C-mole of glucose, using 0.0399 moles of ammonia. Assuming the inlet and outlet streams are maintained at 25◦ C, and that the work input is negligible, what is the heat load on the reactor?

Solution From Table 15.7-2, the heat of combustion of ammonia is 383.0 kJ/mol, that of glucose is 467.8 kJ/C-mole, and ethanol is 684.5 kJ/mol. The heat of combustion of the biomass is unknown, so the energy regularity method will be used: ξ = 4 × 1 + 1 × 1.8 − 2 × 0.56 = 4.68

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 949 Therefore, Δc H = 110.9 · ξ = 110.9 · 4.68 = 519.0

kJ

C-mole

So that, for each C-mole of glucose consumed, YQ/S = 0.235 × 519.0 + 0.451 × 684.5 − 467.8 − 0.0399 × 348.1 = −51.0 kJ

of heat must be removed.

Note It is useful to compare this answer with the more approximate one that is obtained using the generalized degree of reduction for all components: YQ/S = 110.9 (0.235 × 4.68 + 0.451 × 6.00 − 4.00 − 0.0399 × 3) = −34.8 kJ

per C-mole of glucose consumed. The difference between the two results largely arises from the error in the generalized degree of reduction approximation for the heat of combustion of glucose (4 × 110.9 = 443.6 kJ/C-mole, compared with the experimental value of 467.8).

There is one linear combination of the carbon, nitrogen, and oxygen balances that is especially useful. If we take the sum of four times the carbon balance plus the hydrogen balance and subtract twice the oxygen balance (note that the multipliers are the valences, and are the same numbers that appear in the definition of the generalized degree of reduction), we obtain Alternative form of oxygen balance

(4 + HS − 2OS ) + (4 + HN − 2ON ) YN/S − 4YO2 /S = (4 + HB − 2OB ) YB/S + (4 + HP − 2OP ) YP/S or ξS + ξN YN/S − 4YO2 /S = ξB YB/S + ξP YP/S and YO/S =

ξS + ξN YN/S − ξB YB/S − ξP YP/S 4

(15.7-12)

This last equation has several uses, as we will see shortly. Now we move on to the entropy balance. Following the same type of analysis used for the energy balance on a bioreactor, we obtain the following equation for the entropy balance:       0 = (N˙ S )1 − (N˙ S )3 S S + (N˙ N )1 − (N˙ N )3 S N + (N˙ W )1 − (N˙ W )3 S W     + (N˙ O2 )2 − (N˙ O2 )4 S O2 + (N˙ N2 )2 − (N˙ N2 )4 S N2 − (N˙ B )3 S B

Q˙ − (N˙ P )3 S P − (N˙ CO2 )4 S CO2 + + S˙ gen (15.7-13) T For an isothermal system, we can eliminate the heat flow between the energy and entropy balances, and use the partial molar Gibbs energy Gi = H i − T S i to obtain       0 = (N˙ S )1 − (N˙ S )3 GS + (N˙ N )1 − (N˙ N )3 GN + (N˙ W )1 − (N˙ W )3 GW     + (N˙ O2 )2 − (N˙ O2 )4 GO2 + (N˙ N2 )2 − (N˙ N2 )4 GN2 − (N˙ B )3 GB − (N˙ P )3 GP − (N˙ CO2 )4 GCO2 − T S˙ gen

(15.7-14)

950 Chapter 15: Some Additional Biochemical Applications of Thermodynamics The use of this equation is more complicated than using the energy balance because the partial molar Gibbs energy contains both the activity coefficient and the mole fraction of each component (which in the case of biomass is really a complex mixture); that is,

Gi = Gi + RT ln (xi γi )

(15.7-15)

Evaluation of this logarithmic term is very difficult since the mole fractions are not likely to be known (we may know only the elemental compositions of the biomass and perhaps other species, but not their molecular formulae), and little or no information is available on solution nonidealities in the mixtures containing yeast, bacteria, or other incompletely characterized substances. Consequently, we will make what might appear to be a rather severe assumption, that the logarithmic term can be neglected—that is, that we can replace the partial molar Gibbs energies with the pure-component Gibbs energies. Why is such an assumption not unreasonable here? There are four reasons: 1. There are an equal number of inlet and outlet streams, and each one has a term of this form, so there will be some degree of cancellation of this term among the four streams. 2. The contribution of any one species to the Gibbs energy of the stream is Ni (Gi + RT ln xi γi ). When N˙ i is large, the mole fraction of that species will be near 1, so that the logarithmic term will be small. When the mole fraction of the species is very small (so that the logarithmic term is larger), the flow term N˙ i multiplying that term is small. Therefore, in both cases the contribution will not be large. 3. The value of RT that is the multiplier for the logarithmic term has a value of 2.5 kJ/mole, which, while not insignificant, is much smaller in value than the pure-component Gibbs energy terms. 4. The logarithmic terms being neglected here are of great importance in determining the equilibrium state, as was shown in Chapter 13. However, in manufacturing processes chemical reactors, fermenters, and other bioreactors are not generally operated to achieve equilibrium as the residence time required, and therefore the reactor size, would be so large as to be uneconomical. For the reasons stated in points 1, 2, and 3, these logarithmic terms are of less importance away from the equilibrium state. (Note that the comments made here about the contributions of the solution nonidealities and the mixing terms are similar to those made in Sec. 14.5 in discussing calculations involving the availability function.) Therefore, it is reasonable to simplify the equation to       0 = (N˙ S )1 − (N˙ S )3 GS + (N˙ N )1 − (N˙ N )3 GN + (N˙ W )1 − (N˙ W )3 GW     + (N˙ O2 )2 − (N˙ O2 )4 GO2 + (N˙ N2 )2 − (N˙ N2 )4 GN2 − (N˙ B )3 GB − (N˙ P )3 GP − (N˙ CO2 )4 GCO2 − T S˙ gen

(15.7-16)

(where G is a pure-component Gibbs energy on a per–C-mole basis). Now rewriting this equation in terms of the yield factors gives

0 = GS + YN/S GN + YO2 /S GO2 + YW/S GW + YN2 /S GN2 + YB/S GB + YP/S GP + YCO2 /S GCO2 − T S gen

(15.7-17)

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 951 and S gen is the entropy generated per C-mole of substrate consumed. Here, as with the energy balance, we can use Gibbs energies of combustion (see Table 15.7-2) in place of the Gibbs energies and simplify the equation to

c GS + YN/S c GN = YB/S c GB + YP/S c GP + T S gen

Second-law balance

(15.7-18)

Our interest in this equation will be in determining the constraint that thermodynamics (and, in particular, the second law or entropy balance) places on the maximum conversion possible of substrate to product and additional biomass. Based on discussions elsewhere in this textbook, the maximum production of product will occur when the process is operated such that Sgen = 0. However, for real processes, Sgen ≥ 0. Therefore, the form of the equation that will be used is

c GS + YN/S c GN ≥ YB/S c GB + YP/S c GP

Second-law constraint

(15.7-19)

Finally, for the purposes of making an approximate calculation of the maximum possible conversion, we can use the energy regularity correlation and obtain the following constraint based on a C-mole of substrate consumed: Approximate second-law constraint

ξS + ξN YN/S ≥ ξB YB/S + ξP YP/S

(15.7-20)

The interesting and important result of Eq. 15.7-19 (and its simplification, Eq. 15.7-20) is that it provides a thermodynamic limit on the amount of product and biomass that can be produced from a given substrate and nitrogen source based on the Gibbs energy (or availability) of the reactants. One should recognize that the constraint of Eq. 15.7-19 has a simple interpretation, which is that the Gibbs energy of the substrate and nitrogen source consumed must exceed (or at least equal) that of the biomass and products produced. Illustration 15.7-7 Second-Law Limitation on a Fermenter In Illustration 15.7-4, 0.451 C-moles of ethanol and 0.235 C-moles of biomass were produced per C-mole of glucose consumed. How does this compare with the second-law limit on the production of ethanol and biomass?

Solution Using the data in Table 15.7-2, we have per C-mole of glucose consumed 478.7 + 0.0399 × 391.9 ≥ 0.235 × 4.68 × 112 + 0.451 × 659.5 kJ

or 494.3 > 420.6 kJ

So we conclude that the fermentation satisfies the second law of thermodynamics. Further, we can say that 420.6 × 100 = 85.1 percent of the Gibbs energy in the glucose and dissolved ammonia 494.3 appears in the products of ethanol and biomass. Presumably the rest is a result of inefficiencies in the conversion process, and in the respiration and metabolism processes to keep the yeast alive. Since the second-law constraint has been satisfied, the data reported for the production of ethanol from glucose using the yeast Saccharomyces cerevisiae are thermodynamically consistent.

952 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Note It is of interest to examine the simplified version of the second-law constraint in terms of the generalized degrees of reduction. In this case we have 4.00 + 3 × 0.0399 = 4.1197 > 4.68 × 0.235 + 6.00 × 0.451 = 3.8058

so again the second-law constraint is satisfied, though this approximate result gives a somewhat higher thermodynamic efficiency of 3.8058 × 100 = 92.4 percent. 4.1197

Illustration 15.7-8 The Thermodynamic Analysis of Fermentation Data In Illustration 15.7-3 the yeast Saccharomyces cerevisiae was produced anaerobically (without oxygen) from the fermentation of glucose using ammonia as the nitrogen source. In that illustration it was found using the atom balance that YP/S = 0.569

YW/S = −0.063

YB/S = 0.14

and

YN/S = 0.028

Determine the heat load required to keep the fermenter at a constant temperature of 25◦ C, with the reactants entering at that temperature; whether the data satisfy the second-law consistency requirement; and the fraction of the Gibbs energy of the reactants that appears in the products. All calculations should be on a per C-mole of glucose consumed basis.

Solution The heat load on the fermenter is found using the energy balance, Eq. 15.7-8, YQ/S = 0.14 × 532.3 + 0.569 × 684.5 − 467.8 − 0.028 × 348.1 = −13.5 kJ

So 13.5 kJ per C-mole of glucose consumed, or 6 × 14.5 = 81 kJ per mole of glucose consumed, must be removed from the fermenter to keep the temperature constant. Note that if the glucose was merely burned, 467.8 kJ per C-mole of glucose would have to be removed to keep the temperature at 25◦ C. To see if the second-law constraint is satisfied, Eq. 15.7-19 is used: ?

478.7 + 0.028 × 391.9 > 0.14 × 537.6 + 0.569 × 659.5

or 489.6 > 450.2 kJ

Therefore, the second-law constraint is satisfied. Finally, 450.5 × 100 = 92.0 percent of the Gibbs 489.6 energy of the feed appears in the two products, ethanol and the yeast.

A complete thermodynamic analysis of a fermenter involves the following: a. Elemental mass balances for each of the atoms present (usually carbon, hydrogen, nitrogen, and oxygen, but perhaps also including sulfur, phosphorus, and other trace atoms that may be present in proteins and other biomass); b. The energy balance (Eq. 15.7-8 or 15.7-11); and c. The second-law constraint (Eq. 15.7-19 or 15.7-20). All of these can be used together to decide on the effect of using different substrates to carry out a fermentation to produce a specified product and biomass. What we expect is that different amounts of product and biomass will be produced depending on the substrate chosen, and that the thermodynamic analysis in this section can provide some guidance. This is demonstrated in the illustration below.

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 953 Illustration 15.7-9 Determining the Maximum Amount of Product That Can Be Obtained from a Substrate A biologist is trying to genetically engineer bacteria to produce valuable organic chemicals from the inexpensive organic sugar xylose found in a variety of plants and berries. One such organism appears capable of producing 2,3-butanediol. The following data are available. Chemical

Formula

c H , J/mol

2,3-butanediol Xylose

C4 H10 O2 C5 H10 O5

2461.0 2345.2

a. Test the generalized degree of reduction approximation for each of these compounds using the reported heat of combustion data. b. Estimate the maximum amount of 2,3-butanediol that can be be produced per mole of xylose consumed assuming no additional biomass production. c. Assuming that no biomass is produced (and therefore no nitrogen source is needed), determine the amount of carbon dioxide produced and oxygen and water consumed as a function of the amount of 2,3-butanediol produced. d. Determine the amount of heat produced as a function of the amount of 2,3-butanediol produced.

Solution a. The definition of the generalized degree of reduction is ξ=

4·C+H−2·O

C

so that ξX =

4 × 5 + 10 − 2 × 5 =4 5

and

ξB =

4 × 4 + 10 − 2 × 2 = 5.5 4

Using the correlation of Eq. 15.7-10, we have for 2,3-butanediol

c H B = 110.9 × 5.5

kJ

C-mole

×4

C-mole mole

= 2439.8

kJ mole

compared with the measured value of 2461.0 kJ/mole, and for xylose

c H X = 110.9 × 4

kJ

C-mole

×5

C-mole mole

= 2218

kJ mole

compared with the measured value of 2345.2 kJ/mole. So that while the correlation is not exact, the estimates are reasonable. b. Since there is no nitrogen source, no products other than 2,3-butanediol, and no biomass produced, the second-law constraint, Eq. 15.9-20, reduces to ξS ≥ ξP YP/S

so that here

YP/S ≤

4 ξS = 0.727 = ξP 5.5

So there is a limit of 0.727 C-moles of 2,3-butanediol produced per C-mole of xylose consumed. Since xylose has five carbons and a molecular weight of 150, one C-mole of xylose is 30 g. Similarly 2,3-butanediol has four carbon atoms and a molecular weight of 90, so one C-mole is 22.5 g. Therefore, 0.727 C-moles of 2,3-butanediol per C-mole of xylose equals 0.727 × 22.5/30 = 0.545 g 2,3-butanediol per g of xylose, or 545 g/kg of xylose.

954 Chapter 15: Some Additional Biochemical Applications of Thermodynamics c. The overall fermentation reaction can be written as CH2 O + YO2 /S O2 + YW/S H2 O → YP/S CH2.5 O0.5 + YC/S CO2

The carbon mass balance is 1 = YP/S + YC/S

or

YC/S = 1 − YP/S

or

YW/S = 1.25YP/S − 1

The hydrogen mass balance is 2 + 2YW/S = 2.5YP/S

The oxygen mass balance can be developed two different but equivalent ways. The first is 1 + YW/S + 2YO/S = 0.5YP/S + 2YC/S

which, substituting in the carbon and hydrogen balances, reduces to YO/S = 1 − 1.375YP/S

The second method of obtaining the oxygen balance comes from Eq. 15.7-12, which here reduces to YO/S =

ξS − ξP YP/S 4 − 5.5YP/S = = 1 − 1.375YP/S 4 4

The results are shown in Fig. 15.7-2. Note that this figure and the one that follows end at YP/S = 0.727, which is the second-law (or availability) limit for the maximum production of 2,3-butanediol.

1 YC/S 0.5

YO 2 /S

0

–0.5

–1

YW/S

0

0.2

0.4 YP/S

0.6

Figure 15.7-2

d. The energy balance, from Eq. 15.7-11, is



YQ/S = 110.9 YB/S ξB + YP/S ξP − ξS − YN/S ξN = 110.9 5.5YP/S − 4 kJ

The result is shown in Fig. 15.7-3.

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 955 –200

YQ/S kJ/C-mole

–300 –400 –500 –600 –700 0

0.2

0.4 YP/S

0.6

Figure 15.7-3 Heat flow as a function of 2,3-butanediol formed.

Illustration 15.7-10 Choosing a Substrate for a Fermentation An unspecified substrate that does not contain nitrogen, characterized only by its generalized degree of reduction ξS , together with dissolved ammonia as the nitrogen source is to be fermented to produce “average biomass” and no other products. As a function of the generalized degree of reduction of the substrate, determine the following: a. The maximum product yield per C-mole of substrate consumed if the amount of ammonia is limited b. The maximum product yield per C-mole of substrate consumed if the amount of oxygen is limited c. The second-law limitation on the amount of product that can be formed d. The heat that must be removed to keep the fermenter at a constant temperature per C-mole of substrate consumed

Solution This fermentation is written as Substrate + YN/S NH3 + YO2 /S O2 + YW/S H2 O → YB/S CH1.8 O0.5 N0.2 + YC/S CO2

a. The nitrogen balance is YN/S = YB/S · 0.2

or

YB/S = 5 · YN/S

so that if the nitrogen source, ammonia, is limited, there is a mass balance limitation on the amount of biomass (cells) that can be produced. This limit is shown as solid lines in the figure that follows. b. For the oxygen balance, we use Eq. 15.9-12, which here becomes ξS + 35 YB/S − 4.8YB/S ξS + ξN YN/S − ξB YB/S ξS + 3YN/S − 4.8YB/S = = 4 4 4 ξS − 4.2YB/S = 4

YO2 /S =

956 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Values of YB/S as a function of ξS and YO2 /S are also plotted in the figure as dashed-dotted lines. Of special interest is the YO2 /S = 0 line (corresponding to anaerobic fermentation), which is YB/S = 0.2381 · ξS

YO2 /S = 0

We will come back to this result shortly. The carbon balance is 1 = YB/S + YC/S

or

YB/S = 1 − YC/S

Note that the result of this constraint is independent of the properties of the substrate, and merely requires that for the case here in which there is no other product, the carbon in the substrate can be completely converted to biomass if there is no carbon dioxide evolved. If there is CO2 evolution, only a fraction of the substrate will appear as biomass. (We cannot do a hydrogen balance because we do not know the hydrogen content of the substrate; only its degree of reduction is being specified.) c. The last restriction is the second-law constraint, which for the case here of no product, and with the Gibbs energy of the dissolved ammonia known but that of the substrate and biomass unknown, has the form 112 · ξS + YN/S · 391.9 ≥ 112 · ξB · YB/S

Now using the nitrogen balance, the second-law maximum conversion of substrate to biomass is 112 · ξB · YB/S = 112 · ξS + 0.2 · YB/S · 391.9 or (112 · 4.8 − 0.2 · 391.9) YB/S = 112 · ξS

or

YB/S = 0.2439 · ξS

This constraint is also plotted in the following figure. Note that it is almost identical to the YO2 /S = 0 line, and differs from it only to the extent that the dissolved ammonia is not exactly described by the energy regularity principle. The interpretation of this figure is as follows. For any substrate, characterized only by its degree of reduction ξS , the maximum conversion to biomass, YB/S , is governed by whichever is the limiting constraint: 1. The availability of a nitrogen source; 2. The availability of oxygen; or 3. The second law constraint. In particular, assuming complete availability of a nitrogen source, the maximum conversion max of a substrate of a given value of ξS to biomass for this system, YB/S , can be determined from the second-law constraint line. However, since no fermentation or other biological process operates without irreversibilities (i.e., the generation of entropy), the actual max biomass production will be less than the theoretical maximum, that is, YB/S ≥ YB/S . d. Finally, the energy balance on the fermenter (using the nitrogen balance) is YQ/S (kJ/C-mole) = 110.9YB/S ξB − 110.9ξS − 348.1YN/S = 532.3YB/S − 110.9ξS − 0.2 · 348.1YB/S = 462.7YB/S − 110.9ξS kJ

The relationship between YQ/S , YB/S , and ξS is also plotted in the figure as the light dotted line.

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 957 1.2 Carbon limitation 1

0

0.8

Y

Q /S

=–

25

0

YN/S = 0.2

0

2

kJ /C

Y

O

-m

/S

ol

=

Y n,

00 –5 Q

Y

/S

Y

Q

/S

2

Y

O

/S

=

–7

=1 .5

YN/S = 0.05

0.2

50

Y

2

O

/S =

=1 .0

.5 =0 /S 2

O

Y

Y

O

2 /S

2n d

0.4

YN/S = 0.1

=0 .25

lim ita tio

0.6

-la w

YB/S

O

/S

=

YN/S = 0.15

0 0

1

2

3

4 ξS

5

6

7

8

Figure 1 Second-law limitation, oxygen YO2 /S and nitrogen source YN/S consumption, and heat release YQ/S as a function of degree of reduction of the substrate ξS and biomass production YB/S .

Before leaving this section, it is useful to consider two further implications of the second-law constraint on bioreactors. To do this, we will consider the approximate forms of the energy balance and the second-law constraint based on the energy regularity assumption. While the simple conclusions obtained will not be exact, the errors should not be great. For simplicity, we first repeat the oxygen and energy balances and the second-law constraint here. Oxygen balance

YO2 /S =

ξS + ξN YN/S − ξB YB/S − ξP YP/S 4

(15.7-21)

Energy balance

YQ/S = 110.9 (YB/S ξB + YP/S ξP − ξS − YN/S ξN ) kJ/C-mole

(15.7-22)

and the second-law constraint written as

ξS + ξN YN/S ≥ ξB YB/S + ξP YP/S

(15.7-23)

ξS + ξN YN/S − ξB YB/S − ξP YP/S ≥ 0

(15.7-24)

The important thing to notice is that the same term that appears in the second-law constraint and must be greater than or at best equal to zero also appears in the oxygen balance and the energy balance. By comparing these equations, we conclude that as a result of the second law of thermodynamics (and the energy regularity assumption),

YO2 /S ≥ 0

958 Chapter 15: Some Additional Biochemical Applications of Thermodynamics that is, oxygen can only be consumed, and not produced by a fermentation process, and

YQ/S ≤ 0 so that all fermentation processes are exothermic; that is, cooling is needed to keep the temperature constant. Finally, by comparing Eqs. 15.7-11 and 15.7-12, we also conclude that

YQ/S = −443.6YO2 /S

kJ C-mole substrate consumed

(15.7-25)

that is, in a fermentation there is a linear relation between the oxygen consumption and the heat released. In fact, such a linear relationship has been found in laboratory measurements19 . From Eq. 15.7-18 we have

S gen =

( c GS + YN/S c GN − YB/S c GB − YP/S c GP ) T

(15.7-26a)

or more approximately, using the energy regularity principle,

S gen =

YO /S YQ/S (ξS + ξN YN/S − ξB YB/S − ξP YP/S ) =4 2 =− T T 110.9T

(15.7-26b)

These equations can be used to compute the entropy generation of a biochemical process. Illustration 15.7-11 Entropy Generation in a Fermentation The following data have been reported by von Stockar and Birou20 for the fermentation of glucose by K. fragilis at 25◦ C to produce ethanol as the product and biomass of elemental composition CH1.75 O0.52 N0.15 : YB/S = 0.572

YCO2 /S = 0.398

and

YQ/S = −180.9

kJ

C-mole

Compute the following per C-mole of substrate consumed: a. b. c. d. e.

The amount of product formed (YP/S ) The amount of ammonia consumed (YN/S ) The amount of oxygen consumed (YO2 /S ) The amount of heat released (YQ/S ) The amount of entropy generated (Sgen )

Solution To begin, we compute the generalized degree of reduction of the biomass produced: ξB = 4 × C + H − 2 × O = 4 + 1.75 − 2 × 0.52 = 4.71

19 C. 20 U.

L. Cooney, D. I. C. Wang, and R. I. Mateles, Biotech. Bioeng., 11, 269 (1968). von Stockar and B. Birou, Biotech. Bioeng., 34, 86 (1989).

15.7 Thermodynamic Analysis of Fermenters and Other Bioreactors 959 Therefore, for the biomass,

c GB = 112 × ξB = 527.5

kJ

C-mole

and

c HB = 110.9 × ξB = 522.3

kJ

C-mole

The heats and Gibbs energies of all other components are available in Table 15.7-2. a. By the carbon balance, 1 = YB/S + YP/S + YCO2 /S

so that

YP/S = 1 − YB/S − YCO2 /S = 1 − 0.572 − 0.398 = 0.030

b. By the nitrogen balance, YN/S = 0.15 × YN/S = 0.15 × 0.572 = 0.086

c. From Eq. 15.7-12, we have ξS + ξN · YN/S − ξB · YB/S − ξP · YP/S 4 4 + 0.086 × 3 − 4.71 × 0.572 − 0.030 × 6 = = 0.346 4

YO2 /S =

d. From the energy balance, YQ/S = YB/S · c HB + YP/S · c HP − c HS − YN/S · c HN = 0.572 · 522.3 + 0.030 · 684.5 − 467.8 − 0.086 · 348.1 = −178.5

kJ

C-mole

e. From the entropy balance, we have 1 ( c GS + YN/S · c GN − YB/S · c GB − YP/S · c GP ) T (478.7 + 0.086 · 391.9 − 0.572 · 527.5 − 0.030 · 684.5) kJ = = 0.640 298.15 C-mol K

S gen =

Finally, it is of interest to compare the heat release calculated in part (d) above, YQ/S = −178.5 kJ/ C-mole, with the more approximate result from Eq. 15.9-22: YQ/S = −443.6 × YO/S = −443.6 × 0.346 = −153.5

kJ

C-mole

Though the approximate result is not very accurate, it is not unreasonable given the simplicity of the calculation.

The results of this illustration, and previous ones in this section, show that the methods of thermodynamics are just as applicable to biochemical processes as they are to more classical problems in chemical engineering. As a final comment on this section it is interesting to note that Eq. 15.7-22 established a linear relationship between the heat released in fermentation and the oxygen consumed. The following similar linear correlation exists in the mammalian animal world between the average metabolic rate (directly proportional to the energy released as heat, muscular work, and in basic metabolic processes) and the rate of oxygen consumed:

 Metabolic rate

kJ day



= 21 800 × Oxygen consumption

m3 day



960 Chapter 15: Some Additional Biochemical Applications of Thermodynamics This correlation is shown in Figure 15.7-5 for a number of warm-blooded (39◦ C) mammals, including a shrew (average body weight of 0.0048 kg), a cat (2.5 kg), dog (11.7 kg), a human (70 kg), a horse (650 kg), and an elephant (3833 kg). It is amusing to note that this correlation spans animals whose body weight varies by almost a factor of one million! [There are similar but different correlations for cold-blooded (20◦ C) animals, which have lower heat losses and slower metabolic rates.] Using this correlation, since the average human male consumes about 15 liters of oxygen per hour (0.36 m3 per day), we estimate that the average human metabolic energy release is about 7850 kJ/day or 1875 kcal/day, which is close to the recommended daily food allowance for weight control. 106

Metabolic rate, kJ/day

105 104 103 100 10 1 – 10 4

–

0.01 0.1 10 10 3 1 Oxygen consumption, m3/day

100

Figure 15.7-5 Correlation between average metabolic rate and oxygen consumption for warm-blooded animals.

15.8 GIBBS-DONNAN EQUILIBRIUM AND MEMBRANE POTENTIALS In Sec. 11.5 we considered the osmotic pressure that arose between two cells separated by a membrane that was permeable to some components (there the solvent) but not permeable to the solute. Here we revisit the phenomenon of osmotic pressure for a somewhat more complicated case in which the solvent (usually water) contains a strong electrolyte, such as sodium hydroxide

NaOH −→ Na+ + OH− the ions of which can pass through the membrane, and a protein that ionizes to a net charge of Z− P (H)Z ⇐⇒ P Z− + ZH+ that cannot pass through the membrane, though the hydrogen ions produced can. There are also hydrogen and hydroxyl ions present that result from the ionization of water; however, those concentrations are very low and can be neglected. Consequently, the hydrogen ions formed by the ionization of the protein can react with the hydroxyl ions to form water. By charge neutrality the molality of the free hydroxyl ions is related to the molality of the sodium ions and the molality of the ionized protein as follows:

MOH = MNa − ZMP

15.8 Gibbs-Donnan Equilibrium and Membrane Potentials 961 For notational simplicity, we will not use the superscripts + and − to indicate the charges of the ions. The situation we will consider is that in which the water, sodium ions, and hydroxide ions can pass through the membrane, but the protein cannot. We will designate the protein-free side of the membrane with the superscript I and the protein-containing side by II. Each of the compartments must itself be electrically neutral (because considerable energy is required to create a charge imbalance). Therefore, in compartment I, in which there is no protein, we have (using molality for the unit of concentration) I I MNa = MOH (15.8-1) and in compartment II II II MNa = ZMPII + MOH

(15.8-2)

Because both the sodium and hydroxyl ions are free to pass through the membrane, their partial molar Gibbs energies are the same in both compartments:

GINa = GII Na

and GIOH = GII OH

(15.8-3a)

However, here as in Sec. 9.10, the partial molar Gibbs energies of each species cannot be separately determined, so in fact we use II GINa + GIOH = GII Na + GOH

(15.8-3b)

 I 2  II 2 I I II II γ± = MNa γ± MNa MOH MOH

(15.8-3c)

which leads to

For simplicity, we will neglect the activity coefficients (or assume they cancel from both sides of the equation) so that  I 2 I I II II MNa MOH = MNa = MNa MOH (15.8-4) since the sodium ion and hydroxyl ion concentrations are equal in compartment I. Combining Eq. 15.8-2 and 15.8-4, we obtain 2

II I MNa = MNa ·

or

I MOH (M I ) = II Na II MOH MNa − Z · MPII

 II 2  I 2 II − MNa =0 MNa − Z · MPII · MNa

(15.8-5)

which has the solution Gibbs-Donnan equilibrium

II MNa

⎡ ⎤ 2  Z · MPII ⎣ Z · MPII 2 I + = + (MNa ) ⎦ 2 2 

and II MOH =

Z · MPII 2

2 2

I + (MNa ) −

Z · MPII 2

(15.8-6a)

(15.8-6b)

962 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Therefore, if the sodium hydroxide molality in compartment I and the protein molality in compartment II are known, the sodium and hydroxyl ion molalities in the proteincontaining compartment II can be computed. We see from these equations that if the protein is not ionized (i.e., Z = 0), the sodium ion concentrations and the hydroxide ion concentrations will be the same in compartments I and II. Also, if the sodium hydroxide concentration is much greater than the protein concentration (and especially I I = MNaOH  Z · MPII ), the sodium ion concentrations and the hydroxyl ion if MNa concentrations will be the same in compartments I and II. Under any other circumstance, the sodium and hydroxyl ion concentrations will be different in the two compartments. This is the Gibbs-Donnan equilibrium effect. (It is left to the reader to develop the equations for the case in which a positively charged protein is produced on ionization. See Problem 15.25.) In this analysis it has been assumed that the activity coefficients are unity in going from Eq. 15.8-3c to Eq. 15.8-4. Including the activity coefficients would make the analysis and final results more complicated. However, the correction is usually small since the activity coefficients in both compartments may be close in value, and so may cancel in Eq. 15.8-3c. Illustration 15.8-1 Gibbs-Donnan Equilibrium of a Lysozyme Solution One compartment in an osmotic cell contains lysozyme, which has a molecular weight of 14 000 at a concentration of 50 mg/g water. a. The second compartment contains a sodium hydroxide solution that is maintained at 1.585 × 10−3 M, which keeps the pH at 11.2, the isoelectric point of lysozyme. Find the concentration of sodium and hydroxyl ions in the lysozyme-containing compartment. b. The second compartment contains a sodium hydroxide solution that is maintained at 0.1 M, which keeps the pH at 13.0. Find the concentration of sodium and hydroxyl ions in the lysozyme-containing compartment.

Solution a. At a pH = 11.2, which is the isoelectric point of lysozyme, its net charge is zero, so that from the equations above, the sodium and hydroxide ion concentrations in compartment II will be the same as in compartment I, that is, II MNa = 1.585 × 10−3 M

and

II MOH = 1.585 × 10−3 M

b. At pH = 13, from Fig. 15.1-1, the average charge on lysozyme is −9, so that in the equation above, Z = 9. Also, at 50 mg/g water, the protein molality is 50 MP =

mg 1000 g 1g × × mol g water kg 1000 mg = 3.571 × 10−3 M = 3.571 × 10−3 glysozyme kg 14000 mol

I I The solutions to Eqs. 15.7-6a and b (with MNa = MOH = 0.1 M) are II MNa = 0.117 352 M

and

II MOH = 0.085 213 M

15.8 Gibbs-Donnan Equilibrium and Membrane Potentials 963 Comment Note that the sodium ion concentrations in the two compartments are quite different in this last case, as are the hydroxyl ion concentrations. This is the Gibbs-Donnan equilibrium effect.

The fact that each ion partitions unevenly between the two compartments results in a contribution to the osmotic pressure from the ions in addition to that which results from the protein. Also, the fact that neither of the two phases is pure water (or another pure solvent) requires a modification to the derivation of the osmotic pressure of Sec. 11.5. The starting point now is

f Isolvent = f II solvent

(15.8-7)

which leads to I II II xIsolvent γsolvent fsolvent (T, P I ) = xII solvent γsolvent fsolvent (T, P )

(15.8-8)

Next, neglecting the activity coefficients of the solvent (which are usually close to 1 in value unless the solutions are concentrated in the solutes) and using the Poynting pressure correction, we obtain   V solvent (P II − P I ) xIsolvent = xII exp (15.8-9a) solvent RT or

P −P =Π = II

Osmotic pressure in Gibbs-Donnan equilibrium

I

RT V solvent

 ln

xIsolvent xII solvent

(15.8-9b)

Illustration 15.8-2 Osmotic Pressure of a Lysozyme Solution a. One compartment of an osmotic cell contains lysozyme at a concentration of 50 mg/g of water, and the second cell contains an aqueous solution of NaOH maintained at a concentration of 1.585 × 10−3 M, which has a pH of 11.2. Compute the osmotic pressure difference between the two cells. b. One compartment of an osmotic cell contains lysozyme at a concentration of 50 mg/g of water, and the second cell contains an aqueous solution of NaOH maintained at a concentration of 0.1 M, which has a pH of 13. Compute the osmotic pressure difference between the two cells.

Solution a. Since at pH = 11.2, lysozyme is at its isoelectric point and therefore uncharged. Consequently, the concentration of sodium ions and hydroxyl ions in each phase is the same, 1.585 × 10−3 moles per kg of water. Also, from the previous illustration, the lysozyme concentration is 3.517 × 10−3 moles per kg of water, so that the water mole fractions in each cell are xIwater =

55.51 = 0.999 943 2 × 1.585 × 10−3 + 55.51

964 Chapter 15: Some Additional Biochemical Applications of Thermodynamics and xII water =

55.51 = 0.999 879 2 × 1.585 × 10−3 + 55.51 + 3.571 × 10−3

Therefore, the osmotic pressure is Π =



RT V solvent

ln

xIsolvent xII solvent



bar m3 × 298.15 K  0.999 943  mol K ln = 0.881 bar m3 0.999 879 18 × 10−6 mol

8.314 × 10−5 =

b. Using the results from Illustration 15.7-1, the water mole fraction in compartment I is xIwater =

55.51 = 0.996 410 2 × 0.1 + 55.51

the water mole fraction in compartment II is xII water =

55.51 = 0.996 300 0.085 213 + 0.117 352 + 55.51 + 3.571 × 10−3

and the osmotic pressure difference is

Π =

RT ln V solvent



 xIsolvent xII solvent

bar m3 × 298.15 K  0.996 410  mol K ln = 1.520 bar m3 0.996 300 18 × 10−6 mol

8.314 × 10−5 =

Comment If the sodium ions partitioned equally in this last case, the concentrations in compartment II would be II II Mlys = 3.571 × 10−3 M MNa = 0.1 M and

II MOH = 0.1 − 9 × 3.571 × 10−3 = 0.067 861 M

xII water =

55.51 = 0.996 921 0.1 + 0.067 861 + 55.51 + 3.571 × 10−3

and the osmotic pressure difference would be bar m3 × 298.15 K  0.996 410  mol K ln = −7.060 bar 3 0.996 921 −6 m 18 × 10 mol

8.314 × 10−5 Π =

We see from this the importance of including the Gibbs-Donnan equilibrium effect when determining the partitioning of ions across a membrane when there is an ionic species (here the protein) that cannot pass through the membrane. In this case we see that by neglecting this effect, not only is the magnitude of the osmotic pressure changed, but so is its sign.

If there is a difference in ion concentrations across a membrane, an electrical potential difference will arise. To compute the magnitude of that potential difference, we start with the Nernst equation, Eq. 14.6-5, which can be applied to any one of the ions that can pass through the membrane:

G = RT ln

  aII i = W = Zi F E II − E I = Zi F E I ai

(14.6-5)

15.8 Gibbs-Donnan Equilibrium and Membrane Potentials 965

Electrostatic potential in Gibbs-Donnan equilibrium

where ai is the activity of ion species i, Zi is its charge, and F is the Faraday constant. If, as we have frequently done, we neglect the activity coefficients of the ions, this equation reduces to  II  RT M II E − EI = (15.8-10) ln i I Zi F Mi Illustration 15.8-3 Electrostatic Potential of Lysozyme Solutions in an Osmometer a. One compartment of an osmotic cell contains lysozyme at a concentration of 50 mg/g of water, and the second cell contains an aqueous solution of NaOH maintained at a concentration of 1.585 × 10−3 M, which has a pH of 11.2. Compute the electrostatic potential difference between the two cells. b. One compartment of an osmotic cell contains lysozyme at a concentration of 50 mg/g of water, and the second cell contains an aqueous solution of NaOH maintained at a concentration of 0.1 M, which has a pH of 13. Compute the electrostatic potential difference between the two cells.

Solution a. At pH = 11.2, lysozyme is at its isoelectric point and therefore uncharged. Consequently, the concentration of sodium ions and hydroxyl ions in each phase is the same, and there is no electrostatic potential difference between the two compartments. b. From Illustration 15.7-1, we have for the sodium ions that I I MNa = 0.1 M and MNa = 0.117 352 M

Therefore, E II − E I =

25.7 mV 0.117 352 ln = 4.11 mV 1 0.1

As a check on this calculation, we can also calculate the potential difference resulting from the difference in hydroxyl ion concentrations: E II − E I =

25.7 mV 0.085 213 ln = 4.11 mV −1 0.1

which agrees with the calculation based on the sodium ion concentration, as must be the case based on Eq. 15.7-3c or 15.7-4.

Gibbs-Donnan equilibrium determines the concentration difference across simple membranes made of polymers, porous ceramic media, and other ultrafiltration devices. However, the difference of ion concentrations across the membranes of living cells and nerves is more complicated because of the existence of “ion pumps” as a result of carrier-mediated or facilitated diffusion, so that the concentrations of some ions are not in thermodynamic equilibrium. For example, there is a much higher sodium concentration outside cells than there is inside, while the reverse is true for potassium ions. This occurs because there is a carrier (probably a lipoprotein) that binds with a sodium ion inside the cell, transports the ion across membrane, and then releases it into the fluid outside the cell. The carrier is then transformed and binds with a potassium ion, which is then transported into the cell. This mechanism is discussed in courses in biology and physiology. The resulting concentration difference leads to diffusion of the sodium ion across the membrane, so that the steady-state (not equilibrium) sodium ion

966 Chapter 15: Some Additional Biochemical Applications of Thermodynamics concentration gradient is a result of rapid active transport by the ion pump out of the cell, and much slower diffusion into the cell. This pump does not operate equally in both directions, and two to three sodium ions are transported out of the cell for each potassium ion that enters the cell. Also, cell membranes are almost 100 times more permeable to potassium ions than to sodium ions, so that it is reasonable to assume the potassium ion concentration difference across a cell is an equilibrium state. On average, for a resting nerve fiber, the sodium ion concentrations are 10 millimolar (mM) within cells and 142 mM outside the cell, while the potassium ion concentrations are 5 mM outside the cell (in the extracellular fluid, ECF) and 140 mM inside the cell. (Note that the actual situation is further complicated by the fact that the ion channels open and close depending on the physiological situation. For example, during a nerve impulse the permeability of the membrane to sodium ions can increase by a factor of a thousand, almost eliminating the sodium ion concentration difference.) Illustration 15.8-4 Estimate of the Cell Membrane Potential Using the potassium ion concentrations mentioned above, estimate the cell membrane potential in a nerve fiber.

Solution The potassium ion concentrations are approximately 140 mM within the cell and 5 mM in the plasma outside the cell. Consequently, at body temperature of 37◦ C the electrostatic potential across the cell membrane, referred to as the membrane potential, is J · 310.15 K 140 8.314 cell RT MK mol K ln

E = ln ECF = = 0.0891 V = 89.1 mV C J F MK 5 96 485 ·1 mol CV

(At a laboratory temperature of 25◦ C, the electrostatic potential difference is 85.6 mV.)

Chloride ions (and many other anions) diffuse easily through cell membranes, even though there is not a chloride ion pump. Consequently, the chloride ion concentration difference inside and outside of a cell is determined by the electrostatic potential. Illustration 15.8-5 Chloride Ion Concentration within a Nerve Cell The chloride ion concentration in extracellular fluid is approximately 100 mM. Using the cell membrane potential as a result of potassium ions from the illustration above, estimate the chloride ion concentration within a resting nerve cell.

Solution Using the Nernst equation, we have

J · 310.15 K M cell 8.314 cell MCl RT mol K ln Cl ln ECF = −0.0891 V =

E = − C J F MCl 100 96 485 ·1 mol CV cell which has the solution MCl = 5.0 mM, which agrees with reported values of 4 to 5 mM.

This illustration shows how the unequal distribution of a dominant ion, here K+ , across a cell membrane can establish an electrostatic potential difference that then results in the unequal distribution of other ions, many of which are vital to processes in the human body.

15.9 Protein Concentration in an Ultracentrifuge 967

15.9 PROTEIN CONCENTRATION IN AN ULTRACENTRIFUGE While we have generally neglected the effect of kinetic and potential energies in this textbook, there are some cases in which such contributions can be important. One example is the case of an ultracentrifuge, a very-high-speed centrifuge that can be used to concentrate or separate proteins and other biological molecules based on their molecular weights. Figure 15.9-1 is a very simple schematic diagram of a short centrifuge tube in an ultracentrifuge. The steady-state mass, energy, and entropy balance equations will be written for the region of the centrifuge tube between distances r and r + Δr from the axis of rotation of the ultracentrifuge tube. The steady-state mass balance for each species i is     0 = N˙ i r + N˙ i r+ r or N˙ i r+ r = −N˙ i r (15.9-1) The energy balance for this same volume element is     mi ω 2 r2 mi ω 2 r2 ˙ ˙ 0= + + Q˙ Ni H i − Ni H i − 2 2 r r+ r i

(15.9-2)

i

where ω is the angular velocity, mi is the molecular weight of species i (and N˙ i mi is its mass flow rate), and −mi ω 2 r2 /2 is the potential energy per unit mass of species i at a distance r from the center of a centrifuge with a rotational velocity ω . The entropy balance on this same volume element is

0=

 i

r

r + Δr

 ⏐ ⏐ Q˙ N˙ i S i ⏐r + N˙ i S i ⏐r+ r + + S˙ gen T

(15.9-3)

i

Figure 15.9-1 Schematic diagram of a tube containing a protein solution in an ultracentrifuge.

Now recognizing that during operation the temperature is constant, and combining the mass, energy, and entropy balances, we have

   2 2 2 2  ω r ω r 0= H i − T S i − Mi − H i − T S i − Mi − T S˙ gen N˙ i 2 2 r r+ r i

Therefore, at equilibrium

   2 2 2 2  ω r ω r 0= Gi − Mi − Gi − Mi N˙ i 2 2 r r+ r i

(15.9-4)

968 Chapter 15: Some Additional Biochemical Applications of Thermodynamics Since this equation must be satisfied for all values of N˙ i , it then follows that the equilibrium condition for each species is   ω 2 r2 ω 2 r2 Gi − Mi = Gi − Mi (15.9-5a) 2 2 r r+ r where r and r + Δr are any two points along the centrifuge tube. Indeed, to emphasize this we will write the equilibrium relation as   ω 2 r2 ω 2 r2 Gi − Mi = Gi − Mi (15.9-5b) 2 2 1 2 Therefore, the basic equation describing the concentration of each species in an ultracentrifuge at any two points is

Gi,1 (T, P1 , x1 ) − Gi,2 (T, P2 , x2 ) = mi but by definition

 ω2  2 r1 − r22 2



f i,1 (T, P1 , x1 ) Gi,1 (T, P1 , x1 ) − Gi,2 (T, P2 , x2 ) = RT ln f i,2 (T, P2 , x2 )

(15.9-6)

 (15.9-7)

Now using the Poynting factor to correct for the effect of pressure on fugacity, assuming that the partial molar volume is not dependent on concentration or pressure,     xi,1 γi,1 f¯i (T, P1 , x1 ) (P1 − P2 )V i f¯i (T, P1 , x1 ) (P1 − P2 )V i = exp = ¯ exp RT xi,2 γi,2 RT f¯i (T, P2 , x2 ) fi (T, P1 , x2 ) (15.9-8) we obtain  ω2 xi,1 γi,1 RT ln (15.9-9) + (P1 − P2 )V i = mi (r12 − r22 ) xi,2 γi,2 2 As a result of the centrifugal force, the pressure varies at each point in the centrifuge as follows:  ω2  2 r1 − r22 (P1 − P2 ) V = m 2 or  m ω2  2 r1 − r22 (P1 − P2 ) = V 2 (Note that we can derive this equation simply by taking the pure-component limit of Eq. 15.6-9. That is, for a pure component xi = 1 and γi = 1, so the logarithmic term vanishes, and m/V becomes the density of the pure component, which is the solvent.) For the case of water as the solvent, ρ = 1 g/cc, we have  ω2 ω2 xi,1 γi,1 = mi (r12 − r22 ) − ρi V i (r12 − r22 ) RT ln xi,2 γi,2 2 2  2 ρV i ω 2 = mi 1 − (r − r22 ) mi 2 1  ρ ω2 2 (15.9-10) = mi 1 − (r − r22 ) ρi 2 1

15.9 Protein Concentration in an Ultracentrifuge 969 where ρi = mi /V i is the effective mass density of the protein in solution. As a simplification, we will neglect the departures from ideality (either because the system is ideal, or particularly for systems with biochemical compounds, there are so many other uncertainties due to their complexity, that the effects of nonidealities are frequently neglected.) Consequently, the final equation is

 RT ln

Ultracentrifuge equation

xi,1 xi,2



  ρ ω2  2 = mi 1 − r1 − r22 ρi 2

(15.9-11)

Illustration 15.9-1 Use of an Ultracentrifuge to Concentrate a Protein The protein lysozyme has a molecular weight (mi ) of about 14 000 and a partial molar volume (V i ) of 0.75mi , for an effective density (ρi ) of 1.333 g/cc (which is a typical value for proteins). An aqueous solution of lysozyme is placed in an ultracentrifuge tube, the top of which is 2 cm and the bottom of which is 6 cm from the axis of rotation of the ultracentrifuge. The ultracentrifuge is at room temperature and operating at 10 000 revolutions per minute (rpm). What will be the ratio of the lysozyme concentration at the bottom of the tube to that at the top of the tube at equilibrium?

Solution Since the lysozyme is in aqueous solution, the solvent density ρ ≈ 1 g/cc. Also, ω = 2π × (revolutions per second) =

2π × 10 000 = 1047 s−1 60

Therefore,  ln

xi,1 xi,2



  ρ ω2 2 (r − r22 ) mi 1 − ρi 2 1 = RT   g g 1 × 1− × 10472 s−2 × (22 − 62 )cm2 14 000 mol 1.333 cc = J 2 × 8.314 × 298.15 K mol K 14 000 =

 ln

xi,1 xi,2



m2 g × 0.2498 × 1.0962 × 106 s−2 × (22 − 62 )cm2 × 10−4 2 mol cm J kg m2 g 2 × 8.314 × 298.15 K × 1 2 × 1000 mol K s J kg

= −2.023 = −2.023

or

xi,2 = xi,1 e2.023 = 7.56xi,1

Thus, at 10 000 rpm, the lysozyme equilibrium concentration at the bottom of the centrifuge tube should be a factor of 7.56 higher than at the top of the centrifuge tube.

970 Chapter 15: Some Additional Biochemical Applications of Thermodynamics PROBLEMS 15.1 The solubility of creatine in water is 16 g/kg of water. Problem 13.80 provides the structure and dissociation constants for this compound. Determine the solubility of creatine in water as a function of pH. 15.2 The solubility of of the amino acid L-cystine of water is 280 g/L, and its molecular weight is 121.16. Problem 13.83 provides the dissociation constants for this compound. Determine the solubility of L-cystine in water as a function of pH. 15.3 The solubility of oxalic acid in water is 1.350 g/kg of water. Problem 13.81 provides the dissociation constants for this compound. Determine the solubility of oxalic in water as a function of pH. 15.4 The solubility of the amino acid tyrosine in water is 0.453 g/kg of water. Problem 13.85 provides the dissociation constants for this compound. Determine the solubility of tyrosine in water as a function of pH. 15.5 Reading from the graphs in the paper by M. M. Santoro and D. W. Bolen, Biochem. 31, 4901-4907 (1992) the following data are obtained for the Gibbs energy change on chemical denaturation (foldedunfolded transition) of E. coli Thioredoxin separately guanidine hydrochloride and urea as a function of the concentrations of. [M]

Δunf G (kcal/mol)

MGdnHCl = 0 MGdnHCl = 2.5 Murea = 0 Murea = 6.5

7.9 0 9.3 0

15.8 From data in the literature it has been estimated that when the bacteria K. fragilis is used in the aerobic (that is, with added oxygen) fermentation of glucose C6 H12 O6 , using dissolved ammonia as a nitrogen source produces a biomass of atomic composition CH1.75 O0.52 N0.15 and ethanol C2 H5 OH. The following yield factors have been reported: YB/S = 0.569 and YC/S = 0.407. Determine the other yield factors for this fermentation. What fraction of the carbon atoms present in glucose are converted to ethanol? 15.9 When a yeast of average biomass composition is grown on an aqueous solution of methanol with dissolved nitrogen, it is found that 400 g of dry yeast are produced for each kilogram of methanol consumed. No other products are formed. Compute the oxygen consumed and carbon dioxide produced per kilogram of methanol consumed. 15.10 When the bacterium Zymomonas mobilis is used to produce ethanol from glucose in an anaerobic process with ammonia as the nitrogen source, it is found that bacteria biomass is produced to the extent of 0.06 C-moles per C-mole of glucose. The biomass has an elemental composition of CH1.8 O0.5 N0.2 . What are the fractional conversion of glucose to ethanol (on a C-mole basis), the amount of ammonia used, and the amount of carbon dioxide produced per C-mole of glucose consumed? 15.11 The following data have been reported by von Stockar and Birou for biomass production and heat release for the fermentation of glucose by K. fragilis, which also produces ethanol as the product.

Further, with both chemical denaturants, the Gibbs energy of unfolding is found to be a linear function of the molarity of the denaturant. Plot on the same graph the molar extents of the unfolding reaction separately with guanidine hydrochloride and urea as a function of denaturant molarity over the range of 0 to 10 molar. 15.6 M. M. Santoro and D. W. Bolen, Biochem. 31, 49014907 (1992) report the following data for the thermal denaturation (folded-unfolded transition) of E. coli Thioredoxin as a function of NaCl concentration. NaCl [M] 0.1 0.5 1.0 1.5

Tunf (o C) Δunf H (kcal/mol) 87.0 106.9 ± 1.1 87.0 103.5 ± 1.1 87.1 98.1 ± 0.2 88.0 99.8 ± 0.6

Δunf G (kcal/mol) 8.9 ± 0.3 8.4 ± 0.3 7.5 ± 0.3 7.7 ± 0.3

Use these data to compute Δunf S (kcal/mol · K) at each of the salt molarities. 15.7 Derive the equations that replace Eqs. 15.1-3 and 15.3-4b if electrolyte solution nonideality is included.

YQ/S YB/S

YC/S

(kJ/C-mole)

0.569 0.567 0.584 0.552 0.492 0.456 0.345 0.294

0.407 0.403 0.375 0.373 0.339 0.329 0.320 0.307

−186.7 −199.8 −192.0 −166.7 −129.0 −108.9 −83.6 −59.4

Assuming the biomass has the elemental composition CH1.75 O0.52 N0.15 and dissolved ammonia is used as the nitrogen source, compute the following for each data point: a. The ethanol yield factor, YP/S . b. The amount of oxygen consumed, YO2 /S . c. The amount of heat released, YQ/S . Compare the predicted values with those reported in the table above. d. The amount of entropy generated.

Problems 971 15.12 Repeat Illustration 15.7-10 if nitric acid C5 H9 NO4 is used as the nitrogen source instead of ammonia. 15.13 Derive the analogue of Eqs. 15.7-6 for Gibbs-Donnan equilibrium involving a negatively charged protein. 15.14 Glucose and fructose both have the composition C6 H12 O6 , but differ slightly in structure, as shown below, and more so in sweetness. Repeat Illustration 15.7-9 using these sugars instead of xylose. Glucose H

Fructose H

O C C

OH

HO

C

H

H

C

OH

H

C

OH

H

C

H

OH

CH2 O +

C

OH

C

O

HO

C

H

H

C

OH

H

C

OH

H

C

H

H

H

15.16 Succinic acid, which has the structure HOOC— CH2 —CH2 —COOH and on a C-mole basis is represented by CH3/2 O, is useful as a starting material in the manufacture of solvents and polymers. It is typically made by the catalytic oxidation of butane. However, by that route only about 40 percent of the carbon initially present in the butane is converted to succinic acid. It has been suggested instead to use glucose as a starting material and a biochemical pathway with the following proposed reaction stoichiometry:

OH

a. Estimate the maximum amount of 2,3-butanediol that can be produced per kg of glucose and fructose consumed, assuming no additional biomass production. b. Estimate the amount of water and oxygen consumed, and carbon dioxide produced per C-mole of 2,3-butanediol produced assuming no additional biomass production. c. Estimate the amount of heat released to keep the fermenter at 25◦ C per C-mole of 2,3-butanediol produced assuming no additional biomass production. 15.15 The biologist of Illustration 15.7-9 is examining other genetically engineered bacteria to produce 2,3butanediol from lignins, which are derived from abundant and renewable resources such as trees, plants, and agricultural crops. While lignin polymerizes, the repeating unit is believed to have the following general formula: C10 H12 O3 . a. Estimate the maximum amount of 2,3-butanediol that can be produced per kg of lignin consumed, assuming no additional biomass production. b. Estimate the amount of water and oxygen consumed, and carbon dioxide produced per C-mole of 2,3-butanediol produced, assuming no additional biomass production. c. Estimate the amount of heat released to keep the fermenter at 25◦ C per C-mole of 2,3-butanediol produced, assuming no additional biomass production.

1 1 8 CO2 → CH3/2 + H2 O 7 7 7

a. Is this reaction stoichiometry possible? b. Determine the maximum kg of succinic acid that can be produced per kg of glucose. c. Estimate the heat released per C-mole of glucose consumed. d. What fraction of the Gibbs energy of the glucose appears in the succinic acid? 15.17 A considerable amount of methane is produced in Norway from its North Sea oil and gas wells. Some of this methane is converted to methanol by partial oxidation, and then biochemically converted to biomass that is used as animal feed. The reported stoichiometry for the biochemical reaction is CH3 OH + 0.731 · O2 + 0.146 · NH3

→ 0.269 · CO2 + 0.731C-mol biomass a. What is the atomic composition of the biomass produced? b. How much heat is released per mole of methanol consumed? c. What fraction of the Gibbs energy of the reactants is present in the biomass? 15.18 Another possibility is to convert the methane directly to biomass that is used as animal feed. The stoichiometry for this biochemical reaction is CH4 + YO2 /S · O2 + YN/S · NH3 + YW/S · H2 O

→ YB/S · CH1.8 O0.5 N0.2 + YC/S · CO2 It has been found that the consumption of 2.5 m3 of methane (measured at 273 K and 1.013 bar) results in the production of 1 kg of biomass. a. Determine all the yield factors for this reaction. b. How much heat is released per m3 of methane consumed? c. What fraction of the Gibbs energy of the methane is present in the biomass? 15.19 The compound 1,3-propanediol, which has the structure HO—CH2 —CH2 —CH2 —OH, is a monomer

972 Chapter 15: Some Additional Biochemical Applications of Thermodynamics used in the production of polyester fibers and in the manufacture of polyurethanes. 1,3 Propanediol can be produced by chemical processes and also by the fermentation of glycerol in a two-step process using certain bacterial strains. However, neither the chemical nor biological method is well suited for industrial production, because the chemical processes are energy intensive and the biological processes are based on glycerol, an expensive starting material. The DuPont Company has developed a process using a microorganism containing the dehydratase enzyme that can convert other carbon sources, such as carbohydrates or sugars, to 1,3-propanediol. a. What is the maximum number of kg of 1,3propanediol that can be biologically produced from each kg of glycerol, assuming no other energy source is used? What is the stoichiometry of the reaction at the maximum product yield? b. What is the maximum number of kg of 1,3propanediol that can be biologically produced from each kg of glucose, assuming no other energy source is used? What is the stoichiometry of the reaction at the maximum product yield? 15.20 Various different substrates are being considered for the biochemical production of 1,3-propanediol (see previous problem) using dissolved ammonia as the nitrogen source. Determine the maximum amount of 1,3-propanediol that can be produced as a function of the generalized degree of reduction of the substrate if there are no other by-products. Also, when less than the maximum amount of 1,3-propanediol is produced, determine the oxygen and heat flow yield factors as a function of the generalized degree of reduction of the substrate and the 1,3-propanediol yield factor. 15.21 The bacterium Zymomonas mobilis can reproduce while producing ethanol from glucose in an anaerobic (i.e., without added oxygen) continuous process using ammonia as the nitrogen source. It has been found that 0.06 C-moles of bacterium are produced per C-mole of glucose, and that the bacterium can be represented as CH1.8 O0.5 N0.2 . How much ethanol, carbon dioxide and water are produced per C-mole of glucose consumed? 15.22 It has been claimed that coal, which we can consider to be pure carbon, can as a result of microbial action react with water in situ and be converted to methane by the reaction: 2C + 2H2 O → CH4 + CO2

Is this reaction possible without the use of nutrients or another free energy source.

15.23 A proposed chemical composition for anthracite coal is C240 H90 O4 NS. The stoichiometry for a biochemical reaction of anthracite with water is as follows: C240 H90 O4 NS + xH2 O → aCH4 + bCO2 + NO2 + SO2

Determine the maximum thermodynamically allowed value of the extent of conversion of anthracite to methane, and is the value of a, for this reaction if no nutrients or other energy sources are provided. How many moles of carbon dioxide are produced per mole of coal (that is, the value of b)? 15.24 Biogas is produced as in an anaerobic digester that treats farm wastes or energy crops. During the process, an air-tight tank transforms biomass waste into methane, producing renewable energy that can be used for heating, to generate electricity, and or fuel for a gas engines and turbines. The composition of biogas from one such plant has been reported, by volume, to be 62% methane, 8% nitrogen and 30% carbon dioxide. a. Compute the heat that would be produced per 100 m3 of biogas if it was isothermally combusted with a stoichiometric amount of pure oxygen at 298 K. b. Compute the maximum amount of work that could be produced if the biogas was isothermally combusted in fuel cell with a stoichiometric amount of pure oxygen at 298 K. c. Compute the maximum amount of work that could be produced if the biogas was isothermally combusted in a fuel cell with a stoichiometric amount of air (21 mole % oxygen and 79 mole % nitrogen) at 298 K. 15.25 An ultracentrifuge is found to have a lysozyme concentration of 1 mg/g of water 2 cm from the axis of rotation. Determine the equilibrium concentration profile over the range from 2 cm to 4 cm from the rotation axis if the temperature is 25◦ C and a. The rotational speed is 10 000 rpm; b. The rotational speed is 15 000 rpm. 15.26 A centrifuge operating at 5000 rpm and 25◦ C is to being used to partially separate antipneumococcus serum globulin (molecular weight 195 000 and partial molar volume of 0.745 cc/g) from hemoglobin (molecular weight 68 000 and partial molar volume of 0.749 cc/g) in aqueous solution. If the concentrations of hemoglobin and antipneumococcus serum globulin are both 0.1 mg/g of water 2 cm from the centrifuge axis of rotation, what are their concentrations at 4 cm from the axis of rotation?

Appendix A Thermodynamic Data A.I Conversion Factors to SI Units To Convert from:

To:

Multiply by:

atmosphere (standard) bar British thermal unit* BTU/lb BTU/lb ◦ F calorie* cal/g cal/g ◦ C cm of mercury (0◦ C) cm3 dyne dyne cm dyne/cm2 erg foot gallon (U.S., liquid) g/cm3 g/liter horsepower inch inch water (60◦ F) kilogram-force kilowatt hour liter millibar millimeter of mercury (0◦ C) ounce (avoirdupois) pound (avoirdupois) lb/ft3 lb/gal (U.S.) lb-force/ft2 lb-force/in2 (psi) ton (refrigeration) ton (2000 lb) torr (mm Hg, 0◦ C) yard degree Celsius degree Fahrenheit degree Rankine

Pa Pa J J kg−1 J kg−1 K−1 J J kg−1 J kg−1 K−1 Pa m3 N Nm Pa J m m3 kg m−3 kg m−3 W m Pa N J m3 Pa Pa kg kg kg m−3 kg m−3 Pa Pa W kg Pa m K K K

1.013 25 × 105 1 × 105 1.054 × 103 2.324 × 103 4.184 × 103

4.184 4.184 × 103 4.184 × 103 1.33 × 103 1 × 10−6 1 × 10−5 1 × 10−7 1 × 10−1 1 × 10−7 3.048 × 10−1 3.785 × 10−3 1 × 103

1 7.457 × 102 2.540 × 10−2 2.488 × 102

9.807

3.600 × 106 1 × 10−3 1 × 102 1.333 × 102 2.835 × 10−2 4.536 × 10−1 1.602 × 101 1.198 × 102 4.788 × 101 6.895 × 103 3.517 × 103 9.072 × 102 1.333 × 102 9.144 × 10−1 T (K) = T (◦ C) + 273.15 T (K) = [T (◦ F) + 459.67]/1.8 T (K) = T (◦ R)/1.8

∗

Thermochemical unit that is used, for example, in the tables in the Chemical Engineers Handbook. For International Table units use the constant 4.1868 instead of 4.184, and for mean calorie use 4.190 02.

973

974 Appendix A.II: The Molar Heat Capacities of Gases in the Ideal Gas (Zero-Pressure) State

A.II The Molar Heat Capacities of Gases in the Ideal Gas (Zero-Pressure) State*

a

b × 102

c × 105

d × 109

Temperature Range (K)

Paraffinic Hydrocarbons Methane Ethane Propane n-Butane i-Butane n-Pentane n-Hexane

CH4 C2 H6 C3 H8 C4 H10 C4 H10 C5 H12 C6 H14

19.875 6.895 −4.042 3.954 −7.908 6.770 6.933

5.021 17.255 30.456 37.126 41.573 45.398 55.188

1.268 −6.402 −15.711 −18.326 −22.992 −22.448 −28.636

−11.004 7.280 31.716 34.979 49.875 42.259 57.657

273–1500 273–1500 273–1500 273–1500 273–1500 273–1500 273–1500

Monoolefinic Hydrocarbons Ethylene Propylene 1-Butene i-Butene cis-2-Butene trans-2-Butene

C2 H4 C3 H6 C4 H8 C4 H8 C4 H8 C4 H8

3.950 3.151 −1.004 6.904 −7.439 9.791

15.628 23.812 36.193 32.226 33.799 30.209

−8.339 −12.176 −21.381 −16.657 −17.046 −14.239

17.657 24.603 50.502 33.557 33.013 25.398

273–1500 273–1500 273–1500 273–1500 273–1500 273–1500

Cycloparaffinic Hydrocarbons Cyclopentane Methylcyclopentane Cyclohexane Methylcyclohexane

C5 H10 C6 H12 C6 H12 C7 H14

−54.213 −50.686 −66.674 −63.054

54.757 64.352 68.845 79.381

−31.159 −37.301 −38.506 −45.979

68.661 83.808 80.628 100.795

273–1500 273–1500 273–1500 273–1500

Aromatic Hydrocarbons Benzene Toluene Ethylbenzene Styrene Cumene

C6 H6 C7 H8 C8 H10 C8 H8 C9 H12

−36.193 −34.364 −35.138 −24.971 −39.548

48.444 55.887 66.674 60.059 78.184

−31.548 −34.435 −41.854 −38.285 −49.661

77.573 80.335 100.209 92.176 120.502

273–1500 273–1500 273–1500 273–1500 273–1500

Oxygenated Hydrocarbons Formaldehyde Acetaldehyde Methanol Ethanol Ethylene oxide Ketene

CH2 O C2 H4 O CH4 O C2 H6 O C2 H4 O C2 H2 O

22.791 17.531 19.038 19.875 −4.686 17.197

4.075 13.239 9.146 20.946 20.607 12.410

0.713 −2.155 −1.218 −10.372 −9.996 −7.502

−8.695 −15.900 −8.034 20.042 13.176 17.657

273–1500 273–1000 273–1000 273–1500 273–1000 273–1500

Miscellaneous Hydrocarbons Cyclopropane Isopentane Neopentane o-Xylene m-Xylene p-Xylene

C3 H6 C5 H12 C5 H12 C8 H10 C8 H10 C8 H10

−27.117 −9.511 16.172 −15.854 −27.335 −22.318

34.335 52.025 55.670 59.795 62.364 59.498

−23.335 −29.695 −33.548 −34.954 −36.950 −33.406

65.314 66.360 78.787 78.661 83.891 71.255

273–1000 273–1500 273–1500 273–1500 273–1500 273–1500

* Constants

∗ = a + bT + cT 2 + dT 3 , where T is in Kelvins and C ∗ in J(mol K)−1 . are for the equation CP P

Appendix A.II: The Molar Heat Capacities of Gases in the Ideal Gas (Zero-Pressure) State 975 Appendix II (Continued) a

b × 102

c × 105

d × 109

Temperature Range (K)

C3 O2 C3 H6 O C3 H8 O C3 H8 O C3 H6 O

34.322 6.799 3.321 −5.469 2.177

12.858 27.870 35.573 38.640 29.799

−8.707 −15.636 −20.987 −24.268 −17.820

21.682 34.757 48.368 59.163 41.623

273–1500 273–1500 273–1500 273–1500 273–1500

Chloroethenes Chloroethene 1,1-Dichloroethene cis-1,2-Dichloroethene trans-1,2-Dichloroethene Trichloroethene Tetrachloroethene

C2 H3 Cl C2 H2 Cl2 C2 H2 Cl2 C2 H2 Cl2 C2 HCl3 C2 Cl4

10.046 24.682 18.142 23.686 38.494 63.222

17.866 18.339 19.628 17.971 18.900 15.895

−11.511 −13.314 −14.213 −12.644 −15.063 −13.301

28.439 35.632 37.699 33.017 42.259 38.029

273–1500 273–1500 273–1500 273–1500 273–1500 273–1500

Nitrogen Compounds Ammonia Hydrazine Methylamine Dimethylamine Trimethylamine

NH3 N2 H4 CH5 N C2 H7 N C3 H9 N

27.551 16.276 12.534 −1.151 −8.778

2.563 14.870 15.105 27.679 40.246

0.990 −9.640 −6.881 −14.572 −23.217

−6.687 25.063 12.345 29.921 52.017

273–1500 273–1500 273–1500 273–1500 273–1500

Halogens and Halogen Acids Fluorine Chlorine Bromine Iodine Hydrogen fluoride Hydrogen chloride Hydrogen bromide Hydrogen iodide

F2 Cl2 Br2 I2 HF HCl HBr HI

25.586 28.541 33.686 35.582 30.130 30.310 29.996 28.042

2.454 2.389 1.030 0.550 −0.493 −0.762 −0.671 0.190

−1.752 −2.137 −0.890 −0.447 0.659 1.326 1.387 0.509

4.099 6.473 2.680 1.308 −1.573 −4.335 −4.858 −2.014

273–2000 273–1500 273–1500 273–1800 273–2000 273–1500 273–1500 273–1900

Chloromethanes Methyl chloride Methylene chloride Chloroform Carbon tetrachloride Phosgene Thiophosgene

CH3 Cl CH2 Cl2 CHCl3 CCl4 COCl2 CSCl2

12.762 17.573 31.841 51.213 43.305 45.188

10.862 14.305 14.481 14.226 6.916 7.778

−5.205 −9.833 −11.163 −12.531 −3.518 −4.372

9.623 25.389 30.728 36.937

273–1500 273–1500 273–1500 273–1500 273–1000 273–1000

Cyanogens Cyanogen Hydrogen cyanide Cyanogen chloride Cyanogen bromide Cyanogen iodide Acetonitrile Acrylic nitrile

(CN)2 HCN CNCl CNBr CNI CH3 CN CH2 CHCl

41.088 26.527 33.347 36.904 40.544 21.297 19.038

6.217 3.504 4.496 3.801 3.018 11.562 17.171

−2.749 −1.093 −2.203 −1.827 −1.366 −3.812 −7.087

Oxides of Nitrogen Nitric oxide Nitric oxide Nitrous oxide Nitrogen dioxide Nitrogen tetroxide

NO NO N2 O NO2 N2 O4

27.034 29.322 24.092 22.929 33.054

0.987 −0.094 5.859 5.711 18.661

−0.322 0.974 −3.560 −3.519 −11.339

C3 Oxygenated Hydrocarbons

Carbon suboxide Acetone i -Propyl alcohol n -Propyl alcohol Allyl alcohol

273–1000 273–1500 273–1000 273–1000 273–1000 273–1200 273–1000 0.365 −4.184 10.569 7.866

273–3800 273–1500 273–1500 273–1500 273–600

976 Appendix A.II: The Molar Heat Capacities of Gases in the Ideal Gas (Zero-Pressure) State Appendix II (Continued) a

b × 102

c × 105

d × 109

Temperature Range (K)

21.799 17.615 14.812 10.167 −5.398 −1.841

9.208 17.042 24.427 19.636 34.937 43.590

−6.523 −9.172 −11.548 −11.636 −23.356 −28.293

18.197 19.720 20.812 27.130 59.582 70.837

273–1500 273–1500 273–1500 273–1500 273–1500 273–1500

Combustion Gases (Low Temperature Range) Nitrogen N2 Oxygen O2 Air Hydrogen H2 Carbon monoxide CO Carbon dioxide CO2 Water vapor H2 O Ammonia NH3

28.883 25.460 28.088 29.088 28.142 22.243 32.218 24.619

−0.157 1.519 0.197 −0.192 0.167 5.977 0.192 3.75

0.808 −0.715 0.480 0.400 0.537 −3.499 1.055 −0.138

−2.871 1.311 −1.965 −0.870 −2.221 7.464 −3.593

273–1800 273–1800 273–1800 273–1800 273–1800 273–1800 273–1800 300–1500

Combustion Gases (High Temperature Range)† Nitrogen N2 Oxygen O2 Air Hydrogen H2 Carbon monoxide CO Water vapor H2 O

27.318 28.167 27.435 26.879 27.113 29.163

0.623 0.630 0.618 0.435 0.655 1.449

−0.095 −0.075 −0.090 −0.033 −0.100 −0.202

Sulfur Compounds Sulfur Sulfur dioxide Sulfur trioxide Hydrogen sulfide Carbon disulfide Carbonyl sulfide

27.193 25.762 16.393 29.582 30.921 26.034

2.217 5.791 14.573 1.309 6.230 6.427

−1.627 −3.809 −11.193 0.571 −4.586 −4.427

Acetylenes and Diolefins Acetylene Methylacetylene Dimethylacetylene Propadiene 1,3-Butadiene Isoprene

C2 H2 C3 H4 C4 H6 C3 H4 C4 H6 C5 H8

S2 SO2 SO3 H2 S CS2 COS

273-3800 273–3800 273–3800 273–3800 273–3800 273–3800 3.983 8.607 32.402 −3.292 11.548 10.711

273–1800 273–1800 273–1300 273–1800 273–1800 273–1800

√ ∗ †The equation for CO2 in the temperature range of 273–3800 K is CP = 75.464 − 1.872 × 10−4 T − 661.42/ T . Based on data in O. Hougen, K. Watson, and R. A. Ragatz, Chemical Process Principles, Part 1, John Wiley & Sons, New York (1954). Used with permission.

Appendix A.III: The Thermodynamic Properties of Water and Steam 977

A.III The Thermodynamic Properties of Water and Steam1 THERMODYNAMIC PROPERTIES OF STEAM Saturated Steam: Temperature Table Temp (◦ C)

Press. (kPa)

T

P

0.01 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95

Specific Volume Sat. Sat. Liquid Vapor

0.6113 0.8721 1.2276 1.7051 2.339 3.169 4.246 5.628 7.384 9.593 12.349 15.758 19.940 25.03 31.19 38.58 47.39 57.83 70.14 84.55

Internal Energy Sat. Sat. Liquid Vapor Evap.

Enthalpy Sat. Liquid

Entropy

Evap.

Sat. Vapor

Sat. Liquid

Evap.

Sat. Vapor

Vˆ L

Vˆ V

ˆL U

ˆ ΔU

ˆV U

ˆL H

ˆ ΔH

ˆV H

SˆL

ΔSˆ

SˆV

0.001 000 0.001 000 0.001 000 0.001 001 0.001 002 0.001 003 0.001 004 0.001 006 0.001 008 0.001 010 0.001 012 0.001 015 0.001 017 0.001 020 0.001 023 0.001 026 0.001 029 0.001 033 0.001 036 0.001 040

206.14 147.12 106.38 77.93 57.79 43.36 32.89 25.22 19.52 15.26 12.03 9.568 7.671 6.197 5.042 4.131 3.407 2.828 2.361 1.982

0.00 20.97 42.00 62.99 83.95 104.88 125.78 146.67 167.56 188.44 209.32 230.21 251.11 272.02 292.95 313.90 334.86 355.84 376.85 397.88

2375.3 2361.3 2347.2 2333.1 2319.0 2304.9 2290.8 2276.7 2262.6 2248.4 2234.2 2219.9 2205.5 2191.1 2176.6 2162.0 2147.4 2132.6 2117.7 2102.7

2375.3 2382.3 2389.2 2396.1 2402.9 2409.8 2416.6 2423.4 2430.1 2436.8 2443.5 2450.1 2456.6 2463.1 2469.6 2475.9 2482.2 2488.4 2494.5 2500.6

0.01 20.98 42.01 62.99 83.96 104.89 125.79 146.68 167.57 188.45 209.33 230.23 251.13 272.06 292.98 313.93 334.91 355.90 376.92 397.96

2501.3 2489.6 2477.7 2465.9 2454.1 2442.3 2430.5 2418.6 2406.7 2394.8 2382.7 2370.7 2358.5 2346.2 2333.8 2321.4 2308.8 2296.0 2283.2 2270.2

2501.4 2510.6 2519.8 2528.9 2538.1 2547.2 2556.3 2565.3 2574.3 2583.2 2592.1 2600.9 2609.6 2618.3 2626.8 2635.3 2643.7 2651.9 2660.1 2668.1

0.0000 0.0761 0.1510 0.2245 0.2966 0.3674 0.4369 0.5053 0.5725 0.6387 0.7038 0.7679 0.8312 0.8935 0.9549 1.0155 1.0753 1.1343 1.1925 1.2500

9.1562 8.9496 8.7498 8.5569 8.3706 8.1905 8.0164 7.8478 7.6845 7.5261 7.3725 7.2234 7.0784 6.9375 6.8004 6.6669 6.5369 6.4102 6.2866 6.1659

9.1562 9.0257 8.9008 8.7814 8.6672 8.5580 8.4533 8.3531 8.2570 8.1648 8.0763 7.9913 7.9096 7.8310 7.7553 7.6824 7.6122 7.5445 7.4791 7.4159

418.94 440.02 461.14 482.30 503.50 524.74 546.02 567.35 588.74 610.18 631.68 653.24 674.87 696.56 718.33

2087.6 2072.3 2057.0 2041.4 2025.8 2009.9 1993.9 1977.7 1961.3 1944.7 1927.9 1910.8 1893.5 1876.0 1858.1

2506.5 2512.4 2518.1 2523.7 2529.3 2534.6 2539.9 2545.0 2550.0 2554.9 2559.5 2564.1 2568.4 2572.5 2576.5

419.04 440.15 461.30 482.48 503.71 524.99 546.31 567.69 589.13 610.63 632.20 653.84 675.55 697.34 719.21

2257.0 2243.7 2230.2 2216.5 2202.6 2188.5 2174.2 2159.6 2144.7 2129.6 2114.3 2098.6 2082.6 2066.2 2049.5

2676.1 2683.8 2691.5 2699.0 2706.3 2713.5 2720.5 2727.3 2733.9 2740.3 2746.5 2752.4 2758.1 2763.5 2768.7

1.3069 1.3630 1.4185 1.4734 1.5276 1.5813 1.6344 1.6870 1.7391 1.7907 1.8418 1.8925 1.9427 1.9925 2.0419

6.0480 5.9328 5.8202 5.7100 5.6020 5.4962 5.3925 5.2907 5.1908 5.0926 4.9960 4.9010 4.8075 4.7153 4.6244

7.3549 7.2958 7.2387 7.1833 7.1296 7.0775 7.0269 6.9777 6.9299 6.8833 6.8379 6.7935 6.7502 6.7078 6.6663

MPa 100 105 110 115 120 125 130 135 140 145 150 155 160 165 170

0.101 35 0.120 82 0.143 27 0.169 06 0.198 53 0.2321 0.2701 0.3130 0.3613 0.4154 0.4758 0.5431 0.6178 0.7005 0.7917

Vˆ [=] m3 /kg;

0.001 044 0.001 048 0.001 052 0.001 056 0.001 060 0.001 065 0.001 070 0.001 075 0.001 080 0.001 085 0.001 091 0.001 096 0.001 102 0.001 108 0.001 114

1.6729 1.4194 1.2102 1.0366 0.8919 0.7706 0.6685 0.5822 0.5089 0.4463 0.3928 0.3468 0.3071 0.2727 0.2428

ˆ, H ˆ [=] J/g = kJ/kg; U

1 From

Sˆ [=] kJ/kg K

G. J. Van Wylen and R. E. Sontag, Fundamentals of Classical Thermodynamics, S. I. Version. 2nd ed., John Wiley & Sons, New York (1978). Used with permission.

978 Appendix A.III: The Thermodynamic Properties of Water and Steam Saturated Steam: Temperature Table (Continued) Specific Volume Sat. Sat. Liquid Vapor

Internal Energy Sat. Sat. Liquid Vapor Evap.

Temp (◦ C)

Press. (MPa)

T

P

Vˆ L

Vˆ V

ˆL U

ˆ ΔU

175 180 185 190 195 200 205 210 215 220 225 230 235 240 245 250 255 260 265 270 275 280 285 290 295 300 305 310 315 320 330 340 350 360 370 374.14

0.8920 1.0021 1.1227 1.2544 1.3978 1.5538 1.7230 1.9062 2.104 2.318 2.548 2.795 3.060 3.344 3.648 3.973 4.319 4.688 5.081 5.499 5.942 6.412 6.909 7.436 7.993 8.581 9.202 9.856 10.547 11.274 12.845 14.586 16.513 18.651 21.03 22.09

0.001 121 0.001 127 0.001 134 0.001 141 0.001 149 0.001 157 0.001 164 0.001 173 0.001 181 0.001 190 0.001 199 0.001 209 0.001 219 0.001 229 0.001 240 0.001 251 0.001 263 0.001 276 0.001 289 0.001 302 0.001 317 0.001 332 0.001 348 0.001 366 0.001 384 0.001 404 0.001 425 0.001 447 0.001 472 0.001 499 0.001 561 0.001 638 0.001 740 0.001 893 0.002 213 0.003 155

0.2168 0.194 05 0.174 09 0.156 54 0.141 05 0.127 36 0.115 21 0.104 41 0.094 79 0.086 19 0.078 49 0.071 58 0.065 37 0.059 76 0.054 71 0.050 13 0.045 98 0.042 21 0.038 77 0.035 64 0.032 79 0.030 17 0.027 77 0.025 57 0.023 54 0.021 67 0.019 948 0.018 350 0.016 867 0.015 488 0.012 996 0.010 797 0.008 813 0.006 945 0.004 925 0.003 155

740.17 762.09 784.10 806.19 828.37 850.65 873.04 895.53 918.14 940.87 963.73 986.74 1009.89 1033.21 1056.71 1080.39 1104.28 1128.39 1152.74 1177.36 1202.25 1227.46 1253.00 1278.92 1305.2 1332.0 1359.3 1387.1 1415.5 1444.6 1505.3 1570.3 1641.9 1725.2 1844.0 2029.6

1840.0 1821.6 1802.9 1783.8 1764.4 1744.7 1724.5 1703.9 1682.9 1661.5 1639.6 1617.2 1594.2 1570.8 1546.7 1522.0 1496.7 1470.6 1443.9 1416.3 1387.9 1358.7 1328.4 1297.1 1264.7 1231.0 1195.9 1159.4 1121.1 1080.9 993.7 894.3 776.6 626.3 384.5 0

Vˆ [=] m3 /kg;

ˆ, H ˆ [=] J/g = kJ/kg; U

Sˆ [=] kJ/kg K

Enthalpy Sat. Liquid

ˆV U 2580.2 2583.7 2587.0 2590.0 2592.8 2595.3 2597.5 2599.5 2601.1 2602.4 2603.3 2603.9 2604.1 2604.0 2603.4 2602.4 2600.9 2599.0 2596.6 2593.7 2590.2 2586.1 2581.4 2576.0 2569.9 2563.0 2555.2 2546.4 2536.6 2525.5 2498.9 2464.6 2418.4 2351.5 2228.5 2029.6

Entropy

Evap.

Sat. Vapor

Sat. Liquid

Evap.

Sat. Vapor

ˆL H

ˆ ΔH

ˆV H

SˆL

ΔSˆ

SˆV

741.17 763.22 785.37 807.62 829.98 852.45 875.04 897.76 920.62 943.62 966.78 990.12 1013.62 1037.32 1061.23 1085.36 1109.73 1134.37 1159.28 1184.51 1210.07 1235.99 1262.31 1289.07 1316.3 1344.0 1372.4 1401.3 1431.0 1461.5 1525.3 1594.2 1670.6 1760.5 1890.5 2099.3

2032.4 2015.0 1997.1 1978.8 1960.0 1940.7 1921.0 1900.7 1879.9 1858.5 1836.5 1813.8 1790.5 1766.5 1741.7 1716.2 1689.8 1662.5 1634.4 1605.2 1574.9 1543.6 1511.0 1477.1 1441.8 1404.9 1366.4 1326.0 1283.5 1238.6 1140.6 1027.9 893.4 720.5 441.6 0

2773.6 2778.2 2782.4 2786.4 2790.0 2793.2 2796.0 2798.5 2800.5 2802.1 2803.3 2804.0 2804.2 2803.8 2803.0 2801.5 2799.5 2796.9 2793.6 2789.7 2785.0 2779.6 2773.3 2766.2 2758.1 2749.0 2738.7 2727.3 2714.5 2700.1 2665.9 2622.0 2563.9 2481.0 2332.1 2099.3

2.0909 2.1396 2.1879 2.2359 2.2835 2.3309 2.3780 2.4248 2.4714 2.5178 2.5639 2.6099 2.6558 2.7015 2.7472 2.7927 2.8383 2.8838 2.9294 2.9751 3.0208 3.0668 3.1130 3.1594 3.2062 3.2534 3.3010 3.3493 3.3982 3.4480 3.5507 3.6594 3.7777 3.9147 4.1106 4.4298

4.5347 4.4461 4.3586 4.2720 4.1863 4.1014 4.0172 3.9337 3.8507 3.7683 3.6863 3.6047 3.5233 3.4422 3.3612 3.2802 3.1992 3.1181 3.0368 2.9551 2.8730 2.7903 2.7070 2.6227 2.5375 2.4511 2.3633 2.2737 2.1821 2.0882 1.8909 1.6763 1.4335 1.1379 .6865 0

6.6256 6.5857 6.5465 6.5079 6.4698 6.4323 6.3952 6.3585 6.3221 6.2861 6.2503 6.2146 6.1791 6.1437 6.1083 6.0730 6.0375 6.0019 5.9662 5.9301 5.8938 5.8571 5.8199 5.7821 5.7437 5.7045 5.6643 5.6230 5.5804 5.5362 5.4417 5.3357 5.2112 5.0526 4.7971 4.4298

Appendix A.III: The Thermodynamic Properties of Water and Steam 979 Saturated Steam: Pressure Table Press. (kPa)

Temp (◦ C)

P

T

0.6113 1.0 1.5 2.0 2.5 3.0 4.0 5.0 7.5 10 15 20 25 30 40 50 75

Specific Volume Sat. Sat. Liquid Vapor

Internal Energy Sat. Sat. Liquid Vapor Evap.

Enthalpy Sat. Liquid

Entropy

Evap.

Sat. Vapor

Sat. Liquid

Evap.

Sat. Vapor

Vˆ L

Vˆ V

ˆL U

ˆ ΔU

ˆV U

ˆL H

ˆ ΔH

ˆV H

SˆL

ΔSˆ

SˆV

0.01 6.98 13.03 17.50 21.08 24.08 28.96 32.88 40.29 45.81 53.97 60.06 64.97 69.10 75.87 81.33 91.78

0.001 000 0.001 000 0.001 001 0.001 001 0.001 002 0.001 003 0.001 004 0.001 005 0.001 008 0.001 010 0.001 014 0.001 017 0.001 020 0.001 022 0.001 027 0.001 030 0.001 037

206.14 129.21 87.98 67.00 54.25 45.67 34.80 28.19 19.24 14.67 10.02 7.649 6.204 5.229 3.993 3.240 2.217

0.00 29.30 54.71 73.48 88.48 101.04 121.45 137.81 168.78 191.82 225.92 251.38 271.90 289.20 317.53 340.44 384.31

2375.3 2355.7 2338.6 2326.0 2315.9 2307.5 2293.7 2282.7 2261.7 2246.1 2222.8 2205.4 2191.2 2179.2 2159.5 2143.4 2112.4

2375.3 2385.0 2393.3 2399.5 2404.4 2408.5 2415.2 2420.5 2430.5 2437.9 2448.7 2456.7 2463.1 2468.4 2477.0 2483.9 2496.7

0.01 29.30 54.71 73.48 88.49 101.05 121.46 137.82 168.79 191.83 225.94 251.40 271.93 289.23 317.58 340.49 384.39

2501.3 2484.9 2470.6 2460.0 2451.6 2444.5 2432.9 2423.7 2406.0 2392.8 2373.1 2358.3 2346.3 2336.1 2319.2 2305.4 2278.6

2501.4 2514.2 2525.3 2533.5 2540.0 2545.5 2554.4 2561.5 2574.8 2584.7 2599.1 2609.7 2618.2 2625.3 2636.8 2645.9 2663.0

0.0000 0.1059 0.1957 0.2607 0.3120 0.3545 0.4226 0.4764 0.5764 0.6493 0.7549 0.8320 0.8931 0.9439 1.0259 1.0910 1.2130

9.1562 8.8697 8.6322 8.4629 8.3311 8.2231 8.0520 7.9187 7.6750 7.5009 7.2536 7.0766 6.9383 6.8247 6.6441 6.5029 6.2434

9.1562 8.9756 8.8279 8.7237 8.6432 8.5776 8.4746 8.3951 8.2515 8.1502 8.0085 7.9085 7.8314 7.7686 7.6700 7.5939 7.4564

99.63 105.99 111.37 116.06 120.23 124.00 127.44 130.60 133.55 136.30 138.88 141.32 143.63 147.93 151.86 155.48 158.85 162.01 164.97 167.78 170.43 172.96 175.38 177.69

0.001 043 0.001 048 0.001 053 0.001 057 0.001 061 0.001 064 0.001 067 0.001 070 0.001 073 0.001 076 0.001 079 0.001 081 0.001 084 0.001 088 0.001 093 0.001 097 0.001 101 0.001 104 0.001 108 0.001 112 0.001 115 0.001 118 0.001 121 0.001 124

417.36 444.19 466.94 486.80 504.49 520.47 535.10 548.59 561.15 572.90 583.95 594.40 604.31 622.77 639.68 655.32 669.90 683.56 696.44 708.64 720.22 731.27 741.83 751.95

2088.7 2069.3 2052.7 2038.1 2025.0 2013.1 2002.1 1991.9 1982.4 1973.5 1965.0 1956.9 1949.3 1934.9 1921.6 1909.2 1897.5 1886.5 1876.1 1866.1 1856.6 1847.4 1838.6 1830.2

2506.1 2513.5 2519.7 2524.9 2529.5 2533.6 2537.2 2540.5 2543.6 2546.4 2548.9 2551.3 2553.6 2557.6 2561.2 2564.5 2567.4 2570.1 2572.5 2574.7 2576.8 2578.7 2580.5 2582.1

417.46 444.32 467.11 486.99 504.70 520.72 535.37 548.89 561.47 573.25 584.33 594.81 604.74 623.25 640.23 655.93 670.56 684.28 697.22 709.47 721.11 732.22 742.83 753.02

2258.0 2241.0 2226.5 2213.6 2201.9 2191.3 2181.5 2172.4 2163.8 2155.8 2148.1 2140.8 2133.8 2120.7 2108.5 2097.0 2086.3 2076.0 2066.3 2057.0 2048.0 2039.4 2031.1 2023.1

2675.5 2685.4 2693.6 2700.6 2706.7 2712.1 2716.9 2721.3 2725.3 2729.0 2732.4 2735.6 2738.6 2743.9 2748.7 2753.0 2756.8 2760.3 2763.5 2766.4 2769.1 2771.6 2773.9 2776.1

1.3026 1.3740 1.4336 1.4849 1.5301 1.5706 1.6072 1.6408 1.6718 1.7006 1.7275 1.7528 1.7766 1.8207 1.8607 1.8973 1.9312 1.9627 1.9922 2.0200 2.0462 2.0710 2.0946 2.1172

6.0568 5.9104 5.7897 5.6868 5.5970 5.5173 5.4455 5.3801 5.3201 5.2646 5.2130 5.1647 5.1193 5.0359 4.9606 4.8920 4.8288 4.7703 4.7158 4.6647 4.6166 4.5711 4.5280 4.4869

7.3594 7.2844 7.2233 7.1717 7.1271 7.0878 7.0527 7.0209 6.9919 6.9652 6.9405 6.9175 6.8959 6.8565 6.8213 6.7893 6.7600 6.7331 6.7080 6.6847 6.6628 6.6421 6.6226 6.6041

MPa 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.275 0.300 0.325 0.350 0.375 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95

Vˆ [=] m3 /kg;

1.6940 1.3749 1.1593 1.0036 0.8857 0.7933 0.7187 0.6573 0.6058 0.5620 0.5243 0.4914 0.4625 0.4140 0.3749 0.3427 0.3157 0.2927 0.2729 0.2556 0.2404 0.2270 0.2150 0.2042

ˆ, H ˆ [=] J/g = kJ/kg; U

Sˆ [=] kJ/kg K

980 Appendix A.III: The Thermodynamic Properties of Water and Steam Saturated Steam: Pressure Table (Continued) Specific Volume Sat. Sat. Liquid Vapor

Internal Energy Sat. Sat. Liquid Vapor Evap.

Press. (MPa)

Temp (◦ C)

P

T

Vˆ L

Vˆ V

ˆL U

ˆ ΔU

1.00 1.10 1.20 1.30 1.40 1.50 1.75 2.00 2.25 2.5 3.00 3.5 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 22.09

179.91 184.09 187.99 191.64 195.07 198.32 205.76 212.42 218.45 223.99 233.90 242.60 250.40 263.99 275.64 285.88 295.06 303.40 311.06 318.15 324.75 330.93 336.75 342.24 347.44 352.37 357.06 361.54 365.81 369.89 373.80 374.14

0.001 127 0.001 133 0.001 139 0.001 144 0.001 149 0.001 154 0.001 166 0.001 177 0.001 187 0.001 197 0.001 217 0.001 235 0.001 252 0.001 286 0.001 319 0.001 351 0.001 384 0.001 418 0.001 452 0.001 489 0.001 527 0.001 567 0.001 611 0.001 658 0.001 711 0.001 770 0.001 840 0.001 924 0.002 036 0.002 207 0.002 742 0.003 155

0.194 44 0.177 53 0.163 33 0.151 25 0.140 84 0.131 77 0.113 49 0.099 63 0.088 75 0.079 98 0.066 68 0.057 07 0.049 78 0.039 44 0.032 44 0.027 37 0.023 52 0.020 48 0.018 026 0.015 987 0.014 263 0.012 780 0.011 485 0.010 337 0.009 306 0.008 364 0.007 489 0.006 657 0.005 834 0.004 952 0.003 568 0.003 155

761.68 780.09 797.29 813.44 828.70 843.16 876.46 906.44 933.83 959.11 1004.78 1045.43 1082.31 1147.81 1205.44 1257.55 1305.57 1350.51 1393.04 1433.7 1473.0 1511.1 1548.6 1585.6 1622.7 1660.2 1698.9 1739.9 1785.6 1842.1 1961.9 2029.6

1822.0 1806.3 1791.5 1777.5 1764.1 1751.3 1721.4 1693.8 1668.2 1644.0 1599.3 1558.3 1520.0 1449.3 1384.3 1323.0 1264.2 1207.3 1151.4 1096.0 1040.7 985.0 928.2 869.8 809.0 744.8 675.4 598.1 507.5 388.5 125.2 0.0

Vˆ [=] m3 /kg;

ˆ, H ˆ [=] J/g = kJ/kg; U

Sˆ [=] kJ/kg K

Enthalpy Sat. Liquid

ˆV U

2583.6 2586.4 2588.8 2591.0 2592.8 2594.5 2597.8 2600.3 2602.0 2603.1 2604.1 2603.7 2602.3 2597.1 2589.7 2580.5 2569.8 2557.8 2544.4 2529.8 2513.7 2496.1 2476.8 2455.5 2431.7 2405.0 2374.3 2338.1 2293.0 2230.6 2087.1 2029.6

Entropy

Evap.

Sat. Vapor

Sat. Liquid

Evap.

Sat. Vapor

ˆL H

ˆ ΔH

ˆV H

SˆL

ΔSˆ

SˆV

762.81 781.34 798.65 814.93 830.30 844.89 878.50 908.79 936.49 962.11 1008.42 1049.75 1087.31 1154.23 1213.35 1267.00 1316.64 1363.26 1407.56 1450.1 1491.3 1531.5 1571.1 1610.5 1650.1 1690.3 1732.0 1776.5 1826.3 1888.4 2022.2 2099.3

2015.3 2000.4 1986.2 1972.7 1959.7 1947.3 1917.9 1890.7 1865.2 1841.0 1795.7 1753.7 1714.1 1640.1 1571.0 1505.1 1441.3 1378.9 1317.1 1255.5 1193.6 1130.7 1066.5 1000.0 930.6 856.9 777.1 688.0 583.4 446.2 143.4 0.0

2778.1 2781.7 2784.8 2787.6 2790.6 2792.2 2796.4 2799.5 2801.7 2803.1 2804.2 2803.4 2801.4 2794.3 2784.3 2772.1 2758.0 2742.1 2724.7 2705.6 2684.9 2662.2 2637.6 2610.5 2580.6 2547.2 2509.1 2464.5 2409.7 2334.6 2165.6 2099.3

2.1387 2.1792 2.2166 2.2515 2.2842 2.3150 2.3851 2.4474 2.5035 2.5547 2.6457 2.7253 2.7964 2.9202 3.0267 3.1211 3.2068 3.2858 3.3596 3.4295 3.4962 3.5606 3.6232 3.6848 3.7461 3.8079 3.8715 3.9388 4.0139 4.1075 4.3110 4.4298

4.4478 4.3744 4.3067 4.2438 4.1850 4.1298 4.0044 3.8935 3.7937 3.7028 3.5412 3.4000 3.2737 3.0532 2.8625 2.6922 2.5364 2.3915 2.2544 2.1233 1.9962 1.8718 1.7485 1.6249 1.4994 1.3698 1.2329 1.0839 0.9130 0.6938 0.2216 0.0

6.5865 6.5536 6.5233 6.4953 6.4693 6.4448 6.3896 6.3409 6.2972 6.2575 6.1869 6.1253 6.0701 5.9734 5.8892 5.8133 5.7432 5.6772 5.6141 5.5527 5.4924 5.4323 5.3717 5.3098 5.2455 5.1777 5.1044 5.0228 4.9269 4.8013 4.5327 4.4298

Appendix A.III: The Thermodynamic Properties of Water and Steam 981 Superheated Vapor‡ P = 0.010 MPa (45.81) ◦

P = 0.050 MPa (81.33)

P = 0.10 MPa (99.63)

T ( C)

Vˆ

ˆ U

ˆ H

Sˆ

Vˆ

ˆ U

ˆ H

Sˆ

Vˆ

ˆ U

ˆ H

Sˆ

Sat. 50 100 150 200 250 300 400 500 600 700 800 900 1000 1100 1200 1300

14.674 14.869 17.196 19.512 21.825 24.136 26.445 31.063 35.679 40.295 44.911 49.526 54.141 58.757 63.372 67.987 72.602

2437.9 2443.9 2515.5 2587.9 2661.3 2736.0 2812.1 2968.9 3132.3 3302.5 3479.6 3663.8 3855.0 4053.0 4257.5 4467.9 4683.7

2584.7 2592.6 2687.5 2783.0 2879.5 2977.3 3076.5 3279.6 3489.1 3705.4 3928.7 4159.0 4396.4 4640.6 4891.2 5147.8 5409.7

8.1502 8.1749 8.4479 8.6882 8.9038 9.1002 9.2813 9.6077 9.8978 10.1608 10.4028 10.6281 10.8396 11.0393 11.2287 11.4091 11.5811

3.240

2483.9 — 2511.6 2585.6 2659.9 2735.0 2811.3 2968.5 3132.0 3302.2 3479.4 3663.6 3854.9 4052.9 4257.4 4467.8 4683.6

2645.9 — 2682.5 2780.1 2877.7 2976.0 3075.5 3278.9 3488.7 3705.1 3928.5 4158.9 4396.3 4640.5 4891.1 5147.7 5409.6

7.5939 — 7.6947 7.9401 8.1580 8.3556 8.5373 8.8642 9.1546 9.4178 9.6599 9.8852 10.0967 10.2964 10.4859 10.6662 10.8382

1.6940

2506.1 — 2506.7 2582.8 2658.1 2733.7 2810.4 2967.9 3131.6 3301.9 3479.2 3663.5 3854.8 4052.8 4257.3 4467.7 4683.5

2675.5 — 2676.2 2776.4 2875.3 2974.3 3074.3 3278.2 3488.1 3704.7 3928.2 4158.6 4396.1 4640.3 4891.0 5147.6 5409.5

7.3594

— 3.418 3.889 4.356 4.820 5.284 6.209 7.134 8.057 8.981 9.904 10.828 11.751 12.674 13.597 14.521

P = 0.20 MPa (120.23)

Sat. 150 200 250 300 400 500 600 700 800 900 1000 1100 1200 1300

0.8857 0.9596 1.0803 1.1988 1.3162 1.5493 1.7814 2.013 2.244 2.475 2.706 2.937 3.168 3.399 3.630

2529.5 2576.9 2654.4 2731.2 2808.6 2966.7 3130.8 3301.4 3478.8 3663.1 3854.5 4052.5 4257.0 4467.5 4683.2

2706.7 2768.8 2870.5 2971.0 3071.8 3276.6 3487.1 3704.0 3927.6 4158.2 4395.8 4640.0 4890.7 5147.3 5409.3

0.3749 0.4249 0.4744 0.5226 0.5701 0.6173 0.7109 0.8041 0.8969 0.9896 1.0822 1.1747 1.2672 1.3596 1.4521

2561.2 2642.9 2723.5 2802.9 2882.6 2963.2 3128.4 3299.6 3477.5 3662.1 3853.6 4051.8 4256.3 4466.8 4682.5

2748.7 2855.4 2960.7 3064.2 3167.7 3271.9 3483.9 3701.7 3925.9 4156.9 4394.7 4639.1 4889.9 5146.6 5408.6

7.1272 7.2795 7.5066 7.7086 7.8926 8.2218 8.5133 8.7770 9.0194 9.2449 9.4566 9.6563 9.8458 10.0262 10.1982

0.6058 0.6339 0.7163 0.7964 0.8753 1.0315 1.1867 1.3414 1.4957 1.6499 1.8041 1.9581 2.1121 2.2661 2.4201

2543.6 2570.8 2650.7 2728.7 2806.7 2965.6 3130.0 3300.8 3478.1 3662.9 3854.2 4052.3 4256.8 4467.2 4683.0

2725.3 2761.0 2865.6 2967.6 3069.3 3275.0 3486.0 3703.2 3927.1 4157.8 4395.4 4639.7 4890.4 5147.1 5409.0

6.9919 7.0778 7.3115 7.5166 7.7022 8.0330 8.3251 8.5892 8.8319 9.0576 9.2692 9.4690 9.6585 9.8389 10.0110

0.4625 0.4708 0.5342 0.5951 0.6548 0.7726 0.8893 1.0055 1.1215 1.2372 1.3529 1.4685 1.5840 1.6996 1.8151

P = 0.60 MPa (158.85) 6.8213 7.0592 7.2709 7.4599 7.6329 7.7938 8.0873 8.3522 8.5952 8.8211 9.0329 9.2328 9.4224 9.6029 9.7749

0.3157 0.3520 0.3938 0.4344 0.4742 0.5137 0.5920 0.6697 0.7472 0.8245 0.9017 0.9788 1.0559 1.1330 1.2101

2567.4 2638.9 2720.9 2801.0 2881.2 2962.1 3127.6 3299.1 3477.0 3661.8 3853.4 4051.5 4256.1 4466.5 4682.3

2756.8 2850.1 2957.2 3061.6 3165.7 3270.3 3482.8 3700.9 3925.3 4156.5 4394.4 4638.8 4889.6 5146.3 5408.3

— 7.3614 7.6134 7.8343 8.0333 8.2158 8.5435 8.8342 9.0976 9.3398 9.5652 9.7767 9.9764 10.1659 10.3463 10.5183

P = 0.40 MPa (143.63)

2553.6 2564.5 2646.8 2726.1 2804.8 2964.4 3129.2 3300.2 3477.9 3662.4 3853.9 4052.0 4256.5 4467.0 4682.8

2738.6 2752.8 2860.5 2964.2 3066.8 3273.4 3484.9 3702.4 3926.5 4157.3 4395.1 4639.4 4890.2 5146.8 5408.8

6.8959 6.9299 7.1706 7.3789 7.5662 7.8985 8.1913 8.4558 8.6987 8.9244 9.1362 9.3360 9.5256 9.7060 9.8780

P = 0.80 MPa (170.43)

6.7600 6.9665 7.1816 7.3724 7.5464 7.7079 8.0021 8.2674 8.5107 8.7367 8.9486 9.1485 9.3381 9.5185 9.6906

‡Note: Number in parentheses is temperature of saturated steam at the specified pressure. ˆ, H ˆ [=] J/g = kJ/kg; U Sˆ [=] kJ/kg K

Vˆ [=] m3 /kg;

1.6958 1.9364 2.172 2.406 2.639 3.103 3.565 4.028 4.490 4.952 5.414 5.875 6.337 6.799 7.260

P = 0.30 MPa (133.55)

P = 0.50 MPa (151.86)

Sat. 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300

—

0.2404 0.2608 0.2931 0.3241 0.3544 0.3843 0.4433 0.5018 0.5601 0.6181 0.6761 0.7340 0.7919 0.8497 0.9076

2576.8 2630.6 2715.5 2797.2 2878.2 2959.7 3126.0 3297.9 3476.2 3661.1 3852.8 4051.0 4255.6 4466.1 4681.8

2769.1 2839.3 2950.0 3056.5 3161.7 3267.1 3480.6 3699.4 3924.2 4155.6 4393.7 4638.2 4889.1 5145.9 5407.9

6.6628 6.8158 7.0384 7.2328 7.4089 7.5716 7.8673 8.1333 8.3770 8.6033 8.8153 9.0153 9.2050 9.3855 9.5575

982 Appendix A.III: The Thermodynamic Properties of Water and Steam Superheated Vapor (Continued) P = 1.00 MPa (179.91)

P = 1.20 MPa (187.99)

P = 1.40 MPa (195.07)

◦

T ( C)

Vˆ

ˆ U

ˆ H

Sˆ

Vˆ

ˆ U

ˆ H

Sˆ

Vˆ

ˆ U

ˆ H

Sˆ

Sat. 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300

0.194 44 0.2060 0.2327 0.2579 0.2825 0.3066 0.3541 0.4011 0.4478 0.4943 0.5407 0.5871 0.6335 0.6798 0.7261

2583.6 2621.9 2709.9 2793.2 2875.2 2957.3 3124.4 3296.8 3475.3 3660.4 3852.2 4050.5 4255.1 4465.6 4681.3

2778.1 2827.9 2942.6 3051.2 3157.7 3263.9 3478.5 3697.9 3923.1 4154.7 4392.9 4637.6 4888.6 5145.4 5407.4

6.5865 6.6940 6.9247 7.1229 7.3011 7.4651 7.7622 8.0290 8.2731 8.4996 8.7118 8.9119 9.1017 9.2822 9.4543

0.163 33 0.169 30 0.192 34 0.2138 0.2345 0.2548 0.2946 0.3339 0.3729 0.4118 0.4505 0.4892 0.5278 0.5665 0.6051

2588.8 2612.8 2704.2 2789.2 2872.2 2954.9 3122.8 3295.6 3474.4 3659.7 3851.6 4050.0 4254.6 4465.1 4680.9

2784.8 2815.9 2935.0 3045.8 3153.6 3260.7 3476.3 3696.3 3922.0 4153.8 4392.2 4637.0 4888.0 5144.9 5407.0

6.5233 6.5898 6.8294 7.0317 7.2121 7.3774 7.6759 7.9435 8.1881 8.4148 8.6272 8.8274 9.0172 9.1977 9.3698

0.140 84 0.143 02 0.163 50 0.182 28 0.2003 0.2178 0.2521 0.2860 0.3195 0.3528 0.3861 0.4192 0.4524 0.4855 0.5186

2592.8 2603.1 2698.3 2785.2 2869.2 2952.5 3121.1 3294.4 3473.6 3659.0 3851.1 4049.5 4254.1 4464.7 4680.4

2790.0 2803.3 2927.2 3040.4 3149.5 3257.5 3474.1 3694.8 3920.8 4153.0 4391.5 4636.4 4887.5 5144.4 5406.5

6.4693 6.4975 6.7467 6.9534 7.1360 7.3026 7.6027 7.8710 8.1160 8.3431 8.5556 8.7559 8.9457 9.1262 9.2984

P = 1.60 MPa (201.41)

Sat. 225 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300

0.123 80 0.132 87 0.141 84 0.158 62 0.174 56 0.190 05 0.2203 0.2500 0.2794 0.3086 0.3377 0.3668 0.3958 0.4248 0.4538

2596.0 2644.7 2692.3 2781.1 2866.1 2950.1 3119.5 3293.3 3472.7 3658.3 3850.5 4049.0 4253.7 4464.2 4679.9

2794.0 2857.3 2919.2 3034.8 3145.4 3254.2 3472.0 3693.2 3919.7 4152.1 4390.8 4635.8 4887.0 5143.9 5406.0

P = 1.80 MPa (207.15)

6.4218 6.5518 6.6732 6.8844 7.0694 7.2374 7.5390 7.8080 8.0535 8.2808 8.4935 8.6938 8.8837 9.0643 9.2364

0.110 42 0.116 73 0.124 97 0.140 21 0.154 57 0.168 47 0.195 50 0.2220 0.2482 0.2742 0.3001 0.3260 0.3518 0.3776 0.4034

P = 2.50 MPa (223.99)

Sat. 225 250 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300

0.079 98 0.080 27 0.087 00 0.098 90 0.109 76 0.120 10 0.130 14 0.139 98 0.159 30 0.178 32 0.197 16 0.215 90 0.2346 0.2532 0.2718 0.2905

Vˆ [=] m3 /kg;

2603.1 2605.6 2662.6 2761.6 2851.9 2939.1 3025.5 3112.1 3288.0 3468.7 3655.3 3847.9 4046.7 4251.5 4462.1 4677.8

2803.1 2806.3 2880.1 3008.8 3126.3 3239.3 3350.8 3462.1 3686.3 3914.5 4148.2 4387.6 4633.1 4884.6 5141.7 5404.0

2598.4 2636.6 2686.0 2776.9 2863.0 2947.7 3117.9 3292.1 3471.8 3657.6 3849.9 4048.5 4253.2 4463.7 4679.5

2797.1 2846.7 2911.0 3029.2 3141.2 3250.9 3469.8 3691.7 3918.5 4151.2 4390.1 4635.2 4886.4 5143.4 5405.6

P = 2.00 MPa (212.42)

6.3794 6.4808 6.6066 6.8226 7.0100 7.1794 7.4825 7.7523 7.9983 8.2258 8.4386 8.6391 8.8290 9.0096 9.1818

0.099 63 0.103 77 0.111 44 0.125 47 0.138 57 0.151 20 0.175 68 0.199 60 0.2232 0.2467 0.2700 0.2933 0.3166 0.3398 0.3631

P = 3.00 MPa (233.90)

6.2575 6.2639 6.4085 6.6438 6.8403 7.0148 7.1746 7.3234 7.5960 7.8435 8.0720 8.2853 8.4861 8.6762 8.8569 9.0291

ˆ, H ˆ [=] J/g = kJ/kg; U

0.066 68

— 0.070 58 0.081 14 0.090 53 0.099 36 0.107 87 0.116 19 0.132 43 0.148 38 0.164 14 0.179 80 0.195 41 0.210 98 0.226 52 0.242 06

Sˆ [=] kJ/kg K

2604.1 — 2644.0 2750.1 2843.7 2932.8 3020.4 3108.0 3285.0 3466.5 3653.5 3846.5 4045.4 4250.3 4460.9 4676.6

2804.2 — 2855.8 2993.5 3115.3 3230.9 3344.0 3456.5 3682.3 3911.7 4145.9 4385.9 4631.6 4883.3 5140.5 5402.8

2600.3 2628.3 2679.6 2772.6 2859.8 2945.2 3116.2 3290.9 3470.9 3657.0 3849.3 4048.0 4252.7 4463.3 4679.0

2799.5 2835.8 2902.5 3023.5 3137.0 3247.6 3467.6 3690.1 3917.4 4150.3 4389.4 4634.6 4885.9 5142.9 5405.1

6.3409 6.4147 6.5453 6.7664 6.9563 7.1271 7.4317 7.7024 7.9487 8.1765 8.3895 8.5901 8.7800 8.9607 9.1329

P = 3.50 MPa (242.60)

6.1869 — 6.2872 6.5390 6.7428 6.9212 7.0834 7.2338 7.5085 7.7571 7.9862 8.1999 8.4009 8.5912 8.7720 8.9442

0.057 07

— 0.058 72 0.068 42 0.076 78 0.084 53 0.091 96 0.099 18 0.113 24 0.126 99 0.140 56 0.154 02 0.167 43 0.180 80 0.194 15 0.207 49

2603.7 — 2623.7 2738.0 2835.3 2926.4 3015.3 3103.0 3282.1 3464.3 3651.8 3845.0 4044.1 4249.2 4459.8 4675.5

2803.4 — 2829.2 2977.5 3104.0 3222.3 3337.2 3450.9 3678.4 3908.8 4143.7 4384.1 4630.1 4881.9 5139.3 5401.7

6.1253 — 6.1749 6.4461 6.6579 6.8405 7.0052 7.1572 7.4339 7.6837 7.9134 8.1276 8.3288 8.5192 8.7000 8.8723

Appendix A.III: The Thermodynamic Properties of Water and Steam 983 Superheated Vapor (Continued) P = 4.0 MPa (250.40) ◦

P = 4.5 MPa (257.49)

P = 5.0 MPa (263.99)

T ( C)

Vˆ

ˆ U

ˆ H

Sˆ

Vˆ

ˆ U

ˆ H

Sˆ

Vˆ

ˆ U

ˆ H

Sˆ

Sat. 275 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300

0.049 78 0.054 57 0.058 84 0.066 45 0.073 41 0.080 02 0.086 43 0.098 85 0.110 95 0.122 87 0.134 69 0.146 45 0.158 17 0.169 87 0.181 56

2602.3 2667.9 2725.3 2826.7 2919.9 3010.2 3099.5 3279.1 3462.1 3650.0 3843.6 4042.9 4248.0 4458.6 4674.3

2801.4 2886.2 2960.7 3092.5 3213.6 3330.3 3445.3 3674.4 3905.9 4141.5 4382.3 4628.7 4880.6 5138.1 5400.5

6.0701 6.2285 6.3615 6.5821 6.7690 6.9363 7.0901 7.3688 7.6198 7.8502 8.0647 8.2662 8.4567 8.6376 8.8100

0.044 06 0.047 30 0.051 35 0.058 40 0.064 75 0.070 74 0.076 51 0.087 65 0.098 47 0.109 11 0.119 65 0.130 13 0.140 56 0.150 98 0.161 39

2600.1 2650.3 2712.0 2817.8 2913.3 3005.0 3095.3 3276.0 3459.9 3648.3 3842.2 4041.6 4246.8 4457.5 4673.1

2798.3 2863.2 2943.1 3080.6 3204.7 3323.3 3439.6 3670.5 3903.0 4139.3 4380.6 4627.2 4879.3 5136.9 5399.4

6.0198 6.1401 6.2828 6.5131 6.7047 6.8746 7.0301 7.3110 7.5631 7.7942 8.0091 8.2108 8.4015 8.5825 8.7549

0.039 44 0.041 41 0.045 32 0.051 94 0.057 81 0.063 30 0.068 57 0.078 69 0.088 49 0.098 11 0.107 62 0.117 07 0.126 48 0.135 87 0.145 26

2597.1 2631.3 2698.0 2808.7 2906.6 2999.7 3091.0 3273.0 3457.6 3646.6 3840.7 4040.4 4245.6 4456.3 4672.0

2794.3 2838.3 2924.5 3068.4 3195.7 3316.2 3433.8 3666.5 3900.1 4137.1 4378.8 4625.7 4878.0 5135.7 5398.2

5.9734 6.0544 6.2084 6.4493 6.6459 6.8186 6.9759 7.2589 7.5122 7.7440 7.9593 8.1612 8.3520 8.5331 8.7055

P = 6.0 MPa (275.64)

Sat. 300 350 400 450 500 550 600 700 800 900 1000 1100 1200 1300

0.032 44 0.036 16 0.042 23 0.047 39 0.052 14 0.056 65 0.061 01 0.065 25 0.073 52 0.081 60 0.089 58 0.097 49 0.105 36 0.113 21 0.121 06

2589.7 2667.2 2789.6 2892.9 2988.9 3082.2 3174.6 3266.9 3453.1 3643.1 3837.8 4037.8 4243.3 4454.0 4669.6

2784.3 2884.2 3043.0 3177.2 3301.8 3422.2 3540.6 3658.4 3894.2 4132.7 4375.3 4622.7 4875.4 5133.3 5396.0

P = 7.0 MPa (285.88)

5.8892 6.0674 6.3335 6.5408 6.7193 6.8803 7.0288 7.1677 7.4234 7.6566 7.8727 8.0751 8.2661 8.4474 8.6199

P = 9.0 MPa (303.40)

Sat. 325 350 400 450 500 550 600 650 700 800 900 1000 1100 1200 1300

0.020 48 0.023 27 0.025 80 0.029 93 0.033 50 0.036 77 0.039 87 0.042 85 0.045 74 0.048 57 0.054 09 0.059 50 0.064 85 0.070 16 0.075 44 0.080 72

Vˆ [=] m3 /kg;

2557.8 2646.6 2724.4 2848.4 2955.2 3055.2 3152.2 3248.1 3343.6 3439.3 3632.5 3829.2 4030.3 4236.3 4447.2 4662.7

2742.1 2856.0 2956.6 3117.8 3256.6 3386.1 3511.0 3633.7 3755.3 3876.5 4119.3 4364.8 4614.0 4867.7 5126.2 5389.2

0.027 37 0.029 47 0.035 24 0.039 93 0.044 16 0.048 14 0.051 95 0.055 65 0.062 83 0.069 81 0.076 69 0.083 50 0.090 27 0.097 03 0.103 77

2580.5 2632.2 2769.4 2878.6 2978.0 3073.4 3167.2 3260.7 3448.5 3639.5 3835.0 4035.3 4240.9 4451.7 4667.3

2772.1 2838.4 3016.0 3158.1 3287.1 3410.3 3530.9 3650.3 3888.3 4128.2 4371.8 4619.8 4872.8 5130.9 5393.7

P = 8.0 MPa (295.06)

5.8133 5.9305 6.2283 6.4478 6.6327 6.7975 6.9486 7.0894 7.3476 7.5822 7.7991 8.0020 8.1933 8.3747 8.5473

0.023 52 0.024 26 0.029 95 0.034 32 0.038 17 0.041 75 0.045 16 0.048 45 0.054 81 0.060 97 0.067 02 0.073 01 0.078 96 0.084 89 0.090 80

P = 10.0 MPa (311.06)

5.6772 5.8712 6.0361 6.2854 6.4844 6.6576 6.8142 6.9589 7.0943 7.2221 7.4596 7.6783 7.8821 8.0740 8.2556 8.4284

ˆ, H ˆ [=] J/g = kJ/kg; U

0.018 026 0.019 861 0.022 42 0.026 41 0.029 75 0.032 79 0.035 64 0.038 37 0.041 0l 0.043 58 0.048 59 0.053 49 0.058 32 0.063 12 0.067 89 0.072 65 Sˆ [=] kJ/kg K

2544.4 2610.4 2699.2 2832.4 2943.4 3045.8 3144.6 3241.7 3338.2 3434.7 3628.9 3826.3 4027.8 4234.0 4444.9 4460.5

2724.7 2809.1 2923.4 3096.5 3240.9 3373.7 3500.9 3625.3 3748.2 3870.5 4114.8 4361.2 4611.0 4865.1 5123.8 5387.0

2569.8 2590.9 2747.7 2863.8 2966.7 3064.3 3159.8 3254.4 3443.9 3636.0 3832.1 4032.8 4238.6 4449.5 4665.0

2758.0 2785.0 2987.3 3138.3 3272.0 3398.3 3521.0 3642.0 3882.4 4123.8 4368.3 4616.9 4870.3 5128.5 5391.5

5.7432 5.7906 6.1301 6.3634 6.5551 6.7240 6.8778 7.0206 7.2812 7.5173 7.7351 7.9384 8.1300 8.3115 8.4842

P = 12.5 MPa (327.89)

5.6141 5.7568 5.9443 6.2120 6.4190 6.5966 6.7561 6.9029 7.0398 7.1687 7.4077 7.6272 7.8315 8.0237 8.2055 8.3783

0.013 495

— 0.016 126 0.020 00 0.022 99 0.025 60 0.028 01 0.030 29 0.032 48 0.034 60 0.038 69 0.042 67 0.046 58 0.050 45 0.054 30 0.058 13

2505.1 — 2624.6 2789.3 2912.5 3021.7 3125.0 3225.4 3324.4 3422.9 3620.0 3819.1 4021.6 4228.2 4439.3 4654.8

2673.8 — 2826.2 3039.3 3199.8 3341.8 3475.2 3604.0 3730.4 3855.3 4103.6 4352.5 4603.8 4858.8 5118.0 5381.4

5.4624 — 5.7118 6.0417 6.2719 6.4618 6.6290 6.7810 6.9218 7.0536 7.2965 7.5182 7.7237 7.9165 8.0987 8.2717

984 Appendix A.III: The Thermodynamic Properties of Water and Steam Superheated Vapor (Continued) P = 15.0 MPa (342.24) ◦

P = 17.5 MPa (354.75)

P = 20.0 MPa (365.81)

T ( C)

Vˆ

ˆ U

ˆ H

Sˆ

Vˆ

ˆ U

ˆ H

Sˆ

Vˆ

ˆ U

ˆ H

Sˆ

Sat. 350 400 450 500 550 600 650 700 800 900 1000 1100 1200 1300

0.010 337 0.011 470 0.015 649 0.018 445 0.020 80 0.022 93 0.024 91 0.026 80 0.028 61 0.032 10 0.035 46 0.038 75 0.042 00 0.045 23 0.048 45

2455.5 2520.4 2740.7 2879.5 2996.6 3104.7 3208.6 3310.3 3410.9 3610.9 3811.9 4015.4 4222.6 4433.8 4649.1

2610.5 2692.4 2975.5 3156.2 3308.6 3448.6 3582.3 3712.3 3840.1 4092.4 4343.8 4596.6 4852.6 5112.3 5376.0

5.3098 5.4421 5.8811 6.1404 6.3443 6.5199 6.6776 6.8224 6.9572 7.2040 7.4279 7.6348 7.8283 8.0108 8.1840

0.007 920

2390.2 — 2685.0 2844.2 2970.3 3083.9 3191.5 3296.0 3398.7 3601.8 3804.7 4009.3 4216.9 4428.3 4643.5

2528.8 — 2902.9 3109.7 3274.1 3421.4 3560.1 3693.9 3824.6 4081.1 4335.1 4589.5 4846.4 5106.6 5370.5

5.1419 — 5.7213 6.0184 6.2383 6.4230 6.5866 6.7357 6.8736 7.1244 7.3507 7.5589 7.7531 7.9360 8.1093

0.005 834

2293.0 — 2619.3 2806.2 2942.9 3062.4 3174.0 3281.4 3386.4 3592.7 3797.5 4003.1 4211.3 4422.8 4638.0

2409.7 — 2818.1 3060.1 3238.2 3393.5 3537.6 3675.3 3809.0 4069.7 4326.4 4582.5 4840.2 5101.0 5365.1

4.9269 — 5.5540 5.9017 6.1401 6.3348 6.5048 6.6582 6.7993 7.0544 7.2830 7.4925 7.6874 7.8707 8.0442

— 0.012 447 0.015 174 0.017 358 0.019 288 0.021 06 0.022 74 0.024 34 0.027 38 0.030 31 0.033 16 0.035 97 0.038 76 0.041 54

P = 25.0 MPa

375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300

0.001 973 0.006 004 0.007 881 0.009 162 0.011 123 0.012 724 0.014 137 0.015 433 0.016 646 0.018 912 0.021 045 0.023 10 0.025 12 0.027 11 0.029 10

1798.7 2430.1 2609.2 2720.7 2884.3 3017.5 3137.9 3251.6 3361.3 3574.3 3783.0 3990.9 4200.2 4412.0 4626.9

1848.0 2580.2 2806.3 2949.7 3162.4 3335.6 3491.4 3637.4 3777.5 4047.1 4309.1 4568.5 4828.2 5089.9 5354.4

0.001 641 0.001 908 0.002 532 0.003 693 0.005 622 0.006 984 0.008 094 0.009 063 0.009 941 0.011 523 0.012 962 0.014 324 0.015 642 0.016 940 0.018 229

Vˆ [=] m3 /kg;

1677.1 1854.6 2096.9 2365.1 2678.4 2869.7 3022.6 3158.0 3283.6 3517.8 3739.4 3954.6 4167.4 4380.1 4594.3

1742.8 1930.9 2198.1 2512.8 2903.3 3149.1 3346.4 3520.6 3681.2 3978.7 4257.9 4527.6 4793.1 5057.7 5323.5

0.009 942 0.012 695 0.014 768 0.016 555 0.018 178 0.019 693 0.021 13 0.023 85 0.026 45 0.028 97 0.031 45 0.033 91 0.036 36

P = 30.0 MPa

4.0320 5.1418 5.4723 5.6744 5.9592 6.1765 6.3602 6.5229 6.6707 6.9345 7.1680 7.3802 7.5765 7.7605 7.9342

0.001 789 0.002 790 0.005 303 0.006 735 0.008 678 0.010 168 0.011 446 0.012 596 0.013 661 0.015 623 0.017 448 0.019 196 0.020 903 0.022 589 0.024 266

P = 40.0 MPa

375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300

—

1737.8 2067.4 2455.1 2619.3 2820.7 2970.3 3100.5 3221.0 3335.8 3555.5 3768.5 3978.8 4189.2 4401.3 4616.0

1791.5 2151.1 2614.2 2821.4 3081.1 3275.4 3443.9 3598.9 3745.6 4024.2 4291.9 4554.7 4816.3 5079.0 5344.0

P = 35.0 MPa

3.9305 4.4728 5.1504 5.4424 5.7905 6.0342 6.2331 6.4058 6.5606 6.8332 7.0718 7.2867 7.4845 7.6692 7.8432

0.001 700 0.002 100 0.003 428 0.004 961 0.006 927 0.008 345 0.009 527 0.010 575 0.011 533 0.013 278 0.014 883 0.016 410 0.017 895 0.019 360 0.020 815

P = 50.0 MPa

3.8290 4.1135 4.5029 4.9459 5.4700 5.7785 6.0114 6.2054 6.3750 6.6662 6.9150 7.1356 7.3364 7.5224 7.6969

ˆ, H ˆ [=] J/g = kJ/kg; U

0.001 559 0.001 731 0.002 007 0.002 486 0.003 892 0.005 118 0.006 112 0.006 966 0.007 727 0.009 076 0.010 283 0.011 411 0.012 496 0.013 561 0.014 616

Sˆ [=] kJ/kg K

1638.6 1788.1 1959.7 2159.6 2525.5 2763.6 2942.0 3093.5 3230.5 3479.8 3710.3 3930.5 4145.7 4359.1 4572.8

1716.6 1874.6 2060.0 2284.0 2720.1 3019.5 3247.6 3441.8 3616.8 3933.6 4224.4 4501.1 4770.5 5037.2 5303.6

1702.9 1914.1 2253.4 2498.7 2751.9 2921.0 3062.0 3189.8 3309.8 3536.7 3754.0 3966.7 4178.3 4390.7 4605.1

1762.4 1987.6 2373.4 2672.4 2994.4 3213.0 3395.5 3559.9 3713.5 4001.5 4274.9 4541.1 4804.6 5068.3 5333.6

3.8722 4.2126 4.7747 5.1962 5.6282 5.9026 6.1179 6.3010 6.4631 6.7450 6.9886 7.2064 7.4057 7.5910 7.7653

P = 60.0 MPa

3.7639 4.0031 4.2734 4.5884 5.1726 5.5485 5.8178 6.0342 6.2189 6.5290 6.7882 7.0146 7.2184 7.4058 7.5808

0.001 503 0.001 634 0.001 817 0.002 085 0.002 956 0.003 956 0.004 834 0.005 595 0.006 272 0.007 459 0.008 508 0.009 480 0.010 409 0.011 317 0.012 215

1609.4 1745.4 1892.7 2053.9 2390.6 2658.8 2861.1 3028.8 3177.2 3441.5 3681.0 3906.4 4124.1 4338.2 4551.4

1699.5 1843.4 2001.7 2179.0 2567.9 2896.2 3151.2 3364.5 3553.5 3889.1 4191.5 4475.2 4748.6 5017.2 5284.3

3.7141 3.9318 4.1626 4.4121 4.9321 5.3441 5.6452 5.8829 6.0824 6.4109 6.6805 6.9127 7.1195 7.3083 7.4837

Appendix A.III: The Thermodynamic Properties of Water and Steam 985 Compressed Liquid P = 5 MPa (263.99) ◦

T ( C)

Sat. 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340

Vˆ

ˆ U

P = 10 MPa (311.06)

ˆ H

0.001 285 9 1147.8 1154.2 0.000 997 7 0.04 5.04 0.000 999 5 83.65 88.65 0.001 005 6 166.95 171.97 0.001 014 9 250.23 255.30 0.001 026 8 333.72 338.85 0.001 041 0 417.52 422.72 0.001 057 6 501.80 507.09 0.001 076 8 586.76 592.15 0.001 098 8 672.62 678.12 0.001 124 0 759.63 765.25 0.001 153 0 848.1 853.9 0.001 186 6 938.4 944.4 0.001 226 4 1031.4 1037.5 0.001 274 9 1127.9 1134.3

Vˆ

ˆ U

ˆ H

Sˆ

Vˆ

ˆ U

ˆ H

Sˆ

2.9202 0.0001 0.2956 0.5705 0.8285 1.0720 1.3030 1.5233 1.7343 1.9375 2.1341 2.3255 2.5128 2.6979 2.8830

0.001 452 4 0.000 995 2 0.000 997 2 0.001 003 4 0.001 012 7 0.001 024 5 0.001 038 5 0.001 054 9 0.001 073 7 0.001 095 3 0.001 119 9 0.001 148 0 0.001 180 5 0.001 218 7 0.001 264 5 0.001 321 6 0.001 397 2

1393.0 0.09 83.36 166.35 249.36 332.59 416.12 500.08 584.68 670.13 756.65 844.5 934.1 1026.0 1121.1 1220.9 1328.4

1407.6 10.04 93.33 176.38 259.49 342.83 426.50 510.64 595.42 681.08 767.84 856.0 945.9 1038.1 1133.7 1234.1 1342.3

3.3596 0.0002 0.2945 0.5686 0.8258 1.0688 1.2992 1.5189 1.7292 1.9317 2.1275 2.3178 2.5039 2.6872 2.8699 3.0548 3.2469

0.001 658 1 0.000 992 8 0.000 995 0 0.001 001 3 0.001 010 5 0.001 022 2 0.001 036 1 0.001 052 2 0.001 070 7 0.001 091 8 0.001 115 9 0.001 143 3 0.001 174 8 0.001 211 4 0.001 255 0 0.001 308 4 0.001 377 0 0.001 472 4 0.001 631 1

1585.6 0.15 83.06 165.76 248.51 331.48 414.74 498.40 582.66 667.71 753.76 841.0 929.9 1020.8 1114.6 1212.5 1316.6 1431.1 1567.5

1610.5 15.05 97.99 180.78 263.67 346.81 430.28 514.19 598.72 684.09 770.50 858.2 947.5 1039.0 1133.4 1232.1 1337.3 1453.2 1591.9

3.6848 0.0004 0.2934 0.5666 0.8232 1.0656 1.2955 1.5145 1.7242 1.9260 2.1210 2.3104 2.4953 2.6771 2.8576 3.0393 3.2260 3.4247 3.6546

P = 20 MPa (365.81)

Sat. 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 380

0.002 036 0.000 990 4 0.000 992 8 0.000 999 2 0.001 008 4 0.001 019 9 0.001 033 7 0.001 049 6 0.001 067 8 0.001 088 5 0.001 112 0 0.001 138 8 0.001 169 3 0.001 204 6 0.001 246 2 0.001 296 5 0.001 359 6 0.001 443 7 0.001 568 4 0.001 822 6

Vˆ [=] m3 /kg;

1785.6 0.19 82.77 165.17 247.68 330.40 413.39 496.76 580.69 665.35 750.95 837.7 925.9 1016.0 1108.6 1204.7 1306.1 1415.7 1539.7 1702.8

1826.3 20.01 102.62 185.16 267.85 350.80 434.06 517.76 602.04 687.12 773.20 860.5 949.3 1040.0 1133.5 1230.6 1333.3 1444.6 1571.0 1739.3

P = 15 MPa (342.24)

Sˆ

P = 30 MPa

4.0139 0.0004 0.2923 0.5646 0.8206 1.0624 1.2917 1.5102 1.7193 1.9204 2.1147 2.3031 2.4870 2.6674 2.8459 3.0248 3.2071 3.3979 3.6075 3.8772

ˆ, H ˆ [=] J/g = kJ/kg; U

0.000 985 6 0.000 988 6 0.000 995 1 0.001 004 2 0.001 015 6 0.001 029 0 0.001 044 5 0.001 062 1 0.001 082 1 0.001 104 7 0.001 130 2 0.001 159 0 0.001 192 0 0.001 230 3 0.001 275 5 0.001 330 4 0.001 399 7 0.001 492 0 0.001 626 5 0.001 869 1

Sˆ [=] kJ/kg K

0.25 82.17 164.04 246.06 328.30 410.78 493.59 576.88 660.82 745.59 831.4 918.3 1006.9 1097.4 1190.7 1287.9 1390.7 1501.7 1626.6 1781.4

29.82 111.84 193.89 276.19 358.77 441.66 524.93 608.75 693.28 778.73 865.3 953.1 1042.6 1134.3 1229.0 1327.8 1432.7 1546.5 1675.4 1837.5

P = 50 MPa

0.0001 0.2899 0.5607 0.8154 1.0561 1.2844 1.5018 1.7098 1.9096 2.1024 2.2893 2.4711 2.6490 2.8243 2.9986 3.1741 3.3539 3.5426 3.7494 4.0012

0.000 976 6 0.000 980 4 0.000 987 2 0.000 996 2 0.001 007 3 0.001 020 1 0.001 034 8 0.001 051 5 0.001 070 3 0.001 091 2 0.001 114 6 0.001 140 8 0.001 107 2 0.001 203 4 0.001 241 5 0.001 286 0 0.001 338 8 0.001 403 2 0.001 483 8 0.001 588 4

0.20 81.00 161.86 242.98 324.34 405.88 487.65 569.77 652.41 735.69 819.7 904.7 990.7 1078.1 1167.2 1258.7 1353.3 1452.0 1556.0 1667.2

49.03 130.02 211.21 292.79 374.70 456.89 539.39 622.35 705.92 790.25 875.5 961.7 1049.2 1138.2 1229.3 1323.0 1420.2 1522.1 1630.2 1746.6

0.0014 0.2848 0.5527 0.8502 1.0440 1.2703 1.4857 1.6915 1.8891 2.0794 2.2634 2.4419 2.6158 2.7860 2.9537 3.1200 3.2868 3.4557 3.6291 3.8101

986 Appendix A.III: The Thermodynamic Properties of Water and Steam Saturated Solid–Vapor Specific Volume Sat. Sat. Solid Vapor

Internal Energy Sat. Sat. Solid Vapor Evap.

Temp (◦ C)

Press. (kPa)

T

P

Vˆ S × 103

Vˆ V

ˆS U

ˆ ΔU

0.01 0 −2 −4 −6 −8 −10 −12 −14 −16 −18 −20 −22 −24 −26 −28 −30 −32 −34 −36 −38 −40

0.6113 0.6108 0.5176 0.4375 0.3689 0.3102 0.2602 0.2176 0.1815 0.1510 0.1252 0.1035 0.0853 0.0701 0.0574 0.0469 0.0381 0.0309 0.0250 0.0201 0.0161 0.0129

1.0908 1.0908 1.0904 1.0901 1.0898 1.0894 1.0891 1.0888 1.0884 1.0881 1.0878 1.0874 1.0871 1.0868 1.0864 1.0861 1.0858 1.0854 1.0851 1.0848 1.0844 1.0841

206.1 206.3 241.7 283.8 334.2 394.4 466.7 553.7 658.8 786.0 940.5 1128.6 1358.4 1640.1 1986.4 2413.7 2943 3600 4419 5444 6731 8354

−333.40 −333.43 −337.62 −341.78 −345.91 −350.02 −354.09 −358.14 −362.15 −366.14 −370.10 −374.03 −377.93 −381.80 −385.64 −389.45 −393.23 −396.98 −400.71 −404.40 −408.06 −411.70

2708.7 2708.8 2710.2 2711.6 2712.9 2714.2 2715.5 2716.8 2718.0 2719.2 2720.4 2721.6 2722.7 2723.7 2724.8 2725.8 2726.8 2727.8 2728.7 2729.6 2730.5 2731.3

Vˆ [=] m3 /kg;

ˆ, H ˆ [=] J/g = kJ/kg; U

Sˆ [=] kJ/kg K

Enthalpy Sat. Liquid

ˆV U

2375.3 2375.3 2372.6 2369.8 2367.0 2364.2 2361.4 2358.7 2355.9 2353.1 2350.3 2347.5 2344.7 2342.0 2339.2 2336.4 2333.6 2330.8 2328.0 2325.2 2322.4 2319.6

Entropy

Evap.

Sat. Vapor

Sat. Liquid

Evap.

Sat. Vapor

ˆS H

ˆ ΔH

ˆV H

SˆS

ΔSˆ

SˆV

−333.40 −333.43 −337.62 −341.78 −345.91 −350.02 −354.09 −358.14 −362.15 −366.14 −370.10 −374.03 −377.93 −381.80 −385.64 −389.45 −393.23 −396.98 −400.71 −404.40 −408.06 −411.70

2834.8 2834.8 2835.3 2835.7 2836.2 2836.6 2837.0 2837.3 2837.6 2837.9 2838.2 2838.4 2838.6 2838.7 2838.9 2839.0 2839.0 2839.1 2839.1 2839.1 2839.0 2838.9

2501.4 2501.3 2497.7 2494.0 2490.3 2486.6 2482.9 2479.2 2475.5 2471.8 2468.1 2464.3 2460.6 2456.9 2453.2 2449.5 2445.8 2442.1 2438.4 2434.7 2430.9 2427.2

−1.221 −1.221 −1.237 −1.253 −1.268 −1.284 −1.299 −1.315 −1.331 −1.346 −1.362 −1.377 −1.393 −1.408 −1.424 −1.439 −1.455 −1.471 −1.486 −1.501 −1.517 −1.532

10.378 10.378 10.456 10.536 10.616 10.698 10.781 10.865 10.950 11.036 11.123 11.212 11.302 11.394 11.486 11.580 11.676 11.773 11.872 11.972 12.073 12.176

9.156 9.157 9.219 9.283 9.348 9.414 9.481 9.550 9.619 9.690 9.762 9.835 9.909 9.985 10.062 10.141 10.221 10.303 10.386 10.470 10.556 10.644

Appendix A.IV: Enthalpies and Gibbs Energies of Formation 987

A.IV Enthalpies and Gibbs Energies of Formation Standard Enthalpies and Gibbs Energies of Formation at 298.15 K for One Mole of Each Substance from Its Elements ◦

◦

Δf H Δf G State (See Note) kJ/mol of the Substance Formed

Chemical Species Paraffinic Hydrocarbons Methane Ethane Propane n-Butane n-Pentane n-Pentane n-Hexane n-Heptane n-Octane n-Octane

CH4 C2 H6 C3 H8 C4 H10 C5 H12 C5 H12 C6 H14 C7 H16 C8 H18 C8 H18

(g) (g) (g) (g) (g) (l) (g) (g) (g) (l)

−74.5 −83.8 −104.7 −125.8 −146.8 −173.1 −166.9 −187.8 −208.8 −255.1

−50.5 −31.9 −24.3 −16.6 −8.7 −9.2 0.2 8.3 16.3 —

Unsaturated Hydrocarbons Acetylene Ethylene Propylene 1-Butene 1-Pentene 1-Hexene 1-Heptene 1,3-Butadiene

C2 H2 C2 H4 C3 H6 C4 H8 C5 H10 C6 H12 C7 H14 C4 H6

(g) (g) (g) (g) (g) (g) (g) (g)

227.5 52.5 19.7 1.2 −21.3 −42.0 −62.8 109.2

210.0 68.5 62.2 70.3 78.4 86.8 — 149.8

Aromatic Hydrocarbons Benzene Benzene Ethylbenzene Naphthalene Styrene Toluene Toluene

C6 H6 C6 H6 C8 H10 C10 H8 C8 H8 C7 H8 C7 H8

(g) (l) (g) (s) (g) (g) (l)

82.9 49.1 29.9 78.5 147.4 50.2 12.2

129.7 124.5 130.9 — 213.9 122.1 113.6

Cyclic Hydrocarbons Cyclohexane Cyclohexane Cyclopropane Methylcyclohexane Methylcyclohexane Cyclohexene

C6 H12 C6 H12 C3 H6 C7 H14 C7 H14 C6 H10

(g) (l) (g) (g) (l) (l)

−123.1 −156.2 53.3 −154.8 −190.2

31.9 26.9 104.5 27.5 20.6 102.5

Oxygenated Hydrocarbons Acetaldehyde Acetic Acid Acetic Acid 1,2-Ethanediol Ethanol Ethanol Ethyl acetate

C2 H4 O CH3 COOH CH3 COOH C2 H6 O2 C2 H6 O C2 H6 O CH3 COOC2 H5

(g) (l) (aq) (l) (g) (l) (l)

−166.2 −484.5 −486.1 −454.8 −235.1 −277.7 −478.8

−134.4 −389.9 −396.5 −323.1 −168.5 −174.8 −332.4

988 Appendix A.IV: Enthalpies and Gibbs Energies of Formation Appendix A.IV (Continued) State (See Note)

Chemical Species Ethylene oxide Formaldehyde Formic acid Methanol Methanol Phenol Inorganic Compounds Aluminum oxide Aluminum chloride Ammonia Ammonia Ammonium nitrate Ammonium chloride Barium oxide Barium chloride Bromine Bromine Calcium carbide Calcium carbonate Calcium chloride Calcium chloride Calcium chloride Calcium hydroxide Calcium hydroxide Calcium oxide Carbon dioxide Carbon dioxide Carbon disulfide Carbon monoxide Carbon tetrachloride Carbonic acid Hydrochloric acid Hydrochloric acid Hydrogen bromide Hydrogen cyanide Hydrogen cyanide Hydrogen flouride Hydrogen iodide Hydrogen peroxide Hydrogen sulfide Hydrogen sulfide Iodine Iodine Iron (II) Iron (III) oxide (hematite) Iron (II) sulfide Iron sulfide (pyrite) Lead oxide Lead oxide

C2 H4 O CH2 O HCOOH CH4 O CH4 O C6 H5 OH

(g) (g) (l) (g) (l) (s)

Al2 O3 (s, α) AlCl3 (s) NH3 (g) (aq) NH3 NH4 NO3 (s) NH4 Cl (s) BaO (s) (s) BaCl2 Br2 (l) (g) Br2 CaC2 (s) CaCO3 (s) (s) CaCl2 CaCl2 (aq) CaCl2 ·6H2 O (s) (s) Ca(OH)2 Ca(OH)2 (aq) CaO (s) (g) CO2 CO2 (aq) CS2 (l) CO (g) (l) CCl4 (aq) H2 CO3 HCl (g) HCl (aq) HBr (g) HCN (g) HCN (l) HF (g) HI (g) H2 O2 (l) (g) H2 S H2 S (aq) I2 (g) (s) I2 FeO (s) Fe2 O3 (s) FeS (s, α) FeS2 (s) PbO (s, yellow) PbO (s, red)

Δf H ◦

Δ f G◦

kJ/mol of the Substance Formed −52.6 −108.6 −424.7 −200.7 238.7 −165.0

−13.0 −102.5 −361.4 −162.0 −166.3 −50.9

−1675.7 −704.2 −46.1 — −365.6 −314.4 −553.5 −858.6 0 30.9 −59.8 −1206.9 −795.8 — −2607.9 −986.1 — −635.1 −393.5 −413.8 89.7 −110.5 −128.4 −699.7 −92.3 −167.2 −36.4 135.1 108.9 −271.1 26.5 −187.8 −20.6 −39.7 62.3 0 −272.0 −824.2 −100.0 −178.2 −217.3 −219.0

−1582.3 −628.8 −16.5 −26.5 −183.9 −202.9 −525.1 −810.4 0 3.1 −64.9 −1128.8 −748.1 −8101.9 — −898.5 −868.1 −604.0 −394.4 −386.0 65.3 −137.2 −65.2 −623.1 −95.3 −131.2 −53.5 124.7 125.0 −273.2 1.7 −120.4 −33.6 −27.8 19.8 0 — −742.2 −100.4 −166.9 −187.9 −188.9

Appendix A.IV: Enthalpies and Gibbs Energies of Formation 989 Appendix A.IV (Continued) ◦

Chemical Species Lead dioxide Lithium chloride Lithium chloride Lithium chloride Lithium chloride Magnesium oxide Magnesium carbonate Magnesium chloride Mercury (I) chloride Mercury (II) chloride Nitric acid Nitric acid Nitric oxide Nitrogen dioxide Nitrous oxide Nitrogen tetroxide Potassium chloride Silicon dioxide Silver bromide Silver chloride Silver nitrate Sodium bicarbonate Sodium carbonate Sodium carbonate Sodium chloride Sodium chloride Sodium hydroxide Sodium hydroxide Sodium sulfate Sodium sulfate·decahydrate Sulfur Sulfur Sulfur Sulfur dioxide Sulfur trioxide Sulfur trioxide Sulfuric acid Sulfuric acid Water Water Zinc oxide

◦

Δf H Δf G State (See Note) kJ/mol of the Substance Formed

PbO2 LiCl LiCl·H2 O LiCl·2H2 O LiCl·3H2 O MgO MgCO3 MgCl2 Hg2 Cl2 HgCl2 HNO3 HNO3 NO NO2 N2 O N2 O4 KCl SiO2 AgBr AgCl AgNO3 NaHCO3 Na2 CO3 Na2 CO3 ·10H2 O NaCl NaCl NaOH NaOH Na2 SO4 Na2 SO4 ·10H2 O S2 S2 S2 SO2 SO3 SO3 H2 SO4 H2 SO4 H2 O H2 O ZnO

(s) (s) (s) (s) (s) (s) (s) (s) (s) (s) (l) (aq) (g) (g) (g) (g) (s) (s, α) (s) (s) (s) (s) (s) (s) (s) (aq) (s) (aq) (s) (s) (g) (l) (s) (g) (g) (l) (l) (aq) (g) (l) (s)

−277.4 −408.6 −712.6 −1012.7 −1311.3 −601.7 −1095.8 −641.3 −265.2 −224.3 −174.1 — 90.3 33.2 82.1 9.2 −436.8 −910.9 −100.4 −127.1 −124.4 −945.6 −1130.7 −4081.3 −411.2 — −425.6 — −1382.8 −4322.5 129.8 1.1 0 −296.8 −395.7 −441.0 −814.0 — −241.8 −285.8 −348.3

−217.3 — — — — −569.4 −1012.1 −591.8 −210.8 −178.6 −80.7 −111.3 86.6 51.3 104.2 97.9 −409.1 −856.6 −96.9 −109.8 −33.4 −847.9 −1044.4 — −384.1 −393.1 −379.5 −419.2 −1265.2 −3642.3 81.0 0.3 0 −300.2 −371.1 — −690.0 −744.5 −228.6 −237.1 −318.3

Taken from TRC Thermodynamic Tables—Hydrocarbons, Thermodynamics Research Center, Texas A&M University System, College Station, Tex.; “The NBS Tables of Chemical Thermodynamic Properties,” I. Physical and Chemical Reference Data, Vol. 11, Suppl. 2, 1982; and a collection of other sources. Because of the variety of sources used here and in the PROPERTY program, there may be very small differences between the data here and in that program. Note: All standard states are at 25◦ C. The standard state for a gas, designated by (g), is the pure ideal gas at 1 bar. For liquids (l) and solids (s), the standard state is the substance in that state of aggregation at 1 bar and 25◦ C. For solutes in aqueous solution, denoted by (aq), the standard state is the hypothetical ideal 1-molal solution of the solute in water at 1 bar and 25◦ C.

990 Appendix A.V: Heats of Combustion

A.V Heats of Combustion

Compound

Formula

State

Hydrogen Carbon Carbon monoxide Methane Ethane Propane Propane n-Butane n-Butane 2-Methylpropane (isobutane) 2-Methylpropane (isobutane) n-Pentane n-Pentane n-Hexane n-Hexane n-Heptane n-Heptane n-Octane n-Octane n-Nonane n-Nonane n-Decane n-Decane n-Dodecane n-Dodecane n-Hexadecane n-Hexadecane n-Eicosane n-Eicosane Benzene Benzene Methylbenzene (toluene) Methylbenzene (toluene) Ethylbenzene Ethylbenzene 1,2-Dimethylbenzene (o-xylene) 1,2-Dimethylbenzene (o-xylene) 1,3-Dimethylbenzene (m-xylene) 1,3-Dimethylbenzene (m-xylene) 1,4-Dimethylbenzene (p-xylene) 1,4-Dimethylbenzene (p-xylene) Cyclopentane

H2 C CO CH4 C2 H6 C3 H8 C3 H8 C4 H10 C4 H10 C4 H10 C4 H10 C5 H12 C5 H12 C6 H14 C6 H14 C7 H16 C7 H16 C8 H18 C8 H18 C9 H20 C9 H20 C10 H22 C10 H22 C12 H26 C12 H26 C16 H34 C16 H34 C20 H42 C20 H42 C6 H6 C6 H6 C7 H8 C7 H8 C8 H10 C8 H10 C8 H10 C8 H10 C8 H10 C8 H10 C8 H10 C8 H10 C5 H10

(g) (s, graphite) (g) (g) (g) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g) (l) (g)

Heat of Combustion in kJ/mol H2 O (l) H2 O (g) CO2 (g) CO2 (g) 285.840 393.513 282.989 890.35 1559.88 2220.05 2204.06 2878.52 2857.02 2871.65 2851.92 3536.15 3509.54 4194.75 4163.12 4853.48 4816.91 5512.21 5470.71 6170.98 6124.54 6829.71 6778.33 8147.21 8085.96 10782.17 10701.17 13417.13 13316.37 3301.51 3267.62 3947.94 3909.95 4607.13 4564.87 4596.29 4552.86 4594.53 4551.86 4595.25 4552.86 3319.54

241.826 — — 802.32 1427.84 2044.00 2028.00 2658.45 2636.95 2651.58 2631.85 3272.06 3245.45 3886.64 3855.01 4501.36 4464.79 5116.07 5074.56 5730.82 5684.38 6345.58 6294.20 7575.05 7513.79 10033.94 9952.94 12492.84 12392.09 3169.46 3135.57 3771.88 3733.89 4387.05 4344.79 4376.21 4332.78 4374.46 4331.78 4375.17 4332.78 3099.47

Appendix A.V: Heats of Combustion 991 Appendix A.V (Continued)

Compound

Formula

Cyclopentane Methylcyclopentane Methylcyclopentane Cyclohexane Cyclohexane Methylcyclohexane Methylcyclohexane Ethene (ethylene) Propene (propylene) 1-Butene Ethyne (acetylene) Propyne (methylacetylene) 1-Butyne (ethylacetylene) 2-Butyne (dimethylacetylene)

C5 H10 C6 H12 C6 H12 C6 H12 C6 H12 C7 H14 C7 H14 C2 H4 C3 H6 C4 H8 C2 H2 C3 H4 C4 H6 C4 H6

State (l) (g) (l) (g) (l) (g) (l) (g) (g) (g) (g) (g) (g) (g)

Heat of Combustion in kJ/mol H2 O (l) H2 O (g) CO2 (g) CO2 (g) 3290.88 3969.44 3937.73 3953.00 3919.91 4600.68 4565.29 1410.99 2058.47 2718.58 1299.61 1937.65 2597.68 2579.57

3070.80 3705.35 3673.64 3688.91 3655.81 4292.57 4257.18 1322.96 1926.43 2542.53 1255.60 1849.62 2465.64 2447.53

Appendix B Brief Descriptions of Computer Aids for Use with This Book—A more detailed description of the programs and worksheets appear on the Wiley website for this book

Programs and worksheets to enhance the use of this book are on the website www.wiley .com/college/sandler, with instructions on how to use them. These programs are in the folder Appendix B on the website. However, the user is encouraged instead to use the thermodynamics package in Aspen Plus(R). That program has updated data bases and a very nice interface. The most easily understood computational aids are the MATHCAD worksheets because in this computer-algebra package equations are written, in general, in the exact same form as in this book. Consequently, the MATHCAD worksheets are easily changed to accommodate a different equation of state or activity coefficient model. The suite of MATHCAD worksheets allows calculations of the thermodynamic properies and phase behavior for both pure fluids and mixtures, including various flash calculations, activity coefficient fitting and phase behavior calculations, and chemical equilibrium constant and adiabatic flame temperature computations. However, the MATHCAD package is required to use these worksheets. VISUAL BASIC programs are provided in two forms. One is as easy-to-use standalone Windows-based executable programs. The second form is as VISUAL BASIC code that can be modified if the user has access to a VISUAL BASIC compiler. The first VISUAL BASIC program, called PROPERTY, retrieves pure component data from a database of over 600 compounds, and can also be used to calculate vapor pressures and heat capacities at a given temperature or over a range of temperatures. This program is also called by the others in this suite to provide the pure component data needed in those calculations. The pure fluid Peng-Robinson equation of state program can be used to calculate thermodynamic properies and phase behavior at any temperature and pressure. The Peng-Robinson mixture program is for the calculation of mixture vaporliquid equilibrium. The UNIFAC program uses the modified UNIFAC model for vapor-liquid equilibrium calculations with the most recent parameters available at the time of publication of this book. The final program in this set is used to calculate chemical equilibrium constants at a given temperature or over a temperature range. The DOS BASIC programs were provided with the third edition, and are also supplied in stand-alone executable and BASIC code forms. They are a similar suite of VISUAL BASIC programs, except that there is no PROPERTY database. The stand-alone versions run in a DOS window, and have less functionality and are not as user-friendly as the VISUAL BASIC programs. Also, the DOS BASIC UNIFAC program uses the 992

Appendix B: Brief Descriptions of Computer Aids for Use with This Book 993 original version of UNIFAC (as a result of DOS BASIC memory limitations). The BASIC code programs can be modified if a BASIC or QUICKBASIC compiler is available. We also supply three MATLAB programs: a pure fluid Peng-Robinson program, a mixture Peng-Robinson program and a modified UNIFAC program. As there is no property database for the MATLAB programs, all component data must be entered manually. To run these either MATLAB 7.0 (or higher) must be installed on the computer or the file MCRInstaller.exe must first be run (as described in Appendix B.IV). The MCRInstaller is available from MATHWORKS, Inc. The programs and worksheets mentioned here are available on the website for this book. Once again I want to stress that using the Aspen Plus(R) is preferrable to using the programs mentioned here.

Appendix C Aspen Illustration Input Files. These are on the Website for this Book

Illustrations 3.4-1, 3.4-2, and 3.4-4 Illustrations 4.5-1 Illustrations 5.1-1a and b, 5.2-1, 5.2-2, and 5.2-6 Illustrations 8.5-1, and 8.5-2 Illustrations 10.1-1, 10.1-2, 10.1-3, 10.1-4, 10.1-5, 10.1-7, and 10.1-8 Illustrations 10.2-2, 10.2-3, 10.2-4 (both NRTL and UNIQUAC),10.2-5 (NRTL), 10.2-6 (both NRTL, and UNIQUAC) Illustrations 11.2-2, 11.2-5, 11.2-7, 11.2-11, 11.3-1, 11.3-2, and 11.4-1 Illustrations 13.1-1, 13.1-2, 13.1-3, 13.1-4, 13.1-5, 13.1-17, 13.1-8, 13.1-9, 13.2-1, 13.4-1, and 13.4-2 Illustration 14.3-2

994

Appendix D Answers to Selected Problems This appendix contains answers to selected problems so that the user can check his or her work. However, only the final answers are provided, not the complete solution. Therefore, these answers must be used with some discretion. Unlike a simple algebra problem, for which there will be an exact solution, here the user should remember that if the thermodynamic property chart is read somewhat differently than was done here, a slightly different answer will be obtained. Likewise, if you have used one equation of state and I have used another in the solution, or if you have used one activity coefficient model and I have used another, or even if you have fit the parameters in an activity coefficient or equation-of-state model somewhat differently than I have, the final answers will be somewhat different. Also, since there are slight differences in the databases used, there may be R or the using the DOS- and Windows-based programs some differences in the calculated results using AspenPlus I have provided. Therefore, except for the simplest problems (for example, those involving ideal gases), you may not get exactly the same answers as I did. However, the differences should not be large. Finally, because the world is not an ideal gas, equations of state or activity coefficients frequently must be used to describe a real system of interest. This results in considerable mathematical complexity. Consequently, R many problems are best solved using AspenPlus or the computer software I have provided that is discussed in the previous appendix, or with MATHCAD, MATLAB, MATHEMATICA, POLYMATH, or equation-solving software and programs you develop on your own. The solutions presented here were obtained with the software I developed, so the solutions may differ from R as a result of the difference in the thermodynamic databases used.” those obtained with “AspenPlus

Chapter 1 1.2 a. Water is inappropriate as a thermometric fluid between 0◦ C and 10◦ C, since the volume is not a unique function of temperature in this range. b. The thermal expansion of mercury is not linear with temperature, which introduces an error. For example, at 10◦ C the error is −0.068◦ C.

Chapter 2 2.1 a. 0.1 g/m3 . b. 32.35 minutes. 2.4 a. 0.5 moles of NO2 . b. 1.32 hours.

Chapter 3 3.1 a. h = 1.57 m. b. vf = 134.9 m/s, which is much too high because of neglected air and road resistance. 3.5 With steam the final temperature is about 383 K or 656◦ C, while with an ideal gas it is 600◦ C. 3.14 Initial amount of steam in the tank is 194.9 kg. At end of the process there is 215.3 kg of liquid and 67.5 kg of steam. Fraction of contents by weight that is liquid is 0.761.

3.22 At 5 minutes T = 152.5 K, P = 0.69 bar, and the rate of temperature change is −1.15 K/s. 3.25 Final pressure in both tanks is 133.3 bar; the final temperature in tank 1 is 223.4 K; and in tank 2 is 328.0 K. ˆ = 153 3.32 a. Using the pressure-enthalpy diagram, H kJ/kg, leading to T = ∼90 K with 55 percent of the nitrogen being liquid. b. T = 135 K, and no liquid is formed. Why?

Chapter 4 4.1 4.4 4.9 4.12 4.16

a. T = 8.31◦ C. b. S = Sgen = 64.6 J/K. a. T = 308◦ C. b. S gen = 1.62 kJ/kg · K. a. 627 kW. b. S gen = 843 kJ/kg · h. c. 415 kW. a. N = −1768 mol. b. N = −1152 mol. a. W = 9334 J/mol and T = 549.4 K. b. W = 9191 J/mol and T = 520.9 K.

4.22 P1f = P2f = 133.3 bar, T1f = 226 K and T2f = 285 K. 4.23 a. P1f = P2f = 5 bar, T1f = 275.6◦ C and T2f = 497.7◦ C . b. P1f = P2f = 5 bar, T1f = T2f = 366◦ C.

995

996 Appendix D: Answers to Selected Problems

Chapter 5 5.1 a. If the turbine does not drive the compressor, the work is 30 kJ/kg, and C.O.P. = 4.30. b. The compressor work is 34.8 kJ/kg and C.O.P. = 3.18. 5.8 a. Work = 533 kJ/kg methane through compressors. b. Work is 1713 kJ for each kg of LNG, and the fraction of vapor is 0.689. c. Work is 1168 kJ/kg of LNG produced. 5.9 0.535 kg of TNT. 5.12 a. 610 kJ/kg. b. 630 kJ/kg. c. 685 kJ/kg.

Chapter 6 6.1 a. 4.463◦ C/MPa. b. 17.02◦ C/MPa. 0.262. 6.12 3186 J/mol. 6.17 From P-R EOS, V = 1.667 × 10−3 m3 /mol, H = 21, 033 J/mol, S = 7.061 J/mol K; from corresponding states V = 0.967 × 10−3 m3 /mol, H = 7253 J/mol, S = 27.5 J/mol K.

6.21 a. Q = −14088 J/mol, W = 11 540 J/mol. b. Q = −12 136 J/mol, W = 12 319 J/mol. 6.31 a. T = 300◦ C. b. T = 301.9◦ C. c. T = 299.9◦ C. d. T = 267◦ C.

b. fN2 = 1088 bar from corresponding states and 1043 bar from Peng-Robinson program. c. LewisRandall rule with pure component corresponding states fN2 = 761.6, and fN2 = 246.0 bar. d. fN2 = 732.3, and fN2 = 224.9 bar. 9.23 a. γCH4 = 1.080 and γN2 = 1.073. b. 88.8 J/mol. c. 5.984 J/mol K.

Chapter 10 Section 10.1 10.1-1 a. T = 293.7 K, yET = 0.4167, yP = 0.3730, yNB = 0.1610, and yMP = 0.2502. b. T = 314.2 K, yET = 0.0039, yP = 0.0337, yNB = 0.5215, and yMP = 0.4409. c. L = 0.8667, xET = 0.0225, xP = 0.0858, xNB = 0.4258, and xMP = 0.4659. V = 0.1333, yET = 0.2289, yP = 0.1921, yNB = 0.2326, and yMP = 0.3464. 10.1-3 a. Pdew = 0.768 bar, y5 = 0.071, y6 = 0.338, y7 = 0.592. b. Pbub = 1.258 bar, y5 = 0.541, y6 = 0.36, y7 = 0.093. 10.1-6 21.8%.

Section 10.2 Chapter 7 ˆ = −307.8 J/g in both phases. b. At 225◦ C 7.1 a. G ˆ ˆ = −273.8 J/g. G = −314.1 J/g. c. At 210◦ C G 7.13 a. vap H ∼ 42.7 kJ/mol. b. vap H ∼ 313.6 kJ/mol. 7.15 a. f = 15.3 bar. b. f = 69.25 kPa at 100 bar and

129.6 kPa at 1000 bar. 7.19 a. 31.8%. b. 28.3%. 7.24 PTP = 0.483 bar, TTP = 278.7 K. 7.35 a. Using the Peng-Robinson equation of state, by trial and error, T = 225.35 K. b. 47.4 wt % liquid and 52.6 wt % vapor (which corresponds to only 3.1 vol % liquid).

Chapter 8 a. 2. b. 3. c. 3. a. 1. b. 4. c(i) 2, c(ii) 1. a. 5. b. 3. c. 2. a. −59.6 kJ. b. −6.59 kJ. c. −357 kJ. e. H 1 − H 1 = −46.0 kJ/mol and H 2 − H 2 = −2.3 kJ/mol. 8.29 170.5 kg ethylene glycol per minute. 8.35 59.9 kJ/mol of nitrogen entering the reactor. 8.9 8.10 8.13 8.15

Chapter 9 9.12 a. fO2 = 820 bar from corresponding states and 735.1 bar from Peng-Robinson program.

10.2-2 Using the Van Laar expression, xW = 0.075, xFURF = 0.925, yW = 0.867, and yFURF = 0.129. 10.2-5 a. α = 1.075, β = 1.029. 10.2-14 P = 0.6482 bar, y1 = 0.4483, and y2 = 0.5517.

10.2-33 a. 3.658 bar. b. 3.601 bar.

Section 10.3 10.3-1 At

1 bar, T = 231.31 K, yC2 = 0.3444, yC3 = 0.5907, and yC4 = 0.0649. At 20 bar, T = 341.99 K, yC2 = 0.1261, yC3 = 0.6642, and yC4 = 0.2097. At 40 bar, T = 384.56 K, yC2 = 0.0699, yC3 = 0.6135, and yC4 = 0.3168. 10.3-3 At 20 bar, at T = 350 K, xC2 = 0.0258, xC3 = 0.4928, and xC4 = 0.4814. yC2 = 0.0690, yC3 = 0.6304, and yC4 = 0.3007; also V /L = 0.5612/ 0.4388. 10.3-8 KM = 13.95, and KB = 0.009 08.

Chapter 11 Section 11.1 11.1-3 a. xN = 0.0001. b. xN = 0.000 059 1, HN = 1692 bar. 11.1-6 xCO2 = 0.022 15. 11.1-12 a. From the Steam Tables P = 3.169 kPa. b. P = 101.3 kPa, yW = 0.0313.

Appendix D: Answers to Selected Problems 997

Section 11.2 11.2-4 a. One-constant Margules is not consistent with the data; using the liquid compositions of both species gives two different A values. b. δP = 12.0 or 6.4.

Section 11.3 11.3-1 a. γB = 15.24, γB = 0.257, and yW = 0.743. b. γB = 1.524, and γW = 1.60. c. 1.013 bar. 11.3-3 a. xIH = 0.0902, and xII H = 0.9098. b. P = 1.9657 bar, yH = 0.5795, and yEtOH = 0.4205.

12.4-7 Sample result: At 85 K, xN = 0.158, yN = 0.512, PNvap = 3.284 bar, and POvap = 0.5867 bar.

Section 12.5 12.5-3 a. CB,H2 O = 1.05 × 10−3 g/m3 , CB,soil = 1.070 × 10−3 g/m3 , and CB,air = 0.239 × 10−3 g/m3 , CB,fish = 0.707 × 10−3 g/m3 .

Section 12.6 12.6-1 b. x152 = 0.0377, x114 = 0.6025, and x22 = 0.3598.

Section 11.4 CCl4 11.4-3 Sample answer: At xBr = 0.047, γBr = 1.990, H2 O and γBr = 298.8.

Section 11.5 11.5-5 The calculated osmotic pressure is 96.68 bar, which is reasonably close to the rule-of-thumb value of 100 bar.

Chapter 12 Section 12.1 12.1-1 Using the regular solution model, xN is predicted to be 0.2655 in chlorobenzene, 0.256 in benzene, 0.239 in toluene, and 0.221 in CCl4 . 12.1-3 The 49.5◦ C isotherm is fit reasonably well with kCO2 −B = 0.08.

12.1-4 a. yN = 0.006 92. b. yN = 0.002 25. 12.1-7 Heat of fusion is 6.093 kJ/mol

Section 12.3 12.3-2 221.6 g of methanol, 318.6 g of ethanol, or 636.9 g of glycerol, assuming the solutions are ideal. These correspond to 16.5 wt % (vs. 18.1 wt % exp) methanol, 21 wt % (24.4 wt % exp) ethanol, and 35.5 wt % (38.9 wt % exp) glycerol. 12.3-5 a. γW = 0.9445 (87 g), 0.8744 (177 g), and 0.8105 (272 g). b. T (K) = −15.95 (87 g), −33.76 (177 g) and −49.99 (272 g) including the CP term.

Section 12.4 12.4-4 Sample result: At z = 0.5, TS = 393.54 K, TL = 372.61 K, TLLE = 500 K, xI1 = 0.256, xI1 = 0.744.

Chapter 13 (Note that all the calculations were done with the DOSbased CHEMEQ program. Slightly different answers will R be obtained if AspenPlus or the Windows-based program is used to calculate the chemical equilibrium constants.) 94.4%. 14970 bar. a. 0.0253 bar. b. 0.0122 bar. yC2 H5 OH = yH2 ∼ 0, yCH3 CHO = yH2 O ∼ 0.197, and yO2 ∼ 0.049. 13.18 a. Ka,1 = 0.9731, Ka,2 = 0.1191, yC4 H10 = 13.1 13.4 13.6 13.14

0.147, yC4 H10 = 0.308, and yH2 = 0.446.

13.32 rxn Go = 30.2 kJ/mol. 13.50 a. rxn Go (267 K) = 6.65 kJ, rxn Go (255 K) = 5.47 kJ. b. rxn H o = 13.57 kJ, rxn S o = −75.74 kJ/mol. c. 288.8 kPa. 13.78 Sample result: At pH = 5.6 the charge is calculated to be +1.255 × 10−4 .

Chapter 14 14.1 a. Tad = 3553 K, X = 0.659, yH2 = 0.291, yO2 = 0.146, yH2 O = 0.563. b. Tad = 3343 K, X = 0.835, yH2 = 0.104, yO2 = 0.368, yH2 O = 0.528. c. Tad = 1646 K, X = 1.0, yH2 = 0, yO2 = 0.095, yH2 O = 0.190, yN2 = 0.715. 14.3 Using the CHEMEQ program, Ka = 345, X = 0.927, P = 0.5434 bar, and Q = −96.7 kJ.

14.5 11.87 kJ/mol at a body temperature of 310.1 K.

Chapter 15 15.21 YP/S = 0.024, YN/S = 0.08535, YW/S = −0.594, YO/S = 0.3579, and only 2.4% of carbon in glucose is converted to ethanol. 15.25 Sample results: At 4 cm the concentration is 0.649 mg/g. b. 6.763 mg/g.

Index

Absolute entropy, 384 Absolute pressure. See Pressure Absolute temperature scale, 14, 187, 219 Absolute-pressure scale, 12 Acentric factor, 255–256 Acidity of solutions, 799–817 apparent equilibrium constant, 807 buffer, 815–816 effect of added salt on weak acid solution, 811–813 estimating pK of amino acid in solution, 811 Henderson-Hasselbalch equation, 817 strong acid with strong base pH of solution with, 800–804 titration of, 805 weak acid, 807–808 pH of solution with, 807–810 weak acid + strong base, 813–815 pH of solution with, 814 titration of, 814–815 Activity coefficient calculation of from data for the solubility of a solid, 696 from distribution coefficient data, 666–670 consistency test for, 434–435, 557 defined, 429 essentially immiscible mixture, 628 experimental determination of, 541, 552–566 for a partially miscible mixture, 629–630 from the Gibbs energy, 432–433 general liquid-liquid equilibrium relation using, 621 Gibbs-Duhem equation for, 434–435 group contribution model, see UNIFAC infinite dilution, 562–565 interrelating two in a binary mixture, 459 mean ionic, 485 multicomponent excess Gibbs energy models, 493 of a solvent in an electrolyte solution, 495–498 of hemoglobin, 477–478 of proteins, 477 of water in an aqueous sodium chloride solution, 498 one-constant Margules, 448 pressure dependence of, 435 temperature dependence of, 435–436 vapor-liquid equilibrium (γ -φ method), 538–568

998

Activity coefficient models Analytical Solution of Groups (ASOG), 466 correlative liquid mixture, 446–460 Flory-Huggins, 454 in choosing the appropriate thermodynamic model, 490–493 Margules, 449–450 multicomponent excess Gibbs energy, 493–495 NRTL, 453–454 recommended, 492 regular solution, 463–467 to calculate vapor-liquid-liquid equilibrium, 654–659 to predict liquid-liquid equilibrium, 622–626 to predict vapor-liquid-liquid equilibrium, 654–659 UNIFAC, 466–468 UNIQUAC, 455–456 use of to correlate data, 452–453 van Laar, 451 Wilson, 453 Activity coefficient models and species activities at equilibrium, aqueous. See Electrolyte solutions Activity, defined, 477, 742 Adiabatic flame temperature, equations to calculate, 864–867 Adiabatic flash vaporization, 522 Adiabatic mixing of acid and water, temperature change on, 376–377 Adiabatic reaction temperature, 863–867 Adiabatic system, defined, 3, 4 Air stripping, 2, 609–615 Air-water partition coefficient, calculation of, 721–722 Anode, 883 Antifreeze, choice of, 728–729 Antoine equation, 337–339 Apparent equilibrium constant, 807 Aqueous solutions. See Electrolyte solutions Atmospheric pressure, 11, 12 ATP-ADP energy storage and delivery mechanism, 932–937 Availability function, 140, 876, 950 Average biomass, 946–948 Azeotrope(s) and distillation, 548–549 defined, 540–547 maximum boiling, minimum boiling, 540

Azeotropic data, to predict vapor-liquid equilibrium of a binary mixture, 527–530 Azeotropic mixtures, 540 Balance equations. See also Energy balance equations; Mass balance equations applications of, 69–92 change in state variables between fixed initial and final states, 80–83 for a batch reactor, 854–857 for a biochemical reactor, 937–960 for a black-box system, 861, 862, 869 for a chemical reactor, see Secs. 14.1–14.4 for a closed system, 286 for energy, Chapter 3, 372–373 for entropy, Chapter 4, 374–376 for mass, Chapter 2, 371–372 for mixtures, 371–378 for momentum, 93 in a frictionless process, 86 tank-type chemical reactor, 849–857 tubular reactor, 857–860 vapor-liquid explosion, 178–181 vapor-phase explosion, 186–189 overall reactor, Sec. 14.3, 860–869 Batch reactor, 854–857 Batteries, 895–896 Bender equarion, 244 Benedict-Webb-Rubin equation of state, 250 Binary interaction parameter kij , 440, 586–595 table of, 441 Biochemical reactions, 922–925 pH dependence of, 923–925 protein unfolding (denaturation), 925–932 as a function of pressure, 932, 968 as a function of temperature, 925–932 trimerization of a pancreatic hormone, 922–923 yield factors for, 939–941 Bioconcentration, 723 Bioreactors, 937–960 approximate bioreactor energy balance, 945–946 average biomass, properties of, 946–948 carbon balance, 941 data from, thermodynamic analysis of, 952 energy balance, 945–946 on an isothermal fermenter, 948–949

Index 999 entropy generation, 959 generalized degree of reduction, 947 hydrogen balance, 942 mass balance on, 938–943 nitrogen balance, 942 oxygen balance, 942–964 alternative form of, 949–951 production of Saccharomyces cerevisiae, 943–945 second law limitation on, 951–952 substrate choosing, 955–958 maximum amount of product obtainable from, 953–955 yield factors for, 939–941 Biota-water partition coefficient, 722–723 Bjerrium equation, 614 Boiling liquid-evaporating vapor explosions (BLEVE), 189, 192 Boiling-point temperature, 153 Brayton power generation cycle, 174–176 British thermal unit (Btu), 5, 18 Bubble point pressure, 515–516 determination of, 545–547 estimation of, 521 Bubble point temperature, 515–521 estimation of, 519–521, 583 Buffer amino acid as, 826–831 defined, 815–816 Calorie, defined, 18 Carbon balance in a biochemical reaction, 941–945 Carnot cycle, 116–118, 158, 160 efficiency, 118–122 heat pump, 176 schematic diagram of, 117 work, 116–118 Carnot engine, 116 Carnot, Nicolas L´eonard Sadi, 116 Cathode, 883 Celsius temperature scale, 13 Chain rule, 204 Change of state in thermodynamic properties of real substances accompanying, 220–245 of a real gas, 245–250 Chemical equilibrium, 5, 734–843 combined chemical and phase equilibrium, 399–404, 791–799 criteria for, 400–404 in multiple reactions, 400–404 in single reaction, 400–401 for a reaction involving gas and solid, 769–772 in a single-phase system, 735–739 as a function of temperature, 749–755 at high temperatures, 754–756 effect of pressure on an ideal gas-phase, 758–759 effect of pressure on in an ideal gas, 751–753 gas phase chemical equilibrium calculation, 744–745

high-pressure chemical equilibrium, 759–762 ideal gas-phase chemical equilibrium calculation, 745–746 in a liquid mixture, 762–763 ionic dissociation reaction from concentration data, 763–765 ratios, comparing, 757 reactions in gas phase, 739 reactions in ideal gas phase, 739 reactions in ideal liquid phase, 739–742 reactions in liquid phase, 739 nonideal gas/nonideal solution contributions to, 756–759 several reactions occurring in a single phase, 781–791 chemical equilibrium, 781–791 common ion effect in chemical reaction of electrolytes, 786–788 starting point for calculations of, 735–739 Chemical equilibrium constant and species activities at equilibrium, 744 as a function of temperature, 749–751 defined, 742 for ionic dissociation reaction, 763–765 variation of with temperature, 746–747 with ionic strength, simplified equation for, 765–768 Chemical explosions, 869–875 energy released in, estimating liquid, 873–874 solid, 874–875 vapor-phase, 871–872 reaction stoichiometry, 871 Chemical kinetics, 5 Chemical potential, 363–365 Chemical reaction(s) availability and available work in, 882 biochemical, 938, 945, 958 completion, 741 coupled, 932–937 endothermic, 747 equilibria, 3 equilibrium state of, 745 exothermic, 747 heat of and convention for, 378–385 heterogeneous, 768–781 independent, 369 ionic, 763–765, 889–896 maximum work from, 882 notation for, 35, 368, 372 of electrolytes, common ion effect in, 786–788 standard state Gibbs energy change on, 381 standard state heat of, 380–382 third law reference state, 384–385 Chemical reactors availability and available work, 882, 950, 951 bioreactors,, 937–960 black box, 860–866 electrochemical processes, 882–891

overall balance equations and adiabatic reaction temperature, 860–869 tank-type, 849–857 thermodynamics of chemical explosions, 869–875 tubular reactor, 857–860 Chlorinated fluorinated hydrocarbons (CFCs), 729–732 Clapeyron equation, 335 Clausius statement, Second Law of Thermodynamics, 105, 115 Clausius, Rudolf, 106 Clausius-Clapeyron equation, 336, 513, 607 Closed system defined, 3, 4 differential entropy change for, 114 energy balance, 57, 58 entropy balance for, 103–105, 135, 136 Coefficient of performance (C.O.P.) defined, 165 for refrigeration, heat pump, 176–180 of an automobile air conditioner, 167 vapor compression cycle, 177, 178 Coexistence curves, 10, 11, 303–307, 332 Combined equation-of-state and excess Gibbs energy model, 479–482 Combined equilibrium chemical and phase, 399–405, 791–808, 891 chemical and vapor-liquid, 791–794 Combining rule, 440 Completion of a chemical reaction, 769, 773 Compressibility factor, 223, 253–256, 437–442 critical, 251–255 isothermal, 205, 213 Compressor, defined, 75 Concentration equilibrium ratios, 755 Congruent solidification, 718 Conservation of energy, 47–54 Conservation of mass equations, 29–35 Conservation of momentum, 93, 99 Constant-pressure heat capacity, 61 and constant-volume heat capacity, calculating the difference between, 243–245 of liquid water, calculation of, 218–219 Constant-volume heat capacity, 60, 61 Continuous-flow stirred tank reactor, 373–378 Corresponding states, 250–262, 348 acentric factor, 255–256 critical-point conditions, 251–252 enthalpy change from, 258–260 entropy change from, 260–262 for van der Waals equation of state, 251, 253 fugacity coefficient, 315–316 three-parameter, 256–258 two-parameter, 255 Coupled chemical reactions, 932–937 Cricondenbar, 588 Cricondentherm, 588 Critical end point K, 582 Critical locus, 580

1000 Index Critical point, 251–252, 304, 491 Critical properties estimation methods for, 268–272 for selected fluids, 254 of a pure fluid, 271–272 Critical solution temperature, 581 Cubic equations of state, 221–223 binary interaction parameters for, 440 Cubic equations of state. See also specific equation names Cycles: Brayton, 174–176 Carnot, 116–122, 177 Ericsson, 170–174 Rankine, 158–165 Refrigeration, 164–167, 171, 176, 185 Stirling, 169–170 vapor compression refrigeration, 176, 185 Cyclic process, 3, 18, 22 Debye-H¨uckel limiting law, 485–486 Decomposition pressure, 772–774 Degrees of freedom, 330, 404–405 Denaturation of a protien as a function of pressure, 932–935 as a function of temperature, 926–928 defined, 926 Denbigh procedure, 369 Density, defined, 4 Departure functions, 229–230 Derivatives of extensive properties at constant mole number, 202 Dew point pressure, 514–515 determination of, 545–550 estimation of, 521–522 Dew point temperature, 515–521 estimation of, 519–521 Dibasic acid, ionization of, 818–828 Differential entropy change, 114 Diffusion, Fick’s first law of, 23 Dissociation constant, 764, 811–814 Distillation, 1 azeotropes and, 549 calculation of, 529–534 Rayleigh, 535 steam, of turpentine, 659–660 vapor-liquid equilibrium and, 527–529, 551–556 Distillation column graphical design of, 531–535 simple design of, 528–535 Distribution coefficient, 663–677 calculation of activity coefficients from, 664–675 environmental, 614–617 Gibbs energy of transfer of an amino acid, 674–675 liquid-liquid equilibrium relation, 664 octanol-water partition coefficient for a pollutant, 670–675 purification of an antibiotic, 672–674 Ebulliometer, 561–563 Efficiency, engine, 114–116 Ehrenfest equations, 345 Electrical energy, 15

conversion of radiant energy to, 123–124 maximum conversion of solar energy to, 124–125 Electrical neutrality, 483–485 Electrical work, units of, 18, 884–886 Electrochemical cell chemical equilibrium, 888 uses of, 883 voltage, 889–891 Electrochemical processes, 882–891 cell voltage from concentration difference, 889–891 equilibrium constant from standard half-cell potentials, 887–889 standard cell potential, 887 Electrolyte solution(s), 482–490 activity coefficient of a solvent in, 495–506 Debye-H¨uckel limiting law, 485, 486 electrical neutrality in, 483–435 ionic strength, 485–486 mean ionic activity coefficient, 485 mean ionic molality, 485 models, use of, 486–488 molality and molarity, 489–490 Electrostatic potential, in Gibbs-Donnan equilibrium, 963–965 Ellingham diagram, 778 Endothermic chemical reaction, 746, 767, 866 Energy conservation of, 47–54 conversion of heat to, 114–125 electrical, 15 equivalence of work and heat, experiments designed to prove, 16, 17 flow problems, 5 general conservation principle, 17, 18 Gibbs, 111–114, 290, 307–322, 331–334, 361, 398–409, 416–506 Helmholtz, 111, 361 internal, 4, 47–54, 201–203, 207–224, 444 variation with T and V , 60–63, 228 kinetic, 4, 47–53, 73, 84–90 mechanical, 17, 18 potential, 4, 47–54, 84–90, 960, 967 thermal, 17, 18 transfer, 15 Energy balance and calculated heat loads, 862–864 bioreactor, 946–948 closed and open systems, 57–58 commonly used forms of, 51–54 complete, 50–51 difference form of, 53 differential form of, 51 examples of using, 54–59 final result and choice of system, 54–57 for a continuous-flow stirred-tank reactor, 373–377, 784 for a nonreacting system, 377–378 for a reacting system, 372–385 in an isothermal fermenter, 948–949 stirred-tank reactor, 850 Energy flow accompanying mass flow, 47

Engine Carnot, 116–119 efficiency, 117–119 Enhancement factor, 696 Enthalpy, 50 differential expression for, 207 effect of temperature and pressure on, 61–63, 225 gas, 61–63 liquid, 67, 225–231 solid, 67 excess change on mixing, 428–429 experimental determination of, 385–395 of formation, 380–381, 765 of fusion (melting), 335, 689–693, 704 partial molar, 359–360, 387–390 Enthalpy change and departure functions, 229 between two states, 226–227 from corresponding states, 258–260 on mixing, 355 Enthalpy departure equation, 230–231 Enthalpy-entropy diagram, 63–64 Entropy, 99–108 absolute, 384 at equilibrium in a closed, isolated system, 134–135 defined, 101–102 effect of temperature and pressure, 124–126, 225–231, 260 effect of temperature and volume, 126–127, 226–231 excess change on mixing, 429, 454 generation, 102–105, 127, 374–376 heat, work, engines and, 114–125 ideal gas, 125–126 internal generation of, 102–104, 127, 374 of all substances in perfect crystalline or liquid state, 267 partial molar, 359–360, 418–421, 429 Entropy balance and reversibility, 108–114 applications of, 128–140 and change in state variables, 123–125 at equilibrium in a closed, isolated system, 134–137 to determine possibility of a process, 135–137 difference form of, 104–105 differential form of, 104 for a closed system, 102–104 for an open system, 102 rate-of-change form of, 104 Entropy change and departure functions, 229 between two states, 227–228 for a solid or liquid, 126–128 from corresponding states, 260 of an ideal gas, 126 of matter, 125–128 Entropy generation calculation of for a process, 127 in a fermentation, 841, 958–959 Environment. See Sec. 12.5

Index 1001 Equations of change, for a multicomponent system, 370–378 difference forms of, 376 differential forms of, 375 energy balance for a continuous-flow stirred-tank reactor, 373–378 for a reacting system, 372–377 formulation of, 371 species mass balance for a reacting system, 371 total mass balance for a reacting system, 371–372 Equations of change, for a pure fluid difference forms of, 32, 53, 104–107 differential forms of, 32, 51, 104 Equations of state, 20–21 Bender, 224 Benedict-Webb-Rubin, 250 Clausius, 222 combining rule, 440 computation of vapor pressure from, 322–330 corresponding states, 250–267 fundamental, 215–216 generalized, 263–268 ideal gas, 247–248 in choosing appropriate thermodynamic model, 490 mixing rule, 439, 440, 479–482 models, recommended, 493 Peng-Robinson, 221–223, 231, 233–236, 250, 263–266, 268, 314–315, 323, 329, 341–344 Soave-Redlich-Kwong, 223, 266–267 thermal, 20 to calculate liquid-liquid equilibrium, 631–632 to calculate vapor-liquid-liquid equilibrium, 654–660 to construct thermodynamics properties chart, 231–232, 236–238, 325–330 to predict solubility of a gas in a liquid, 606–609 to predict solubility of a solid in supercritical fluid (SCF), 696–698 van der Waals, 245–253, 314, 462 vapor-liquid equilibrium (φ-φ method), 510, 578–596 virial, 221–234, 313–314, 438, 695–696 volumetric, 20, 220–225, 250 Equilibrium chemical, 286–294, 309–310, 734–843 combined chemical and phase, 402–405, 791–809 criteria for, 286–296, 309–310 in design of separation and purification processes, 1 liquid-liquid, 617–653 local, 21 metastable, unstable, 8–9, 307 osmotic, 677–686, 960–965 phase, 10–12, 300–307, 322–330 reaction in a single phase, 735–768 solid-liquid, 689–694 solid-supercritical fluid, 696–699 stable, 8–9

vapor-liquid, 10, 299–307, 322–330 vapor-liquid-liquid, 652–663 Equilibrium constant and species-activities at equilibrium, relation between, 744 calculation of from standard half-cell potentials, 887–893 variation of with temperature, 747, 780 general equation, 749 simplified equation, 749 Equilibrium partial pressure, calculation of for oxidation of a metal, 778–780 Equilibrium state, 7–10, 15 defined, 5, 8, 22 evolution from nonequilibrium state, 10, 22 for the low pressure, gas-phase reaction, 735 of a chemically reacting system, 735 specification of, 18–19, 404–408 stable, 18 Ericsson cycle, 170–174 Excess Gibbs energy models for binary mixtures, 446–460 for multicomponent mixtures, 493, 495 Excess mixing property, 429 Excess properties, Gibbs-Duhem equation for, 435 Exothermic chemical reaction, 746 Exothermic mixing process, energy release of, 362 Explosions chemical, 869–875 energy release, 185–197, 872–873 mechanical, 185–197 Extensive variables, 19 Extent of reaction, molar, 37–39, 737 External energy, defined, 4 Fahrenheit temperature scale, 13 Fermentation choosing a substrate for, 955–958 data, thermodynamic analysis of, 952 entropy generation in, 958–962 isothermal, energy balance in, 948–952 second-law limitations, 951, 952 Fermenters. See Bioreactors Fick’s first law of diffusion, 23 First Law of Thermodynamics, 50–52, 105, 267–268 First Law of Thermodynamics. See also Energy balance First-order phase transition, 340 Flame temperature, adiabatic, 863 Flash calculations, 521–527, 584–587, 595 defined, 523 Flory-Huggins equations, 454–455 interaction parameter, 454 Flow compressor process, energy balance equation, 57 Fluid mechanics, 5, 45 Fourier’s law of heat conduction, 23 Fractional ligand coverage, 917–918 Freezing point of liquid mixtures, 704–710 depression and eutectic point, calculating, 706–708

of blood, 709 of water, 705 Freezing-point depression of a solvent due to solute presence, 704–710 general equation, 704–705 simplified equation, 705 Fuel cell, calculation of energy produced in, 867–869 Fugacity approximation of, 318–319 computing from density data, 311–312 defined, 308 developing expressionsfor, from the Gibbs energy, 432–434 equality of, as starting point of all phase equilibrium calculations, 689 for a hydrocarbon mixture, 444–446 in phase equilibrium, 424–425 of a gas using the Peng-Robinson equation of state, 314–315 using the virial equation, 314–315 of a liquid using the Peng-Robinson equation of state, 317–318 of a pure gas, 310 of a pure liquid, 316–318 of a solid, 320–322 of a subcooled liquid, 471 of ice from density data, 320–321 of species in mixtures, 422 gaseous mixtures, 436–443 in nonsimple mixtures, 468–478 liquid mixtures, 443–445 saturated steam, calculation of, 313 using the Lewis-Randall Rule, 436–438 using the Peng-Robinson equation of state, 314–315, 440–445 using the virial equation of state, 438–439 Fugacity coefficient corresponding states, 315–316 defined, 308–309 equation, 310–313 for a species in a mixture, 422–424 from Peng-Robinson equation of state, 314–318, 440–443 from van der Waals equation of state, 314 from virial equation of state, 313–314, 438–439 of gases and vapors, 471 Functional groups, 456 Fundamental equations of state, 215–216 Fusion, Gibbs energy of, 689–691, 704 γ -φ method for the calculation of

vapor-liquid equilibrium, Sec. 11.2, 509–510 Gas, chemical reaction in, 736, 739 constant, 13 heat capacity, 60–63, 205, 225–226 ideal, 12, 61, 125, 126, 417–421 solubility, 600–617 Gaseous mixtures, fugacity of species in, 436–443 Generalized degree of reduction, 946–948

1002 Index Generalized equations of state, 263–267 Gibbs-Duhem equation, 363–367, 387, 401, 459, 495, 557, 559 Peng-Robinson, 264–266 Soave-Redlich-Kwong, 266–267 Gibbs energy, 111–114, 290, 307–322, 335, 361, 385, 416–506 developing expressions for activity coefficient and special fugacities from, 432–434 of formation, 379 selected oxides, 778 of mixing, 355–363, 419, 420, 428, 470 of transfer of an amino acid, 674–675 partial molar, 363–367, 372, 416–506 pure component, 738 to determine whether a mixture is ideal, 431–432 Gibbs phase rule applications of, 406 defined, 410 for a one-component system, 330–334 multicomponent, multiphase system, 404–415 use of, 333–334 Gibbs, Josiah Willard, 215 Gibbs-Donnan equilibrium, 960–966 electrostatic potential in, 965–966 of a lysozyme solution, 963–965 osmotic pressure in, 963–965 Gibbs-Duhem equation at constant T and P , 365–367 for activity coefficients, 435–436, 488 for binary mixture at constant T and P , 367 for excess properties, 434 generalized, 363–367 Group contribution method, 456 Group contribution method. See also Joback, UNIFAC and UNIQUAC Half-cell reaction, 884–887 Harlecher-Braun equation, 338 Heat conversion of mechanical energy or work, 114 flow of, 15, 21 in conservation of energy, 47–54 of formation, 379, 764 of fusion, 335, 689, 704 of mixing, 355–359, 389, 420, 428 of reaction, 378–385, 747–749, 849, 852–854 standard states, 380–384, 747–749 of vaporization, 336, 463 sign convention for, 48 Heat capacity, 60–61, 225, 250 constant pressure, 61 constant volume, 60–61 of an ideal gas, 61–63, 226, 250 Heat conduction, Fourier’s law of, 23 Heat engine, 114–125, 158–180 Heat flow, by radiation, 107–108 Heat flux, 22 Heat load on a reactor, 851–863, 944 Heat of fusion, 335, 691–693, 704

Heat of mixing, 361 Heat of reaction and convention for the thermodynamic properties of reacting mixtures, 378–385 standard state, 380–384 Heat pump, 176–180 Helmholtz energy, 111, 360 Henderson-Hasselbalch equation, 817 Henry’s law based on molality, 475–477 based on mole fraction, 474–475 constant, 473, 603 standard state, 478–479 Heterogeneous azeotropy, 659 Heterogeneous chemical reactions, 768–781 calculation of chemical equilibrium for a reaction involving a gas and a solid, 769–772 equilibrium partial pressures for the oxidation of a metal, 777–780 solubility product from solubility data, 775–777 the decomposition pressure of a solid, 772–774 components that appear in the same phase, 768 determining whether an oxidation reaction will occur, 780–781 solubility product, definition of, 774 High-pressure chemical equilibrium, 759–762 High-pressure vapor-liquid equilibrium using equations of state (φ-φ method), 578–598 Hildebrand, Joel H., 463 Hilsch-Ranque vortex tube, 136 Horsepower, 5 Hydrocarbon mixture, species fugacity calculation for, 444–446 Hydrogen balance for a biochemical reactor, 942 Ice point, 14 Ideal gas equation of state, 60–63, 226–229, 246–274 Ideal gas heat capacity, 61–63, 226, 250 Ideal gas mixing properties, 499 Ideal gas mixture, 417–421 defined, 419 determining whether a gas mixture is ideal, 421 hypothetical, 418 partial molar properties for, 418–421 properties of, 420 Ideal gas problem entropy change to solve, 132 mass and energy balances to solve, 69–92 Ideal gas(es), 60–63, 226–229 defined, 12, 60, 417 effect of pressure on chemical equilibrium of, 751–753

entropy change, 126 mass and energy balances, use of with, 74–75 pressure and heat and work flows in computing, 90–92 temperature scale, 14, 219–220 thermometer, 14 Ideal gas-phase chemical equilibrium, 739 calculation, 745–746 effect of pressure on, 758–759 Ideal mixture(s) defined, 426–428 Gibbs energy of, 431–433 vapor-liquid equilibrium in, 510–538 Ideal solution equilibrium relations, 603–604 Idealized incompressible fluid, 67–69 Incompressible fluid, 67 Independent chemical reactions, 368 example, 369, 375 Inert diluent, effect of, 745–746 Infinite-dilution activity coefficients, determined from ebulliometry, 562–565 Insecticides, 720, 726 Integral mass balance, 31–35 Intensive variables, 19 Internal energy, defined, 4 Internal generation, 28–30, 99–103, 125, 373–376 International Association for the Properties of Water and Steam (IAPWS), 224 International System (SI) of units, 6 International Table calorie, 18 Intrinsic stability, 293–300, 398 Inversion temperature, 279 Ion activity coefficient. See Electrolyte Solutions Ion dissociation reaction, chemical equilibrium constant for, 763–765 Ionic strength, 485 Ionization constant, 764 Ionization of biochemicals, 817–831 charge on an amino acid as function of pH, 822–826 dibasic acid, 818–831 pH of solution containing amino acid as buffer, 826–831 Irreversible processes, 109–110, 139. See also Entropy generation Isenthalpic expansion, 58–59. See also Joule-Thomson expansion Isentropic process, 123. See also Reversible processes Isolated system, defined, 3–4 Isotherm, defined, 917 Isothermal compressibility, 205, 213 Isothermal enthalpy change of mixing, calculation of, 395 Isothermal heat of reaction, 374 Isothermal partial vaporization (flash), 521–527, 584–586 Joback group contribution method, 269–271 Joule, 7, 18

Index 1003 Joule’s experiments, 16–18 Joule, James Prescott, 16, 23 Joule-Thomson coefficient of steam, calculation of, 216–217 Joule-Thomson expansion, 59, 70–71, 77, 93, 128, 153–155, 165 inversion temperature, 279 Kelvin temperature scale, 13–15 Kelvin, Lord, 106 Kelvin-Planck statement, Second Law of Thermodynamics, 105–106, 115 Kinetic energy, 4, 47–54, 73–74, 83–89 Lagrange multiplier, 400 Langmuir isotherm, 917 Lattice vacancies, 713–718 LeChatelier and Braun, Principle of, 768, 836 Lever rule, 63–67 Lewis-Randall rule, 436–438, 510, 600 Ligand binding to a substrate, 917–921 fractional ligand coverage, 917–918 hemoglobin + oxygen binding, 918–921 warfarin, binding to human albumin, 921 Limit of stability, 293–300 Linde liquefaction process, 154–158 Liquefaction, 153–158 Liquid mixtures activity coefficient of a species in, 429, 446–460, 493 fugacity of species in, 443–445 models, 446–460 nonsimple, 446 recommended equation-of-state models, 493 simple, 446 Liquid(s) at low pressure, 67 critical point of, 304 effect of pressure on enthalpy, 239–245 entropy, 241–242 internal energy, 240–241 volume, 239–242 energy required to pressurize, 242–243 entropy change for, 125–128 fugacity of, 316–317 solubility of a gas in, 606–609 solubility of a liquid in, 617–652 subcooled, 307 superheated, 307 Liquid-liquid equilibrium, 652–665 application to liquid-liquid extraction, 640–646 calculation of using an equation of state, 631–632 for a partially miscible mixture, 629–630 for an essentially immiscible mixture, 628–629 general relation using activity coefficients, 621–626 mass balance calculation on a triangular diagram, 636–640 measurement of, 620–621 polymer recycling, 632–635

prediction of using an activity coefficient model, 622–626 Liquid-liquid extraction, 2, 640–646 Liquid-liquid-vapor equilibrium, 654–659 Liquid-solid equilibrium, 689 Liquid-vapor equilibrium, 322–330, 335, 507–598 Liquid-vapor interface, surface tension at, 342 Local equilibrium, 21 Low-pressure gas-phase reaction, equilibrium state for, 735 Low-pressure vapor-liquid equilibrium equation, 510 in nonideal mixtures, 538–578 Lower consulate/critical solution temperature, 620, 626 Lower critical end point, 582 Mass balance, 2–3 calculation of on a triangular diagram, 635–640 difference form, 31–32 differential form, 31–32 for a nonreacting system, 377–378 for liquid mixture with a reversible reaction, 40–42 for mixture with chemical reaction, 38–40, 860–864 for simultaneous reactions, 781–783 modeling of simple environmental problem, 42–43 molar extent of reaction notation, 37–38 on a bioreactor, 940–941 on a black-box reactor, 855–857 rate-of-change form, 2, 30–35, 38 species, for a reacting system, 371 stirred-tank reactor, 850 total, for a reacting system, 371–372 tubular reactor, 857–860 using biochemical yield factors, 938–941 vapor-liquid equilibrium, 521–522 Mass flux, 22 Mass transfer, 5 Matter, thermodynamic properties of, 59–69 constant-pressure heat capacity, 61 constant-volume heat capacity, 60–61 entropy changes of, 125–128 ideal gas heat capacity, 61–63 idealized incompressible fluid or solid, 67–69 lever rule, 63, 66–67 solids or liquids at low pressure, 67 Maximum boiling azeotrope, 540 Maximum useful work, Sec. 14.5, 875–881 Maxwell relations, 207–208 Mayer, J. R., 23 Mean calorie, 18 Mean ionic activity coefficient, 485–487 Mean ionic molality, 485 Mechanical contact, 4 Mechanical energy, 17–18, 119 conversion of radiant energy to, 123–124

maximum conversion of solar energy to, 122–123 Mechanical explosions, 185–193 boiling liquid-evaporating vapor explosions (BLEVEs), 191–193 compressed air tank, 188–190 steam tank, 187–188 steam-water, energy released from, 190–192 two-phase, energy released from, 193 vapor-phase, 186–187 Mechanical stability, 626 Mechanical stability criterion, 297–299 Mechanical work, units of, 18 Membrane potentials chloride ion concentration within a nerve cell, 932–936 estimate of, 966 Metal oxides, 777–780 Metastable region, 300–307 Methane, thermodynamic properties of, 66 Minimum boiling azeotrope, 540 Minimzation of Gibbs energy, 290, 396–404, 735–739, 797 Mixing at constant T and P , 385–395, 417–420 at constant T and V , 499 enthalpy change on, 355, 426, 430 entropy change on, 429, 454 Gibbs energy change on, 354–363, 419–421, 470 Helmholtz energy change on, 361, 396–397 of ideal gases, 416–421 volume change on, 355–359, 385–395 Mixing rules for equations of state, 439–440, 479–482 Mixture(s) azeotropic, 540–549 Clapeyron equation for, 335–336 critical point of, 491 enthalpy change on mixing, 355 equations of change for, 370–378 excess property, 429 exothermic mixing process, energy release of, 362–363 fugacity of a species by two methods, 442–443 gaseous, 436–446 hydrocarbon mixture, 444–445 liquid, 443–445 solid, 445–446 generalized Gibbs-Duhem equation, 363–367 Gibbs energy, analysis of, 431–432 heat of mixing, 361–363 ideal gas, 417–421 ideal mixture, 425–436 isothermal enthalpy change of mixing, calculation of, 395 liquid, 493 mass and energy balances, nonreacting system, 377–378 multicomponent, 493 nonsimple, 446, 468–478 partial molar Gibbs energy, 363–367 partial molar properties, 359–360

1004 Index polymer-solvent, 565 pure component thermodynamic properties, 359 Redlich-Kister expansion, 388–391 Solid, 445–446, 714–718 species concentration in, 354, 357–358 temperature change on adiabatic mixing of acid and water, 362–363 thermodynamic description of, 354–363 thermodynamic properties at constant T and P , 444 estimating, 426–427 thermodynamic theory of, 3 two-phase, properties of, 302 van Laar constants for selected binary, 451 vapor, 446–535 vapor-liquid equilibrium in, 2, 507–594 volume change on mixing, 355–357, 361 Moderation, Principle of, 767 Molality, 475–477, 489–490, 903 mean ionic, 485 Molar extent of reaction, 37–38, 736 Molar Gibbs energy, 307–322 Molar properties, 20 Molarity, 489–490, 799 Mole fractions, 365, 436 Henry’s law based on, 473–474 Molecular motion, 4 Molecules, classes of, 250 Mollier diagram, 63–64, 70–73 Momentum, conservation of, 93, 99 Montreal protocol, 731 Multicomponent excess Gibbs energy (activity coefficient) models, 493 Multicomponent system, equations of change for, 370–378 Multiphase system, equilibrium state, 404–406 Multiple chemical reactions, 367–370, 788 Natural flows, 8 Natural processes, unidirectional character of, 99 Nernst equation, 885 Newton’s law of viscosity, 23 Nitrogen balance in a biochemical reaction, 942 Nitrogen, thermodynamic properties of, 67 Nonconserved properties, 26–30, 102, 373–375 Nonequilibrium state, evolution from equilibrium state to, 10, 22 Nonideal gases. See Equations of state Nonideal mixtures, low-pressure vapor-liquid equilibrium in, 538–578 Nonreacting system, mass and energy balances on, 377–378 Nonsimple mixture fugacity of species in, 468–478 liquid, 446 NRTL model, 453–454 Ocean Thermal Energy Conversion (OTEC) project, 146

Octanol-water partition coefficient, 670, 721–726 One-constant Margules activity coefficient, 448 One-constant Margules equation, 448–449, 452 Open system defined, 4 differential entropy change for, 114 energy balance for, 57–58 entropy balance for, 102 Osmotic equilibrium, 677–684 molecular weight and osmotic viral coefficients of a protein, 681–683 molecular weight of a polymer, 680 molecular weight of a protein, 680–681 Osmotic pressure defined, 678 equation, 677–679 estimation of in human blood, 683–684 in Gibbs-Donnan equilibrium, 965 Ostwald ripening, 344 Overall reactor balance equations and the adiabatic reaction temperature, 860–869 adiabatic flame temperature calculation, 864–867 calculated load and overall/specific energy balances compared, 862–864 energy produced in a fuel cell, calculation of, 867–869 maximum work from a fuel cell at constant T and P , 867 overall steady-state mass and energy balances for a reactor, 861–864 Oxidation reaction, 780 Oxidation, defined, 883 Oxygen balance in a biochemical reaction, 942–945, 949–951 pK , 800 Partial molar enthalpy, calculation of from experimental data, 394–395 heat of mixing, 361–363 infinite dilution from experimental data, 394–395 Partial molar entropy, 359–361, 418–421, 429 Partial molar Gibbs energy, 363–367, 421–425 Partial molar thermodynamic property, 359–361 at infinite dilution, 388–390, 474–475, 607 for an ideal gas mixture, 418–421 general equation relating to pure component property, 391–394 method of intercepts, 385–395 Partial molar volume equation for calculation of, 387–388 experimental determination of, 385–395 volume change on mixing, 361 Partial pressure, 419, 510, 777 Partial vaporization, calculation of, 521–527 Partition chromatography, 665

Partitioning of a solid solute between two liquid phases, 701–703 of a solute among two coexisting liquid phases, 665–675, see also Distribution coefficient Path-dependent properties, 10, 81–83, 111–114, 132–134 PCBs, 723, 726 Peng-Robinson equation of state, 221–225, 250, 264–266 binary interaction parameters for, 440 enthalpy and entropy changes for, 231, 236 fugacity coefficient from, 314–317, 440–442 generalized, 263–264 to calculate thermodynamic properties of oxygen, 234–235 to calculate vapor pressure, 322–330 to solve a real gas problem, 265–266 Performance, coefficient of, 165–168, 177–180 pH, 800 Phase, defined, 4 Phase behavior modeling of chemicals in the environment, 720–726 air-water partition coefficient, 722–723 concentration of pollutant in different environmental compartments, 724–726 distribution of PCBs, 723–724 Phase behavior of solid mixtures, 710–720 estimating lattice vacancies, 713–718 Phase diagrams, 303–307, 472, 512–516, 538, 544, 573, 578–594, 618, 654–659 Phase equilibrium, 5 application of equilibrium and stability criteria to the equation of state, 300–307 calculations, starting point for all, 511, 582–586, 600–602, 621, 666, 688–690, 701–704 combined chemical and phase equilibrium, 399–404 fugacity in, 424–425 in multicomponent systems, criteria for, 396–399 first criterion, 397–398 second criterion, 398 third criterion, 398–399 involving solids, 688–733 under all constraints, important criteria for, 299–300 Phase rule, 330–334, 404–408 Phase stability, 293–300, 399–404 Phase transitions first-order, 340 second order, 340 thermodynamic properties of, 334–340 φ-φ method for the calculation of vapor-liquid equilibrium, 578–594 Piston-and-cylinder process, energy balance equation, 57 Pitzer, Kenneth, 255, 764 Plait point, 637 Planck, Max, 106–107 pOH, 800–806

Index 1005 Polymer solutions, 454, 565–567, 632–635 Polymerization reaction, 2 Polymers, recycling of, 632–635 Polytropic process, work in, 122–123 Potential energy, 3, 47–54, 84–90, 967–969 Power generation and refrigeration cycles, 164–167, 171–176, 178–190, 196 Brayton cycle, 174–176 coefficient of performance (C.O.P.), 165–167 Ericsson cycle, 170–174 heat pump, 176–180 Rankine power generation cycle, 161–165 Rankine refrigeration cycle, 164–169 Stirling cycle, 169–171, 173–174 Poynting pressure correction, 318 Prausnitz, John M., 224, 254, 256, 437, 453, 455, 463, 464, 601–602 Prausnitz-Shair correlation, 606 Predictive activity coefficient models, 460–468 Pressure, 10–12 -composition diagram, 511–513, 518, 538–594, 655 absolute, 12 critical, 251–252 decomposition, 772 explicit equation of state, 220–245 partial, 419, 510, 780 reduced, 253 sublimation, 320–322, 338–339 vapor, 303–307, 317–319, 322–330, 335–336 Pressure gauge, 11 Pressure-enthalpy diagram for methane, 66 for nitrogen, 67 Principle of corresponding states, 250–262, 348 Process design, 726–732 antifreeze, choice of, 727 drop-in replacement for the antifreeze ethylene glycol, 727 refrigerant, choice of, 729–732 solvent, choice of, 732 Process engineering, 268 Product design, 726–732 Product engineering, 268 Protein unfolding. See Denaturation P -T -x data, predicting vapor-phase compositions from, 560–562 Pure component properties of, Joback group contributions, 269–271 Pure fluid-phase equilibrium, computation of vapor pressure from an equation of state, 322–330 Purification of an antibiotic, 672–674 processes, 1 Quality, of steam, 63 Quasistatic process, 103 Radiant energy, conversion to mechanical and electrical energy, 123–124

Radiation, heat flow by, 107, 108 Rankine power generation cycle, 161–165 Rankine refrigeration cycle, 164–167 Rankine temperature scale, 13, 15 Raoult’s law, 510–512, 519 deviations from, 538–565 Rayleigh distillation, 534 Rayleigh equation, 534–535 Reacting system energy balance for, 372–385 regular, see Regular solutions species mass balance for, 371 total mass balance for, 371–372 Reaction stoichiometry, 737, 871–872 Redlich-Kister expansion, 388–389 Redlich-Kwong equation of state, 223, 266–267, 280 Reduced pressure, 253 Reduced temperature, 253 Reduced volume, 253 Reduction, generalized degree of, 946–948 Reference state, 62, 478 Refrigerant, choice of, 729–732 Refrigeration cycles, 164–167, 171, 176, 185 Regular solutions defined, 463 model, activity coefficients, 463–464, 541 Retrograde behavior defined, 589 of the first kind, 589 of the second kind, 589 Reversible chemical reactions, 740 Reversible heat, 104–117, 139 Reversible process(es), 108–114, 139, 374, 867 Reversible work, 105–117, 139 Riedel equation, 337 Salting in/out, 613 Salts, dissociation of a weak, 763, 773–774 Saturated liquid, 156, 157. See also Coexistence curves Saturated vapor. See Coexistence curves Second Law of Thermodynamics, 102, 105, 267, 286–289, 951. See also Entropy balance Clausius statement of, 106, 115 constraint, 951 Kelvin-Planck statement of, 106–107, 115 limitation on a fermenter, 951–952 Second order phase transition, 340 Sediment-water partition coefficient, 723–724 Separation processes. See Distillation and Liquid-liquid extraction Shaft work, 47–48, 110–113 Shock wave, 185, 869–870 SI unit system, 6–7 Simple heat engine, 115 Simple liquefaction, 153–157 Simple liquid mixture, 446 Simultaneous reactions, mass balance equations for, 781–783

Soave-Redlich-Kwong equation of state, 223, 266–267 Soil-water partition coefficient, 723–724 Solar energy, conversion of to mechanical or electrical energy, 124–125 Solid(s) at low pressure, 67 decomposition pressure, calculation of, 772–774 effect of pressure on enthalpy, 239–240 entropy, 241–242 internal energy, 239–240 volume, 238–239 entropy change for, 126–127 fugacity of, 320 idealized incompressible, 67–69 mixtures, fugacity of species in, 445–446 phase behavior of, 710–718 Solid-liquid phase transition, 689–701, 715–718 Solubilities as function of pH, 900–910 weak acids, 901–910 weak amino acid, 907–908 weak base, 901–910 Solubility of a gas in a liquid, 600–615 air stripping of radon from groundwater, 609–613 conversion formulas of various expressions of gas solubility, 612 estimation of, 604–606 ideal solution equilibrium relation, 603–604 mole fraction solubility of gases in water, 610–611 prediction of using an equation of state, 606–609 solubility of oxygen in an aqueous salt solution, blood, and seawater, 613–615 solvent equilibrium relation, 603–604 Solubility of a liquid in a liquid, 652–665 Solubility of a solid, 617–621 calculation of activity coefficient from data for, 694 in a gas, 695–696 estimating using the virial equation of state, 695–696 in a liquid, 690–696 equation for, 690 using UNIFAC, 693 in supercritical fluid (SCF), 696–698 determination of the heat of fusion of insulin, 699 using an equation of state, 696–698 Solubility parameter, 463 Solubility product calculation of from solubility data, 775–777 defined, 774 simplified expression for the variation of with ionic strength, 774 Solution aqueous, see Electrolyte solutions polymer, 454, 565–547, 632–635 regular, see Regular solutions

1006 Index Solution acidity. See Acidity of solutions Solvent equilibrium relation, 602–603 Solvent partial pressure, computing, 565–567 Solvent, choice of, 732 Species activity, based on different choices of standard-state activity, 743 Species mass balance for a reacting system, 38–40, 371, 860–863 Specific heat. See Heat capacity Specification of the equilibrium state, 18–21, 330–334, 404–405, 412, 521 Spite, Principle of, 768 Stability, 293–300 first or thermal criterion, 296, 297 intrinsic, 293–300 limit of, 293–300 mutual, 300 of chemical equilibrium, 767 phase equilibria criteria for, 299–300 second or mechanical criterion, 297 Stable equilibrium state, 18, 22 Stable system, 7–10 Standard cell potential, computation of, 887 Standard Gibbs energy change on reaction, 381–382, 744–745 Standard half-cell potentials, equilibrium constant from, 886–889 Standard heat of combustion, 382 Standard heat of reaction, 381–382 calculation of as a function of temperature, 383–384 calculation of at 25◦ C, 382–383 Standard state, 478–479, 742 defined, 381, 742 species activity based on, 743 State variable(s), 19–21, 354–363, 366 change in between fixed initial and final states, 130, 131 entropy balance in change of, 131–133 reversible path to calculate change in, 123 State, defined, 4 Static cell, 558 Steady states, 8 Steady-state system, defined, 3 Steam distillation, of turpentine, 659–660 Mollier diagram, 64 thermodynamic properties of, 64–65, 973–991 Steam power cycle, calculating the efficiency of, 161–162 Steam-water explosion, energy released from, 190–193 Stefan-Boltzmann law, 107 Stirling cycle, 169–174 Stirred-tank reactor design equations for, 851 mass and energy balances for, 851 simplified energy balance, two forms of, 851–854 steady-state design of, 852–854 steady-state mass balance for, 851 Stoichiometric coefficient, 36, 368, 938

Subcooled liquid fugacity, estimate of, 471 Subcooled vapor, 307 Subcooling, 341–344 Sublimation pressure, 320–322, 338–339 Substrate binding of a ligand to, 917–921 choosing, for fermentation, 955–956 maximum of product obtainable from, 953–955 Supercritical fluid (SCF), solubility of a solid in, 696–698 Superheated liquid, 307, 341–344 Surroundings, defined, 4 Syst´eme International d’Unit´es. See SI unit system System adiabatic, 3–4 choice of, 54–59, 103, 127–128 closed, 3–4 constraints imposed upon, 5 defined, 4 driven, 8 System dynamics, 5 System of units, 5 Tank-type chemical reactor, 849–857 batch reactor, 854–857 design equations for a stirred-tank reactor, 851 design of a steady-state stirred-tank reactor, 851–852 mass and energy balances for a stirred-tank reactor, 850–851 steady-state mass balance for a stirred-tank reactor, 851–852 two forms of simplified energy balance for stirred-tank reactor, 852–854 Temperature, 12–15 defined, 12 scales of, 12–13, 219–220 Temperature-composition diagram, 513–514, 519, 545 Temperature-entropy diagram for steam, 65 Ternary systems, 493–495, 635 Thermal contact, 4 Thermal energy, 17–18 conversion of to mechanical energy, 119 Thermal equation of state, 20 Thermal equilibrium, 12. See also Temperature Thermal stability criterion, 296–297 Thermochemical calorie, 18 Thermodynamic consistency relation, 366, 556–561 Thermodynamic model, criteria for choosing, 490–493 Thermodynamic partial derivatives definitions and identities, 209 evaluation of, 205–219 with different state variables held constant, different values, 211–213 with different state variables held constant, not equal, 208–211 Thermodynamic properties effect of pressure on volume of liquids and solids, 238–239

estimation of, 2 of matter, 59–69 of phase transitions, 334–340 of reacting mixtures, 378–385 of real substances changes in, accompanying a change of state, 220–245 estimation methods for, 268–272 example, involving change of state of a real gas, 245–250 generalized equations of state, 263–267 heat capacity data, 225–231 ideal gas and absolute temperature, 219–220 partial derivatives, evaluation of, 205–219 principle of corresponding states, 250–263 volumetric equation-of-state data, 220–225 of small systems, 341–344 of water and steam, 977–986 two-phase mixture, 63, 66–67 Thermodynamic properties chart, construction of, 231–238, 325–329 Thermodynamic systems, stability of, 293–300 Thermodynamics development of, 23 First Law of, 50–51, 105, 267 Second Law of, 102, 105–107, 267, 286–289 Third Law of, 267–268, 384–385 Thermometer, calibration of, 15 Thermometric property, 14–15 Third Law of Thermodynamics, 267–268, 384–385 reference state, 384–385 Tie line, 512–514, 579, 618, 637–641 Time-invariant state, 5, 8, 22 Triangular diagrams, 635–636 mass balance calculation in, 636–640 Triple point, 306 Triple product rule, 204 Trivial solution, 324 Tubular reactor, 857–860 design of, 859–860 mass and energy balances for, 859–860 uses of, 857–858 Turbine, efficiency of, 146 Two-constant Margules expansion, 449–450 Two-phase explosion, simplified equation for estimating the energy released in, 193 Two-phase thermodynamic mixture, properties of, 63, 66–67 Ultracentrifuge equation, 969 protein concentration in, 967–969 Unidirectional character of natural processes, 10, 101–107 UNIFAC (UNIquac Functional-group Activity Coefficient) model, 466–468, 709

Index 1007 UNIQUAC equation, 455 computation of volume and surface area fractions for use in, 456–460 Universal temperature scale, 13 Unstable system, 8–9, 307 Upper consulate/upper critical solution temperature, 620, 626 Upper critical end point, 582 van der Waals equation of state, 221, 245–253, 314, 462 van der Waals loops, 303, 626 van der Waals, J. D., 221 van Laar equations, 450–453, 460 constants for, 451 for a ternary system, 493–495 van’t Hoff equation, 746 Vapor compression cycle, 176–179, 185 Vapor mixtures, recommended equation-of-state models, 493 Vapor pressure, 303 computation of from equation of state, 322–330 Vapor-liquid equilibrium. See Chapter 10 activity coefficient (γ -φ method), 509–510 measurement of, 551 coexistence pressure, 303–307 composition (x-y) diagram, 512–514 data, correlation of, 551–560 equation-of-state (φ-φ method), 508 high-pressure, using equations of state (φ-φ method), 578–598 binary interaction parameter kij , 586–598 measurement of, 582 in ideal mixtures, 510–535 bubble point pressure and bubble point pressure, 514–521 development of diagrams for a mixture that obeys Raoult’s law, 516–519 dew point temperature and dew point temperature, 514–521 distillation, 527–531, 549 distillation column, design of, 527–532, 549 low-pressure vapor-liquid equilibrium equation, 510 partial vaporization of, 522–525

Raoult’s law, 510–512 Rayleigh distillation, 535 Rayleigh equation, 534 low-pressure, in nonideal mixtures, 538–567, 579 azeotropes and distillation, 547–550 azeotropic mixture, 540 composition diagram (x-y), construction of, 512–514, 543–545 computing the solvent partial pressure above a polymer-solvent mixture, 565–567 construction of vapor-liquid equilibrium diagrams for a nonideal system, 543–545 correlation of vapor-liquid data, 551–556 determination of dew point and bubble point pressures, 545–546 integral form of thermodynamic consistency relation, 556–559 measurement of vapor-liquid equilibrium data, 554 of a binary mixture, predicting from azeotropic data, 540–543 mass balances for, 521–522 predicting from infinite dilution activity coefficients, 562–565 predicting vapor-phase compositions from P -T -x data, 560–562 relevance of to distillation, 527, 529 Vapor-liquid explosion (BLEVE), balance equations for, 189–193 Vapor-liquid-liquid equilibrium, 652–665 application to steam distillation, 659–660 calculation using an activity coefficient model, 634–635 calculation using an equation of state, 634–659 Vapor-phase compositions, predicting from P -T -x data, 560–562 Vapor-phase explosion, 871–872 balance equations for, 186–187 examples, 189–190 Variables extensive, 18–21, 202–204, 354–358 intensive, 18–21, see also State variables state, 19–21, 113, 132–133, 226–228, 354, 852, 864

Virial equation of state, 223–225, 313–314, 439, 695–696 coefficients, 275, 279–281, 439 Viscosity, Newton’s law of, 23 Volume critical, 251–252 partial molar, 385–395 at infinite dilution, 394–396 reduced, 253 Volume change on melting, 335 on mixing, 355, 361 Volumetric equation-of-state, 20, 220–225, 250 Water and steam, thermodynamic properties of, 977–986 Weak acid, 807–810 + strong base, 813–815 pH of solution of, 808–810 solubility of, 901–910 Weak base, solubility of, 901–910 Well-mixed assumption, 850 Wilson equation, 453, 494 Wohl equation, 450 Wohl, Kurt, 450 Wong-Sandler mixing rules, 481 Work and conservation of energy, 47–48 Carnot cycle, 116–118 conversion of heat to, 114 defined, 15 flow of, 21 in a polytropic process, 122–123 maximum useful, Sec. 14.5, 875–881 of a flowing fluid against pressure, 48–50 reversible, 110 x-y (vapor-liquid equilibrium composition)

diagram, 512–514 Yield factors, biochemical, 856–857 Zwitterions, 821 γ -φ method for the calculation of

vapor-liquid equilibrium, Sec. 10.2 φ-φ method for the calculation of

vapor-liquid equilibrium, Sec. 10.3

SUPERSCRIPTS

SUBSCRIPTS Symbol

Designates

Symbol

Designates

A, B, C, . . . , AB, D ad conf c EOS eq i imp in j

species dissociated electrolyte AB adiabatic process configurational critical property equation of state equilibrium state ith species, i = 1, . . . , C impurities inlet conditions generally denotes j th reaction, j = 1,. . ., M kth flow stream, k = 1, . . . , K mixture property mixing or mixture reference property reaction property along a two-phase coexistence line coordinate direction vacancy

I, II calc ex exp fus

phase calculated property excess property on mixing measured property property change on melting or fusion initial and final states, respectively ideal property ideal gas property ideal gas mixture property ideal mixture property maximum residual reversible process property along vapor–liquid coexistence line property change on sublimation vapor, liquid, and solid phase, respectively property change on vaporization charge on an ionic species

k m mix R rxn sat x, y , z vac

i, f ID IG IGM IM max res rev sat sub V, L, S vap z+ , z−

Abbreviated list of SI units Physical Quantity Basic units Length Mass Time Electric current Temperature Amount of substance Luminous intensity

SI Unit

Symbol

meter kilogram second ampere kelvin mole candela

m kg s A K mol cd

Derived units with assigned names Energy joule Force newton Power watt Electric charge coloumb Electric potential difference volt Pressure pascal Frequency hertz

J N W C V Pa Hz

Derived units without assigned names Area square meter Volume cubic meter Density cubic meter kilogram per Molar heat capacity joule per mole Kelvin Specific heat capacity joule per kilogram Kelvin Concentration mole per cubic meter

m2 m3 kg m−1 J mol−1 K−1 J kg−1 K−1 mol m−3

Units and conversion factors Length: Mass: Energy:

Pressure:

Molar Heat Capacity: Specific Heat Capacity:

Power:

Relation to Other SI Units

J = kg m2 s−2 N = kg m s−2 = J m−1 W = kg m2 s−3 = J s−1 C = As V = kg m2 s−3 A−1 = J A−1 s−1 Pa = kg m−1 s−2 = N m−2 s−1

The Gas Constant 1 m = 3.281 feet 1 kg = 2.2046 pounds 1 J = 1 watt-second = 107 ergs = 107 dyne-cm = 0.2390 cal = 0.948 × 10−3 BTU = 0.7376 ft-lb = 0.9868 × 10−2 liter-atm 1 Pa = 1 newton m−2 = 1 joule m−3 = 0.9869 × 10−5 atmospheres = 0.750 × 10−2 millimeters of energy = 1 × 10−5 bars = 1.450 × 10−4 psi

1 J mol−1 K−1 = 0.2390 cal(mol K)−1 = 0.2390 BTU(lb-mol ◦ F)−1 1 J g−1 K−1 = 0.2390 cal(g K)−1 = 0.2390 BTU(lb ◦ F)−1 = 1 kJ (kg K)−1 1 J sec−1 = 1 watt = 0.1341 × 10−2 horsepower (hp) = 0.7356 ft-lbf /sec = 0.948 × 10−3 BTU/sec

R = 8.314 = 8.314 = 8.314 × 10−3 = 8.314 × 10−5 = 8.314 × 10−2 = 8.314 × 10−6

J/mol K N m/mol K kPa m3 /mol K bar m3 /mol K bar m3 /kmol K MPa m3 /mol K

TABLES OF THERMODYNAMIC DATA The Gas Constant, Table 1.4-1 Parmeters for the van der Waals Equation of State, Table 6.4-1 Molar Volume, Coefficient of Thermal Expansion, and Coefficient of Isothermal Compressiblity at 20◦ C for Some Condensed Phases, Table 6.4-5 The Critical and Other Constants for Selected Fluids, Table 6.6-1 Binary Interaction Parameters k12 for the Peng-Robinson Equation of State, Table 9.4-1 The van Laar Constants for Some Binary Mixtures, Table 9.5-1 The Group Volume and Surface Area Parameters, R and Q, for Use with the UNIQUAC and UNIFAC Models, Table 9.5-2 Molar Liquid Volumes and Solubility Parameters of Some Nonpolar Liquids, Table 9.6-1 Values of the Parameters in the Equations for γ± for Aqueous Solutions, Table 9.10-1 “Liquid” Volumes and Solubility Parameters for Gaseous Solutes at 25◦ C, Table 11.1-1 Mole Fraction Solubility of Gases in Water at 1.013 bar Partial Pressure as a Function of Temperature, ln x = A + B/T + C ln T + DT + ET 2 ; (T in K), Table 11.1-2 Octanol-Water Partition Coefficients and Other Properties of Some Organic Chemicals, Herbicides, and Pesticides, Table 11.4-1 The Heats and Gibbs Energies of Formation for Ions in an Ideal 1-Molal Solution at 25◦ C, Table 13.1-4 Standard Half-Cell Potentials at 25◦ C, Table 14.6-1 Yield Factors Based on Substrate Consumed, Table 15.7-1. The Degree of Reduction ξ , Gibbs Energy, and Heat of Combustion of Biochemicals and Hydrocarbons on a Carbon Mole Basis (kJ/C-mole)a,b , Table 15.7-2 Conversion Factors to SI Units, Appendix A.I The Molar Heat Capacities of Gases in the Ideal Gas (Zero-Pressure) State*, Appendix A.II The Thermodynamic Properties of Water and Steam, Appendix A.III Enthalpies and Gibbs Energies of Formation, Appendix A.IV Heats of Combustion, Appendix A.V

13 221 238 254 441 451 457 464 486 602 610 671 764 886 939 947 973 974 977 987 990

DIAGRAMS OF THERMODYNAMIC DATA Enthalpy-Entropy (Mollier) Diagram for Steam, Figure 3.3-1a Temperature-entropy diagram for steam, Figure 3.3-1b Pressure-enthalpy diagram for methane, Figure 3.3-2 Pressure-enthalpy diagram for nitrogen, Figure 3.3-3 Pressure-enthalpy diagram for HFC-134a, Figure 3.3-4 Generalized Compressibility Factor Diagram, Figure 6.6-3 Generalized Enthalpy Departure Diagram, Figure 6.6-4 Generalized Entropy Departure Diagram, Figure 6.6-5 Enthalpy-concentration diagram for aqueous sulfuric acid at 0.1MPa., Figure 8.1-1 Enthalpy-Concentration Diagram for the Ethanol + Water System Chemical equilibrium constants as a function of temperature, Figure 13.1-2 Standard Gibbs energies of formation of selected oxides of Ellingham diagram, Figure 13.2-3

64 65 66 67 68 257 259 261 356 574 748 778

All the thermodynamic data diagrams that are listed above appear as ADOBE PDF files on the website for this book, and may be enlarged and printed for easier reading and for use in solving problems.

Notation Standard, generally accepted notation has been used throughout this text. This list contains the important symbols, their definition, and, when appropriate, the page of first occurrence (where a more detailed definition is given). Symbols used only once, or within only a single section are not listed. In a few cases it has been necessary to use the same symbol twice. These occurrences are rare and widely separated, so it is hoped no confusion will result. SPECIAL NOTATION Symbol

Designates

ˆ ˆ (caret as in H) (overbar as in H i ) (underscore as in H ) ± (as in M± ) ∗ (as in G∗i )

property per unit mass (enthalpy per unit mass) partial molar property (partial molar enthalpy) property per mole (enthalpy per mole) mean ionic property (mean ionic molality) property (Gibbs energy) in hypothetical pure component state extrapolated from infinite dilution behavior ideal unit molal property (Gibbs energy) extrapolated from infinite dilution behavior standard state

 (as in G i ) ◦

GENERAL NOTATION Symbol A ai a, b, c, . . . Aˆ, Bˆ

B(T), C(T), . . . C ◦ C Ci CV , CP ∗ CV , CP∗ ∂, d, D D

F ◦ F f f¯i Ffr g G Δfus G, Δrxn G, Δmix G Δf G◦i H H, O , N , C Hi , Hi Δfus H , Δmix H, Δrxn H

Designates Helmholtz energy (111) activity of species i (742) constants in heat capacity equation, equation of state, etc. availability (141) virial coefficients (223) number of components (354) degrees Celsius (13) concentration of species i (667) constant-volume and constant-pressure heat capacities (60) ideal gas heat capacity (61) partial, total, and substantial derivative symbols diffusion coefficient (22) degrees of freedom (331) degrees Fahrenheit (13) pure component fugacity (308) fugacity of a species in a mixture (422) frictional forces (84) acceleration of gravity (11) Gibbs energy (111) Gibbs free energy changes on fusion (472), reaction (381), and mixing (420) molar Gibbs free energy of formation of species i (363) enthalpy (50) stoichiometric coefficient in a biochemical compound (888) Henry’s law constants (474, 475) enthalpy changes on fusion (335), mixing (355), and reaction (380)

Symbol

Designates

Δsub H, Δvap H Δc Hi◦ Δf Hi◦ Δs H I K

enthalpy changes on sublimation (338) and vaporization (336) standard heat of combustion of species i (382) molar heat of formation of species i (380) molar integral heat solution (410) ionic strength (485) in Chapters 9–15 number of flow streams (28) degrees Kelvin (13) distribution coefficients (523, 665, 666) chemical equilibrium constant (742) chemical equilibrium ratios (755) in Chapter 13 ideal solution ionization (762) and solubility (774) products coefficient of sliding friction (86) K −factor, yi /xi (523) octanol-water position coefficient (723) solubility product (742) product of activity coefficients (756) product of fugacity coefficients (756) equation of state binary interation parameter (440) moles of liquid phase (522) mass (30) mass flow rate (30) molecular weight (30) number of independent chemical reactions (70) amount of mass that entered from kth flow stream (31) mass of system in state 1 (31) number of moles (30) initial number of moles of species i (370) pressure (11) number of phases (331) vapor (335), sublimation (338), and saturation pressures (335) atmospheric pressure (12) critical pressure (251) measure of solution acidity (800) -log(Ka ) (808) triple point pressure (305) partial pressure of species i (510) reduced pressure (251) reaction pressure (752) heat flow (22) and heat flow rate (47) volumetric flow rate (850) gas constant (13) drop radius (341) or specific reaction rate (370) solubility (902) total solubility (902) entropy (101) entropy generated (102) and entropy generation rate (102) second partial derivatives of entropy (294) temperature (13) critical temperature (251) mixture freezing temperature (708) lower and upper consolute temperatures (620) melting temperature (708)

K K , Kc , Kx Ka Kc , Kp , Kx , Ky Kc◦ , Ks◦ kfr Ki KOW Ks Kγ Kν kij L M ˙ M

mw M ΔMk M1 N Ni,0 P

P P vap , P sub , P sat Patm Pc

pH pKa Pt Pi Pr Prxn Q, Q˙ q R r SO ST S Sgen , S˙ gen SUV , SUN , etc. T Tc Tf Tlc , Tuc Tm

Symbol

Designates

Tr Tt U V Δfus V v Vc Δmix V Vr wi wI , wII ˙ W, W NET W Wfr ˙s Ws , W X

reduced temperature (252) triple point temperature (305) internal energy (47) volume (12) (also moles of vapor in Secs. 10.1 and 10.2) volume change on melting or fusion (690) velocity (23) molar critical volume (251) volume change on mixing (355) reduced volume (252) mass fraction of species i (355) fraction of mass in a phase (63) work (48) and rate at which work is supplied to system (48) net work supplied from surroundings (86) work against function (87) shaft work (48) and rate of shaft work (48) any thermodynamic variable in Chapter 6, molar extent of chemical reaction (37) elsewhere coordinate direction mole fraction of species i in vapor or liquid phase (354) any thermodynamic variable (203) coordinate direction vapor-phase mole fraction (436) C-moles of X produced per C-mole substrate consumed (939) Heat flow per C-mole substrate consumed (945) compressibility factor (231) or charge on a protein (960) coordinate direction critical compressibility (252) ionic valence (484) coefficient of thermal expansion (205) (in Chapter 6) coefficients in van Laar and Debye-H¨uckel equations (Chapters 9–15) specific heat ratio (188) activity coefficients of species i (447, 474, 475) mean ionic activity coefficient (485) solubility parameter (463) general thermodynamic variable (28) adiabatic compressibility (275) isothermal compressibility (205) viscosity (23), Joule-Thomson coefficient (275), chemical potential (398) stoichiometric coefficient (36) ionic stoichiometric coefficient (483) mass density (22) osmotic pressure (678) surface tension (341) volume fraction (455, 463) generalized degree of reduction (946) potential energy (47) fugacity coefficient of pure species i (308), and species i in a mixture (422) acentric factor (255)

x xi Y y yi YX/S YQ/S Z z Zc z+ , z− α α, β γ γi , γi∗ , γi γ± δ θ κS κT μ ν ν+, ν− ρ Π σ φ, Φ ξ ψ φi , φ¯i ω

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