Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics 9783030491970, 9783030491987

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Daniel Blankschtein

Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics

Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics

Daniel Blankschtein

Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics

Daniel Blankschtein Department of Chemical Engineering Massachusetts Institution of Technology Cambridge, MA, USA

ISBN 978-3-030-49197-0 ISBN 978-3-030-49198-7 https://doi.org/10.1007/978-3-030-49198-7

(eBook)

© Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Cover illustration: Schematic pressure (P) - molar volume (V) phase diagram of a pure substance depicting the Binodal, the Spinodal, and the Critical Point (CP). The sigmoidal dashed curve corresponds to an isotherm at a temperature (T) which is lower than the critical temperature Tc. Outside the Binodal, the system is thermodynamically Stable. Between the Binodal and the Spinodal, the system is thermodynamically Metastable. Inside the Spinodal, the system is thermodynamically Unstable. For complete details, please refer to Lecture 17. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To my wife Anna, for her unconditional love, loyalty, and support

Contents

Lecture 1

Part I

Introduction to the Book . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Motivation and Scope of the Book . . . . . . . . . . . . . 1.2 Organization of the Book . . . . . . . . . . . . . . . . . . . . 1.3 Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . .

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1 1 2 2

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7 7 7

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9 9 10 10 10 11

Fundamental Principles and Properties of Pure Fluids

Lecture 2

Lecture 3

Useful Definitions, Postulates, Nomenclature, and Sample Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Useful Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Postulates I and II (Adapted from Appendix A in T&M) . . . . . . . . . . . . . . . . . . . . . . . 2.4 Sample Problem 2.1 . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Sample Problem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The First Law of Thermodynamics for Closed Systems: Derivation and Sample Problems . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Work Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Sample Problem 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Specific Types of Work Interactions . . . . . . . . . . . . . 3.5 Sample Problem 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Sample Problem 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Postulate III (Adapted from Appendix A in T&M) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13 13 13 14 14 15 16 16 16 17 20 vii

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3.8 3.9 3.10 Lecture 4

Lecture 5

Lecture 6

Energy Decomposition . . . . . . . . . . . . . . . . . . . . . . . Heat Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . The First Law of Thermodynamics for Closed Systems . . . . . . . . . . . . . . . . . . . . . . . . . .

The First Law of Thermodynamics for Closed Systems: Thermal Equilibrium, the Ideal Gas, and Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Thermal Equilibrium and the Directionality of Heat Interactions . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Postulate IV (Adapted from Appendix A in T&M) . . 4.4 Ideal Gas Properties . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Equation of State (EOS) . . . . . . . . . . . . . . 4.4.2 Internal Energy . . . . . . . . . . . . . . . . . . . . 4.4.3 Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.4 Other Useful Relationships . . . . . . . . . . . . 4.5 Sample Problem 4.1: Problem 3.1 in T&M . . . . . . . 4.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . The First Law of Thermodynamics for Closed Systems: Sample Problem 4.1, Continued . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Sample Problem 4.1: Problem 3.1 in T&M, Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Solution 1: System III-Atmosphere (a) . . . . . . . . . . . 5.4 Solution 2: System I-Gas (g) . . . . . . . . . . . . . . . . . . 5.5 Food for Thought . . . . . . . . . . . . . . . . . . . . . . . . . . The First Law of Thermodynamics for Open Systems: Derivation and Sample Problem . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 The First Law of Thermodynamics for Open Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2 Derivation . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Sample Problem 6.1: Problem 3.9 in T&M . . . . . . . 6.3.1 Solution Strategy . . . . . . . . . . . . . . . . . . . 6.3.2 Well-Mixed Gas Model System . . . . . . . . 6.3.3 Pressurization Step: Well-Mixed Gas Model System . . . . . . . . . . . . . . . . . . . . . 6.3.4 Layered or Stratified Gas Model System . .

20 20 21

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23 23

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23 24 25 25 25 25 26 26 27

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31 31

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31 32 35 40

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41 41

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41 41 42 43 44 45

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46 48

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6.3.5 6.3.6 Lecture 7

Pressurization Step: Stratified Gas Model System . . . . . . . . . . . . . . . . . . . . . . Emptying Step: Well-Mixed Gas Model System . . . . . . . . . . . . . . . . . . . . . .

The Second Law of Thermodynamics: Fundamental Concepts and Sample Problem . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Natural Processes . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Statement (1) of the Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Sample Problem 7.1 . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Statement (1a) of the Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Heat Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 Efficiency of a Heat Engine . . . . . . . . . . . . . . . . . . 7.8 Reversible Process . . . . . . . . . . . . . . . . . . . . . . . . .

Lecture 8

48 48

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53 53 53

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56 56 57

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57 58 59 60

Heat Engine, Carnot Efficiency, and Sample Problem . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Heat Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Theorem of Carnot . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Corollary to Theorem of Carnot . . . . . . . . . . . . . . . 8.5 Sample Problem 8.1 . . . . . . . . . . . . . . . . . . . . . . . . 8.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Theorem of Clausius . . . . . . . . . . . . . . . . . . . . . . .

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63 63 63 65 65 66 67 70

Lecture 9

Entropy and Reversibility . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Reversible Process . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Irreversible Process . . . . . . . . . . . . . . . . . . . . . . . .

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71 71 71 77 77

Lecture 10

The Second Law of Thermodynamics, Maximum Work, and Sample Problems . . . . . . . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 A More General Statement of the Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . 10.3 Heat Interactions Along Reversible and Irreversible Paths (Closed System) . . . . . . . . . 10.4 Work Interactions Along Reversible and Irreversible Paths (Closed System) . . . . . . . . . 10.5 Sample Problem 10.1 . . . . . . . . . . . . . . . . . . . . . . 10.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Sample Problem 10.2 . . . . . . . . . . . . . . . . . . . . . . 10.6.1 Solution . . . . . . . . . . . . . . . . . . . . . . . .

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81 81

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84 85 86 87 87

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10.7 10.8 Lecture 11

Criterion of Equilibrium Based on the Entropy . . . . . Sample Problem 10.3 . . . . . . . . . . . . . . . . . . . . . . . 10.8.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . .

88 89 89

The Combined First and Second Law of Thermodynamics, Availability, and Sample Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Closed, Single-Phase, Simple System . . . . . . . . . . 11.3 Open, Single-Phase, Simple System . . . . . . . . . . . . 11.4 Sample Problem 11.1 . . . . . . . . . . . . . . . . . . . . . . 11.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Sample Problem 11.2 . . . . . . . . . . . . . . . . . . . . . . 11.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . .

. 93 . 93 . 93 . 94 . 96 . 96 . 99 . 100

Lecture 12

Flow Work and Sample Problems . . . . . . . . . . . . . . . . . 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Sample Problem 12.1 . . . . . . . . . . . . . . . . . . . . . . 12.2.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Sample Problem 12.2: Problem 4.3 in T&M . . . . . . 12.3.1 Solution: Assumptions . . . . . . . . . . . . . .

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103 103 103 103 106 107

Lecture 13

Fundamental Equations and Sample Problems . . . . . . . . . 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Thermodynamic Relations for Simple Systems . . . . . 13.3 Fundamental Equation . . . . . . . . . . . . . . . . . . . . . . 13.4 The Theorem of Euler in the Context of Thermodynamics (Adapted from Appendix C in T&M) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5 Sample Problem 13.1 . . . . . . . . . . . . . . . . . . . . . . . 13.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . 13.6 Sample Problem 13.2 . . . . . . . . . . . . . . . . . . . . . . . 13.6.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . 13.7 Variable Transformations and New Fundamental Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.8 Sample Problem 13.3 . . . . . . . . . . . . . . . . . . . . . . . 13.8.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . 13.9 Sample Problem 13.4 . . . . . . . . . . . . . . . . . . . . . . . 13.9.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . .

117 117 117 119

Lecture 14

Manipulation of Partial Derivatives and Sample Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Two Additional Restrictions on the Internal Energy Fundamental Equation . . . . . . . . . . . . . . . . 14.3 Corollary to Postulate I . . . . . . . . . . . . . . . . . . . . .

121 122 122 123 123 124 126 126 127 127

. 129 . 129 . 129 . 130

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14.4 14.5 14.6 14.7

14.8 14.9 14.10 Lecture 15

Lecture 16

Reconstruction of the Internal Energy Fundamental Equation . . . . . . . . . . . . . . . . . . . . . . Manipulation of Partial Derivatives of Thermodynamic Functions . . . . . . . . . . . . . . . . . . . Internal Energy and Entropy Fundamental Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Useful Rules to Calculate Partial Derivatives of Thermodynamic Functions . . . . . . . . . . . . . . . . . 14.7.1 The Triple Product Rule . . . . . . . . . . . . . . 14.7.2 The Add Another Variable Rule . . . . . . . . 14.7.3 The Derivative Inversion Rule . . . . . . . . . 14.7.4 Maxwell’s Reciprocity Rule . . . . . . . . . . . Sample Problem 14.1 . . . . . . . . . . . . . . . . . . . . . . . 14.8.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . Jacobian Transformations . . . . . . . . . . . . . . . . . . . . 14.9.1 Properties of Jacobians . . . . . . . . . . . . . . . Sample Problem 14.2 . . . . . . . . . . . . . . . . . . . . . . . 14.10.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . .

Properties of Pure Materials and Gibbs Free Energy Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Gibbs Free Energy Fundamental Equation . . . . . . . 15.3 Derivation of the Gibbs-Duhem Equation . . . . . . . . 15.4 Relating the Gibbs Free Energy to Other Thermodynamic Functions . . . . . . . . . . . . . . . . . . 15.5 First-Order and Second-Order Partial Derivatives of the Gibbs Free Energy . . . . . . . . . . . . . . . . . . . 15.6 Determining Which Data Set Has the Same Thermodynamic Information Content as the Gibbs Free Energy Fundamental Equation . . . . . . . . . . . . Evaluation of Thermodynamic Data of Pure Materials and Sample Problems . . . . . . . . . . . . . . . . . . . 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Summary of Changes in Entropy, Internal Energy, and Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2.1 Using T and P as the Two Independent Intensive Variables . . . . . . . . . . . . . . . . 16.2.2 Using T and V as the Two Independent Intensive Variables . . . . . . . . . . . . . . . . 16.2.3 The Ideal Gas Limit . . . . . . . . . . . . . . . . 16.2.4 Relation between CP and CV . . . . . . . . . . 16.3 Sample Problem 16.1 . . . . . . . . . . . . . . . . . . . . . . 16.3.1 Solution . . . . . . . . . . . . . . . . . . . . . . . .

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130 131 132 132 132 133 133 133 134 134 134 135 136 136 137 137 137 138

. 139 . 139

. 142 . 147 . 147 . 147 . 148 . . . . .

148 148 149 149 149

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Contents

16.4 16.5

Lecture 17

Lecture 18

Sample Problem 16.2 . . . . . . . . . . . . . . . . . . . . . . 16.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . Evaluation of Changes in the Thermodynamic Properties of Pure Materials . . . . . . . . . . . . . . . . . 16.5.1 Calculation of the Entropy Change . . . . . 16.5.2 Strategy I . . . . . . . . . . . . . . . . . . . . . . . 16.5.3 Strategy II . . . . . . . . . . . . . . . . . . . . . . . 16.5.4 Strategy III . . . . . . . . . . . . . . . . . . . . . .

Equations of State of a Pure Material, Binodal, Spinodal, Critical Point, and Sample Problem . . . . . . . . 17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 Equations of State of a Pure Material . . . . . . . . . . . 17.3 Examples of Equations of State (EOS) . . . . . . . . . . 17.3.1 The Ideal Gas EOS . . . . . . . . . . . . . . . . 17.3.2 The van der Waals EOS . . . . . . . . . . . . . 17.4 Sample Problem 17.1 . . . . . . . . . . . . . . . . . . . . . . 17.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 17.5 Pressure-Explicit Form of the Isotherm P ¼ f (V, T) of a Pure Material . . . . . . . . . . . . . . . 17.6 Stable, Metastable, and Unstable Equilibrium . . . . . 17.7 Mechanical Analogy of Stable, Metastable, and Unstable Equilibrium States . . . . . . . . . . . . . . 17.8 Mathematical Conditions for Stability, Metastability, and Instability . . . . . . . . . . . . . . . . . 17.9 Mathematical Conditions for the Spinodal and the Critical Point . . . . . . . . . . . . . . . . . . . . . . The Principle of Corresponding States and Sample Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Examples of Additional Equations of State (EOS) . 18.2.1 The Redlich-Kwong (RK) EOS . . . . . . . 18.2.2 Sample Problem 18.1 . . . . . . . . . . . . . . . 18.2.2.1 Solution . . . . . . . . . . . . . . . . . 18.2.3 Sample Problem 18.2 . . . . . . . . . . . . . . . 18.2.3.1 Solution . . . . . . . . . . . . . . . . . 18.2.4 The Peng-Robinson (PR) EOS . . . . . . . . 18.2.5 The Virial EOS . . . . . . . . . . . . . . . . . . . 18.2.6 Sample Problem 18.3 . . . . . . . . . . . . . . . 18.2.6.1 Solution . . . . . . . . . . . . . . . . . 18.3 The Principle of Corresponding States . . . . . . . . . . 18.3.1 The Compressibility Factor . . . . . . . . . . . 18.3.2 Sample Problem 18.4 . . . . . . . . . . . . . . . 18.3.2.1 Solution . . . . . . . . . . . . . . . . .

. 150 . 151 . . . . .

151 151 152 152 152

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159 159 159 160 160 160 161 161

. 162 . 166 . 167 . 168 . 168 . . . . . . . . . . . . . . . .

171 171 171 171 172 172 173 173 174 174 175 175 177 177 180 180

Contents

Lecture 19

Lecture 20

xiii

Departure Functions and Sample Problems . . . . . . . . . . 19.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2 Sample Problem 19.1 . . . . . . . . . . . . . . . . . . . . . . 19.2.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 19.3 Departure Functions . . . . . . . . . . . . . . . . . . . . . . . 19.4 Calculation of the Entropy Departure Function, DS (T, P) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.5 Calculation of the Helmholtz Free Energy Departure Function, DA(T, V) . . . . . . . . . . . . . . . 19.6 Calculation of the Entropy Departure Function, DS (T, V) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.7 Sample Problem 19.2 . . . . . . . . . . . . . . . . . . . . . . 19.7.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 19.8 Important Remark . . . . . . . . . . . . . . . . . . . . . . . .

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187 187 187 187 191

. 192 . 193 . . . .

194 196 196 197

Review of Part I and Sample Problem . . . . . . . . . . . . . . . 20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 Basic Concepts, Definitions, and Postulates . . . . . . . 20.3 Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.4 The First Law of Thermodynamics for Closed Systems . . . . . . . . . . . . . . . . . . . . . . . . . 20.5 The First Law of Thermodynamics for Open, Simple Systems . . . . . . . . . . . . . . . . . . . . 20.6 The First Law of Thermodynamics for Steady-State Flow Systems . . . . . . . . . . . . . . . . 20.7 Carnot Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.8 Entropy of a Closed System . . . . . . . . . . . . . . . . . . 20.9 The Second Law of Thermodynamics . . . . . . . . . . . 20.10 The Combined First and Second Law of Thermodynamics for Closed Systems . . . . . . . . . 20.11 Entropy Balance for Open Systems . . . . . . . . . . . . . 20.12 Maximum Work, Availability . . . . . . . . . . . . . . . . . 20.13 Fundamental Equations . . . . . . . . . . . . . . . . . . . . . . 20.14 Manipulation of Partial Derivatives . . . . . . . . . . . . . 20.15 Manipulation of Partial Derivatives Using Jacobian Transformations . . . . . . . . . . . . . . . . . . . . 20.16 Maxwell’s Reciprocity Rules . . . . . . . . . . . . . . . . . . 20.17 Important Thermodynamic Relations for Pure Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.18 Gibbs-Duhem Equation for a Pure Material . . . . . . . 20.19 Equations of State (EOS) . . . . . . . . . . . . . . . . . . . . 20.20 Stability Criteria for a Pure Material . . . . . . . . . . . . 20.21 Sample Problem 20.1 . . . . . . . . . . . . . . . . . . . . . . . 20.21.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . .

199 199 199 199 200 200 200 201 201 201 201 202 202 202 203 204 204 205 206 206 207 208 208

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Part II

Contents

Mixtures: Models and Applications to Phase and Chemical Reaction Equilibria

Lecture 21

Lecture 22

Lecture 23

Extensive and Intensive Mixture Properties and Partial Molar Properties . . . . . . . . . . . . . . . . . . . . . 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Extensive and Intensive Differentials of Mixtures . . 21.3 Choose Set 1: {T, P, N1, . . ., Nn} and Analyze B ¼ B (T, P, N1, . . ., Nn) . . . . . . . . . . 21.4 Important Remarks . . . . . . . . . . . . . . . . . . . . . . . . 21.5 Choose Set 2: {T, P, x1, . . ., xn-1, N} and Analyze B ¼ B (T, P, x1, . . ., xn-1, N) . . . . . . . 21.6 Choose Set 1: {T, P, N1, . . ., Nn} and Analyze B ¼ B (T, P, N1, . . ., Nn) . . . . . . . . . . 21.7 Choose Set 2: {T, P, x1, . . ., xn-1, N} and Analyze B ¼ B (T, P, x1, . . ., xn-1, N) . . . . . . . Generalized Gibbs-Duhem Relations for Mixtures, Calculation of Partial Molar Properties, and Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Partial Molar Properties . . . . . . . . . . . . . . . . . . . . 22.3 Useful Relations Between Partial Molar Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.4 How Do We Calculate Bi ? Cases 1, 2, and 3 . . . . . 22.5 Sample Problem 22.1 . . . . . . . . . . . . . . . . . . . . . . 22.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 22.6 Generalized Gibbs-Duhem Relations for Mixtures .

. 213 . 213 . 214 . 215 . 216 . 216 . 218 . 219

. 223 . 223 . 223 . . . . .

Mixture Equations of State, Mixture Departure Functions, Ideal Gas Mixtures, Ideal Solutions, and Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.2 Sample Problem 23.1 . . . . . . . . . . . . . . . . . . . . . . . 23.2.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . 23.3 Equations of State for Gas Mixtures . . . . . . . . . . . . 23.3.1 Ideal Gas (IG) Mixture EOS . . . . . . . . . . . 23.3.2 van der Waals (vdW) Mixture EOS . . . . . . 23.3.3 Peng-Robinson (PR) Mixture EOS . . . . . . 23.3.4 Virial Mixture EOS . . . . . . . . . . . . . . . . . 23.4 Calculation of Changes in the Thermodynamic Properties of Gas Mixtures . . . . . . . . . . . . . . . . . . . 23.4.1 Mixture Attenuated State Approach . . . . . . 23.4.2 Mixture Departure Function Approach . . .

224 225 228 228 230

233 233 233 234 235 235 235 236 237 238 239 240

Contents

xv

23.5

Lecture 24

Lecture 25

Lecture 26

Ideal Gas Mixtures and Ideal Solutions . . . . . . . . . 23.5.1 One Component (Pure, n ¼ 1) Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.5.2 Ideal Gas Mixture: For Component i . . . . 23.5.3 Ideal Solution: For Component i . . . . . . .

. 241 . 241 . 242 . 242

Mixing Functions, Excess Functions, and Sample Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.2 Sample Problem 24.1 . . . . . . . . . . . . . . . . . . . . . . 24.2.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 24.3 Sample Problem 24.2 . . . . . . . . . . . . . . . . . . . . . . 24.3.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 24.4 Sample Problem 24.3 . . . . . . . . . . . . . . . . . . . . . . 24.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 24.5 Other Useful Relations for an Ideal Solution . . . . . 24.6 Summary of Results for an Ideal Solution . . . . . . . 24.7 Mixing Functions . . . . . . . . . . . . . . . . . . . . . . . . . 24.7.1 The Mixing B and Reference States . . . . 24.7.2 Pure Component Reference State for Component j . . . . . . . . . . . . . . . . . . . . . 24.7.3 Useful Relations for Mixing Functions . . 24.8 Mixing Functions: Mixing of Three Liquids at Constant T and P . . . . . . . . . . . . . . . . . . . . . . . . . 24.9 Ideal Solution Mixing Functions . . . . . . . . . . . . . . 24.10 Excess Functions . . . . . . . . . . . . . . . . . . . . . . . . .

. 252 . 254 . 255

Ideal Solution, Regular Solution, and Athermal Solution Behaviors, and Fugacity and Fugacity Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.2 Ideal Solution Behavior . . . . . . . . . . . . . . . . . . . . 25.3 Regular Solution Behavior . . . . . . . . . . . . . . . . . . 25.4 Athermal Solution Behavior . . . . . . . . . . . . . . . . . 25.5 Fugacity and Fugacity Coefficient . . . . . . . . . . . . . 25.5.1 Variations of bf i and f i with Pressure . . . . 25.5.2 Variations of bf i and f i with Temperature . 25.6 Other Relations Involving Fugacities . . . . . . . . . . . 25.7 Calculation of Fugacity . . . . . . . . . . . . . . . . . . . . . 25.8 The Lewis and Randall Rule . . . . . . . . . . . . . . . . .

. . . . . . . . . . .

259 259 259 260 261 261 263 263 265 265 268

Activity, Activity Coefficient, and Sample Problems . . . . 26.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.2 Activity and Activity Coefficient . . . . . . . . . . . . . . 26.3 Pure Component Reference State . . . . . . . . . . . . . . 26.4 Calculation of Activity . . . . . . . . . . . . . . . . . . . . .

. . . . .

271 271 272 273 275

. . . . . . . . . . . .

243 243 243 244 245 246 247 248 248 249 249 250

. 251 . 251

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26.5

. . . .

. . . .

276 277 279 279

Lecture 27

Criteria of Phase Equilibria, and the Gibbs Phase Rule . 27.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.2 Use of Other Reference States . . . . . . . . . . . . . . . . 27.3 Phase Equilibria: Introduction . . . . . . . . . . . . . . . . 27.4 Criteria of Phase Equilibria . . . . . . . . . . . . . . . . . . 27.4.1 Thermal Equilibrium . . . . . . . . . . . . . . . 27.4.2 Mechanical Equilibrium . . . . . . . . . . . . . 27.4.3 Diffusional Equilibrium . . . . . . . . . . . . . 27.5 The Gibbs Phase Rule . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . .

281 281 281 282 283 284 285 285 286

Lecture 28

Application of the Gibbs Phase Rule, Azeotrope, and Sample Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.2 The Gibbs Phase Rule for a Pure Substance . . . . . . 28.3 Sample Problem 28.1 . . . . . . . . . . . . . . . . . . . . . . 28.3.1 Solution . . . . . . . . . . . . . . . . . . . . . . . .

. . . . .

291 291 291 293 293

Differential Approach to Phase Equilibria, Pressure-Temperature-Composition Relations, Clausius-Clapeyron Equation, and Sample Problem . . . 29.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29.2 Sample Problem 29.1 . . . . . . . . . . . . . . . . . . . . . . 29.2.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 29.3 Simplifications of Eqs. (29.21) and (29.22) . . . . . .

. . . . .

295 295 295 296 300

26.6

Lecture 29

Lecture 30

Sample Problem 26.1 . . . . . . . . . . . . . . . . . . . . 26.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . Sample Problem 26.2 . . . . . . . . . . . . . . . . . . . . 26.6.1 Solution . . . . . . . . . . . . . . . . . . . . . .

. . . .

Pure Liquid in Equilibrium with Its Pure Vapor, Integral Approach to Phase Equilibria, Composition Models, and Sample Problems . . . . . . . . . . . . . . . . . . . . . 30.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.2 From Lecture 29 . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.3 Pure Liquid in Equilibrium with Its Pure Vapor . . . . 30.4 Integral Approach to Phase Equilibria . . . . . . . . . . . 30.4.1 Sample Problem 30.1 . . . . . . . . . . . . . . . . 30.4.2 Solution Strategy . . . . . . . . . . . . . . . . . . . 30.4.3 Calculation of Vapor Mixture Fugacities . . 30.4.4 Calculation of Liquid Mixture Fugacities . . 30.4.5 Calculation of Pure Component i Liquid Fugacity . . . . . . . . . . . . . . . . . . . . . . . . . 30.4.6 Simplifications of Eq. (30.29) . . . . . . . . . . 30.4.7 Models for the Excess Gibbs Free Energy of Mixing of n ¼ 2 Mixtures . . . . .

305 305 306 306 310 310 310 311 312 313 315 316

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30.4.8 Lecture 31

Lecture 32

Lecture 33

Models for the Excess Gibbs Free Energy of Mixing of n > 2 Mixtures . . . . . 318

Chemical Reaction Equilibria: Stoichiometric Formulation and Sample Problem . . . . . . . . . . . . . . . . . . 31.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Contrasting the Calculation of Changes in Thermodynamic Properties With and Without Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . 31.2.1 Case I: Closed Binary System of Inert Components 1 and 2 . . . . . . . . . . . . . . . . 31.2.2 Case II: Closed Binary System of Components 1 and 2 Undergoing a Dissociation Reaction . . . . . . . . . . . . . . . . 31.3 Stoichiometric Formulation . . . . . . . . . . . . . . . . . . . 31.4 Important Remark . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 Sample Problem 31.1 . . . . . . . . . . . . . . . . . . . . . . . 31.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . Criterion of Chemical Reaction Equilibria, Standard States, and Equilibrium Constants for Gas-Phase Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Derivation of the Criterion of Chemical Reaction Equilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.3 Derivation of the Equilibrium Constant for Chemical Reaction r . . . . . . . . . . . . . . . . . . . . 32.4 Derivation of the Equilibrium Constant for a Single Chemical Reaction . . . . . . . . . . . . . . . 32.5 Discussion of Standard States for Gas-Phase, Liquid-Phase, and Solid-Phase Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.6 Comments on the Standard-State Pressure . . . . . . . 32.7 Decomposition of the Equilibrium Constant into Contributions from the Fugacity Coefficients, the Gas Mixture Mole Fractions, and the Pressure . . Equilibrium Constants for Condensed-Phase Chemical Reactions, Response of Chemical Reactions to Temperature, and Le Chatelier’s Principle . . . . . . . . . 33.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33.2 Derivation of the Equilibrium Constant for a Condensed-Phase Chemical Reaction . . . . . . . 33.3 Determination of the Standard Molar Gibbs Free Energy of Reaction . . . . . . . . . . . . . . . . . . . . 33.4 Response of Chemical Reactions to Changes in Temperature and Pressure . . . . . . . . . . . . . . . . .

321 321

321 322

322 323 326 326 327

. 329 . 329 . 330 . 331 . 334

. 334 . 336

. 337

. 341 . 341 . 341 . 343 . 348

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Contents

33.5 33.6 Lecture 34

Lecture 35

Lecture 36

Lecture 37

How Does a Chemical Reaction Respond to Temperature? . . . . . . . . . . . . . . . . . . . . . . . . . . . 348 Le Chatelier’s Principle . . . . . . . . . . . . . . . . . . . . . 350

Response of Chemical Reactions to Pressure, and Sample Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 34.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34.2 How Does a Chemical Reaction Respond to Pressure? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34.3 Sample Problem 34.1 . . . . . . . . . . . . . . . . . . . . . . 34.3.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 34.4 Sample Problem 34.2 . . . . . . . . . . . . . . . . . . . . . . 34.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . The Gibbs Phase Rule for Chemically-Reacting Systems and Sample Problem . . . . . . . . . . . . . . . . . . . . . 35.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.2 Sample Problem 35.1 . . . . . . . . . . . . . . . . . . . . . . 35.2.1 Solution Strategy . . . . . . . . . . . . . . . . . . 35.2.2 Selection of Standard States . . . . . . . . . . 35.2.3 Remarks . . . . . . . . . . . . . . . . . . . . . . . . 35.2.4 Evaluation of Fugacities . . . . . . . . . . . . . 35.2.5 Calculation of the Equilibrium Constant . 35.2.6 Comment on the Standard-State Pressure . . . . . . . . . . . . . . . . . . . . . . . . Effect of Chemical Reaction Equilibria on Changes in Thermodynamic Properties and Sample Problem . . . . 36.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36.2 Sample Problem 36.1: Production of Sulfuric Acid by the Contact Process . . . . . . . . . . . . . . . . . 36.3 Solution Strategy . . . . . . . . . . . . . . . . . . . . . . . . . 36.4 Evaluation of K(T) . . . . . . . . . . . . . . . . . . . . . . . . 36.5 Derivation of the Second Equation Relating T and ξ . . . . . . . . . . . . . . . . . . . . . . . . . . Review of Part II and Sample Problem . . . . . . . . . . . . . 37.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37.2 Partial Molar Properties . . . . . . . . . . . . . . . . . . . . 37.3 Generalized Gibbs-Duhem Relations for Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37.4 Gibbs-Helmholtz Relation . . . . . . . . . . . . . . . . . . . 37.5 Mixing Functions . . . . . . . . . . . . . . . . . . . . . . . . . 37.6 Ideal Gas Mixtures . . . . . . . . . . . . . . . . . . . . . . . . 37.7 Ideal Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .

. 353 . 353 . . . . .

353 355 355 358 358

. . . . . . . .

367 367 367 368 369 370 370 370

. 373 . 375 . 375 . 375 . 376 . 379 . 382 . 389 . 389 . 389 . . . . .

390 390 390 391 391

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37.8 37.9 37.10 37.11 37.12 37.13 37.14 37.15 37.16 37.17 37.18 37.19 37.20 37.21 37.22 37.23 37.24 37.25 37.26 37.27 37.28 37.29 37.30 37.31

Part III

Excess Functions . . . . . . . . . . . . . . . . . . . . . . . . . . Fugacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Variation of Fugacity with Temperature and Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Generalized Gibbs-Duhem Relation for Fugacities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Fugacity Coefficient . . . . . . . . . . . . . . . . . . . . . . . . Lewis and Randall Rule . . . . . . . . . . . . . . . . . . . . . Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Activity Coefficient . . . . . . . . . . . . . . . . . . . . . . . . Variation of Activity Coefficient with Temperature and Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Generalized Gibbs-Duhem Relation for Activity Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Conditions for Thermodynamic Phase Equilibria . . . Gibbs Phase Rule . . . . . . . . . . . . . . . . . . . . . . . . . . Differential Approach to Phase Equilibria . . . . . . . . Dependence of Fugacitities on Temperature, Pressure, and Mixture Composition . . . . . . . . . . . . . Integral Approach to Phase Equilibria . . . . . . . . . . . Pressure-Temperature Relations . . . . . . . . . . . . . . . . Stoichiometric Formulation for Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Equilibrium Constant . . . . . . . . . . . . . . . . . . . . . . . Typical Reference States for Gas, Liquid, and Solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Equilibrium Constant for Gases Undergoing a Single Chemical Reaction . . . . . . . . . . . . . . . . . . Equilibrium Constants for Liquids and Solids . . . . . . Calculation of the Standard Molar Gibbs Free Energy of Reaction . . . . . . . . . . . . . . . . . . . . . . . . . Variation of the Equilibrium Constant with Temperature and Pressure . . . . . . . . . . . . . . . . . . . . Sample Problem 37.1 . . . . . . . . . . . . . . . . . . . . . . . 37.31.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . .

392 393 394 394 394 395 395 396 396 396 396 397 397 397 398 398 399 399 399 399 400 400 401 401 402

Introduction to Statistical Mechanics

Lecture 38

Statistical Mechanics, Canonical Ensemble, Probability and the Boltzmann Factor, and Canonical Partition Function . . . . . . . . . . . . . . . . . . . 411 38.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 38.2 Canonical Ensemble and the Boltzmann Factor . . . . 412

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38.3

38.4 Lecture 39

Lecture 40

Lecture 41

Probability That a System in the Canonical Ensemble Is in Quantum State j with Energy Ej(N, V) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 Physical Interpretation of the Canonical Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416

Calculation of Average Thermodynamic Properties Using the Canonical Partition Function and Treatment of Distinguishable and Indistinguishable Molecules . . . . 39.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39.2 Calculation of the Average Energy of a Macroscopic System . . . . . . . . . . . . . . . . . . . 39.3 Calculation of the Average Heat Capacity at Constant Volume of a Macroscopic System . . . . 39.4 Calculation of the Average Pressure of a Macroscopic System . . . . . . . . . . . . . . . . . . . 39.5 Canonical Partition Function of a System of Independent and Distinguishable Molecules . . . . . . 39.6 Canonical Partition Function of a System of Independent and Indistinguishable Molecules . . . . . 39.7 Decomposition of a Molecular Canonical Partition Function into Canonical Partition Functions for Each Degree of Freedom . . . . . . . . . . . . . . . . . 39.8 Energy States and Energy Levels . . . . . . . . . . . . . . Translational, Vibrational, Rotational, and Electronic Contributions to the Partition Function of Monoatomic and Diatomic Ideal Gases and Sample Problem . . . . . . . 40.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.2 Partition Functions of Ideal Gases . . . . . . . . . . . . . 40.3 Translational Partition Function of a Monoatomic Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.4 Electronic Contribution to the Atomic Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.5 Sample Problem 40.1 . . . . . . . . . . . . . . . . . . . . . . 40.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 40.6 Average Energy of a Monoatomic Ideal Gas . . . . . 40.7 Average Heat Capacity at Constant Volume of a Monoatomic Ideal Gas . . . . . . . . . . . . . . . . . . 40.8 Average Pressure of a Monoatomic Ideal Gas . . . . . 40.9 Diatomic Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . .

. 419 . 419 . 419 . 421 . 421 . 422 . 424

. 425 . 427

. 429 . 429 . 429 . 430 . . . .

432 433 433 434

. 435 . 435 . 435

Thermodynamic Properties of Ideal Gases of Diatomic Molecules Calculated Using Partition Functions and Sample Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439 41.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439 41.2 Vibrational Partition Function of a Diatomic Molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 440

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41.3 41.4 41.5 41.6 41.7 41.8 41.9 41.10 Lecture 42

Lecture 43

Sample Problem 41.1 . . . . . . . . . . . . . . . . . . . . . . 41.3.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . Rotational Partition Function of a Diatomic Molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Average Rotational Energy of an Ideal Gas of Diatomic Molecules . . . . . . . . . . . . . . . . . . . . . Average Rotational Heat Capacity at Constant Volume of an Ideal Gas of Diatomic Molecules . . . Fraction of Diatomic Molecules in the Jth Rotational Level . . . . . . . . . . . . . . . . . . . . . . . . . . Rotational Partition Functions of Diatomic Molecules Contain a Symmetry Number . . . . . . . . Total Partition Function of a Diatomic Molecule . . Sample Problem 41.2 . . . . . . . . . . . . . . . . . . . . . . 41.10.1 Solution . . . . . . . . . . . . . . . . . . . . . . . .

Statistical Mechanical Interpretation of Reversible Mechanical Work, Reversible Heat, and the First Law of Thermodynamics, the Micro-Canonical Ensemble and Entropy, and Sample Problem . . . . . . . . . . . . . . . . . 42.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Statistical Mechanical Interpretation of Reversible Mechanical Work, Reversible Heat, and the First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . 42.3 Micro-Canonical Ensemble and Entropy . . . . . . . . 42.4 Relating Entropy to the Canonical Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.5 Sample Problem 42.1 . . . . . . . . . . . . . . . . . . . . . . 42.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 42.6 Relating the Statistical Mechanical Relation, S ¼ kBlnW, to the Thermodynamic Relation, dS ¼ δQrev =T . . . . . . . . . . . . . . . . . . . . . . . . . . . Statistical Mechanical Interpretation of the Third Law of Thermodynamics, Calculation of the Helmholtz Free Energy and Chemical Potentials Using the Canonical Partition Function, and Sample Problems . . . . . . . . . . . 43.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43.2 The Third Law of Thermodynamics and Entropy . . 43.3 Calculation of the Helmholtz Free Energy of a Pure Material Using the Canonical Partition Function . . . 43.4 Sample Problem 43.1 . . . . . . . . . . . . . . . . . . . . . . 43.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . .

. 442 . 442 . 443 . 444 . 445 . 445 . . . .

446 447 447 447

. 449 . 449

. 449 . 452 . 453 . 456 . 456

. 457

. 459 . 459 . 459 . 461 . 462 . 462

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43.5 43.6 Lecture 44

Lecture 45

Lecture 46

Sample Problem 43.2 . . . . . . . . . . . . . . . . . . . . 43.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . Sample Problem 43.3 . . . . . . . . . . . . . . . . . . . . 43.6.1 Solution . . . . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

Grand-Canonical Ensemble, Statistical Fluctuations, and Sample Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 44.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44.2 Grand-Canonical Ensemble . . . . . . . . . . . . . . . . . . 44.3 Statistical Fluctuations . . . . . . . . . . . . . . . . . . . . . 44.4 Sample Problem 44.1 . . . . . . . . . . . . . . . . . . . . . . 44.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 44.5 Sample Problem 44.2 . . . . . . . . . . . . . . . . . . . . . . 44.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 44.6 Fluctuations in the Number of Molecules . . . . . . . . 44.7 Sample Problem 44.3 . . . . . . . . . . . . . . . . . . . . . . 44.7.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 44.8 Equivalence of All the Ensembles in the Thermodynamic Limit . . . . . . . . . . . . . . . . . . . . .

. . . .

463 463 464 465

. . . . . . . . . . .

467 467 467 471 474 474 475 475 478 480 480

. 482

Classical Statistical Mechanics and Sample Problem . . . . 45.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45.2 Classical Statistical Mechanics . . . . . . . . . . . . . . . . 45.3 Classical Molecular Partition Function . . . . . . . . . . . 45.4 Classical Partition Function of an Atom in an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45.5 Classical Partition Function of a Rigid Rotor . . . . . . 45.6 Classical Partition Function of a System Consisting of N Independent and Indistinguishable Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45.7 Classical Partition Function of a System Consisting of N Interacting and Indistinguishable Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45.8 Sample Problem 45.1 . . . . . . . . . . . . . . . . . . . . . . . 45.8.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . 45.9 Simultaneous Treatment of Classical and Quantum Mechanical Degrees of Freedom . . . . . 45.10 Equipartition of Energy . . . . . . . . . . . . . . . . . . . . . . Configurational Integral and Statistical Mechanical Derivation of the Virial Equation of State . . . . . . . . . . . . . 46.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46.2 Modeling Gases at Number Densities Approaching Zero . . . . . . . . . . . . . . . . . . . . . . . . . 46.3 Modeling Gases at Higher Number Densities . . . . . .

485 485 485 486 487 488

489

490 491 491 492 493 495 495 495 497

Contents

xxiii

46.4 Lecture 47

Lecture 48

Lecture 49

Derivation of the Virial Equation of State Using the Grand-Canonical Partition Function . . . . . 497

Virial Coefficients in the Classical Limit, Statistical Mechanical Derivation of the van der Waals Equation of State, and Sample Problem . . . . . . . . . . . . . 47.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47.2 Virial Coefficients in the Classical Limit . . . . . . . . 47.3 Spatial Dependence of the Two-Body Interaction Potential Including Its Long-Range Asymptotic Behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47.4 Sample Problem 47.1: Calculate the Second Virial Coefficient Corresponding to the Hard-Sphere Interaction Potential . . . . . . . . . . . . . . . . . . . . . . . 47.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . 47.5 Calculating the Second Virial Coefficient Corresponding to an Interaction Potential Consisting of a Hard-Sphere Repulsion and a van der Waals Attraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47.6 Important Remarks About the Behavior of Interaction Potentials . . . . . . . . . . . . . . . . . . . . 47.7 Derivation of the van der Waals Equation of State Using Statistical Mechanics . . . . . . . . . . . Statistical Mechanical Treatment of Chemical Reaction Equilibria and Sample Problem . . . . . . . . . . . . 48.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48.2 Expressing the Equilibrium Constant Using Partition Functions . . . . . . . . . . . . . . . . . . . 48.3 Relating the Pressure-Based and the Number Density-Based Equilibrium Constants for Ideal Gas Mixtures . . . . . . . . . . . . . . . . . . . . . 48.4 Sample Problem 48.1 . . . . . . . . . . . . . . . . . . . . . . 48.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . Statistical Mechanical Treatment of Binary Liquid Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49.2 Modeling Binary Liquid Mixtures Using a Statistical Mechanical Approach . . . . . . . . . . . . . 49.3 Calculating ΔSmix Using the Micro-Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49.4 Range of Validity of the Lattice Description of Binary Liquid Mixtures . . . . . . . . . . . . . . . . . . 49.5 Lattice Theory Calculation . . . . . . . . . . . . . . . . . .

. 505 . 505 . 505

. 508

. 509 . 509

. 511 . 512 . 513 . 515 . 515 . 515

. 519 . 519 . 520 . 523 . 523 . 523 . 524 . 525 . 526

xxiv

Contents

49.6 49.7 Lecture 50

Calculation of Chemical Potentials . . . . . . . . . . . . . 528 Molecular Characteristics of Ideal Solutions . . . . . . . 529

Review of Part III and Sample Problem . . . . . . . . . . . . . . 50.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.2 Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . 50.3 Average Properties in the Canonical Ensemble . . . . . 50.4 Calculation of the Canonical Partition Function . . . . 50.5 Molecular Partition Functions of Ideal Gases . . . . . . 50.6 Summary of Thermodynamic Functions of Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.7 Grand-Canonical Ensemble . . . . . . . . . . . . . . . . . . . 50.8 Average Properties in the Grand-Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.9 Micro-Canonical Ensemble . . . . . . . . . . . . . . . . . . . 50.10 Average Entropy in the Micro-Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.11 Classical Statistical Mechanics . . . . . . . . . . . . . . . . 50.12 Calculation of Virial Coefficients . . . . . . . . . . . . . . . 50.13 Statistical Mechanical Treatment of Chemical Reaction Equilibria . . . . . . . . . . . . . . . . . . . . . . . . . 50.14 Statistical Mechanical Treatment of Binary Liquid Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.15 Useful Constants in Statistical Mechanics . . . . . . . . . 50.16 Useful Relations in Statistical Mechanics . . . . . . . . . 50.17 Sample Problem 50.1 . . . . . . . . . . . . . . . . . . . . . . . 50.17.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . .

531 531 531 532 533 534 536 536 537 538 538 538 539 540 540 540 540 541 541

Solved Problems for Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545 Solved Problems for Part II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 639 Solved Problems for Part III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706

Lecture 1

Introduction to the Book

1.1

Motivation and Scope of the Book

This book is based on lectures that I delivered in the one semester, graduate-level course Chemical Engineering Thermodynamics (10.40) in the Department of Chemical Engineering at the Massachusetts Institute of Technology (MIT). For my teaching of 10.40, I was awarded the Outstanding Teaching Award by the graduate students nine times. Encouraged and motivated by repeated requests from my 10.40 students, in 2018, I finally decided to write this book which is based on my lecture notes, supplemented by many solved sample problems which help crystallize the material taught. Including this lecture, the book consists of 50 lectures and is primarily designed for graduate students and senior undergraduate students in Chemical Engineering. The book is also suitable for graduate students in Mechanical Engineering, Chemistry, and Materials Science. It focuses on developing the ability of the reader to solve a broad range of challenging problems in Classical Thermodynamics, by applying the fundamental principles and concepts taught to new and often unusual scenarios. In addition, the book introduces readers to the fundamentals of Statistical Mechanics, including demonstrating how the microscopic properties of atoms and molecules, as well as their interactions, can be accounted for to calculate various practically relevant average thermodynamic properties of macroscopic systems. Again, solved sample problems are presented to help the reader better understand the material taught. My lecture notes, and therefore this book, are inspired by concepts, principles, methods, and applications that I distilled and adapted from a number of books (see below), as well as by my own in-depth understanding of the material taught. The book provides a pedagogical presentation of the fundamentals of Classical Thermodynamics, with an introduction to Statistical Mechanics, including applying these fundamentals to the solution of illuminating and practically relevant sample problems which are dispersed throughout the lectures. © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_1

1

2

1.2

1 Introduction to the Book

Organization of the Book

The book is organized as follows: Lecture 1 provides an introduction to the book. Part I discusses Fundamental Principles and Properties of Pure Fluids and consists of Lectures 2–20. Part II discusses Mixtures: Models and Applications to Phase and Chemical Reaction Equilibria and consists of Lectures 21–37. Part III presents an Introduction to Statistical Mechanics and consists of Lectures 38–50. Illuminating solved sample problems are presented throughout the book. In addition, following Lecture 50, the book contains solved problems pertaining to Part I (10 solved problems), Part II (10 solved problems), and Part III (15 solved problems). These problems are challenging and will assist readers to crystalize the material taught. For complete details about the organization of the book, readers are referred to the comprehensive Table of Contents. The material on Classical Thermodynamics presented in Lectures 2–20 of Part I, as well as in Lectures 21–37 of Part II, is adapted from Thermodynamics and Its Applications, third edition, by Jefferson W. Tester and Michael Modell, Prentice Hall International Series in the Physical and Chemical Sciences, Upper Saddle River, NJ (1996). Hereafter, the names of the authors will be abbreviated as T&M. In addition, some of the material presented in Lectures 9 and 12 of Part I is adapted from The Principles of Chemical Thermodynamics, fourth edition, by Kenneth Denbigh, Cambridge University Press, London (1981). Hereafter, the name of the author will be abbreviated as Denbigh. The material on Introduction to Statistical Mechanics presented in Lectures 38– 43 of Part III is adapted from Molecular Thermodynamics, by Donald A. McQuarrie and John D. Simon, University Science Books, Sausalito CA (1999). Hereafter, the names of the authors will be abbreviated as M&S. The material on Introduction to Statistical Mechanics presented in Lectures 44–48 of Part III is adapted from Statistical Mechanics, by Donald A. McQuarrie, Harper & Row, New York (1976). Hereafter, the name of the author will be abbreviated as McQuarrie. The material presented in Lectures 49 and 50 of Part III is adapted from my lecture notes. Starting with Lecture 2, each lecture begins with an introduction section which summarizes the material that will be discussed in the lecture, including solved sample problems. These introductions will serve as useful road maps for the 49 lectures and, as such, will help readers to more readily navigate the material taught.

1.3

Acknowledgments

I am grateful to the students who attended my 10.40 lectures over the years for challenging me with insightful questions which helped me improve the quality of my lectures. Many thanks to all the teaching assistants who worked with me over the years, and who helped the students crystallize the material taught in class through fruitful interactions during office hours and one-on-one meetings, including

1.3 Acknowledgments

3

preparing outstanding solutions to homework and exam problems. In particular, I am indebted to Nancy Zoeller, Daniel Kamei, Ahmed Ismail, Henry Lam, Srinivas Moorkanikkara, Fei Chen, Amanda Engler, Jaisree Iyer, Michael Stern, Bomy Lee Chung, Sven Schlumpberger, Jennifer Lewis, Manish Shetty, Ananth Govind Rajan, Ran Chen, Tzyy-Shyang Lin, and Dimitrios Fraggedakis for their efforts and intellectual contributions to 10.40. Many thanks to Vishnu Sresht and Rahul Prasanna Misra for their valuable insights on various aspects of Classical Thermodynamics and Statistical Mechanics. I am also grateful to my colleagues, Jefferson Tester, who introduced me to the teaching of Chemical Engineering Thermodynamics, and Jonathan Harris, Bernhardt Trout, and Bradley Olsen, who co-taught 10.40 with me and with whom I discussed many challenging aspects of the material taught. I am indebted to Dimitrios Fraggedakis for creating all the beautiful graphical figures, and to Tzyy-Shyang Lin for creating all the tables. I am also indebted to my assistant, Cindy Welch, for her immense dedication and skill in typing the majority of the lectures in the book. Many thanks to our department head, Paula Hammond, and to our Executive Officer, Martin Bazant, for facilitating writing of the book. I am most grateful to Steven Elliot, Senior Publishing Editor, Cambridge University Press, for his insightful advise and encouragement about publishing engineering books, and for introducing me to Michael Luby, Senior Publishing Editor at Springer, who eventually oversaw the publication of my book. In fact, I am indebted to Michael Luby, Brian Halm, Sathya Stephen, Brinda Megasyamalan, and Mario Gabriele from Springer for their dedication and guidance with all aspects of my book project. Finally, my greatest gratitude and deepest love go to my parents, Samuel and Julia, for instilling in me the importance of education and hard work, to my wife, Anna, for her love, patience, and understanding, and to my daughters, Suzan and Dana, for their love and continued support.

Part I

Fundamental Principles and Properties of Pure Fluids

Lecture 2

Useful Definitions, Postulates, Nomenclature, and Sample Problems

2.1

Introduction

The material presented in this lecture is adapted from Chapter 2 in T&M. First, we will introduce useful definitions which will be utilized throughout Parts I, II, and III of this book. Second, we will discuss two (I and II) out of the four (I, II, III, and IV) postulates which serve as the pillars on which the edifice of thermodynamics is erected. These postulates can be viewed as truisms which cannot be proven from first principles, but that are consistent with our experimental observations, and have withstood the test of time. Third, we will present the nomenclature which will be utilized throughout this book. Fourth, we will solve Sample Problem 2.1 to determine if a system is simple or composite. Finally, we will solve Sample Problem 2.2 to shed light on the liquid-vapor monovariant line in a pressure (P)-temperature (T) phase diagram.

2.2

Useful Definitions

System – Subject of the experiment that we carry out. Well-defined region in terms of spatial coordinates. Boundary – Surface enclosing the system. Can be real or imaginary. Environment – Region of space external to the system and sharing a common boundary with the system. Work and heat interactions occur at the boundary. Primitive Property – Property of the system which can be measured at a particular time without perturbing the system and whose value does not depend on the history of the system. Examples include temperature, pressure, and volume. Event – An occurrence where at least one primitive property changes. Interaction – A simultaneous event which occurs in the system and its environment and would change if the environment (or the system) was changed. © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_2

7

8

2 Useful Definitions, Postulates, Nomenclature, and Sample Problems

Closed System – System whose boundary is impermeable to mass flow. Open System – System whose boundary is permeable to mass flow, through at least one point, of at least one component of the system. Adiabatic Boundary – Prevents the transfer of heat. Prevents temperature equilibration. Diathermal Boundary – Allows the transfer of heat. Leads to instantaneous temperature equilibration. Rigid Boundary – Prevents changes in the volume of the system. Movable Boundary – Permits changes in the volume of the system. External Constraints – Set of boundaries: permeable versus impermeable to mass flow, adiabatic versus diathermal, rigid versus movable. Isolated System – System enclosed by impermeable, adiabatic, and rigid boundaries. Simple System – System devoid of any internal impermeable, rigid, and adiabatic boundaries, and not acted upon by external force fields (e.g., gravitational, electric, or inertial forces), which can change the energy (e.g., gravitational, electric, kinetic) of the system. Composite System – System composed of two or more simple systems. No restrictions apply to the type of internal boundaries separating the various simple systems. Phase – Region within a simple system having uniform properties. Restraints – Barriers within a system (simple or composite) that prevent some changes from occurring within the time span of interest. Internal boundaries in a composite system which are impermeable, adiabatic, or rigid are considered restraints. State of the System – Identified by the values of the properties needed to reproduce the system. Stable Equilibrium State – State whose properties do not vary with time. Postulate I postulates the existence of stable equilibrium states and indicates the number of properties that need to be specified to unambiguously characterize such states (see below). Postulate II postulates what conditions are required to attain a stable equilibrium state (see below). Change of State – Identified by a change in the value of at least one property. Path – Describes all the states that the system traverses during a change of state. Quasi-static Path – A path for which all the intermediate states are equilibrium states. Derived Property – Property which exists only for stable equilibrium states, and is not measurable. A derived property is defined in terms of changes in the state of the system between initial and final stable equilibrium states. As such, a derived property is a state function (e.g., energy, enthalpy, entropy). Extensive Property – Property which depends on the size (mass) of the system (e.g., volume, energy, number of molecules). Intensive Property – Property which is independent of the size (mass) of the system (e.g., pressure, temperature, chemical potential).

2.4 Sample Problem 2.1

2.3

9

Postulates I and II (Adapted from Appendix A in T&M)

I. For closed, simple systems with given internal restraints, there exist stable equilibrium states that can be characterized completely by two independently variable properties in addition to the masses of the particular chemical species initially charged. – Postulates the existence of stable equilibrium states, but does not indicate when they exist – As we will show in Part II, is consistent with the Gibbs Phase Rule II. In processes for which there is no net effect on the environment, all systems (simple and composite) with given internal restraints will change in such a way that they approach one and only one stable equilibrium state for each simple subsystem. In the limiting condition, the entire system is said to be at equilibrium. – Describes the natural tendency of an isolated system to approach a state of stable equilibrium characterized by minimal internal energy and maximal entropy – If the set of internal restraints changes, then so will the state of stable equilibrium to which the system tends

2.4

Sample Problem 2.1

A closed container is filled with water coexisting with its vapor at room temperature

Fig. 2.1

(see Fig. 2.1). Is the system simple or composite?

10

2.4.1

2 Useful Definitions, Postulates, Nomenclature, and Sample Problems

Solution

The system in Fig. 2.1 consists of two equilibrated phases (water and water vapor), separated by an interface which is diathermal, movable, and open. In addition, the change in the gravitational potential energy of the molecules in the vapor phase is negligible and is zero for the fixed center of mass of the liquid phase. As a result, according to the requirements imposed by Postulate I, the system in Fig. 2.1 is indeed simple.

2.5

Sample Problem 2.2

Fig. 2.2

Discuss the liquid/vapor equilibrium line in the pressure (P)-temperature (T) phase diagram shown in Fig. 2.2, where TP and CP denote the Triple Point and the Critical Point, respectively.

2.5.1

Solution

The variation of pressure (P) with temperature (T) along the L/V equilibrium line in Fig. 2.2 is governed by the Clapeyron equation, which we will derive in Part II, and is given by: 

dP dT

 ¼ L=V

HV  HL ΔHvap
V L ¼ TΔVvap T V V

ð2:1Þ

2.6 Nomenclature

11

In the last term in Eq. (2.1), the numerator is equal to the molar enthalpy of vaporization, and the denominator is equal to the absolute temperature times the molar volume of vaporization. Figure 2.2 and Eq. (2.1) show that along the liquidvapor (L/V) equilibrium line, pressure and temperature are not independent. As a result, only one intensive variable can be specified, for example, the temperature, which uniquely determines the pressure. As we will show in Part II, this result is consistent with the celebrated Gibbs Phase Rule.

2.6

Nomenclature Variable

Extensive Intensive

General case, B Energy Internal energy Enthalpy Entropy Gibbs free energy Volume Mole number Temperature Pressure Chemical potential

B E U H S G V N __ __ __

B E U H S G V __ T P μ

In the table above and throughout this book, we will utilize the underbar to denote extensive variables like V, U, H, and S. The corresponding molar (intensive) variables like V, U, H, and S will not carry the underbar.

Lecture 3

The First Law of Thermodynamics for Closed Systems: Derivation and Sample Problems

3.1

Introduction

The material presented in this lecture is adapted from Chapter 3 in T&M. First, we will discuss mechanical work done on a rigid body. We will also introduce other types of work, including surface, electric, and magnetic. Second, we will extend the concept of mechanical work done on a rigid body to that associated with a thermodynamic system, for example, work done by a gas expanding against the atmosphere in a cylinder-piston assembly in the presence of friction. Third, we will discuss a key result that the work done by a system on the environment is equal to minus the work done by the environment on the system. We will stress that work is a mode of energy transfer which exists only at the boundary between a system and the environment. In the system or in the environment, there is only energy, which we will introduce through Postulate III as work done on the system under adiabatic conditions. Fourth, we will introduce heat absorbed by the system as the difference between the work done on the system under adiabatic conditions and the actual work done on the system. Like work, heat is a form of energy transfer which exists solely at the boundary between the system and the environment. Fifth, after we introduce work, energy, and heat, the First Law of Thermodynamics for a closed system will emerge naturally. Finally, we will solve Sample Problems 3.1, 3.2, and 3.3 to help crystallize the material taught.

3.2

Work Interactions

The differential mechanical work associated with the movement of a rigid body is defined as follows:

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_3

13

14

3 The First Law of Thermodynamics for Closed Systems: Derivation and Sample. . .

δW ¼ |{z} Differential mechanical work

X !  ! Fs • |{z} dr |fflfflfflffl{zfflfflfflffl} Differential Sum of all the forces acting on the surface or boundary of the rigid body at a point where there is a differential displacement of the boundary

ð3:1Þ

displacement of the boundary

The symbol δ denotes path-dependent differentials of functions which are not state variables, such as work. The integral of such functions yields the value of the function. For example: State ð 2

δW ¼ W

ð3:2Þ

State 1

and depends on the path connecting states 1 and 2.

3.3

Sample Problem 3.1

Calculate the work done by a gas expanding isothermally (see Fig. 3.1).

3.3.1

Solution

Given N moles of gas expanding isothermally, the gas pressure, Pg, is a function of the gas volume, Vg (see Fig. 3.1). The work done by the expanding gas is given by (see Fig. 3.1): V2g

State ð 2

Wg ¼

δWg ¼ State 1



ð

Pg dVg ¼ V1g

Fig. 3.1

Area beneath the path; depends on the path!

 ð3:3Þ

3.4 Specific Types of Work Interactions

15

The symbol d denotes total differentials of functions which are state variables. The integral of such functions yields the change in the value of the function. For example: State ð 2

!

!

!

dr ¼ r2  r1

ð3:4Þ

State 1

and is independent of the path connecting states 1 and 2. The total mechanical work is given by (see Eq. (3.1)): !

!

ðr 2

W¼

δW ¼ !

r1

ðr 2 X

 ! Fs • d r

!

ð3:5Þ

!

r1

For a list of specific types of work interactions, see below. Body (b) forces, or forces associated with external fields, act on molecules in the ! system. They are denoted by F b . Examples include gravitational, inertial, centrifugal, and Coulombic forces.

3.4

Specific Types of Work Interactions

As discussed above, all work interactions are path-dependent and are defined to occur at the boundary of a system. We saw that the symbol δ is used here to designate a path-dependent property. Further, we saw that, in general, differential work can be ! represented by the dot product of a boundary (surface, s) force, F s , and a differential ! displacement, d r . Specifically, !

!

δW ¼ F s • d r

The following specific types of work are encountered in nature: Pressure: – PdV (P ¼ pressure, V ¼ volume) Surface: σda (σ ¼ surface tension, a ¼ area) !

! !

!

Electric: E • dD (E ¼ electric field strength, D ¼ electric flux density) !

! !

!

Magnetic: H • dB (H ¼ magnetic field strength, B ¼ magnetic flux density) !

! !

!

Frictional: F f • d r (F f ¼ frictional force, r ¼ displacement)

ð3:6Þ

3 The First Law of Thermodynamics for Closed Systems: Derivation and Sample. . .

16

3.5

Sample Problem 3.2

Calculate the balance of forces on a weight suspended by a string rising in the þbz direction in a gravitational field in the absence of viscous forces (see Fig. 3.2).

Fig. 3.2

3.5.1

Solution

Newton’s Second Law of Motion states that: X! X! ! Fs þ Fb ¼ m a

ð3:7Þ

!

Considering the inertial force,  m a , as a body force, it follows that:   X !  X! The sum of all surface and body ! Fb  m a ¼ 0 Fs þ forces acting on a rigid body is equal to 0

ð3:8Þ

3.6

Sample Problem 3.3

Calculate the work done by a gas on its environment as it expands against the atmosphere in a cylinder-piston assembly in the presence of friction (see Fig. 3.3).

3.6 Sample Problem 3.3

17

Fig. 3.3

3.6.1

Solution

Figure 3.3 shows a gas expanding against the atmosphere in a cylinder-piston assembly: • The gas is a simple, closed system with one moving boundary • Assume a slow, quasi-static gas expansion According to our definition of mechanical work, the work done by the gas on its environment is given by: !

!

δWg ¼ F g • d z ¼ Fg dz

ð3:9Þ

Fg ¼ Pg A ) δWg ¼ Pg ðAdzÞ

ð3:10Þ

dVg ¼ Adz ) δWg ¼ Pg dVg ð3:11Þ 9 8 To calculate Wg , we must know > > V2g > > zð2 zð2 > > ð = < the path, P ¼ P ð z Þ or P ¼ P g g g g Pg dVg Wg ¼ δWg ¼ Pg Adz ¼ > >   > > > z1 z1 ; : Vg , connecting states 1 and 2 > V1g ð3:12Þ Note that all the forces depicted in Fig. 3.4 are colinear (along þbz or bz). As a result, Newton’s Second Law of Motion implies that: Fg  Pa A  Ff  mg  ma ¼ 0

ð3:13Þ

Fg ¼ Pa A þ Ff þ mg þ ma

ð3:14Þ

or

18

3 The First Law of Thermodynamics for Closed Systems: Derivation and Sample. . .

Fig. 3.4

dv dv dz ¼ ¼ ðdv=dzÞv dt dz dt

ð3:15Þ

  dv dz Pa A þ Ff þ mg þ mv dz

ð3:16Þ

Recall that: a ¼ Accordingly, δWg ¼ Fg dz ¼

δWg ¼ Pa ðAdzÞ þ Ff dz þ mgdz þ mvdv   1 δWg ¼ Pa dVg þ dðmgzÞ þ d mv2 þ Ff dz |fflfflfflfflfflffl{zfflfflfflfflfflffl} |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflffl2ffl{zfflfflfflfflfflfflfflfflfflffl} Work done by 



Work done by the gas to push back the atmosphere

Work done by the gas to increase the potential energy of the piston

Work done by the gas to increase the kinetic energy of the piston

ð3:17Þ ð3:18Þ

the gas on the cylinder walls to overcome friction

Equation (3.18) shows that work done by the gas can be computed through its effect on the gas environment (the atmosphere, the piston, and the cylinder walls). To calculate Wg, we simply integrate δWg in Eq. (3.18) along a path connecting states 1 and 2. For simplicity, if Ff is assumed to be constant, then: zð2

Wg ¼

δWg

ð3:19Þ

z1

Wg ¼ Pa Aðz2  z1 Þ þ mgðz2  z1 Þ þ

1 2 m v 2  v 1 2 þ Ff ð z 2  z 1 Þ 2

ð3:20Þ

Equation (3.20) shows that if we know Ff, we can measure (Pa, A, m, z1, z2, v1, and v2) and calculate Wg in terms of its effects on the gas environment. Following a similar procedure, we can also calculate:

3.6 Sample Problem 3.3

19

• Wa – Work done by the atmosphere • Wp – Work done by the piston • Ww – Work done by the cylinder walls on their respective environments. We can then show that Wg + Wa + Wp + Ww ¼ 0, or that: WSystem on Environment

¼

WEnvironment on System

ð3:21Þ

Equation (3.21) is a key result that we will utilize to solve many problems involving the calculation of mechanical work done by a system on its environment. It is always possible to measure mechanical work in terms of the change in the potential energy of a mass in a gravitational field. Mechanical work is a form of energy transfer. For example, work done by the gas was transferred into potential and kinetic energies of the piston. By carrying out mechanical work, a system can transfer energy to its environment. There is a different (nonmechanical) mode of energy transfer known as heat (see Fig. 3.5). Energy is transferred as heat “Hot Body”

“Cold Body”

Diathermal Boundary

Fig. 3.5

To introduce heat in a rigorous manner, we invoke the notion of an adiabatic work interaction associated with a purely mechanical process (no heat can be transferred across adiabatic boundaries). Simply put, work and heat are two modes of energy transfer. Once in the system, only energy exists. Later in Part I, we will see that there is a distinction in the quality and efficiency of work and heat. The concept of the adiabatic work interaction which is always possible between stable equilibrium states is introduced through Postulate III (see below). Because the adiabatic (a) work is only a function of the end states, it is a derived property of the system, which we will refer to as the Energy, E, of the system. By convention, the energy of the system increases when work is done on the system by its environment, that is: Ef ‐Ei ¼ þWai!f

ð3:22Þ

The adiabatic work, Wai!f , is independent of the path connecting states i and f and is therefore a state function.

3 The First Law of Thermodynamics for Closed Systems: Derivation and Sample. . .

20

3.7

Postulate III (Adapted from Appendix A in T&M)

For any states, (1) and (2), in which a closed system is at equilibrium, the change of state represented by (1) ! (2) and/or the reverse change (2) ! (1) can occur by at least one adiabatic process, and the adiabatic work interaction between this system and its surroundings is determined uniquely by specifying the end states (1) and (2). * States that adiabatic work interactions of any type in closed systems are state functions * Implies that the First Law of Thermodynamics is, in fact, a restatement of the Law of Energy Conservation, because the adiabatic form of the First Law of Thermodynamics states that dE ¼ δQ þ δW, with δQ ¼ 0

3.8

Energy Decomposition

The total energy, E, can be decomposed into three main contributions: (i) Internal energy, U, associated with microscopic energy storage at the molecular level (ii) Potential energy, EPE, associated with the gravitational force (iii) Kinetic energy, EKE, associated with the inertial force Adding up (i), (ii), and (iii) above yields: E ¼ U þ EPE þ EKE

ð3:23Þ

EPE ¼ 0, EKE ¼ 0 ) E ¼ U

ð3:24Þ

For a simple system:

3.9

Heat Interactions

The energy difference between two states can always be determined by measuring the work in an adiabatic process connecting the two states (Postulate III, see above). With the same initial and final states, we can visualize any process (adiabatic or nonadiabatic) connecting the two states. Because energy is a state function, the energy difference is the same as that found for the adiabatic process. However, if the process is not adiabatic, the work interaction will be different than that for the adiabatic process. Nevertheless, it is always possible to measure work as discussed above.

3.10

The First Law of Thermodynamics for Closed Systems

21

One can then define heat, Q, as the difference between the energy change and the actual work carried out (see Fig. 3.6). 

Q ¼ ðEf  Ei Þ  Wi!f



Ef  Ei ¼ þWai!f ∴ Q ¼ Wai!f  Wi!f

3.10

ð3:25Þ ð3:26Þ

The First Law of Thermodynamics for Closed Systems

The First Law of Thermodynamics for closed systems is a restatement of the Law of Energy Conservation and the manner in which energy (associated with the bulk of the system) and work and heat (associated with the boundaries of the system) are interconverted.

Fig. 3.6

In integral form, the First Law of Thermodynamics for closed systems is written as follows: ΔE ¼ Ef  Ei ¼ Q þ W

ð3:27Þ

where Q is the total amount of heat absorbed by the system and W is the total amount of work done on the system. In differential form, the First Law of Thermodynamics for closed systems is written as follows:

22

3 The First Law of Thermodynamics for Closed Systems: Derivation and Sample. . .

dE ¼ δQ þ δW

ð3:28Þ

For a closed system interacting with its environment, the composite of [system (S) + environment (E)] can always be considered as a “new system” of constant volume surrounded by an adiabatic and impermeable boundary (see Fig. 3.7):

Fig. 3.7

An examination of Fig. 3.7 shows that: ΔESþE ¼ ΔES þ ΔEE

ð3:29Þ

ΔESþE ¼ Q þ W ¼ 0 þ 0 ¼ 0

ð3:30Þ

fΔES ¼ ΔEE ; WS ¼ W E g

ð3:31Þ

∴ QS ¼ QE

ð3:32Þ

Lecture 4

The First Law of Thermodynamics for Closed Systems: Thermal Equilibrium, the Ideal Gas, and Sample Problem

4.1

Introduction

The material presented in this lecture is adapted from Chapter 3 in T&M. First, we will discuss thermal equilibrium, Postulate IV, and the directionality of heat flow. Second, we will discuss the thermodynamic properties of an ideal gas. Third, we will utilize the ideal gas as a model fluid which will allow us to obtain mathematically simple solutions when we solve problems involving the First Law of Thermodynamics. Fourth, we will begin solving Sample Problem 4.1, an illuminating problem which will allow us to select various possible systems, including ascertaining which one leads to the most challenging, or to the simplest, solution. It will become apparent that the engineer, or the scientist, needs to develop a facility to select the optimal system which will lead to the simplest solution. This, of course, will require experiense and practice. Finally, we will present a four-step strategy that can be used to solve problems involving the First Law of Thermodynamics.

4.2

Thermal Equilibrium and the Directionality of Heat Interactions

Subsystems A and B which are at equilibrium across a diathermal boundary are in thermal equilibrium. If there is a heat interaction between subsystems A and B of the isolated, composite system, it follows from Postulate IV (see below) that this intereaction must eventually cease because each subsystem, as well as the composite system, will approach equilibrium. If a heat interaction occurs, it follows that QA + QB ¼ 0 (see Fig. 4.1). The equations below describe the isolated composite system (A + B) shown in Fig. 4.1.

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_4

23

4 The First Law of Thermodynamics for Closed Systems: Thermal Equilibrium. . .

24

Fig. 4.1

ΔEAþB ¼ QAþB þ WAþB ðFirst Law of ThermodynamicsÞ

ð4:1Þ

ΔEAþB ¼ 0 ðIsolated SystemÞ

ð4:2Þ

WAþB ¼ 0 ðRigid BoundaryÞ

ð4:3Þ

∴QAþB ¼ QA þ QB ¼ 0 ðAdiabatic BoundaryÞ

ð4:4Þ

What is the direction of a heat interaction? Postulate IV (see below) helps us answer this question. Postulate IV is often referred to as the Zeroth Law of Thermodynamics.

4.3

Postulate IV (Adapted from Appendix A in T&M)

If the sets of systems A,B and A,C each have no heat interaction when connected across nonadiabatic walls, then, there will be no heat interaction if systems B and C are also so connected. – This postulate is sometimes referred to as the “Zeroth Law of Thermodynamics” – States that temperature differences are required for heat transfer to occur – Introduces the concept of thermal equilibrium in the absence of heat interactions When sytems A and B undergo a purely heat interaction, that is, WA ¼ 0 and WB ¼ 0, such that: ΔEA ¼ QA þ WA ¼ QA ¼ ðΔEB ¼ QB þ WB ¼ QB Þ

ð4:5Þ

QA ¼ QB

ð4:6Þ

or

then, heat is transferred from system A to system B. We can also state that QA!B > 0.

4.4 Ideal Gas Properties

25

The thermometric temperature, θ, can be used to rank systems with respect to the direction of heat interactions. If for the three systems, A, B, and C, QA!B > 0 (heat is transferred from A to B) and QB!C > 0 (heat is transferred from B to C), then, θA > θB > θC

ð4:7Þ

By convention, when θA > θB, the heat interaction is such that EA decreases and EB increases or: dE >0 dθ

4.4 4.4.1

Ideal Gas Properties Equation of State (EOS)

PV ¼ NRT (Extensive form) PV ¼ RT (Intensive form)

4.4.2

Internal Energy

o U is only R o a function of temperature: dU ¼ Cv dT U ¼ Cv dT þ U0 ; U ¼ NU U0 ¼ Reference-state constant Cov ¼ Ideal gas heat capacity at constant volume Cov ¼ ð∂U=∂TÞV Cov ¼ gðTÞ ¼ a þ bT þ cT2 þ . . .

where a, b, c, etc. are fitted empirical constants.

4.4.3

Enthalpy

H is only a function of temperature: dH ¼ Cop dT R H ¼ U þ PV ¼ Cop dT þ H0 ; H ¼ NH ¼ U þ PV H0 ¼ Reference-state constant Cop ¼ Ideal gas heat capacity at constant pressure

ð4:8Þ

26

4 The First Law of Thermodynamics for Closed Systems: Thermal Equilibrium. . .

Cop ¼ ð∂H=∂TÞP Cop ¼ f ðTÞ ¼ a* þ b* T þ c* T2 þ . . . where a*, b*, c*, etc. are fitted empirical constants and a*  a ¼ R, b ¼ b*, c ¼ c*, etc. (see Section 4.4.2).

4.4.4

Other Useful Relationships Cop  Cov ¼ R dV=V ¼ dT=T  dP=P dV=V þ dP=P ¼ dT=T þ dN=N

4.5

Sample Problem 4.1: Problem 3.1 in T&M

A small well-insulated cylinder and piston assembly (see Fig. 4.2) contains an ideal gas at 10.13 bar and 294.3 K. A mechanical lock prevents the piston from moving. The length of the cylinder containing the gas is 0.305 m, and the piston crosssectional area is 1.858  102 m2. The piston, which weighs 226 kg, is tightly fitted, and when allowed to move, there are indications that considerable friction is present. When the mechanical lock is released, the piston moves in the cylinder until it impacts and is engaged by another mechanical stop; at this point, the gas volume has just doubled. The heat capacity of the ideal gas is 20.93 J/mol K, independent of temperature and pressure. Consider the heat capacities of the piston and the cylinder walls to be negligible.

Fig. 4.2

4.5 Sample Problem 4.1: Problem 3.1 in T&M

27

(a) As an engineer, can you estimate the temperature and pressure of the gas after such an expansion? Clearly state any assumptions. (b) Repeat the calculations if the cylinder was rotated 90 and 180 before tripping the mechanical lock.

4.5.1

Solution

Fig. 4.3

To solve this problem, we will make use of the following four steps: 1. Sketch Given Configuration (see Fig. 4.3) 2. Summarize Given Data and Information • Gas: Ideal Initial Condition: Ti ¼ 294.3 K Pi ¼ 10.13 bar zi ¼ 0.305 m Cv ¼ 20.93 J/mol K Final Condition: Tf ¼?; Pf ¼? zf ¼ 2zi ¼ 0.610 m • Atmosphere:

4 The First Law of Thermodynamics for Closed Systems: Thermal Equilibrium. . .

28

Pa ¼ 1.013 bar Infinite reservoir • Piston: Area – Ap ¼ 1.858  102 m2 Mass – Mp ¼ 226 kg Negligible heat capacity, Cv ¼ 0 Well insulated • Cylinder Walls: Negligible heat capacity, Cv ¼ 0 Well insulated • Considerable friction is present between the tightly fitted piston and the cylinder walls! Find Tf and Pf of the gas if the cylinder is: (i) Upright, as in the sketch in Fig. 4.3. (ii) Tilted 90 or 180 before removing Stop 1. 3. Identify Critical Issues for Solution (i) What system and boundaries should we select? (ii) How do we deal with the friction? 4. Make Physically Reasonable Approximations (i) Due to the friction, the gas expansion is slow and the path is quasi static.

Fig. 4.4

(ii) Gravitational effects on the gas and the atmosphere are negligible ) both systems are simple (Figs. 4.4, 4.5, and 4.6).

4.5 Sample Problem 4.1: Problem 3.1 in T&M

Fig. 4.5

Fig. 4.6

29

Lecture 5

The First Law of Thermodynamics for Closed Systems: Sample Problem 4.1, Continued

5.1

Introduction

The material presented in this lecture is adapted from Chapter 3 in T&M. First, we will choose one of the systems that we introduced in Lecture 4, where friction does not need to be accounted for, and show that it will lead to a relatively simple solution of Sample Problem 4.1 (denoted as Solution 1). Second, we will choose another system where friction will need to be accounted for. Recall that in the Statement of Sample Problem 4.1, no information is provided about the friction! Third, the need to deal explicitly with the friction will lead to a more challenging solution (denoted as Solution 2). Nevertheless, through a creative derivation, we will show that the mechanical work done by the gas to overcome friction at the cylinder walls is recovered 100% as heat absorbed by the gas, and therefore, it cancels out in the context of the First Law of Thermodynamics for the chosen system. Finally, as expected, we will show that the simpler Solution 1 and the more challenging Solution 2 yield identical results.

5.2

Sample Problem 4.1: Problem 3.1 in T&M, Continued

For completeness, below, we again present the Statement of Sample Problem 4.1. A small well-insulated cylinder and piston assembly (see Fig. 5.1) contain an ideal gas at 10.13 bar and 294.3 K. A mechanical lock prevents the piston from moving. The length of the cylinder containing the gas is 0.305 m, and the piston cross-sectional area is 1.858
102 m2. The piston, which weighs 226 kg, is tightly fitted, and when allowed to move, there are indications that considerable friction is present. When the mechanical lock is released, the piston moves in the cylinder until it impacts and is engaged by another mechanical stop; at this point, the gas volume has just doubled. The heat © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_5

31

32

5 The First Law of Thermodynamics for Closed Systems. . .

capacity of the ideal gas is 20.93 J/mol K, independent of temperature and pressure. Consider the heat capacity of the piston and cylinder walls to be negligible.

Fig. 5.1

(a) As an engineer, can you estimate the temperature and pressure of the gas after such an expansion? Clearly state any assumptions. (b) Repeat the calculations if the cylinder were rotated 90 and 180 before tripping the mechanical lock.

5.3

Solution 1: System III-Atmosphere (a)

Fig. 5.2

We begin by solving System III (the Atmosphere, a) introduced in Lecture 4, and for completeness, shown again in Fig. 5.2. System III is a simple, closed system, with an adiabatic boundary that has one movable part (the piston). Accordingly,

5.3 Solution 1: System III-Atmosphere (a)

33

δQa ¼ 0

ð5:1Þ

A First Law of Thermodynamics analysis of the closed system yields: dEa ¼ δQa þ δWa

ð5:2Þ

Combining Eqs. (5.1) and (5.2) yields: dEa ¼ δWa

ð5:3Þ

¼ Pa dVa ¼ Pa dVg |fflfflfflffl{zfflfflfflffl}

ð5:4Þ

We also know that: δWa |ffl{zffl} Work done on

Work done by

the atmosphere

the atmosphere

where we have used the fact that dVa ¼ dVg . To compute dEa in Eq. (5.2), we consider the composite system (x + a) where system x consists of the gas + piston + cylinder (see Fig. 5.2). Note that system (x + a) is isolated by the imaginary boundary at 1 (see the outer dashed boundary in Fig. 5.2). The imaginary boundary at 1 is impermeable, adiabatic, and rigid by choice. In that case, a First Law of Thermodynamics analysis of system (x + a) yields: dExþa ¼ dEx þ dEa dExþa ¼

δQxþa |fflffl{zfflffl} Adiabatic boundary!0

þ

δWxþa |fflfflffl{zfflfflffl}

¼0

Rigid boundary!0

∴ dEa ¼ dEx

ð5:5Þ

Because system x ¼ gas (g) + piston (p) + cylinder walls (w), it follows that its total energy is given by: Ex ¼ Eg þ Ep þ Ew

ð5:6Þ

Eg ¼ Ug ðThe gas is a simple systemÞ

ð5:7Þ

We also know that:

1 Mp v2p Ep ¼ Up þ Mp gz þ 2|fflfflffl{zfflfflffl} |fflffl{zfflffl}

ð5:8Þ

Piston Piston Potential Energy Kinetic Energy

where Up is the internal energy of the piston. Because Cvp ¼ 0 by choice, Up ¼ 0.

34

5 The First Law of Thermodynamics for Closed Systems. . .

Ew ¼ Uw ðThe walls are a simple systemÞ

ð5:9Þ

Because Cvw ¼ 0 by choice, it follows that: Ew ¼ 0

ð5:10Þ

Using Eqs. (5.7), (5.8), and (5.10) in Eq. (5.6) yields: 1 Ex ¼ Ug þ Mp gz þ Mp v2p 2

ð5:11Þ

Taking the differential of Eq. (5.11), including using it in Eq. (5.5), yields:     1 dEa ¼ dEx ¼ dUg  d Mp gz  d Mp v2p 2

ð5:12Þ

Because the gas is ideal and Ng is constant, it follows that: dUg ¼ Ng Cvg dT

ð5:13aÞ

Combining Eqs. (5.12) and (5.13a) yields:     1 dEa ¼ Ng Cvg dT  d Mp gz  d Mp vp 2 2

ð5:13bÞ

Using Eq. (5.13b) in Eq. (5.3), with δWa given in Eq. (5.4), yields:     1 Ng Cvg dT  d Mp gz  d Mp V2p ¼ Pa dVg 2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflffl{zfflffl} δW

ð5:14Þ

    1 Ng Cvg dT ¼ Pa dVg þ d Mp gz þ d Mp v2p 2

ð5:15Þ

dEa

a

or

In Eq. (5.15), Ng, Cvg, Pa, Mp, and g are all known. Therefore, Eq. (5.15) can be integrated directly from [Ti , Ngi , Vgi , zi , vpi ¼ 0 (initially, the piston is at rest) to Tf , Ngf ¼ Ngi , Vgf ¼ 2Vgi (chosen), zf, vpf ¼ 0 (finally, the piston is at rest)].

5.4 Solution 2: System I-Gas (g)

35

We will integrate Eq. (5.15) later after we again derive Eq. (5.15) using System I (The Gas) introduced in Lecture 4. We encourage the readers to undertake the solution of System II (System x – Gas + Piston + Cylinder Walls), which is similar to that of System III discussed above. System I (Gas, g), although the most natural to choose, will lead to the most challenging solution. Of course, we will again obtain Eq. (5.15). For completeness, Fig. 5.3 depicting System I is shown again below. As discussed in Lecture 4, System I (Gas, g) is a simple, closed system, with one moving boundary (the piston), and heat is generated internally due to the friction of the piston with the cylinder walls (as a result, the dashed boundary surrounding the gas in Fig. 5.3 is diathermal).

Fig. 5.3

5.4

Solution 2: System I-Gas (g)

If we choose System I, then, δQg 6¼ 0, and the system is simple. Unlike System III and System II, first introduced in Lecture 4, in which there was no friction, the challenge with System I involves dealing with the friction, about which we have no information. The first step is to apply the First Law of Thermodynamics to System I (Gas, g). Specifically, dEg ¼ dUg ¼ |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} Simple system

δQg |{z} Heat absorbed by the gas due to the friction

þ

δWg |ffl{zffl}

ð5:16Þ

Work done on the gas by the environment

We know that the source of the heat, δQg, is the friction generated at the cylinder walls by the work done on the cylinder walls by the moving piston against the friction. This suggests that to calculate δQg, we should focus on the cylinder walls as a system and carry out a First Law of Thermodynamics analysis on the cylinder walls. Note that the boundary of the cylinder walls is impermeable, rigid, and diathermal. Nevertheless, work is done on the cylinder walls by the moving piston against the friction. In other words, although there is no PdV-type work, there is friction work. It then follows that:

5 The First Law of Thermodynamics for Closed Systems. . .

36

dEw ¼ dUw ¼ |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} Simple system

δQw |{z}

þ

δWw |ffl{zffl}

ð5:17Þ

Work done on the cylinder walls to overcome friction

Heat absorbed by the cylinder walls

As discussed in Lecture 4, because the heat capacity of the cylinder walls, Cvw, is zero according to the Problem Statement, the cylinder walls have no capacity to change their internal energy. In other words: dUw ¼ 0

ð5:18Þ

Using Eq. (5.18) in Eq. (5.17) yields: δQw |fflffl{zfflffl}

¼

Heat released by the cylinder walls

δWw |ffl{zffl}

ð5:19Þ

Work done on the cylinder walls to overcome friction

Next, we relate δQw to δWg. Intuitively, we expect that the heat released by the cylinder walls should be completely absorbed by the gas. This follows because according to the Problem Statement, neither the cylinder walls (w) nor the piston (p) can absorb the heat released by the cylinder walls (both Cvw and Cvp are zero). In addition, the heat released by the cylinder walls, δQw, cannot be released to the atmosphere, because the gas is adiabatically enclosed by the cylinder walls and the piston. To prove that, we can choose as our system the (Gas + Cylinder Walls). Because this system is adiabatically enclosed, it follows that: δQg þ δQw ¼ 0

ð5:20Þ

or that δQg |{z}

¼

δQw |fflffl{zfflffl}

ð5:21Þ

Heat released by the cylinder walls

Heat absorbed by the gas

Using Eq. (5.21) in Eq. (5.19) yields: δQg |{z} Heat absorbed by the gas

¼

δWw |ffl{zffl} Work done on the cylinder walls

ð5:22Þ

5.4 Solution 2: System I-Gas (g)

37

Equation (5.22) is a key result. It shows that the heat generated by the frictional work at the cylinder walls, δWw, is transmitted 100% back to the gas which absorbs it (δQg). Otherwise, the problem would be much more challenging, and we would require additional information and assumptions in order to solve it. Next, we go back to the First Law of Thermodynamics and deal with the work term δWg. In Lecture 4, we saw that: W ðSystem on EnvironmentÞ ¼ W ðEnvironment on SystemÞ

ð5:23Þ

Using Eq. (5.23) where System ¼ Gas, and Environment ¼ (Cylinder Walls + Atmosphere + Piston) yields: δWg |fflffl{zfflffl}

¼

Work done by the gas on the three elements of its environment

δWw |ffl{zffl} Work done on the cylinder walls to overcome friction

þ

δWa |ffl{zffl}

þ

Work done on the atmosphere to push it back

δWp |ffl{zffl}

ð5:24Þ

Work done on the piston to increase its potential and kinetic energies

where δWa ¼ Pa dVa ¼ Pa dVg

ð5:25Þ

    1 δWp ¼ d Mp gz þ d Mp v2p 2

ð5:26Þ

and

Combining Eqs. (5.24), (5.25), and (5.26) yields:     1 δWg ¼ Pa dVg  δWw  d Mp gz  d Mp v2p 2

ð5:27Þ

Next, we can return to the First Law of Thermodynamics for the Gas ((Eq. 5.16)), along with Eq. (5.22), which yields: dUg ¼ δQg þ δWg ¼ δWw þ δWg

ð5:28Þ

Using Eq. (5.27) for δWg in Eq. (5.28) yields:     1 dUg ¼ δWw  Pa dVg  δWw  d Mp gz  d Mp v2p 2 or

ð5:29Þ

38

5 The First Law of Thermodynamics for Closed Systems. . .

    1 dUg ¼ Pa dVg  d Mp gz  d Mp v2p 2

ð5:30Þ

Equation (5.30) is a very interesting result which shows that the gas expands in a manner where the work done on the cylinder walls to overcome friction (δWw) and the heat absorbed by the gas (δQg ¼ δWw) cancel each other out! Indeed, the internal energy of the gas (see Eq. (5.30)) decreases because, as it expands, the gas works against the atmosphere as well as to increase the potential and the kinetic energies of the piston. The frictional work done on the cylinder walls is recovered 100% in the form of heat reabsorbed by the gas! Note that if the cylinder walls where such that dUw 6¼ 0, contrary to what was assumed in Eq. (5.18) based on the Problem Statement, and if the cylinder walls as well as the piston could absorb some heat (an effect neglected here based on the Problem Statement), then, only part of the frictional work lost on the cylinder walls would be recovered! In that case, the internal energy of the gas would decrease to a greater extent, and we can anticipate a lower final temperature of the gas! Because the gas is ideal, it follows that (recall that Ng ¼ constant): dUg ¼ Ng Cvg dT

ð5:31Þ

Using Eq. (5.31) in Eq. (5.30) yields the desired result:     1 Ng Cvg dT ¼ Pa dVg þ d Mp gz þ d Mp v2p 2

ð5:32Þ

Note that, as expected, Eq. (5.32) is identical to Eq. (5.15)! of Vg. Integrating Finally, we can  solve Eq. (5.32)  to find Tf as a function  Eq. (5.32) from Ti , Vgi , zi , and vpi to Tf , Vgf , zf , and vpf , we obtain:   Ng Cvg ðTf  Ti Þ ¼ Pa Vgf  Vgi þ Mp gðzf  zi Þ h   2 i 1 2 þ Mp • vpf  vpi 2 |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} Note that vpi and vpf

ð5:33Þ

are both zero, because the piston starts at rest and stops at rest!

Setting the last term in Eq. (5.33) to zero, including rearranging, we obtain:

5.4 Solution 2: System I-Gas (g)

Tf ¼ Ti 

39

 9 8

< þ1, Up I ¼ 0, Sideways > : ‐1, Down

9 > = > ;

ð5:38Þ

Using Eq. (5.38) in Eq. (5.34) for Tf, including accounting for the direction of the moving piston, Eq. (5.34) can be generalized as follows:

5 The First Law of Thermodynamics for Closed Systems. . .

40

Tf ¼ Ti 

 9 8 Vi! If the original state of the system (N, T, Vi) is restored, so that the system does not change its state, as required by Statement (1) of the Second Law of Thermodynamics, then energy from the environment in the form of work is needed to compress the gas back from (N, T, Vf) to (N, T, Vi). At the same time, energy in the form of heat will be transferred from the system to the environment to maintain constant temperature. Therefore, the reverse gas compression process requires exactly the same amount of work produced by the gas expansion, and therefore, no net work is produced! The solution to sample Problem 7.1 clearly shows that Statement (1) of the Second Law of Thermodynamics can be expressed in the following alternative way.

7.5

Statement (1a) of the Second Law of Thermodynamics

“It is impossible by a cyclic process to convert the heat absorbed by a system completely (100%) into work.” The word cyclic in Statement (1a) of the Second Law of Thermodynamics requires that the system be restored periodically to its

58

7 The Second Law of Thermodynamics: Fundamental Concepts and Sample Problem

original state. In the gas example discussed in Sample Problem 7.1, the expansion and compression of the gas back to its original state constitute a complete cycle. If the process is repeated, it becomes a cyclic process. The restriction to a “cyclic process” in Statement (1a) of the Second Law of Thermodynamics provides the same limitations as the restriction “only effect” in Statement (1) of the Second Law of Thermodynamics. The Second Law of Thermodynamics does not prohibit the production of work from heat, but it does place a limit on the fraction of heat that may be converted into work in a cyclic process [Statement (1a) of the Second Law of Thermodynamics]. The partial conversion of heat into work is the basis for most commercial generation of power. Next, we introduce a heat engine and derive a quantitative expression for the conversion efficiency.

7.6

Heat Engine

A heat engine is a device (or a machine) which produces work from heat in a cyclic process. It is useful to illustrate the operation of a heat engine with a steam power cycle (Rankine cycle, see Fig. 7.2).

Fig. 7.2

7.7 Efficiency of a Heat Engine

59

The Rankine cycle consists of the following four steps: 1. Liquid water at ambient temperature is pumped, WP, into a boiler 2. Heat, QH, from a fuel is transferred in the boiler to the water, converting it to steam at high temperature and pressure 3. Useful work is obtained by expanding the steam to a low pressure in a turbine, WT 4. Exhaust steam from the turbine is condensed by the transfer of heat, QC, to cooling water, thus completing the cycle

7.7

Efficiency of a Heat Engine

Essential to the operation of a heat engine are (see Fig. 7.3): – Absorbing heat from a hot reservoir – Doing work – Rejecting heat to a cold reservoir

Fig. 7.3

In Fig. 7.3, |QH|, |QC|, and |WE| denote the magnitudes of the “vectors,” while the arrows for Q and ⟹ for W denote directions. In Lecture 8, we will use the vectorial representation of heat and work to model an ideal engine, known as the Carnot engine. The efficiency of a heat engine is defined as follows: ηE ¼

Work done by the engine Heat absorbed by the engine from the hot reservoir

ð7:11Þ

Using our work and heat conventions, Eq. (7.11) can be expressed as follows: ηE ¼

WE QH

ð7:12Þ

60

7 The Second Law of Thermodynamics: Fundamental Concepts and Sample Problem

where WE is the work done on the engine and QH is the heat absorbed by the engine from the hot reservoir. In all allowable processes, either WE or QH will be < 0, but not both, so that ηE > 0 in Eq. (7.12). To make the equations independent of the sign conventions for W and Q, we can use |WE| and |QH|, with the directions indicated by the arrows for Q and ⟹ for W. In that case: ηE ¼

jWE j jQH j

ð7:13Þ

As stressed above, ηE < 1! If this is the case, what is it that limits the engine efficiency? From a practical viewpoint, factors such as friction and other resistances that dissipate energy will certainly decrease the value of ηE. However, even if one could construct an ideal engine, “free of friction,” which could operate fully reversibly, one would still find that there is a Second Law of Thermodynamics limit of the efficiency of a heat engine, so that: ηideal 1 ) QH > 0, heat is absorbed by the Carnot engine from the hot reservoir. 2. Calculation of |QC| : Isothermal (T = TC) Compression from c→d Once again, along this isotherm: δQ ¼

RTC dV V

ð8:14Þ

Integrating Eq. (8.14) from Vc to Vd yields: V ðd

V ðd

δQ ¼ QC ¼ Vc

  RTC V dV ¼ RTC ln c V Vd

ð8:15Þ

Vc

Note that because Vc/Vd > 1 ) QC < 0, heat is rejected by the Carnot engine to the cold reservoir. Note that QH being positive in Eq. (8.13) and QC being negative in Eq. (8.15) are consistent with the directions of the arrows ( ) in the Carnot cycle (see Fig. 8.3).

8.5 Sample Problem 8.1

69

Using Eqs. (8.13) and (8.15), it follows that: T lnðVb =Va Þ jQH j ¼ H jQC j TClnðVc =Vd Þ

ð8:16Þ

We can show (left as an exercise) that: lnðVb =Va Þ ¼ lnðVc =Vd Þ

ð8:17Þ

by analyzing the two adiabatic portions (δQ ¼ 0), b!c and d!a, using Eq. (8.9). Combining Eqs. (8.16) and (8.17) yields: jQ H j jT H j ¼ jQC j jTC j

ð8:18Þ

Equation (8.18) can be rewritten without the absolute magnitude signs as follows: Q QH þ C ¼0 TH TC

ð8:19Þ

ψðTH Þ jQH j ¼ jQ C j ψðT C Þ

ð8:20Þ

However, because

a comparison of Eqs. (8.20) and (8.18) clearly shows that: ψðTÞ ¼ T

ð8:21Þ

ψðTH Þ  ψðTC Þ ψð T H Þ

ð8:22Þ

Because ηC ¼

using Eq. (8.21) for TH and TC in Eq. (8.22) yields: ηC ¼

TH  TC TH

ð8:23Þ

Equation (8.23) is a central result. It clearly shows that only if TC ¼ 0 K (273  C) or if TH ! infinity can ηE ¼ 1! Cold reservoirs (e.g., the atmosphere, lakes, rivers, oceans) have TC values of ~300 K. Hot reservoirs (e.g., fuel combustion, nuclear reactors) have TH values of ~600 K. As a result, ηC  0.5. Real irreversible heat engines have ηE values which rarely exceed 0.35!

70

8.6

8 Heat Engine, Carnot Efficiency, and Sample Problem

Theorem of Clausius

“Given any reversible process in which the temperature changes in any prescribed manner, it is always possible to find a reversible zigzag process consisting of adiabatic-isothermal-adiabatic steps such that the heat interaction in the isothermal step is equal to the heat interaction in the original process” (see Fig. 8.4):

Fig. 8.4

Lecture 9

Entropy and Reversibility

9.1

Introduction

The material presented in this lecture is adapted from Chapter 4 in T&M, as well as from Chapter 1 in Denbigh. First, we will introduce a new thermodynamic function, the entropy S, which strictly is only defined for a reversible process. Second, we will show that the entropy S is a function of state, first for a reversible Carnot cycle and then for a reversible arbitrary cycle. Fourth, after showing that the entropy is a function of state, we will show that if two states can be bridged both along reversible and irreversible paths, then, the entropy change along the irreversible path, although not defined, can nevertheless be equated to that along the reversible path. This will allow us to calculate entropy changes for irreversible processes. Interestingly, we will show that some states cannot be bridged both along reversible and irreversible paths. Finally, we will show that the entropy change for a closed and adiabatic system is always positive for an irreversible (natural) process and is zero for a reversible one. This statement is often referred to as the Second Law of Thermodynamics.

9.2

Entropy

The differential of the new thermodynamic function, the Entropy, denoted as S, is defined as follows:
 δQ dS ¼ T rev

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_9

ð9:1Þ

71

72

9 Entropy and Reversibility

where δQ is the differential amount of heat absorbed by the system at temperature, T, as it undergoes a reversible change of state. First, we would like to show that S is a function of state. To this end, we need to show that: þ dS ¼ 0

ð9:2Þ

where the circle symbol in the integral indicates that the integration is carried out around the reversible cycle, such that the system begins and ends at the same state. Figure 9.1 depicts going reversibly from state i to state f along Path α and then returning reversibly from state f to state i along Path β.

Fig. 9.1

Indeed, if we can prove that: þ

ðf dS ¼

ði ðdSÞPath α þ

i

ðdSÞPath β ¼ 0

ð9:3Þ

f

it would then follow that: ðf

ði

ðf

ðdSÞPath α ¼  ðdSÞPath i

f

β

¼ þ ðdSÞPath

β

ð9:4Þ

i

or that: ðΔSi!f ÞPath

α

¼ ðΔSi!f ÞPath

β

ð9:5Þ

Equation (9.5) clearly shows that the change in the entropy of the system is the same along any path connecting states i and f. The change is determined entirely by i and f, and therefore, S is a function of state. With the above in mind, we would like to prove that for a reversible cyclic process:

9.2 Entropy

73

þ dS ¼ 0

ð9:6Þ

To this end, we will first prove this result for a special reversible cyclic process, the Carnot cycle, introduced in Lecture 8, which consists of two isotherms (one at a constant hot temperature, TH, and the other at a constant cold temperature, TC) and two adiabats. We will use a (P-V) representation of the Carnot cycle (see Fig. 9.2):

Fig. 9.2

Using the (P-V) phase diagram in Fig. 9.2, it follows that: þ dS ¼ ΔSa!b!c!d ¼ ΔSa!b þ ΔSb!c þ ΔSc!d þ ΔSd!a

ð9:7Þ

where again the circle symbol in the integral indicates that the integration is carried out around the reversible Carnot cycle. Next, we evaluate separately each of the four entropy contributions in Eq. (9.7). Specifically,

ΔSa!b

 ðb
ðb δQH 1 Q ¼ ¼ ðδQH Þrev ¼ H ðIsothermÞ TH TH rev TH

ð9:8Þ

 ðc
δQb!c ¼ ¼ 0 ðAdiabatÞ T rev

ð9:9Þ

a

ΔSb!c

a

b

74

9 Entropy and Reversibility

ðd
ΔSc!d ¼ c

δQC TC

 ¼ rev

1 TC

ðd ðδQc Þrev ¼

QC ðIsothermÞ TC

ð9:10Þ

c

ða ΔSd!a ¼ d


 δQd!a ¼ 0 ðAdiabatÞ T rev

ð9:11Þ

Using Eqs. (9.8), (9.9), (9.10), and (9.11) in Eq. (9.7) it follows that: þ dS ¼

Q QH þ C TH TC

ð9:12Þ

In Lecture 8, we showed that the sum in Eq. (9.12) is zero! Therefore, þ dS ¼ 0 ðFor a reversible Carnot cycleÞ

ð9:13Þ

Next, we need to prove that the same result applies to any type of reversible cycle, not necessarily consisting of two isotherms and two adiabats. The key to the proof is to replace the arbitrary reversible cycle by a series of Carnot cycles (this is possible due to the Theorem of Clausius that we proved in Lecture 8) and then to use the results just obtained for each Carnot cycle. One can then show that: þ dSs ¼ 0 ðFor a reversible arbitrary cycleÞ

ð9:14Þ

Note that in Eq. (9.14), the subscript s in Ss has been added to stress the fact that dSs is the entropy change of the system which absorbed heat, δQ, in a reversible process at temperature, T, and it does not include the entropy change of the heat reservoir that supplied the heat. Because the entropy, S, is a function of state, if two states, A and B, can be bridged by both reversible and irreversible paths (see Fig. 9.3):

Fig. 9.3

9.2 Entropy

75

it follows that: ðΔSA!B Þrev ¼ ðΔSA!B Þirrev

ð9:15Þ

This implies that although we do not know how to directly compute (ΔS)irrev, because dS was only defined for a reversible process, we can nevertheless compute this quantity by equating it with (ΔS)rev, because S is a function of state! We will use this important result to solve various problems involving irreversible processes, where one is asked to compute (ΔS)irrev. It is important to recognize that for some processes, it will be impossible to bridge the same two states along irreversible and reversible paths! For example, if states A and B can be bridged by an adiabatic, irreversible path, then, they cannot be bridged by an adiabatic, reversible path (see Fig. 9.4):

Fig. 9.4

To prove this, one assumes that B0 ¼ B and then shows that this leads to an inconsistency (see Fig. 9.5):

Fig. 9.5

Because the adiabatic path is reversible, one can come back from B to A! Clearly, the cyclic process A!B!A would be irreversible. However, the system returned to its original state (A), and no heat interaction occurred (the process A!B!A is

76

9 Entropy and Reversibility

adiabatic). This implies that the environment did not change its state either. Therefore, the system and its environment were restored to their original states. This would correspond to the process A!B!A being reversible. However, this is inconsistent with the original statement that the process A!B!A is irreversible. Hence, B0 6¼ B! If a process is such that B0 ¼ B, then: ðΔSA!B Þirrev ¼ ðΔSA!B Þrev

ð9:16Þ

However, we will soon see that the heat and work interactions along the reversible and the irreversible paths are different. If one has a collection of subsystems 1, 2, . . ., n, the total entropy of the composite system is the sum of the entropies of each subsystem, because the entropy, S, is extensive (see Fig. 9.6).

Fig. 9.6

“The entropy change of a closed and adiabatic system is always positive for an irreversible (natural) process and zero for a reversible one.” This statement is often referred to as the Second Law of Thermodynamics. Let us examine this statement, first for a reversible process (see Fig. 9.7) and then for an irreversible one.

Fig. 9.7

9.4 Irreversible Process

9.3

77

Reversible Process

ðB
¼ A


 δQ • dS ¼ T rev

ð9:17Þ

• ðΔSA!B Þrev ¼ ðSB  SA Þrev

ð9:18Þ

 δQ ¼ 0 ðBecause δQ ¼ 0 along the AB adiabatic pathÞ T rev

ð9:19Þ

We have therefore shown that the entropy change for a closed and adiabatic system is zero for a reversible process.

9.4

Irreversible Process

Because the process A!B is irreversible (see Fig. 9.8), the defining equation, dS ¼ (δQ/T)rev, cannot be used to calculate (ΔSA!B)irrev. Therefore, we will assume that after the original change of state, A!B, has taken place irreversibly and adiabatically, the reverse change of state, B!A, is carried out reversibly. Note that, in general, the return path, B!A, cannot be carried out adiabatically. Indeed, because the overall cyclic process from A to B to A is irreversible, it cannot be completed without leaving some change in the environment, because the system itself returns to its original state at A. If the process could be completed adiabatically, then, the entire process, A!B!A, would be reversible, which is not the case.

Fig. 9.8

We can design the reversible return path, B!A, in such a way that any heat interaction occurring along the path, QB!A, occurs isothermally. Recall that this is possible according to the Theorem of Clausius that we proved in Lecture 8. Specifically, we choose a reversible return path consisting of the following three steps:

78

9 Entropy and Reversibility

(i) The adiabatic step, B!C (ii) The isothermal step, C!D (iii) The adiabatic step, D!A The chosen return path B!C!D!A is such that: QB!C!D!A ¼ QB!A ¼ QC!D

ð9:20Þ

because QB!C and QD!A are both zero along the adiabats BC and DA, respectively. Figure 9.9 shows the (P-V) phase diagram corresponding to the entire process.

Fig. 9.9

We seek to calculate (SBSA)irrev. Because S is a function of state, it follows that: þ 0¼ A!A

dS ¼ ðSB  SA Þirrev þ ðSA  SB Þrev

ð9:21Þ

or ðSB  SA Þirrev ¼ ðSA  SB Þrev

ð9:22Þ

We know that along the reversible return path B!A ¼ B!C!D!A, one has: dSB!A ¼ dSB!C þ dSC!D þ dSD!A

ð9:23Þ

9.4 Irreversible Process

79

Because along the two adiabatic paths, δQB!C ¼ 0 and δQD!A ¼ 0 and, therefore, dSB!C ¼ 0 and dSD!A ¼ 0, it follows that:
dSB!A ¼ dSC!D ¼

δQC!D T

 ¼ rev


 δQB!A T rev

ð9:24Þ

In Eq. (9.24), T is the same along the isothermal step (C!D), and δQC!D ¼ δQB!A according to the Theorem of Clausius. Integrating Eq. (9.24) from B to A, we obtain: ðA dSB!A B

 ðA
δQB!A ¼ SA  SB ¼ T rev

ð9:25Þ

B

Because T is constant, it follows that: 1 SA  SB ¼ T

ðA ðδQB!A Þrev ¼

ðQB!A Þrev T

ð9:26Þ

B

or ðSA  SB Þrev ¼

ðQB!A Þrev T

ð9:27Þ

Next, we need to show that (QB!A)rev < 0, that is, that the system releases heat along the reversible return path from B to A. To do this, we consider the closed system undergoing the cyclic process A!B!A and apply the First Law of Thermodynamics. Specifically, ΔUA!B!A ¼ 0 ¼ QA!B þ QB!A þ WA!B þ WB!A

ð9:28Þ

Because QA!B is zero along the adiabat AB, rearranging Eq. (9.28), we obtain: QB!A ¼ ðWA!B þ WB!A Þ

ð9:29Þ

Note that – (WA!B + WB!A) is the total work done by the system during the cyclic process A!B!A. Next, we will show that QB!A must be negative. Indeed, (i) If QB!A ¼ 0, then, the environment would not change its state, and because the system does not change its state either, the cyclic process A!B!A would be reversible, contrary to what we know (the process is irreversible)

80

9 Entropy and Reversibility

(ii) If QB!A > 0, then, this would correspond to the complete conversion of heat, QB!A, absorbed by the system into work in a cyclic process, which is not possible according to Statement 1(a) of the Second Law of Thermodynamics In view of (i) and (ii) above, it follows that: QB!A < 0

ð9:30Þ

Equation (9.30) reveals that the system releases heat to the environment during its reversible return path from B to A. Because we showed that: ðSB  SA Þirrev ¼ ðSA  SB Þrev ¼ 

QB!A T

ð9:31Þ

and that QB!A < 0, it follows that: ðSB  SA Þirrev > 0

ð9:32Þ

ðΔSA!B Þirrev ¼ ðSB  SA Þirrev > 0

ð9:33Þ

or that:

For the system to return to A, the entropy created along the irreversible path A!B must be decreased by releasing heat to the environment along the reversible return path B!A!

Lecture 10

The Second Law of Thermodynamics, Maximum Work, and Sample Problems

10.1

Introduction

The material presented in this lecture is adapted from Chapter 4 in T&M. First, we will present a more general statement of the Second Law of Thermodynamics than the one presented in Lecture 9. To this end, we will introduce the concept of entropy created in a process (zero for a reversible process and greater than zero for an irreversible one). Second, we will show that, as discussed in Lecture 9, although the entropy changes along reversible and irreversible paths are equal (because entropy is a function of state), more heat is produced along the irreversible path and more work is produced along the reversible path. Third, we will solve Sample Problem 10.1 to illustrate the calculation of maximum work for a particular reversible process. Fourth, we will solve Sample Problem 10.2 to prove that heat always flows from a hot body to a cold body, including showing that as long as heat transfer takes place (an irreversible process), entropy is created. Fifth, we will derive a criterion of equilibrium based on the entropy, which states that the entropy attains its maximum value at thermodynamic equilibrium. Finally, we will solve Sample Problem 10.3 to calculate the entropy change of a closed, simple system undergoing a reversible process, including when the process is isobaric.

10.2

A More General Statement of the Second Law of Thermodynamics

Continuing with the Second Law of Thermodynamics discussed in Lecture 9, we will present a more general statement of this law. Indeed, because the entire “universe” ¼ (system + environment) is an isolated (adiabatic + closed + rigid) system, the Second Law of Thermodynamics can also be stated as follows:

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_10

81

82

10 The Second Law of Thermodynamics, Maximum Work, and Sample Problems

“The entropy of the universe must increase in any irreversible (natural) process, and is conserved (remains constant) in any reversible process”. In other words: dSuniverse > 0, for an irreversible process dSuniverse ¼ 0, for a reversible process It is useful to convert the inequality in the Second Law of Thermodynamics for an irreversible process into an equality. To this end, we introduce the quantity, σ, the total entropy created in the irreversible process. One can then rewrite the Second Law of Thermodynamics as follows: dSS ¼ dσ

ð10:1Þ

where the subscript S in S denotes system, and:

10.3

dσ ¼ 0, For a reversible process

ð10:2Þ

dσ > 0, For an irreversible process

ð10:3Þ

Heat Interactions Along Reversible and Irreversible Paths (Closed System)

In Lecture 9, we showed that (ΔSA!B)irrev ¼ (ΔSA!B)rev. Next, we will show that the heat interaction is different for each process. Consider system (S) undergoing a purely heat interaction with a heat reservoir (R). A heat reservoir is an ideal body which acts solely as a donor or an acceptor of heat while maintaining constant volume (PdV- type work ¼ 0). The heat reservoir can absorb or reject an infinite amount of heat while maintaining constant temperature. This absorption or rejection of heat results in a unique change of state of the reservoir. Indeed, a First Law of Thermodynamics analysis of the heat reservoir as a closed, simple, rigid, and diathermal system (see Fig. 10.1) yields:

Fig. 10.1

10.3

Heat Interactions Along Reversible and Irreversible Paths (Closed System)

83

In other words, the heat, QR, absorbed (QR > 0) or released (QR < 0) by the heat reservoir is equal to the change of the state function, ΔUR. Therefore, the change of state of the heat reservoir is uniquely determined by the amount of heat transferred, δQR, irrespective of whether δQR is transferred reversibly or irreversibly. As a result, in defining dSR, there is no need to indicate (δQR)rev or (δQR)irrev, because they lead to the same change of state of the heat reservoir. That is, dSR ¼ δQR/TR. If the heat reservoir (R) releases heat, δQR, to system (S) at temperature, TR, then, the following entropy changes occur (see Fig. 10.2):

Fig. 10.2

In Fig. 10.2, the composite (t) of [heat reservoir (R) + system (S)] is a closed + adiabatic system surrounded by the dashed boundary, for which the Second Law of Thermodynamics applies. Specifically, dSt ¼ dSS þ dSR ¼ dσ

ð10:4Þ

δQ δQR ¼  S TR TR

ð10:5Þ

For the heat reservoir: dSR ¼

In Eq. (10.5), δQR is the heat absorbed by the heat reservoir (R), and δQS is the heat absorbed by the system (S), where δQS > 0 and δQR > 0. Combining Eqs. (10.4) and (10.5) yields: δQS þ dσ TR

ð10:6Þ


 δQS dSS ¼ TR rev

ð10:7Þ

dSS ¼ However,

84

10 The Second Law of Thermodynamics, Maximum Work, and Sample Problems

Accordingly, Eqs. (10.6) and (10.7) show that:
 δQS δQS ¼ þ dσ TR rev TR

ð10:8Þ

Equation (10.8) shows that:

 δQS δQS > , for dσ > 0 TR rev TR irrev

ð10:9Þ

ðδQS Þrev > ðδQS Þirrev

ð10:10Þ

or that:

The inequality in Eq. (10.10) shows that more heat is absorbed by the system in a reversible process. Alternatively, more heat is released by the system in an irreversible process. In other words, ðδQS Þirrev > ðδQS Þrev

10.4

ð10:11Þ

Work Interactions Along Reversible and Irreversible Paths (Closed System)

Fig. 10.3

Given states A and B that can be bridged both along reversible and irreversible paths (see Fig. 10.3), the First Law of Thermodynamics for a closed system states that:

10.5

Sample Problem 10.1

85

ΔEA!B ¼ QA!B þ WA!B

ð10:12Þ

Because E is a function of state, it follows that: ðΔEA!B Þrev ¼ ðΔEA!B Þirrev

ð10:13Þ

We have just shown that: ðQA!B Þrev > ðQA!B Þirrev

ð10:14Þ

Equations (10.12), (10.13), and (10.14) show that: ∴ðWA!B Þrev > ðWA!B Þirrev ðWork done by the systemÞ

ð10:15Þ

The inequalities in Eqs. (10.15) and (10.11) indicate that in any reversible change of state, A!B, of a closed system, a greater amount of work and a smaller amount of heat are produced relative to the corresponding irreversible change of state. This, of course, is consistent with the expected dissipative nature of an irreversible process. In a reversible process, one can obtain maximum work.

10.5

Sample Problem 10.1

In a given process, a closed system absorbs heat, |δQ|, from a heat reservoir at the constant temperature, TR, of the hot reservoir and performs work (see Fig. 10.4). Calculate the maximum work.

Fig. 10.4

86

10 The Second Law of Thermodynamics, Maximum Work, and Sample Problems

10.5.1 Solution The composite system (t) of system (S) + heat reservoir (R) is closed + adiabatic, and therefore, the Second Law of Thermodynamics applies. Specifically, dSt ¼ dSS þ dSR ¼ dσ

ð10:16Þ

jδQj δQR ¼  TR TR

ð10:17Þ

In Eq. (10.16), dSR ¼

where – |δQ| is the differential heat released by the heat reservoir (see Fig. 10.4). Combining Eqs. (10.16) and (10.17) yields: jδQj ¼ TR dSS  TR dσ

ð10:18Þ

where σ is the entropy created in the process, with dσ > 0 for an irreversible process and dσ ¼ 0 for a reversible process. Next, we focus on system (S) and carry out a First Law of Thermodynamics analysis on it. The system is closed and simple and has a diathermal, movable boundary. Accordingly: dUS ¼ jδQj þ δW

ð10:19Þ

δW ¼ jδQj  dUS

ð10:20Þ

or

where –δW is the work done by the system. Using Eq. (10.18) in Eq. (10.20) yields: δW  d ½TR SS  US   TR dσ

ð10:21Þ

where dσ is either zero for a reversible process or greater than zero for an irreversible process. Accordingly, we can express Eq. (10.21) as follows: δW  d½TR SS  US  |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}

ð10:22Þ

State Function

Integrating Eq. (10.22) from state A to state B, we obtain:    B  A WA!B  TR SBS  SA S  US  US

ð10:23Þ

10.6

Sample Problem 10.2

87

where the two terms on the right-hand side of Eq. (10.23) are independent of the path connecting states A and B. Equation (10.23) shows that (WA!B) cannot be larger than the quantity on the right-hand side, which we will denote as Wmax, where:    B  A Wmax ¼ TR SBS  SA S  US  US

ð10:24Þ

is the maximum work done by the system for the process considered and corresponds to the reversible case (dσ ¼ 0). If the process considered were irreversible (dσ > 0), then: ðWA!B Þirrev < Wmax

ð10:25Þ

Note that if a different process is considered, a different expression for Wmax will result.

10.6

Sample Problem 10.2

Calculate the differential total entropy change for a process involving heat transfer from a heat reservoir at T2 to another heat reservoir at T1. Figure 10.5 depicts the composite system (t) under consideration.

10.6.1 Solution

Fig. 10.5

Applying the Second Law of Thermodynamics to the composite system (t) in Fig. 10.5 yields:

88

10 The Second Law of Thermodynamics, Maximum Work, and Sample Problems

dSt ¼ dS1 þ dS2 ¼ dσ dS1 ¼

jδQj jδQj , dS2 ¼  T1 T2

ð10:26Þ ð10:27Þ

Combining Eqs. (10.26) and (10.27) yields: ∴ dSt ¼ jδQj ðT2  T1 Þ=T1 T2 ¼ dσ

ð10:28Þ

Because for any allowable process dσ  0, Eq. (10.28) indicates that T2  T1. Therefore, heat can only be transferred from the “hotter” to the “colder” body, as observed in nature. Note that if T2 < T1, it would follow that dσ < 0, which would violate the Second Law of Thermodynamics! As long as T2 > T1, entropy is created (dσ > 0) as heat is transferred from “hot” body 2 to “cold” body 1. Thermal equilibrium is eventually attained when T2 ¼ T1 and dSt ¼ 0. Note that as long as T2 > T1, the heat transfer process is irreversible. When T2 – T1 ¼ dT, heat can be transferred reversibly, because: dSt ¼ dσ ¼

jδQjdT  0 ðDifferential of the second orderÞ T22

ð10:29Þ

This is, in fact, what is assumed for the reversible heat transfer process in a Carnot engine. If a system is open and not adiabatically enclosed, the Second Law of Thermodynamics does not have to apply.

10.7

Criterion of Equilibrium Based on the Entropy

As we already saw, the only changes of state that can occur within a closed + adiabatic boundary are those for which the entropy either increases (irreversible process) or remains constant (reversible process). The same holds true for an isolated system, which in addition to being closed (N ¼ constant) and adiabatic (Q ¼ 0) is also surrounded by a rigid (V ¼ constant) boundary. For an isolated system, the First Law of Thermodynamics indicates that ΔE ¼ Q + W ¼ 0 or that E ¼ constant. It then follows that whenever an isolated system (E, V, and N are constant) can change from a state of lower entropy to one of higher entropy, it is possible for this change of state to occur according to the Second Law of Thermodynamics. Because the isolated system must eventually reach an equilibrium state whose properties (in particular, the entropy) are constant, it follows that S must be a maximum at equilibrium. Mathematically, this implies that (see Fig. 10.6):

10.8

Sample Problem 10.3

89

Fig. 10.6

10.8

Sample Problem 10.3

Calculate the entropy of a closed, simple system undergoing a reversible process.

10.8.1 Solution Consider states A and B connected along a reversible path (see Fig. 10.7):

Fig. 10.7

Figure 10.7 indicates that: ðB ðΔSA!B Þrev ¼

ðdSA!B Þrev A

According to the definition of entropy:

ð10:30Þ

90

10 The Second Law of Thermodynamics, Maximum Work, and Sample Problems

ðdSA!B Þrev ¼

ðδQA!B Þrev T

ð10:31Þ

We can compute (δQA!B)rev by invoking the First Law of Thermodynamics, that is: ðdEA!B Þrev ¼ ðdUA!B Þrev ¼ ðδQA!B Þrev þ ðδWA!B Þrev |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

ð10:32Þ

Simple

Rearranging Eq. (10.32) yields: ðδQA!B Þrev ¼ ðdUA!B Þrev  ðδWA!B Þrev

ð10:33Þ

Using Eq. (10.33) in Eq. (10.31) yields: ðdSA!B Þrev ¼

ðdUA!B Þrev ðdWA!B Þrev  T T

ð10:34Þ

If only PdV- type work is involved, it follows that: ðδWA!B Þrev ¼ PdV

ð10:35Þ

where P varies with V along path A to B. Combining Eqs. (10.35) and (10.34) yields: ðdSA!B Þrev ¼

ðdUA!B Þrev P þ dV T T

ð10:36Þ

Using Eq. (10.36) in Eq. (10.30), and integrating, yields: ðB ðΔSA!B Þrev ¼ A

ðdUA!B Þrev þ T

ðB

P dV T

ð10:37Þ

A

If the process is irreversible and one can bridge states A and B along a reversible path, Eq. (10.37) can be used to compute (ΔSA!B)irrev, because S is a function of state. If in addition to being reversible the process is isobaric (P is constant), Eq. (10.36) can be expressed as follows: ðdSA!B Þrev,P ¼

1 d ðUA!B þ PVÞrev,P T

Recalling that U + PV ¼ H, Eq. (10.38) can be expressed as follows:

ð10:38Þ

10.8

Sample Problem 10.3

91

ðdSA!B Þrev,P ¼

ðdHA!B Þrev,P T

ð10:39Þ

Equation (10.39) will be particularly useful to compute entropy changes in the case of reversible processes which occur isobarically. The product, TS, has units of energy. Typically, if we choose [T] ¼ Kelvin (K) and [E] ¼ Joule (J), it follows that [S] ¼ J/K.

Lecture 11

The Combined First and Second Law of Thermodynamics, Availability, and Sample Problems

11.1

Introduction

The material presented in this lecture is adapted from Chapter 4 in T&M. First, we will derive the Combined First and Second Law of Thermodynamics for both a closed and an open, single-phase, simple system. Second, we will solve Sample Problem 11.1 which will help us crystallize the material presented in Lecture 10, including calculating changes in entropy and changes in energy, heat, and work when a gas expands isothermally in a cylinder-piston assembly, without friction, in one case, reversibly, and in another case, irreversibly. Finally, we will solve Sample Problem 11.2 to calculate the maximum work done by an open system, consisting of shaft work and Carnot work, including introducing a new thermodynamic function of state, the Availability or Exergy.

11.2

Closed, Single-Phase, Simple System

For this system, the First Law of Thermodynamics states that: dE ¼ dU ¼ δQ þ δW

ð11:1Þ

For an internally reversible, quasi-static process with only PdV-type work, it follows that: δQ ¼ δQrev ¼ TdS

ð11:2Þ

δW ¼ δWrev ¼ PdV

ð11:3Þ

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_11

93

94

11 The Combined First and Second Law of Thermodynamics, Availability, and Sample. . .

Using Eqs. (11.2) and (11.3) in Eq. (11.1) yields: dU ¼ TdS  PdV

11.3

ð11:4Þ

Open, Single-Phase, Simple System

For an internally reversible, quasi-static process with a one-component stream entering (in) and leaving (out) the system (see Fig. 11.1), all the intensive properties must remain the same. In other words:

Fig. 11.1

A First Law of Thermodynamics analysis of the system bounded by the σ-surface in Fig. 11.1 yields: dE ¼ dU ¼ δQrev þ δWrev þ ðUin þ Pin Vin Þ δnin  ðUout þ Pout Vout Þ δnout ð11:5Þ Because the intensive properties of the entering (in) and leaving (out) streams are the same and δnin  δnout ¼ dN, where N is the total number of moles in the system bounded by the σ-surface, Eq. (11.5) can be rewritten as follows: dE ¼ dU ¼ δQrev þ δWrev þ ðU þ PVÞdN

ð11:6Þ

A Second Law of Thermodynamics entropy balance on the system bounded by the σ-surface (see Fig. 11.1) yields: dS ¼ δQrev =T þ Sin δnin  Sout δnout

ð11:7Þ

¼ δQrev =T þ Sðδnin  δnout Þ

ð11:8Þ

11.3

Open, Single-Phase, Simple System

95

dS ¼ δQrev =T þ SdN

ð11:9Þ

Rearranging Eq. (11.9), we obtain: δQrev ¼ TdS  TSdN

ð11:10Þ

δWrev ¼ PdV

ð11:11Þ

For PdV-type work:

Combining Eqs. (11.6), (11.10), and (11.11) yields the Combined First and Second Law of Thermodynamics, that is: dU ¼ TdS  PdV þ ðU þ PV  TSÞdN ¼ TdS  PdV þ μdN

ð11:12Þ

In Eq. (11.12), μ is the molar Gibbs free energy, or the chemical potential, defined as follows: μ ¼ G ¼ U þ PV  TS ¼ H  TS

ð11:13Þ

Equation (11.12) can be generalized for a multi-component, single-phase system that traverses a quasi-static path. Specifically: dU ¼ TdS  PdV þ

n X

μi dNi

ð11:14Þ

i¼1

where U is a continuous function of its (n + 2) independent extensive variables, that is, U ¼ f (S, V, N1, N2, . . ., Nn), whose differential can be expressed as follows: dU ¼ ð∂U=∂SÞV,N dS þ ð∂U=∂VÞS,N dV þ

n X

ð∂U=∂Ni ÞS,V,NjðiÞ dNi

ð11:15Þ

i¼1

In Eq. (11.15), the subscript N is a short-hand notation for N1, N2, . . ., Nn are kept constant, and the subscript Nj(i) indicates that all the mole numbers j except i are kept constant. A comparison of Eqs. (11.14) and (11.15) shows that: ð∂U=∂SÞV,N ¼ T; ð∂U=∂VÞS,N ¼ P; and ð∂U=∂Ni ÞS,V,NjðiÞ ¼ μi

ð11:16Þ

96

11 The Combined First and Second Law of Thermodynamics, Availability, and Sample. . .

11.4

Sample Problem 11.1

Two moles of an ideal gas expand isothermally and without friction in a cylinderpiston assembly which is in contact with a heat reservoir maintained at a constant temperature of 300 K. The initial state of the gas is (0.5 m3, 300 K), and the final state of the gas is (5 m3, 300 K). Assuming that the piston is massless, calculate: (1) The entropy change of the gas (g), (2) the entropy change of the heat reservoir (R), and (3) the entropy change of the universe (U). Carry out the calculation when (a) the gas expansion is reversible and (b) the gas expansion is irreversible.

11.4.1 Solution (a) The gas expansion is reversible and quasi-static, such that Pg ¼ Pext + dP at all times. Figure 11.2 illustrates the various elements of this problem.

Fig. 11.2

ΔUg ¼ 0 (Ideal gas, closed + isothermal) 0 ¼ ΔUg ¼ Qg + Wg (First Law of Thermodynamics analysis of the gas) Qg ¼  Wg (Heat absorbed by the gas from the heat reservoir ¼ work done by the gas) The entropy change of the gas results from the heat interaction with the heat reservoir and is given by: ΔSg ¼

Qg Wg ¼ TR TR

where TR is the temperature of the heat reservoir (300 K).

ð11:17Þ

11.4

Sample Problem 11.1

97

The work done by the gas, Wg, is given by: V ðf

Wg ¼

V ðf

Pg dVg ¼ Vi

Ng RTg

dVg Vg

ð11:18Þ

Vi

or   V Wg ¼ Ng RTgln f Vi

ð11:19Þ

The values of the gas properties are given in the Problem Statement and are summarized below for completeness: Ng ¼ 2 mol R ¼ 8.314 J/mol K Tg ¼ 300 K Vi ¼ 0.5 m3 Vf ¼ 5 m3 Using the gas property values above in Eq. (11.19) yields: 

Wg

 rev

  ¼ Qg ¼ 1:15  104 J

ð11:20Þ

Wg 1:15  104 J ¼ 300 K TR

ð11:21Þ

¼ þ38:3 J=K

ð11:22Þ

Accordingly: ΔSg ¼  or   ΔSg

rev

Because the heat reservoir lost precisely an amount of heat, QR ¼  Qg, it follows that: ðΔSR Þrev ¼ 38:3 J=K

ð11:23Þ

Consistent with the Second Law of Thermodynamics for a reversible process, it follows that the change in the entropy of the universe (U) is given by (see Eqs. (11.22) and (11.23)): ðΔSU Þrev ¼ ΔSg þ ΔSR ¼ 0

ð11:24Þ

98

11 The Combined First and Second Law of Thermodynamics, Availability, and Sample. . .

(b) Because Pg > Pext, the gas expansion is irreversible and not quasi-static. Let us consider the most extreme irreversibility which corresponds to Pext ¼ 0, where the gas expands against vacuum. Because Pg is no longer uniform, we cannot use the expression PgdVg to compute (δWg)! Instead, we can use the central result, presented in Lecture 3, that the work done by the gas (g) is equal to the work done on the three elements of its environment (in this problem, the vacuum, the friction (f) of the cylinder walls, and the piston (p)). Specifically: Wg ¼ Wvacuum þ Wf þ Wp

ð11:25Þ

As per the Problem Statement, there is no friction with the cylinder walls (Wf ¼ 0), and the piston is massless (Wp ¼ 0). Therefore, Eq. (11.25) reduces to: Wg ¼ Wvacuum ¼ 0

ð11:26Þ

or 

Wg

 irrev

¼0

ð11:27Þ

Because we already saw that, in general: Qg ¼ Wg

ð11:28Þ

it follows that: 

Qg

 irrev

¼0

ð11:29Þ

Comparing Eqs. (11.20) and (11.29), we obtain:     Qg rev ¼ 1:15  104 J > Qg irrev ¼ 0

ð11:30Þ

Further, comparing Eqs. (11.20) and (11.27), we obtain: 

Wg

 rev

  ¼ 1:15  104 J > Wg irrev ¼ 0

ð11:31Þ

Equations (11.30) and (11.31) are consistent with the observations that we made in Lecture 10 about heat and work for reversible and irreversible processes. Because S is a function of state, it follows that:   ΔSg

irrev

  ¼ ΔSg

rev

¼ þ38:3 J=K

ð11:32Þ

11.5

Sample Problem 11.2

99

Because the heat reservoir did not deliver any heat, that is, QR ¼  Qg ¼ 0, it did not change its state, and therefore: ðΔSR Þirrev ¼ 0

ð11:33Þ

Consistent with the Second Law of Thermodynamics for an irreversible process and using Eqs. (11.32) and (11.33), it then follows that: ðΔSU Þirrev ¼

11.5

  ΔSg

irrev

þ ðΔSR Þirrev ¼ þ38:3 J=K

ð11:34Þ

Sample Problem 11.2

Calculate the maximum work done by an open system undergoing the processes, depicted in Fig. 11.3.

Fig. 11.3

The following assumptions can be made: (1) reversible, quasi-static processes, (2) steady-state operation, indicating that: dN ¼ δnin  δnout ¼ 0 ) δnin ¼ δnout  δn

ð11:35Þ

dE ¼ dðNEÞ ¼ dU ¼ dðNUÞ ¼ 0

ð11:36Þ

dS ¼ dðNSÞ ¼ 0

ð11:37Þ

,and (3) the heats δQS and δQR are absorbed and rejected isothermally by the Carnot engine. The Carnot engine is used to produce Carnot work out of these heat interactions.

100 11 The Combined First and Second Law of Thermodynamics, Availability, and Sample. . .

11.5.1 Solution (1) To calculate the maximum work, we will assume that all the processes considered are reversible. First, we will calculate the maximum value of the shaft work, δWS, by doing a First Law of Thermodynamics analysis of the open system (see Fig. 11.4):

Fig. 11.4

dE ¼ dU ¼ 0 ðSteady stateÞ ¼ δQS  δWS þ Hin δnin  Hout δnout ∴

δWmax S

ð11:38Þ

dN ¼ 0 ) δnin ¼ δnout  δn ðSteady stateÞ

ð11:39Þ

¼ δQS þ ðHout  Hin Þ δn ðMaximum shaft workÞ

ð11:40Þ

(2) To calculate the Carnot work, which is the maximum work that the Carnot engine produces, we do a First Law of Thermodynamics analysis of the Carnot engine (see Fig. 11.5):

Fig. 11.5

dU ¼ 0 ðSteady stateÞ ¼ δQS  δQR  δWC

ð11:41Þ

δWmax ¼ δQS þ δQR ðMaximum Carnot workÞ C

ð11:42Þ

or

11.5

Sample Problem 11.2

101

Therefore, the total maximum work is given by the sum of the maximum shaft work (Eq. 11.40) and the maximum Carnot work (Eq. 11.42), that is, by: ðδWS þ δWC Þmax ¼ δQR þ ðHout  Hin Þδn

ð11:43Þ

(3) It is convenient to express δQR in Eq. (11.43) in terms of a function of state. To this end, we carry out a Second Law of Thermodynamics entropy balance on the composite system of (system + Carnot engine; see Fig. 11.6):

Fig. 11.6

dS ¼ 0 ðSteady stateÞ ¼ 

δQR þ Sin δnin  Sout δnout To

ð11:44Þ

Because, at steady state, δnin ¼ δnout  δn, Eq. (11.44) can be expressed as follows: δQR ¼ To ðSout  Sin Þ δn

ð11:45Þ

ðδWS þ δWC Þ ¼ δWnet ½Total ðnetÞ work obtained

ð11:46Þ

Denoting

where the three differential work contributions in Eq. (11.46) are positive. Using Eq. (11.45) in Eq. (11.43) yields: δWnet max ¼ fðHout  Hin Þ  To ðSout  Sin Þg δn

ð11:47Þ

It is convenient to define a new derived thermodynamic property, the Availability or Exergy, B, as follows:

102 11 The Combined First and Second Law of Thermodynamics, Availability, and Sample. . .

B ¼ H  To S

ð11:48Þ

where To is the constant temperature of the cold reservoir. It then follows that: δWnet max ¼ ðBout  Bin Þ δn ¼ ΔBin!out δn

ð11:49Þ

δWnet max ¼ ðBout  Bin Þ ¼ ΔBin!out δn

ð11:50Þ

or that:

Equation (11.50) shows that the maximum work per mole done by the system is equal to –ΔBin!out, which is a state function. Further, the maximum power is given by: δWnet max _ max ¼ ΔBin!out δn ¼W net δt δt

ð11:51Þ

_ net ¼ ΔBin!out n_ W max

ð11:52Þ

or

Lecture 12

Flow Work and Sample Problems

12.1

Introduction

In this lecture, we will solve Sample Problem 12.1 to calculate the reversible work of expansion or compression in flow systems (adapted from Chapter 2 in Denbigh). Second, we will solve Sample Problem 12.2 to analyze the operation of a Hilsch vortex tube (see below), which will allow us to crystallize material presented in Lectures 10, 11, and 12.

12.2

Sample Problem 12.1

Reversible work of expansion or compression in flow systems. Figure 12.1 depicts the process under consideration:

Fig. 12.1

12.2.1 Solution First, we will carry out a First Law of Thermodynamics analysis on δn moles of fluid as they flow from state 1 to state 2 (denoted hereafter as 1!2). Specifically,

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_12

103

104

12

Flow Work and Sample Problems

ΔE1!2 ¼ Q1!2  W1!2

ð12:1Þ

In Eq. (12.1), Q1!2 is the heat absorbed by the δn moles of fluid as they flow from 1!2, and W1!2 is the work done by the δn moles of fluid as they flow from 1!2, where: ΔE1!2 ¼ ðE2  E1 Þ δn 0

Q1!2 ¼ Q δn

ð12:2Þ ð12:3Þ

and W1!2 ¼ W0u δn þ

P2 ðV2 δnÞ  P1 ðV1 δnÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Work done by the δn moles on the fluid which lies ahead to change its volume by V2 δn

Work done on the δn moles by the fluid which lies behind to change its volume by V1 δn

ð12:4Þ

where W1!2 is the total work done by the δn moles of fluid as they flow from 1!2, and W0u is the useful (shaft) work done by one mole of fluid as it flows from 1!2. Using Eqs. (12.2), (12.3), and (12.4) in Eq. (12.4), we obtain:   ðE2  E1 Þ δn ¼ Q0  W0u þ P1 V1  P2 V2 δn

ð12:5Þ

E2  E1 ¼ Q0  W0u þ ðP1 V1  P2 V2 Þ

ð12:6Þ

or

where Q0 is the heat absorbed per mole, W0u is the useful (shaft) work done per mole, and (P1V1P2V2) represents the flow work done on a mole of fluid. In the absence of external fields, e.g., gravitational and inertial, the system is simple. In that case, E ¼ U, and the left-hand side in Eq. (12.6) can be expressed as follows: E2  E1 ¼ U2  U1 ¼ ðH2  P2 V2 Þ  ðH1  P1 V1 Þ

ð12:7Þ

Combining Eqs. (12.6) and (12.7) yields: H2  H1 þ P1 V1  P2 V2 ¼ Q0  W0u þ P1 V1  P2 V2 Cancelling the P1V1 and P2V2 terms in Eq. (12.8) yields:

ð12:8Þ

12.2

Sample Problem 12.1

105

H2  H1 ¼ ΔH1!2 ¼ Q0  W0u

ð12:9Þ

dH ¼ δQ0  δW0u

ð12:10Þ

or, in differential form:

Because the process is reversible, it follows that: δQ0 ¼ TdS

ð12:11Þ

In addition, the following well-known thermodynamic relation applies: dH ¼ TdS þ VdP

ð12:12Þ

Combining Eqs. (12.10), (12.11), and (12.12) yields: TdS þ VdP ¼ TdS  δW0u

ð12:13Þ

δW0u ¼ VdP

ð12:14Þ

or

Integrating Eq. (12.14) from P1 to P2 yields: W0u

Pð2

¼

ð12:15Þ

VdP P1

In general, dH ¼ dðU þ PVÞ ¼ dU þ PdV þ VdP

ð12:16Þ

If the process is reversible, such that the δn moles of fluid constitute a closed, simple system traversing a quasi-static path, then, the Combined First and Second Law of Thermodynamics applies. Specifically, dU ¼ TdS  PdV

ð12:17Þ

Equation (12.15) indicates that: W0u

δWu ¼ ¼ δn

Pð2

VdP P1

ð12:18Þ

106

12

Flow Work and Sample Problems

Integrating Eq. (12.18) with respect to n yields: ð Pð2 Wu ¼  VdP δn

ð12:19Þ

n P1

where Wu is the useful (shaft) work done by the δn moles of fluid as they flow from 1!2. Note that the limits of integration with respect to pressure may vary with the number of moles of fluid (n) flowing through the device. Dividing Eq. (12.4) by δn yields: W 01!2 ¼ W 0u þ ½P2 V 2  P1 V 1 

ð12:20Þ

In addition, the pressure integral in Eq. (12.18) can be rewritten as follows: W0u

Pð2

¼

Pð2

VdP ¼  P1

½dðPVÞ  PdV

ð12:21Þ

P1

or W0u

V ð2

¼  ½P2 V2  P1 V1  þ

PdV

ð12:22Þ

V1

Using Eq. (12.22) in Eq. (12.20), and then cancelling the P1V1 and P2V2 terms, yields: W01!2

V ð2

¼

PdV

ð12:23Þ

V1

12.3

Sample Problem 12.2: Problem 4.3 in T&M

A Hilsch vortex tube (Hvt) for sale commercially is fed with air at 300 K and 5 bar into a tangential slot near the center (point A in Fig. 12.2). Stream B leaves at the left end at 1 bar and 250 K, and stream C leaves at the right end at 1 bar and 310 K. These two streams then act as a sink and a source for a Carnot engine, and both streams leave the engine at 1 bar and TD. Assume ideal gases that have a constant heat capacity Cp ¼ 29.3 J/mol K.

12.3

Sample Problem 12.2: Problem 4.3 in T&M

107

If stream A flows at 1 mol/s, what are the flow rates of streams B and C? What is TD? What is the Carnot power output per mole of stream A? What is the entropy change of the overall process per mole of stream A? What is the entropy change of the Hilsch vortex tube (Hvt) per mole of stream A? (f) What is the maximum power that one could obtain by any process per mole of stream A if all the heat were rejected or absorbed from an isothermal reservoir at TD?

(a) (b) (c) (d) (e)

Fig. 12.2

12.3.1 Solution: Assumptions • • • •

Ideal gases, constant Cp The Hilsch vortex tube operates adiabatically Steady-state operation The gases are well mixed in the Hilsch vortex tube

(a) If stream A flows at ṅA ¼ 1 mol/s, what are the flow rates of streams B and C?

108

12

Flow Work and Sample Problems

Choose the Hilsch vortex tube as the system (see Fig. 12.3):

Fig. 12.3

A First Law of Thermodynamics analysis of the system in Fig. 12.3 yields: dU ¼ δQ þ δW þ HA δnA  HB δnB  HC δnC

ð12:24Þ

Dividing Eq. (12.24) by δt, and using the fact that dU ¼ 0 (Steady state), δQ ¼ 0 (Adiabatic boundary), and δW ¼ 0 (Rigid boundary), we obtain: HA n_ A  HB n_ B  HC n_ C ¼ 0

ð12:25Þ

δni ¼ n_ i ði ¼ A, B, CÞ δt

ð12:26Þ

dN δnA δnB δnC ¼   ¼ 0 ðSteady stateÞ dt δt δt δt

ð12:27Þ

where

and

Combining Eq. (12.27) with Eq. (12.26) for i ¼ A, B, and C yields: n_ A ¼ n_ B þ n_ C

ð12:28Þ

Combining Eqs. (12.25) and (12.28) yields: ðHA  HB Þn_ B ¼ ðHC  HA Þn_ C Because the gases are ideal, it follows that:

ð12:29Þ

12.3

Sample Problem 12.2: Problem 4.3 in T&M

109

HA  HB ¼ Cp ðTA  TB Þ

ð12:30Þ

HC  HA ¼ Cp ðTC  TA Þ

ð12:31Þ

and

Therefore, combining Eqs. (12.29), (12.30), and (12.31) yields: Cp ðTA  TB Þn_ B ¼ Cp ðTC  TA Þn_ C

ð12:32Þ

  TC  TA n_ B ¼ n_ TA  TB C

ð12:33Þ

or

As per the Problem Statement, we have: TA ¼ 300 K TB ¼ 250 K TC ¼ 310 K Using these three temperature values in Eq. (12.33) yields: ∴ n_ B ¼

1 n_ 5 C

ð12:34Þ

At steady state, we know that: n_ B þ n_ C ¼ n_ A ¼ 1 mol= sec

ð12:35Þ

Solving Eqs. (12.34) and (12.35) yields: n_ B ¼

1 n_ ¼ 0:167 mol= sec ða1Þ 6 A

ð12:36Þ

n_ B ¼

5 n_ ¼ 0:833 mol= sec ða2Þ 6 A

ð12:37Þ

and

(b) What is TD?

110

12

Flow Work and Sample Problems

Fig. 12.4

Consider the reversible Carnot engine. It absorbs heat from the “Hot” stream C and rejects heat to the “Cold” stream B at temperatures which are changing (see Fig. 12.4). For the reversible Carnot engine in Fig. 12.4, we have: δQC δQB þ ¼ 0 TB TC

ð12:38Þ

Dividing Eq. (12.38) by δt, we obtain: ðδQB =δtÞ ðδQC =δtÞ þ ¼0 TB TC

ð12:39Þ

Q_ Q_ B þ C ¼0 TB TC

ð12:40Þ

or

Because the flows in streams B and C are isobaric, we can express Q_ in terms of dH as follows: δQSt ¼ dHSt ¼ δn Cp dT

ð12:41Þ

where the subscript St denotes stream. Dividing Eq. (12.41) by δt, we obtain: ðδQSt =δtÞ ¼ Q_ St ¼ ðδn=δtÞ Cp dT

ð12:42Þ

Q_ St ¼ n_ Cp dT ðIsobaric flowÞ

ð12:43Þ

or

12.3

Sample Problem 12.2: Problem 4.3 in T&M

111

However, Q_ Engine ¼ Q_ Stream , and therefore: Q_ C ¼ n_ C Cp dT

ð12:44Þ

Q_ B ¼ n_ B Cp dT

ð12:45Þ

and

Using Eqs. (12.44) and (12.45) in Eq. (12.40), we obtain: n_ B

dTB dT  n_ C C ¼ 0 TB TC

ð12:46Þ

Integrating Eq. (12.46) from TBi !TD and from TCi !TD yields: TðD

n_ B T1B

dTB  n_ C TB

TðD

dTC ¼0 TC

ð12:47Þ

TiC

where we expect that TCi > TD and TBi < TD (see Fig. 12.4). Carrying out the two integrations in Eq. (12.47), we obtain: n_ B ln

   i TC TD _ ln þ n ¼0 C i TD TB

ð12:48Þ

In Eq. (12.48), the first term represents the heat rejected (0) by the Carnot engine from stream C. Defining x ¼ n_ B =n_ C , and rearranging Eq. (12.48), we obtain:   x  1 TD ¼ TiC TiB 1þx

ð12:49Þ

Using TCi¼ 310 K, TBi ¼ 250 K, and x ¼ n_ B =n_ C ¼ 1=5 in Eq. (12.49) yields: TD ¼ 299:1K ðbÞ

ð12:50Þ

(c) What is the Carnot power output per mole of stream A? Because no work is done in the Hilsch vortex tube, and no work is done by the flows after they exit the Hilsch vortex tube (dP ¼ 0), we expect that: Wtotal ¼ WC

ð12:51Þ

112

12

Flow Work and Sample Problems

where Wtotal (>0) is the total work done by the system, and WC (>0) is the work done by the Carnot engine (see Fig. 12.4). We can therefore choose a system consisting of the Carnot engine and the Hilsch vortex tube and compute the work done by this system, Wtotal, which should be equal to the Carnot work, WC. Alternatively, we can calculate WC directly by choosing the Carnot engine as the system Here, we will utilize the first approach to calculate WC. Figure 12.5 illustrates the system in question.

Fig. 12.5

At steady state, a First Law of Thermodynamics analysis of the system depicted in Fig. 12.5 yields: dU δQ δWtotal δn_ δn_ ¼ 0 ¼  þ HA A  HD D dt δt δt δt δt

ð12:52Þ

where δQ ¼ 0 (Adiabatic boundary). Rearranging Eq. (12.52) yields: _ total ¼ HA n_ A  HD n_ D W

ð12:53Þ

n_ A ¼ n_ D

ð12:54Þ

At steady state,

Combining Eqs. (12.53) and (12.54) yields: _ total ¼ n_ A ðHA  HD Þ ¼ n_ A Cp ðTA  TD Þ W

ð12:55Þ

Using the data provided in the Problem Statement in Eq. (12.55) yields: _ total ¼ W _ C ¼ 26:4 J= sec ðcÞ W

ð12:56Þ

12.3

Sample Problem 12.2: Problem 4.3 in T&M

113

(d) What is the entropy change of the overall process per mole of stream A? To answer this question, it is convenient to consider a small element of fluid flowing from the initial condition of (5 bar, 300 K) to the final condition of (1 bar, 299.1 K) along a reversible path. We can then use the Combined First and Second Law of Thermodynamics for a closed system (the element of fluid). Specifically, dS ¼

dU P þ dV T T

ð12:57Þ

Because δn is constant, Eq. (12.57) can be expressed as follows: dS ¼

dU P þ dV T T

ð12:58Þ

For an ideal gas: P dT R dV ¼ R  dP, and dU ¼ Cv dT T T P

ð12:59Þ

Using the results in Eq. (12.59) in the dS relation in Eq. (12.58), and rearranging, we obtain: dS ¼ ðCv þ RÞ

dT dP R T P

ð12:60Þ

or dS ¼ Cp

dT dP R T P

ð12:61Þ

Integrating Eq. (12.61) from (TA, PA) ! (TD, PD) yields: 

ΔSA!D

T ¼ Cp ln D TA



  P  Rln D PA

ð12:62Þ

Using the data provided in the Problem Statement in Eq. (12.62) yields: ΔSA!D ¼ 13:3 J=mol K ðdÞ

ð12:63Þ

Note that because ΔSR ¼ 0, it follows that: ΔSU ¼ ΔSA!D þ ΔSR ¼ 13:3 J=mol K ðIrreversible processÞ

ð12:64Þ

(e) What is the entropy change of the Hilsch vortex tube (Hvt) per mole of stream A?

114

12

Flow Work and Sample Problems

Because ΔSC ¼ 0, it follows that: ΔSHvt ¼ ΔSA!D ¼ 13:3 J=mol K ðeÞ

ð12:65Þ

We can obtain the result in (e) above in an alternative manner by passing n_ B moles of stream B from (TA,PA) to (TB,PB) reversibly, and n_ C moles of stream C from (TA, PA) to (TC,PC) reversibly, calculating the entropy changes associated with each stream per mole of stream A and then adding up these two changes. This yields:    

T P n_ B Cp ln B  Rln B ¼ 1:35J=molK TA PA n_ A

ð12:66Þ

   

TC P n_ C Cp ln  Rln C ¼ 11:95J=molK ΔSC ¼ TA PA n_ A

ð12:67Þ

ΔSB ¼ and

Adding Eqs. (12.66) and (12.67), we obtain the desired result: ΔSHvt ¼ ΔSB þ ΔSC ¼ 13:3 J=mol K ¼ ΔSA!D ðeÞ

ð12:68Þ

(f) What is the maximum power that one could obtain by any process per mole of stream A if all the heat were rejected or absorbed from an isothermal reservoir at TD? Figure 12.6 depicts the system of interest:

Fig. 12.6

where n_ A ¼ n_ D ¼ 1 mol= sec TA ¼ 300 K; TD ¼ 299:1 K PA ¼ 5 bar; PD ¼ 1 bar

12.3

Sample Problem 12.2: Problem 4.3 in T&M

115

First, we carry out a First Law of Thermodynamics analysis of “The Perfect Engine”: dU ¼ 0 ðSteady stateÞ ¼ δQ  δW þ HA δnA  HD δnD

ð12:69Þ

where δnD ¼ δnA, and therefore, Eq. (12.69) can be expressed as follows: δW ¼ ðHA  HD Þδn  δQ

ð12:70Þ

Next, for this reversible process, it is convenient to express δQ in terms of entropies. To this end, we carry out a Second Law of Thermodynamics entropy balance on “The Perfect Engine.” This yields (see Fig. 12.6): dS ¼ 0 ðSteady stateÞ ¼ 

δQ þ SA δnA  SD δnD TD

ð12:71Þ

Because, at steady state, δnA ¼ δnD ¼ δn, rearranging Eq. (12.71) yields: δQ ¼ TD ðSA  SD Þ δn

ð12:72Þ

where TD is the temperature of the heat sink. Using Eq. (12.72) for δQ in Eq. (12.70) for δW yields the following expression for the differential maximum work done by “the Perfect Engine”: δWmax ¼ fðHA  TD SA Þ  ðHD  TD SD Þg δn

ð12:73Þ

Using the availability introduced in Lecture 11, we can express Eq. (12.73) as follows: δWmax ¼ ðBD  BA Þδn ¼ ΔBA!D δn

ð12:74Þ

Dividing Eq. (12.74) by δt, we obtain: δWmax δn ¼ ΔBA!D δt δt

ð12:75Þ

_ max ¼ ΔBA!D n_ δW

ð12:76Þ

or

_ max is the maximum power produced by “The Perfect Engine” per mole of where W stream A. Using the data provided in the Problem Statement in Eq. (12.76), we obtain: _ max ¼ 4000 W ðf Þ W

ð12:77Þ

Lecture 13

Fundamental Equations and Sample Problems

13.1

Introduction

The material presented in this lecture is adapted from Chapter 5 in T&M. First, we will motivate the need to calculate partial derivatives of thermodynamic properties. Second, we will discuss the internal energy and the entropy fundamental equations. Third, we will introduce the celebrated Theorem of Euler in the context of thermodynamics, including solving Sample Problems 13.1 and 13.2 to illustrate its use. We will also discuss a method that we refer to as “the Euler integration,” including demonstrating its usefulness. Fourth, we will discuss how to systematically transform from one set of independent thermodynamic variables to another, each associated with a different fundamental equation, in a manner that ensures that the original and the new fundamental equations possess the same thermodynamic information content. Fifth, we will solve Sample Problems 13.3 and 13.4 to illustrate how to carry out these variable transformations. Finally, we will derive the celebrated Gibbs-Duhem equation which relates the differential changes in temperature, pressure, and n chemical potentials and shows that these (n + 2) intensive variables are not independent. This conclusion is consistent with the Corollary to Postulate I that we will discuss in Lecture 14.

13.2

Thermodynamic Relations for Simple Systems

In this lecture, and in Lecture 14, we will discuss material that is somewhat mathematical. Nevertheless, the results presented are essential for the calculation of changes in thermodynamic properties of systems which are not necessarily ideal. For example, suppose that we are given a one-component system (n ¼ 1), whose extensive equilibrium thermodynamic state is characterized by the two independent intensive variables, T and V, plus the number of moles, N, initially charged, © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_13

117

118

13 Fundamental Equations and Sample Problems

consistent with Postulate I, first introduced in Lecture 2, which requires that n + 2 ¼ 1 + 2 ¼ 3. We are then asked to calculate the change in the molar entropy of the system as it evolves from state 1 (characterized by T1,V1) to state 2 (characterized by T2,V2). Having chosen T and V as the two independent intensive variables, it follows that the molar entropy, S, can be expressed as S ¼ S (T,V). Therefore, the differential of S is given by:
dS ¼

∂S ∂T






∂S ∂V

dT þ V

 dV

ð13:1Þ

T

Because S in Eq. (13.1) is a function of state, we can use a convenient constant volume (isochoric) + constant temperature (isothermal) two-step path to carry out the integration from state 1 to state 2. The (V-T) phase diagram in Fig. 13.1 shows the real path and the two-step path involved in connecting state 1 and state 2:

Fig. 13.1

Integrating Eq. (13.1) from state 1 to state 2 along the isochoric-isothermal path shown in Fig. 13.1 yields: ð2

Tð2

ΔS1!2 ¼ dS ¼ 1

T1




∂S ∂T



V ð2

dT þ VjV1

V1




 ∂S dV ∂V TjT2

ð13:2Þ

In Eq. (13.2), the notation |V1 indicates that the temperature partial derivative of the molar entropy is evaluated at V ¼ V1, and the notation |T2 indicates that the molar volume partial derivative of the molar entropy is evaluated at T ¼ T2. In order to carry out the temperature integration in Eq. (13.2), we need to know the partial derivative, (∂S/∂T)V. In order to carry out the molar volume integration in Eq. (13.2), we need to know the partial derivative, (∂S/∂V)T. We will soon show that (∂S/∂T)V ¼ Cv/T and that (∂S/∂V)T ¼ (∂P/∂T)V. With the above in mind, in this lecture, we will discuss various methods to calculate the required partial derivatives. To this end, we will relate partial

13.3

Fundamental Equation

119

derivatives to other partial derivatives or to quantities that can be evaluated in terms of measurable properties. As stated above and as we will show, (∂S/∂T)V can be determined using the heat capacity at constant volume, CV. In addition, we will show that (∂S/∂V)T can be determined using an appropriate equation of state, P ¼ f (T,V).

13.3

Fundamental Equation

We have seen that the properties {S, V, N1, N2, . . ., Nn} form a set of (n + 2) natural independent extensive variables for the internal energy, U. This is consistent with Postulate I first introduced in Lecture 2. Specifically, U ¼ U ðS, V, N1 , N2 , . . . , Nn Þ

ð13:3Þ

Equation (13.3) is referred to as the Internal Energy Fundamental Equation (FE). By solving the U expression in Eq. (13.3) for S, we can express S in terms of the (n + 2) independent extensive variables (U, V, N1, N2, . . ., Nn), specifically, S ¼ S ðU, V, N1 , N2 , . . . , Nn Þ

ð13:4Þ

Equation (13.4) is referred to as the Entropy Fundamental Equation (FE). Geometrically, the FE is a surface in (n + 3) dimensional space. For example, the U surface as a function of the (n + 2) independent extensive variables (S, V, N1, N2, . . ., Nn) is plotted schematically in Fig. 13.2:

Fig. 13.2

120

13 Fundamental Equations and Sample Problems

The points on the (n + 3) FE surface represent stable equilibrium states of the simple system. Quasi-static processes can be represented by curves on the surface. Processes that are not quasi-static are not identified with curves on the surface. Given U ¼ U (S, V, N1, . . ., Nn), the differential of U is given by: 

 n
X ∂U ∂U ∂U dU ¼ dS þ dV þ dNi ∂S V,N ∂V S,N ∂Ni S,V,Nj½i i¼1

ð13:5Þ

In Eq. (13.5), N is a shorthand notation for {N1, . . ., Nn} are constant, and Nj[i] is a shorthand notation for all Njs except Ni are constant. In Lecture 11, we saw that the Combined First and Second Law of Thermodynamics for an open, simple, n-component system undergoing a reversible process is given by: dU ¼ TdS  PdV þ

n X

μi dNi

ð13:6Þ

i¼1

Recall that Eq. (13.6) accounts solely for the reversible PdV-type work done on the system. A comparison of Eqs. (13.5) and (13.6) for dU shows that:





∂U ∂Ni

 ∂U ¼ T ¼ gT ðS, V, N1 , . . . , Nn Þ ∂S V,N

ð13:7Þ


 ∂U ¼ P ¼ gP ðS, V, N1 , . . . , Nn Þ ∂V S,N

ð13:8Þ

 ¼ μi ¼ gi ðS, V, N1 , . . . , Nn Þ, i ¼ 1, . . . , n

ð13:9Þ

S,V,Nj½i

In Eqs. (13.7), (13.8), and (13.9), S and T, V and –P, and Ni and μi (i ¼ 1, . . ., n) are (n + 2) conjugate variables. Because S, V, and Ni (i ¼ 1, . . ., n) are all extensive, it follows that the (n + 2) conjugate variables, T, P, and μi (i ¼ 1, . . ., n), are all intensive. The (n + 2) first-order partial derivatives of U are referred to as “Equations of State.” We will see shortly that only (n + 1) of these equations of state are independent (Corollary to Postulate I).

13.4

The Theorem of Euler in the Context of Thermodynamics. . .

The following results apply to an n-component system: n X 1. U ¼ U ðS, V, N1 , . . . , Nn Þ, dU ¼ TdS  PdV þ μi dNi

121

ð13:10Þ

i¼1

where in the dU expression above, only PdV-type work is accounted for. 2. There are (n + 1) independent first-order partial derivatives of the FE, U. When n ¼ 1, there are two: (∂U/∂S)V,N ¼ T and (∂U/∂V)S,N ¼ P. Note that, as we will show later, the first-order partial derivative of U with respect to N is given by (∂U/∂N)S,V ¼ μ (T,P) and, therefore, depends on the other two first-order partial derivatives of U. 3. There are (n + 1)(n + 2)/2 independent second-order partial derivatives of U. When n ¼ 1, there are three: (∂2U/∂S2)V,N, (∂2U/∂V2)S,N, and (∂2U/ ∂S∂V)N ¼ (∂2U/∂V∂S)N, because the order of differentiation in the mixed second-order partial derivative is immaterial. The set of (n + 1) independent first-order partial derivatives of U, and the set of (n + 1)(n + 2)/2 independent second-order partial derivatives of U, is very important, because it is possible to prove that any other partial derivative of U can be expressed in terms of this set of partial derivatives. If the FEs (U, S, etc.) were known, they would allow us to obtain all the (n + 2) equations of state by differentiation with respect to their independent variables. Unfortunately, FEs depend explicitly on the material that they describe. In other words, there is no single, universal FE governing the properties of all materials. As such, the calculation of a FE is outside the scope of Thermodynamics. Instead, given information about the molecules comprising a material, and the intermolecular forces operating in the material, we can use Statistical Mechanics to calculate a FE of interest. In Part III, we will use Statistical Mechanics to illustrate the calculation of several FEs. Thermodynamics does not specify the complete form of the FE, but it imposes some restrictions on its functional form. We know that U and S are extensive properties which are first-order in mass or (moles). A well-known mathematical theorem due to Euler can be implemented in the context of Thermodynamics to obtain a very powerful result which we present next.

13.4

The Theorem of Euler in the Context of Thermodynamics (Adapted from Appendix C in T&M)

The Theorem of Euler applies to all smoothly varying, homogeneous functions f (a, b, . . ., x, y, . . .), where a, b, . . . are intensive variables homogeneous to zero-order in mass (or moles), and x, y, . . . are extensive variables homogeneous to first-order in

122

13 Fundamental Equations and Sample Problems

mass (or moles). Further, df (total differential) can be integrated directly, where df is an exact differential, that is, it is not path dependent. It then follows that if Y ¼ ky and X ¼ kx, we obtain: f ða, b, . . . , X, Y, . . .Þ ¼ kf ða, b, . . . , x, y, . . .Þ

ð13:11Þ

and x ð∂f=∂xÞa,b,...,y... þ y ð∂f=∂yÞa,b,...,x... þ . . . ¼ ð1Þ f ða, b, . . . , x, y, . . .Þ ð13:12Þ In Eq. (13.12), only the extensive variables x, y, . . . appear on the left-hand side of the equality.

13.5

Sample Problem 13.1

U ¼ U (S, V, N1, . . ., Nn) is homogeneous to first-order in all its (n + 2) independent extensive variables {S, V, N1, . . ., Nn}. Apply the Theorem of Euler to U.

13.5.1 Solution According to the Theorem of Euler, it follows that: UðkS, kV, kN1 , . . . , kNn Þ ¼ kUðS, V, N1 , . . . , Nn Þ

ð13:13Þ



 n
X ∂U ∂U ∂U Sþ Vþ N i ¼ ð 1Þ U ∂S V,N ∂V S,N ∂Ni S,V,Nj½i i¼1

ð13:14Þ

and

Note that because the (n + 2) independent variables in U are all extensive, they all appear on the left-hand side of Eq. (13.14). In Eq. (13.14), the first-order partial derivatives multiplying S, V, and Ni are equal to gT ¼ T, gP ¼ P, and gi ¼ μi, respectively (see Eqs. (13.7), (13.8), and (13.9)). Therefore, Eq. (13.14) can be expressed as follows: TS  PV þ

n X i¼1

μi Ni ¼ U

ð13:15Þ

13.6

Sample Problem 13.2

123

The Theorem of Euler enables us to reconstruct U, if we know the set of (n + 2) intensive first-order partial derivatives {gT, gP, and gi ¼ μi (i ¼ 1, . . ., n)}. In fact, below, we will show that only (n + 1) of these (n + 2) intensive variables are independent. As shown in Lecture 11, the Combined First and Second Law of Thermodynamics states that: TdS  PdV þ

n X

μi dNi ¼ dU

ð13:16Þ

i¼1

If we “Euler Integrate” (EI) the Combined First and Second Law of Thermodynamics in Eq. (13.16), that is, if we replace: d ðExtensive PropertyÞ EI ! Extensive Property

ð13:17Þ

d ðIntensive PropertyÞ EI ! 0

ð13:18Þ

and

we recover: TS  PV þ

n X

μi N i ¼ U

ð13:19Þ

i¼1

because all the differentials in Eq. (13.16) are of extensive properties. The “Euler Integration” can be carried out on other differential forms of fundamental equations, as illustrated in the problem below.

13.6

Sample Problem 13.2

Apply the Theorem of Euler to the enthalpy, H.

13.6.1 Solution The enthalpy, H, is given by: H ¼ H ðS, P, N1 , . . . , Nn Þ

ð13:20Þ

124

13 Fundamental Equations and Sample Problems

Because the variable P in H is intensive, Eq. (13.20) shows that only (n + 1) variables out of the (n + 2) independent variables in H are extensive. Therefore, use of the Theorem of Euler yields: HðkS, P, kN1 , . . . , kNn Þ ¼ kHðS, P, N1 , . . . , Nn Þ

ð13:21Þ


 n
X ∂H ∂H Sþ Ni ¼ ð1Þ H ∂S P,N ∂Ni S,P,Nj½i i¼1

ð13:22Þ

and

In Eq. (13.22), the first-order partial derivatives multiplying S and Ni are equal to gT ¼ T and gi ¼ μi, respectively. Note that the intensive variable, P, does not appear on the left-hand side of Eq. (13.22).

13.7

Variable Transformations and New Fundamental Equations

From the Combined First and Second Law of Thermodynamics emerged naturally the FE, U, whose natural variables {S, V, N1, . . ., Nn} are all extensive. However, practically, S (the entropy) is not a convenient variable to work with. Indeed, we do not know how to precisely control the entropy S, and no devices are available to measure entropy. On the other hand, the variable conjugate to the entropy, T ¼ (∂U/∂S)V,N, the temperature, is simple to control using a water bath and to measure using a thermometer. Therefore, for practical reasons, it is often convenient to work with other variables. When we change a variable to its conjugate variable, we introduce a new FE. Further, when we change variables, we need to ensure that the new FE obtained contains the same thermodynamic information content as the FE from which it was derived. Typically, Legendre transforms (LTs) are used to ensure that this is the case. However, here, we will not discuss LTs. Readers who are interested in learning more about LTs, please refer to Chapter 5 in T&M. Instead, we will explore a simpler procedure that can be used to ensure that the thermodynamic information content of each FE is preserved. As a first example, consider the FE, U {S, V, N1, . . ., Nn}, and replace the variable, V, by its conjugate variable, P. We know that: dU ¼ TdS  PdV þ

n X i¼1

μi dNi

ð13:23Þ

13.7

Variable Transformations and New Fundamental Equations

125

Let us add the quantity d (PV) to both sides of the equality in Eq. (13.23) to create the new FE, H: dU þ dðPVÞ ¼ d ðU þ PVÞ ¼ dH ¼ TdS  PdV þ

n X

μi dNi þ PdV þ VdP

ð13:24Þ

i¼1

Cancelling the two PdV terms on the right-hand side of the equality in Eq. (13.24) yields: dH ¼ TdS þ VdP þ

n X

μi dNi

ð13:25Þ

i¼1

If we EI Eq. (13.25), recalling that H, S, and Ni (i ¼ 1, . . ., n) are all extensive variables, and that P is an intensive variable, we obtain: H ¼ TS þ 0 þ

n X

μi N i

ð13:26Þ

i¼1

Subtracting and adding PV to the right-hand side of the equality in Eq. (13.26), and rearranging, we obtain: H¼

TS  PV þ

n X

! μi dNi

þ PV

ð13:27Þ

i¼1

Recognizing that the expression in parenthesis in Eq. (13.27) is the internal energy, U, it follows that: H ¼ U þ PV

ð13:28Þ

Recall that H ¼ H (S, P, N1, . . ., Nn). Note also that, strictly, the intensive variable appearing in H in Eq. (13.28) should be –P, which is the conjugate variable of V. However, for all practical purposes, not showing explicitly the – sign is fine. Taking the differential of H ¼ H (S, P, N1, . . ., Nn), we obtain: 

 n
X ∂H ∂H ∂H dH ¼ dS þ dP þ dNi ∂S P,N ∂P S,N ∂Ni S,P,Nj½i i¼1

ð13:29Þ

126

13 Fundamental Equations and Sample Problems

Comparing Eqs. (13.25) and (13.29) for dH, we obtain:


∂H ∂S

 ¼ T, P,N



 ∂H ∂H ¼ V, ¼ μi ði ¼ 1, . . . , nÞ ∂P S,N ∂Ni S,P,Nj½i

ð13:30Þ

We also know that:


∂U ∂S






∂U ∂Ni

¼ T, V,N

 ¼ μi ði ¼ 1, . . . , nÞ

ð13:31Þ

S,V,Nj½i

Because T, derived from U in Eq. (13.31) and from H in Eq. (13.30), are the same, and μi, derived from U in Eq. (13.31) and from H in Eq. (13.30), are the same, it follows that U (S, V, N1, . . ., Nn) and H (S, P, N1, . . ., Nn) possess the same thermodynamic information content, although they depend on different sets of (n + 2) independent variables: {S, V, N1, . . ., Nn} for U, and {S, P, N1, . . ., Nn} for H. To further crystallize the material taught, below, we will solve Sample Problems 13.3 and 13.4.

13.8

Sample Problem 13.3

Transform U to A (Helmholtz Free Energy).

13.8.1 Solution We first replace S in U(S, V, N1, . . ., Nn) by its conjugate variable:
 ∂U T¼ ∂S V,N

ð13:32Þ

When we do, we will find that: A ¼ AðT, V, N1 , . . . , Nn Þ

ð13:33Þ

and dA ¼ SdT  PdV þ

n X i¼1

μi dNi

ð13:34Þ

13.9

13.9

Sample Problem 13.4

127

Sample Problem 13.4

Transform U to G (Gibbs Free Energy).

13.9.1 Solution We first replace S by its conjugate variable: T¼


 ∂U ∂S V,N

ð13:35Þ

and then V by its conjugate variable: P ¼


 ∂U ∂V S,N

ð13:36Þ

When we do, we will find that G ¼ G (T, P, N1, . . ., Nn) and that: dG ¼ SdT þ VdP þ

n X

μi dNi

ð13:37Þ

i¼1

The transformation of variables presented above suggests the following “Rule of Thumb”: To transform a variable to its conjugate variable in the differential form of a fundamental equation, we simply: 1. Flip the variable with its conjugate variable 2. Multiply the result by 1 For example: TdS in dU ! SdT in dA PdV in dU ! þVdP in dH and dG If we transform all the (n + 2) original extensive variables {S, V, N1, . . ., Nn} in the FE, U, to their (n + 2) conjugate intensive variables, that is,


 ∂U ∂U ∂U , P¼ , μ ¼ ði ¼ 1, . . . , nÞ T¼ ∂S V,N ∂V S,N i ∂Ni S,V,Nj½i

ð13:38Þ

128

13 Fundamental Equations and Sample Problems

we obtain a very useful relation between the (n + 2) intensive conjugate variables known as the Gibbs-Duhem Equation. To derive the Gibbs-Duhem Equation, we begin with the differential form of the Combined First and Second Law of Thermodynamics, that is: dU ¼ TdS  PdV þ

n X

μi dNi

ð13:39Þ

i¼1

If we Euler Integrate Eq. (13.39), we obtain: U ¼ TS  PV þ

n X

μi Ni

ð13:40Þ

i¼1

If we then differentiate the Euler Integrated form of U in Eq. (13.40), we obtain: dU ¼ TdS þ SdT  PdV  VdP þ

n X i¼1

μi dNi þ

n X

Ni dμi

ð13:41Þ

i¼1

Because dU in Eq. (13.39) and dU in Eq. (13.41) are the same, it follows that: TdS  PdV þ

n X

μi dNi ¼ TdS þ SdT  PdV  VdP þ

i¼1

n X

μi dNi

i¼1

þ

n X

Ni dμi

ð13:42Þ

i¼1

Cancelling the equal terms on both sides of the equality in Eq. (13.42), we obtain: SdT  VdP þ

n X

Ni dμi ¼ 0 ðThe Gibbs‐Duhem EquationÞ

ð13:43Þ

i¼1

The Gibbs-Duhem Equation (GDE) is a relation between the (n + 2) intensive variables {T, P, μ1, . . ., μn} and shows that only (n + 1) of these are independent, a statement that was made earlier in this lecture, and that we have now proven.

Lecture 14

Manipulation of Partial Derivatives and Sample Problems

14.1

Introduction

The material presented in this lecture is adapted from Chapter 5 in T&M. Continuing with the material introduced in Lecture 13, first, we will discuss two additional restrictions which need to be imposed on the internal energy fundamental equation, U, including presenting the Corollary to Postulate I. Second, we will discuss how to reconstruct the fundamental equation, U, if we know the (n + 2) first-order partial derivatives of U. Third, we will motivate the need to manipulate partial derivatives of thermodynamic functions in order to calculate changes in these thermodynamic functions when the system evolves from an initial state i to a final state f. To this end, we will discuss how to devise useful integration paths which connect state i with state f. Fourth, we will discuss various useful methods to relate an unknown partial derivative to other partial derivatives which can be calculated or determined experimentally. Fifth, among the useful methods to relate a desired partial derivative to others, we will discuss the triple product rule, the add another variable rule, the derivative inversion rule, and Maxwell’s reciprocity rule. In addition, we will discuss Jacobian transformations. Finally, we will solve Sample Problems 14.1 and 14.2 to illustrate the use of some of the methods presented to calculate partial derivatives.

14.2

Two Additional Restrictions on the Internal Energy Fundamental Equation

In addition to the restrictions imposed by the Theorem of Euler on the form of the fundamental equation, U, the following two restrictions need to be imposed: (i) U is a single-valued function of its (n + 2) independent variables S, V, N1, . . ., Nn (ii) (∂U/∂S)V,N ¼ T is nonnegative (0) © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_14

129

130

14.3

14

Manipulation of Partial Derivatives and Sample Problems

Corollary to Postulate I

“For a single-phase, simple system of n components, any intensive property can be defined by the values of any other (n + 1) independent intensive properties.”

14.4

Reconstruction of the Internal Energy Fundamental Equation

We saw that given U ¼ U (S, V, N1, . . ., Nn), it follows that:


 ∂U ¼ T ¼ gT ðS, V, N1 , . . . , Nn Þ ∂S V,N


 ∂U ¼ P ¼ gP ðS, V, N1 , . . . , Nn Þ ∂V S,N


∂U ∂Ni

ð14:1Þ

ð14:2Þ

 ¼ μi ¼ gi ðS, V, N1 , . . . , Nn Þ, i ¼ 1, 2, . . . , n

ð14:3Þ

S,V,Nj½i

Alternatively, if we know the (n + 2) “Equations of State”: fgT ¼ T, gP ¼ P, g1 ¼ μ1 , . . . , gn ¼ μn g

ð14:4Þ

we can reconstruct U using the Euler integrated form. Specifically, U ¼ TS  PV þ

n X

μi Ni

ð14:5Þ

iþ1

Because the (n + 2) variables, gT, gP, g1, . . ., gn, are all intensive, according to the Corollary to Postulate I, only (n + 1) of them are independent. Therefore, given (n + 1) of them, we can determine the remaining one to within an arbitrary constant of integration. Accordingly, only (n + 1) “Equations of State” are needed to determine the Fundamental Equation, U, to within an arbitrary constant of integration. Because only ΔU has physical significance, this constant of integration is not important.

14.5

14.5

Manipulation of Partial Derivatives of Thermodynamic Functions

131

Manipulation of Partial Derivatives of Thermodynamic Functions

Suppose that we want to calculate ΔS1!2 for a one-component (n ¼ 1) system and that we choose T and P as the n + 1 ¼ 1 + 1 ¼ 2 independent intensive variables. Consistent with the Corollary to Postulate I, it then follows that another intensive variable, for example, the molar entropy, S, can be uniquely expressed as a function of T and P, that is, S ¼ S ðT, PÞ

ð14:6Þ

The differential of S in Eq. (14.6) is given by:
dS ¼

∂S ∂T




dT þ

P

 ∂S dP ∂P T

ð14:7Þ

Integrating Eq. (14.7) from state 1 to state 2 yields: ð2 ΔS1!2 ¼

dS1!2

ð14:8Þ

1

In order to carry out the integration in Eq. (14.8), it is useful to consider the schematic (P-T) phase diagram in Fig. 14.1.

Fig. 14.1

Because S is a function of state, it is convenient to choose an isobaric (constant pressure)-isothermal (constant temperature) two-step path to go from state 1 to state 2 in order to carry out the integration (see Fig. 14.1). Specifically,

132

14 Tð2

ΔS1!2 ¼ T1

Manipulation of Partial Derivatives and Sample Problems




Pð2
  ∂S ∂S dT þ dP ∂T PjP1 ∂P TjT2

ð14:9Þ

P1

To carry out the temperature integration in Eq. (14.9), we need to know (∂S/ ∂T)P ¼ CP/T. To carry out the pressure integration in Eq. (14.9), we need to know (∂S/∂P)T ¼  (∂V/∂T)P.

14.6

Internal Energy and Entropy Fundamental Equations

The Fundamental Equations U and S can be obtained via an Euler integration of the Combined First and Second Law of Thermodynamics. Specifically, U can be expressed as follows: U ¼ f u ½S, V, N1 , . . . , Nn  ¼ TS  PV þ

n X

μi N i

ð14:10Þ

ðμi =TÞNi

ð14:11Þ

i¼1

Further, by solving Eq. (14.10) for S, we obtain: S ¼ f s ½U, V, N1 , . . . , Nn  ¼ U=T þ ðP=TÞV 

n X i¼1

14.7

Useful Rules to Calculate Partial Derivatives of Thermodynamic Functions

A number of useful rules are available to calculate partial derivatives of thermodynamic functions. These include:

14.7.1 The Triple Product Rule


 ∂F ∂x ∂y Fðx, yÞ ) ¼ 1 ∂x y ∂y F ∂F x

Example :

HðT, PÞ ) ð∂H=∂TÞP ð∂T=∂PÞH ð∂P=∂HÞT ¼ 1

ð14:12Þ

ð14:13Þ

14.7

Useful Rules to Calculate Partial Derivatives of Thermodynamic Functions

133

14.7.2 The Add Another Variable Rule ð∂F=∂yÞx ¼

Example :

ð∂F=∂ϕÞx ð∂y=∂ϕÞx

ϕ ¼ T ) ð∂S=∂HÞP ¼

Cp =T ð∂S=∂TÞP 1 ¼ ¼ T Cp ð∂H=∂TÞP

ð14:14Þ

ð14:15Þ

14.7.3 The Derivative Inversion Rule ð∂F=∂yÞx ¼ 1=ð∂y=∂FÞx Example :

ð∂T=∂SÞP ¼ 1=ð∂S=∂TÞP ¼ T=CP

ð14:16Þ ð14:17Þ

14.7.4 Maxwell’s Reciprocity Rule Given a function F (x, y):

 ∂F ∂F dx þ dy dF ¼ ∂x y ∂y x "

ð14:18Þ


# 
 ∂ ∂F ∂ ∂F ¼ ∂y ∂x y ∂x ∂y x y

ð14:19Þ

Fyx ¼ Fxy

ð14:20Þ

x

or

For a smoothly varying function, the order of differentiation is immaterial.

134

14.8

14

Manipulation of Partial Derivatives and Sample Problems

Sample Problem 14.1

Derivation of a Maxwell reciprocity relation starting with U(S, V).

14.8.1 Solution For a pure material (n ¼ 1): U ¼ UðS, VÞ

ð14:21Þ

dU ¼ TdS  PdV

ð14:22Þ




 ∂U ¼T ∂S V

ð14:23Þ

 ∂U ¼ P ∂V S

ð14:24Þ


 
 ∂ ∂U ∂ ∂U ¼ ∂V ∂S V S ∂S ∂V S V

ð14:25Þ




However, 

or



 ∂T ∂P ¼ ∂V S ∂S V

ð14:26Þ

Equation (14.26) is an example of a Maxwell reciprocity relation.

14.9

Jacobian Transformations

Given f(x,y) and g(x,y), the Jacobian is defined in terms of the following determinant:

14.9

Jacobian Transformations

135



   ∂f ∂f     ∂x ∂y ∂ðf, gÞ  y x  

  Jacobian ¼ ∂ðx, yÞ  ∂g ∂g    ∂x ∂y x  y
Jacobian ¼

∂f ∂x


y

∂g ∂y






x

∂f ∂y


x

∂g ∂x

ð14:27Þ

 ð14:28Þ y

14.9.1 Properties of Jacobians 1. Transposition ∂ðf, gÞ ∂ðg, f Þ ¼  ∂ðx, yÞ ∂ðx, yÞ

ð14:29Þ

1 ∂ðf, gÞ ¼ ∂ ðx, yÞ ∂ðx, yÞ ∂ ðf, gÞ

ð14:30Þ

∂ðf, gÞ ∂ðf, gÞ ∂ðz, wÞ ¼ ∂ðx, yÞ ∂ðz, wÞ ∂ðx, yÞ

ð14:31Þ

2. Inversion

3. Chain Rule Expansion

A simplification occurs if we need to compute (∂f/∂z)g, where f ¼ f (z,g). First, we recognize that:


∂f ∂z

 ¼ g

∂ðf, gÞ ∂ðz, gÞ

Implementing Properties 3 and 2 above in Eq. (14.32), we obtain:

ð14:32Þ

136

14




14.10

Manipulation of Partial Derivatives and Sample Problems

∂f ∂z



∂ðf, gÞ ∂ðz, gÞ = ∂ðx, yÞ ∂ðx, yÞ

¼ g

ð14:33Þ

Sample Problem 14.2

Calculate (∂T/∂P)H for a one-component fluid.

14.10.1

Solution

Choosing T and P as the two independent intensive variables, it follows that H ¼ H (T,P). We can therefore use Eq. (14.33), with f ¼ T, g ¼ H, and z ¼ P, that is,


∂T ∂P

 H

∂ ðT, HÞ ∂ ðx, yÞ ¼ ∂ ðP, HÞ ∂ ðx, yÞ

ð14:34Þ

If we substitute x ¼ T and y ¼ P in Eq. (14.34), we obtain:


∂ ðT, HÞ  ∂ ðT, PÞ ∂T ¼ ∂P H ∂ ðP, H Þ

ð14:35Þ

∂ ðT, PÞ Using Property 1 in Eq. (14.35), we obtain:


∂T ∂P



 ¼ H



∂ ðH, TÞ ∂ ðP, TÞ





∂ ðH, PÞ ∂ ðT, PÞ



¼

∂H

∂P T  ∂H ∂T P

¼

∂H

∂P T

CP

ð14:36Þ

Note that we could have utilized the triple product rule directly to obtain the result in Eq. (14.36).

Lecture 15

Properties of Pure Materials and Gibbs Free Energy Formulation

15.1

Introduction

The material presented in this lecture is adapted from Chapter 5 in T&M. In this lecture, we will consider pure materials (n ¼ 1). First, we will discuss the Gibbs Free Energy Fundamental Equation, G. Second, we will again derive the Gibbs-Duhem Equation in a complementary way. Third, we will relate the Gibbs Free Energy to other thermodynamic functions. Fourth, we will calculate the first-order and secondorder partial derivatives of G with respect to its independent variables T, P, and N. In particular, we will show that the three independent second-order partial derivatives of G are related to three widely used fluid properties – the heat capacity at constant pressure, the isothermal compressibility, and the coefficient of thermal expansion. Finally, we will determine which data set has the same thermodynamic information content as the Fundamental Equation, G.

15.2

Gibbs Free Energy Fundamental Equation

For a single phase, simple system of a pure material (n ¼ 1), the independent variables for the Internal Energy Fundamental Equation (FE), U, are {S, V, N}. From a practical point of view, it is often useful to transform S and V into their conjugate variables, T ¼ (∂U/∂S)V,N and –P ¼ (∂U/∂V)S,N, respectively. The new variables {T, P, N} correspond to the new FE, G ¼ G(T, –P, N). Note that the – sign in P is not needed unless we use Legendre transforms, which we will not discuss. Therefore, hereafter, we will write: G ¼ G(T, P, N). Beginning with the Combined First and Second Law of Thermodynamics, dU ¼ TdS –PdV + μdN, and following the “Rule of Thumb” that we discussed in Lecture 13, we obtain:

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_15

137

138

15

Properties of Pure Materials and Gibbs Free Energy Formulation

þTdS in dU ! SdT in dG

ð15:1Þ

PdV in dU ! þVdP in dG

ð15:2Þ

dG ¼ SdT þ VdP þ μdN

ð15:3Þ

and

It then follows that:

Because G and N are extensive variables, and T and P are intensive variables, if we Euler integrate Eq. (15.3), we obtain: G ¼ 0 þ 0 þ μN

ð15:4Þ

G ¼ μN

ð15:5Þ

For n ¼ 1 ) μ ¼ G=N ¼ G

ð15:6Þ

or

Key Result :

Equation (15.6) shows that, for a pure material (n ¼ 1), the chemical potential, μ, is equal to the molar Gibbs Free Energy, G ¼ G/N. As we will see in Part II, knowing the chemical potential, μ, is essential to solve phase equilibria and chemical reaction equilibria problems.

15.3

Derivation of the Gibbs-Duhem Equation

Recalling that S ¼ NS, V ¼ NV, and G ¼ NG, and using these relations in Eq. (15.3), we obtain: dG ¼ dðNGÞ ¼ dðNμÞ ¼ Ndμ þ μdN ¼ SdT þ VdP þ μdN

ð15:7Þ

Cancelling the equal terms in Eq. (15.7), and rearranging, yields: Ndμ ¼ NSdT þ NVdP Dividing Eq. (15.8) by N, we obtain:

ð15:8Þ

15.5

First-Order and Second-Order Partial Derivatives of the Gibbs Free Energy

dμ ¼ SdT þ VdP

139

ð15:9Þ

Equation (15.9) is, in fact, the Gibbs-Duhem Equation (GDE) derived in Lecture 13 for n ¼ 1, and clearly shows that μ ¼ μ(T, P) for a pure (n ¼ 1) material. The GDE is also consistent with the Corollary to Postulate I which we presented in Lecture 14. Indeed, for n ¼ 1, T and P are (n + 1) ¼ (1 + 1) ¼ 2 independent intensive variables on which the intensive variable, μ, depends.

15.4

Relating the Gibbs Free Energy to Other Thermodynamic Functions

We also know that: dU ¼ TdS  PdV þ μdN EI U ¼ TS  PV þ μN !

ð15:10Þ

Using μN ¼ G in the expression for U in Eq. (15.10), we obtain: U ¼ TS  PV þ G

ð15:11Þ

G ¼ U þ PV  TS

ð15:12Þ

G ¼ ðU þ PVÞ  TS ¼ H  TS

ð15:13Þ

G ¼ ðU  TSÞ þ PV ¼ A þ PV

ð15:14Þ

or

or

or

15.5

First-Order and Second-Order Partial Derivatives of the Gibbs Free Energy

Similar to U ¼ U(S, V, N), for n ¼ 1, the FE, G ¼ G(T, P, N), for n ¼ 1, also has three first-order partial derivatives or “Equations of State” (EOS). Specifically, beginning with:

140

15

Properties of Pure Materials and Gibbs Free Energy Formulation

dG ¼ SdT þ VdP þ μdN

ð15:15Þ

it follows that: • ð∂G=∂TÞP,N ¼ S ¼ g1 ðT, P, NÞ ¼ Ng1 ðT, PÞ ! SN ¼ Ng1 ðT, PÞ ! S ¼ g1 ðT, PÞ

ð15:16Þ

• ð∂G=∂PÞT,N ¼ V ¼ g2 ðT, P, NÞ ¼ Ng2 ðT, PÞ ! VN ¼ Ng2 ðT, PÞ ! V ¼ g2 ðT, PÞ • ð∂G=∂NÞT,P ¼ G ¼ μ ¼ g3 ðT, PÞ ! μ ¼ g3 ðT, PÞ

ð15:17Þ ð15:18Þ

Note that the set of (n + 2) ¼ (1 + 2) ¼ 3 “EOS,” {g1(T,P), g2(T,P), and g3(T,P)}, which are all intensive, is not independent. As discussed earlier, only (n + 1) ¼ (1 + 1) ¼ 2 are independent. This is fully consistent with the Corollary to Postulate I that we presented in Lecture 14. We will utilize this important result below. Similar statements to those made earlier about the second-order partial derivatives of U can be made about the second-order partial derivatives of G. In particular, there are [(n + 2)(n + 1)]/2 independent second-order partial derivatives of G. Specifically, for n ¼ 1, there are [(1 + 2)(1 + 1)]/2 ¼ 3, given by:
 2 (1) ∂ G=∂T2 ) ð∂S=∂TÞP,N ¼ g11 ðT, P, NÞ ð15:19Þ P,N

or Nð∂S=∂TÞP ¼ Ng11 ðT, PÞ

ð15:20Þ

ð∂S=∂TÞP ¼ g11 ðT, PÞ

ð15:21Þ

or

(2)




2

∂ G=∂P2

 T,N

) ð∂V=∂PÞT,N ¼ g22 ðT, P, NÞ

ð15:22Þ

or Nð∂V=∂PÞT ¼ Ng22 ðT, PÞ or

ð15:23Þ

15.5

First-Order and Second-Order Partial Derivatives of the Gibbs Free Energy

ð∂V=∂PÞT ¼ g22 ðT, PÞ and (3)




2

∂ G=∂T∂P

 N

141

ð15:24Þ


 2 ) ð∂S=∂PÞT,N ¼ ∂ G=∂P∂T

N

ð15:25Þ

¼ ð∂V=∂TÞP,N ¼ g12ðT, P, NÞ ¼ Ng12 ðT, PÞ ¼ g21ðT, P, NÞ ¼ Ng21ðT, PÞ

ð15:26Þ

Accordingly, Nð∂V=∂TÞP ¼ Nð∂S=∂PÞT ¼ Ng12 ðT, PÞ ¼ Ng21 ðT, PÞ

ð15:27Þ

ð∂S=∂PÞT ¼ ð∂V=∂TÞP ¼ g12 ðT, PÞ ¼ g21 ðT, PÞ

ð15:28Þ

or

There are three additional second-order partial derivatives of G involving differentiations with respect to N. However, these partial derivatives are either zero or redundant with the previous partial derivatives of G. For example:
 2 ∂ G=∂N2

T,P

¼ ð∂μ=∂NÞT,P ¼ 0 ðRecall that μ ¼ μðT, PÞÞ

ð15:29Þ

For additional practice, interested readers may want to calculate the two mixed partial derivatives of G with respect to T and N, as well as with respect to P and N, that is, to calculate:


2

∂ G=∂T∂N

 P


 2 and ∂ G=∂P∂N

T

ð15:30Þ

The three independent, second-order partial derivatives of G, that is, (∂S/∂T)P, (∂V/∂P)T, and (∂V/∂T)P, are related to the following three widely used fluid properties: (1) (2) (3)

CP ¼ Tð∂S=∂TÞP , Heat capacity at constant pressure 1 ð∂V=∂PÞT , Isothermal compressibility V

ð15:32Þ

1 ð∂V=∂TÞP , Coefficient of thermal expansion V

ð15:33Þ

κT ¼  αP ¼

ð15:31Þ

142

15

Properties of Pure Materials and Gibbs Free Energy Formulation

Recall that: ð∂S=∂TÞP ¼ g11 ðT, PÞ

(i)

ð15:34Þ

ð∂V=∂PÞT ¼ g22 ðT, PÞ

ð15:35Þ

ð∂V=∂TÞP ¼ g12 ðT, PÞ ¼ g21 ðT, PÞ

ð15:36Þ

where g1, g2, and g3 are the three “EOS” associated with G. (ii) Knowing κT and αP is equivalent to knowing the P-V-T volumetric equation of state of the pure (n ¼ 1) fluid (iii) CP, κT, and αP are experimentally accessible (iv) To prove that CP ¼ T (∂S/∂T)P, we begin from the basic definition CP ¼ (∂H/ ∂T)P. Because H ¼ H (S,P) and dH ¼ TdS + VdP, it follows that (∂H/∂T)P ¼ T (∂S/∂T)P + 0 ¼ CP

15.6

Determining Which Data Set Has the Same Thermodynamic Information Content as the Gibbs Free Energy Fundamental Equation

Clearly, because: GðT, P, NÞ ¼ NμðT, PÞ ¼ Ng3 ðT, PÞ

ð15:37Þ

it is sufficient to know g3(T,P) ¼ μ(T,P) to reconstruct G when n ¼ 1. However, to derive μ(T,P) ¼ g3(T,P) to within an arbitrary constant of integration, we need to know the two independent “EOS,” g1(T,P) and g2(T,P). To show this more clearly, we recall that: dG ¼ dμ ¼ SdT þ VdP ðThe GDE for n ¼ 1Þ

ð15:38Þ

where S ¼ S(T,P) and V ¼ V(T,P). Recall that: SðT, PÞ ¼ g1 ðT, PÞ

ð15:39Þ

VðT, PÞ ¼ g2 ðT, PÞ

ð15:40Þ

and

15.6

Determining Which Data Set Has the Same Thermodynamic Information. . .

143

Therefore, dμ ¼ SðT, PÞ dT þ VðT, PÞdP

ð15:41Þ

dμ ¼ g1 ðT, PÞ dT þ g2 ðT, PÞdP

ð15:42Þ

or

To integrate Eq. (15.42), we recall that μ ¼ G is a function of state, and therefore, we can choose a convenient integration path connecting the initial state (i) to the final state ( f ). Specifically, in a (P-T) phase diagram, we can choose a two-step isobaric (constant pressure)  isothermal (constant temperature) path to go from state i (Po,To) to state f (P,T). The (P-T) phase diagram in Fig. 15.1 shows the chosen two-step path. Carrying out the integration of dμ in Eq. (15.42) from (To,Po) to (T,P) yields: ðμ

ðT dμ ¼ μ  μo ¼

μo

ðP g1 ðT, PÞjPo dT þ

To

g2 ðT, PÞjT dP

ð15:43Þ

Po

Fig. 15.1

where in Eq. (15.43) and Fig. 15.1, the temperature integration corresponds to the isobaric path, and the pressure integration corresponds to the isothermal path. Equation (15.43) can be expressed as follows:

144

15

Properties of Pure Materials and Gibbs Free Energy Formulation

ðT μðT, PÞ ¼ GðT, PÞ ¼ μo þ

ðP g1 ðT, PÞjPo dT þ

To

g2 ðT, PÞjT dP

ð15:44Þ

Po

where μo is an arbitrary constant of integration. When calculating the difference between μ(T,P), or G(T,P), in states 2 and 1, μo cancels out. Therefore, to compute μ (or G) to within an arbitrary constant of integration, μo, we need to know the two “Equations of State,” g1(T,P) and g2(T,P). Note that this is fully consistent with the Corollary to Postulate I that we discussed in Lecture 13. Next, what do we do if we do not know g1(T,P) and g2(T,P)? How can we nevertheless reconstruct the FE for G (or μ) in that case? To this end, we can first evaluate g1(T,P) and g2(T,P) from knowledge of CP, κT, and αP, or equivalently, from knowledge of CP and the P-V-T volumetric equation of state. Recall that: 

   ∂S ∂S dT þ dP ∂T P ∂P T

g1 ðT, PÞ ¼ SðT, PÞ ) dS ¼

ð15:45Þ

where ð∂S=∂TÞP ¼ CP =T, and ð∂S=∂PÞT ¼ ð∂V=∂TÞP ¼ Vαp

ð15:46Þ

and  g2 ðT, PÞ ¼ VðT, PÞ ) dV ¼

∂V ∂T





∂V dT þ ∂P P

 dP

ð15:47Þ

T

where ð∂V=∂TÞP ¼ Vαp , and ð∂V=∂PÞT ¼ VκT

ð15:48Þ

Let us rewrite Eq. (15.45) for dS and Eq. (15.47) for dV in terms of CP, κT, and αP. Specifically,
 CP dT  ðVαP ÞdP T

ð15:49Þ

dV ¼ ðVαP ÞdT  ðVκT ÞdP

ð15:50Þ

dS ¼ and

To integrate Eqs. (15.49) and (15.50), we again utilize a convenient isobaricisothermal two-step path. Specifically, we integrate from an initial state i at (Po,To),

15.6

Determining Which Data Set Has the Same Thermodynamic Information. . .

145

characterized by S ¼ So (arbitrary) and V ¼ Vo(To,Po) (not arbitrary), to a final state f at (P,T). The (P-T) phase diagram in Fig. 15.2 illustrates the various paths involved. Integrating Eqs. (15.49) and (15.50) yields:

Fig. 15.2

SðT, PÞ ¼ So þ

ðT
 ðP CP jPo dT  ðVαP ÞjT dP T To

Po

ðT

ðP

VðT, PÞ ¼ Vo þ

ðVαP ÞjPo dT  To

ðVκT ÞjT dP

ð15:51Þ

ð15:52Þ

Po

Note that in Eqs. (15.51) and (15.52), the temperature integrations correspond to the isobaric path, and the pressure integrations correspond to the isothermal path (see Fig. 15.2). Because dμ ¼ S(T,P)dT + V(T,P)dP, we can use the expressions for S(T,P) in Eq. (15.51) and for V(T,P) in Eq. (15.52) in the expression for dμ and then integrate from (Po,To) to (P,T) to calculate μ(T,P) ¼ g3(T,P). To this end, the integration will be carried out along the same isobaric-isothermal two-step path shown in Fig. 15.2. The result of the integration is presented below: μðT, PÞ ¼ μo  ÐT ÐT CP  ÐP  So þ ðVαP Þj T dP dT T j Po dT  To

þ

ÐP Po



To

VoðTo, PoÞ þ

Po

ÐT To

ðVαP Þj Po dT 

Po

ÐP Po

ð15:53Þ



ðVκT Þj T dP dP T

where in Eq. (15.53), μo and So are both arbitrary constants of integration, and the pressure integration between Po and P, when P ¼ Po, is equal to zero. Rearranging Eq. (15.53) for μ(T,P), we obtain:

146

15

Properties of Pure Materials and Gibbs Free Energy Formulation

ðT μðT, PÞ ¼ ðμo  PoVoÞ  SoðT  ToÞ þ VoP  To

ðP þ Po

2T 3 ð
 CP 4 j dT5jPo dP T Po To

2T 3 ð ðP 4 ðVαP Þj Po dT  ðVκT Þj T dP5jT dP To

ð15:54Þ

Po

where in Eq. (15.54), μo –PoVo ¼ Ao, which is the Helmholtz free energy at (To, Po). In addition, in Eq. (15.54), Ao and So are arbitrary constants, while Vo(To,Po) is not. Finally, the thermodynamic information content of the Fundamental Equation, G ¼ μ (for n ¼ 1), is contained in the data set of CP, αP, and κT, or alternatively, of CP and the P-V-T volumetric equation of state, in addition to one value of the molar volume Vo in the reference state characterized by (Po,To). However, we can only determine Δμ ¼ ΔG to within the arbitrary constant of integration, So. In other words: ΔG1!2 ¼ Δμ1!2 ¼ SoðT2  T1 Þ þ VoðP2  P1 Þ þ Integrals

ð15:55Þ

In the special case of an isothermal process, Eq. (15.38) shows that: dμ ¼ dG ¼ SdT þ VdP ¼ VdP ðAt constant temperatureÞ

ð15:56Þ

Integrating Eq. (15.56) for dG from P1 to P2 at constant T yields: Pð2

ΔG1!2 jT ¼

VjT dP

ð15:57Þ

P1

Equation (15.57) shows that ΔG1!2 jT does not depend on any arbitrary constant of integration and, as shown earlier in Part I, represents the negative of the shaft work associated with a flowing n ¼ 1 fluid undergoing a reversible, isothermal process.

Lecture 16

Evaluation of Thermodynamic Data of Pure Materials and Sample Problems

16.1

Introduction

The material presented in this lecture is adapted from Chapter 8 in T&M. In this lecture, we will continue discussing pure materials (n ¼ 1). First, we will present a summary of the differentials of S, U, and H, expressed in terms of two sets of (n + 1) ¼ 2 intensive variables: (T, P) and (T, V). We will also consider the ideal gas limit and derive an expression relating the heat capacities at constant pressure and volume. Second, we will solve Sample Problem 16.1 to calculate the variation of the heat capacity at constant pressure with pressure at a given temperature. Third, we will solve Sample Problem 16.2 to calculate the variation of the heat capacity at constant volume with volume at a given temperature. Fourth, we will show how to calculate changes in thermodynamic properties given a mathematical expression relating P, T, and V (referred to as the equation of state, EOS) and different types of heat capacity data. In particular, we will present three strategies that can be used depending on the type of heat capacity data available to us. Fifth, we will present a three-step integration method, using either a (P-T) or a (V-T) phase diagram, which can be used when ideal gas heat capacity data is available (i.e., in the attenuated state, corresponding to the pressure approaching zero, or to the molar volume approaching infinity). Finally, we will demonstrate the use of this three-step integration method, referred to as the attenuated state approach, to calculate entropy changes when a pure material evolves from an initial state 1 to a final state 2, using either a (P-T) or a (VT) phase diagram.

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_16

147

148

16.2

16

Evaluation of Thermodynamic Data of Pure Materials and Sample Problems

Summary of Changes in Entropy, Internal Energy, and Enthalpy

16.2.1 Using T and P as the Two Independent Intensive Variables dS ¼

 
 CP ∂V dP dT  T ∂T P

         ∂V ∂V ∂V þ P dU ¼ CP  P dT  T dP ∂T P ∂T P ∂P T    ∂V dP dH ¼ CP dT þ V  T ∂T P

ð16:1Þ

ð16:2Þ

ð16:3Þ

16.2.2 Using T and V as the Two Independent Intensive Variables  
 CV ∂P dS ¼ dV dT þ T ∂T V

ð16:4Þ

    ∂P  P dV dU ¼ CV dT þ T ∂T V

ð16:5Þ

       ∂P ∂P ∂P dH ¼ CV þ V þV dT þ T dV ∂T V ∂T V ∂V T 

ð16:6Þ

16.2.3 The Ideal Gas Limit     ∂V R , ∂V RT PV ¼ RT ) ¼ ¼  2 P ∂T P ∂P T P

ð16:7Þ

16.3

Sample Problem 16.1

149

 PV ¼ RT )

∂P ∂T

 ¼ V

R, V



 ∂P RT ¼  2 ∂V T V

ð16:8Þ

16.2.4 Relation between CP and CV    CP ∂V dS ¼ dP dT  T ∂T P


 ∂S ¼ However, ∂T V rearranging, yields:

CV T

    ∂S C ∂V ∂P ¼ P T ∂T V ∂T P ∂T V

ð16:9Þ

ð16:10Þ

: Using this result in Eq. (16.10), multiplying by T, and

CP  CV ¼ T

    ∂V ∂P ∂T P ∂T V

ð16:11Þ

For an ideal gas (IG), Eqs. (16.7) and (16.8) show that:     ∂V R , ∂P R ¼ ¼ ∂T P P ∂T V V

ð16:12Þ

Using the results in Eq. (16.12) in Eq. (16.11) for an IG, we obtain: ðCP  CV ÞIG ¼ T

16.3

R R TR2 TR2 ¼ ¼ ¼R P V PV RT

ð16:13Þ

Sample Problem 16.1

Calculate how CP varies with P at constant T.

16.3.1 Solution Imagine that CP was measured as a function of temperature at a pressure, Po, and that we would like to know CP(T,P) at a different pressure P, without having to carry out any additional measurements. For this purpose, if we could calculate:

150

16

Evaluation of Thermodynamic Data of Pure Materials and Sample Problems

  ∂CP ∂P T

ð16:14Þ

it would follow that: ðP  CP ðT, PÞ ¼ CP ðT, PoÞ þ Po

∂CP ∂P

 dP

ð16:15Þ

T

To compute the desired partial derivative of CP with respect to P, at constant T, in the integrand in Eq. (16.15), we begin with: 

 ∂S CP ¼ T ∂T P

ð16:16Þ

Differentiating Eq. (16.16) with respect to P, at constant T, and using the fact that the order of differentiation is immaterial, we obtain:         ∂CP ∂ ∂S ∂ ∂S ¼ T ¼ T ∂P T ∂P ∂T P T ∂T ∂P T P

ð16:17Þ

Equation (16.1), repeated below for completeness, shows that:      
 CP ∂V ∂S ∂V dP ) ¼ dT  dS ¼ T ∂T P ∂P T ∂T P

ð16:18Þ

Using the result on the right-hand side of the arrow in Eq. (16.18) in Eq. (16.17), we obtain:    2  ∂CP ∂ V ¼ T ∂P T ∂T2 P

ð16:19Þ

Equation (16.19) shows that, for an ideal gas, for which V ¼ RT/P, it follows that: ð∂CP =∂PÞT ¼ 0

16.4

Sample Problem 16.2

Calculate how CV varies with V at constant T.

16.5

Evaluation of Changes in the Thermodynamic Properties of Pure Materials

151

16.4.1 Solution Given CV(T,Vo), in order to calculate CV(T,V) at another volume V, we begin with: ðV  CV ðT, VÞ ¼ CV ðT, VoÞ þ Vo

∂CV ∂V

 dV

ð16:20Þ

T

We know that:  CV ¼ T

 ∂S ∂T V

ð16:21Þ

and therefore that: 

∂CV ∂V

 T

∂ ¼T ∂V



∂S ∂T

  V T

∂ ¼ T ∂T



∂S ∂V

  ð16:22Þ T V

We also know that (see Eq. (16.4)):      
 CV ∂P ∂S ∂P dS ¼ dT þ dV ) ¼ T ∂T V ∂V T ∂T V

ð16:23Þ

Using the result on the right-hand side of the arrow in Eq. (16.23) in Eq. (16.22), we obtain:  2    ∂CV ∂ P ¼T ∂V T ∂T2 V

ð16:24Þ

Equation (16.24) shows that, for an ideal gas, for which P ¼ RT/V, it follows that: ð∂CV =∂VÞT ¼ 0

16.5

Evaluation of Changes in the Thermodynamic Properties of Pure Materials

16.5.1 Calculation of the Entropy Change Choosing the n + 1 ¼ 2 independent intensive variables, T and P, we know that: S ¼ S(T, P), and that (see Eq. (16.1)):

152

16

Evaluation of Thermodynamic Data of Pure Materials and Sample Problems

dS ¼




   CP ∂V dP dT  T ∂T P

ð16:25Þ

16.5.2 Strategy I If CP(T,P) is known, and the volumetric P-V-T equation of state (EOS) is known, as we did in Lecture 15, we can integrate Eq. (16.25) directly using an isobaricisothermal two-step path as follows: T ð2

ΔS1!2 ¼ T1






Pð2  



CP

∂V

dT  dP

T

∂T P

P1

P1

ð16:26Þ

T2

In Eq. (16.26), the temperature integration corresponds to the isobaric path at P1, and the pressure integration corresponds to the isothermal path at T2.

16.5.3 Strategy II If CP(T,Po) is known, and the volumetric P-V-T EOS is known, we can first calculate CP(T,P) by integrating Eq. (16.19). Specifically, ðP CP ðT, PÞ ¼ CP ðT, PoÞ 

T Po

 2  ∂ V dP ∂T2 P

ð16:27Þ

Following that, we can use Strategy I.

16.5.4 Strategy III If we know the ideal gas heat capacity at constant pressure, CoP ¼ a* + b*T + c*T2 + . . ., (see Lecture 4) corresponding to P!P*!0, referred to as an attenuated state, as well as the volumetric P-V-T EOS, we can choose the following three-step path in a (P-T) phase diagram to evaluate ΔS1!2 when the system evolves from an initial state 1 (at P1, T1) to a final state 2 (at P2, T2). The schematic (P-T) phase diagram in Fig. 16.1 illustrates the three-step path involved:

16.5

Evaluation of Changes in the Thermodynamic Properties of Pure Materials

153

Fig. 16.1

Following the three-step path shown in Fig. (16.1) yields: ΔS1!2 ¼ ΔS1!a þ ΔSa!b þ ΔSb!2

ð16:28Þ

Next, we will calculate each of the three entropy contributions on the right-hand side of the equal sign in Eq. (16.28). 1. Path 1!a: Isothermal P*¼0 ð 

ΔS1!a ¼  P1



∂V

dP ∂T P T1

ð16:29Þ

In Eq. (16.29), when P* ¼ 0, V ¼ RT/P, (∂V/∂T)P ¼ R/P, and there is a log (P* ¼ 0) divergence. Further, the partial derivative in the integrand in Eq. (16.29) can be evaluated if a volumetric P-V-T EOS in available. 2. Path a!b: Isobaric T ð2 

ΔSa!b ¼



CoP

dT T P*

ð16:30Þ

T1

where the temperature integration is only carried out in the attenuated state, and where CPo is known. Further, holding P* constant in the CPo/T term in Eq. (16.30) is redundant, because CPo is independent of pressure.

154

16

Evaluation of Thermodynamic Data of Pure Materials and Sample Problems

3. Path b!2: Isothermal Pð2

ΔSb!2 ¼  P*¼0

 

∂V

dP ∂T P T2

ð16:31Þ

In Eq. (16.31), when P* ¼ 0, V ¼ RT/P, (∂V/∂T)P ¼ R/P, and there is a log (P* ¼ 0) divergence. Further, the partial derivative in the integrand in Eq. (16.31) can be evaluated if a volumetric P-V-T EOS is available. 4. How do we deal with the divergences that exist at P* ¼ 0 when carrying out the pressure integrals isothermally along path 1!a (see Eq. (16.19)) and path b!2 (see Eq. (16.31))? Because the system behaves ideally when P* ¼ 0, it is possible to express the singularity in analytical form as follows: (a) In Eq. (16.29), we add and subtract: P*¼0 ð




R

P

P1

dP

ð16:32Þ

T1

which yields: P*¼0 ð

ΔS1!a ¼  P1



    ∂V R P*  dP  R ln P1 ∂T P P T1

ð16:33Þ

In the limit when P* ! 0: V!

RT ) P

  ∂V R ! P ∂T P

ð16:34Þ

Eq. (16.34) in the integrand in Eq. (16.33), it follows that h  Using i ∂V R  ! 0 when P* ! 0, thereby removing the singularity from the integral. P ∂T P (b) In Eq. (16.31), we add and subtract: Pð2

P*¼0

which yields:




R

dP P

T2

ð16:35Þ

16.5

Evaluation of Changes in the Thermodynamic Properties of Pure Materials Pð2



ΔSb!2 ¼  P*¼0

∂V ∂T

  P

R P

 dP  Rln T2




P2 P*



155

ð16:36Þ

Like in Eq. (16.33), the integrand in Eq. (16.36) !0 in the limit P*!0 and the singularity is removed from the integral. When we calculate ΔS1!a + ΔSb!2 (see Eqs. (16.33) and (16.36)), the two ln (P*!0) contributions cancel out. Specifically, we obtain: P*¼0 ð

ΔS1!a þ ΔSb!2 ¼  P1 Pð2

 P*¼0

     ∂V R P  dP  Rln 2 P P1 ∂T P T1    ∂V R  dP ∂T P P T2

ð16:37Þ

In Eq. (16.37), the first pressure integration corresponds to the real gas state to the ideal gas state transition at T1, and the second pressure integration corresponds to the ideal gas state to the real gas state transition at T2. If the two independent intensive variables are V and T, we can use a (V-T) phase diagram to calculate ΔS1!2. In this case, the ideal gas attenuated state corresponds to V* ¼ 1, and we can choose the convenient three-step path shown in Fig. 16.2:

Fig. 16.2

In the attenuated state at V* ¼ 1, the gas behaves ideally with a heat capacity at constant volume given by: CVo ¼ a + bT + cT2 + ... (see Lecture 4). If CVo and the volumetric P-V-T EOS are known, we can calculate ΔS1!2 as follows (see Fig. 16.2):

156

16

Evaluation of Thermodynamic Data of Pure Materials and Sample Problems

ΔS1!2 ¼ ΔS1!a þ ΔSa!b þ ΔSb!2

ð16:38Þ

Next, we calculate each of the three entropy contributions on the right-hand side of the equal sign in Eq. (16.38). 1. Path 1!a: Isothermal 

∂P

dV ∂T V T1

V*ð ¼1 

ΔS1!a ¼ V1

ð16:39Þ

 , ∂P ¼ R , and there is a log (V* ¼ 1) In Eq. (16.39), when V* ¼ 1, P ¼ RT V V ∂T V divergence. In addition, the partial derivative in the integrand in Eq. (16.39) can be evaluated if a volumetric P-V-T EOS is available. 2. Path a!b: Isochoric Tð2

ΔSa!b ¼

 o 

Cv

dV T V*

ð16:40Þ

T1

The temperature integration in Eq. (16.40) is only carried out in the attenuated state, where CVo is known. In addition, holding V* constant in the (CVo/T) term in Eq. (16.40) is redundant, because CVo is independent of the molar volume. 3. Path b!2: Isothermal V ð2

ΔSb!2 ¼ V* ¼ 1



∂P ∂T







dV

ð16:41Þ

V T2

In Eq. (16.41), when V* ¼ 1, P ¼ RT/V, (∂P/∂T)V ¼ R/V, and there is a log (V* ¼ 1) divergence. In addition, the partial derivative in the integrand in Eq. (16.41) can be evaluated if a volumetric P-V-T EOS is available. In order to deal with the divergences in ΔS1!a and ΔSb!2 when V* ¼ 1 (see Eqs. (16.39) and (16.40)), we add and subtract: V*¼1 ð

V1

and


 R dV for ΔS1!a V T1

ð16:42Þ

16.5

Evaluation of Changes in the Thermodynamic Properties of Pure Materials V ð2


 R dV for ΔSb!2 V T2

157

ð16:43Þ

V*¼1

This yields: V*¼1 ð



ΔS1!a þ ΔSb!2 ¼ V1 V ð2

þ V*¼1

∂P ∂T





R  V V

∂P ∂T







T1

R  V V

V2 dV þ Rln V1



 dV

ð16:44Þ

T2

In Eq. (16.44), the first molar volume integration corresponds to the real gas state to the ideal gas state transition at T1, and the second molar volume integration corresponds to the ideal gas state to the real gas state transition at T2.

Lecture 17

Equations of State of a Pure Material, Binodal, Spinodal, Critical Point, and Sample Problem

17.1

Introduction

The material presented in this lecture is adapted from Chapters 8 and 7 in T&M. In this lecture, we will continue discussing pure materials (n ¼ 1). First, we will discuss the mathematical relation between P, V, and T, referred to as the volumetric equation of state, or in short, the equation of state (EOS). In particular, we will discuss the ideal gas EOS and the van der Waals EOS, including providing an underlying molecular interpretation for both EOS. Second, we will solve Sample Problem 17.1 to calculate the excluded volume between two spheres of equal radius. Third, we will examine the various forms of the isotherms in a pressure (P)-volume (V) phase diagram at temperatures which are high, equal, or low relative to the critical temperature. Fourth, we will discuss the coexistence curve, the spinodal curve, and the critical point, including providing mathematical criteria to calculate them. Finally, we will discuss stability, metastability, and instability, including providing mathematical criteria to characterize these behaviors, as well as useful mechanical analogies to rationalize what each behavior entails.

17.2

Equations of State of a Pure Material

In Lecture 16, we saw that knowledge of the relation between P, V, and T is essential for the calculation of thermodynamic properties of a pure (one-component, n ¼ 1) system. The P-V-T relation is known as the volumetric equation of state, or in short, the equation of state (EOS), and is available as a: Pressure  Explicit EOS : P ¼ f ðT, VÞ

ð17:1Þ

or © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_17

159

160

17

Equations of State of a Pure Material, Binodal, Spinodal, Critical Point. . .

Volume  Explicit EOS : V ¼ f ðT, PÞ

ð17:2Þ

Equations of state may be very simple, such as the ideal gas EOS which contains no parameters, or very complex, such as the Martin-Hou EOS which contains 21 parameters. The choice of EOS depends on (1) the desired accuracy and (2) the endurance of the user. The various parameters which appear in the EOS are evaluated by fitting the EOS to experimental P-V-T data. Therefore, an EOS can never be more accurate than the experimental P-V-T data that it describes.

17.3

Examples of Equations of State (EOS)

17.3.1 The Ideal Gas EOS Applicable when P is low (~5 atm), or V is high, or ρ ¼ 1/V is low. The ideal gas EOS can be written as follows: Intensive Form : PV ¼ RT

ð17:3Þ

Extensive Form : PV ¼ NRT

ð17:4Þ

and

In Eqs. (17.3) and (17.4), the gas constant R ¼ 8.314 J/mol K.

17.3.2 The van der Waals EOS The ideal gas EOS assumes that: (i) Molecules have no volume, such that the entire system volume, V, is available to each of the N molecules comprising the system (ii) No interactions operate between the N molecules comprising the system Assumptions (i) and (ii) result in Eqs. (17.4) and (17.3). van der Waals relaxed assumptions (i) and (ii) as follows: (1) Every molecule excludes a volume, b, from every other molecule, such that the free volume available in the system is (V-Nb)

17.4

17.4

Sample Problem 17.1

161

Sample Problem 17.1

Evaluate the excluded volume between two hard spheres of equal radius, Ro, (Fig. 17.1).

Fig. 17.1

17.4.1 Solution As Fig. 17.1 shows, the distance of closest approach between the centers, 0 and 00, of the two hard spheres is equal to 2Ro. The volume excluded by the sphere of radius Ro, centered at 0, from the center of the second sphere of radius Ro, centered at 00, is given by: Vexcluded ¼


 4π 4π 3 ð2 RoÞ3 ¼ 8 Ro 3 3

ð17:5Þ

Note that: b ¼ 0.5Vexcluded (per sphere/molecule). The assumption of additive excluded volumes is reasonable at low densities, where there is no overlap of excluded volumes (see Fig. 17.2):

Fig. 17.2

162

17

Equations of State of a Pure Material, Binodal, Spinodal, Critical Point. . .

However, as Fig. 17.2 shows, at high densities, the excluded volume calculated assuming that the excluded volumes are independent is overpredicted. (2) Because liquids and solids exist in nature, van der Waals recognized the inevitable existence of attractive interactions between molecules, which he described as pairwise and proportional to the number density squared. If each pair of molecules experiences an attraction of strength, a, the resulting attractive contribution to the system pressure is given by:  2 N aρ ¼ a V 2

ð17:6Þ

van der Waals (vdW) used assumption (1) to replace V in Eq. (17.4) by (V-Nb), and assumption (2) to replace the pressure P in Eq. (17.4) by (P + a (N/V)2), and proposed the celebrated vdW EOS given by: "

 2 # N ½V  Nb ¼ NRT Pþa V

ð17:7Þ

Equation (17.7) can be expressed in the following two forms: P ¼

P ¼

NRT aN2  ðPressure‐explicit extensive form of the vdW EOSÞ V  Nb V2 ð17:8Þ

RT a  2 ðPressure‐explicit intensive form of the vdW EOSÞ ð17:9Þ Vb V

Note that when V ! 1, Eq. (17.8) reduces to Eq. (17.4), and when V ! 1, Eq. (17.9) reduces to Eq. (17.3). Finally, as we will show in Lecture 18, the two parameters, a and b, in the vdW EOS can be determined using the experimental critical point conditions.

17.5

Pressure-Explicit Form of the Isotherm P = f (V, T) of a Pure Material

Figure 17.3 shows schematically the family of isotherms in a (P-V) phase diagram for a pure (n ¼ 1) material (e.g., carbon dioxide).

17.5

Pressure-Explicit Form of the Isotherm P ¼ f (V, T) of a Pure Material

163

Fig. 17.3

The following observations can be made about Fig. 17.3: (1) At sufficiently high temperatures, each isotherm is a continuous curve, where as V increases, P decreases. In other words, on isotherms such as 1 and 2, (∂P/ ∂V)T < 0. (2) At sufficiently low temperatures, each isotherm is discontinuous and consists of three sections. For example, in isotherm 5, the first section of the curve at high pressures corresponds to the liquid state, while the third section of the curve at low pressures corresponds to the vapor (gas) state. (3) The two curves (sections) in (2) are joined by a horizontal line which corresponds to the simultaneous presence of two phases, liquid and vapor (gas), coexisting at thermodynamic equilibrium. (4) Isotherm 3 corresponds to the transition between isotherms corresponding to the vapor phase only (like 1 and 2) and those which include a horizontal section and correspond to liquid-vapor equilibrium (like 5). In isotherm 3, the horizontal

164

17

Equations of State of a Pure Material, Binodal, Spinodal, Critical Point. . .

section has contracted to a single point of inflection, which is the critical point of the system, denoted as CP. Isotherm 3 is referred to as the critical isotherm, for which T ¼ TC. Mathematically, therefore, the CP is characterized by the conditions for the existence of a point of inflection with a horizontal tangent (see below). (5) The curve LL’CP gives the molar volumes of the liquid phases in equilibrium with the vapor phases, at various temperatures below TC, as a function of P. Similarly, the curve VV’CP gives the molar volumes of the vapor phases in equilibrium with the liquid phases, at various temperatures below TC, as a function of P. The complete curve LL’CPV’V is referred to as the L/V coexistence curve, the binodal, or the saturation curve. (6) The molar volume of the vapor, VV, decreases, or equivalently, the vapor number density, ρV ¼ 1/VV, increases as T increases toward TC from below. On the other hand, the molar volume of the liquid, VL, increases, or equivalently, the liquid number density, ρL ¼ 1/VL, decreases as T increases toward TC from below. As a result, at the critical point, CP, the molar volumes of the vapor and the liquid are equal to VC, and the corresponding pressure is PC. (7) In general, a critical state is characterized by the fact that the two coexisting phases (in the present case, liquid and vapor) are identical. Above the critical point, that is, for T > TC, the pure substance (n ¼ 1) can exist in true equilibrium only in the vapor state. In other words, the critical temperature, TC, is the highest temperature at which the liquid and the vapor can coexist in true equilibrium. The existence of a critical point makes it possible to pass from one physical state to another without ever observing the appearance of a new phase. Figure 17.4 illustrates this point.

Fig. 17.4

17.5

Pressure-Explicit Form of the Isotherm P ¼ f (V, T) of a Pure Material

165

Indeed, starting with vapor at A, increasing T, at constant V, along the isochoric path AB to T > TC, then compressing the system while maintaining T > TC (T changes) from B to B0 , and finally decreasing T to below TC, at constant V, from B0 to D, we end with liquid at D. Therefore, we can pass in a continuous manner from the vapor state at A to the liquid state at D. The continuity of state, shown in Fig. 17.4, which results from the existence of a critical point, indicates that, in a certain sense, the vapor and liquid states are two different manifestations of the same physical state (a fluid state). This was recognized by Kelvin who suggested that the segments DL and AV in Fig. 17.4, as well as in the (P-V) phase diagram in Fig. 17.5, are really parts of the single continuous curve DLMNVA.

Fig. 17.5

The idea of a continuous isotherm for T < TC, shown in Fig. 17.5, was subsequently picked up by van der Waals, and developed further into the celebrated vdW EOS, which is cubic in V. Specifically,  

 RT 2 a ab V þ V ¼0 V3  b þ P P P

ð17:10Þ

The (P-V) phase diagram in Fig. 17.6 shows the form of the isotherm of the vdW EOS for T < TC.

166

17

Equations of State of a Pure Material, Binodal, Spinodal, Critical Point. . .

Fig. 17.6

Figure 17.6 shows that, for T < TC, the isotherm has the required sigmoidal shape. At a given value of P (see the horizontal dashed line), there are three solutions: the smallest one at 1 corresponds to VL, the middle one at 2 is unphysical, and the largest one at 3 corresponds to VV. In addition, Fig. 17.5 shows that, for T > TC, there is a single real solution corresponding to VV for every P value. The resulting isotherm satisfies (∂P/ ∂V)T < 0. Finally, for T ¼ TC, there are three identical solutions corresponding to the CP located at PC and VC.

17.6

Stable, Metastable, and Unstable Equilibrium

As shown in the (P-V) phase diagram in Fig. 17.5, the various portions of the DLMNVA continuous curve have distinct physical significance. First, between M (a minimum) and N (a maximum), we have: 

∂P ∂V

 >0

ð17:11Þ

T

which indicates that the states between M and N are mechanically unstable (that is, when P increases, V increases as well). As a result, such states are not realizable in practice. The portion VN corresponds to an overcompressed (supersaturated) vapor, which can exist in a metastable state which will disappear spontaneously if condensation nuclei are introduced in the system. This is precisely what happens in clouds (large masses of supersaturated vapor), which upon seeding with silver halide particles, produce liquid drops which subsequently fall under gravity as artificial rain drops.

17.7

Mechanical Analogy of Stable, Metastable, and Unstable Equilibrium States

167

Similarly, the portion LM corresponds to an overexpanded liquid, which again is metastable. The points M and N are therefore boundary points between metastable and unstable states of the system. For every isotherm having T < TC, there are pairs of points like M and N. If we join all the M points and all the N points with the curve aMCPNb (see Fig. 17.7), the resulting curve is referred to as the spinodal, a curve which separates unstable states, for which (∂P/∂V)T > 0, from metastable states, for which (∂P/∂V)T < 0. Clearly, the mathematical condition for the spinodal is given by: 

∂P ∂V

 ¼0

ð17:12Þ

T

The CP is located at the maximum of the spinodal where it touches the binodal. Therefore, the binodal and the spinodal divide the (P-V) phase diagram into stable, metastable, and unstable regions. The (P-V) phase diagram in Fig. 17.7 illustrates the various features discussed above:

Fig. 17.7

17.7

Mechanical Analogy of Stable, Metastable, and Unstable Equilibrium States

Figure 17.8 provides mechanical analogs of a stable equilibrium state (a ball resting at the bottom of a valley), an unstable equilibrium state (a ball resting at the top of a hill), and a metastable equilibrium state (a ball resting at the bottom of a shallow local minimum separated from a deeper global minimum by an energy barrier).

168

17

Equations of State of a Pure Material, Binodal, Spinodal, Critical Point. . .

Fig. 17.8

17.8

Mathematical Conditions for Stability, Metastability, and Instability 

 ∂P 2Ro

Using statistical mechanics, we can show that (see Part III):

ð18:20Þ

176

18

The Principle of Corresponding States and Sample Problems

N B ¼ A ð4πÞ 2

1 ð



 Uij r2 dr 1  exp  kB T

ð18:21Þ

0

where NA is Avogadro’s number and kB is the Boltzmann constant. Using the expression for Uij in Eq. (18.20) in the expression for B in Eq. (18.21) yields:

B ¼

8 2Ro < ð

NA ð4πÞ 2 :

1 ð

½1  0 r2 dr þ

0

9 = ½1  1 r2 dr ;

ð18:22Þ

2Ro

In Eq. (18.22), the first integration yields 8Ro3, and the second integration yields zero. Using these results in Eq. (18.22) yields: B ¼

h i NA 4π Ro3 8 3 2

ð18:23Þ

16π Ro3 NA 3

ð18:24Þ

or B ¼

In Fig. 18.2, the volume of the dashed sphere corresponds to the volume excluded by a sphere of radius Ro from the center of a second sphere of radius Ro which touches the first sphere and is given by 8(4π/3)Ro3.

Fig. 18.2

As discussed in Lecture 17, the parameter, b, in the van der Waals EOS reflects the excluded-volume interactions between two molecules. Therefore, b should be related to B calculated above. To show this, we begin with the vdW EOS, repeated here for completeness (see Eq. (18.5)):

18.3

The Principle of Corresponding States

P ¼

177

RT a  V  b V2

ð18:25Þ

We then assume that no attractive interactions operate between the molecules by setting a ¼ 0 in Eq. (18.25). Following that, we expand Eq. (18.5), with a ¼ 0, in powers of 1/V to second order. This yields: P ¼

RT bRT þ 2 V V

ð18:26Þ

Comparing the virial EOS in Eq. (18.17), truncated to second order in 1/V, with Eq. (18.26) shows that: b¼B ¼

16π Ro3 NA 3

ð18:27Þ

The important result in Eq. (18.27) shows that, if we know the chemical structure of the molecule, we can estimate Ro and then compute b at the molecular level.

18.3

The Principle of Corresponding States

“All fluids in corresponding states (same Pr and Tr) have the same reduced volume, Vr.”

18.3.1 The Compressibility Factor

Z ¼

PV PV ¼ NRT RT

ð18:28Þ

Equation (18.28) shows that: (i) for an ideal gas (IG), ZIG ¼ 1, (ii) knowledge of Z ¼ Z (T,P) or Z ¼ Z (T,V) is equivalent to knowing the EOS. Recalling that P ¼ PrPC, V ¼ VrVC, and T ¼ TrTC, and then using these P, V, and T expressions in Eq. (18.28), we obtain: Z ¼ ZC where

Pr Vr Tr

ð18:29Þ

178

18

The Principle of Corresponding States and Sample Problems

ZC ¼

PC VC RTC

ð18:30Þ

Table 18.1 shows predictions of equations of state for the critical compressibility factor. Based on the expression for Z in Eq. (18.29), we can ask: “Can all the fluids in nature be correlated from knowledge of their Pr and Tr values,” as predicted by the Principle of Corresponding States (POCS)? According to the POCS, Vr is a universal function of Pr and Tr. As a result, for all the fluids with the same ZC, Z should be universal too. Table 18.1 Predictions of equations of state for ZC Equation of state Ideal van der Waals (vdW) Redlich-Kwong (RK) Redlich-Kwong-Soave (RKS) Peng-Robinson (PR) “Best Fit” Martin

ZC 1 0.375 0.333 0.333 0.307 0.25

Table 18.2 Experimental values of ZC for various fluids Substance He H2 C02 S02 C6H6 NH3 H20

ZC 0.3141 0.3049 0.2869 0.2774 0.2663 0.2420 0.2290

Table 18.2 shows that for various fluids, 0.22 < ZC < 0.32. Because the experimental ZC values are not identical, the experimental Z (Pr, Tr) values are not universal, contrary to the prediction made by the POCS (see Fig. 18.3).

18.3

The Principle of Corresponding States

179

1.1

1.0

Tr = 2.00

0.9

Compressibility Factor, Z = PV/NRT

Tr = 1.50 0.8

Tr = 1.30

0.7

0.6

Tr = 1.20

0.5

Tr = 1.10 0.4

Tr = 1.00

0.3

0.2

Methane

Iso-pentane

Ethylene

n-Heptane

Ethane

Nitrogen

Propane

Carbo n dioxide

n-Butane

Water

Average curve based on data on hydrocarbons

0.1 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 [Reprinted from Yunus A. Cengel & Michael A. Boles, 2005, McGraw Hill, Figure 3-51, p 141, based on data from Gouq-Jen Su, 1946, Ind. Eng. Chem. 38, p 803.]

Fig. 18.3 Gas compressibilities as a function of reduced pressure and temperature

Because the original two-parameter (Pr,Tr) POCS cannot reproduce the experimental data in Fig. 18.3 for all the fluids considered, it is convenient to introduce a third differentiating parameter. A natural first choice is ZC. In terms of Pr, Tr, and ZC, we can generalize the original POCS as follows: Z ¼ Z ðTr , Pr , ZC Þ

ð18:31Þ

Another choice for the third differentiating parameter is Pitzer’s acentric factor, ω, defined as follows: h i ω ¼ 1:0  log 10 ðPr sat ÞTr ¼0:7 ð18:32Þ where Prsat is the reduced vapor pressure and is evaluated at Tr ¼ 0.7. In Sample Problem 18.4, we present a derivation of Eq. (18.32).

180

18

The Principle of Corresponding States and Sample Problems

18.3.2 Sample Problem 18.4 Derive Eq. (18.32) for ω.

18.3.2.1

Solution

The Clausius-Clapeyron equation, which we will derive in Part II, states that:   d lnPvap ΔHvap ð18:33Þ ¼ R d ð1=TÞ where Pvap is the fluid vapor pressure, and ΔHvap is the fluid molar enthalpy of vaporization. If we assume that ΔHvap is independent of temperature, Eq. (18.33) can be expressed as follows:   d lnPvap ΔHvap dðlnPr sat Þ ¼ a ¼ constant ð18:34Þ ¼ ¼ R d ð1=TÞ d ð1=Tr Þ where a is the slope of ln (Prsat) vs. 1/Tr. If the two-parameter principle of corresponding states was generally valid, the slope a would be the same for all the pure (n ¼ 1) fluids. In practice, we find that each fluid has its own characteristic value of a, which can therefore serve as a third differentiating parameter. Pitzer recognized that all the vapor pressure data for the simple fluids (SF: Ar, Kr, Xe) lie on the same line when plotted as log10 (Prsat) vs. (1/Tr) and that the line passes through log10 (Prsat) ¼ 1.0 at Tr ¼ 0.7. Data for other fluids define other lines whose location can be fixed in relation to the line corresponding to the simple fluids by the difference: ½ log 10 ðPr sat ðSFÞÞ  log 10 ðPr sat ÞTr ¼0:7

ð18:35Þ

The Pitzer acentric factor, ω, is defined as this difference, that is, ω ¼ log 10 ðPr sat ðSFÞÞ jTr ¼0:7  log 10 ðPr sat ÞjTr ¼0:7

ð18:36Þ

The first term on the right-hand side of the equal sign in Eq. (18.36) is equal to 1.0, and therefore, Eq. (18.36) can be expressed as follows: ω ¼ 1:0  log 10 ðPr sat Þ jTr ¼0:7

ð18:37Þ

Equation (18.37) completes our derivation. To actually determine ω, we need to know TC, PC, and a single vapor pressure value at Tr ¼ 0.7. By definition, ω ¼ 0 for the simple fluids (Ar, Kr, Xe). Values of ω, as well as critical point coordinates, for other fluids, are listed in Table 18.3.

18.3

The Principle of Corresponding States

181

Table 18.3 Critical point coordinates and Pitzer’s acentric factors

It is noteworthy that, if we do not have access to an EOS, the ability to calculate Z at a given T and P enables us to calculate V ¼ RTZ/P. Of course, even if an EOS is available, as shown next, we can use the generalized three-parameter Pitzer correlation approach to compute Z.

182

18

The Principle of Corresponding States and Sample Problems

It is possible to express Z as an expansion in powers of ω truncated to linear order. Specifically, Z ¼ Z0 ðTr , Pr Þ þ ωZ1 ðTr , Pr Þ

ð18:38Þ

Interestingly, Z0 and Z1 are available graphically (see Figs. 18.4, 18.5, 18.6, and 18.7). In Lecture 19, we will utilize Eq. (18.38), along with Figs. 18.4 and 18.5, to solve an interesting sample problem.

Fig. 18.4 Generalized correlation for Z0, Pr < 1.0

18.3

The Principle of Corresponding States

Fig. 18.5 Generalized correlation for Z1, Pr < 1.0

183

184

18

The Principle of Corresponding States and Sample Problems

Fig. 18.6 Generalized correlation for Z0, Pr > 1.0

18.3

The Principle of Corresponding States

Fig. 18.7 Generalized correlation for Z1, Pr > 1.0

185

Lecture 19

Departure Functions and Sample Problems

19.1

Introduction

The material presented in this lecture is adapted from Chapter 8 in T&M. In this lecture, we will continue discussing pure materials (n ¼ 1). First, we will solve Sample Problem19.1 to calculate the molar volume of n-butane at a given temperature and pressure. For this purpose, we will first use the ideal gas EOS and then can use more realistic EOS. In addition, we will use the Generalized Pitzer Correlation Approach presented in Lecture 18. Second, we will discuss the departure function (DF) approach to calculate changes in thermodynamic properties, including the entropy departure function, DS, and the Helmholtz free energy departure function, DA. Third, we will solve Sample Problem 19.2 to calculate the internal energy departure function, DU. Finally, in order to calculate isothermal variations of A or G, we will show that it is simpler to directly integrate the differentials of A or G, instead of using departure functions.

19.2

Sample Problem 19.1

Calculate the molar volume of n-butane at 510 K and 25 bar. You can assume that, for n-butane, Tc ¼ 425.2 K, Pc ¼ 38 bar, and ω ¼ 0.193.

19.2.1 Solution The (P-T) phase diagram in Fig. 19.1 shows the L/V equilibrium line, the CP, and the operating conditions (the black circle). © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_19

187

188

19

Departure Functions and Sample Problems

Fig. 19.1

Because the temperature and the pressure of interest are known, we can use an EOS to calculate the molar volume. For example, we can: (a) Use the ideal gas EOS V ¼

RT ¼ 1696:1 cm3 =mol P

ð19:1Þ

where we have used the gas constant R, T ¼ 510 K, and P ¼ 25 bar. (b) Use other EOS For example, the vdW EOS, the RK EOS, or the PR EOS discussed in Lecture 18. To estimate the various parameters in these EOS, we can use the given values of TC, PC, and ω as needed. We would then need to solve a cubic equation for V, where only one solution will be physical in the gas phase (25 bar, 510 K). Alternatively, if an EOS is not available, we can use the Generalized Pitzer Correlation Approach discussed in Lecture 18, as shown in (c) below. (c) Generalized Pitzer Correlation Approach We know that: V ¼

RTZ P

ð19:2Þ

where Z ¼ Z0ðTr , Pr Þ þ ω Z1ðTr , Pr Þ

ð19:3Þ

To use the graphs for Z0 and Z1 presented in Lecture 18, we first calculate Tr and Pr corresponding to the T and P values given in the Problem Statement. Specifically, Tr ¼ T=TC ¼ 510=425:2 ¼ 1:198

ð19:4Þ

19.2

Sample Problem 19.1

189

Pr ¼ P=PC ¼ 25=38 ¼ 0:658

ð19:5Þ

Because Pr < 1.0, we need to use the Generalized Correlation Graphs for Z0 and Z corresponding to Pr < 1.0, first introduced in Lecture 18, and shown again in Fig. 19.1 and Fig. 19.3, respectively, for completeness. 1

Fig. 19.2 Generalized correlation for Z0, Pr < 1.0

190

19

Departure Functions and Sample Problems

Fig. 19.3 Generalized correlation for Z1, Pr < 1.0

19.3

Departure Functions

191

• Using the graph for Z0 in Fig. 19.2, at Tr ¼ 1.198 and Pr ¼ 0.658, yields: Z0 ¼ 0:865 ) Vo ¼

Z0 RT cm3 ’ 1467 P mol

ð19:6Þ

• Using the graph for Z1 in Fig. 19.3, at the same Tr and Pr values, yields: Z1 ’ 0:038

ð19:7Þ

Using the calculated values for Z0 and Z1 in Eq. (19.6) and Eq. (19.7), respectively, as well as ω ¼ 0.193, in Eq. (19.3) yields: Z ¼ Z0 þ ω Z1 ’ 0:872

ð19:8Þ

Using the calculated Z value in Eq. (19.8), along with the T and P values given in the Problem Statement, in Eq. (19.2) for V yields the desired result: V¼

ZRT cm3 ’ 1479 P mol

ð19:9Þ

Interestingly, the predicted molar volume of n-butane in Eq. (19.9) is in remarkably good agreement with the experimental value given below: Vexp ¼ 1480:7 cm3 =mol

19.3

ð19:10Þ

Departure Functions

In Lecture 16, we discussed the three-step integration approach which can be used to calculate a thermodynamic property change. This approach combines two isothermal property changes from the real to an attenuated, ideal gas (IG) state (P* ¼ 0 or V* ¼ 1) and a temperature variation in the attenuated state. As we will show in this lecture, the isothermal variation can be more formally defined in terms of a so-called departure function of a property, defined as follows: “A Departure Function is the difference between the property of interest in its real state at a specified (T,P) or (T,V), and in an Ideal Gas State at the same T and P.” Therefore, there are two equivalent formulations of departure functions:

192

19

Departure Functions and Sample Problems

ð1Þ BðT, PÞ  BIG ðT, PÞ ¼ BðT, PÞ  Bo ðT, PÞ

ð19:11Þ

and ð2Þ BðT, VÞ  BIG ðT, Vo Þ ¼ BðT, VÞ  Bo ðT, Vo Þ, where Vo ¼

RT P

ð19:12Þ

In Eqs. (19.11) and (19.12), B is any derived property of the system (S, H, U, A, G, etc.). Hereafter, we will abbreviate departure function as DF.

19.4

Calculation of the Entropy Departure Function, DS(T, P)

If S ¼ S(T,P), then:    CP ∂V dT  dP dS ¼ T ∂T P


ð19:13Þ

At constant temperature, integrating Eq. (19.13) with respect to pressure, from the ideal gas state at P!0 to the real state at P, yields: ðP   ∂V SðT, PÞ  S ðT, P! 0Þ ¼  dP ∂T P o

ð19:14Þ

0

Further, at constant temperature, integrating Eq. (19.13) with respect to pressure, from the ideal gas state at P!0 to an ideal gas state at a low P value, yields: ðP
 R S ðT, PÞ  S ðT, P! 0Þ ¼  dP P o

o

ð19:15Þ

0

In Eqs. (19.14) and (19.15), the superscript o in S denotes an ideal gas when P ¼ 0. In Eq. (19.15), because we are connecting two ideal gas states, V ¼ RT/P and (∂V/∂T)P ¼ R/P in the integrand. Subtracting Eq. (19.15) from Eq. (19.14), cancelling the two equal terms on the left-hand side, and combining the two integrals, yields the entropy departure function, DS(T,P), given by:

19.5

Calculation of the Helmholtz Free Energy Departure Function, DA(T, V)

ðP  DSðT, PÞ ¼ SðT, PÞ  S ðT, PÞ ¼  o

0

∂V ∂T

  P

 R dP P

193

ð19:16Þ

Note that, in Lecture 16, we also obtained the integral in Eq. (19.16) when we removed the apparent divergence in the limit P*!0. In essence, we used a DF to do that. Departure functions can always be computed using a P-V-T EOS. Because most P-V-T EOS are pressure explicit (with V and T as the independent variables), this suggests that we should first calculate the departure function of the molar Helmholtz free energy A(T,V), from which we should be able to calculate all the other departure functions (see below).

19.5

Calculation of the Helmholtz Free Energy Departure Function, DA(T, V)

Recall that: A ¼ AðT, VÞ ) dA ¼ SdT  PdV

ð19:17Þ

At constant T, integrating dA in Eq. (19.17) with respect to V from the ideal gas state at V ¼ 1 to the real state at V yields: ðV A ðT, VÞ  A ðT, 1Þ ¼  o

ð19:18Þ

PdV 1

In Eq. (19.18), Ao(T, 1) carries the superscript o to indicate that the system is an ideal gas when V ¼ 1. Because the behavior between V ¼ Vo and V ¼ 1 corresponds to that of an ideal gas (IG), it follows that:    Ao ðT, 1Þ  Ao ðT, Vo Þ ¼  PdV  o 1 ð

V

ð19:19Þ IG

Adding Eqs. (19.18) and (19.19), cancelling the two equal terms on the left-hand side, and combining the two integrals, yields:

194

19

Departure Functions and Sample Problems

ðV AðT, VÞ  A ðT, V Þ ¼  o

1 ð

PdV 

o

1

RT dV V

ð19:20Þ

o

V

where in Eq. (19.19), PdV|IG is equal to (RT/V)dV. In Eq. (19.20), there is an apparent divergence in the two integrals in the limit V ¼ 1. To remove the apparent divergence in Eq. (19.20), we add and subtract the following integral: ðV
1

 RT dV V

ð19:21Þ

Rearranging the left-hand side of the resulting equation, we obtain the Helmholtz free energy departure function, DA(T, V). Specifically,  o ðV h i RT V DAðT, VÞ ¼ AðT, VÞ  A ðT, V Þ ¼  dV þ RT ln ð19:22Þ P V V o

o

1

When V ! 1, the integrand in Eq. (19.22) is proportional to 1/V2. Therefore, the integral over dV scales as 1/V and is equal to 0 when V ¼ 1, thereby eliminating the apparent divergence. By using any pressure-explicit EOS (vdW, RK, PR, etc.), DA(T,V) can be computed using Eq. (19.22). Once DA(T,V) is known, we can readily compute the other DFs. As an illustration, below, we calculate DS(T, V).

19.6

Calculation of the Entropy Departure Function, DS(T, V)

To calculate DS(T,V), we begin from the definition of S in terms of A. Specifically,   ∂A S¼ ∂T V

ð19:23Þ

Next, we can calculate DS(T,V) by differentiating DA(T,V) in Eq. (19.22) with respect to T, at constant V. This yields:

19.6



Calculation of the Entropy Departure Function, DS(T, V)

195

2V 3  o  o   ð
 ∂A ∂A ∂ 4 RT V ∂lnVo dV5 þ Rln þ RT  ¼ P V V ∂T V ∂T V ∂T ∂T V 

1

V

ð19:24Þ where (∂A/∂T)V ¼  S. Using the fact that Vo ¼ (RT/P), the last term in Eq. (19.24) can be expressed as follows:  RT

∂ ln Vo ∂T

 ¼ V

   o RT ∂Vo ∂V ¼ P Vo ∂T V ∂T V

ð19:25Þ

In addition, dAo ¼  SodT  PdVo, which upon differentiation with respect to T, at constant V, yields:  o  o ∂A ∂V o ¼ S  P ∂T V ∂T V

ð19:26Þ

Combining Eqs. (19.24), (19.25), and (19.26), including cancelling the equal terms, yields the desired expression for DS(T,V) ¼ S(T,V)  So (T,Vo). Specifically, 2V 3  o ð
 RT V o o 4 5 ð19:27Þ DS ðT, VÞ ¼ S ðT, VÞ  S ðT, V Þ ¼ P dV  Rln V V 1

V

From knowledge of DA(T,V) and DS(T,V), we can calculate the departure function of any other thermodynamic property, B(T,V), abbreviated as DB(T,V). For example, below, we provide expressions for the internal energy departure function, DU(T,V), the enthalpy departure function, DH(T,V), and the Gibbs free energy departure function, DG(T,V). Specifically, ! DUðT, VÞ ¼ UðT, VÞ  U ðT, V Þ ¼ o

o

A ffl Affl} |fflfflffl {zfflfflffl

! SS þ T |fflfflffl{zfflfflffl}

o

o

DA

ð19:28Þ

DS

! DHðT, VÞ ¼ HðT, VÞ  H ðT, V Þ ¼ o

o

U ffl Uffl} |fflfflffl {zfflfflffl o

! þ

PV  |ffl{zffl} RT

! DGðT, VÞ ¼ G ðT, VÞ  G ðT, V Þ ¼ o

o

H ffl Hffl} |fflfflffl {zfflfflffl o

DH

ð19:29Þ

PVo

DU

! T

SS |fflfflffl{zfflfflffl} o

DS

ð19:30Þ

196

19.7

19

Departure Functions and Sample Problems

Sample Problem 19.2

Use the DF approach to calculate ΔU1!2 as a function of T and V.

19.7.1 Solution

Fig. 19.4

The (V-T) phase diagram in Fig. 19.4 illustrates the initial state (1), the final state (2), and the three paths connecting states 1 and 2. Recall that the superscript o denotes properties of the ideal gas state. In Fig. 19.4, Vo1 ¼

RT1 RT2 and Vo2 ¼ P1 P2

ð19:31Þ

where P1 and P2 are the actual pressures in states 1 and 2, respectively. Utilizing the DF approach as it applies to the three-path process depicted in Fig. 19.4 yields:

 UðT2 , V2 Þ  UðT1 , V1 Þ ¼ ΔU1!2 ¼ UðT2 , V2 Þ  Uo T2 , Vo2 Þ





 þ Uo T2 , Vo2  Uo T1 , Vo1 Þ  UðT1 , V1 Þ  Uo T1 , Vo1 Þ

ð19:32Þ

In Eq. (19.32), the first bracketed term corresponds to DU at T2, the third bracketed term corresponds to DU at T1, and the second bracketed term is evaluated in the ideal gas state (o) as follows:

19.8

Important Remark

197

o

U



T2 , Vo2

U

o



T1 , Vo1

Tð2

¼

Cov dT

ð19:33Þ

T1

Recall that for an ideal gas, Uo ¼ f(T) only. In order to compute DU in the first and the third bracketed terms in Eq. (19.32), we only require a pressure-explicit EOS. The same procedure can be used to compute ΔS1!2, ΔH1!2, ΔA1!2, ΔG1!2, etc. In all cases, we only require a pressure-explicit EOS + C0v data. Note the similarity between the DF approach, which uses the ideal gas state with Vo ¼ RT/ P, and the attenuated state approach which uses V* ¼ 1.

19.8

Important Remark

To compute isothermal variations of G or A, it is more convenient to work directly with the equations for dG or dA derived earlier, rather than to use DFs. Recall that: Pð2

dGjT ¼ VdP ) GðT, P2 Þ  GðT, P1 Þ ¼

VðT, PÞ dP

ð19:34Þ

P1

where V(T, P) is a volume-explicit EOS. Similarly: Vð2

dAjT ¼ PdV ) AðT, V2 Þ  AðT, V1 Þ ¼ 

PðT, V Þ V1

where P (T,V) is a pressure-explicit EOS.

ð19:35Þ

Lecture 20

Review of Part I and Sample Problem

20.1

Introduction

In this lecture, first, we will present a comprehensive review of the material discussed in Part I. Following that, we will solve Sample Problem 20.1 to demonstrate that the expression for the efficiency of a Carnot engine that we derived in Lecture 8 using an ideal gas as the engine working fluid is in fact generally valid for any working fluid.

20.2

Basic Concepts, Definitions, and Postulates

Definitions of systems and boundaries, states and paths, work and heat, and the four postulates of thermodynamics.

20.3

Ideal Gas PV ¼ NRT

ð20:1Þ

dU ¼ CV dT

ð20:2Þ

dH ¼ CP dT

ð20:3Þ

CP ¼ CV þ R

ð20:4Þ

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_20

199

200

20.4

20.5

20

Review of Part I and Sample Problem

The First Law of Thermodynamics for Closed Systems dE ¼ δQ þ δW

ð20:5Þ

E ¼ U þ EPE þ EKE

ð20:6Þ

For a simple system, E ¼ U

ð20:7Þ

The First Law of Thermodynamics for Open, Simple Systems dU ¼ δQ þ δW þ

X

Hin,i δnin,i 

X Hout,j δnout,j

i

ð20:8Þ

j

If the system is not simple, then, dU ! dE ¼ dU + dEKE + dEPE. There can also be KE and PE terms in the “in/out” stream terms in Eq. (20.8). Mass Balance for an Open System X X δnin,i  δnout,j

dN ¼

i

20.6

ð20:9Þ

j

The First Law of Thermodynamics for Steady-State Flow Systems _ þ Q_ þ W

X

Hin,i n_ in,i 

i

X i

X

Hout,j n_ out,j ¼ 0

ð20:10Þ

j

n_ in,i 

X j

n_ out,j ¼ 0

ð20:11Þ

20.10

20.7

The Combined First and Second Law of Thermodynamics for Closed Systems

Carnot Engine jδQH j jδQC j ¼ TH TC

ð20:12Þ

δQH δQC þ ¼0 TH TC

ð20:13Þ

ηc ¼

20.8

W TH  TC ¼ QH TH

ð20:14Þ

Entropy of a Closed System dS ¼

20.9

201

  δQ T rev

ð20:15Þ

The Second Law of Thermodynamics

For a closed system, the entropy dS: (1) Is greater than 0 for an irreversible adiabatic process (2) Is equal to 0 for a reversible adiabatic process

20.10

The Combined First and Second Law of Thermodynamics for Closed Systems dU ¼ TdS  PdV

ð20:16Þ

In addition, choosing a convenient path to compute a change in a state function, e.g., choosing a reversible path to compute Δ S.

202

20.11

20

Entropy Balance for Open Systems dS ¼

20.12

Review of Part I and Sample Problem

  X X δQ þ Sin,i δnin,i  Sout,j δnout,j þ δσ T rev i j

ð20:17Þ

Maximum Work, Availability ΔB ¼ ΔH  T0 ΔS

ðB ¼ AvailabilityÞ

ð20:18Þ

For an open system, δWmax ¼ δnΔB

ð20:19Þ

where δWmax is the maximum work done on the system. For a reversible, steady-state process with all heat interactions at To, _ max ¼ nΔB _ W

ð20:20Þ

For a non-reversible, steady-state, constant volume, adiabatic process, _ net ¼ nΔB _ W þ To σ_

20.13

ð20:21Þ

Fundamental Equations Internal Energy :

dU ¼ TdS  PdV þ

X

μi dNi

ð20:22Þ

i

Enthalpy :

dH ¼ TdS þ VdP þ

X i

μi dNi

ð20:23Þ

20.14

Manipulation of Partial Derivatives

203

Helmholtz Free Energy: dA ¼ SdT  PdV þ

X μi dNi

ð20:24Þ

i

Gibbs Free Energy: dG ¼ SdT þ VdP þ

X μi dNi

ð20:25Þ

i

For a Single-Component (Pure) System:

Also useful :

dU ¼ TdS  PdV

ð20:26Þ

dH ¼ TdS þ VdP

ð20:27Þ

dA ¼ SdT  PdV

ð20:28Þ

dG ¼ SdT þ VdP

ð20:29Þ

G ¼ U þ PV  TS ¼ H  TS ¼ μ

ð20:30Þ

In addition, the Theorem of Euler, and Euler integration to reconstruct Fundamental Equations.

20.14

Manipulation of Partial Derivatives  Inversion Rule :

Chain Rule :

Add Another Variable Rule :

∂X ∂Y

 Z

 1 ∂Y ¼ ∂X Z

ð20:31Þ

      ∂X ∂X ∂Z ¼ ∂Y W ∂Z W ∂Y W

ð20:32Þ

∂X    ∂X W ¼ ∂Z ∂Y ∂Y W ∂Z

ð20:33Þ

W

204

20

 Triple Product Rule :

20.15

∂X ∂Y



 ¼ Z

∂Y ∂Z

 1 ∂Z ∂X Y

 X

ð20:34Þ

Manipulation of Partial Derivatives Using Jacobian Transformations 

∂f ∂z

 ¼ g

∂ðf, gÞ ∂ðz, gÞ

The Jacobian :

 

∂f

∂ðf, gÞ ∂x y ¼  ∂ðx, yÞ ∂g

∂x y

Transposition :

∂ðf, gÞ ∂ðg, f Þ ¼ ∂ðx, yÞ ∂ðx, yÞ

ð20:35Þ 

∂f ∂y x



∂g

∂y x

∂ðf, gÞ 1 ¼ ∂ðx, yÞ ∂ðx, yÞ ∂ðf, gÞ

Inversion :

∂ðf, gÞ ∂ðf, gÞ ∂ðz, wÞ ¼ ∂ðx, yÞ ∂ðz, wÞ ∂ðx, yÞ

Chain Rule Expansion :

20.16

Review of Part I and Sample Problem

ð20:36Þ

ð20:37Þ

ð20:38Þ

ð20:39Þ

Maxwell’s Reciprocity Rules  for dU

∂T ∂V

 S

  ∂P ¼ ∂S V

ð20:40Þ

20.17

Important Thermodynamic Relations for Pure Materials

for dH

    ∂T ∂V ¼ ∂P S ∂S P 

for dA

∂S ∂V

 for dG



∂S ∂P

 ¼

T

ð20:41Þ

 ∂P ∂T V

ð20:42Þ

  ∂V ∂T P

ð20:43Þ

 ¼ T

205

In addition, the fundamental equations and their partial derivatives.

20.17

Important Thermodynamic Relations for Pure Materials

Change in thermodynamic properties using T and P as variables:   CP ∂V dP dT  dS ¼ T ∂T P         ∂V ∂V ∂V þP dU ¼ CP  P dT  T dP ∂T P ∂T P ∂P T   ∂V dP dH ¼ CP dT þ V  T ∂T P

ð20:44Þ

ð20:45Þ



ð20:46Þ

Change in thermodynamic properties using T and V as variables: dS ¼

  CV ∂P dT þ dV T ∂T V

    ∂P dU ¼ CV dT þ T  P dV ∂T V

ð20:47Þ

ð20:48Þ

206

20



Review of Part I and Sample Problem

      ∂P ∂P ∂P dH ¼ CV þ V þV dT þ T dV ∂T V ∂T V ∂V T 

ð20:49Þ

Relation between CP and CV:    ∂V ∂P CP  CV ¼ T ∂T P ∂T V

ð20:50Þ

Variations of CP and CV at constant T: ðP  2  ∂ V CP ðT, PÞ ¼ CP ðT, P0 Þ  T dP ∂T2 P

ð20:51Þ

P0

ðV  CV ðT, VÞ ¼ CV ðT, V0 Þ þ T V0

2

∂ P ∂T2

 dV

ð20:52Þ

V

In addition, the attenuated state and the departure function approaches.

20.18

Gibbs-Duhem Equation for a Pure Material SdT þ VdP  Ndμ ¼ 0

20.19

ð20:53Þ

Equations of State (EOS)

The van der Waals EOS: P¼

RT a  V  b V2

ð20:54Þ

20.20

Stability Criteria for a Pure Material

207

The Redlich-Kwong EOS: RT a  pffiffiffi Vb TVðV þ bÞ

ð20:55Þ

aðω, TÞ RT  V  b VðV þ bÞ þ bðV  bÞ

ð20:56Þ

RT BRT CRT þ 2 þ 3 V V V

ð20:57Þ

P¼ The Peng-Robinson EOS: P¼ The Virial EOS:

P¼

The Pitzer correlation for the compressibility factor: Z¼

PV ¼ Z0 ðTr , Pr Þ þ ωZ1 ðTr , Pr Þ RT

ð20:58Þ

In addition, the principle of corresponding states and the evaluation of EOS parameters based on the critical point conditions.

20.20

Stability Criteria for a Pure Material

Spinodal Condition: 

∂P ∂V

 ¼0

ð20:59Þ

T

Critical Point Conditions: 

∂P ∂V



 ¼ 0; TC

2

∂ P ∂V2

 ¼0

ð20:60Þ

TC

In addition, stable, metastable, and unstable equilibrium states and the binodal and spinodal curves in the (P-V) phase diagram of a pure material (n ¼ 1).

208

20.21

20

Review of Part I and Sample Problem

Sample Problem 20.1

Demonstrate that, for any working fluid, the thermal efficiency of a Carnot engine is given by: ηC ¼

20.21.1

TH  TC T ¼ 1 C TH TH

Solution

As discussed in Lecture 8, the Carnot cycle can be viewed as consisting of the following four steps: (1) A reversible, isothermal addition of heat to the working fluid at a high temperature, TH, as it expands from state 1 to state 2. (2) A reversible, adiabatic expansion of the working fluid to a low temperature, TC, as it flows from state 2 to state 3. (3) A reversible, isothermal rejection of heat from the working fluid at a low temperature, TC, as it contracts from state 3 to state 4.

Fig. 20.1

(4) A reversible, adiabatic compression of the working fluid to a high temperature, TH, as it flows from state 4 to state 1 in order to complete the cycle and return to its original state 1. When we first solved this problem in Lecture 8, we used an ideal gas as the engine’s working fluid. In addition, we used a (P-V) phase diagram to represent the

20.21

Sample Problem 20.1

209

Carnot cycle. However, this required using an equation of state to express the dependence of P on V. To simplify the algebra, in Lecture 8, we chose the ideal gas equation of state (EOS) given by: P ¼

NRT V

and claimed, without proof, that the solution would not change if the working fluid was not ideal. As per the Sample Problem 20.1 Statement, in order to deal with any working fluid without being limited to any particular EOS, it is convenient to represent the Carnot cycle using a temperature-entropy (T-S) phase diagram, instead of using a pressure-volume (P-V) phase diagram (see Fig. 20.1). Let us carry out a First Law of Thermodynamics analysis of the Carnot cycle by following an element of the working fluid (closed system) as it flows around the Carnot cycle from state1!state2!state3!state4!state1 (see Fig. 20.1). Specifically, ΔUj cycle ¼ 0 ¼ Qj cycle þ Wj cycle

ð20:61Þ

where the first term on the right-hand side of the second equal sign in Eq. (20.61) is the total heat absorbed by the element of fluid as it completes the Carnot cycle, and the second term on the right-hand side of the second equal sign in Eq. (20.61) is the total work done on the element of fluid as it completes the Carnot cycle. Rearranging Eq. (20.61) yields: Wj cycle ¼ Qj cycle

ð20:62Þ

The heat interactions along the Carnot cycle are as follows (see Fig. 20.1):   ðaÞ ðδQÞrev,a ¼ TH ðdSÞrev,a ) Qa ¼ TH S2  S1

ð20:63Þ

ðbÞ ðδQÞrev,b ¼ 0 ) Qb ¼ 0

ð20:64Þ

  ðcÞ ðδQÞrev,c ¼ TC ðdSÞrev,c ) QC ¼ TC S4  S3

ð20:65Þ

ðdÞ ðδQÞrev,d ¼ 0 ) Qd ¼ 0

ð20:66Þ

Adding up Eqs. (20.63), (20.64), (20.65), and (20.66) yields Qjcycle, the total amount of heat absorbed by the element of fluid as it flows from 1!2!3!4!1. Specifically,

210

20

Review of Part I and Sample Problem

Qj cycle ¼ THðS2  S1 Þ þ 0 þ TC ðS4  S3 Þ þ 0

ð20:67Þ

However, the (T-S) phase diagram in Fig. 20.1 shows that: S4 ¼ S1 , and S3 ¼ S2

ð20:68Þ

Using the two equalities in Eq. (20.68) in Eq. (20.67) yields: Qj cycle ¼ ðTH  TC ÞðS2  S1 Þ

ð20:69Þ

Using Eq. (20.69) in Eq. (20.62) yields: WjE ¼ WE ¼ Q j cycle ¼ ðTH  TC ÞðS2  S1 Þ

ð20:70Þ

Recall that the efficiency of a Carnot engine (the element of working fluid in this case) is defined as follows: ηC ¼ 

ðWork Done by the EngineÞ  Heat Absorbed by the Engine from the Hot Reservoir at TH

ð20:71Þ

or, using Eq. (20.70) and Eq. (20.63) in Eq. (20.71), yields: ηC ¼

Wjcycle ðTH  TC ÞðS2  S1 Þ ¼ Qa TH ðS2  S1 Þ

ð20:72Þ

Cancelling the S2  S1 terms in Eq. (20.72) yields the desired result, that is, ηC ¼

ðTH  TC Þ T ¼ 1 C TH TH

ð20:73Þ

Part II

Mixtures: Models and Applications to Phase and Chemical Reaction Equilibria

Lecture 21

Extensive and Intensive Mixture Properties and Partial Molar Properties

21.1

Introduction

The material presented in this lecture is adapted from Chapter 9 in T&M. First, we will discuss extensive and intensive differentials of mixture thermodynamic properties, B (extensive) and B (intensive), using two sets of (n + 2) independent variables. The first set consists of the temperature, the pressure, and the n mole numbers of all the components comprising the mixture. The second set consists of the temperature, the pressure, the mixture composition, and the total number of moles of the components comprising the mixture. Second, we will introduce partial molar properties of thermodynamic properties of mixtures. Finally, we will show how to “assemble” an extensive thermodynamic property of a mixture comprising n components by adding up the products of the partial molar property of component i times the number of moles of component i. In Part I, we considered pure materials (n ¼ 1). However, there are many systems of fundamental and practical importance where several components (n > 1) are present, for example, mixtures of: (i) solvent (say, water) and salt (say, NaCl), water and polymer (say, polyethylene glycol), water and protein (say, ovalbumin), water and colloids (say, silica particles), or water and surfactant (say, sodium dodecyl sulfate), (ii) mixtures of gases (say, methane and carbon dioxide), or mixtures of liquids (say, water and methanol), just to mention a few. The new important feature in all these mixtures is that the composition of the mixture, in addition to its temperature (T) and pressure (P), controls the thermodynamic behavior of the mixture. In other words, while for a pure material (n ¼ 1), the intensive (molar) properties depend only on T and P; for mixtures (n > 1), these properties depend on T, P, and the mixture composition (hereafter, referred to as composition). This will add a new dimension to the calculation of changes in the thermodynamic properties of mixtures which we did not encounter in Part I. With the above in mind, in Lectures 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, and 37 of Part II, in addition to introducing several new concepts, © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_21

213

214

21

Extensive and Intensive Mixture Properties and Partial Molar Properties

including partial molar properties, mixing functions excess functions, mixture fugacities and fugacity coefficients, mixture activities and activity coefficients, an important challenge in going from n ¼ 1 to n > 1 will involve the more complex notation of multiple indices that we need to keep track of. First, we will consider single-phase, multi-component (n > 1) mixtures. Subsequently, we will consider multi-phase, multi-component (n > 1) mixtures. Initially, we will assume that the various components comprising the mixture are all inert, postponing the treatment of chemical reactions to the conclusion of Part II (for details, please refer to the Table of Contents).

21.2

Extensive and Intensive Differentials of Mixtures

Postulate I states that any thermodynamic property can be expressed as a function of (n + 2) independent variables. For a single-phase, simple system, there are no restrictions regarding the choice of these (n + 2) variables. On the other hand, for a mixture (n > 1), we will choose two convenient sets of (n + 2) independent variables to express an extensive mixture property, B, or an intensive mixture property, B. Specifically, (i) Set 1 : fT, P, N1 , N2 , . . . , Nn g In terms of Set 1, we have: B ¼ BðT, P, N1 , N2 , . . . , Nn Þ B ¼ BðT, P, N1 , N2 , . . . , Nn Þ where Ni is the number of moles of component i (1, 2, . . ., n). (ii) Set 2 : fT, P, x1 , x2 , . . . , xn‐1 , Ng In terms of Set 2, we have: B ¼ BðT, P, x1 , x2 , . . . , xn‐1 , NÞ B ¼ BðT, P, x1 , x2 , . . . , xn‐1 , NÞ where xi is the mole fraction of component i (1, 2, . . ., n) and is given by:

21.3

Choose Set 1: {T, P, N1, . . ., Nn} and Analyze. . .

xi ¼ Ni =N, N ¼

215

n X

ð21:1Þ

Ni

i¼1

It is noteworthy that instead of using T and P, we can use the two general variables, Y1 and Y2, which can be extensive or intensive. However, consideration of such variables is not necessary here. Next, we will analyze separately how B and B vary as a function of the variables in Set 1 and Set 2, respectively.

21.3

Choose Set 1: {T, P, N1, . . ., Nn} and Analyze B = B (T, P, N1, . . ., Nn)

We begin with the differential of B, given by:  dB ¼

∂B ∂T





∂B dT þ ∂P P,Ni

 dP þ T,Ni

 n  X ∂B i¼1

∂Ni

dNi

ð21:2Þ

T,P,Nj ½i

Recall that, in Eq. (21.2), the subscript Ni in the partial derivatives of B with respect to temperature and pressure indicates that all the Nis remain constant. In addition, the subscript Nj[i] in the partial derivative of B with respect to Ni indicates that all the Njs, except for Ni, remain constant. The partial derivative of B with respect to Ni, at constant T,P,Nj[i], appears so frequently in the study of mixtures, that it was given its own name and symbol. Specifically, 

∂B ∂Ni

 ¼ Bi ðPartial molar B of component iÞ

ð21:3Þ

T,P,Nj ½i

where Bi is an intensive mixture property. Recall that we have already seen that when B ¼ G, Bi ¼ Gi ¼ μi , the chemical potential of component i. We will discuss partial molar properties in more detail in Lecture 22, including assigning physical significance to them. If we Euler integrate the dB relation in Eq. (21.2), recalling that T and P are intensive variables, we obtain: B ¼ 0þ0þ

n X i¼1

or

Bi Ni

ð21:4Þ

216

21

Extensive and Intensive Mixture Properties and Partial Molar Properties

B¼

n X

Bi Ni

ð21:5Þ

i¼1

Equation (21.5) is a central result, which indicates that any extensive property, B, of  a mixture can  be “assembled” from the set of n partial molar properties, B1 , B2 , . . . , Bn , using the mole numbers, {N1, N2, . . ., Nn}, as the weighting factors.

21.4

Important Remarks

(i) In a mixture (n > 1), the partial molar property, Bi , plays a role which is analogous to that played by the molar property, B, in a pure (one-component, n ¼ 1) system. In other words: • For n ¼ 1 ) B ¼ NBðT, PÞ n X • For n > 1 ) B ¼ Ni Bi

ð21:6Þ ð21:7Þ

i¼1

where Bi ¼ Bi (T, P, x1, x2, . . ., xn-1), and x1, x2, . . ., xn-1 is the composition. (ii) When n ¼ 1 ) Bi ¼ Bi ðT, PÞ:

21.5

Choose Set 2: {T, P, x1, . . ., xn-1, N} and Analyze B = B (T, P, x1, . . ., xn-1, N)

Note that in Set 2, we eliminated the mole fraction: xn ¼ 1 

n1 X

xi

ð21:8Þ

i¼1

We begin with the differential of B, given by:     n1  X ∂B ∂B ∂B dT þ dP þ dxi ∂T P,x,N ∂P T,x,N ∂xi T,P,x½i,n,N i¼1   ∂B þ dN ∂N T,P,x

 dB ¼

ð21:9Þ

21.5

Choose Set 2: {T, P, x1, . . ., xn-1, N} and Analyze. . .

217

In Eq. (21.9), the following shorthand notation was used in the partial derivatives: x – Indicates that all the xis are kept constant x[i,n] – Indicates that, when we change xi, we also change the eliminated mole fraction xn, because dxn ¼  dxi. Accordingly, x[i,n] indicates that all the xs, except for xi and xn, are kept constant In Eq. (21.9) for dB, the variables, B and N, are extensive, while the variables, T, P, and xi, are intensive. Accordingly, if we Euler integrate Eq. (21.9), we obtain:  B ¼ 0þ0þ0þ

∂B ∂N

 N

ð21:10Þ

T,P,x

or 

 ∂B B ¼ ¼ B N ∂N T,P,x

ð21:11Þ

We can relate the partial derivative, (∂B/∂xi)T,P,x[i,n],N, in Eq. (21.9) to the partial molar properties, Bi and Bn , as follows. Starting with B ¼ B (T, P, N1, . . ., Nn), we write the differential of B as follows:     n X ∂B ∂B dT þ dP þ Bj dNj dB ¼ ∂T P,Ni ∂P T,Ni j¼1

ð21:12Þ

Differentiating dB in Eq. (21.12) with respect to xi, keeping T, P, x[i,n], and N constant, yields: 

∂B ∂xi

 T,P,x½i,n,N

¼ 0 þ 0 þ

n X j¼1

  ∂Nj Bj ∂xi T,P,x½i,n,N

ð21:13Þ

Because Nj ¼ xjN, it follows that: 9 8 > 0, if j 6¼ i and n >     = < ∂Nj ∂xj ¼N ¼ N • þ1, if j ¼ i > > ∂xi T,P,x½i,n,N ∂xi x ½i,n ; : 1, if j ¼ n

ð21:14Þ

Using Eq. (21.14) on the right-hand side of Eq. (21.13), we obtain: n X j¼1

Bj

    ∂Nj ¼ 0 þ NBi  NBn ¼ N Bi  Bn ∂xi T,P,x½i,n,N

ð21:15Þ

218

21

Extensive and Intensive Mixture Properties and Partial Molar Properties

Using Eq. (21.15) in Eq. (21.3) yields the desired result: 

∂B ∂xi

21.6

 T,P,x ½i,j,N

  ¼ N Bi  Bn

ð21:16Þ

Choose Set 1: {T, P, N1, . . ., Nn} and Analyze B = B (T, P, N1, . . ., Nn)

We begin with the differential of B, given by:  dB ¼

∂B ∂T





∂B dT þ ∂P P,Ni

 dP þ T,Ni

 n  X ∂B ∂Ni

i¼1

dNi

ð21:17Þ

T,P,Nj ½i

Because B, T, and P are intensive, and the Nis {i ¼ 1, 2, . . ., n} are extensive, if we Euler integrate the dB expression in Eq. (21.17), we obtain: 0¼0þ0þ

 n  X ∂B i¼1

∂Ni

Ni

ð21:18Þ

T,P,Nj ½i

or  n  X ∂B ∂Ni

i¼1

Ni ¼ 0

ð21:19Þ

T,P,Nj ½i

Equation (21.19) shows that the n Nis satisfy a constraint, and therefore, cannot be varied independently. In Eq. (21.19): 

∂B ∂Ni

 6¼ Bi

ð21:20Þ

T,P,Nj ½i

The readers are encouraged to always include the underbar to highlight an extensive property. One can also show that (see Appendix F in T&M): 

∂B ∂Ni

or

 ¼ T,P,Nj ½i

 1 B B N i

ð21:21Þ

21.7

Choose Set 2: {T, P, x1, . . ., xn-1, N} and Analyze. . .

 Bi ¼ B þ N

∂B ∂Ni

219

 ð21:22Þ T,P,Nj ½i

Equation (21.22) for Bi is particularly useful to compute Bi when B ¼ B (T, P, N1, . . ., Nn) is known.

21.7

Choose Set 2: {T, P, x1, . . ., xn-1, N} and Analyze B = B (T, P, x1, . . ., xn-1, N)

Note that B is intensive, and the (n + 1) variables {T, P, x1, . . ., xn-1} are also intensive. Accordingly, and consistent with the Corollary to Postulate I, we will show that B does not depend on N. We begin with the differential of B (T, P, x1, . . ., xn-1, N), given by: dB ¼

     n1  X ∂B ∂B ∂B dT þ dP þ dxi ∂T P,x,N ∂P T,x,N ∂xi T,P,x½i,n,N i¼1   ∂B þ dN ∂N T,P,x

ð21:23Þ

In Eq. (21.23) for dB only N is extensive, and therefore, if we Euler integrate the dB expression, we obtain: 

 ∂B 0¼0þ0þ0þ N ∂N T,P,x

ð21:24Þ

or 

∂B ∂N

 N¼ 0

ð21:25Þ

T,P,x

Because N 6¼ 0, it follows that: 

 ∂B ¼ 0 ∂N T,P,x

ð21:26Þ

Equation (21.26) shows that the intensive property, B, depends on the (n + 1) intensive properties {T, P, x1, . . ., xn-1}, as required by the Corollary to Postulate I.

220

21

Extensive and Intensive Mixture Properties and Partial Molar Properties

We can also show that (see Appendix F in T&M): 

∂B ∂xi

 T,P,x½i,n

¼ Bi  Bn

ð21:27Þ

where Eq. (21.27) follows from Eq. (21.16). Specifically, 

∂B ∂xi





T,P,x½i,n,N

¼N

∂B ∂xi

 T,P,x½i,n

  ¼ N Bi  Bn

ð21:28Þ

Cancelling the two Ns in Eq. (21.28) yields: 

∂B ∂xi

 T,P,x½i,n

¼



Bi  Bn



ð21:29Þ

We can derive another useful relation for dB as follows. We have just shown that: B ¼ BðT, P, x1 , x2 , . . . , xn1 Þ

ð21:30Þ

Taking the differential of B yields: dB ¼

     ðX n1Þ  ∂B ∂B ∂B dT þ dP þ dxi ∂T P,x ∂P T,x ∂xi T,P,x½i,n i¼1

ð21:31Þ

Using Eq. (21.29) in the last sum in Eq. (21.31) yields: ðX n1Þ

ðX n1Þ ðX n1Þ   Bi  Bn dxi ¼ Bi dxi  Bn dxi

i¼1

i¼1

ð21:32Þ

i¼1

The second sum on the right-hand side of Eq. (21.32) can be expressed as follows: ðX n1Þ

Bn

! dxi

¼ Bn ðdxn Þ ¼ Bn dxn

ð21:33Þ

i¼1

Using Eq. (21.33) in the sum on the left-hand side of Eq. (21.32) yields: ðX n1Þ i¼1





Bi  Bn dxi ¼

ðX n1Þ i¼1

Bi dxi þ Bn dxn

ð21:34Þ

21.7

Choose Set 2: {T, P, x1, . . ., xn-1, N} and Analyze. . .

221

or ðX n1Þ i¼1



n X  Bi  Bn dxi ¼ Bi dxi

ð21:35Þ

i¼1

Using Eq. (21.29) in the last term in Eq. (21.31), and then using Eq. (21.35), we obtain:     n X ∂B ∂B dB ¼ dT þ dP þ Bi dxi ∂T P,x ∂P T,x i¼1

ð21:36Þ

Lecture 22

Generalized Gibbs-Duhem Relations for Mixtures, Calculation of Partial Molar Properties, and Sample Problem

22.1

Introduction

The material presented in this lecture is adapted from Chapter 9 in T&M. First, we will provide physical insight into the concept of a partial molar property by considering the volume of the system as the property of interest. Second, we will introduce the partial molar operator and use it to derive useful relations between partial molar properties. Third, we will consider three cases to illustrate how to calculate partial molar properties of (i) an extensive thermodynamic property which depends on T, P, and the n mole numbers, (ii) an intensive (molar) thermodynamic property which depends on T, P, and the n mole numbers, and (iii) an intensive (molar) thermodynamic property which depends on T, P, and the mixture composition. Fourth, we will solve Sample Problem 22.1 to calculate the partial molar Bs of components 1 and 2 in a binary mixture, given the mixture molar property B as a function of T, P, and the mole fraction of component 2. Fifth, we will provide a useful geometrical interpretation of the results derived in item four above. Finally, we will derive the generalized Gibbs-Duhem relations for mixtures.

22.2

Partial Molar Properties

To gain physical insight into the concept of a partial molar property, including how it differs from a molar property, it is convenient to consider the volume of the system as the property of interest. Pure Water (w) at T = 25
C and P = 1 bar. What is the volume occupied by one mole of water at 25
C and 1 bar? In other words, what is the molar volume of water, Vw (25
C, 1 bar)? In pure water, the measured value is:

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_22

223

224

22

Generalized Gibbs-Duhem Relations for Mixtures, Calculation of Partial Molar. . .

Vw ðT ¼ 25o C, P ¼ 1 barÞ ¼ 18 cm3 =mol

ð22:1Þ

On average, in pure water, at this T and P, the water molecules are a distance d (w, pure) from each other. A 50:50 Mixture of Water (w) and Methanol (m) at T = 25
C and P = 1 bar. What is the volume occupied by one mole of water at these conditions? In other words, what is the partial molar volume of water, Vw (25
C, 1 bar, xw ¼ 0.5)? Recall that xw + xm ¼ 1. In the binary water (w)-methanol (m) mixture, the measured value is: Vw ðT ¼ 25o C, P ¼ 1 bar, xw ¼ 0:5Þ ¼ 14 cm3 =mol

ð22:2Þ

A comparison of Eq. (22.2) and Eq. (22.1) reveals that, in the binary watermethanol mixture, water contracts! In other words, the volume occupied by one mole of water in the binary water-methanol mixture is smaller than that occupied by one mole of water in pure water at the same T and P. That is, Vw ð25o C, 1 bar, xw ¼ 05Þ < Vw ð25o C, 1 barÞ

ð22:3Þ

Molecularly, due to the presence of the methanol molecules in the mixture with water, the average distance between the water molecules, d (w, mixture), decreases, that is, d (w, mixture) < d (w, pure). It turns out that depending on the water mixture: < Vw > Vw ¼

ð22:4Þ

at the same T and P. Understanding the relation between Vw and Vw is not trivial and requires a deep understanding of the intermolecular forces operating between the various components comprising the water mixture.

22.3

Useful Relations Between Partial Molar Properties

In Lecture 21, we saw that:
Bi ¼ The operator

∂B ∂Ni

 ð22:5Þ T,P,Nj½i

22.4

How Do We Calculate Bi ? Cases 1, 2, and 3




∂ ∂Ni

225

 ð22:6Þ T,P,Nj½i

is referred to as the partial molar operator. Clearly, if the partial molar operator in Eq. (22.6) operates on an extensive property B which is multiplied by either T or P, it follows that:


∂ ∂Ni

 TB ¼ TBi

ð22:7Þ

PB ¼ PBi

ð22:8Þ

T,P,Nj½i

and


∂ ∂Ni

 T,P,Nj½i

The relations in Eqs. (22.7) and (22.8) imply that: H ¼ U þ PV ) Hi ¼ Ui þ PVi

ð22:9Þ

A ¼ U  TS ) Ai ¼ Ui  TSi

ð22:10Þ

and

Recall that Bi is an intensive mixture property, and according to the Corollary to Postulate I, it can be expressed in terms of (n + 1) independent intensive properties. For example: Bi ¼ Bi ðT, P, x1 , . . . , xn‐1 Þ

ð22:11Þ

Taking the differential of Bi in Eq. (22.11) yields:
dBi ¼

∂Bi ∂T




  ðX n1Þ
∂Bi ∂Bi dT þ dP þ dxj ∂P T,x ∂xj T,P,x½j,n P,x j¼1

ð22:12Þ

In Eq. (22.12), xn was eliminated in the last term.

22.4

How Do We Calculate Bi ? Cases 1, 2, and 3

If experimental data, or analytical expressions, of B or B are available, Bi can be calculated directly. Consider three possible cases:

226

22

Generalized Gibbs-Duhem Relations for Mixtures, Calculation of Partial Molar. . .

Case 1: B ¼ B(T, P, N1, . . ., Nn) is known. In this case, Bi is calculated using the definition of the partial molar B of component i, which we repeat below for completeness:
Bi ¼

∂B ∂Ni

 ð22:13Þ T,P,Nj½i

Case 2: B ¼ B(T, P, N1, . . ., Nn) is known. In this case, (i) We can first multiply B by N to obtain B ¼ NB(T, P, N1, . . ., Nn) and then use Case 1. or (ii) We can first obtain (∂B/∂Ni)T,P,Nj[i] directly from the given data and then use Eq. (21.22) presented in Lecture 21, which we repeat below for completeness:


∂B ∂Ni

 T,P,Nj½i


  1  ∂B ¼ Bi  B ) Bi ¼ B þ N N ∂Ni T,P,N

ð22:14Þ

j½i

where the partial derivative of B with respect to Ni can be calculated directly using the given data. Case 3: B ¼ B(T, P, x1, . . ., xi-1, xi+1, . . ., xn) is known. In this case, xi was eliminated from the set of n mole fractions, and as a result, B depends only on the (n-1) independent mole fractions: fx1 , . . . , xi1 , xiþ1 , . . . , xn g

ð22:15Þ

In addition, (i) Similar to Case 2 (i) above, we can first multiply the given B by N to obtain B ¼ NB (T, P, x1, . . ., xi-1, xi+1, . . ., xn) and then use Case 1. or (ii) We can first obtain (∂B/∂xj)T,P,x[j,i] directly from the given data and then use this partial derivative in one of the expressions presented above. For example, we know that:


∂B Bi ¼ B þ N ∂Ni

 ð22:16Þ T,P,Nj½i

22.4

How Do We Calculate Bi ? Cases 1, 2, and 3

227

Accordingly, let us relate (∂B/∂Ni)T,P,Nj[i] in Eq. (22.16) to (∂B/∂xj)T,P,x[j,i]. Because B ¼ B(T, P, x1, . . ., xi-1, xi+1, . . ., xn), where xi was eliminated, the differential of B is given by:
dB ¼

∂B ∂T

 dT þ P,x


 X
∂B  ∂B dP þ dxj ∂P T,x ∂xj T,P,x½j,i j6¼i

ð22:17Þ

Differentiating dB in Eq. (22.17) with respect to Ni, at constant T, P, Nj[i], yields:


∂B ∂Ni

 ¼

X
∂B 

∂xj • ∂xj T,P,x½j,i ∂Ni

j6¼i

T,P,Nj½i




 ð22:18Þ T,P,Nj½i

where


∂xj ∂Ni



N

¼ ¼

∂ Nj ∂Ni



! ¼

N

∂Nj ∂Ni



 Nj



∂N ∂Ni



2

N

¼

0  Nj Nj ¼ 2 2 N N

xj N

ð22:19Þ

Using Eq. (22.19) in Eq. (22.18) yields:


∂B ∂Ni

 ¼ T,P,Nj½i

X
∂B  j6¼i

 x j  N ∂xj T,P,x½j,i

ð22:20Þ

Using Eq. (22.20) in Eq. (22.16) and rearranging, we obtain: X
∂B  Bi ¼ B  xj ∂xj T,P,x½j,i j6¼i

ð22:21Þ

It is noteworthy that Eq. (22.21) for Bi should only be used for a data set in which the mole fraction, xi, was eliminated, that is, when B is known as a function of: fT, P, x1 , . . . , xi1 , xiþ1 , . . . , xn g

ð22:22Þ

If we would like to obtain a partial molar Bk , where k is in the set of variables on which B depends (i.e., where xk was not eliminated), we need to use a different relation (see T&M, Section 9.3, pages 331 and 332). Specifically,

228

22

Generalized Gibbs-Duhem Relations for Mixtures, Calculation of Partial Molar. . .


Bk ¼ B þ

∂B ∂xk

  T,P,x½k,i

X
∂B  xj ∂xj T,P,x½j,i j6¼i

ð22:23Þ

where the mole fraction, xi, was eliminated. Note that Eq. (22.23) for Bk should only be used for all the xks which are in the set of variables on which B depends.

22.5

Sample Problem 22.1

Consider a binary mixture of components 1 and 2, where B(T, P, x2) is known. Calculate B1 (T, P, x2) and B2 (T, P, x2) and provide a geometrical interpretation of your results.

22.5.1 Solution Clearly, x1 ¼ 1 – x2 was eliminated from the set of the two mole fractions. In order to compute B1 , we use Eq. (22.21) for a mole fraction that was eliminated (x1 in this case). Specifically, X
∂B  B1 ¼ B  xj ∂xj T,P,x½j,i j6¼1

ð22:24Þ

or, because j ¼ 2,
B1 ¼ B  x2

∂B ∂x2

 ¼ B1 ðT, P, x2 Þ

ð22:25Þ

T,P

To compute B2 , we use Eq. (22.23), where k ¼ 2 is in the set of mole fractions on which B depends. Specifically, for k ¼ 2, i ¼ 1, and j ¼ 2, we obtain:
B2 ¼ B þ

∂B ∂x2

  T,P

n X


x2

26¼1



∂B ∂x2

ð22:26Þ T,P

where in Eq. (22.26), the sum is redundant. In other words:


∂B B2 ¼ B þ ∂x2 or




 x2

T,P

∂B ∂x2

 ð22:27Þ T,P

22.5

Sample Problem 22.1

229


B2 ¼ B þ ð1  x2 Þ

∂B ∂x2

 ¼ B2 ðT, P, x2 Þ

ð22:28Þ

T,P

The expressions for B1 and B2 in Eq. (22.25) and Eq. (22.28), respectively, have a very nice geometrical interpretation that can be used to graphically compute B1 and B2 : Suppose that we are given B as a function of x2, at constant T and P, in graphical format (see Fig. 22.1). In Fig. 22.1, as well as in the derivation below, we have replaced x by X to enhance visualization.

Fig. 22.1

Let us consider triangles I and II in Fig. 22.1: Triangle I: Using Trigonometry  
 B T, P, X2  d1 ∂B ¼ X2  0 ∂X2 T,P j X

ð22:29Þ

2

where j X2 indicates that the partial derivative is evaluated at X2 ¼ X2 : Equation (22.29) can be rearranged as follows (see Fig. 22.1):   d1 ¼ B T, P, X2  X2




∂B ∂X2

 T,P jX2

ð22:30Þ

230

22

Generalized Gibbs-Duhem Relations for Mixtures, Calculation of Partial Molar. . .

Triangle II: Using trigonometry  
 d2  B T, P, X2 ∂B ¼ 1  X2 ∂X2 T,P j X

ð22:31Þ

2

Rearranging Eq. (22.31) yields (see Fig. 22.1):   d2 ¼ B T, P, X2 þ ð1  X2 Þ




∂B ∂X2

 T,P j X2

ð22:32Þ

A comparison of Eq. (22.30) for d1 with Eq. (22.25) for B1 reveals that:   d1 ¼ B1 T, P, X2

ð22:33Þ

A comparison of Eq. (22.32) for d2 with Eq. (22.28) for B2 reveals that:   d2 ¼ B2 T, P, X2

ð22:34Þ

Accordingly, for any 0 < X2 < 1, a tangent to the B vs. X2 curve at X2 intercepts the X2 ¼ 0 axis at B1 (T, P, X2) and the X2 ¼ 1 axis at B2 (T, P, X2) (see Fig. 22.1). This provides a very nice geometrical interpretation. As the number of mole fractions increases, the geometrical depiction of our results becomes increasingly challenging, because it requires a multidimensional space, where tangent lines become tangent hyper planes in n dimensional space.

22.6

Generalized Gibbs-Duhem Relations for Mixtures

For an n-component system, there are n partial molar quantities representing any extensive property B: B1 , B2 , B3 , . . . , Bn : These n intensive properties, along with T and P, form a set of (n + 2) intensive properties. According to the Corollary to Postulate I, only (n + 1) of these intensive properties are independent. The mathematical equations relating these (n + 2) intensive properties are called the generalized Gibbs-Duhem relations. To find the relation between B1 , B2 , . . . , Bn , T and P, we proceed as follows: B ¼ BðT, P, N1 , . . . , Nn Þ

ð22:35Þ

22.6

Generalized Gibbs-Duhem Relations for Mixtures


dB ¼

∂B ∂T




dT þ

P,Ni

∂B ∂P

231

 dP þ T,Ni

n X

Bi dNi

ð22:36Þ

i¼1

where Ni is a shorthand notation which indicates that N1, N2, . . ., Nn are all constant. Euler integrating Eq. (22.36) yields: B¼0þ0þ

n X

Bi Ni

ð22:37Þ

i¼1

or B¼

n X

ð22:38Þ

Bi Ni

i¼1

The differential of Eq. (22.38) is given by: dB ¼

n X

Bi dNi þ

i¼1

n X

Ni dBi

ð22:39Þ

i¼1

Equating dB in Eq. (22.36) and Eq. (22.39), and cancelling the equal terms, yields: n X


Ni dBi ¼

i¼1

Dividing Eq. (22.40) by N ¼

n P

∂B ∂T




 ∂B dT þ dP ∂P T,Ni P,Ni

ð22:40Þ

Ni yields:

i¼1 n X i¼1


xi dBi ¼


 ∂B ∂B dT þ dP ∂T P,x ∂P T,x

ð22:41Þ

where, in Eq. (22.41), x is a shorthand notation which indicates that x1, x2, . . ., xn are all constant. Equations (22.40) and (22.41) are known as  the generalized Gibbs-Duhem relations. Clearly, the (n + 2) intensive properties T, P, B1 , B2 , . . ., Bn are related, as shown in Eqs. (22.40) and (22.41). We note that Eqs. (22.40) and (22.41) are particularly useful when T and P are constant. In that case, the right-hand sides in Eqs. (22.40) and (22.41) are both equal to zero, which yields:

232

22 n X i¼1

Generalized Gibbs-Duhem Relations for Mixtures, Calculation of Partial Molar. . .

Ni dBi ¼ 0, or

n X

xi dBi ¼ 0 ðAt constant T and PÞ

ð22:42Þ

i¼1

Equation (22.42) shows that, at constant T and P, given (n-1) Bi s, we can calculate the nth one by integration. Note that when B ¼ G, Eq. (22.36) can be expressed as follows: dB ¼ dG ¼ SdT þ VdP þ

n X

Gi dNi

ð22:43Þ

i¼1

Equation (22.43) shows that:
 ∂G ¼ μi • Bi ¼ Gi ¼ ∂Ni T,P,Nj½i

ð22:44Þ





 ∂B ∂G ¼ ¼ S • ∂T P,Ni ∂T P,Ni

ð22:45Þ



 ∂B ∂G ¼ ¼ V ∂P T,Ni ∂P T,Ni

ð22:46Þ

•

Using Eqs. (22.44), (22.45), and (22.46) in Eq. (22.40) yields: n X

Ni dμi ¼  SdT þ VdP

ð22:47Þ

i¼1

Equation (22.47) is the celebrated Gibbs-Duhem equation for a mixture of n components.

Lecture 23

Mixture Equations of State, Mixture Departure Functions, Ideal Gas Mixtures, Ideal Solutions, and Sample Problem

23.1

Introduction

The material presented in this lecture is adapted from Chapter 9 in T&M. First, we will solve Sample Problem 23.1 to calculate the partial molar enthalpy of component 2, and the molar enthalpy of a binary liquid mixture of components 1 and 2, given the partial molar enthalpy of component 1 as a function of T, P, and the mole fraction of component 1. Second, we will discuss various equations of state (EOS) for gas mixtures, including the ideal gas mixture EOS, the van der Waals mixture EOS, the Peng-Robinson mixture EOS, and the virial mixture EOS. As needed, we will present several composition-dependent mixing rules to relate the parameters of the mixture EOS to those of the EOS corresponding to the various components comprising the mixture. We will see that, as expected, when the mixture reduces to a single component, the mixture EOS reduces to the EOS of the single component. Third, we will discuss how to calculate changes in the thermodynamic properties of gas mixtures by generalizing the attenuated state approach, first presented in Lecture 16, and the departure function approach, first presented in Lecture 19, from a single component (n ¼ 1) to several components (n > 1). Specifically, the isothermal variations will be calculated using a mixture EOS, and the temperature variations will be calculated using mixture heat capacity data, obtained through a composition average of the heat capacities of the various components comprising the gas mixture. Finally, we will define an ideal gas, an ideal gas mixture, and an ideal solution in terms of the chemical potential of component i in each case.

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_23

233

234

23.2

23

Mixture Equations of State, Mixture Departure Functions, Ideal Gas Mixtures,. . .

Sample Problem 23.1

For a binary liquid mixture of components 1 and 2, if the partial molar enthalpy H1 is known as a function of T, P, and x1, calculate H2 and H. The data H1 ¼ f ðx1 Þ is available at constant T and P.

23.2.1 Solution We begin by writing the generalized Gibbs-Duhem relation for B ¼ H at constant T and P, derived in Lecture 22, which for components 1 and 2, is given by: x1 dH1 þ x2 dH2 ¼ 0

ð23:1Þ

Because we are given H1 ¼ H1 ðT, P, x1 Þ, we differentiate Eq. (23.1) with respect to x1, at constant T and P. This yields:
x1

∂H1 ∂x1

 þ x2 T,P


 ∂H2 ¼ 0 ∂x1 T,P

ð23:2Þ

where x2 ¼ 1 – x1. Rearranging Eq. (23.2) yields:

 ∂H2 x1 ∂H1 ¼ 1  x1 ∂x1 T,P ∂x1 T,P

ð23:3Þ

where the partial derivative of H1 with respect to x1, at constant T and P, is known as a function of x1. Integrating Eq. (22.3) from x1 ¼ 0 (pure component 2) to x1 yields: xð1

0

xð1
   ∂H2 dx1 ¼ dH2 T,P ¼ H2 ðT, P, x1 Þ  H2 ðT, P, 0Þ ∂x1 T,P 0

xð1


x1 1  x1

¼ 0


 ∂H1 dx1 ∂x1 T,P

ð23:4Þ

where H2 ðT, P, 0Þ ¼ H2 ðT, PÞ is the molar enthalpy of pure component 2. Rearranging Eq. (23.4) yields:

23.3

Equations of State for Gas Mixtures

235 xð1


x1 1  x1

H2 ðT, P, x1 Þ ¼ H2 ðT, PÞ 




0

∂H1 ∂x1

 dx1

ð23:5Þ

T,P

Finally, given H1 (T, P, x1), and having calculated H2 (T, P, x1) using Eq. (23.5), we can “assemble” H as follows: H ¼ x1 H1 þ x2 H2

ð23:6Þ

Equation (23.6) shows that if we know (n1) Bi s, we can compute the nth partial molar B and then assemble B. In this example, B ¼ H, and n ¼ 2.

23.3

Equations of State for Gas Mixtures

In this lecture, and in all the coming ones dealing with mixtures, the notation y will be used to denote gas-phase mole fractions, and the notation x will be used to denote condensed (liquid or solid)-phase mole fractions.

23.3.1 Ideal Gas (IG) Mixture EOS

PV ¼ NRT; N ¼

n X

Ni ðExtensive formÞ

ð23:7Þ

i¼1

PV ¼ RTðIntensive formÞ

ð23:8Þ

All mixture EOS must approach the IG Mixture EOS in the limits: P ! 0, V ! 1, or ρ ! 0

ð23:9Þ

23.3.2 van der Waals (vdW) Mixture EOS Extensive Form P¼

NRT N2 am  V  Nbm V2

where bm and am are mixture parameters. Specifically,

ð23:10Þ

236

23

Mixture Equations of State, Mixture Departure Functions, Ideal Gas Mixtures,. . .

Nbm ¼

n X

N i bi ; N 2 a m ¼

i¼1

n X

!2 ð23:11Þ

Ni ai 1=2

i¼1

where bi and ai are the pure component parameters. Intensive Form P¼

RT a  m2 V  bm V

ð23:12Þ

where bm ¼

n X

yi bi ðArithmetic meanÞ

ð23:13Þ

i¼1

am 1=2 ¼

n X

  yi ai 1=2 “ Geometric” mean

ð23:14Þ

i¼1

In Eqs. (23.13) and (23.14), yi ¼ Ni/N is the mole fraction of component i, and bi and ai are pure component parameters.

23.3.3 Peng-Robinson (PR) Mixture EOS

P ¼

NRT N 2 am  V  Nbm V ðV þ Nbm Þ þ Nbm ðV  Nbm Þ

ðExtensive formÞ ð23:15Þ

where bm ¼

n X

ðArithmetic meanÞ

ð23:16Þ

ðComposition‐weighted averageÞ

ð23:17Þ

yi bi

i¼1

am ¼

n X n X i¼1 j¼1

yi yj aij

23.3

Equations of State for Gas Mixtures

237

  1=2 aij ¼ 1  δij aii ajj , for i 6¼ j

ð23:18Þ

In Eq. (23.18), δij is a binary interaction parameter and is typically a small number (0.1). In addition, aii ¼ ai ¼ ai ðTci Þ αi ðωi , Tri Þ; Tri ¼ T=Tci

ð23:19Þ

where Tci and Tri are the critical temperature and the reduced temperature of pure component i, respectively. In Eq. (23.19), h  i2 R2 Tci 2 αi ¼ 1  κi 1  Tri 1=2 ; ai ðTci Þ ¼ 0:045724 Pci

ð23:20Þ

where κi ¼ f ðωi Þ ¼ A þ Bωi þ Cωi 2

ð23:21Þ

In Eq. (23.16), bi ¼ 0:07780

RTci Pci

ð23:22Þ

The parameters aii, αi, κi, and bi are pure component parameters. In addition, when δij ¼ 0, it follows that (am)PR ¼ (am)vdW.

23.3.4 Virial Mixture EOS

P¼

NRT B N2 RT C N3 RT þ m 3 þ ... þ m 2 V V V

P¼

RT B RT C RT þ m2 þ m3 þ ... V V V

ðExtensive formÞ

ðIntensive formÞ

ð23:23Þ

ð23:24Þ

where Bm ¼

n X n X i¼1 j¼1

and

yi yj Bij

ðMixture second viral coefficientÞ

ð23:25Þ

238

23

Mixture Equations of State, Mixture Departure Functions, Ideal Gas Mixtures,. . .

For i ¼ j ) Bii ¼ Bi

ðSecond virial coefficient of pure component iÞ ð23:26Þ

 1=2  1=2 For i 6¼ j ) ðiÞ Bij ¼ Bii Bjj ¼ Bi Bj

ð23:27Þ

 1=2  1=2 ðiiÞ Bij ¼ 1  sij Bi Bj

ð23:28Þ

In Eq. (23.28), sij is a binary interaction parameter which is different than δij.

23.4

Calculation of Changes in the Thermodynamic Properties of Gas Mixtures

For a closed, multi-component gas mixture, or for an open, multi-component gas mixture at steady state, in the absence of chemical reactions, it follows that: (i) The mole number, Ni, of component i is fixed for every i ¼ 1, 2, . . ., n in the gas mixture. (ii) Or, equivalently, the gas-phase mole fraction, yi, is fixed for every i ¼ 1, 2, . . ., n in the gas mixture. In other words, in the absence of chemical reactions, the gas mixture composition, {y1, y2, . . ., yn1}, is fixed as the mixture evolves from state 1 to state 2. According to the Corollary to Postulate I, we can characterize the equilibrium intensive state of the gas mixture by the set of (n + 1) independent, intensive variables: {T, P, y1, y2, . . ., yn1}. To evaluate the change, ΔB1!2, we can proceed exactly as we did in the (n ¼ 1) case, albeit using a gas mixture EOS (at fixed composition) to evaluate the isothermal variations and Copm ðor Covm Þ to evaluate the temperature variations. It is noteworthy that Copm and Covm are the mixture heat capacities at constant pressure and volume in the ideal gas (or attenuated) state, respectively. The mixture heat capacities in the ideal gas (or attenuated) state are related to the pure component ideal gas heat capacities by a simple arithmetic-mean mixing rule. Specifically: Copm ¼

n X i¼1

yi Copi

ð23:29Þ

23.4

Calculation of Changes in the Thermodynamic Properties of Gas Mixtures

Covm ¼

n X

yi Covi

239

ð23:30Þ

i¼1

where Copi and Covi are the pure component i ideal gas heat capacities at constant pressure and volume, respectively. For gas mixtures, we can use the mixture attenuated state approach, first introduced for a pure material in Lecture 16, as follows:

23.4.1 Mixture Attenuated State Approach Imagine that we need to compute the change in the mixture molar property B when the mixture evolves from state 1, characterized by the (n + 1) independent intensive variables (T1, P1, y1, y2, . . ., yn1), to state 2, characterized by the (n + 1) independent intensive variables (T2, P2, y1, y2, . . ., yn1). To calculate ΔB1!2, we can choose the (P-T) phase diagram shown on the left-hand side of Fig. 23.1 and then use the attenuated state approach that we discussed in Lecture 16 for a pure component fluid. Similarly, imagine that we need to compute the change in the mixture molar property B when the mixture evolves from state 1, characterized by the (n + 1) independent intensive variables (T1, V1, y1, y2, . . ., yn1), to state 2, characterized by the (n + 1) independent intensive variables (T2, V2, y1, y2, . . ., yn1). To calculate ΔB1!2, we can choose the (V-T) phase diagram shown on the right-hand side of Fig. 23.1 and then use the attenuated state approach that we discussed in Lecture 16 for a pure component fluid. The key observation is that in the two phase diagrams shown in Fig. 23.1, the mixture composition (y1, y2, . . . ., yn1) remains constant as the mixture evolves from state 1 to state 2. Consequently, the mixture heat capacities, Copm and Covm, as well as the parameters in the mixture EOS, can be evaluated at the fixed mixture composition, and do not change when the mixture evolves from state 1 to state 2.

Fig. 23.1

240

23

Mixture Equations of State, Mixture Departure Functions, Ideal Gas Mixtures,. . .

It then follows (see the phase diagrams in Fig. 23.1), that: ΔB1!2 ¼ ΔB1!a þ ΔBa!b þ ΔBb!2

ð23:31Þ

where (i) To compute ΔB1!a and ΔBb!2 (isothermal steps), we use a mixture EOS. (ii) To compute ΔBa!b (T-variation), we use Copm or Covm data.

23.4.2 Mixture Departure Function Approach If the mixture composition, {y1, y2, . . ., yn1}, is fixed, then, the departure function of B for the mixture, (DB)m, is defined as in the pure component case. Because most EOS are pressure explicit, it is convenient to define the mixture departure function using V, rather than P, as one of the independent intensive variables. Specifically, ðDBÞm ¼ BðT, V, y1 , y2 , . . . , yn1 Þ  Bo ðT, Vo , y1 , y2 , . . . , yn1 Þ

ð23:32Þ

where Vo ¼

RT P

ð23:33Þ

In Eq. (23.32), the superscript o in Bo and Vo denotes an ideal gas mixture state. In addition, P corresponds to the actual mixture pressure.

Fig. 23.2

23.5

Ideal Gas Mixtures and Ideal Solutions

241

To calculate ΔB1!2 of the mixture using the departure function approach, we follow the three-step path shown in Fig. 23.2, where the mixture composition remains fixed as the mixture evolves from state 1 to state 2. This yields: ΔB1!2 ¼  ðDBÞm1 þ ΔBa!b þ ðDBÞm2

ð23:34Þ

where (i) To compute (DB)m1 and (DB)m2, we use a mixture EOS. (ii) To compute ΔBa!b, we use Covm and the ideal gas mixture EOS.

23.5

Ideal Gas Mixtures and Ideal Solutions

We saw that an ideal gas satisfies the following two requirements: (1) It obeys the ideal gas EOS: PV ¼ NRT or PV ¼ RT (2) U ¼ U(T) and H ¼ H(T) In an ideal gas mixture, requirement (2) is replaced by: Ui ¼ Ui ðTÞ and Hi ¼ Hi ðTÞ

ð23:35Þ

Next, we will define: (i) an ideal gas, (ii) an ideal gas mixture, and (iii) an ideal solution, by first defining the corresponding chemical potentials. We will then show that (i) and (ii) above satisfy requirements (1) and (2) above.

23.5.1 One Component (Pure, n = 1) Ideal Gas μ¼

G ¼ G ¼ λðTÞ þ RTlnP N

ð23:36Þ

where μ is the chemical potential, G is the molar Gibbs free energy, λ is only a function of temperature, and P is the gas pressure.

242

23

Mixture Equations of State, Mixture Departure Functions, Ideal Gas Mixtures,. . .

For a non-ideal, one component gas, we will see that P in Eq. (23.36) will be replaced by a new thermodynamic function – the fugacity of component i in the non-ideal gas, f(T, P).

23.5.2 Ideal Gas Mixture: For Component i μi ¼ Gi ¼ λi ðTÞ þ RTlnPi

ð23:37Þ

where μi is the chemical potential of component i, Gi is the partial molar Gibbs free energy of component i, λi is only a function of temperature and is specific to component i, and: Pi ¼ yi P

ð23:38Þ

is the partial pressure of component i, P is the gas mixture pressure, and yi is the mole fraction of component i in the gas mixture. For a non-ideal gas mixture, we will see that Pi in Eq. (23.37) will be replaced by a new thermodynamic function – the fugacity of component i in the non-ideal gas mixture, bf i (T, P, y1, . . ., yn1).

23.5.3 Ideal Solution: For Component i μi ¼ Gi ¼ Λi ðT, PÞ þ RTlnxi

ð23:39Þ

where Λi is a function of T and P and is specific to component i, and xi is the mole fraction of component i in the condensed (liquid or solid) phase. Note that when xi ¼ 1 (pure component i), Eq. (23.39) can be expressed as follows: μi ¼ Gi ¼ Λi ðT, PÞ þ 0 ) Λi ðT, PÞ ¼ Gi ðT, PÞ

ð23:40Þ

Using the result in Eq. (23.40) in Eq. (23.39) yields: μi ¼ Gi ¼ Gi ðT, PÞ þ RTlnxi

ð23:41Þ

For a non-ideal solution, we will see that xi in Eq. (23.41) will be replaced by a new thermodynamic function – the activity of component i in the non-ideal solution, ai(T, P, x1, . . ., xn  1).

Lecture 24

Mixing Functions, Excess Functions, and Sample Problems

24.1

Introduction

The material presented in this lecture is adapted from Chapter 9 in T&M. First, we will show that if the expressions for the chemical potentials presented in Lecture 23 apply, then, pure ideal gas behavior (see Sample Problem 24.1), ideal gas mixture behavior (see Sample Problem 24.2), and ideal solution behavior (see Sample Problem 24.3) are realized. Second, we will present expressions for G, S, H, U, and V of an ideal solution as a function of the solution composition. Third, we will show that it is often advantageous to calculate the deviation of a thermodynamic property from its value in a suitably chosen reference state, referred to as the mixing function, and then to compute the desired thermodynamic property by combining the mixing function with the value of the thermodynamic property in the reference state. Fourth, we will discuss reference states, with particular emphasis on the pure component reference state. Fifth, we will provide a physical interpretation of a mixing function when mixing three pure liquids to create a ternary liquid mixture. Sixth, we will discuss ideal mixing functions. Finally, we will show that it is often convenient to calculate the deviation of a thermodynamic property from its ideal solution value, at the same T, P, and composition, referred to as the excess function, and then to compute the desired thermodynamic property by combining the excess function with the value of the thermodynamic property in the ideal solution.

24.2

Sample Problem 24.1

Show that if μ ¼ G(T, P) ¼ λ(T) + RTlnP for a one component gas, then, the gas is ideal because it satisfies the two requirements below: (i) PV ¼ RT (ii) U ¼ U(T) and H ¼ H(T) © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_24

243

244

24

Mixing Functions, Excess Functions, and Sample Problems

24.2.1 Solution Choosing T and P as the two independent intensive variables for a one component gas, it follows that G ¼ G(T,P). In addition, it follows that dG ¼  SdT + VdP, and therefore, that:
 ∂G ¼ V ðGeneral result for n ¼ 1Þ ∂P T

ð24:1Þ

Using μ ¼ G (T, P) ¼ λ(T) + RT lnP in Eq. (24.1), it follows that:


∂G ∂P

 ¼0þ T

RT RT ¼ P P

ð24:2Þ

Equating Eqs. (24.1) and (24.2), including rearranging, yields: PV ¼ RT ðRequirement ðiÞ is satisfiedÞ

ð24:3Þ

Next, we show that H ¼ H(T). Recall that: G ¼ H  TS )

G H ¼ S T T

ð24:4Þ

Taking the temperature partial derivative, at constant pressure, of the expression on the right-hand side of the arrow in Eq. (24.4) yields:    
 ∂ G ∂ H ∂S ¼  ∂T T P ∂T T P ∂T P

ð24:5Þ

The first partial derivative on the right-hand side of Eq. (24.5) is given by:

   ∂ H 1 ∂H H ¼  T ∂T P ∂T T P T2

ð24:6Þ

The second partial derivative on the right-hand side of Eq. (24.5) is given by:



 ∂S CP 1 ∂H ¼ ¼ T ∂T P T ∂T P

ð24:7Þ

Using Eqs. (24.6) and (24.7) in Eq. (24.5), including cancelling the equal terms, yields:

24.3

Sample Problem 24.2

245

  ∂ G H ¼ 2 ∂T T P T

ð24:8Þ

Equation (24.8) is known as the Gibbs-Helmholtz equation, and is a general relation between G and H for n ¼ 1. If we know G(T,P), Eq.(24.8) allows us to compute H(T,P) by differentiation. Alternatively, if we know H(T,P), Eq. (24.8) allows us to compute G(T,P) by integration. In extensive form, the Gibbs-Helmholtz equation is expressed as follows:   H ∂ G ¼ 2 ∂T T P,N T

ð24:9Þ

According to the Problem Statement, for a one component (n ¼ 1) Ideal Gas, μ ¼ G(T,P) ¼ λ(T) + RTlnP, and therefore: λðTÞ G þ RlnP ¼ T T

ð24:10Þ

Taking the partial derivative of Eq. (24.10) with respect to temperature, at constant pressure, yields:
   ∂ G ∂ λðTÞ þ 0 ¼ ∂T T P ∂T T

ð24:11Þ

Using Eq. (24.11) in Eq. (24.8), including rearranging, yields:
   ∂ G 2 ∂ λðTÞ ¼ T H ¼ T ∂T T P ∂T T 2

ð24:12Þ

Equation (24.12) shows that H is only a function of temperature. In addition, because U ¼ H – PV ¼ H(T) – RT, it follows that U ¼ U(T). To derive the last result, we used the fact that, for an ideal gas, PV ¼ RT.

24.3

Sample Problem 24.2

Show that if μi ¼ Gi ¼ λi ðTÞ þ RTln ðyi PÞ , the gas mixture is ideal, because it satisfies the following requirements: (i) Vi ¼ RT P

ðIndependent of yi Þ

Equation (24.13) implies that:

ð24:13Þ

246

24

V¼

n X

Mixing Functions, Excess Functions, and Sample Problems

n RT X RT y ¼ ð1Þ ¼ RT=P ðIdeal gas EOSÞ P i¼1 i P

yi V i ¼

i¼1

(ii) Hi ¼ Hi ðTÞ and Ui ¼ Ui ðTÞ ðIndependent of yi Þ

ð24:14Þ

ð24:15Þ

24.3.1 Solution In general,



  ∂G ∂Gi ¼V ) ¼ Vi ∂P T,y ∂P T,Ni

ð24:16Þ

Recall that Eq. (24.16) is a general result for component i in the gas mixture. According to the Problem Statement, in an ideal gas mixture, Gi ¼ λi ðTÞ þ RTlnðyi PÞ, where yiP is the partial pressure, Pi, of component i. Using this result in the expression on the right-hand side of the arrow in Eq. (24.16) yields:
 ∂Gi RT RT ¼ ¼ 0þ P P ∂P T,y

ð24:17Þ

Equating Eqs. (24.16) and (24.17) yields: Vi ¼

RT P

ðIndependent of yi Þ

ð24:18Þ

It then follows that: V¼

n X i¼1

RT yi Vi ¼ P

n X

! yi

¼

i¼1

RT ) PV P

¼ RT ðIdeal gas mixture EOSÞ

ð24:19Þ

Next, starting with the Gibbs-Helmholtz equation in extensive form, repeated below for completeness (see Eq. (24.9)), it follows that:
   ∂ G H ∂ Gi H ¼ 2 ) ¼  2i ∂T T P,Ni ∂T T P,y T T

ð24:20Þ

The relation on the right-hand side of the arrow in Eq. (24.20) applies to component i in the mixture. Recall that in Eq. (24.9), the subscript Ni is a shorthand notation indicating that every mole number Ni is kept constant, and in Eq. (24.20),

24.4

Sample Problem 24.3

247

the subscript y is a shorthand notation indicating that every mole fraction yi is kept constant. According to the Problem Statement, Gi in an ideal gas mixture is given by:   Gi ¼ λi ðTÞ þ RTln yi P

ð24:21Þ

Dividing Eq. (24.21) by T, and then taking the partial derivative of the resulting expression with respect to T, at constant P and mixture composition, yields:

   λi ðTÞ Gi ∂ Gi ∂ λi ðTÞ þ Rln yi P ) þ0 ¼ ¼ T T ∂T T P,y ∂T T

ð24:22Þ

Using Eq. (24.22) in the expression on the right-hand side of the arrow in Eq. (24.20), including rearranging, yields:

 ∂ Gi 2 ∂ λi ðTÞ Hi ¼ T ¼ T ∂T T P,y ∂T T 2

ð24:23Þ

Equation (24.23) shows that in an ideal gas mixture, Hi is only a function of T, and therefore, is equal to Hi(T). Because Ui ¼ Hi  PVi, and in an ideal gas mixture PVi ¼ RT, and Hi ¼ Hi ðTÞ, it follows that: Ui ðTÞ ¼ Hi ðTÞ  RT

ð24:24Þ

Equation (24.24) shows that, in an ideal gas mixture, Ui is only a function of T, and therefore, is equal to Ui(T).

24.4

Sample Problem 24.3

As discussed in Lecture 23, in an ideal solution, μi ¼ Gi ¼ Gi ðT, PÞ þ RTlnxi

ð24:25Þ

Show that: (i) Vi is not a function of composition, that is, Vi ¼ Vi ðT, PÞ

ð24:26Þ

(ii) Hi is not a function of composition, that is, Hi ¼ Hi ðT, PÞ

ð24:27Þ

(iii) Ui is not a function of composition, that is, Ui ¼ Ui ðT, PÞ

ð24:28Þ

248

24

Mixing Functions, Excess Functions, and Sample Problems

In other words, show that the partial molar volume, the partial molar enthalpy, and the partial molar internal energy of component i in an ideal solution are equal to the molar values at the same T and P.

24.4.1 Solution To prove (i), (ii), and (iii) above, we proceed as follows. In general, in a solution of n components, it follows that:
 ∂Gi ¼ Vi ∂P T,x

ð24:29Þ

Recall that the subscript x in Eq. (24.29) indicates that all the mole fractions xi are kept constant. Recall also that xi denotes the mole fraction of component i in a condensed (liquid or solid) phase. On the other hand, yi denotes the mole fraction of component i in the gas phase. According to the Problem Statement, in an ideal solution, Gi ¼ Gi ðT, PÞ þ RTln xi : Using this expression for Gi in Eq. (24.29) yields:


∂Gi ∂P




¼

T,x

∂Gi ∂P

 þ 0 ¼ Vi ) Vi ¼ Vi ðT, PÞ

ð24:30Þ

T

In addition, using the expression for Gi given in the Problem Statement in the Gibbs-Helmholtz equation for a mixture yields: ∂ ∂T


   Gi ∂ Gi H ¼ ¼  2i ) Hi ¼ Hi ðT, PÞ T P,x ∂T T P T

ð24:31Þ

Having shown that, in an ideal solution, Vi ¼ Vi ðT, PÞ and Hi ¼ Hi ðT, PÞ , it follows that: Ui ¼ Hi  PVi ¼ Hi  PVi ¼ Ui ) Ui ¼ Ui ðT, PÞ

24.5

Other Useful Relations for an Ideal Solution Gi ¼ Gi þ RT lnxi

and

ð24:32Þ

ð24:33Þ

24.7

Mixing Functions

249

G ¼ H  TS ) Gi ¼ Hi  TSi

ðComponent i in the mixtureÞ

Gi ¼ Hi  TSi ðPure component iÞ

ð24:34Þ ð24:35Þ

Subtracting Eq. (24.35) from Eq. (24.34), including rearranging, yields:     Gi  Gi ¼ Hi  Hi  T Si  Si ) Si ¼ Si ðT, PÞ  Rln xi

ð24:36Þ

where in Eq. (24.36), we have used the facts that, in an ideal solution, Gi  Gi ¼ RTln xi , and Hi  Hi ¼ 0: The expression on the right-hand side of the arrow in Eq. (24.36) corresponds to the partial molar entropy of component i in an ideal solution.

24.6

Summary of Results for an Ideal Solution G¼

n X

xi G i ¼

n X

i¼1

S¼

xi Gi ðT, PÞ þ RT

i¼1

n X

xi Si ¼

i¼1

H¼

n X

xi Si ðT, PÞ  R

xi H i ¼

i¼1

U¼

n X

n X i¼1

24.7

ð24:37Þ

n X

xilnxi

ð24:38Þ

i¼1 n X

xi Hi ðT, PÞ

ð24:39Þ

xi Ui ðT, PÞ

ð24:40Þ

xi Vi ðT, PÞ

ð24:41Þ

i¼1

xi U i ¼

i¼1

V¼

xi lnxi

i¼1

i¼1 n X

n X

n X i¼1

xi V i ¼

n X i¼1

Mixing Functions

Sometimes, it is convenient to relate a mixture property, B, to the value of that property in some reference state (RS) that can be real or hypothetical (for example, pure component RS, dilute mixture RS, etc.). The difference between B and the value of B in the RS, denoted as B{, is referred to as the Mixing B, and denoted as ΔBmix .

250

24

Mixing Functions, Excess Functions, and Sample Problems

24.7.1 The Mixing B and Reference States The defining equation is given by: ΔBmix ¼ BðT, P, N1 , . . . , Nn Þ 

n X

  { Nj Bj T{ , P{ , x{j , . . . , x{n‐1

ð24:42Þ

j¼1 {

where Nj is the actual number of moles of component j in the mixture, and Bj is the {

partial molar B of component j in the RS. Further, in Eq. (24.42), Bj is not necessarily equal to Bj . In molar form, the mixing B can be expressed as follows: ΔBmix ¼ BðT, P, x1 , . . . , xn‐1 Þ 

n X

  { xj Bj T{ , P{ , x{j , . . . , x{n‐1

ð24:43Þ

j¼1 {

where xj is the actual mole fraction of component j in the mixture, and Bj is the {

partial molar B of component j in the RS. Again, recall that Bj is not necessarily equal to Bj : Moreover, ΔBmix ðor ΔBmix Þ is only specified when the RS of each component j has been specified, in addition to T{ and P{. Clearly, for a mixing function to be useful, it should depend on the properties (T, P, x1, . . ., xn-1) of the mixture that it is supposed to describe. In other words, we would like: ΔBmix ¼ f ðT, P, x1 , . . . , xn1 Þ

ð24:44Þ

Equation (24.44) imposes restrictions on the RS variables. Three potential restrictions are discussed below: {

{

1. T{ ¼ T, P{ ¼ P, x{j ¼ xj ) Bj ¼ Bj ðT, P, x1 , . . . , xn1 Þ and varies as the actual mixture conditions change. 2. T{ ¼ To , P{ ¼ Po , x{j ¼ xjo, where the subscript o denotes constant values in the  { o RS. In this case, Bj ¼ Bj To , Po , x10 , . . . , xn1o is constant for every j, independent of variations in the actual mixture conditions. 3. In practice, the most common RS is the pure component RS at the same T, P, and state of aggregation (vapor, liquid, or solid) of the mixture, which we discuss next.

24.7

Mixing Functions

251

24.7.2 Pure Component Reference State for Component j The pure component reference state is characterized by: • Mixture state of aggregation • T{ ¼ T, P{ ¼ P

ð24:45Þ

{ { • xj ¼ 1, xi ¼ 0, for i 6¼ j

ð24:46Þ

{

• Bj ¼ Bj ðT, PÞ

ð24:47Þ

24.7.3 Useful Relations for Mixing Functions Because a mixing function is a thermodynamic property of the mixture, all the relations derived in previous lectures for B (or B) of a mixture also apply to ΔBmix ðor ΔBmix Þ. For example,
 ∂ΔBmix { ΔBj ¼ ¼ Bj  Bj ðPartial molar mixing B of component jÞ ∂Nj T,P,N i½j

ð24:48Þ

ΔBmix ¼

n X

Nj ΔBj

ð24:49Þ

xj ΔBj

ð24:50Þ

j¼1

ΔBmix ¼

n X j¼1

Further, because ΔBmix ¼ ΔBmix ðT, P, x1 , . . . , xn1 Þ

ð24:51Þ

it follows that:



 ∂ðΔBmix Þ ∂ðΔBmix Þ dT þ dP dðΔBmix Þ ¼ ∂T ∂P P,x T,x  n‐1
X ∂ðΔBmix Þ dxj þ ∂xj T,P,x½j,n j¼1

ð24:52Þ

252

24

Mixing Functions, Excess Functions, and Sample Problems

In addition, the generalized Gibbs-Duhem relations apply as follows: n X i¼1

  xi d ΔBi ¼




∂ðΔBmix Þ ∂T

 dT þ P,x


 ∂ðΔBmix Þ dP ∂P T,x

ð24:53Þ

At constant T and P, the generalized Gibbs-Duhem relations in Eq. (24.53) simplify to: 
 n X ∂ ΔBi xi ¼0 ð24:54Þ ∂xj T,P,x½j,k i¼1 As before, given (n-1) partial molar mixing Bs, one can calculate the remaining one, to within an arbitrary constant of integration, by integrating Eq. (24.54).

24.8

Mixing Functions: Mixing of Three Liquids at Constant T and P

Fig. 24.1

The partial molar properties Hi , Vi , and Si (i ¼ 1, 2, 3) depend on the nature of the liquids i being mixed, as well as on the mixture composition (x1, x2), the temperature (T), and the pressure (P). Recall that x3 ¼ 1x1x2. The molar properties Hi,Vi, and Si (i ¼ 1, 2, 3) depend on the nature of liquid i, as well as on the temperature (T) and the pressure (P). In general: Hi ðT, P, x1 , x2 Þ 6¼ Hi ðT, PÞ

ð24:55Þ

24.8

Mixing Functions: Mixing of Three Liquids at Constant T and P

253

Vi ðT, P, x1 , x2 Þ 6¼ Vi ðT, PÞ

ð24:56Þ

Si ðT, P, x1 , x2 Þ 6¼ Si ðT, PÞ

ð24:57Þ

where i ¼ 1, 2, and 3. The following three observations can be made (see Fig. 24.1): 1. The enthalpy of mixing, ΔHmix , simply measures the difference between Hmix (in this example, the enthalpy of the ternary mixture) and the sum of the enthalpies of the three pure liquids (1, 2, and 3) which were mixed to create the ternary mixture, that is, ΔHmix ¼ Hmix  ðH1 þ H2 þ H3 Þ

ð24:58Þ

where H1 þ H2 þ H3 is the pure component reference state enthalpy. Expanding Eq. (24.58) in terms of the partial molar enthalpies and molar enthalpies of components 1, 2, and 3 yields: ΔHmix ¼ N1 H1 þ N2 H2 þ N3 H3  ðN1 H1 þ N2 H2 þ N3 H3 Þ

ð24:59Þ

Combining the N1, N2, and N3 terms, we obtain:       ΔHmix ¼ N1 H1  H1 þ N2 H2  H2 þ N3 H3  H3

ð24:60Þ

or ΔHmix ¼ N1 ΔH1 þ N2 ΔH2 þ N3 ΔH3

ð24:61Þ

where ΔHi is the partial molar enthalpy of mixing of component i (1, 2, and 3). 2. Similarly, the volume of mixing is given by: ΔVmix ¼ Vmix  ðV1 þ V2 þ V3 Þ

ð24:62Þ

where V1 þ V2 þ V3 is the pure component reference state volume. Similar to the enthalpy calculation presented above, Eq. (24.62) can be expressed as follows:       ΔVmix ¼ N1 V1  V1 þ N2 V2  V2 þ N3 V3  V3 or

ð24:63Þ

254

24

Mixing Functions, Excess Functions, and Sample Problems

ΔVmix ¼ N1 ΔV1 þ N2 ΔV2 þ N3 ΔV3

ð24:64Þ

where ΔVi (i ¼ 1, 2, 3) is the partial molar mixing volume of component i. 3. Finally, the entropy of mixing is given by: ΔSmix ¼ Smix  ðS1 þ S2 þ S3 Þ

ð24:65Þ

where S1 þ S2 þ S3 is the pure component reference state entropy. Similar to the enthalpy and entropy of mixing calculations presented above, it follows that:       ΔSmix ¼ N1 S1  S1 þ N2 S2  S2 þ N3 S3  S3

ð24:66Þ

Equation (24.66) can also be expressed as follows: ΔSmix ¼ N1 ΔS1 þ N2 ΔS2 þ N3 ΔS3

ð24:67Þ

where ΔSi (i ¼ 1, 2, and 3) is the partial molar entropy of mixing of component i.

24.9

Ideal Solution Mixing Functions

For an ideal solution, we choose the pure component RS. Specifically, 8 9 < T{ ¼ T, P{ ¼ P, x{ ¼ 1, x{ ¼ 0 ðfor i 6¼ jÞ, same = j i : aggregation state as the mixture ! B{ ¼ Bj ðT, PÞ ; j

ð24:68Þ

Therefore, in an ideal solution, the Mixing B is given by: ID ΔBID mix ¼ B 

n X

xj Bj ðT, PÞ

ð24:69Þ

j¼1

Expressions for five ideal mixing functions follow: ΔGID mix ¼ RT

n X j¼1

ID

xj lnxj ; ΔGj ¼ RTlnxj

ð24:70Þ

24.10

Excess Functions

255

ΔSID mix ¼ RT

n X

ID

xj lnxj ; ΔSj ¼ Rlnxj

ð24:71Þ

j¼1 ID

ð24:72Þ

ID

ð24:73Þ

ID

ð24:74Þ

ΔHID mix ¼ 0; ΔHj ¼ 0 ΔVID mix ¼ 0; ΔVj ¼ 0 ΔUID mix ¼ 0; ΔUj ¼ 0

24.10

Excess Functions

The deviation of a mixture property, B, from its ideal solution value at the same T, P, and composition (x1, . . ., xn-1) as in the original mixture, BID, is referred to as the Excess B, and is given by: BEX ¼ B  BID

ð24:75Þ

We know that: B¼

n X

n X

Nj Bj ; BID ¼

j¼1

ID

Nj Bj

ð24:76Þ

j¼1

Using Eq. (24.76) in (24.75), we obtain: BEX ¼ B  BID ¼

n X

  ID Nj Bj  Bj

ð24:77Þ

j¼1

Further, because BEX ¼

n X

EX

Nj Bj

ð24:78Þ

j¼1

Equations (24.78) and (24.77) show that: EX

Bj

ID

¼ Bj  Bj ðPartial molar excess B of component jÞ

ð24:79Þ

256

24

Mixing Functions, Excess Functions, and Sample Problems

Clearly, every relation derived earlier for B (or B) applies to BEX (or BEX). In particular, we can define excess mixing functions, ΔBEX mix , as follows: ID ΔBEX mix ¼ ΔBmix  ΔBmix

ð24:80Þ

Expanding Eq. (24.80) in terms of partial molar properties yields: n X

EX

Nj ΔBj

j¼1

¼

n X

Nj ΔBj 

j¼1

n X

ID

Nj ΔBj

ð24:81Þ

j¼1

Equation (24.81) shows that: EX

ΔBj

ID

¼ ΔBj  ΔBj ðPartial molar excess mixing B of component jÞ

ð24:82Þ

One can show that unlike ΔBmix or ΔBID mix , an excess function is independent of ID the RS, provided that the same RS is used for ΔBmix and ΔBID mix : Because for ΔBmix we always choose the pure component RS, the same RS will also be used for ΔBmix , unless specified otherwise, for the calculation of BEX . The proof of this last statement is presented below. Choosing the pure component reference state yields: ΔBj ¼ Bj  Bj

ð24:83Þ

and ID

ID

ΔBj ¼ Bj  Bj

ð24:84Þ

Subtracting Eq. (24.84) from (Eq. 24.83), including cancelling the two equal terms, yields: ID

EX

ΔBj  ΔBj ¼ ΔBj

ID

EX

¼ Bj  Bj ¼ Bj

ð24:85Þ

or EX

ΔBj

EX

¼ Bj

ð24:86Þ

24.10

Excess Functions

257

Equation (24.86) shows that the partial molar excess mixing B of component j is equal to the partial molar excess B of component j. The equality in Eq. (24.86) ensures that: EX ΔBEX ðThe excess mixing B is equal to the excess BÞ mix ¼ B

ð24:87Þ

Lecture 25

Ideal Solution, Regular Solution, and Athermal Solution Behaviors, and Fugacity and Fugacity Coefficient

25.1

Introduction

The material presented in this lecture is adapted from Chapter 9 in T&M. First, we will discuss ideal solution behavior, including presenting a molecular interpretation of ideality, followed by discussing regular solution and athermal solution behaviors. Second, we will introduce the concept of fugacity of component i in a non-ideal gas mixture, including showing that when the pressure approaches zero, the fugacity of component i approaches the partial pressure of component i. We will also define the ratio of the mixture fugacity of component i and the partial pressure of component i as the fugacity coefficient of component i. Clearly, in an ideal gas mixture, the fugacity of component i is equal to the partial pressure of component i, and the fugacity coefficient of component i is equal to unity. For a one component gas, the fugacity is equal to the pressure, and their ratio is equal to the fugacity coefficient of the one component gas, which is equal to unity when the one component gas is ideal. Second, we will derive expressions for the variations of the fugacity with pressure and temperature, both for a pure component gas and for a gas mixture. These expressions will be used when we discuss the differential approach to phase equilibria. Third, we will discuss how to calculate fugacities using an EOS approach, both for a pure component gas and for a gas mixture. Fourth, we will derive the generalized Gibbs-Duhem relation for the fugacities, which is particularly useful when T and P are constant, and allows us to calculate an unknown fugacity if we know the remaining (n-1) independent fugacities. Finally, we will derive the Lewis and Randal Rule, which relates the fugacity of component i in an ideal solution to the product of the fugacity of pure component i, at the same T and P as those in the ideal solution, and the mole fraction of component i in the ideal solution.

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_25

259

25 Ideal Solution, Regular Solution, and Athermal Solution Behaviors, and Fugacity. . .

260

25.2

Ideal Solution Behavior

For an ideal solution, where the pure component reference state is used for every component, all the excess functions are zero, that is, BID EX ¼ BID  BID ¼ 0: In Fig. 25.1, we present a simple molecular model of an ideal binary solution where both the excess entropy and the excess enthalpy are zero. Indeed, when the “colored blind” black and gray billiard balls of equal radius are mixed, “they cannot differentiate if they are in a mixture or by themselves,” which corresponds to ideal mixing.

Fig. 25.1

25.3 EX

(a) Sj

Regular Solution Behavior ¼ 0, for every component j in the solution. Accordingly,

SEX ¼

n X

EX

Nj Sj

¼0

ð25:1Þ

j¼1

Equation (25.1) shows that the excess entropy of a regular solution is zero. In other words, from the entropy point of view, the solution behaves ideally. This typically occurs when the various components in the solution are similar in size.

25.5

Fugacity and Fugacity Coefficient EX

(b) Hj

261

6¼ 0, for every component j in the solution. Therefore, it follows that: HEX 6¼ 0

ð25:2Þ

The result in Eq. (25.2) shows that the excess enthalpy of a regular solution is non-zero. In other words, from the enthalpy (interactions) point of view, an athermal solution is not ideal. Using Eqs. (25.1) and (25.2), it follows that: (c) GEX ¼ HEX  TSEX ¼ HEX ¼ ΔHEX mix

25.4

ð25:3Þ

Athermal Solution Behavior EX

(a) Hj

¼ 0, for every component j in the solution. Therefore, it follows that: HEX ¼ 0

ð25:4Þ

Equation (25.4) shows that from the enthalpy (interactions) point of view, an athermal solution behaves ideally. EX

(b) Sj

6¼ 0, for every component j in the solution. Therefore, it follows that: SEX 6¼ 0

ð25:5Þ

Equation (25.5) shows that the excess entropy of an athermal solution is non-zero. In other words, from the entropy point of view, an athermal solution is not ideal. Using Eqs. (25.4) and (25.5), it follows that: (c) GEX ¼ HEX  TSEX ¼  TSEX ¼ TΔSEX

25.5

ð25:6Þ

Fugacity and Fugacity Coefficient

We know that for a one component (n ¼ 1) ideal gas (IG) i: μIG i ðT, PÞ ¼ Gi ðT, PÞ ¼ λi ðTÞ þ RTlnP

ð25:7Þ

262

25 Ideal Solution, Regular Solution, and Athermal Solution Behaviors, and Fugacity. . .

In Eq. (25.7), the units are implicitly specified by the ideal gas (IG) state condition contained in the λi(T) term. For example: λi ðTÞ ¼ μIG i ðT, P ¼ 1 bar or 1 atmÞ

ð25:9Þ

If gas i is not ideal, we introduce a new thermodynamic function which replaces P, and is called fugacity, fi(T,P). The fugacity of pure component i is an intensive property and, according to the Corollary to Postulate I, depends on the (n + 1) ¼ (1 + 1) ¼ 2 independent intensive variables T and P. For a non-ideal gas i, it follows that: μi ðT, PÞ ¼ Gi ðT, PÞ ¼ λi ðTÞ þ RTlnf i ðT, PÞ

ð25:10Þ

Clearly, in the limit when P ! P* 0, the gas must behave ideally, and therefore: lim P!0

  f i ðT, PÞ ¼ 1 ðFor pure gas iÞ P

ð25:11Þ

We also saw that for gas i in an ideal gas mixture (IGM), it follows that: μIGM ðT, P, yi Þ ¼ Gi ðT, P, yi Þ ¼ λi ðTÞ þ RTlnPi i

ð25:12Þ

where Pi ¼ yiP (Partial pressure of i). In Eq. (25.12), μIGM depends only on yi of gas i. In principle, it should depend on i {y1, y2, . . ., yn1}, not only on yi. If the gas mixture is not ideal, we replace Pi by the fugacity, bf i ðT, P, y1 , . . . , yn1 Þ, of gas i in the mixture. It then follows that: μi ðT, P, y1 , . . . , yn1 Þ ¼ Gi ðT, P, y1 , . . . , yn1 Þ ¼ λi ðTÞ þ RTlnbf i ðT, P, y1 , . . . , yn1 Þ

ð25:13Þ

In the limit when P ! P* 0, the gas mixture must behave ideally, and therefore: lim P!0

  bf i ¼ 1 ðFor gas i in the mixtureÞ Pi

ð25:14Þ

Next, we will derive useful relations involving the variations of bf i and f i with pressure and temperature.

25.5

Fugacity and Fugacity Coefficient

263

25.5.1 Variations of bf i and f i with Pressure We know that, in general,  G ¼ G ðT, P, N1 , . . . , Nn Þ )

∂G ∂P



 ¼V) T,Ni

∂Gi ∂P

 ¼ Vi

ð25:15Þ

T,y

We also know that: Gi ¼ λi ðTÞ þ RTlnbf i

ð25:16Þ

Using Eq. (25.16) in Eq. (25.15) yields: 

∂Gi ∂P



 ¼ 0 þ RT

T,y

∂ ln bf i ∂P

 ¼ Vi

ð25:17Þ

T,y

Rearranging Eq. (25.17) yields:   ∂ ln bf i Vi ¼ RT ∂P T,y

ðFor component i in the mixtureÞ

ð25:18Þ

Recall that both bf i and Vi are intensive variables which depend on {T, P, y1, . . ., yn1}. In a similar manner, we can show that for pure gas i:   ∂lnf i Vi ¼ RT ∂P T

ðFor pure component iÞ

ð25:19Þ

Recall that both fi and Vi are intensive variables which depend on T and P.

25.5.2 Variations of bf i and f i with Temperature We saw that, in general, component i in a mixture must satisfy the Gibbs-Helmholtz equation, given by:    ∂ Gi =T H ¼  2i ∂T T P,y

ð25:20Þ

264

25 Ideal Solution, Regular Solution, and Athermal Solution Behaviors, and Fugacity. . .

We know that: Gi ¼ λi ðTÞ þ RTlnbf i

ð25:21Þ

In order to use Eq. (25.21) in Eq. (25.20), we need to compute:   ∂λi ðTÞ ∂T P,y

ð25:22Þ

where keeping P and y constant in the partial derivative is redundant, because λi depends only on temperature. Because we have no information about λi (T), it would be convenient to eliminate λi (T) altogether. The “trick” is to calculate Gi for an ideal gas mixture, denoted by o o Gi , and then to subtract Gi from Gi . Specifically, o

Gi ¼ λi ðTÞ þ RTlnPi ; Pi ¼ yi P

ð25:23Þ

Subtracting Eq. (25.23) from Eq. (25.21), including cancelling the equal terms, we obtain:   bf Gi  Gi ¼ RTln i ðFor component i in the mixtureÞ Pi o

ð25:24Þ

Dividing Eq. (25.24) by RT, and then taking the partial derivative with respect to T, at constant P and y, yields: 1 0   o ∂ln bf i =yi P  Hi  Hi ∂  o A Gi  Gi =RT P,y ¼  ¼@ ∂T ∂T RT2  ¼

∂lnbf i ∂T

P,y



ð25:25Þ P,y

Equating the second and the last terms in Eq. (25.25), including rearranging, yields: !    o Hi  Hi ∂lnbf i ¼ ∂T P,y RT2

ð25:26Þ

P,y

o

Recall that in an ideal gas mixture, Hi ¼ Hi o (T), independent of y and P. Accordingly, Eq. (25.26) can be expressed as follows:

25.7

Calculation of Fugacity

265

    Hi  Hi o ðTÞ ∂lnbf i ¼ ∂T P,y RT2

ðFor component i in the mixtureÞ

ð25:27Þ

Recall that both bf i and Hi in Eq. (25.27) are intensive variables which depend on {T, P, y1, . . ., yn1}. In a similar manner, we can show that for pure gas i: 

∂lnf i ∂T

 ¼ P

ðHi  Hi o ðTÞÞ RT2

ðFor pure component iÞ

ð25:28Þ

Recall that both fi and Hi in Eq. (25.28) are intensive variables which depend on T and P.

25.6

Other Relations Involving Fugacities

n o The set of (n + 2) intensive variables, T, P, bf 1 , . . . , bf n , is not independent. Indeed, according to the Corollary to Postulate I, only (n + 1) intensive variables are independent. Therefore, these (n + 2) intensive variables are related by a generalized Gibbs-Duhem relation. Specifically, n X i¼1

"  # n X Hi  Hi o b xi d ln f i ¼  xi dT þ RT2 i¼1 

! n X xi V i dP RT i¼1

ð25:29Þ

At constant T and P, Eq. (25.29) yields: n X i¼1

  ∂lnbf i xi ¼ 0 ∂xj T,P,x½j,k

ð25:30Þ

Equation (25.30) shows that, given (n-1) of the bf i s, the nth one can be calculated by integration, to within an arbitrary constant of integration.

25.7

Calculation of Fugacity

The calculation of bf i or f i makes use of an EOS, where this is not restricted to gases, if the EOS used can accurately describe the volumetric behavior of liquids. Multiplying both sides of Eq. (25.18) by dP yields:

266

25 Ideal Solution, Regular Solution, and Athermal Solution Behaviors, and Fugacity. . .

  

∂lnbf i

dP ¼ d ln bf i

∂P T,y

 ¼ T,y

 Vi dP RT

ð25:31Þ

Integrating the second and the third terms in Eq. (25.31) with respect to P from P!0 to P yields: ðP P !0



d ln bf i





bf i ¼ ln yi P

ðP 

 ¼

T,y

P !0

 Vi dP RT

ð25:32Þ

In the limit P!0, the mixture behaves ideally, and: Vi !

RT P

ð25:33Þ

In that case, the pressure integral in Eq. (25.32) diverges logarithmically when P!0. To take care of this divergence, as we have done in the (n ¼ 1) case, we subtract: ðP 

 dP P ¼ ln  P P

ð25:34Þ

P !0

from the second and the third terms in Eq. (25.32). This yields:       ðP   bf i bf i bf i P Vi 1 ln dP ¼ ln   ln  ¼ ln ¼ P yi P  yi P Pi RT P 

ð25:35Þ

0

where the integral in Eq. (25.35) is well-behaved when P ¼ 0. The ratio, bf i =Pi , in Eq. (25.35) measures deviations from the ideal mixture behavior and is known as the fugacity coefficient of component i in the mixture. Specifically, b b bi ¼ f i ¼ f i ϕ P i yi P

ð25:36Þ

b iðT, P, y1 , . . . , yn1 Þ, and ϕ b IGM ¼ 1: Using Eq. (25.36) in the third bi ¼ ϕ where ϕ i term in Eq. (25.35), including equating it with the last term in Eq. (25.35) and then multiplying both terms by RT, yields:

25.7

Calculation of Fugacity

267

  ðP  bf i RT b i ¼ RTln dP ¼ RTln ϕ Vi  P yi P

ð25:37Þ

0

We can use Eq. (25.37) if we have access to a volume-explicit EOS, that is, if we are given: V ¼ V ðT, P, N1 , . . . , Nn Þ

ð25:38Þ

from which we can compute: Vi ¼ ð∂V=∂Ni ÞT,P,Nj½i

ð25:39Þ

Unfortunately, most EOS are pressure explicit, that is, they provide: P ¼ P (T, V, N1, . . ., Nn). In that case, we can use a different equation, which relies on a pressureb i . Alternatively, we can use the triple-product rule to explicit EOS, to compute ϕ calculate Vi in terms of the pressure-explicit EOS. Specifically,  Vi ¼

∂V ∂Ni







∂P ∂Ni T,V,Nj½i

T,P,Nj½i

¼ 

∂P ∂V T,N i

ð25:40Þ

It is important to recognize that, in Eq. (25.40), the parameters in the mixture EOS depend on {N1, . . ., Nn} when calculating the partial derivative (∂P/∂Ni)T,V,Nj[i]. If we have access to a pressure-explicit EOS, that is, to P ¼ P (T, V, N1, . . ., Nn), b i : Specifically, we can derive a different equation to compute ϕ #    ðV " bf i ∂P RT b RTlnϕi ¼ RTln ¼  RTlnZ  dV  yi P V ∂Ni T,V,Nj½i

ð25:41Þ

1

In Eq. (25.41), Z is the compressibility factor, and is given by PV/NRT. Again, we recognize that the mixture EOS parameters depend on {N1, N2, . . ., Nn}, when calculating: 

∂P ∂Ni

 T,P,Nj½i

in Eq. (25.41). In a similar manner, we can show that for pure gas i:

ð25:42Þ

268

25 Ideal Solution, Regular Solution, and Athermal Solution Behaviors, and Fugacity. . .

  ðP  fi RT dP ¼ RTlnϕi ¼ RTln Vi  P P

ð25:43Þ

0

In Eq. (25.43), ϕi is the fugacity coefficient of pure gas i. Recall that for a pure ideal gas, ϕIG i ¼ 1. Further, Eq. (25.43) is particularly useful when we have access to V ¼ V (T, P, N), that is, to a volume-explicit EOS. Finally, recall that both ϕi and Vi are intensive variables which depend on T and P. If we have access to P ¼ P (T,V,N), that is, to a pressure-explicit EOS, we can show that: ðV " RTlnϕi ¼  RTlnZ  1

∂P ∂N

# RT  dV V T,V



ð25:44Þ

In Eq. (25.44), Z is the compressibility factor, given by Z ¼ PV/NRT.

25.8

The Lewis and Randall Rule

As stated earlier, in principle, the fugacity can also be used to describe a liquid mixture. In particular, using Eq. (25.21) for an ideal solution yields: ID ID Gi ¼ λi ðTÞ þ RTlnbf i

ð25:45Þ

For pure component i, it follows that (see Eq. (25.10)): Gi ðT, PÞ ¼ λi ðTÞ þ RTlnf i ðT, PÞ

ð25:46Þ

Subtracting Eq. (25.46) from Eq. (25.45), including cancelling the equal terms and rearranging, yields: ID ∴Gi 

bf ID i Gi ðT, PÞ ¼ RTln f i ðT, PÞ

! ð25:49Þ

For an ideal solution, it follows that: ID

Gi  Gi ðT, PÞ ¼ RTlnxi

ð25:50Þ

25.8

The Lewis and Randall Rule

269

A comparison of Eqs. (25.49) and (25.50) shows that: bf ID ¼ f i ðT, PÞxi i

ð25:51Þ

The relation in Eq. (25.51) is known as the Lewis and Randall Rule. It simply states that the fugacity of component i in an ideal solution can be obtained by multiplying the pure component i fugacity at the same T and P as in the ideal solution, by the mole fraction of component i in the ideal solution. Interestingly, bf ID in Eq. (25.51) depends only on xi, and not on the mole fractions of the other i components present in the ideal solution!

Lecture 26

Activity, Activity Coefficient, and Sample Problems

26.1

Introduction

The material presented in this lecture is adapted from Chapter 9 in T&M. First, we will introduce a new thermodynamic function known as the activity of component i, which will describe deviations from the ideal solution behavior. Second, we will define the product of RT and the natural logarithm of the activity of component i as the difference between the partial molar Gibbs free energy of component i in the actual non-ideal solution and the partial molar Gibbs free energy of component i in a reference state having the same temperature as that of the non-ideal solution. Third, we will revisit the pure component reference state, originally introduced in Lecture 24, and use it in this lecture and beyond. Fourth, we will show that if we choose the pure component reference state, when the solution approaches ideality, the activity of component i is equal to the mole fraction of component i. We will also define the ratio of the activity of component i and the mole fraction of component i as the activity coefficient, which will approach unity when the solution approaches ideality. Fifth, we will discuss how to calculate the activity of component i using a mathematical model for the excess Gibbs free energy of mixing. Sixth, we will solve Sample Problem 26.1 to calculate the activity coefficients of components 1 and 2 in a non-ideal binary solution for which a mathematical model for the excess Gibbs free energy of mixing is known, including discussing the model properties when the solution composition varies from 0 (pure component 1) to 0.5 (equimolar composition) to 1 (pure component 2). We will also discuss, mathematically and graphically, deviations from ideality as reflected in the activity coefficients of components 1 and 2 as a function of the solution composition. Specifically, we will show that the activity coefficient is equal to 1 in an ideal solution, exhibits positive deviations from ideality when it is larger than 1, and exhibits negative deviations from ideality when it is smaller than 1. Finally, we will solve Sample Problem 26.2 to calculate the activity coefficient of component 2 in a non-ideal binary solution as a function of T,

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_26

271

272

26

Activity, Activity Coefficient, and Sample Problems

P, and the solution composition, given the activity coefficient of component 1 as a function of T, P, and the solution composition.

26.2

Activity and Activity Coefficient

Recall that to calculate the fugacity of component i in the mixture, bf i, one can use the equations derived earlier which make use of a pressure-explicit or a volume-explicit mixture EOS. Unfortunately, EOS are best suited to describe the volumetric behavior of gas mixtures. To deal with liquid (or even solid) mixtures, it is convenient to develop a new approach, which is based on describing the deviations from the ideal solution behavior using a new thermodynamic function known as the activity. The role played by the activity in a non-ideal solution is analogous to that played by the fugacity in a non-ideal gas mixture. We introduce the activity of component i in the solution, ai, as the difference in the partial molar Gibbs free energy of component i between the real state and a reference state. The reference state (RS) is denoted by {, and as will be explained below, we choose T{ ¼ T of the actual solution, while x{i 6¼ xi (for every i), and P{ 6¼ P. Specifically,   { RTlnai ¼ Gi ðT, P, x1 , . . . , xn1 Þ  Gi T, P{ , x{1 , . . . , x{n1

ð26:1Þ

{

Recall that we can also express Gi and Gi in terms of fugacities as follows: Gi ¼ λi ðTÞ þ RTlnbf i

ð26:2Þ

{ { Gi ¼ λi ðTÞ þ RTlnbf i

ð26:3Þ

Subtracting Eq. (26.3) from Eq. (26.2), cancelling the equal terms and rearranging, yields: Gi 

{ Gi

bf ¼ RTln {i bf i

Using Eq. (26.4) in Eq. (26.1) yields:

! ð26:4Þ

26.3

Pure Component Reference State

bf RTln {i bf

273

! ¼ RTlnai ) ai ¼

i

bf i bf {

ð26:5Þ

i

The last term in Eq. (26.5) shows that the activity of component i in the solution is equal to the ratio of the fugacities of component i in the real state and in the RS, denoted by {. Clearly, the choice of RS determines the activity ai. A number of RSs are available. We will deal mainly with the pure component RS, originally introduced in Lecture 24, and discussed again for completeness next.

26.3

Pure Component Reference State

Recall that the pure component reference state has the following attributes: • Same state of aggregation as the solution • P{ ¼ P (Recall that T{ ¼ T) • x{i ¼ 1, x{j ¼ 0, for j 6¼ i { { • bf i ¼ f i ðT, PÞ and Gi ¼ Gi ðT, PÞ

Therefore, it follows that for this RS: ai ¼

bf i ðT, P, x1 , . . . , xn1 Þ f i ðT, PÞ

ð26:6Þ

or, rearranging Eq. (26.6), that: bf i ¼ f i ai

ð26:7Þ

Equation (26.7) shows that if we know ai, we can calculate bf i . In addition, combining Eqs. (26.4), for the pure component RS, with Eq. (26.6) yields: RTlnai ¼ Gi  Gi ¼ ΔGi

ðPartial molar mixing Gibbs free energy of iÞ

ð26:8Þ

where ai ¼ ai(T, P, x1, . . ., xn-1). Equation (26.6) shows that the pure component RS is symmetric for every component i, in the sense that:

274

26

Activity, Activity Coefficient, and Sample Problems

f ðT, PÞ lim ai ¼ i ¼1 f i ðT, PÞ xi!1

ðFor every iÞ

ð26:9Þ

There are also asymmetric RSs, like the infinite-dilution RS, for which this is different (see below). For an Ideal (ID) Solution, we saw that: ID

ID

ΔGi ¼ Gi  Gi ¼ RTlnxi ¼ RTlnaID i

ð26:10Þ

Comparison of the last two terms in Eq. (26.10) shows that the activity of component i in an ideal solution is equal to xi. When the solution is not ideal, the ratio, ai/xi, is not unity. Therefore, the deviation of this ratio from unity serves as a quantitative measure of the solution non-ideality. The ratio is known as the activity coefficient of component i, and is given by: γi ¼

bf i bf i ai ; ai ¼ ) γi ¼ xi f i ðT, PÞ xi f i ðT, PÞ

ð26:11Þ

The expressions in Eq. (26.11) are all based on the pure component RS. The last term in Eq. (26.11) shows that, for pure component i, xi ¼ 1, bf i ¼ f i (T,P), and γ i (pure i) ¼ 1. In other words, for the pure component RS: limγi ¼ 1 xi !1

ð26:12Þ

For an Ideal (ID) Solution, the last term in Eq. (26.11) is given by: γID i ¼

bf ID i xi f i ðT, PÞ

ð26:13Þ

According to the Lewis and Randall Rule introduced in Lecture 25, it follows that: bf ID ¼ xi f i ðT, PÞ i

ð26:14Þ

Using Eq. (26.14) in Eq. (26.13) shows that for an ideal solution: γID i ¼ 1

ð26:15Þ

26.4

26.4

Calculation of Activity

275

Calculation of Activity

When the pure component RS is used, the activity coefficient can be related to the excess Gibbs free energy of mixing. To show this, we begin with (see the first term in Eq. (26.11)): a i ¼ γ i xi

ð26:16Þ

Taking the natural logarithm of Eq. (26.16), including multiplying the resulting expression by RT, and using Eq. (26.8), yields: RTlnai ¼ RTlnðγi xi Þ ¼ Gi  Gi ¼ ΔGi

ð26:17Þ

For an Ideal (ID) Solution, Eq. (26.17) can be expressed as follows:  ID  ID ID RTlnaID i ¼ RTln γi xi ¼ Gi  Gi ¼ ΔGi

ð26:18Þ

Subtracting Eq. (26.18) from Eq. (26.17), including rearranging, yields: 

γ RTln IDi γi



ID

EX

¼ ΔGi  ΔGi ¼ ΔGi

ID

EX

¼ Gi  Gi ¼ Gi

ð26:19Þ

In the derivation of Eq. (26.19), we have used relations presented in Lecture 25. Recalling that γID i ¼ 1, Eq. (26.19) can be written as follows: EX

RTlnγi ¼ ΔGi

EX

¼ Gi

ð26:20Þ

Equation (26.20) shows that RT times the natural logarithm of the activity coefficient of component i in the solution is equal to the partial molar excess Gibbs free energy of mixing of component i, which is equal to the partial molar excess Gibbs free energy of component i. Equation (26.20) is very useful, because it permits calculation of γi (and, therefore, of ai ¼ γixi) for a non-ideal solution if either GEX or ΔGEX are known. Specifically, given: EX

EX EX (i) ΔG ¼ ΔG ðT, P, N1 , . . . , Nn Þ ) RTlnγi ¼ ΔGi   ∂ΔGEX ¼ ∂Ni T,P,Nj½i

ð26:21Þ

276

26

Activity, Activity Coefficient, and Sample Problems EX

EX EX (ii) G ¼ G ðT, P, N1 , . . . , Nn Þ ) RTlnγi ¼ Gi  EX  ∂G ¼ ∂Ni T,P,Nj½i

EX

Because ΔGi

¼ RTlnγi and ΔGEX ¼

n P i¼1

ð26:22Þ

EX

Ni ΔGi , it follows that:

n X ΔGEX ¼ GEX ¼ RT Ni lnγi

ð26:23Þ

i¼1

The set of (n + 2) intensive variables, {T, P, γ1, . . ., γn}, satisfies the following generalized Gibbs-Duhem relation (gGDr) at constant T and P: n X

xi dlnγi ¼ 0

ðConstant T and PÞ

ð26:24Þ

i¼1

Equation (26.24) shows that, given (n-1) γis at constant T and P, the gGDr can be used to compute the nth γi by integration, to within an arbitrary constant of integration. One can also show that:   ∂lnγi ΔHi ¼ ∂T P,x RT2   ∂lnγi ΔVi ¼ RT ∂P T,x

ð26:25Þ ð26:26Þ

Typically, for liquids and solids, ΔHi and ΔVi are small, and therefore, γi is a weak function of T and P.

26.5

Sample Problem 26.1

For a binary solution of components 1 and 2, it is known that: ΔGEX ¼ NCx1x2 RT

ð26:27Þ

26.5

Sample Problem 26.1

277

where C is a parameter which does not depend on composition (it may depend on T and P), and N ¼ N1 + N2. (i) Show that ΔGEX is symmetric around x1 ¼ x2 ¼ 0.5. (ii) Calculate γ1(T, P, x2) and γ2(T, P, x2), and discuss the type of nonideality.

26.5.1 Solution (i) Equation (26.27) shows that ΔGEX vanishes when x1 ¼ 0 (pure 2) and when x2 ¼ 0 (pure 1). This is a manifestation of us choosing the pure component RS to calculate ΔGEX . Further, ΔGEX attains its maximum value when x1 ¼ x2 ¼ 0.5, and is symmetric in x1 (or x2) around 0.5 (see Fig. 26.1). In Fig. 26.1, we replaced x by X for better visualization.

Fig. 26.1

In fact, any mixing function whose RS is the pure components must vanish in the limit xi¼1 for each component i. (ii) To calculate γ1 and γ2, we use the expression derived above which we repeat below for completeness: EX

RTlnγi ¼ ΔGi

¼

  ∂ΔGEX ∂Ni T,P,Nj½i

ð26:28Þ

278

26

Activity, Activity Coefficient, and Sample Problems

It is useful to first rewrite ΔGEX in Eq. (26.28) as an explicit function of N1 and N2:    ΔGEX N N2 CN1 N2 ¼ ¼ CNx1 x2 ¼ CN 1 RT N N N

ð26:29Þ

ΔGEX CN1 N2 ¼ RT ðN1 þ N2 Þ

ð26:30Þ

or

Using Eq. (26.30) in the last term in Eq. (26.28) for i ¼ 1, we obtain:  EX   2 ΔG N2 RTlnγ1 ¼ ¼ RTC ¼ RTCx22 N1 þ N2 ∂N1 T,P,N2

ð26:31Þ

or 2

γ1 ¼ eCx2

ð26:32Þ

Equation (26.32) shows that when x2 ¼ 0 (pure 1), γ1 (x1 ¼ 1) ¼ 1, as required when we use the pure component RS. Similarly,  EX   2 ΔG N1 RTlnγ2 ¼ ¼ RTC ¼ RTCx21 N1 þ N2 ∂N2 T,P,N1

ð26:33Þ

or γ2 ¼ eCð1‐x2 Þ

2

ð26:34Þ

Equation (26.34) shows that when x2 ¼ 1 (pure 2), γ2 (x2 ¼ 1) ¼ 1, as required when we use the pure component RS Because x2 varies between 0 and 1 (as does x1 ¼ 1  x2), Eqs. (26.32) and (26.34) show that, at fixed T and P, γ1 and γ2 are both >1 over the entire composition range. We refer to this behavior as a positive deviation from ideality, because: γID i ¼ 1 ðFor i ¼ 1 and 2Þ

ð26:35Þ

26.6

Sample Problem 26.2

279

If γi Tob ðPÞ

ð27:3Þ

where in Eqs. (27.1), (27.2), and (27.3), the variables on the right-hand side of the inequality signs denote the vapor pressure, the freezing temperature, and the boiling temperature of the solvent (water), respectively. • Osmotic Pressure : To be discussed below

ð27:4Þ

The four solution properties in Eqs. (27.1), (27.2), (27.3), and (27.4) are known as colligative properties, namely, properties which in dilute solution are proportional to the solute concentration. Because in the dilute sugar-water solution there is a clear asymmetry between the solvent (x1!1), and the solute (x2!0), it makes sense to use an asymmetric reference state (RS), rather than to use the pure component RS for both the solute and the solvent (as we did in previous lectures). Indeed, we will continue to use the pure component RS for the solvent, but will use the infinite-dilution RS for the solute. Therefore, the resulting RS is asymmetric, as shown in Fig. 27.1. In the last entry of the left column of Fig. 27.1, the dilute solution behaves ideally (solvent wise), that is, γ1 ¼ 1 when x1!1. On the other hand, in the last entry of the right column of Fig. 27.1, the dilute solution behaves ideally (solute wise), that is, γ2 ¼ 1 when x2!0.

27.3

Phase Equilibria: Introduction

Next, we will discuss scenarios in which two or more phases are in equilibrium. Phase equilibria are important in many industrial applications of relevance to chemical engineers, including distillation, absorption, and extraction, where two

27.4

Criteria of Phase Equilibria

283

Fig. 27.1

phases are in contact. In some cases, three phases are in contact, for example, at the solid/liquid/vapor equilibrium observed at the triple point. The coexisting phases can be: Liquid/Vapor (L/V), Liquid/Liquid (L/L), Solid/Vapor (S/V), Solid/Liquid (S/L), and Solid/Solid (S/S). In the remainder of this lecture, and in Lectures 28, 29, and 30, we will discuss: 1. The general thermodynamic criteria of phase equilibria 2. The Gibbs Phase Rule (GPR) 3. The differential approach to phase equilibria 4. The integral approach to phase equilibria 5. Sample Problems which illustrate the implementation of the new material in 1 to 4 above

27.4

Criteria of Phase Equilibria

Consider π phases in equilibrium, where each phase contains n components. All the internal boundaries are open, diathermal, and movable. The entire composite system is therefore simple. Further, the entire composite system is surrounded by a closed, adiabatic, and rigid boundary, and therefore, is isolated.

284

27

Criteria of Phase Equilibria, and the Gibbs Phase Rule

Fig. 27.2

With each phase α, we can associate the following set of (n + 2) intensive variables:
α α α α  T , P , μ1 , μ2 , . . . , μαn

ð27:5Þ

which, as we know, are not independent, because they are related by the GibbsDuhem equation in phase α. Figure 27.2 illustrates the phases and variables characterizing the system under consideration. For the simple, composite, multi-phase, multi-component system depicted in Fig. 27.2, the following equilibrium conditions apply.

27.4.1 Thermal Equilibrium T1 ¼ T2 ¼ . . . ¼ Tα ¼ . . . ¼ Tπ

ð27:6Þ

Equation (27.6) indicates that thermal equilibrium implies no temperature gradients, and therefore, no heat transfer between the coexisting phases.

27.4

Criteria of Phase Equilibria

285

27.4.2 Mechanical Equilibrium P1 ¼ P2 ¼ . . . ¼ Pα ¼ . . . ¼ Pπ

ð27:7Þ

Equation (27.7) indicates that mechanical equilibrium implies no pressure gradients, and therefore, no volume transfer between the coexisting phases.

27.4.3 Diffusional Equilibrium μi 1 ¼ μi 2 ¼ . . . ¼ μi α ¼ . . . ¼ μi π ði ¼ 1, . . . , nÞ

ð27:8Þ

Equation (27.8) indicates that diffusional equilibrium implies no chemical potential gradients for every component i, and therefore, no mass transfer of every component i between the coexisting phases. What happens to the conditions of thermodynamic equilibrium in Eqs. (27.6), (27.7), and (27.8) if the multi-phase system is not simple? As an illustration, consider a binary water (•) + sugar (x) solution coexisting with pure water (•) across a membrane which only allows free passage of water. The resulting osmometry cell is shown in Fig. 27.3. In Fig. 27.3, the resulting two-phase system is not simple because the membrane is rigid (no pressure equality is possible), and is impermeable to the sugar molecules (no sugar chemical potential equality is possible). Accordingly, only two conditions of thermodynamic equilibrium apply. Specifically,

Fig. 27.3

286

27

Criteria of Phase Equilibria, and the Gibbs Phase Rule

TPure Water ¼ TSugar Solution Water Solution μPure ¼ μSugar w w

ð27:9Þ ð27:10Þ

Because the membrane is rigid, no pressure equality can be established at equilibrium. In fact, we can show that the pressure in the sugar solution compartment exceeds the pressure in the pure water compartment by an amount known as the osmotic pressure, π, which is the fourth colligative property discussed earlier. Specifically, PSugar Solution > PPure Water ) PSugar Solution  PPure Water ¼ π

27.5

ð27:11Þ

The Gibbs Phase Rule

We would like to determine how many independent intensive variables need to be specified to fully describe the intensive thermodynamic equilibrium state of a simple composite system consisting of π phases, each carrying n components, in thermodynamic equilibrium. The answer is embodied in the celebrated Gibbs Phase Rule. The following derivation of the Gibbs Phase Rule involves four steps: 1. Characterize phase alpha (α) using the following set of (n + 2) intensive variables:
α α α α  T , P , μ1 , μ2 , . . . , μαn

ð27:12Þ

2. Impose the Gibbs-Duhem equation in phase α: Sα dTα  Vα dPα þ

n X

Nαi dμαi ¼ 0 ðα ¼ 1, 2, . . . , πÞ

ð27:13Þ

i¼1

3. Impose the conditions of thermal, mechanical, and diffusional equilibrium: T1 ¼ T2 ¼ . . . Tα ¼ . . . ¼ Tπ  T ðThermal equilibriumÞ

ð27:14Þ

27.5

The Gibbs Phase Rule

287

P1 ¼ P2 ¼ . . . ¼ Pα ¼ . . . ¼ Pπ  P ðMechanical equilibriumÞ μ1i ¼ μ2i ¼ . . . ¼ μαi ¼ . . . ¼ μπi  μi ði ¼ 1, 2, . . . , nÞ ðDiffusional equilibriumÞ

ð27:15Þ ð27:16Þ

4. Relate the set of (n + 2) intensive variables fT, P, μ1 , . . . , μn g

ð27:17Þ

by π Gibbs-Duhem equations, one for each phase α. Specifically, Sα dT  Vα dP þ

n X

Nαi dμi ¼ 0

ðα ¼ 1, 2, . . . , πÞ

ð27:18Þ

i¼1

In other words, out of the (n + 2) intensive variables, only (n + 2  π) are independent. This number is referred to as the variance, and is denoted by L, where, L¼nþ2π

ð27:19Þ

Equation (27.19) is the Gibbs Phase Rule (GPR) and was originally derived by J.W. Gibbs in 1875. Note that if there are additional constraints (e.g., chemical reactions), L is decreased further by the number of additional constraints, r, that is, L ¼ n + 2  π  r. A simpler and illuminating derivation of the Gibbs Phase Rule, which is particularly useful if the systems considered are not simple, is presented next. However, to better understand the derivation for a non-simple system, we will first deal with a simple system, in which all the internal boundaries are open, diathermal, and movable. Using the Corollary to Postulate I, we select (n + 1) independent intensive variables to characterize each of the π coexisting phases of the simple system. As discussed in Part II, these π(n + 1) intensive variables are not independent. Indeed, the π(n + 1) intensive variables are related by the conditions of thermal equilibrium (TE), mechanical equilibrium (ME), and diffusional equilibrium (DE) which apply. If the multi-phase system is simple, all the conditions of TE, ME, and DE apply. It then follows that: (i)

TE : T1 ¼ T2 ¼ . . . ¼ Tπ , Imposes ðπ  1Þ constraints

ð27:20Þ

(ii)

ME : P1 ¼ P2 ¼ . . . ¼ Pπ , Imposes ðπ  1Þ constraints

ð27:21Þ

(iii)

μ1i ¼ μ2i ¼ . . . ¼ μαi ¼ . . . ¼ μπi , Imposes ðπ  1Þn constraints

ð27:22Þ

288

27

Criteria of Phase Equilibria, and the Gibbs Phase Rule

The total number of constraints relating the π(n + 1) intensive variables is therefore given by the sum of all the constraints in Eqs. (27.20), (27.21), and (27.22), that is, by: Total number of TE, ME, and DE constraints ¼ ðπ  1Þ þ ðπ  1Þ þ ðπ  1Þn ¼ ðπ  1Þðn þ 2Þ

ð27:23Þ

It then follows that the number of independent intensive variables, L, is given by: L ¼ πðn þ 1Þ  ðπ  1Þðn þ 2Þ

ð27:24Þ

Rearranging Eq. (27.24), including cancelling the equal terms, yields: L¼ nþ2π

ðGibbs Phase Rule for a simple systemÞ

ð27:25Þ

As expected, Eq. (27.25) is identical to Eq. (27.19). If the multi-phase system is not simple, then, not all the (TE + ME + DE) constraints apply, because some of the internal boundaries are closed, adiabatic, or rigid. Accordingly, L will be determined by: L ¼ π ðn þ 1Þ  ðNumber of constraints which applyÞ

ð27:26Þ

Equation (27.26) shows that, because there is no universal expression for the number of constraints which apply, there is no universal expression for L in the case of a multi-phase, non-simple system. Each case needs to be carefully examined to determine the number of constraints which apply. Returning to a simple system, Postulate I requires that we specify (n + 2) independent properties. If we choose L of these (n + 2) to be intensive, then, the number of extensive ones, E, is given by: E ¼ ðn þ 2Þ  L ¼ ðn þ 2Þ  ðn þ 2  π  rÞ ¼ n þ 2  n  2 þ π þ r ð27:27Þ or E¼πþr

ðFor a simple systemÞ

ð27:28Þ

27.5

The Gibbs Phase Rule

289

For example, for: 9 8 > =

r ¼0 ) E¼πþ0¼π > > ; : π

ð27:29Þ

If π ¼ 1, then E ¼ 1, and a natural choice of extensive variable is the number of moles, N. In addition, L ¼ 1 + 2–1 ¼ 2, and a natural choice of the two independent intensive variables is T and P.

Lecture 28

Application of the Gibbs Phase Rule, Azeotrope, and Sample Problem

28.1

Introduction

The material presented in this lecture is adapted from Chapter 15 in T&M. First, we will utilize the Gibbs Phase Rule derived in Lecture 27 to elucidate the (P-T) phase diagram of a one component (n ¼ 1) substance, in the absence of chemical reactions (r ¼ 0), which can exist in Solid (S), Liquid (L), and Vapor (V) phases. In each of the three phases, T and P can vary independently, and as a result, the system is referred to as divariant. Second, Solid and Vapor, Solid and Liquid, and Liquid and Vapor can coexist along the S/V, S/L, and L/V equilibrium lines, respectively. Along each of these lines, P ¼ f(T), and the system is referred to as monovariant. The S/V, S/L, and L/V monovariant lines meet at a special point, known as the triple point, which is invariant. The L/V monovariant line begins at the triple point and ends at another special point, known as the critical point, which is invariant. Finally, we will qualitatively solve Sample Problem 28.1 to examine the boiling of a binary liquid mixture coexisting with its binary vapor mixture. Specifically, we will utilize the Gibbs Phase Rule to understand qualitatively how the mixture vapor pressure changes as a function of the solution composition at constant temperature, including discussing the pressure-composition phase diagram, and the azeotropic point.

28.2

The Gibbs Phase Rule for a Pure Substance

As discussed in Lecture 27, for a pure substance (n ¼ 1) existing in one phase (π ¼ 1), for example, Solid (S), Liquid (L), or Vapor (V), in the absence of chemical reactions (r ¼ 0), the variance is given by: L ðn ¼ 1, π ¼ 1, r ¼ 0Þ ¼ n þ 2  π  r ¼ 2

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_28

ð28:1Þ

291

292

28

Application of the Gibbs Phase Rule, Azeotrope, and Sample Problem

In other words, L ¼ 2, and therefore, we can vary at most two independent intensive variables, for example, T and P, in the (P-T) phase diagram. The system is said to be divariant. For a pure substance (n ¼ 1) coexisting in two phases (π ¼ 2), for example, Liquid/Vapor (L/V), Solid/Liquid (S/L), or Solid/Vapor (S/V), in the absence of chemical reactions (r ¼ 0), the variance is given by: L ðn ¼ 1, π ¼ 2, r ¼ 0Þ ¼ n þ 2  π  r ¼ 1

ð28:2Þ

In other words, L ¼ 1, and therefore, we can only vary one intensive variable along the L/V, S/L, or S/V coexistence lines, that is, P ¼ P(T), in a (P-T) phase diagram, and the system is said to be monovariant. For a pure substance (n ¼ 1) coexisting in three phases (π ¼ 3), for example, Solid/Liquid/Vapor (S/L/V), in the absence of chemical reactions (r ¼ 0), the variance is given by: L ðn ¼ 1, π ¼ 3, r ¼ 0Þ ¼ n þ 2  π  r ¼ 0

ð28:3Þ

In other words, L ¼ 0, which corresponds to the invariant triple point (TP) in the (P-T) phase diagram. The L/V coexistence line begins at the triple point (TP) and terminates at the critical point (CP). Therefore, the Gibbs Phase Rule helps us understand qualitatively the typical (P-T) phase diagram of a pure substance (see Fig. 28.1).

Fig. 28.1

Interestingly, as discussed in Part I, at the CP of a pure substance, two constraints s = 2, where s denotes the number of additional constraints which need to be enforced. Specifically,

28.3

Sample Problem 28.1

293



∂P ∂V

 T



2

∂ P ¼ 0 and ∂V2

 ¼0

ð28:4Þ

T

Accordingly, at the critical point of a pure substance, it follows that L (n ¼ 1, π ¼ 1, s ¼ 2) ¼ n + 2 – π – s ¼ 0, indicating that the critical point is invariant.

28.3

Sample Problem 28.1

Boiling of a binary liquid mixture coexisting with its binary vapor mixture. Examine qualitatively how the mixture vapor pressure varies with the mixture composition at constant temperature.

28.3.1 Solution For the liquid-vapor equilibrium of a binary mixture, it follows that: n ¼ 2, π ¼ 2, r ¼ 0 ⟹ L ¼ n + 2 – π – r ¼ 2. Accordingly, we need to specify two independent intensive variables, and any other intensive variable should be uniquely determined. In this Sample Problem, we are asked to examine how the vapor pressure of the binary (liquid) mixture varies with T and x1 (the liquid mole fraction of component 1) or with T and y1 (the vapor mole fraction of component 1). To be specific, we will assume that, at temperature T, the vapor pressure of o o component 1 [Pvp1 (T)] is larger than the vapor pressure of component 2 [Pvp2 (T)]. Without yet solving the liquid-vapor equilibrium problem quantitatively (we will learn how to do that in Lectures 29 and 30), let us examine how Pvp varies with x1 or y1 at constant T. Fig. 28.2 depicts the P versus x1 or P versus y1 phase diagram at a given temperature (T).

Fig. 28.2

294

28

Application of the Gibbs Phase Rule, Azeotrope, and Sample Problem

o o Figure 28.2 shows that for Pvp2 (T) < P0 < Pvp1 (T), the system is unstable, and 0 separates into a liquid phase at x1(T, P ) coexisting with a vapor phase at y1(T, P0 ). At each T and P0 , there is a unique value of x1(T, P0 ) and y1(T, P0 ), as required by the Gibbs Phase Rule, because L ¼ 2. Clearly, as shown in Fig. 28.2, the Gibbs Phase Rule helps us understand qualitatively the pressure-composition phase diagram of this binary mixture. There is a special temperature, known as the azeotropic temperature, at which T ¼ Taz and P ¼ Paz, where the liquid and vapor mole fractions are equal, that is:

At Taz , Paz )

az xaz 1 ¼ y1 az x2 ¼ yaz 2

ð28:5Þ

At the azeotrope, the functions P(T,x1) or P(T,y1) are no longer monotonic. The corresponding phase diagram is shown in Fig. 28.3.

Fig. 28.3

If we choose (Taz and P0 ) as the two independent intensive variables (as required by the Gibbs Phase Rule, where L ¼ 2), we do not know if we are located on Lobe I or on Lobe II (see Fig. 28.3). Therefore, a better choice in this case is (Taz, xI1) or (Taz, yI1), or (Taz, xII1 ) or (Taz, yII1 ), which locates us on Lobes I or II. This then enables us to unambiguously characterize the intensive thermodynamic equilibrium state of the system. This example involving the azeotrope shows that great care is sometimes required when one applies the Gibbs Phase Rule.

Lecture 29

Differential Approach to Phase Equilibria, Pressure-Temperature-Composition Relations, Clausius-Clapeyron Equation, and Sample Problem

29.1

Introduction

The material presented in this lecture is adapted from Chapter 15 in T&M. First, we will solve Sample Problem 29.1 to discuss how to model the phase equilibria of a binary liquid mixture coexisting with its binary vapor mixture. Recall that, in Lecture 28, we discussed various aspects of this problem qualitatively in the context of the Gibbs Phase Rule. To solve this problem quantitatively, we can pursue two modeling approaches: (i) the Differential Approach to Phase Equilibria, which equates the differentials of the natural logarithms of the fugacities of every component present in the binary liquid mixture and in the binary vapor mixture, and (ii) the Integral Approach to Phase Equilibria, which equates the fugacities of every component present in the binary liquid mixture and in the binary vapor mixture. In this lecture, we will discuss approach (i) in great detail, and reserve our discussion of approach (ii) to Lecture 30. Specifically, to implement approach (i), we will expand the differentials of the natural logarithms of the fugacities as a function of temperature, pressure, and mixture composition, using the expressions derived in Lecture 25. Doing so, we will derive two equations relating the four unknowns, dT, dP, the differential of the liquid mole fraction of component 1, and the differential of the vapor mole fraction of component 1. According to the Gibbs Phase Rule, for this two-phase, two-component system in the absence of chemical reactions, the number of independent intensive variables is L ¼ 2. This implies that by solving the two equations simultaneously, given two intensive variables, for example, T and P, we will be able to uniquely determine the liquid and the vapor mole fractions of component 1. Third, we will discuss eight simplifications of the two equations which will lead to simple solutions. Finally, we will solve the two equations simultaneously, including discussing the types of predictions that we can make.

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_29

295

296

29.2

29

Differential Approach to Phase Equilibria, Pressure-Temperature-Composition. . .

Sample Problem 29.1

Model the phase equilibria of a binary liquid mixture coexisting with its binary vapor mixture.

29.2.1 Solution To solve this phase equilibria problem, we will undertake the following steps: 1. Draw the system and specify the independent intensive variables in each phase using the Gibbs Phase Rule, as shown in Fig. 29.1.

Fig. 29.1

2. Determine the variance of the two-phase system: L¼nþ2πr

ð29:1Þ

n ¼ 2, π ¼ 2 ðLiquid þ VaporÞ, r ¼ 0 ) L ¼ 2

ð29:2Þ

According to the Gibbs Phase Rule, L ¼ 2 (see Eq. (29.2)), and therefore, if we choose two independent intensive variables, all the other intensive variables must be uniquely determined. Clearly, out of the six intensive variables, {TV, PV, y1, TL, PL, x1}, in Fig. 29.1, only two are independent. 3. Impose the conditions of phase equilibria: (a) Thermal Equilibrium

29.2

Sample Problem 29.1

297

TL ¼ TV ¼ T

ð29:3Þ

PL ¼ PV ¼ P

ð29:4Þ

(b) Mechanical Equilibrium

(c) Diffusional Equilibrium bL bV bL bV μ1L ¼ μV 1 ) λ1 ðTÞ þ RTln f 1 ¼ λ1 ðTÞ þ RTln f 1 ) f 1 ¼ f 1

ð29:5Þ

bL bV bL bV μL2 ¼ μV 2 ) λ2 ðTÞ þ RTln f 2 ¼ λ2 ðTÞ þ RTln f 2 ) f 2 ¼ f 2

ð29:6Þ

Recall that: bf L ¼ bf L ðT, P, x1 Þ; bf L ¼ bf L ðT, P, x2 Þ 1 1 2 2

ð29:7Þ

bf V ¼ bf V ðT, P, y1 Þ; bf V ¼ bf V ðT, P, y2 Þ 1 1 2 2

ð29:8Þ

and

where in Eqs. (29.7) and (29.8), x2 ¼ 1 – x1, and y2 ¼ 1 – y1. 4. Solve the fugacity equations: (a) Use the Integral Approach In the Integral Approach, we equate the fugacity of component i (1 and 2) in each coexisting phase (L and V). Specifically, bf L ðT, P, x1 Þ ¼ bf V ðT, P, y1 Þ 1 1

ð29:9Þ

bf L ðT, P, x2 Þ ¼ bf V ðT, P, y2 Þ 2 2

ð29:10Þ

where x2 ¼ 1 – x1, and y2 ¼ 1 – y1. L L As we discussed in Part II, to model the liquid fugacities, bf 1 and bf 2 , we use an EX activity coefficient approach based on a model for ΔG . In addition, to model the V V vapor fugacities, bf 1 and bf 2 , we use a fugacity coefficient approach based on an Equation of State.

298

29

Differential Approach to Phase Equilibria, Pressure-Temperature-Composition. . .

In this interesting Sample Problem, there are four unknowns: T, P, x1, and y1, which are related by the two fugacity equalities in Eqs. (29.9) and (29.10). This indicates that in order to obtain a unique solution, we must specify two out of the four unknown intensive variables, as required by the Gibbs Phase Rule, which indicated that L ¼ 2. For example, if we choose T and P, we can calculate the liquid composition x1 and the vapor composition y1. (b) Use the Differential Approach

 L   V In the differential approach, we equate d ln bf i with d ln bf i for i ¼ 1 and

2, that is:  L  V d lnbf 1 ¼ d ln bf 1

ð29:11Þ

 L  V d lnbf 2 ¼ d lnbf 2

ð29:12Þ

and

 α and we then expand each d lnbf i term (for i ¼ 1 and 2, and α ¼ L and V) in terms V

V

of T, P, and composition. As we discussed in Part II, because bf i ¼ bf i ðT, P, yi Þ for i ¼ 1 and 2, it follows that:  V d lnbf i ¼

V!

∂ln bf i ∂T

P,yi

V ∂lnbf i dT þ ∂P

! T,yi

V ∂lnbf i dP þ ∂yi

! dyi

ð29:13Þ

T,P

We should recognize that if n > 2, additional dyi terms will appear in Eq. (29.13). L L Further, because bf i ¼ bf i ðT, P, xi Þ for i ¼ 1 and 2, it follows that: L  L ∂lnbf i b d ln f i ¼ ∂T

! P,xi

L ∂lnbf i dT þ ∂P

! T,xi

L ∂lnbf i dP þ ∂xi

! dxi

ð29:14Þ

T,P

Again, if n > 2, additional dxi terms will appear in Eq. (29.14). As shown in Part II, V ∂lnbf i ∂T

!

V

P,yi

Hi  Hoi ðTÞ ¼ RT2

! ð29:15Þ

29.2

Sample Problem 29.1

299 L ∂lnbf i ∂T

!

L

P,xi

V ∂lnbf i ∂P

L ∂lnbf i ∂P

Hi  Hoi ðTÞ ¼ RT2 !

V

¼

¼

Vi RT

!

L

T,xi

ð29:16Þ

!

Vi RT

T,yi

!

ð29:17Þ

! ð29:18Þ

 V  L Using Eqs. (29.15)–(29.18) in d lnbf i ¼ d lnbf i ,

for i ¼ 1 and 2 (see

Eqs. (29.13) and (29.14)) yields the following equation: ! ! V! V V Hi  Hoi ðTÞ Vi ∂lnbf i dyi ¼ dP þ dT þ  RT ∂yi RT2 T,P ! ! L! L L o b Hi  Hi ðTÞ Vi ∂ln f i  dxi dP þ dT þ RT ∂xi RT2

ð29:19Þ

T,P

Combining the dT and dP terms in Eq. (29.19) yields: V

L

Hi  Hi  RT2

L ∂lnbf i  ∂xi

¼0

!

V

dT þ !

L

Vi  Vi RT

!

V ∂lnbf i dP þ ∂yi

! dyi T,P

dxi T,P

ð29:20Þ

where i ¼ 1 and 2, dy2 ¼  dy1 and dx2 ¼  dx1. Equation (29.20), for both i ¼ 1 and i ¼ 2, relates the four unknown intensive variables T, P, x1, and y1. Therefore, to obtain a unique solution, we must specify two of these four unknowns in order to determine the remaining two unknowns. This is fully consistent with the Gibbs Phase Rule which indicated that L ¼ 2. Next, we write Eq. (29.20) separately for i ¼ 1 and i ¼ 2 in terms of the four unknowns: T, P, x1, and y1. This yields:

300

29

Differential Approach to Phase Equilibria, Pressure-Temperature-Composition. . .

! V L H1  H1  dT þ RT2 L! ∂lnbf 1  dx1 ∂x1

! V! V L V1  V1 ∂lnbf 1 dP þ RT ∂y1

dy1

T,P

T,P

¼0

ð29:21Þ

and ! V L H2  H2  dT þ RT2 L! ∂lnbf 2  dx1 ∂x1

! V! V L V2  V2 ∂lnbf 2 dP þ RT ∂y1

dy1

T,P

T,P

¼0

ð29:22Þ

where in Eq. (29.22), we replaced dy2 by -dy1, ∂y2 by -∂y1, dx2 by -dx1, and ∂x2 by -∂x1. Before we discuss the general solution of Eqs. (29.21) and (29.22), we first consider several simplifications.

29.3

Simplifications of Eqs. (29.21) and (29.22)

1. Only component 1 is volatile (that is, evaporates). In that case, the vapor phase is pure component 1, so that only Eq. (29.21) applies, where: V

V

V y1 ¼ 1, dy1 ¼ 0, H1 ¼ HV 1 ðT, PÞ, and V1 ¼ V1 ðT, PÞ

ð29:23Þ

2. Only component 2 is volatile (that is, evaporates). In that case, the vapor phase is pure component 2, so that only Eq. (29.22) applies, where: V

V

V y2 ¼ 1, dy2 ¼ 0, H2 ¼ HV 2 ðT, PÞ, and V2 ¼ V2 ðT, PÞ

ð29:24Þ

3. Components 1 and 2 are volatile (that is, both evaporate), but the vapor mixture composition is constant. In that case, both Eqs. (29.21) and (29.22) apply, where:

29.3

Simplifications of Eqs. (29.21) and (29.22)

301

y1 ¼ const:, y2 ¼ 1  y1 ¼ const: ) dy1 ¼ 0, dy2 ¼ 0 4. Components 1 and 2 are volatile (that is, both evaporate), but the liquid mixture composition is constant. Once again, both Eqs. (29.1) and (29.2) apply, where: x1 ¼ const:, x2 ¼ 1  x1 ¼ const: ) dx1 ¼ 0, dx2 ¼ 0

ð29:25Þ

5. Pressure is kept constant, so that dP ¼ 0 in Eqs. (29.21) and (29.22). 6. Temperature is kept constant, so that dT ¼ 0 in Eqs. (29.21) and (29.22). 7. The vapor mixture is ideal. In that case: V  V ID ∂lnbf i bf ¼ Pi ¼ yi P, so that i ∂yi

!ID ¼ T,P

1 , for i ¼ 1 and 2 yi

ð29:26Þ

If the vapor mixture is not ideal, the dependence of the partial derivative in Eq. (29.26) on composition is more complex, and as discussed in Part II, can be calculated using an EOS for the vapor mixture. 8. The liquid mixture is ideal. In that case, the Lewis and Randall Rule introduced in Part II applies and states that: L  L ID ∂lnbf i L bf ¼ xi f i ðT, PÞ, so that i ∂xi

!ID ¼ T,P

1 , for i ¼ 1 and 2 xi

ð29:27Þ

If the liquid mixture is not ideal, the dependence of the partial derivative in Eq. (29.27) on composition is more complex, and as discussed in Part II, can be calculated using an excess Gibbs free energy of mixing approach to evaluate the liquid mixture fugacities. Recall that, in general: bf L ¼ γL xi f i ðT, PÞ; RT ln γL ¼ i i i

  ∂ΔGEX ∂NLi T,P,NL

ð29:28Þ

j½i

where γLi depends on T, P, and x1, ..., xn-1. In the various problems that we will encounter, some (and if we are fortunate, all) of the simplifications (1) to (8) above may apply. Next, let us return to Eqs. (29.21) and (29.22), and combine them to eliminate one of the mole fractions. For example, imagine that we want to calculate how the vapor mixture composition, y1, varies with T and P (recall that, according to the Gibbs Phase Rule, L ¼ 2). To obtain an equation relating dT, dP, and dy1, we proceed as follows:

302

29

Differential Approach to Phase Equilibria, Pressure-Temperature-Composition. . .

• Multiply Eq. (29.21) by x1 • Multiply Eq. (29.22) by x2 • Add the resulting equations This yields:



  9 8  V L 1) vapor mixture. To this end, we will utilize the equalities of the fugacities of each of the n components present in each phase to model the conditions of diffusional equilibrium. Specifically, gas-phase fugacities will be modeled using a fugacity-coefficient approach based on an EOS. Further, liquid-phase fugacities will be modeled using an activitycoefficient approach based on a model for the excess Gibbs free energy of mixing. Third, we will show how to relate the pure component fugacity at temperature T and pressure P to the pure component fugacity at temperature T and the vapor pressure of the same component at temperature T. The resulting relation will involve introducing the Poynting correction. Fourth, using the results above, we will derive n equations modeling the diffusional equilibrium of the n components, where these equations relate the 2n unknowns, T, P, the (n-1) liquid mixture mole fractions, and the (n-1) vapor mixture mole fractions. In addition, using the Gibbs Phase Rule, we will show that L ¼ n. This implies that if we specify n of the 2n unknowns, we should be able to uniquely calculate the n remaining ones by simultaneously solving the n equations discussed above. Fifth, we will discuss four simplifications of the n equations which will lead to simpler solutions, including the celebrated Raoult’s law. Finally, we will

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_30

305

306

30

Pure Liquid in Equilibrium with Its Pure Vapor, Integral Approach to Phase. . .

present several mathematical models, known as composition models, that can be used to evaluate the excess Gibbs free energy of mixing.

30.2

From Lecture 29

For completeness, below, we present again Eqs. (29.21) and (29.22), derived in the context of the differential approach to phase equilibria in Lecture 29, which relate the four unknowns, dT, dP, dx1, and dy1. Specifically, ! ! V L V L H1  H1 V1  V1  dP þ dT þ RT RT2 L! ∂lnbf 1  dx1 ∂x1

V ∂lnbf 1 ∂y1

! dy1 T,P

T,P

¼0

ð29:21Þ

and ! ! V L V L H2  H2 V2  V2 dP þ  dT þ RT RT2 L! ∂lnbf 2  dx1 ∂x1

V ∂lnbf 2 ∂y1

! dy1 T,P

T,P

¼0

ð29:22Þ

where in Eq. (29.22), dy2 ¼  dy1, dx2 ¼  dx1, ∂y2 ¼  ∂y1, and ∂x2 ¼  ∂x1.

30.3

Pure Liquid in Equilibrium with Its Pure Vapor

First, let us consider a practically relevant application of Eqs. (29.21) and (29.22), that is, a pure liquid in equilibrium with its pure vapor. Because we are dealing with a pure (n ¼ 1) system, we can select to work with either Eq. (29.21) or Eq. (29.22). If we select Eq. (29.21), then, y1 ¼ 1, x1 ¼ 1, V1 ¼ V1 (molar volume of pure component 1), and H1 ¼ H1 (molar enthalpy of pure component 1). In that case, Eq. (29.21) yields:

30.3

Pure Liquid in Equilibrium with Its Pure Vapor

 V L  V L H1 ‐H1 V1 ‐V1  dP þ 0 ¼ 0 dT þ 2 RT RT

307

ð30:1Þ

where the numerator of the term multiplying dT is ΔHvap, minus the Molar Enthalpy of Vaporization of pure component 1, and the numerator of the term multiplying dP is ΔVvap, the Molar Volume of Vaporization of pure component 1. Accordingly, Eq. (30.1) can be expressed as follows: 

    ΔHvap ΔVvap dP ¼ 0 dT þ RT RT2

ð30:2Þ

Rearranging Eq. (30.2), we obtain an equation which describes how P varies with T along the L/V equilibrium monovariant (L ¼ 1) line in a Pressure (P)-Temperature (T) phase diagram (see Fig. 30.1). This equation is known as the Clapeyron Equation. The Clapeyron Equation also describes how P varies with T along the Solid/ Liquid [S/L] and the Solid/Vapor [S/V] equilibrium monovariant lines. Specifically:

Fig. 30.1



 dP ΔHfus ¼ dT ½S=L TΔVfus

ð30:3Þ

where ΔHfus is the Molar Enthalpy of Fusion, and ΔVfus is the Molar Volume of Fusion. In addition, 

dP dT

 ½S=V

¼

ΔHsub TΔVsub

ð30:4Þ

308

30

Pure Liquid in Equilibrium with Its Pure Vapor, Integral Approach to Phase. . .

where ΔHsub is the Molar Enthalpy of Sublimation, and ΔVsub is the Molar Volume of Sublimation. The (P-T) phase diagram depicted in Fig. 30.2 shows the various phases (L ¼ 2), equilibrium lines (L ¼ 1), the triple point (TP, L ¼ 0)), and the critical point (CP, L ¼ 0).

Fig. 30.2

In some cases, the Clapeyron Equation along the Liquid/Vapor [L/V] equilibrium line can be simplified. Specifically, if vaporization takes place at low pressures, one can assume that (1) the vapor phase is ideal. In addition, it is reasonable to assume that (2) the molar volume of the vapor, VV, is much larger than that of the liquid, VL. Assumption (1) implies that: VV ¼

RT ðIdeal gas behavior of the vaporÞ P

ð30:5Þ

Assumption (2) implies that: ΔVvap ¼ VV  VL  VV ¼

RT P

ð30:6Þ

where we have used Eq. (30.5). Utilizing assumptions (1) and (2) above in the [L/V] Clapeyron Equation (see Fig. 30.1) yields:

30.3

Pure Liquid in Equilibrium with Its Pure Vapor



309

 ΔHvap dP ¼ ¼ dT ½L=V TΔVvap

ΔHvap   RT T P

ð30:7Þ

Rearranging Eq. (30.7), we obtain:  )

dP=P dT=T2

 ½L=V

  ΔHvap ΔHvap dðlnPÞ ¼ ¼ ) R R dð1=TÞ ½L=V

ð30:8Þ

or 

dðlnPÞ dð1=TÞ

 ½L=V

¼

ΔHvap R

ð30:9Þ

It is noteworthy that the vapor pressure is often denoted as PL/V, Pvp, or Psat. Equation (30.9) is known as the Clausius-Clapeyron equation, and can be plotted as shown in Fig. 30.3.

Fig. 30.3

Figure 30.3 and Eq. (30.9) show that the tangent to the line at any value of 1/T, say, at 1/T* (the dashed line), corresponds to the value of ΔHvap/R at T*. We should recognize that while the Clapeyron equation is general, the ClausiusClapeyron equation is an approximation.

310

30.4

30

Pure Liquid in Equilibrium with Its Pure Vapor, Integral Approach to Phase. . .

Integral Approach to Phase Equilibria

30.4.1 Sample Problem 30.1 Discuss all aspects of the phase equilibria of a multi-component (n > 1) liquid mixture coexisting with its multi-component (n > 1) vapor mixture in the absence of chemical reactions (r ¼ 0) (see Fig. 30.4).

Fig. 30.4

30.4.2 Solution Strategy For the multi-component (n > 1), two-phase (liquid/vapor; π ¼ 2) system in the absence of chemical reactions (r ¼ 0) shown in Fig. 30.4, the variance is given by 



L¼nþ2πr ¼nþ 2 20¼n

ð30:10Þ

In other words, out of the 2(n + 1) intensive variables, {TV, PV, y1, . . ., yn-1; TL, PL, x1, . . ., xn-1}, only n are independent. Next, we impose the conditions of phase equilibria: (a) Thermal Equilibrium TL ¼ TV  T

ð30:11Þ

PL ¼ P V  P

ð30:12Þ

(b) Mechanical Equilibrium

30.4

Integral Approach to Phase Equilibria

311

(c) Diffusional Equilibrium μLi ¼ μV i , i ¼ 1, 2, . . . , n

ð30:13Þ

As discussed in Part II, Eq. (30.13) can be expressed using fugacities instead of chemical potentials. This yields: L V λi ðTÞ þ RTlnbf i ¼ λi ðTÞ þ RTlnbf i , i ¼ 1, 2, . . . , n

ð30:14Þ

Cancelling the λi (T)s in Eq. (30.14) yields: bf L ¼ bf V , i ¼ 1, 2, . . . , n i i

ð30:15Þ

L L V V where bf i ¼ bf i ðT, P, x1 , . . . , xn1 Þ and bf i ¼ bf i ðT, P, y1 , . . . , yn1 Þ. Given L ¼ n of the 2n intensive variables, {T, P, x1, . . ., xn1; y1, . . ., yn1}, the n fugacity equations in Eq. (30.15) can be solved simultaneously to calculate the remaining n intensive variables.

30.4.3 Calculation of Vapor Mixture Fugacities In the integral approach to phase equilibria, we calculate the vapor mixture fugacV ities, bf i , using the fugacity-coefficient approach based on a reliable mixture EOS (see Lecture 25). Specifically: bf V V V bV b ϕi ¼ i ) bf i ¼ Pyi ϕ i Pi

ð30:16Þ

bf V ¼ yi ϕ b V ðT, P, y1 , . . . , yn1 Þ P i i

ð30:17Þ

or, more explicitly:

b V in Eqs. (30.16) and (30.17) can be calculated using a VolumeRecall that ϕ i Explicit (VE) or a Pressure-Explicit (PE) mixture EOS.

312

30

Pure Liquid in Equilibrium with Its Pure Vapor, Integral Approach to Phase. . .

30.4.4 Calculation of Liquid Mixture Fugacities In the integral approach to phase equilibria, we calculate the liquid mixture fugacL ities, bf i , using the activity-coefficient approach based on a model for the excess Gibbs free energy of mixing, ΔGEX, of the liquid mixture (see Lecture 26). Specifically, γL1 ¼

bf L =f L ðT, PÞ aLi ¼ i 1 xi xi

ð30:18Þ

where f Li(T, P) is the pure component i liquid fugacity. Below, we will show how to calculate these fugacities using a creative strategy. Equation (30.18) can be expressed as follows: bf L ¼ xi γ L ðT, P, x1 , . . . , xn1 Þ f L ðT, PÞ i i i

ð30:19Þ

Recall that γL1 can be calculated using the relation: RTlnγLi

¼

∂ΔGEX L ∂Ni

! ð30:20Þ T,P,NLj½i

V L Using the expressions for bf i and bf i in Eqs. (30.17) and (30.19), respectively, in Eq. (30.15) yields the desired n equations of the integral approach. Specifically,

b ðT, P, y1 , . . . , yn1 ÞP ¼ xi γL ðT, P, x1 , . . . , xn1 Þ f L ðT, PÞ, ði ¼ 1, 2, . . . , nÞ yi ϕ i i i V

ð30:21Þ where the 2n unknowns, {T, P, y1, . . ., yn1; x1, . . ., xn1}, are related by the n equations given in Eq. (30.21). The Gibbs Phase Rule indicates that L ¼ n. Therefore, if we specify L ¼ n of these unknowns, we can uniquely determine the remaining n by simultaneously solving the n equations in Eq. (30.21). In fact, the selection of the n variables that we specify determines the types of phase equilibria calculations that we can carry out. For example, if we specify {T, y1, . . ., yn1},we can calculate {P, x1, . . ., xn1}.

30.4

Integral Approach to Phase Equilibria

313

30.4.5 Calculation of Pure Component i Liquid Fugacity Typically, EOS for liquids are not reliable. Consequently, it is not recommended to calculate f Li ðT, PÞ using a pure component fugacity coefficient approach based on an EOS. Instead, we recommend to replace f Li ðT, PÞ by a vapor fugacity, which can be readily calculated if we have access to a reliable EOS for the vapor. We can do that at conditions where pure liquid i is at equilibrium with pure vapor i. This occurs at temperature, T, and a pressure which is equal to the vapor pressure, Pvpi(T), of pure liquid i (see Fig. 30.5, including the boxed equation).

Fig. 30.5

However, we need to calculate f Li at P 6¼ Pvpi ðTÞ, which requires relating:   f Li ðT, PÞ to f Li T, Pvpi ðTÞ ¼ f V i T, Pvpi ðTÞ

ð30:22Þ

We can readily do so by recalling that: 

∂lnf Li ∂P

 T

 L  L

VLi Vi

¼ ) d lnf i ¼ dP RT RT T

ð30:23Þ

where VLi is the molar volume of pure liquid i. Integrating the differential relation to the right of the arrow in Eq. (30.23), from Pvpi (T) to P, yields:

314

30

Pure Liquid in Equilibrium with Its Pure Vapor, Integral Approach to Phase. . .

"

# f Li ðT, PÞ ¼ ln L f i T, Pvpi ðTÞ

ðP 

 VLi dP RT

ð30:24Þ

Pvpi ðTÞ

or 2

ðP 

 6 f Li ðT, PÞ ¼ f Li T, Pvpi ðTÞ exp 4

VLi RT

3  7 dP5

ð30:25Þ

Pvpi ðTÞ

  Because f Li T, Pvpi ðTÞ is equal to f V i T, Pvpi ðTÞ , we can express Eq. (30.25) as follows: 2  6 f Li ðT, PÞ ¼ f V i T, Pvpi ðTÞ exp 4

ðP 

VLi RT

3  7 dP5

ð30:26Þ

Pvpi ðTÞ

where the exponential term in  Eq. (30.26) is known as the Poynting correction. Next, we can calculate f V T, P ð T Þ using the fugacity coefficient approach for vpi i pure vapor i based on an EOS. Specifically, ϕV i ¼

fV i ðT, PÞ , for P ¼ Pvpi ðTÞ P

ð30:27Þ

Rearranging Eq. (30.27), we obtain:   V fV i T, Pvpi ðTÞ ¼ ϕi T, Pvpi ðTÞ Pvpi ðTÞ

ð30:28Þ

Using Eq. (30.28) in Eq. (30.26), and then using the resulting equation in Eq. (30.21), we obtain the set of n equations relating the 2n intensive variables, {T, P, y1, . . ., yn1; x1, . . ., xn1}, in the context of the integral approach to phase equilibria. Specifically,  b V ðT, P, y1 , . . . , yn1 ÞP ¼ xi γL ðT, P, x1 , . . . , xn1 ÞϕV T, Pvpi ðTÞ Pvpi ðTÞCi yi ϕ i i i ð30:29Þ where

30.4

Integral Approach to Phase Equilibria

2 6 Ci ¼ exp 4

ðP 

315

3  7 5dP ðPoynting correction for i ¼ 1, 2, . . . , nÞ RT VLi

ð30:30Þ

Pvpi ðTÞ

Typically, Ci  1, but we must always check that this is indeed the case. For example, we can first assume that Ci ¼ 1, calculate P, and then carry out the integration in Eq. (30.30) to verify that it is indeed  1.

30.4.6 Simplifications of Eq. (30.29)  ID (1) The vapor mixture is ideal (ID): ϕV ¼ 1, for every i  L ID i ¼ 1, for every i (2) The liquid mixture is ideal: γ i  V ID (3) Pure gas i is ideal: ϕi ¼ 1, for every i (4) The Poynting correction is unity: Ci ¼ 1, for every i If simplifications (1)–(4) all apply, Eq. (30.29) reduces to: yiP ¼ Pi ¼ xi Pvpi(T), i ¼ 1, 2, . . ., n (Raoult0 s law)

(30.31)

If only simplifications (1), (3), and (4) apply, Eq. (30.29) reduces to: yi P ¼ xi γLi ðT, P, x1 , . . . , xn1 Þ Pvpi ðTÞ, i ¼ 1, 2, . . . , n ðModified Raoult0 s lawÞ ð30:32Þ To calculate γLi in Eqs. (30.29) and (30.32), we require mathematical models for ΔGEX , that can be used as follows: RTlnγLi

¼

∂ΔGEX L ∂Ni

! ð30:33Þ T,P,NLj½i

Several mathematical models for ΔGEX , for both binary mixtures (n ¼ 2) and multi-component mixtures (n > 2), are presented in the following tables (all adapted from T&M).

316

30

Pure Liquid in Equilibrium with Its Pure Vapor, Integral Approach to Phase. . .

30.4.7 Models for the Excess Gibbs Free Energy of Mixing of n = 2 Mixtures Type and name First-order polynomial Two-suffixa Margules

Three-suffixa Margules

van Laar

Operating equations ΔGEX ¼ Ax1x2 Binary parameters ¼ A RTlnγ1 ¼ Ax22 RTlnγ2 ¼ Ax21 ΔGEX ¼ x1x2 [A + B(x1  x2)] Binary parameters ¼ A, B RTlnγ1 ¼ ðA þ 3BÞx22  4Bx32 RTlnγ2 ¼ ðA  3BÞx21  4Bx31 Ax1 x2 ΔGEX ¼ x1 ðA=B Þþx2

Binary parameters ¼ A, B  2 1 RTlnγ1 ¼ A 1 þ Ax Bx2  2 Bx2 RTlnγ2 ¼ B 1 þ Ax 1 Four suffixa Margules

General Redlich-Kister

ΔGEX ¼ x1x2 [A + B(x1  x2) + C(x1  x2)2] Binary parameters ¼ A, B, C RTlnγ1 ¼ ðA þ 3B þ 5CÞx22  4ðB þ 4CÞx32 þ 12Cx42 RTlnγ2 ¼ ðA  3B þ 5CÞx21  4ðB  4CÞx31 þ 12Cx41 h i 2 ΔGEX RT ¼ x1 x2 B þ Cðx1  x2 Þ þ Dðx1  x2 Þ þ . . . Binary parameters ¼ B, C, D, . . .

Volume-fraction based Hildebrand-Scatchard

2

ΔGEX ¼  ðδ1 δ2 Þ  1 1 V1 x1 þV2 x2

V1, V2 ¼ molar volumes of pure 1 and 2 Pure component parameters ¼ δ1, δ2 2 RTlnγ1 ¼ V1 Φ2 2 ð δ1  δ2 Þ 2 RTlnγ2 ¼ V2 Φ1 ðδ1  δ2 Þ2  i xi Φi ¼ V1 xV1 þV i ¼ 1, 2 2 x2 Flory-Huggins

Φ1 Φ2 ΔGEX RT ¼ x1 ln x1 þ x2 ln x2 1 2 Φ1 ¼ N1 NþrN Φ2 ¼ N1rN þrN2 2

þ

χ Φ1 Φ2 ðN1 þrN2 Þ ðN1 þN2 Þ

r ¼ chain length of polymer ¼ number of monomer units Binary parameters ¼ χ; 2(1  solvent/monomer; 2  polymer) (continued)

30.4

Integral Approach to Phase Equilibria

Type and name General Wohl expansion

Local-composition based Wilson

TK-Wilsonf

317

Operating equations  ΔGEX ¼ ðx1 q1 þ x2 q2 Þ 2a12 z1 z2 þ 3a112 z21 z2 þ 3a122 z1 z22 þ RT 4a1112 z31 z2 þ 4a1222 z1 z32 þ 6a1122 z21 z22 þ . . . 9 x1 q1 > z1  = x1 q1 þ x2 q2 effective volume fraction x2 q2 > z2  ; x1 q1 þ x2 q2 qi ¼ volume fraction size of pure i (i ¼ 1, 2 Binary parameters ¼ a12, a112, a122, . . . ΔGEX RT

¼ ‐x1 ln ðx1 þ Λ12 x2 Þ  x2 ln ðx2 þ Λ21 x1 Þ Binary parameters ¼ Λ12, Λ21 ln γ1 ¼ ln(x1 + Λ12x2) + βx2 ln γ2 ¼ ln(x2 + Λ21x1) + βx1   Λ12 β ¼ x1 þΛ  Λ21Λx121þx2 12 x2 ΔGEX RT

¼ x1 ln

ðx1 þV2 x2 =V1 Þ ðx1 þΛ12 x2 Þ

þ x2 ln

ðV1 x1 =V2 þx2 Þ ðΛ21 x1 þx2 Þ

V1, V2 ¼ molar volumes of pure 1 and 2 β same as for Wilson model Binary parameters ¼ Λ12, Λ21 2 x2 =V 1 Þ ln γ1 ¼ ln ðxð1xþV þ ðβ  β Þx2 1 þΛ12 x2 Þ V2 =V1 =V2 β ¼ ðx1 þV  ðV1 xV11=V 2 x2 =x1 Þ 2 þx2 Þ   τ21 G21 τ12 G12 ΔGEX ¼ x x þ 1 2 x1 þx2 G21 RT x2 þx1 G12

b

NRTL

Δg21 12 where τ12 ¼ Δg RT , τ21 ¼ RT ln G12 ¼ α12τ12, ln G21 ¼ α12τ21 Binary parameters ¼ Δg12, Δg21, α12c    2 G21 τ12 G12 lnγ1 ¼ x22 τ21 x1 þx þ 2 G 2 21 ðx2 þx1 G12 Þ    2 G12 2 21 G21 lnγ2 ¼ x1 τ12 x2 þx1 G12 þ ðx τþx 2 G Þ 1

UNIQUAC

d

ΔG

2

21

¼ ΔG (combinatorial) + ΔGEX(residual)  Φ Φ ¼ x1 ln x11 þ x2 ln x22 þ 2z q1 x1 ln Φθ1 þ q2 x2 ln 1  0  0 ΔGEX ðresidualÞ 0 0 0 ¼ q x ln θ þ θ τ x ln θ þ θ01 τ12  q 1 1 2 2 1 2 21 2 RT EX

EX

ΔGEX ðcombinatorialÞ RT

θ1 Φ2



x q0

q1 1 r1 , θ1  x1 qx1þx θ01  x1 q0 1þx12 q0 Φ1  x1 rx1 þx 2 r2 2q 1

2

Δu12 21 lnτ21   Δu RT ,lnτ12   RT

1

2

(continued)

318

30

Pure Liquid in Equilibrium with Its Pure Vapor, Integral Approach to Phase. . .

Type and name

Operating equations r, q, and q0 are pure component parameters and coordination number z ¼ 10 Binary parameters ¼ Δu12 and Δu21e     Φ z θ r lnγi ¼ ln i þ qi ln i þ Φj li  i lj  q0i ln θ0i þ θ0j τji xi 2 Φi rj ! τji τij þθ0j q0i 0  θi þ θ0j τji θ0i þ θ0i τij where i ¼ 1, j ¼ 2 or i ¼ 2, j ¼ 1    li ¨ 2z ðri  qi Þ  ðri  1Þ; lj ¨ 2z rj  qj  rj  1

Two-suffix signifies that the expansion for ΔGEX is quadratic in mole fraction. Three-suffix signifies a third-order, and four-suffix signifies a fourth-order equation. b NRTL ¼ non-random two-liquid model. c Δg12 ¼ g12  g22; Δg21 ¼ g21  g11 d UNIQUAC ¼ universal quasi-chemical activity coefficient model. e Δu12 ¼ u12  u22; Δu21 ¼ u21  u11. f works for liquid-liquid systems. Sources: Prausnitz et al. (1986), 2nd ed, Chapter 6 and Walas (1985), Chapter 4 where the Margules, Redlich-Kister, van Laar, Wilson, TK-Wilson, Wohl, Hildebrand-Scatchard, NRTL, and UNIQUAC equations are discussed. a

30.4.8 Models for the Excess Gibbs Free Energy of Mixing of n > 2 Mixtures Model Two-Suffix Margules

ln γi n X n h X k¼1 j¼1

Wilson

1 Aki  Akj xk xj 2

ðbinary interactions onlyÞ 3 2 " # n n P P 6 ½Λki xk  7 Λij xj  1  ln 5 4P n j¼1

"

TK-Wilson  ln

n P

# Λij xj

" þ ln

n P

2

j¼1 n

P k¼1

Gki xk

3

Λij and Λji are defined as in Wilson model

Λkj xj

k¼1

3

2

j¼1

# Vj xj =Vi þ

n P

6 ðVi =Vk Þxk 7 5 4P n ðVj =Vk Þxj j¼1 32 P 3 n

k¼1

2 τji Gji xj

j¼1

n P 6 ½Λki xk  7  5 4P n

j¼1 n P

Λii ¼ Λjj ¼ 1   V λ Λij ¼ Vji exp RTij

Λkj xj

k¼1

j¼1

NRTL

Parameters required Ajj ¼ Akk ¼ 0 Akj ¼ Ajk

i

τmj Gmj xm n P 7 6 Gij xj 76 m¼1 þ 4P 54τij ‐ P 5 n n j¼1

Gkj xk

k¼1

Gkj xk

k¼1

g g

τji ¼ jiRT ii τii ¼ τjj ¼ 0 Gji ¼ exp(αji τji) Gii ¼ Gjj ¼ 1.0 (continued)

30.4

Integral Approach to Phase Equilibria

Model UNIQUAC

319

ln γi lnγi ¼ lnγCi þ lnγRi

Parameters required   ðuji uii Þ τji ¼ exp  RT

n τii ¼ τjj ¼ 1.0 Φ Φ P lnγCi ¼ ln xii þ 2z qi ln Φθi þ li ‐ xii xj lj i j¼1 2 33 2 " # li ¼ 2z ðri  qi Þ  ðri  1Þ n n P P 6 6 θj τij 77 R lnγi ¼ qi 41  ln θj τji  4P 55 n j¼1

Φi

θk τkj

j¼1

k¼1

¼P n

ri xi

k¼1

rk xk

qi xi

andθi ¼ P n k¼1

qk xk

z ¼ 10 (usually)

Lecture 31

Chemical Reaction Equilibria: Stoichiometric Formulation and Sample Problem

31.1

Introduction

The material presented in this lecture is adapted from Chapter 16 in T&M. So far in Part II, we have discussed multi-component (n > 1) and multi-phase (π > 1) systems under the assumption that the various chemical species comprising the system do not participate in any chemical reactions. In Lectures 31, 32, 33, 34, 35, and 36, we will allow the various chemical species to participate in chemical reactions, including generalizing the thermodynamic formulation accordingly. In this lecture, we will first contrast the calculation of changes in the thermodynamic properties of a binary mixture consisting of two inert components with that of a binary mixture where the two components participate in a dissociation reaction. Second, we will discuss the stoichiometric formulation of chemical reaction equilibria. Specifically, we will introduce stoichiometric numbers, stoichiometric coefficients, and the extent of reaction. Finally, we will solve Sample Problem 31.1, where 2 moles of methane and 3 moles of water are introduced into a closed chemical reactor, where they undergo two chemical reactions simultaneously to produce carbon monoxide, carbon dioxide, and hydrogen. Given the initial mole numbers of the reactants (methane and water), we will calculate the two extents of reaction, the final mole numbers of methane, water, carbon monoxide, carbon dioxide, and hydrogen, and their corresponding mole fractions.

31.2

Contrasting the Calculation of Changes in Thermodynamic Properties With and Without Chemical Reactions

When chemical reactions are involved, it is desirable to calculate equilibrium conversions, including the effect of temperature, pressure, and reactant composition on these conversions. We should stress that the calculation of reaction rates is © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_31

321

322

31 Chemical Reaction Equilibria: Stoichiometric Formulation and Sample Problem

outside the realm of thermodynamics. It is also practically important to determine the effects of temperature, pressure, and ratio of reactants on the equilibrium conversion of chemical reactions in order to optimize them. Finally there is an important need to calculate changes in thermodynamic properties when chemical reactions are involved in the thermodynamic process. A comparison of Case I and Case II below will illustrate this important need.

31.2.1 Case I: Closed Binary System of Inert Components 1 and 2 In Case I, in the set of variables, T, P, No1 , No2 , the initial mole numbers, No1 and No2 , remain fixed when the binary mixture evolves from the initial state i to the final state f. As a result, the change in the extensive property B as the binary mixture evolves from the initial state i to the final state f is given by:

 ΔBi!f ¼ Bf Tf , Pf , No1 , No2  Bi Ti , Pi , No1 , No2

ð31:1Þ

where 

No1 , No2

 i

  ¼ No1 , No2 f

ð31:2Þ

31.2.2 Case II: Closed Binary System of Components 1 and 2 Undergoing a Dissociation Reaction In this case, in the set of variables, T, P, N1, and N2, the initial mole numbers, No1 and No2 , change as the binary mixture evolves from the initial state i to the final state f. For example, suppose that a dissociation reaction takes place as follows: I2 ⇄ 2I, where I2 ¼ 1, and I ¼ 2. As we vary T and P, No1 and No2 will change to N1 and N2, where as we will show:

 N1 ¼ N1 T, P, No1 , No2 and N2 ¼ N2 T, P, No1 , No2

ð31:3Þ

As a result, the change in the extensive property B as the binary mixture evolves from the initial state i to the final state f is given by:

31.3

Stoichiometric Formulation

323


 ΔBi!f ¼ Bf ðTf , Pf , N1 , N2 Þ  Bi Ti , Pi , No1 , No2

ð31:4Þ

N1  Nf1 6¼ No1 , and N2  Nf2 6¼ No2

ð31:5Þ

where

31.3

Stoichiometric Formulation

As an illustration, we begin with the following simple, stoichiometrically-balanced chemical reaction: CH4 þ H2 0 ⇄ CO þ 3H2

ð31:6Þ

where the coefficients multiplying the various species i, including a sign convention, are referred to as stoichiometric numbers, and are denoted by υi. The convention for the sign of υi is as follows: ð1Þ υi > 0, For products ð2Þ υi < 0, For reactants

ð31:7Þ

ð3Þ υi ¼ 0, For inert species Specifically, for the chemical reaction in Eq. (31.6), it follows that: υCH4 ¼ 1, υH2 O ¼ 1, For the two reactants

ð31:8Þ

υCO ¼ þ1, υH2 ¼ þ3, For the two products

ð31:9Þ

For the chemical reaction in Eq. (31.6), the changes in the numbers of moles of the species present are in direct proportion to the stoichiometric numbers. For example, if 0.5 moles of CH4 disappear by reaction, then, 0.5 moles of H2O must also disappear. In addition, simultaneously, 0.5 moles of CO and 1.5 moles of H2 are formed by reaction, where: ΔNCH4 ΔNH2 O ΔNCO ΔNH2 ¼ ¼ ¼ υCH4 υH2 O υCO υH2 or

ð31:10Þ

324

31 Chemical Reaction Equilibria: Stoichiometric Formulation and Sample Problem

0:5 0:5 þ0:5 þ1:5 ¼ ¼ ¼ 1 1 þ1 þ3

ð31:11Þ

For a general chemical reaction: j υ1 j A1 þ j υ2 j A2 þ . . . ⇄ j υ3 j A3 þ j υ4 j A4 þ . . . where the |υi|s are the stoichiometric coefficients, and the Ais stand for the chemical formulas of the species. As shown in Eq. (31.7), the υis themselves (with the sign included) are referred to as stoichiometric numbers. For a stoichiometrically-balanced chemical reaction, if we consider a differential amount of reaction, we can write: dN1 dN2 dN3 dNn ¼ ¼ ¼ ... ¼ υ1 υ2 υ3 υn

ð31:12Þ

where each term is related to an amount, or an extent, of reaction, as represented by a change in the number of moles of a chemical species. Because all the n ratios in Eq. (31.12) are equal, they can be identified collectively with a single quantity, dξ, where the variable ξ is referred to as the extent of reaction, or the reaction coordinate. In other words, dNj ¼ υj dξðj ¼ 1, 2, . . . , nÞ

ð31:13Þ

In Eq. (31.13), n is the number of reactive species, and includes inert species, for which υj ¼ 0. If several chemical reactions take place simultaneously, then, for reaction r, there is an extent of reaction ξr. In that case, component j participating in chemical reaction r will satisfy: dNjr ¼ υjr dξr ðr ¼ 1, 2, . . . , mÞ

ð31:14Þ

where m is the number of independent chemical reactions. Clearly, if we add up the changes in the dNjrs over all the chemical reactions in which component j participates, we obtain the total change in the number of moles of component j. Specifically, dNj ¼

m X r¼1

dNjr ¼

m X

υjr dξr

ð31:15Þ

r¼1

where Eq. (31.14) was used. In Eq. (31.15), the sum is over the m independent chemical reactions, and (j ¼ 1, 2, . . ., n), including inert species for which υjr ¼ 0. Integrating Eq. (31.15) from (ξr ¼ 0, Nj ¼ Njo) to some (ξr, Nj), we obtain:

31.3

Stoichiometric Formulation

325

Nj ¼ Nj ðInitialÞ þ

m X

υjr ξr

ð31:16Þ

r¼1

where Nj (Initial) ¼ Njo, and j ¼ 1, 2, . . ., n. In Eq. (31.16), Nj represents the number of moles of component j in the equilibrium mixture, and depends on the number of moles of component j initially charged, Njo, as modified by all the m chemical reactions in which component j participates. Summing over all the species in Eq. (31.16), including the inert ones for which υjr ¼ 0, we obtain: N¼

n X

Nj ¼

j¼1

n X

Njo þ

m X

! υjr ξr

ð31:17Þ

r¼1

j¼1

or N¼

n X

Njo þ

n X m X

υjr ξr

ð31:18Þ

j¼1 r¼1

j¼1

Let us first deal with the double summation in Eq. (31.18), that is, with: n X m X

υjr ξr ¼

j¼1 r¼1

Denoting

n P

m n X X r¼1

! υjr ξr

ð31:19Þ

υr ξr

ð31:20Þ

j¼1

υjr ¼ υr , it follows that:

j¼1 n m X X r¼1

! υjr ξr ¼

n X r¼1

j¼1

Using Eq. (31.20) in Eq. (31.18) yields: N ¼ No þ

m X

υ r ξr

ð31:21Þ

r¼1

where N¼

n X j¼1

Nj , No ¼

n X j¼1

Njo , υr ¼

n X j¼1

υjr

ð31:22Þ

326

31.4

31 Chemical Reaction Equilibria: Stoichiometric Formulation and Sample Problem

Important Remark

Equation (31.22) shows that although the system is closed, N changes with respect to No. This follows because we are considering a balance on the components, instead of a balance on the individual atoms. If we did consider a balance on the individual atoms, then, for a closed system, NInitial Atom,i ¼ Constant

ð31:23Þ

irrespective of any chemical reactions. The mole fraction of component j in the mixture is obtained by taking the ratio of Nj in Eq. (31.16) and N in Eq. (31.21). Specifically, 0 yj ¼

Nj B ¼B N @

Njo þ No þ

m P r¼1 m P

υjr ξr υ r ξr

1 C C A

ð31:24Þ

r¼1

31.5

Sample Problem 31.1

Consider a reactor in which we initially charge 2 moles of methane (CH4) and 3 moles of water (H2O). It is known that the following two chemical reactions take place simultaneously: CH4 þ H2 O ⇄ CO þ 3H2 , r ¼ 1

ð31:25Þ

CH4 þ 2H2 O ⇄ CO2 þ 4H2 , r ¼ 2

ð31:26Þ

If initially NoCH4 ¼ 2 moles, NoH2 O ¼ 3 moles, and NoCO ¼ NoCO2 ¼ NoH2 ¼ 0

ð31:27Þ

Determine the dependence of the five Njs and yjs on ξ1 and ξ2 as the two chemical reactions proceed.

31.5

Sample Problem 31.1

327

31.5.1 Solution It is useful to construct a stoichiometric table to keep track of the various υjr's corresponding to the two simultaneous chemical reactions. Specifically, r j CH4  1 H2O  2 CO  3 CO2  4 H2  5 5 P υr ¼ υjr

Reaction 1, r ¼ 1, υj1 υ11 ¼ 1 υ21 ¼ 1 υ31 ¼ +1 υ41 ¼ 0 υ51 ¼ +3 υ1 ¼ +2

Reaction 2, r ¼ 2, υj2 υ12 ¼ 1 υ22 ¼ 2 υ32 ¼ 0 υ42 ¼ +1 υ52 ¼ +4 υ2 ¼ +2

j¼1

We also know that: NoCH4 ¼ 2 moles

ð31:28Þ

NoH2 O ¼ 3 moles

ð31:29Þ

¼0

ð31:30Þ

NoCO2 ¼ 0

ð31:31Þ

NoH2 ¼ 0

ð31:32Þ

NoCO

Using the stoichiometric table above, and the definitions of Nj and yj above, we obtain the following result:

328

31 Chemical Reaction Equilibria: Stoichiometric Formulation and Sample Problem

It is noteworthy that both Nj and yj depend on ξ1 and ξ2. Therefore, as the two chemical reactions proceed forward, both Nj and yj change for each j.

Lecture 32

Criterion of Chemical Reaction Equilibria, Standard States, and Equilibrium Constants for Gas-Phase Chemical Reactions

32.1

Introduction

The material presented in this lecture is adapted from Chapter 16 in T&M. First, we will derive the criterion of chemical reaction equilibria, which establishes a relation between the chemical potentials of every component i (1, 2, . . ., n) participating in chemical reaction r (1, 2, . . ., m). Second, expressing the chemical potentials in terms of fugacities, and using these expressions in the criterion of chemical reaction equilibria, we will derive an expression for the equilibrium constant associated with chemical reaction r, expressed in terms of the product of the ratios of the fugacities of component i in the actual mixture and in a suitably chosen reference state, both at the same temperature, raised to the power of the stoichiometric number associated with component i in chemical reaction r. Third, we will define the standard molar Gibbs free energy of reaction r and show that the equilibrium constant associated with chemical reaction r can be expressed as the exponential of minus the standard molar Gibbs free energy of reaction r divided by RT. Fourth, we will focus on a single chemical reaction (m ¼ 1), and discuss the selection of reference states to model gas-phase, liquid-phase, and solid-phase chemical reactions, where the last two are collectively referred to as condensed-phase chemical reactions. In particular, we will discuss the selection of the reference state pressure, including its effect on the equilibrium constant. Fifth, for a gas-phase chemical reaction, we will decompose the fugacity of component i into a product of the fugacity coefficient of component i in the gas mixture, the mixture pressure, and the mole fraction of component i in the gas mixture. Using this fugacity decomposition, we will show that the equilibrium constant can be expressed as a product of contributions from the fugacity coefficients, the gas mixture mole fractions, and the pressure. Finally, we will consider the limit of an ideal gas mixture, including deriving an expression for the equilibrium constant.

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_32

329

330

32.2

32 Criterion of Chemical Reaction Equilibria, Standard States, and Equilibrium. . .

Derivation of the Criterion of Chemical Reaction Equilibria

Consider a multi-component (n > 1), single-phase (π ¼ 1) system consisting of {N1, N2, . . ., Nn} moles at T and P. The Gibbs free energy, G, is a function of the (n + 2) variables fT, P, N1 , N2 , . . . , Nn g, i:e:, G ¼ GðT, P, N1 , N2 , . . . , Nn Þ, and its differential is given by: dG ¼ SdT þ VdP þ

n X

μj dNj

ð32:1Þ

j¼1

In Eq. (32.1), in the absence of chemical reactions, the various mole numbers, {N1, N2, . . ., Nn}, are free to vary in any manner that we choose, except for the possible constraint that: n X

dNj ¼ 0

ð32:2Þ

j¼1

if the overall system is closed. However, in the presence of m independent chemical reactions, this is no longer the case, and the various dNjs are all related to each other through the extents of reaction, ξr. Specifically, as shown in Lecture 31, m X

dNj ¼

υjr dξr

ð32:3Þ

r¼1

Using Eq. (32.3) in Eq. (32.1) yields: dG ¼ SdT þ VdP þ

m n X X r¼1

! υjr μj dξr

ð32:4Þ

j¼1

Equation (32.4) shows that G ¼ GðT, P, ξ1 , . . . , ξm Þ: As a result, the variations of the various ξrs in dG are independent. At equilibrium, at constant T and P, the Gibbs free energy must attain its minimum value. This requires that the first derivative of G with respect to ξs, keeping all the other ξrs 6¼ ξs, T, and P constant, be zero, that is: 

∂G ∂ξs

 ¼0 T,P,ξr6¼s

ð32:5Þ

32.3

Derivation of the Equilibrium Constant for Chemical Reaction r

331

Figure 32.1 helps us understand Eq. (32.5):

Fig. 32.1

Using Eq. (32.4) in Eq. (32.5), it follows that: 

∂G ∂ξs

 ¼ T,P,ξr6¼s

n X

υjs μj

ð32:6Þ

j¼1

Equations (32.5) and (32.6) show that, at equilibrium, we have: n X

υjr μj ¼ 0

ð32:7Þ

j¼1

where we replaced the dummy index s by r. Equation (32.7) is the condition of chemical reaction equilibria, a central result which relates the chemical potentials of the n components participating in chemical reaction r.

32.3

Derivation of the Equilibrium Constant for Chemical Reaction r

Next, we can express μj in Eq. (32.7) using its relation to the fugacity of component j in the mixture, bf j . Specifically, as we showed in Part II,

32 Criterion of Chemical Reaction Equilibria, Standard States, and Equilibrium. . .

332

μj ¼ Gj ¼ λj ðTÞ þ RTlnbf j

ð32:8Þ

Because we have no information about λj(T) in Eq. (32.8), we would like to eliminate it. To this end, we select a reference state, denoted by (o), for which To ¼ T. In addition, we select the reference state to be a pure component reference state, for which xoj¼ 1. Note that the reference state pressure, Po, is left arbitrary for now, that is, Po 6¼ P. In this reference state (RS), we have: xoj ¼ 1, To ¼ T, Po , f oj ¼ f oj ðT, Po Þ

ð32:9Þ

Therefore, using the reference state in Eq. (32.9) in Eq. (32.8) yields: μoj ¼ Goj ðT, Po Þ ¼ λj ðTÞ þ RTln f oj

ð32:10Þ

Because λj(T) appears in both Eq. (32.8) for μj and Eq. (32.10) for μoj , by subtracting Eq. (32.10) from Eq. (32.8), the two λj(T)s cancel out, and we obtain: μj ¼

Goj ðT, Po Þ

b  f j ðT, P, CompositionÞ þ RTln f oj ðT, Po Þ

ð32:11Þ

Next, we can use Eq. (32.11) for μj in the condition of chemical reaction equilibria, Eq. (32.7). This yields: n X

υjr

Goj

j¼1

bf j þ RTln o fj

!! ¼0

ð32:12Þ

Rearranging Eq. (32.12) yields: 

n X

! υjr Goj

¼ RT

j¼1

n X j¼1

bf j υjr ln o fj

! ð32:13Þ

First, let us work on the right-hand side of Eq. (32.13). Specifically,

RT

n X j¼1

bf j υjr ln o fj

! ¼ RT

n X j¼1

" ln

bf j f oj

!υjr #

( ¼ RT ln

n b Y fj fo j¼1 j

!υjr ) ð32:14Þ

32.3

Derivation of the Equilibrium Constant for Chemical Reaction r

333

where we used the equality: n X

ð ln Þ ¼ ln

j¼1

! n Y

ð32:15Þ

j¼1

In Eq. (32.14), the product of the fugacity ratios for chemical reaction r is referred to as the Equilibrium Constant for Chemical Reaction r, Kr, that is: n b Y f j ðT, P, CompositionÞ Kr ¼ f oj ðT, PÞ j¼1

!υjr ð32:16Þ

where Composition denotes the mole fractions of the various components in the mixture, as related through the extents of reaction. Second, let us work on the left-hand side of Eq. (32.13). The summation over j from 1 to n represents the molar Gibbs free energy difference between the products and the reactants in the RS, or the Standard State (o), and is referred to as the Standard Molar Gibbs Free Energy of Reaction, given by: ΔGor ¼

n X

υjr Goj ðT, Po Þ

ð32:17Þ

j¼1

Using Eqs. (32.17) and (32.16) in Eq. (32.13), we obtain: ΔGor ¼ RTlnKr

ð32:18Þ

In summary:  Kr ¼ exp

ΔGor ¼

n X

ΔGor RT

 ð32:19Þ

υjr Goj ðT, Po Þ

ð32:20Þ

j¼1

n b Y f j ðT, P, CompositionÞ Kr ¼ f oj ðT, Po Þ j¼1

!υjr

where r ¼ 1, 2, . . ., m (Number of independent chemical reactions).

ð32:21Þ

334

32 Criterion of Chemical Reaction Equilibria, Standard States, and Equilibrium. . .

In Eqs. (32.19)–(32.21), Goj , f oj , and bf j are constant throughout the mixture, irrespective of the chemical reactions in which component j participates. As a result, these three quantities do not depend on r.

32.4

Derivation of the Equilibrium Constant for a Single Chemical Reaction

For the special case of a single chemical reaction, we can eliminate the index r, and express Eqs. (32.19)–(32.21) as follows: 

ΔGo K ¼ exp RT

ΔG ¼

n X



υj Goj ðT, Po Þ

ð32:22Þ

ð32:23Þ

j¼1

! υj n b Y f j ðT, P, CompositionÞ K¼ f oj ðT, Po Þ j¼1

32.5

ð32:24Þ

Discussion of Standard States for Gas-Phase, Liquid-Phase, and Solid-Phase Chemical Reactions

In Eqs. (32.22)–(32.24), ΔGo and K depend on the choice of standard state, particularly on the choice of Po. Next, we will also assume that all the stoichiometric numbers are known. In order to calculate the standard molar Gibbs free energy of reaction, ΔGo ¼

n X j¼1

or the chemical potential of component j,

υj Goj

ð32:25Þ

32.5

Discussion of Standard States for Gas-Phase, Liquid-Phase, and Solid-Phase. . .

μj ¼

Goj

  bf þ RTln oi fj

335

ð32:26Þ

we need to know the values of Goj for all the non-inert components, for which υj 6¼ 0. Therefore, the standard state (o) needs to be specified precisely. As we saw earlier, the standard-state temperature, To, was chosen to be equal to the system temperature, T, in order to eliminate the unknown function λj(T). In addition, the standard-state pressure, Po, the standard-state composition (the various xoj s in the standard state), and the state of aggregation of the standard state (gas, liquid, or solid) may be chosen for convenience. It is common to denote the state of aggregation of the standard state as follows:     υj Aj ðgÞ þ jυk jAk ðlÞ þ υq Aq ðsÞ ¼ 0

ð32:27Þ

where g (gas), l (liquid), and s (solid) designate the state of aggregation of the standard state. Below, we discuss each standard state separately: 1. (g): Indicates the following standard state (o) Pure Ideal Gas j, xoj ¼ 1 To ¼ T (System temperature) Po ¼ 1 bar ) f oj ¼ 1 bar (Unit fugacity standard state) 2. (l): Indicates the following standard state (o) Pure Liquid k, xok ¼ 1 To ¼ T (System temperature) Po ¼ P (System pressure), or Pvpk (T), or 1 bar where vp denotes vapor pressure. 3. (s): Indicates the following standard state (o) Pure Solid q, xoq ¼ 1, at the most stable solid state To ¼ T (System temperature) Po ¼ P (System pressure), or Pvpq (T), or 1 bar Typically, we will choose Po to be either 1 bar or Pvpj (T) for component j, so that the standard-state properties, ΔGo ðT, Po Þand f oj ðT, Po Þ, depend only on temperature, and not on pressure. However, in some cases, we can choose Po ¼ P, such that, when the system pressure, P, changes, Po will also change. When Po ¼ 1 bar or Pvpj(T), it follows that K ¼ exp (ΔGo(T, Po) / RT) depends solely on the temperature, T. On the other hand, when Po ¼ P, it follows that K ¼ exp (ΔGo(T, P) / RT) depends on both the temperature, T, and the pressure, P.

336

32.6

32 Criterion of Chemical Reaction Equilibria, Standard States, and Equilibrium. . .

Comments on the Standard-State Pressure

As we have just shown, the standard-state pressure, Po, is intimately connected with the pressure dependence of the equilibrium constant, K. Indeed, 

ΔGoðTo , Po Þ KðT , P Þ ¼ exp RT o



o

ð32:28Þ

is strictly a property of the standard state (o). As discussed, we typically choose To ¼ T and Po ¼ 1 bar. In that case, K depends only on T. However, sometimes, Po is chosen to be Pvpj(T) in the case of liquids or solids. In that case, one needs to use  Goj T, Pvpj ðTÞ to calculate the molar Gibbs free energy of component j in the standard state. If only GojðT, 1 barÞ is known, then, we need to calculate how   Goj T, Pvpj ðTÞ is related to GojðT, 1 barÞ using a relation derived in Part I. Specifically,  o  ∂Gj ðT, PÞ ¼ Voj ðT, PÞ ) dGoj ¼ Voj ðT, PÞdP ∂P T

ð32:29Þ

Integrating the differential relation to the right of the arrow in Eq. (32.29), from P ¼ 1 bar to P ¼ Pvpj(T), yields: 

Goj



T, Pvpj ðTÞ ¼

PvpjððTÞ

GojðT, 1

barÞ þ

Voj ðT, PÞ dP

ð32:30Þ

1 bar

Recall that in Eqs. (32.29) and (32.30), Voj (T, P) is the molar volume of component j in the standard state (o) at T and P. If for some of the n components we choose Po ¼ P, then, we need to compute GojðT, PÞ in the standard state in terms of Goj(T, 1 bar) using Eq. (32.30), where Pvpj(T) is replaced by P, that is, ðP Goj ðT, PÞ ¼ Goj ðT, 1 barÞ þ

Voj ðT, PÞdP

ð32:31Þ

1bar

If ΔGo is known at 1 bar, the simplest approach is to choose Po ¼ 1 bar for every component j in the standard state.

32.7

32.7

Decomposition of the Equilibrium Constant into Contributions from the Fugacity. . .

337

Decomposition of the Equilibrium Constant into Contributions from the Fugacity Coefficients, the Gas Mixture Mole Fractions, and the Pressure

In the case of a single chemical reaction, we showed that (see Eq. (32.24), repeated below for clarity): n Y bf j ðT, P, CompositionÞ K¼ f ojðT, Po Þ j¼1

!υj ð32:32Þ

It is convenient to express K in Eq. (32.32) in terms of the mole fractions of the various components participating in the chemical reaction. To this end, it is useful to express the mixture fugacity of component j, bf j, as a function of concentrations using the fugacity-coefficient approach based on an equation of state which we discussed in Part II. This approach is particularly useful when dealing with gas-phase chemical reactions. On the other hand, when dealing with liquid-phase or solid-phase chemical reactions, it is convenient to express bf j as a function of concentrations using the activity-coefficient approach based on a model for the excess Gibbs free energy of mixing, ΔGEX, which we also discussed in Part II. We will carry out this calculation in Lecture 33. Recall that if component j is present in two or more phases, at thermodynamic equilibrium, we can use the fugacity of component j in any of the coexisting phases, because they are all equal. Focusing on a single gas (vapor)-phase chemical reaction, we can express the fugacity of component j in the gas mixture as follows: bf j ¼ bf j ðT, P, y1 , y2 , . . . , yn1 Þ ¼ ϕ b j ðT, P, y1 , y2 , . . . , yn1 ÞPyj

ð32:33Þ

In Eq. (32.33), the (n-1) gas mole fractions, {y1, y2, . . ., yn1}, are dependent, b j is the fugacity because they are determined by the extent of reaction, ξ, and ϕ coefficient of component j. To simplify the notation, we can rewrite Eq. (32.33) as follows: bf j ¼ ϕ b j Pyj

ð32:34Þ

bf o ¼ 1 bar j

ð32:35Þ

We also know that:

Dividing Eq. (32.34) by Eq. (32.35) we obtain:

338

32 Criterion of Chemical Reaction Equilibria, Standard States, and Equilibrium. . .

b j Pyj bf j ϕ o ¼ 1 bar fj

ð32:36Þ

where in Eqs. (32.34) and (32.36), P is in units of bar. From the definition of the equilibrium constant, K, in Eq. (32.32), it follows that: n b Y fj K¼ o f j¼1 j

!υj ¼

n

υj Y bf j Pyj

ð32:37Þ

j¼1

or K ¼ Kϕ Ky KP

ð32:38Þ

where Kϕ ¼

n Y

b j υj ϕ

ð32:39Þ

j¼1

where Kϕ reflects the non-idealities of mixing in the gas mixture. Recall that if the b ID ¼ 1 and KID ¼ 1. gas mixture is ideal, then, ϕ j ϕ Further, in Eq. (32.38): Ky ¼

n Y

yj υ j

ð32:40Þ

j¼1

Recall that the yj’s are all related to a single extent of reaction, ξ, and are therefore dependent. Finally, in Eq. (32.38): KP ¼ Pυ , where υ ¼

n X

υj

ð32:41Þ

j¼1 o where P is given in units of bar, because we have used the fact that bf j ¼ 1 bar. In addition, KyKP ¼ KyPυ, which we can express as follows:

Ky Pυ ¼

n Y j¼1

yj P

υj

¼

n Y j¼1

Pj υj

ð32:42Þ

32.7

Decomposition of the Equilibrium Constant into Contributions from the Fugacity. . .

339

where Pj is the partial pressure of gas j in the gas mixture. Accordingly, K ¼ Kϕ

n Y

Pj υj

ð32:43Þ

j¼1

If the gas mixture is ideal, it follows that: b o ¼ 1 ) KID ¼ 1 ϕ j ϕ

ð32:44Þ

Using the result to the right of the arrow in Eq. (32.44) in Eq. (32.43), we obtain: KID ¼

n Y j¼1

Pj υj

ð32:45Þ

Lecture 33

Equilibrium Constants for Condensed-Phase Chemical Reactions, Response of Chemical Reactions to Temperature, and Le Chatelier’s Principle

33.1

Introduction

The material presented in this lecture is adapted from Chapter 16 in T&M. In addition, Tables 33.1 and 33.2 are adapted from Introduction to Chemical Engineering Thermodynamics, Fourth Edition, by J.M. Smith and H.C. Van Ness, McGraw-Hill Book Company, NY (1987). First, we will complete our discussion of equilibrium constants that we began in Lecture 32. Specifically, we will discuss the equilibrium constant of a condensed (liquid or solid)-phase chemical reaction. Second, we will discuss how to determine the standard molar Gibbs free energy of reaction, including decomposing it into enthalpic and entropic contributions which can be calculated using results presented in Part I. Knowledge of the standard molar Gibbs free energy of reaction will then allow us to determine the equilibrium constant of the chemical reaction. Third, we will discuss how a chemical reaction responds to temperature, including classifying it as exothermic or endothermic. Finally, we will discuss Le Chatelier’s Principle.

33.2

Derivation of the Equilibrium Constant for a Condensed-Phase Chemical Reaction

As discussed in Part II, the fugacity of component j in a condensed (liquid or solid) mixture is given by: bf j ðT, P, x1 , x2 , . . . , xn
1 Þ ¼ γjðT, P, x1 , x2 , . . . , xn
1 Þxj f jðT, PÞ

ð33:1Þ

where the mole fractions, {x1, x2, . . ., xn
1}, are all related to a single extent of reaction. Further, in the standard state (o), we have: © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_33

341

342

33

Equilibrium Constants for Condensed-Phase Chemical Reactions, Response of. . .

f oj ¼ f oj ðT, Po Þ, where Po ¼ P, Pvpj ðTÞ, or 1 bar

ð33:2Þ

Equations (33.1) and (33.2) indicate that: bf j γj ðT, P, x1 , x2 , xn
1 Þ xj f j ðT, PÞ ¼ f oj f oj ðT, Po Þ

ð33:3Þ

In Eq. (33.3), we can relate fj(T, P) to f oj (T, Po) using results presented in Part II. Specifically, 2P 3   ð  ∂lnf j Vj Vj ¼ ) f jðT, PÞ ¼ f ojðT, Po Þ exp4 dP5 RT RT ∂P T P

ð33:4Þ

o

In Eq. (33.4), Vj (T, P) is the molar volume of component j, and the exponential term is referred to as the Poynting correction. Using the definition of the equilibrium constant, K, introduced in Lecture 32, and Eq. (33.3), we obtain: n b υj Y fi K¼ ¼ Kγ Kx KP o f j j¼1

ð33:5Þ

where Kγ ¼

n Y

γ j υj

ð33:6Þ

j¼1

where Kγ reflects the nonidealities of mixing in a liquid or solid (condensed) phase. If the mixture is ideal (ID), then, γID j ¼ 1, and Kγ ¼ 1. It also follows that Kx in Eq. (33.5) is given by: Kx ¼

n Y

xj υ j

ð33:7Þ

j¼1

where all the xjs in Eq. (33.7) are related to a single extent of reaction. In addition, the xjs affect both Kx and Kγ. Finally, KP in Eq. (33.5) is given by:

KP ¼

n Y j¼1

2 exp4υj

3  Vj dP5 RT

ðP  Po

ð33:8Þ

33.3

Determination of the Standard Molar Gibbs Free Energy of Reaction

343

Recall that K is intimately related to the standard state (o), because: K ¼ exp ½
ΔGoðT, Po Þ=RT

ð33:9Þ

Equation (33.9) indicates that in order to determine the equilibrium constant K, we need to determine the standard molar Gibbs free energy of reaction. Below, we will discuss how to do that.

33.3

Determination of the Standard Molar Gibbs Free Energy of Reaction

Because direct experimental data for K is only available for very few simple chemical reactions, it is more typical to determine K using Eq. (33.9) from available data on: ΔGoðT, Po Þ ¼

n X

υj Goj ðT, Po Þ

ð33:10Þ

j¼1

Alternatively, ΔGo(T, Po) can be obtained if enthalpy and entropy data in the standard state (o) is available. Specifically, as discussed in Part I, Goj ¼ Hoj
TSoj ) υj Goj ¼ υj Hoj
Tυj Soj )

n X

υj Goj  ΔGo

ð33:11Þ

j¼1

It then follows that: ΔGo ¼

n X

υj Hoj
T

j¼1

n X

υj Soj ) ΔGoðT, Po Þ

j¼1

¼ ΔH ðT, P Þ
TΔSoðT, Po Þ o

o

ð33:12Þ

In Eq. (33.12), ΔGo(T, Po): ΔHo(T, Po): ΔSo:

Standard molar Gibbs free energy of reaction Standard molar enthalpy of reaction, also known as the standard heat of reaction, Qr Standard molar entropy of reaction

Because it is not practical to measure ΔGo (or ΔHo and ΔSo) for every chemical reaction, tables are available for a large number of compounds reporting the standard molar Gibbs free energy of formation and the standard molar enthalpy of formation,

344

33

Equilibrium Constants for Condensed-Phase Chemical Reactions, Response of. . .

or the standard heat of formation, of the species from the elements. In these tables, the function ΔGo becomes, for each species j, ΔGofj , and similarly, ΔHo for each species j, becomes ΔHofj , where f denotes formation. To obtain ΔGo and ΔHo from the given ΔGofj and ΔHofj , respectively, we simply utilize: ΔGo ¼

n X

υj ΔGofj

and ΔHo ¼

j¼1

n X

υj ΔHofj

ð33:13Þ

j¼1

Usually, tables for ΔGofj and ΔHofj are available only at 298 K. If ΔGo or ΔHo are needed at other temperatures, we can compute these using temperature integrations (see below). For pure elements, by convention, we choose: ΔGofj ¼ 0, ΔHofj ¼ 0, at all Ts

ð33:14Þ

For pure elements that are solids at the temperature of interest, the crystal form must be specified. For example, the standard state of carbon is based on graphite, and only for graphite are ΔGof ¼ 0 and ΔHof ¼ 0: Should other forms of carbon be present in the system, then, ΔGof and ΔHof for these forms of carbon are not zero. Tabulations are available that list ΔGofi and ΔHofi for several components (usually at 25  C or 298 K), see Tables 33.1 and 33.2 below. If ΔHo is known at one temperature, it can be found at any other temperature by integration using heat capacity data. Specifically, as discussed in Part I, dHoj ¼ Copj dT

ð33:15Þ

where Copj is the heat capacity at constant pressure of pure component j in the standard state (o) at T and Po. It then follows that: υj dHoj

n n     X X ¼ d υj Hoj ¼ υj Copj dT ) d υj Hoj ¼ d υj Hoj

)

n X

j¼1

! υj Copj dT

!

j¼1

ð33:16Þ

j¼1

where the first summation over j in the large brackets is the standard molar enthalpy of reaction (ΔHo), and the second summation over j in the large brackets is the standard molar heat capacity of reaction at constant pressure ΔCop : Equating the last two terms in Eq. (33.16) yields: dðΔHo Þ ¼ ΔCop dT

ð33:17Þ

33.3

Determination of the Standard Molar Gibbs Free Energy of Reaction

345

Table 33.1 Standard Gibbs Free Energy of reaction at 298 K (Joules per mole of the substance formed) Chemical species Paraffins: Methane Ethane Propane n-Butane n-Pentane n-Hexane n-Heptane n-Octane 1-Alkenes: Ethylene Propylene 1-Butene 1-Pentene 1-Hexene Miscellaneous organics: Acetaldehyde Acetic acid Acetylene Benzene Benzene 1,3-Butadiene Cyclohexane Cyclohexane 1,2-Ethanediol Ethanol Ethanol Ethylbenzene Ethylene oxide Formaldehyde Methanol Methanol Methylcyclohexane Methylcyclohexane Styrene Toluene Toluene

State (Note 2)

ΔGϕf20g

CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18

g g g g g g g g


50,460
31,855
24,290
16,570
8,650 150 8,260 16,260

C2H4 C3H6 C4H8 C5H10 C6H12

g g g g g

68,460 62,205 70,340 78,410 86,830

C2H4O C2H4O2 C2H2 C6H6 C6H6 C4H6 C6H12 C6H12 C2H6O2 C2H6O C2H6O C8H10 C2H4O CH2O CH4O CH4O C7H14 C7H14 C8H8 C7H8 C7H8

g l g g l g g l l g l g g g g l g l g g l


128,860
389,900 209,970 129,665 124,520 149,795 31,920 26,850
323,080
168,490
174,780 130,890
13,010
102,530
161,960
166,270 27,480 20,560 213,900 122,050 113,630

346

33

Equilibrium Constants for Condensed-Phase Chemical Reactions, Response of. . .

Table 33.2 Standard heats of formation at 25  C (Joules per mole of the substance formed) Chemical species Paraffins: Methane Ethane Propane n-Butane n-Pentane n-Hexane n-Heptane n-Octane 1-Alkenes: Ethylene Propylene 1-Butene 1-Pentene 1-Hexene 1-Heptene Miscellaneous organics: Acetaldehyde Acetic acid Acetylene Benzene Benzene 1,3-Butadiene Cyclohexane Cyclohexane 1,2-Ethanediol Ethanol Ethanol Ethylbenzene Ethylene oxide Formaldehyde Methanol Methanol Methylcyclohexane Methylcyclohexane Styrene Toluene Toluene

State

ΔHo298

CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18

g g g g g g g g


74,520
83,820
104,680
125,790
146,760
166,920
187,780
208,750

C2H4 C3H6 C4H8 C5H10 C6H12 C7H14

g g g g g g

52,510 19,710
540
21,280
41,950
62,760

C2H4O C2H4O2 C2H2 C6H6 C6H6 C4H6 C6H12 C6H12 C2H6O2 C2H6O C2H6O C8H10 C2H4O CH2O CH4O CH4O C7H14 C7H14 C8H8 C7H8 C7H8

g l g g l g g l l g l g g g g l g l g g l


166,190
484,500 227,480 82,930 49,080 109,240
123,140
156,230
454,800
235,100
277,690 29,920
52,630
108,570
200,660
238,660
154,770
190,160 147,360 50,170 12,180

33.3

Determination of the Standard Molar Gibbs Free Energy of Reaction

347

Integrating Eq. (33.17) from T1 to T yields: ðT ΔH ðT, P Þ ¼ ΔH ðT1 , P Þ þ o

o

o

ΔCopðT, PoÞdT

o

ð33:18Þ

T1

Typically, T1 in Eq. (33.18) is 298 K. If ΔHo is known as a function of temperature, we can obtain ΔGo as a function of temperature using the Gibbs-Helmholtz equation discussed in Part I. Recall that:   d G H ¼
2 dT T Po T

ð33:19Þ

  d ΔGo ΔHo ¼
2 dT T Po T

ð33:20Þ

and therefore,

We can first calculate how ΔHo varies with T using Eq. (33.18). Subsequently, we can use ΔHo(T,Po) so deduced in Eq. (33.20) to compute how ΔGo varies with T. Specifically, integrating Eq. (33.20) between T1 and T, we obtain: ðT T1

  ðT  o  ΔGoðT, Po Þ ΔGo ðT1 , Po Þ d ΔGo ΔG
dT ¼ d ¼ T T1 dT T Po T Po T1

ðT ¼




ΔHo dT T2

ð33:21Þ

T1

or ΔGoðT, Po Þ ΔGoðT1 , Po Þ
¼ T T1

ðT 

 ΔHo dT T2

ð33:22Þ

T1

Using Eq. (33.22) in Eq. (33.9) we can then calculate: KðT, Po Þ ¼ exp

 
ΔGoðT, Po Þ RT

ð33:23Þ

348

33.4

Equilibrium Constants for Condensed-Phase Chemical Reactions, Response of. . .

33

Response of Chemical Reactions to Changes in Temperature and Pressure

We would like to know how a change in temperature, or in pressure, affects the equilibrium conversion of reactants into products. In other words, does an increase in temperature (T) or in pressure (P) shift the chemical reaction in the direction of the products (the desired outcome), or does it shift the chemical reaction in the direction of the reactants (the undesired outcome)? This information is, of course, essential from a practical viewpoint, because we need to know if we should operate at a higher T or P in order to convert more reactants into products. We will first discuss the effect of T on the equilibrium conversion. The effect of P on the equilibrium conversion will be discussed in Lecture 34.

33.5

How Does a Chemical Reaction Respond to Temperature?

We know that in the standard state (o), we have:  
ΔGo ΔGo ΔG ¼ ΔG ðT, P Þ; K ¼ exp ; and lnK ¼
RT RT o

o

o

ð33:24Þ

It then follows that: 

∂lnK ∂T

 ¼
Po

     1 ∂ ΔGo 1 ΔHo ¼ R ∂T T R T2 Po

ð33:25Þ

where Eq. (33.20) was used. Accordingly, 

∂lnK ∂T

 ¼ Po

ΔHo RT2

ð33:26Þ

or 

∂lnK ∂ð1=TÞ

 ¼
Po

ΔHo R

ð33:27Þ

Equation (33.27) shows that if we plot lnK versus 1/T, at constant Po, the slope of the resulting curve at any T is equal to (
ΔHo/R) at that T.

33.5

How Does a Chemical Reaction Respond to Temperature?

349

Chemical reactions are classified as follows: (a) Exothermic, if ΔHo < 0 (b) Endothermic, if ΔHo > 0 In Part I, we showed that the First Law of Thermodynamics for a closed system is given by: dU ¼ δQ þ δW

ð33:28Þ

where dU is the differential change in the internal energy of the system, δQ is the differential heat absorbed by the system, and δW is the differential work done on the system. Assuming PdV-type work, it follows that: δW ¼
PdV

ð33:29Þ

Using Eq. (33.29) in Eq. (33.28) yields: dU ¼ δQ
PdV

ð33:30Þ

dU þ PdV ¼ δQ

ð33:31Þ

or

If we assume a constant pressure (isobaric) process, Eq. (33.31) can be written as follows: dðU þ PVÞ ¼ δQjP

ð33:32Þ

δQjP ¼ dH

ð33:33Þ

or

Equation (33.33) helps us understand that, at constant pressure, the standard molar enthalpy of reaction, ΔHo, is equal to the standard heat of reaction, Qor . This result shows that: (i) If ΔHo < 0, this implies that Qor < 0: According to our definition of heat flow in Part I, this indicates that the system rejects heat, Qor , to the environment. This explains why the chemical reaction in this case is referred to as exothermic. (ii) If, on the other hand, ΔHo > 0, this implies that Qor > 0: According to our definition of heat flow in Part I, this indicates that the system absorbs heat, Qor , from the environment. This explains why the chemical reaction in this case is referred to as endothermic. To summarize, for an exothermic reaction, ΔHo < 0, and therefore, the slope of lnK versus 1/T is positive. On the other hand, for an endothermic reaction, ΔHo > 0,

350

33

Equilibrium Constants for Condensed-Phase Chemical Reactions, Response of. . .

Fig. 33.1

and therefore, the slope of lnK versus 1/T is negative. Figure 33.1 depicts these two lnK versus 1/T behaviors. Interestingly, only if ΔHo is independent of temperature, the lnK versus 1/T curves are straight lines. Equation (33.27), as well as Fig. 33.1, indicate that when ΔHo < 0 (exothermic reaction), an increase in temperature, that is, a decrease in 1/T, brings about a decrease in lnK, and hence, in K. Therefore, for an exothermic reaction, an increase in temperature shifts the reaction in the direction of the reactants. The opposite occurs for an endothermic reaction (ΔHo > 0), for which an increase in temperature, or a decrease in 1/T, brings about an increase in lnK, and hence, in K. This shifts the reaction in the direction of the products.

33.6

Le Chatelier’s Principle

This principle states that: “A system at equilibrium, when subjected to a perturbation, responds in a manner that tends to eliminate the effect of the perturbation.” Therefore, for an endothermic reaction, when we perturb the system by increasing the temperature, the system responds by shifting the reaction toward the products whose enthalpy is higher than that of the reactants (ΔHo > 0). As a result, some of the heat input goes to the reaction, thereby cooling down the system and lowering its temperature. On the other hand, for an exothermic reaction, when we perturb the system by increasing the temperature, the system responds by shifting the reaction towards the reactants, whose enthalpy is higher than that of the products (ΔHo < 0). As a result, some of the heat provided to increase the temperature goes into the reaction, thereby cooling down the system and lowering its temperature.

33.6

Le Chatelier’s Principle

351

Returning to Eq. (33.27), we can obtain another useful result. Specifically, 

∂lnK ∂ð1=TÞ

 PO

ΔHo ¼
) R

Tð2

T1

   ∂lnK 

T ð2 

¼


ΔHoðT, Po Þ dð1=TÞ R

T1

PO

or 2 6 KðT2 , Po Þ ¼ KðT1 , Po Þ exp4


Tð2

3  ΔH ðT, P Þ 7 dð1=TÞ5 R o

o

ð33:34Þ

T1

where in Eq. (33.34), T1 is typically equal to 298 K. In addition, Eq. (33.34) enables a direct evaluation of K(T2, Po) if ΔHo(T, Po) is known. If ΔHo is a constant independent of T, then, Eq. (33.34) can be readily integrated to yield:   
ΔHo 1 1 KðT2 , P Þ ¼ KðT1 , P Þ exp
T2 T1 R o

o

ð33:35Þ

Lecture 34

Response of Chemical Reactions to Pressure, and Sample Problems

34.1

Introduction

The material presented in this lecture is adapted from Chapter 16 in T&M. First, we will discuss how a chemical reaction responds to pressure. Second, we will solve Sample Problem 34.1 to calculate the effect of pressure on the equilibrium gas mixture composition. Finally, we will solve Sample Problem 34.2 to calculate the equilibrium conversion for a gas mixture undergoing a dissociation reaction.

34.2

How Does a Chemical Reaction Respond to Pressure?

In Lecture 33, we saw that the equilibrium constant, K, depends on the standard molar Gibbs free energy of reaction, ΔGo, which is a function of To and Po, where the reference-state temperature, To, was chosen to be equal to the system temperature, T. If the reference-state pressure, Po, is chosen to be 1 bar or Pvpj (T), the vapor pressure of component j at temperature T, then, ΔGo is not a function of pressure. In that case: lnK ¼ 

ΔGo ðT, Po Þ ) RT




∂lnK ∂P

 ¼ T


 1 ∂ΔGo ¼0 RT ∂P T

ð34:1Þ

However, if Po is chosen to be equal to the system pressure, P, then, ΔGo ¼ ΔGo (T, P) depends explicitly on P, and therefore, K will also depend on P. Specifically: ΔGo ðT, PÞ ) lnK ¼  RT




∂lnK ∂P

 T


 1 ∂ΔGo ¼ RT ∂P T

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_34

ð34:2Þ

353

354

34

Response of Chemical Reactions to Pressure, and Sample Problems

Recall that: ΔGoðT, PÞ ¼

n X

υj Goj ðT, PÞ

ð34:3Þ

j¼1

and, therefore, that:
o 
 n X ∂Gj ðT, PÞ ∂ΔGoðT, PÞ ¼ υj ∂P ∂P T T j¼1

ð34:4Þ

  In Eq. (34.4), ∂Goj ðT, PÞ=∂P ¼ Voj ðT, PÞ, the molar volume of component j T

in the standard state (o) at T and P. Using Eq. (34.4) in Eq. (34.2) yields:
 n ∂lnK 1 X ¼ υ Vo ðT, PÞ RT j¼1 j j ∂P T

ð34:5Þ

where the sum over j is equal to ΔVo(T, P), the standard molar volume of reaction. We can therefore express Eq. (34.5) as follows:
 ΔVoðT, PÞ ∂lnK ¼ RT ∂P T

ð34:6Þ

Equation (34.6) shows that: (i) If ΔVo < 0, then, if P increases, K also increases, and the products are favored. (ii) If ΔVo > 0, then, if P increases, K decreases, and the reactants are favored. Figure 34.1 illustrates behaviors (i) and (ii) above. Recall that in Fig. 34.1, T is constant.

Fig. 34.1

34.3

34.3

Sample Problem 34.1

355

Sample Problem 34.1

When the standard-state pressure Po is 1 bar or Pvpj (T), the vapor pressure of component j at temperature T, we showed that K is independent of P. Nevertheless, there can still be a pressure dependence of the equilibrium mixture composition for the various components participating in the chemical reaction. With this in mind, consider the following gas-phase chemical reaction involving the four gasses A, B, C, and D: jυA jAðgÞ þ jυB jBðgÞ ! jυC jCðgÞ þ jυD jDðgÞ

ð34:7Þ

for which the mixture of gases is ideal. Calculate the variation of the equilibrium constant Ky with pressure P, including plotting your result.

34.3.1 Solution First, we can use the fugacity-coefficient formulation discussed earlier to obtain: bf j ¼ ϕ b j yj P, where ϕ bj ¼ ϕ b ID ¼ 1 ) bf j ¼ bf ID ¼ yj P, for j ¼ A, B, C, and D ð34:8Þ j j For this gas-phase chemical reaction: To ¼ T, Po ¼ 1 bar ) f oj ¼ 1 bar,

for j ¼ A, B, C, and D

ð34:9Þ

By combining Eqs. (34.8) and (34.9), we obtain: bf j f oj

!

  P , for j ¼ A, B, C, and D ¼ yj 1 bar

ð34:10Þ

The equilibrium constant for the gas-phase chemical reaction considered here is given by:  jυC j jυD j bf C =f o bf D =f o C D K¼  jυA j jυB j bf A =f o bf B =f o A B

ð34:11Þ

356

34

Response of Chemical Reactions to Pressure, and Sample Problems

Using Eq. (34.10) for A, B, C, and D in Eq. (34.11) yields: K¼

½yC ðP=1 barÞjυC j ½yD ðP=1 barÞjυD j

½yA ðP=1 barÞjυA j ½yB ðP=1 barÞjυB j

ð34:12Þ

Equation (34.12) can also be expressed as follows: K ¼ Ky KP

ð34:13Þ

where, because the mixture of gases A, B, C, and D is ideal, Kϕ ¼ 1, and does not appear on the right-hand side of Eq. (34.13). In addition, ½ΔGoðT, 1 barÞ RT  jυ j jυ j  C yC yD D Ky ¼ ½ y A j υA j y B j υB j 

K ¼ exp

ð34:14Þ ð34:15Þ

and KP ¼ ðP=1 barÞΔυ , where Δυ ¼ jυC j þ jυD j  jυA j  jυB j

ð34:16Þ

Because Po ¼ 1 bar, we know that ΔGo, and therefore, K, are independent of P (see Eq. (34.14)). Accordingly:


 ∂K ¼0 ∂P T

ð34:17Þ

However, for Eq. (34.17) to be valid, Eq. (34.13) indicates that Ky must depend on P to counterbalance the dependence of KP on P (see Eq. (34.16)). Differentiating Eq. (34.13) with respect to P, at constant T, yields:


∂K ∂P




 ¼ T


 ∂Ky ∂KP KP þ Ky ∂P T ∂P T

ð34:18Þ

Differentiating Eq. (34.16) with respect to P yields:
   ∂KP P Δυ‐1 ¼ Δυ 1 bar ∂P T where holding T constant is redundant.

ð34:19Þ

34.3

Sample Problem 34.1

357

Combining Eqs. (34.17), (34.18), and (34.19) yields:
     ∂Ky P Δυ P Δυ‐1 0¼ þKy Δυ 1 bar ∂P T 1 bar

ð34:20Þ

  
 P Δυ‐1 ∂Ky P þΔυKy 1 bar ∂P T 1 bar

ð34:21Þ

or 0¼



Equation (34.21) shows that because P is not zero, the two terms inside the curly brackets must add up to zero, that is:
  ∂Ky P þΔυKy ¼ 0 ∂P T 1bar

ð34:22Þ

Rearranging Eq. (34.22) yields:

 ∂Ky ∂P þ Δυ ¼ 0 P T Ky T

ð34:23Þ



∂lnKy T þ ð∂lnPÞT Δυ ¼ 0

ð34:24Þ

or

or


∂lnKy ∂lnP

 ¼ Δυ

ð34:25Þ

T

where Δυ is a constant number, and therefore, the slope of the lnKy versus lnP curve is constant, giving rise to a straight line. Figure 34.2 below helps visualize the predicted behavior. Figure 34.2 shows that when P increases, the equilibrium constant Ky increases if Δυ < 0, thus favoring the products. On the other hand, if Δυ > 0, when P increases, the equilibrium constant Ky decreases, thus favoring the reactants. Note that if Δυ ¼ 0, Ky is independent of P.

358

34

Response of Chemical Reactions to Pressure, and Sample Problems

Fig. 34.2

34.4

Sample Problem 34.2

Consider a closed reactor which is charged initially with 1 mole of pure I2(g), and which is maintained at 800  C and a very low pressure, P. It is known that the following dissociation reaction occurs: I2ðgÞ ! 2IðgÞ

ð34:26Þ

where – NoI2  No1 ¼ 1 mole – NoI  No2 ¼ 0 mole – T ¼ 800o C ¼ 1073 K – P ¼ is known, and is very low

 ΔHo ¼ 156:6 kJ mole – ΔSo ¼ 108:4 J=mole K – We can therefore calculate ΔGo from : ðΔHo  TΔSo Þ: Compute the mole fractions of I2 and I at equilibrium, as well as the equilibrium extent of reaction, ξ, and the mole numbers, NI2 and NI, at equilibrium.

34.4.1 Solution Because the dissociation process involves a gas-phase chemical reaction, we choose the following standard state (o): * To ¼ T

34.4

Sample Problem 34.2

359

* Po ¼ 1 bar * Pure gases [j ¼ 1 (I2) and j ¼ 2 (I)] in the ideal gas state, for which, f oj ¼ 1 bar (for j ¼ 1 and 2) Next, we invoke the Gibbs Phase Rule given by: L¼nþ2πr

ð34:27Þ

n ¼ 2, π ¼ 1, r ¼ 1 ) L ¼ 2 þ 2  1  1 ¼ 2

ð34:28Þ

where

The Gibbs Phase Rule indicates that if we fix two independent intensive variables (L ¼ 2), for example, T and P (both known in Sample Problem 34.2), we should be able to calculate any other intensive variables, such as, y1 and y2. In other words, we should be able to calculate: y1 ¼ y1ðT, PÞ and y2 ¼ y2ðT, PÞ

ð34:29Þ

Regarding the extent of reaction, ξ, it is an extensive property as we defined it. Overall, for this simple dissociation reaction, according to Postulate I, we need to specify a total of (n + 2) independent variables. If we specify L of these to be intensive, then, the remaining (n + 2 – L) variables should be extensive. In this case, n ¼ 2, L ¼ 2, and therefore, we need to specify 2 + 2 – 2 ¼ 2 extensive variables, to fully characterize both the intensive and the extensive equilibrium thermodynamic state of the system. In particular, to calculate the extensive properties ξ, N1, and N2, in addition to T and P, two intensive properties, we also specify the two extensive o properties, No2 : Accordingly, the set of (n + 2) ¼ 4 independent variables  N1 and o includes T, P, N1 , No2 , which will allow us to calculate any other intensive as well as extensive property of interest. In Sample Problem 34.2, we are asked to calculate: * * * * *

y1 ¼ y1ðT, PÞ y2 ¼ y2ðT, PÞ

ξ ¼ ξ T, P, No1 , No2 N1 ¼ N1 T, P, No1 , No2 N2 ¼ N2 T, P, No1 , No2

In order to calculate y1(T, P) and y2(T, P), we use the expression which relates K to Kϕ, Ky, and KP. Specifically,  ΔGoðTÞ ¼ Kϕ Ky KP KðTÞ ¼ exp RT

ð34:30Þ

360

34

Response of Chemical Reactions to Pressure, and Sample Problems

In Eq. (34.30), ΔGo(T)¼ΔHo(T)TΔSo(T), and we have assumed that Po ¼1 bar, and therefore, K, ΔHo, ΔSo, and ΔGo do not depend on P. Because we know ΔHo(T) and ΔSo(T), we also know ΔGo(T), and therefore, we also know K(T). For the given dissociation reaction: I2 ðgÞ ! 2IðgÞ

ð34:31Þ

υ1 ¼ 1, υ2 ¼ þ2, Δυ ¼ υ1 þ υ2 ¼ 1 þ 2 ¼ 1

ð34:32Þ

we have:

Using Δυ ¼ 1 (see Eq. (34.32)) in the defining equation for KP (see Eq. (34.16)) yields: KP ¼ PΔυ ¼ P

ð34:33Þ

where P is in units of bars. Because we are told that the pressure, P, is very low, we can model the binary gas mixture as being ideal. Therefore, as we assumed earlier: 8 9 b1 ¼ ϕ b ID ¼ 1 = > 1, Eq. (34.53) shows that ξ < At T ¼ 800 C ¼ 1073K, ΔHo ¼ 156:6 KJ=mole, > : o ΔS ¼ 0:1084 KJ=moleK,

365

9 > = ∴ΔGo ¼ ΔHo  TΔSo ¼ 40:29 KJ=mole ðΔGo =RTÞ ¼ 4:51 > ; KðT ¼ 1073K, Po ¼ 1 barÞ ¼ e4:51 ð34:57Þ

Lecture 35

The Gibbs Phase Rule for Chemically-Reacting Systems and Sample Problem

35.1

Introduction

In this lecture, we will solve Sample Problem 35.1, an illuminating problem that deals with the partial decomposition of calcium carbonate solid into calcium oxide solid and carbon dioxide gas in a closed chemical reactor. We are asked to calculate the equilibrium pressure, the extent of reaction, and the equilibrium mole numbers of the three species present in the reactor. For this purpose, we will follow the chemical reaction equilibria approach, including formulating the Gibbs Phase Rule for chemically reacting systems.

35.2

Sample Problem 35.1

A closed chemical reactor (see Fig. 35.1), which is initially evacuated, is loaded with CaCO3(s). It is known that CaCO3(s) partially decomposes into CaO(s) and CO2 (g) according to the following chemical reaction: CaCO3ðsÞ ! CaOðsÞ þ CO2 ðgÞ

ð35:1Þ

You are asked to calculate: 1. The equilibrium gas pressure, P. 2. The extent of reaction, ξ. 3. The equilibrium mole numbers of the three species, Ns CaCO3 , Ns CaO , and Ng CO2 : We are given: (i) ΔGo(T) – Standard Molar Gibbs Free Energy of Reaction (ii) V sCaCO – Molar volume of CaCO3 solid 3 © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_35

367

368

35

The Gibbs Phase Rule for Chemically-Reacting Systems and Sample Problem

s – Molar volume of CaO solid CaO (iv) An EOS for the gas

(iii) V

Fig. 35.1

For the decomposition reaction in Eq. (35.1), it follows that: 8 9 < υCaCO3 ¼ 1 = υCaO ¼ þ1 Δυ ¼ 1 þ 1 þ 1 ¼ þ1 : ; υCO2 ¼ þ1

ð35:2Þ

35.2.1 Solution Strategy To solve this Sample Problem, we will use the Chemical Reaction Equilibria Approach, where the overall system is simple, Postulate I applies, and the conventional Gibbs Phase Rule can be used. In this case, we treat the system as consisting of n ¼ 3 components, π ¼ 3 phases (two solid and one gas), where a single decomposition reaction takes place (r ¼ 1). The conventional Gibbs Phase Rule indicates that: n ¼ 3, π ¼ 3, r ¼ 1 ) L ¼ n þ 2  π  r ¼ 3 þ 2  3  1 ) L ¼ 1

ð35:3Þ

Consistent with Eq. (35.3), it is convenient to specify T as the single intensive variable. It should then be possible to compute the equilibrium pressure, P, which is another intensive variable. Treating this system as simple, Postulate I applies and indicates that we need to specify (n + 2) ¼ (3 + 2) ¼ 5 independent variables to fully characterize both the intensive and the extensive equilibrium thermodynamic state of

35.2

Sample Problem 35.1

369

the system. Because L ¼ 1, we need to specify (n + 2) – L ¼ (3 + 2) – 1 ¼ 5 – 1 ¼ 4 independent extensive variables. A convenient choice includes the initial moles of CaCO3 solid, the initial moles of CaO solid, the initial moles of CO2 gas, and the total volume of the reactor, that is, N

so so go ,N , N , and V CaCO3 CaO CO2

ð35:4Þ

The set of five independent variables  T, N

 so so go ,N , N ,V CaCO3 CaO CO2

ð35:5Þ

will then allow us to compute what the Sample Problem requests: P (Intensive), s g s (Extensive), N (Extensive), and N (Extensive). ξ (Extensive), N CaO CO2 CaCO3

35.2.2 Selection of Standard States (i) For CO2(g) • Pure CO2(g) in an ideal gas state • To ¼ T go • Po ¼ 1 bar, f ¼ 1 bar CO2 (ii) For CaCO3(s) • Pure CaCO3(s) • To ¼ T so so • Po ¼ 1 bar, f ¼ f ð T, 1 barÞ CaCO3 CaCO3 (iii) For CaO(s) • Pure CaO(s) • To ¼ T • Po ¼ 1 bar, f

so so ¼ f ð T, 1 barÞ CaO CaO

370

35

The Gibbs Phase Rule for Chemically-Reacting Systems and Sample Problem

35.2.3 Remarks We know that ΔGo ¼ ΔGo(To, Po). If we choose Po ¼ 1 bar, then, ΔGo(To, 1 bar) is usually available in tabular form for To ¼ 298 K. If Po ¼ P, then, we need to calculate ΔGo(To, P), where if ΔGo(To, 1 bar) is known, we need to implement a pressure correction to relate ΔGo(To, 1 bar) to ΔGo(To, P), as discussed in Part I. Later in this lecture, we will discuss in more detail what happens if we choose Po ¼ P for the three components under consideration: CaCO3(s), CaO(s), and CO2(g).

35.2.4 Evaluation of Fugacities In this Sample Problem, three pure (n ¼ 1) phases are involved: CaCO3(s), CaO(s), and CO2(g). The fugacity of each phase will therefore depend on (n + 1) ¼ (1 + 1) ¼ 2 independent, intensive variables, consistent with the Corollary to Postulate I. These two variables are T and P. We can therefore write: f

g s s s s g ðT, PÞ, f ¼f ¼ f ðT, PÞ, f ¼ f ðT, PÞ ð35:6Þ CO2 CaO CaCO3 CaCO3 CaO CO2

35.2.5 Calculation of the Equilibrium Constant Having determined the three fugacities at T and P, as well as in the reference state at T and 1 bar, we can write down the expression for the Equilibrium Constant, K, as follows: K ¼ K ðTÞ ¼

Y

f j =f o

 υj

ð35:7Þ

j¼1, 2, 3

where 1 ¼ CaOðsÞ, 2 ¼ CO2ðgÞ, 3 ¼ CaCO3ðsÞ

ð35:8Þ

υ1 ¼ 1, υ2 ¼ 1, υ3 ¼ 1

ð35:9Þ

Using the expressions for f j and f oj (j ¼ 1, 2, and 3) in the expression for K in Eq. (35.7) yields:

35.2

Sample Problem 35.1

371

 KðTÞ ¼

1 f sCaO ðT, PÞ so f CaO ðT, 1 barÞ 

g

1

f CO ðT, PÞ 2

1 bar

1

f sCaCO ðT, PÞ 3 f so CaCO ðT, 1 bar Þ

ð35:10Þ

3

where if all the components involved in the chemical reaction are pure, then, Eq. (35.10) shows that ξ does not appear in the expression for K, because it is independent of composition. In that case, the K expression provides a unique relation between T and P. The fugacity ratio of each pure solid can be evaluated from the Poynting correction, where we need to know the molar volume of each solid as a function of T and P. Specifically, s ð P   ðT, PÞ CaO s V ðT, PÞ=RTÞ dP ¼ exp CaO so 1 bar f ðT, 1 barÞ CaO

ð35:11Þ

s ð P   ðT, PÞ CaCO3 s V ðT, PÞ=RTÞ dP ¼ exp CaCO3 so 1 bar ðT, 1 barÞ f CaCO3

ð35:12Þ

f

and f

The fugacity of CO2(g) is evaluated using the fugacity coefficient approach that we discussed in Part II. Specifically, f

g g ðT, PÞ ¼ ϕ ðT, PÞP CO2 CO2

where we can compute the fugacity coefficient, ϕ

ð35:13Þ

g ðT, PÞ, given a suitable CO2

volumetric EOS for CO2(g). Using Eqs. (35.11), (35.12), and (35.13) in Eq. (35.10), we obtain:

372

35

The Gibbs Phase Rule for Chemically-Reacting Systems and Sample Problem

1 9 s s > >   BV CaOðT, PÞ  V CaCO3ðT, PÞ C = g P B C f dP KðTÞ ¼ exp ð T, P Þ @ A > CO2 RT 1 bar > 1 bar > > ; : 8 > >
> þV N þV N

s so > : ðRT=PÞ þ V V CaO CaO3

9 > > = > > ;

ð35:23Þ

where T, V, N

go so so s s ,N , N ,V , and V CO2 CaO CaCO3 CaO CaCO3

ð35:24Þ

are all known. Knowing ξ, we can readily calculate: N

s so s so g go ¼N þ ξ, N ¼N  ξ, and N ¼N þξ CaO CaO CaCO3 CaCO3 CO2 CO2

ð35:25Þ

374

35

The Gibbs Phase Rule for Chemically-Reacting Systems and Sample Problem

35.2.6 Comment on the Standard-State Pressure As we have seen, the standard-state pressure, Po, is intimately connected with the pressure dependence of the equilibrium constant, K(To, Po), that is,

 ΔGo ðTo , Po Þ KðT , P Þ ¼ exp RT o

o

ð35:26Þ

As stressed earlier, Eq. (35.26) clearly shows that K(To, Po) is indeed a property of the standard state (o). As we have seen, we typically choose To ¼ T (the system temperature), and Po ¼ 1 bar. As a result, ΔGo(T, 1 bar), and K ¼ K(T) as shown in Eq. (35.26) above. However, sometimes Po is chosen to be Pvpj(T) for component j. In that case, we need to compute Goj (T, Pvpj(T)) in order to calculate: ΔGo ðTÞ ¼

n X

  υj Goj T, Pvpj

ð35:27Þ

j¼1

If we know Goj (T, 1 bar), we can use the relation derived in Part I relating the variation of Goj(T, P) with P, at constant T, to the molar volume of component j, that is,  o ∂Gj ðT, PÞ ¼ Voj ðT, PÞ ∂P T

ð35:28Þ

Integration of Eq. (35.28) with respect to P, at constant T, then yields: Goj



 T, Pvpj ðTÞ ¼ Goj ðT, 1 barÞ þ

ð Pvpj ðTÞ 1 bar

Voj ðT, P0 ÞdP0

ð35:29Þ

If Po ¼ P for some components j (say, solids), then, Eq. (35.29) is modified as follows: Goj ðT, PÞ ¼ Goj ðT, 1 barÞ þ

ðP 1 bar

Voj ðT, P0 ÞdP0

ð35:30Þ

Lecture 36

Effect of Chemical Reaction Equilibria on Changes in Thermodynamic Properties and Sample Problem

36.1

Introduction

The material presented in this lecture is adapted from Chapter 16 in T&M. • An important goal in Parts I and II of the book is to learn how to calculate changes in thermodynamic properties of pure component (n ¼ 1) and multi-component (n > 1) systems as these evolve from some initial state (i) to some final state (f). That is, to learn how to calculate: ΔUi!f , ΔHi!f , ΔSi!f , ΔGi!f , etc. So far, we have accomplished this goal in the absence of chemical reactions. • Here, we will consider cases where some of the components in the system undergo chemical reactions. As a result, the system equilibrium composition is no longer constant and has to be determined as a function of the system initial composition. In addition, the thermodynamic properties of the system have to be calculated as they respond to the composition changes. To illustrate how to carry out such calculations, in this lecture, we will solve an interesting Sample Problem which integrates concepts and equations presented in Parts I and II of the book.

36.2

Sample Problem 36.1: Production of Sulfuric Acid by the Contact Process

In the manufacture of sulfuric acid by the contact process, elemental sulfur is burned with air (assumed to be a mixture of O2 and N2) to form SO2, which is then further oxidized to form SO3. The SO3 then reacts with water to produce sulfuric acid (H2SO4). Assume that the product gas stream from the sulfur burner contains 9 mole % of SO2, 80 mole % of nitrogen, and 11 mole % of oxygen. The product gas stream is © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_36

375

376

36

Effect of Chemical Reaction Equilibria on Changes in Thermodynamic. . .

subsequently cooled to 723 K and passed over a catalyst bed to convert SO2 into SO3 according to the following chemical reaction: SO2 ðgÞ þ 0:502 ðgÞ⇄SO3 ðgÞ

ð36:1Þ

(a) Choose 1 mole of the SO2-O2-N2 product gas stream from the sulfur burner as a basis, and derive a relation between the equilibrium constant and the extent of reaction. Assume that the pressure is maintained constant at one bar, and that the gas mixture is ideal. (b) Using the data given in Table 36.1, derive a relation between the equilibrium constant and the temperature. Table 36.1

SO3(g) SO2(g) O2(g) N2(g)

ΔHof ð298KÞ ðJ=molÞ

ΔGof ð298KÞ ðJ=molÞ

Cop ðJ=molKÞ

3.954  105 2.970  105 0 0

3.705  105 3.005  105 0 0

60.19 45.21 29.96 29.96

Recall that Cop is the heat capacity at constant pressure and can be assumed to be constant. (c) If the SO2-O2-N2 mixture is fed to the SO3 reactor at 723 K and if this reactor is adiabatic and operates at steady state, derive a relation between the outlet temperature and the extent of reaction. Indicate how you would calculate the outlet temperature, the extent of reaction, and the mole % of SO2 left. Assume ideal gas behavior of the inlet and the outlet gas mixtures.

36.3

Solution Strategy

• When we solve problems of this type, the first question that we should ask is: Do we have sufficient information to solve the problem? Specifically, we are searching for two unknowns, Tout and ξ, and in order to find them, we require two equations that we can solve simultaneously. The first equation is the equilibrium constant relation, and the second equation is the First Law of Thermodynamics for an Open System. We therefore know, a priori, that we will be able to solve this problem!

36.3

Solution Strategy

377

In this Sample Problem, there are two basic steps, as depicted in Figure 36.1

Fig. 36.1

• Because we are dealing with a gas-phase chemical reaction, we choose the following standard state (o): – To ¼ T – Po ¼ 1 bar – Pure ideal gas for each gas (j ¼ O2, N2 (inert), SO2, SO3) and f oj ¼ 1 bar for each j It then follows that ΔGo ¼ ΔGo(T) and K ¼ K(T), where, for simplicity, we omitted the 1 bar notation in ΔGo and K. • For the gas-phase chemical reaction in Eq. (36.1), the equilibrium constant is given by: KðTÞ ¼ Kϕ Ky KP

ð36:2Þ

• Because the gas mixture is ideal (ID), it follows that: b ID ¼ 1, ϕ b ID ¼ 1, ϕ b ID ¼ 1 ) Kϕ ¼ KID ¼ 1 ϕ SO2 O2 SO3 ϕ

ð36:3Þ

378

36

Effect of Chemical Reaction Equilibria on Changes in Thermodynamic. . .

• Given the gas-phase chemical reaction in Eq. (36.1), it follows that:  Ky ¼ 

ySO3 1 

ySO2

1 yO 2

1

ð36:4Þ

Recall that N2 is an inert component, that is, νN2 ¼ 0, and therefore, it does not appear in Eq. (36.4) for Ky. • Given the gas-phase chemical reaction in Eq. (36.1), we know that: ySO3 ¼

NSO3 ξ  ¼ N 1  12 ξ

NSO2 ð0:09  ξÞ  ¼  N 1  12 ξ   0:11  12 ξ NO2  ¼ ¼  N 1  12 ξ

ð36:5Þ

ySO2 ¼

ð36:6Þ

yO2

ð36:7Þ

yN 2 ¼

NN2 0:80  ¼ N 1  12 ξ

ð36:8Þ

• We also know that: KP ¼



P 1 bar

Δν

with Δν ¼ ð1=2Þ

ð36:9Þ

However, because P ¼ 1 bar, Eq. (36.9) yields: KP ¼ 1

ð36:10Þ

• Using Eqs. (36.3) and (36.10)) in Eq. (36.2), we obtain: KðTÞ ¼ Ky

ð36:11Þ

• Using Eq. (36.4) for Ky, along with Eqs. (36.5), (36.6), and (36.7) for ySO3, ySO2, and yO2, respectively, in Eq. (36.11) yields:   1 ξ= 1  12 ξ KðTÞ ¼ Ky ¼   1     ð0:09  ξÞ= 1  12 ξ 0:11  12 ξ = 1  12 ξ 1=2 or

36.4

Evaluation of K(T)

379

KðTÞ ¼

ðξÞð1  0:5ξÞ1=2 ð0:09  ξÞð0:11  0:5ξÞ1=2

ð36:12Þ

Equation (36.12) provides the first needed relation between the two unknowns, T and ξ. However, to utilize Eq. (36.12), we first need to evaluate K(T). Fortunately, we can do that using some of the data provided in the Problem Statement, including Table 36.1.

36.4

Evaluation of K(T)

We begin from the expression for K(T) given in terms of ΔGo(T), where:  ΔGoðTÞ KðTÞ ¼ exp  RT

ð36:13Þ

In Table 36.1, we are given data at T ¼ 298 K. Using this data, we can calculate ΔGo(298 K), and then, using Eq. (36.13), we can calculate K(298 K). However, we need to calculate K(T) at T 6¼ 298 K. To this end, we can utilize the standard-state molar enthalpy of reaction, ΔHo(T), which we can calculate using the data given in Table 36.1 (see below). • First, we calculate ΔGo(298 K) as follows: X

ΔGo ð298 KÞ ¼

νi ΔGofi ð298 KÞ

ð36:14Þ

i¼SO3 , SO2 , O2 , N2

Using the data in Table 36.1 in Eq. (36.14), we obtain: ΔGoð298 KÞ ¼ 9 8 > > > > =

> |fflfflffl ffl {zfflfflffl ffl } |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl {zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl } |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl {zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl } > > |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} ; : N2 ðInertÞ SO SO 3

2

O2

ð36:15Þ or

380

36

Effect of Chemical Reaction Equilibria on Changes in Thermodynamic. . .

ΔGo ð298 KÞ ¼ 7  104 J=mol

ð36:16Þ

Using Eq. (36.16) and T ¼ 298 K in Eq. (36.13) yields:  ΔGoð298 KÞ Kð298 KÞ ¼ exp  Rð298 KÞ or Kð298 KÞ ¼ exp ð28:253Þ

ð36:17Þ

To find K(T) given K(298 K), we use the following relation derived in Lecture 33: 

∂ln K ∂T

¼ P

ΔHo RT2

ð36:18Þ

Multiplying both sides of Eq. (36.18) by dT0 , integrating from 298 K to T, and exponentiating yields: ð T KðTÞ ¼ Kð298 KÞ exp



298 K

ΔHo ðT0 Þ dT0 2 RT0

ð36:19Þ

Note that in Eq. (36.17), holding P constant in (∂lnK/∂T)P is redundant, because K ¼ K(T, Po) and does not depend on P. Equation (36.19) shows that if we know ΔHo(T), we can calculate K(T). To this end, we can first compute ΔHo(298 K) using the data given in Table 36.1 and, then, calculate ΔHo(T) using the temperature derivative of ΔHo(T) given by:  ∂ΔHo ¼ ΔCoP ∂T P

ð36:20Þ

where, again, holding P constant in Eq. (36.20) is redundant, because ΔHo ¼ ΔHo(T, Po), and does not depend on P. Multiplying both sides of Eq. (36.20) by dT0 and then integrating from 298 K to T yields (recall that according to Table 36.1, CoP and, therefore, ΔCoP , do not depend on T): ΔHo ðTÞ ¼ ΔHo ð298 KÞ þ

ðT 298 K

ΔCoP dT0

ð36:21Þ

In the Problem Statement, we are told that all the standard-state heat capacities at constant pressure (CoPi, for i ¼ SO3, SO2, O2, and N2) are independent of temperature and given by their values at 298 K (see Table 36.1).

36.4

Evaluation of K(T)

381

As a result, in Eqs. (36.20) and (36.21), the standard-state molar heat capacity of reaction at constant pressure, ΔCoP , is also independent of temperature and can be calculated as follows: ΔCoP ¼ ΔCoP ð298 KÞ ¼

X

νi CoPi ð298 KÞ

ð36:22Þ

i¼SO3 , SO2 , O2 , N2

Using the values of CoPi ð298 KÞ and νi given in the Problem Statement, including Table 36.1, in Eq. (36.22) yields: 9 8 O2 > > SO3 SO2 zfflfflfflfflfflfflfflfflffl ffl }|fflfflfflfflfflfflfflfflffl ffl { > > =

|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}> > > : N2 ‐Inert ; ð36:23Þ or ΔCoP ¼ 0!

ð36:24Þ

Equations (36.21) and (36.24) show that ΔHo is independent of temperature and is given by its value at 298 K, which we calculate below using the data given in the Problem Statement, including Table 36.1: X

ΔHo ð298 KÞ ¼

νi Hoif ð298 KÞ

ð36:25Þ

i¼SO3 , SO2 , O2 , N2

ΔHo ð298 KÞ ¼ 9 8 > > > > = <      1 J 4 4 ðþ1Þ 39:54  10 þ ð1Þ 29:7  10 þ  ð0Þ þ ð0Þð0Þ 2 mol > {zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflffl{zfflfflfflfflfflffl} |fflffl{zfflffl}> > > :|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflSO N2 ‐Inert ; SO 3

2

O2

ð36:26Þ or ΔHoð298 KÞ ¼ 9:84  104

J mol

ð36:27Þ

Note that Eq. (36.27) corresponds to an exothermic reaction. Using Eq. (36.17) in Eq. (36.19), along with Eq. (36.27), and rearranging yields:

382

36

Effect of Chemical Reaction Equilibria on Changes in Thermodynamic. . .

1:1835  104 K KðTÞ ¼ exp 11:463 þ Tðin KÞ

 ð36:28Þ

Using Eq. (36.28) for K(T) in Eq. (36.12) yields:

 h   i  1:1835  104 K 1 1=2 ¼ ðξÞð1  0:5ξÞ1=2 = ð0:09  ξÞ 0:11  ξ exp 11:463 þ 2 TðinKÞ ð36:29Þ Equation (36.29) provides an equation relating the two desired unknowns, T and ξ, where all the inputs are known. Clearly, we need a second independent equation involving the two unknowns, T and ξ, that we can solve along with Eq. (36.29) to uniquely determine T and ξ at the outlet.

36.5

Derivation of the Second Equation Relating T and ξ

According to Postulate I, to fully characterize the system under consideration which is simple, we need to specify (n + 2) independent variables, which in this problem, where n ¼ 4 (SO3, SO2, O2, and N2), is equal to (4 + 2) ¼ 6. Based on the description in the Sample Problem, it is convenient to choose the following six independent variables: n o T, P, NoSO3 , NoSO2 , NoO2 , NoN2

ð36:30Þ

The inlet (in) stream is then fully characterized in terms of: n o o in o in o in o Tin , Pin , Nin SO2 ¼ NSO2 , NSO3 ¼ NSO3 ¼ 0, NO2 ¼ NO2 , NN2 ¼ NN2

ð36:31Þ

Note that all the inputs in Eq. (36.31) are known! The outlet (out) stream is fully characterized in terms of: n

o 1 o out out o out o Tout ¼ ?, Pout ¼ Pin ,Nout ¼ N  ξ , N ¼ ξ , N ¼ N  , N ¼ N ξ out out SO2 SO2 SO3 O2 O2 N2 2 out N2 ð36:32Þ

Note that in Eq. (36.32), the only unknown inputs are Tout and ξout. As stressed above, in addition to Eq. (36.29) which relates the two unknowns T ¼ Tout and ξ ¼ ξout, we need to write down a second independent equation which

36.5

Derivation of the Second Equation Relating T and ξ

383

also relates Tout and ξout. This equation is provided by the First Law of Thermodynamics describing the operation of the steady-state, open, adiabatic reactor. First Law of Thermodynamics Analysis of the Gas Mixture in the Reactor

Fig. 36.2

In addition, as per the Problem Statement, the reactor operates isobarically, that is, Pout ¼ Pin. Next, we carry out a First Law of Thermodynamics analysis of the gas mixture in the reactor (see Fig. 36.2). The gas mixture in the SO3 reactor is well-mixed at all times, and therefore, it occupies the entire volume of the reactor. As a result, no PdV-type work is incurred (δW ¼ 0). In addition, the reactor operates adiabatically, and therefore, the gas mixture has no heat interactions (δQ ¼ 0). The gas mixture is a simple, open system, and therefore, the differential form of the First Law of Thermodynamics for an Open System applies. Specifically:  0

dU þ δW ¼ δQ þ dHin  dHout |{z} |{z}   |{z}  Well Steady 0 Adiabatic 0 Mixed State

Accordingly, after rearranging, Eq. (36.33) becomes:

ð36:33Þ

384

36

Effect of Chemical Reaction Equilibria on Changes in Thermodynamic. . .

dHin ¼ dHout

ð36:34Þ

In integral form, Eq. (36.34) is given by:     in in in out out out out Hin Tin , Pin , Nin , N , N , N H T , P , N , N , N , N ¼ out out out SO3 SO2 O2 N2 SO3 SO2 O2 N2 ð36:35Þ the (n + 2) ¼ (4 + 2) ¼ 6 variables n Recall that in Eq. (36.35), o in in in Tin , Pin , Nin , N , N , N Hin are all known. Accordingly, the determining SO3 SO2 O2 N2 left-hand side of Eq. (36.35) is known! On the other hand, Hout on the right-hand side of Eq. (36.35) is determined by Tout in (not known), Pout (known), Nout SO3 (determined by NSO3 (known) and ξout (unknown)), out in in NSO2 (determined by NSO2 (known) and ξout (unknown)), Nout O2 (determined by NO2 in (known) and ξout (unknown)), and Nout N2 (determined by NN2 (known) and ξout (unknown)). In other words, Hout on the right-hand side of Eq. (36.35) depends on the two desired unknowns, Tout and ξout! We will next calculate Hin and Hout and then use them in Eq. (36.35). This will result in an equation relating Tout and ξout. This equation, when solved simultaneously with Eq. (36.29), will allow us to uniquely determine the two unknown quantities, Tout and ξout. In general, for an n component mixture, we know that: H¼

n X

Ni Hi

ð36:36Þ

i¼1

Because we are told that the inlet and outlet gas mixtures are ideal, it follows that the partial molar enthalpy of component i is equal to the molar enthalpy of component i and depends solely on the temperature. Specifically, Hi ¼ Hi ðTÞ

ð36:37Þ

Using Eq. (36.37) for i ¼ SO3, SO2, O2, and N2 in Eq. (36.36) for the inlet gas mixture, we obtain: Hin ¼

X

Nin i Hi ðTin Þ

ð36:38Þ

i¼SO2 , SO2 , O2 , N2

where Nin Nin Nin and Nin SO2 ¼ 0:09 moles, SO3 ¼ 0 moles, O2 ¼ 0:11 moles, N2 ¼ 0:80 moles. Using these four mole numbers in Eq. (36.38), we obtain:

36.5

Derivation of the Second Equation Relating T and ξ

Hin ðTin Þ ¼ 0:09HSO2 ðTin Þ þ 0:11HO2 ðTin Þ þ 0:80HN2 ðTin Þ

385

ð36:39Þ

Carrying out a similar analysis for Hout yields: Hout ¼

X

Ni out Hi ðTout Þ

ð36:40Þ

i¼SO2 , SO2 , O2 , N2

  out out 1 where Nout SO2 ¼ ð0:99  ξout Þ moles, NSO3 ¼ ξout moles, NO2 ¼ 0:11  2 ξout moles, and Nout N2 ¼ 0:80 moles: Using these four mole numbers in Eq. (36.40) yields: Hout ðTout Þ ¼ ð0:09  ξout ÞHSO2 ðTout Þ þ ξout HSO3ðTout Þ   1 þ 0:11  ξout HO2ðTout Þ þ 0:80 HN2ðTout Þ 2

ð36:41Þ

In Eq. (36.41), combining all the terms which depend on ξout and all the terms which do not depend on ξout, we obtain: HoutðTout Þ ¼ ½0:09HSO2 þ 0:11HO2 þ 0:80HN2  h i 1 þ ξout HSO3  HSO2  HO2 2

ð36:42Þ

Note that in Eq. (36.42), HSO3 ¼ HSO3ðTout Þ, HSO2 ¼ HSO2ðTout Þ, HO2 ¼ HO2ðTout Þ, and HN2 ¼ HN2ðTout Þ. In addition, note that in Eq. (36.42), 1 HSO3ðTout Þ  HSO2ðTout Þ  HO2ðTout Þ 2 ¼ ΔHo ðTout Þ ¼ ΔHoð298 KÞ

ð36:43Þ

where ΔH (298 K) is the standard molar enthalpy of reaction, shown above to be independent of temperature and evaluated at 298 K using the given data. In addition, a comparison of the first term on the right-hand side of Eq. (36.42) with Eq. (36.39) shows that: ½0:09 HSO2ðTout Þ þ 0:11HO2ðTout Þ þ 0:80 HN2ðTout Þ ¼ HinðTout Þ

ð36:44Þ

Using Eqs. (36.44) and (36.43) in Eq. (36.42) yields: HoutðTout Þ ¼ HinðTout Þ þ ξout ΔHoð298 KÞ An examination of Eq. (36.45) shows that: (i) If there is no chemical reaction, that is, if ξout ¼ 0, then,

ð36:45Þ

386

36

Effect of Chemical Reaction Equilibria on Changes in Thermodynamic. . .

Hout ðTout Þ ¼ Hin ðTout Þ

ð36:46Þ

We also know that (see Eq. (36.35)): Hout ðTout Þ ¼ Hin ðTin Þ

ð36:47Þ

A comparison of Eqs. (36.46) and (36.47) shows that in the absence of the chemical reaction, Tout ¼ Tin! (ii) On the other hand, in the presence of the chemical reaction, that is, if ξout 6¼ 0, and because we have shown that ΔHo(298 K) < 0 (exothermic reaction), a comparison of Eqs. (36.45) and (36.47) shows that: Hout ðTout Þ ¼ Hin ðTout Þ þ ξout ΔHo ð298 KÞ ¼ Hin ðTin Þ |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}

ð36:48Þ

Tin! This result is expected, because the exothermic chemical reaction releases heat. Next, we rewrite Eq. (36.48) as follows: Hin ðTin Þ  Hin ðTout Þ ¼ ξout ΔHo ð298 KÞ

ð36:49Þ

where Hin(Tin) – Hin(Tout) in Eq. (36.49) is given by: Hin ðTin Þ  HinðTout Þ ¼ 0:09½HSO2ðTin Þ  HSO2ðTout Þ þ0:11½HO2ðTin Þ  HO2ðTout Þ þ 0:80½HN2ðTin Þ  HN2ðTout Þ

ð36:50Þ

For each pure gas i (SO2, O2, and N2) in Eq. (36.50), we can compute the quantity, [Hi(Tin) – Hi(Tout)], using the constant Copi values given in Table 36.1. Specifically: Hi ðTin Þ  Hi ðTout Þ ¼

ð Tin Tout

Copi dT ¼ Copi ½Tin  Tout 

ð36:51Þ

Using Eq. (36.51) for i ¼ SO3, O2, and N2 in Eq. (36.50) yields: Hin ðTin Þ  Hin ðTout Þ ¼ ðTin  Tout Þ   • 0:09CoPSO2 þ 0:11CoPO2 þ 0:80CoPN2 Using Eq. (36.52) in Eq. (36.49) and then solving for Tout yields:

ð36:52Þ

36.5

Derivation of the Second Equation Relating T and ξ

 Tout ¼ Tin 

0:09CoPSO2

387

ΔHo ð298 KÞ ξ þ 0:11CoPO2 þ 0:80CoPN2 out

ð36:53Þ

Using Tin ¼ 723 K, ΔHo (298 K) given in Eq. (36.27) and the Copi values given in Table 36.1 in Eq. (36.53) yields: Tout ¼ ð723 þ 3140:5 ξout ÞK

ð36:54Þ

As expected for an exothermic chemical reaction, Eq. (36.54) shows that Tout > Tin ¼ 723 K. Equation (36.54) can now be solved simultaneously with Eq. (36.29) to obtain the following equation for ξout: ðξout Þð10:5ξout Þ1=2 ð0:09ξout Þð0:110:5ξout Þ

1=2

    ¼ exp 11:463þ 1:1835104 =ð723þ3140:5ξout Þ ð36:55Þ

Solving Eq. (36.55), for example, by iteration, yields: ξout ¼ 0:0555

ð36:56Þ

Using Eq. (36.56) in Eq. (36.54) yields: 2

3

6 7 Tout ¼ 4723 þ ð3140:5Þð0:0555Þ5K |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}

ð36:57Þ

174:30

or Tout ¼ 897:3 K

ð36:58Þ

Again, as expected, Tout > Tin = 723K. Finally, the % of SO2 left is given by:  ð0:09  0:0555Þ • 100% ’ 38:3% 0:09

ð36:59Þ

Lecture 37

Review of Part II and Sample Problem

37.1

Introduction

In this lecture, we will first review the topics covered in Part II and then solve Sample Problem 37.1, an interesting problem which will help crystallize many of the concepts and methodologies presented in Part II.

37.2

Partial Molar Properties

 X
∂B  ∂B ∂B dB ¼ dT þ dP þ dNi ∂T P,Ni ∂P T,Ni ∂Ni T,P,Nj½i i
Bi 

∂B ∂Ni

B¼

ð37:1Þ



X

ð37:2Þ T,P,Nj½i

Ni Bi

ð37:3Þ

i

B¼

X

xi Bi

ð37:4Þ

i

In addition, equations of state (EOS) for mixtures, including mixing rules for the EOS parameters, mixture heat capacities, and the attenuated state and departure function approaches for mixtures. © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_37

389

390

37.3

37

Generalized Gibbs-Duhem Relations for Mixtures X


Ni dBi ¼

i

37.4

37.5

Review of Part II and Sample Problem


 ∂B ∂B dT þ dP ∂T P,Ni ∂P T,Ni

ð37:5Þ

Gibbs-Helmholtz Relation   ∂ G H ¼ 2 ∂T T P T

ð37:6Þ

  ∂ G H ¼ 2 ∂T T P,N T

ð37:7Þ

Mixing Functions ΔBmix ¼ B 

X X { Ni Bi ¼ Ni ΔBi i

ΔBmix ¼ B 

X

{

xi Bi ¼

i


ΔBi ¼

∂ðΔBmix Þ ∂Ni

X

xi ΔBi

ð37:9Þ

i



Reference States {

1. Pure Component: Bi ¼ Bi ðT, PÞ {

ð37:8Þ

i

2. Infinite Dilution: Bi ¼ Bi ðT, P, xi ! 0Þ

{

¼ Bi  Bi T,P,Nj½i

ð37:10Þ

37.7

37.6

Ideal Solutions

391

Ideal Gas Mixtures Gi ¼ λi ðTÞ þ RTln ðyi PÞ

Vi ¼

RT P

ð37:11Þ

ð37:12Þ

Ui ¼ UiðTÞ

ð37:13Þ

Hi ¼ HiðTÞ

ð37:14Þ

ΔHmix ¼ 0

ðPure component reference stateÞ

ð37:15Þ

ΔVmix ¼ 0

ðPure component reference stateÞ

ð37:16Þ

X xi ln xi

ðPure component reference stateÞ

ð37:17Þ

ΔSi ¼ Rln xi

ðPure component reference stateÞ

ð37:18Þ

X xi ln xi

ðPure component reference stateÞ

ð37:19Þ

ðPure component reference stateÞ

ð37:20Þ

ΔSmix ¼ R

i

ΔGmix ¼ RT

i

ΔGi ¼ RTln xi

37.7

Ideal Solutions Gi ¼ ΛiðT, PÞ þ RTln xi ¼ GiðT, PÞ þ RTln xi

ð37:21Þ

392

37

Review of Part II and Sample Problem

Vi ¼ Vi ðT, PÞ

ð37:22Þ

Ui ¼ Ui ðT, PÞ

ð37:23Þ

Hi ¼ Hi ðT, PÞ

ð37:24Þ

ΔHmix ¼ ΔHID mix ¼ 0

ðPure component reference stateÞ

ð37:25Þ

ΔVmix ¼ ΔVID mix ¼ 0

ðPure component reference stateÞ

ð37:26Þ

X ΔSmix ¼ ΔSID xilnxi mix ¼ R

ðPure component reference stateÞ

ð37:27Þ

i

ID

ΔSi ¼ ΔSi ¼ Rlnxi

ðPure component reference stateÞ

X ΔGmix ¼ ΔGID xilnxi mix ¼ RT

ðPure component reference stateÞ

ð37:28Þ

ð37:29Þ

i

ID

ΔGi ¼ ΔGi ¼ RTlnxi

37.8

ðPure component reference stateÞ

ð37:30Þ

BEX ¼ B  BID

ð37:31Þ

BEX ¼ B  BID

ð37:32Þ

Excess Functions

EX

Bi

ID

¼ Bi  Bi

ð37:33Þ

37.9

Fugacity

393 ID ΔBEX mix ¼ ΔBmix  ΔBmix

EX

ΔBi

ID

¼ ΔBi  ΔBi

EX ΔBEX mix ¼ B

EX

ΔBi

EX

ð37:34Þ

ð37:35Þ ð37:36Þ

¼ Bi

ð37:37Þ

ΔSEX mix ¼ 0

ð37:38Þ

ΔHEX mix ¼ 0

ð37:39Þ

Regular Solution

Athermal Solution

37.9

Fugacity Gi ¼ λi ðTÞ þ RTlnf i Gi ¼ λi ðTÞ þ RTln fbi

ðFor pure component iÞ

ð37:40Þ

ðFor component i in a mixtureÞ

ð37:41Þ

Limits of Ideality
 f lim i ¼ 1 P!0 P


fbi lim P!0 yi P

ðFor pure component iÞ

ð37:42Þ

 ¼1

ðFor component i in a mixtureÞ

ð37:43Þ

394

37.10

37

Review of Part II and Sample Problem

Variation of Fugacity with Temperature and Pressure

 ∂ðGi =RTÞ ∂lnf i V ¼ ¼ i RT ∂P ∂P T,Ni T,Ni

ð37:44Þ



 ∂ Gi =RT ∂ln fbi V ¼ ¼ i RT ∂P ∂P T,yi T,yi

ð37:45Þ

 

 ∂ Gi  G0i =RT Hi  H0i ∂lnf i ¼ ¼ ∂T ∂T P,Ni RT2 P,Ni


37.11

∂ln fbi ∂T

 P,yi

 1 0  0 ∂ Gi  Gi =RT A ¼@ ∂T

¼

Hi  H0i RT2

ð37:47Þ

P,yi

Generalized Gibbs-Duhem Relation for Fugacities X X
Hi  H0  X
Vi  i dP xi dln fbi ¼  xi xi dT þ RT RT2 i i i

37.12

ð37:46Þ

ð37:48Þ

Fugacity Coefficient ϕi 

fi P

fb ϕbi  i yi P

ð37:49Þ

ð37:50Þ

37.14

Activity

395

RTlnϕi ¼

ðP 

Vi 

 RT dP P

ð37:51Þ

Vi 

 RT dP P

ð37:52Þ

0

RTln ϕbi ¼

ðP  0




ðV RTlnϕi ¼ RTln Z  1

RTln ϕbi ¼ RTlnZ 

ðV
1

37.13

!  ∂P RT dV  V ∂N T,V

∂P ∂Ni

 T,V,Nj½i

! RT  dV V

ð37:54Þ

Lewis and Randall Rule bf ID ¼ f i ðT, PÞxi i

37.14

ð37:53Þ

ð37:55Þ

Activity {

RTln ai ¼ Gi  Gi ¼ ΔGi

ai 

fbi b f {i

ð37:56Þ

ð37:57Þ

For an Ideal Solution (Pure Component Reference State) a i ¼ xi

ð37:58Þ

396

37.15

37

Review of Part II and Sample Problem

ai fb ¼ i xi xi f { i

ð37:59Þ

Activity Coefficient

γi 

EX

RTln γi ¼ ΔGi

37.16

EX

¼ Gi

Variation of Activity Coefficient with Temperature and Pressure 1 0  EX
 ∂ ΔG =RT i ∂lnγi A ¼@ ∂P ∂P T,X 1 0  EX
 ∂ ΔG =RT i ∂ln γi A ¼@ ∂T ∂T P,X

37.17

ð37:60Þ

¼

ΔVi RT

T,X

¼

ΔHi RT2

ð37:62Þ

X,P

Generalized Gibbs-Duhem Relation for Activity Coefficients
   X ΔHmix ΔVmix dP xi dlnγi ¼  dT þ 2 RT RT i

37.18

ð37:61Þ

ð37:63Þ

Conditions for Thermodynamic Phase Equilibria Thermal Equilibrium : Tα ¼ Tβ ¼ Tγ ¼ . . . ¼ Tπ

ð37:64Þ

37.21

Dependence of Fugacitities on Temperature, Pressure, and Mixture Composition

397

Mechanical Equilibrium : Pα ¼ Pβ ¼ Pγ ¼ . . . ¼ Pπ

ð37:65Þ

Diffusional Equilibrium : μαi ¼ μβi ¼ μγi ¼ . . . ¼ μπi ði ¼ 1, 2, . . . , nÞ

ð37:66Þ

37.19

Gibbs Phase Rule L¼nþ2πrs

ð37:67Þ

where n is the number of components, π is the number of phases, r is the number of independent chemical reactions, and s is the number of additional constraints. In addition, understanding phase diagrams.

37.20

Differential Approach to Phase Equilibria bf α ¼ bf β ¼ . . . ¼ bf π ) d ln bf α ¼ d ln bf β ¼ . . . ¼ d ln bf π i i i i i i

37.21

ð37:68Þ

Dependence of Fugacitities on Temperature, Pressure, and Mixture Composition

! n1 α X b ∂ln f i dlnbf αi ¼ dT þ dP þ dxαi ∂xαi α α α i¼1 P,xi T,xi T,P,xj½i,n !

α  n 1 α X ∂lnbf H  H0 Vαi i ) dlnbf αi ¼  i 2 i dT þ dxαi dP þ α RT ∂x RT i α i¼1 ∂lnbf αi ∂T

!

∂lnbf αi ∂P

!

T,P,xj½i,n

ð37:69Þ

398

37

37.22

Review of Part II and Sample Problem

Integral Approach to Phase Equilibria bf α ¼ bf β ¼ . . . ¼ bf π i i i

ð37:70Þ

When only two phases (π = 2) are in thermodynamic equilibrium, say, α = Vapor (V) and β = Liquid (L), each containing n components, we can compute the vapor fugacity of component i using an EOS approach, and the liquid fugacity of component i using an Excess Gibbs Free Energy model. In that case, Eq. (37.70) with α = V and β = L is given by:   b VðT, P, y1 , . . . , yn1 ÞP ¼ xi γL ðT, P, x1 , . . . , xn1 ÞϕV T, Pvpi ðTÞ Pvpi ðTÞCi yi ϕ i i i ð37:71Þ where the Poynting correction Ci is given by: 2 6 Ci ¼ exp 4

ðP


VLi

RT

3  7 5dP ðPoynting correction for i ¼ 1, 2, . . . , nÞ

ð37:72Þ

Pvpi ðTÞ

In addition, knowledge of how to simplify Eq. (37.72) under various equilibrium conditions.

37.23

Pressure-Temperature Relations

Clapeyron Equation


 ΔHvap dP ¼ dT ½L=V TΔVvap

ð37:73Þ

Clausius-Clapeyron Equation  ΔHvap dð ln PÞ ¼ R dð1=TÞ ½L=V

ð37:74Þ

Recall that the Clausius-Clapeyron Equation assumes that: (1) the vapor phase is ideal, and (2) the molar volume of the vapor is much larger than that of the liquid.

37.27

37.24

Equilibrium Constant for Gases Undergoing a Single Chemical Reaction

399

Stoichiometric Formulation for Chemical Reactions

dni,r dnj,r ¼ ¼ . . . ¼ dξr for all species in all independent chemical reactions, r νi,r νj,r ð37:75Þ

37.25

Equilibrium Constant

Criteria of Chemical Reaction Equilibria X νj,r μj ¼ 0

ðr ¼ 1, 2, . . . , mÞ

ð37:76Þ

j

Y fbi ðT, P, y , y , ::, y Þ 1 2 n1 KðT, P Þ ¼ o o ð T, P Þ f i i o

!νi


 ΔGo ðT, Po Þ ¼ exp  RT

ð37:77Þ

Note that all the terms in Eq. (37.77) are evaluated at the system temperature. Furthermore, the reference-state fugacity and standard molar Gibbs free energy of reaction are evaluated at the same reference pressure.

37.26

Typical Reference States for Gas, Liquid, and Solid

Gas: Pure ideal gas at the system temperature and at 1 bar pressure. Liquid: Pure liquid at the system temperature and at 1 bar pressure, or at its vapor pressure, or at the system pressure. Solid: Pure solid in its most stable crystal state at the system temperature and 1 bar pressure, or at its vapor pressure, or at the system pressure.

37.27

Equilibrium Constant for Gases Undergoing a Single Chemical Reaction

KðT, 1 barÞ ¼

Y νi b ϕ i

i

!

! Y Y P νi νi yi ¼ Kϕ Ky KP 1 bar i i

ð37:78Þ

400

37.28

37

Review of Part II and Sample Problem

Equilibrium Constants for Liquids and Solids

If the reference-state pressure Po is chosen to be 1bar or Pvpi(T), then:

KðT, Po Þ ¼

Y xi ν i

!

Y γ i νi

i

i

0 BX ¼ Kx Kγ exp @ νi i

0

! Y

B exp @νi

i

1 Vi C dPA RT

P0

1

ðP

ðP

Vi C dPA RT

ð37:79Þ

P0

If the reference-state pressure Po is chosen to be the system pressure P, then: KðT, PÞ ¼

Y x i νi

!

i

37.29

Y γ i νi

! ¼ Kx Kγ

ð37:80Þ

i

Calculation of the Standard Molar Gibbs Free Energy of Reaction

(i) Using the standard Gibbs free energy of formation: ΔGo ðT, Po Þ ¼

X νi ΔGof,i ðT, Po Þ

ð37:81Þ

i

(ii) Using: ΔGo(T, Po), ΔHof,i ðT , Po Þ, and CoP,i ðT, Po Þ ΔGo ðT, Po Þ ΔGo ðT , Po Þ ¼  T T

ðT


 ΔHo ðT, Po Þ dT T2

ð37:82Þ

T

ΔHo ðT, Po Þ ¼

X νi ΔHof,i ðT, Po Þ

ð37:83Þ

i

ΔHof,i ðT, Po Þ

¼

ΔHof,i ðT , Po Þ

ðT þ

CoP,i ðT, Po ÞdT 

T

ð37:84Þ

37.31

37.30

Sample Problem 37.1

401

Variation of the Equilibrium Constant with Temperature and Pressure
 ∂lnK ΔHo ¼ ∂T P RT2
 ∂lnK ¼0 ∂P T


37.31

∂lnK ∂P

if

 ¼ T

ΔV RT

ð37:85Þ

P0 ¼ 1 bar or Pvpi

ð37:86Þ

if P0 ¼ P

ð37:87Þ

Sample Problem 37.1

For the solution of this problem, you can assume that: 1. Liquid and vapor mixtures are ideal. 2. Liquid hydrocarbons are completely miscible. 3. Liquid hydrocarbon and water are completely immiscible. (a) Calculate at what pressure will a liquid mixture consisting of droplets of benzene and toluene dispersed in water begin to boil at 40  C (see Fig. 37.1). The following information is provided: (i) The liquid mixture consists of 2 moles of water, 0.5 moles of benzene, and 1 mole of toluene (ii) The pure component vapor pressures at 40  C are: – Water, 55.3 mmHg – Benzene, 181.1 mmHg – Toluene, 59.1 mmHg (iii) The pure component molar volumes at 40  C are: – Water, 18 cm3/mol – Benzene, 89.4 cm3/mol – Toluene, 106.5 cm3/mol

402

37.31.1

37

Review of Part II and Sample Problem

Solution

Fig. 37.1

1. We are dealing with a three-phase system, which is not simple, because the oil mixture (B + T) in the oil droplets and the continuous water (w) phase are fully immiscible (see Fig. 37.1)! We can represent and model the three-phase system in the following useful way (see Fig. 37.2):

Fig. 37.2

37.31

Sample Problem 37.1

403

2. Use the Gibbs Phase Rule in each simple phase to determine the number of independent intensive variables in each phase: Ternary Vapor Mixture n ¼ 3, π ¼ 1, r ¼ 0 ) L ¼ n þ 2  π  r ¼ 3 þ 2  1  0 ¼ 4

ð37:88Þ

We choose these four intensive variables to be: {Tv, Pv, yB, yw}. where, v ¼ vapor (see Fig. 37.2). Pure Water Phase n ¼ 1, π ¼ 1, r ¼ 0 ) L ¼ n þ 2  π  r ¼ 1 þ 2  1  0 ¼ 2

ð37:89Þ

We choose these two variables to be: {Tw, Pw} (see Fig. 37.2). Binary Hydrocarbon Mixture n ¼ 2, π ¼ 1, r ¼ 0 ) L ¼ n þ 2  π  r ¼ 2 þ 2  1  0 ¼ 3

ð37:90Þ

 We choose these three variables to be: To , Po , XoB (see Fig. 37.2). 3. Determine the variance (L) of the phase equilibrium system using the generalized Gibbs approach for composite systems. The system is not simple because there is an impermeable barrier between the water (w) and oil (o) phases. Therefore, we need to use the generalized Gibbs approach to determine L. Specifically, (a) Total number of intensive variables in the three phacses: Tv , Pv , yB , yT |fflfflfflfflfflfflfflffl ffl{zfflfflfflfflfflfflfflfflffl} ;

To , Po , XoB |fflfflfflfflfflffl{zfflfflfflfflfflffl} ;

Vapor phase, L¼4

Oil phase, L¼3

Tw , Pw |fflfflffl{zfflfflffl} Water phase, L¼2

ð37:91Þ

(b) Total conditions of thermodynamic equilibrium which apply: • T.E.: Tv ¼ To ¼ Tw (2 conditions) • M.E.: Pv ¼ Po ¼ Pw (2 conditions) bv bo bv bo bv • D.E.: f w w ¼ f w ; f B ¼ f B ; f T ¼ f T (3 conditions) It then follows that: ð 2 þ 2 þ 3Þ ð 4 þ 3 þ 2Þ L ¼ ðaÞ  ðbÞ ¼ |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}  |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} ¼ 2 9

7

ð37:92Þ

404

37

Review of Part II and Sample Problem

or L¼2

ð37:93Þ

4. We next solve the three fugacity equations: b fw w ¼ fw

v

ð37:94Þ

bf o ¼ bf v B B

ð37:95Þ

bf o ¼ bf v T T

ð37:96Þ

Because of the information provided in the Problem Statement, it is convenient to use the Integral Approach to Phase Equilibria. Specifically,

 w bv bv fw w ¼ f w T, Pvpw ðTÞ Cw ¼ f w ¼ yw ϕw P

ð37:97Þ

where the Poynting correction for water is given by: Cw ¼ exp

"ð P Pvpw ðTÞ



 Vw dP RT

# ð37:98Þ

Because at equilibrium,



 v fw w T, Pvpw ðTÞ ¼ f w T, Pvpw ðTÞ

ð37:99Þ

Equation (37.97) can be rewritten as follows:

 bvP f vw T, Pvpw ðTÞ Cw ¼ yw ϕ w

ð37:100Þ

 b v Pvpw ðTÞ f vw T, Pvpw ðTÞ ¼ ϕ w

ð37:101Þ

We also know that:

Using Eqs. (37.101) and (37.100) yields: bvP b v Pvpw ðTÞCw ¼ yw ϕ ϕ w w

ð37:102Þ

37.31

Sample Problem 37.1

405

Because the vapor mixture is ideal, it follows that:  v ID b bv ¼ ϕ ¼1 ϕ w w

ð37:103Þ

b is the fugacity coefficient of pure water vapor at T ¼ 40  C ¼ 313.15 K Because ϕ w and Pvpw (313.15 K) ¼ 55.3 mmHg, and this vapor pressure is low, we can assume that the pure water vapor phase also behaves ideally, that is, that: v

bv ¼ 1 ϕ w

ð37:104Þ

Using Eqs. (37.103) and (37.104) in Eq. (37.102) then yields: Pvpw ðTÞCw ¼ yw P

ð37:105Þ

We will also assume, and check later, that Cw ¼ 1. In that case, Eq. (37.105) yields: Pvpw ðTÞ ¼ yw P

ðRaoult’s law for waterÞ

ð37:106Þ

We next deal with the phase equilibria of benzene and toluene in the binary oil (o) phase and the ternary vapor phase. For this purpose, we use an activity coefficient approach for benzene and toluene in the binary oil (o) phase, and a fugacity coefficient approach for benzene and toluene in the ternary vapor (v) phase. For benzene we obtain: bf o ¼ Xo γo f o ðT, PÞ B B B B bf v ¼ yB ϕ bv B B

ð37:107Þ

o v bvP ∴bf B ¼ bf B ) XoB γoB f oB ðT, PÞ ¼ yB ϕ B

Similarly, for toluene, we obtain: bf o ¼ Xo γo f o ðT, PÞ ¼ f v ¼ yT ϕ v P T T T T T T

ð37:108Þ

Like in the case for water, we can use the Poynting corrections for benzene and toluene, that is:   f oi ðT, PÞ ¼ f oi T, Pvpi ðTÞ Ci , i ¼ B and T

ð37:109Þ

Using Eq. (37.109), for i ¼ B, in Eq. (37.107) yields:   b vP XoB γoB f oB T, PvpB ðTÞ CB ¼ yB ϕ B

ð37:110Þ

406

37

Review of Part II and Sample Problem

Using Eq. (37.109), for i ¼ T, in Eq. (37.108) yields:   b vP XoT γoT f oT T, PvpT ðTÞ CT ¼ yT ϕ T

ð37:111Þ

Like in the case of water,     f oi T, Pvpi ðTÞ ¼ f vi T, Pvpi ðTÞ ¼ ϕvi Pvpi ðTÞ, i ¼ B and T

ð37:112Þ

Using Eq. (37.112), for i ¼ B, in Eq. (37.110) yields: b P XoB γoB ϕvB PvpB ðTÞCB ¼ yB ϕ B v

ð37:113Þ

Using Eq. (37.112), for i ¼ T, in Eq. (37.111) yields: v

b P XoT γoT ϕvT PvpT ðTÞCT ¼ yT ϕ T

ð37:114Þ

Like in the case for water, we can assume that: b v ¼ 1, ϕ b v ¼ 1, CB ¼ 1, and CT ¼ 1 ϕvB ¼ 1, ϕvT ¼ 1, ϕ B T

ð37:115Þ

In addition, the (B + T) binary mixture is ideal, and therefore, γoB ¼ 1 and γoT ¼ 1. Using all these simplifications in Eqs. (37.113) and (37.114) yields: XoB PvpB ðTÞ ¼ yB P

ð37:116Þ

XoT PvpT ðTÞ ¼ yT P

ð37:117Þ

Adding up Eqs. (37.106), (37.116), and (37.117) we obtain: Pvpw ðTÞ þ XoB PvpB ðTÞ þ XoT PvpT ðTÞ ¼ Pðyw þ yB þ yT Þ

ð37:118Þ

where XoB þ XoT ¼ 1: Equation (37.118) shows that given T and XoB or T and XoT, two independent intensive variables, we can compute the dependent intensive variable, P, consistent with the generalized Gibbs Phase Rule Approach which indicated that L ¼ 2! At the conditions given (see Eq. (37.119)), and using the pure component vapor pressure values provided in the Problem Statement (see Eq. (37.120)):

37.31

Sample Problem 37.1

407


For T ¼ 40o C ¼ 313:15 K,

XoB ¼

 0:5 , 0:5 þ 1

XoT ¼



 1 , 0:5 þ 1

Pvpw ¼ 55:3 mmHg, PvpB ¼ 181:1 mmHg, and PvpT ¼ 50:1 mmHg,

ð37:119Þ

ð37:120Þ

(a) We predict that the mixture will begin to boil at a pressure P given by: 




   0:5 1 P ¼ 55:3 þ 181:1 þ 50:1 mmHg, or P 0:5 þ 1 0:5 þ 1 ¼ 155:1 mmHg

ð37:121Þ

We can next check that, as assumed, the Poynting corrections CW, CB, and CT are close to unity. Indeed, the interested reader is encouraged to show that: CW ¼ 1.00007 CB ¼ 0.99988 CT ¼ 1.00050 Food for Thought Calculate the composition of the first vapor bubble that forms.

Part III

Introduction to Statistical Mechanics

Lecture 38

Statistical Mechanics, Canonical Ensemble, Probability and the Boltzmann Factor, and Canonical Partition Function

38.1

Introduction

The material presented in this lecture is adapted from Chapter 3 in M&S. First, we will discuss the Canonical ensemble and the Boltzmann factor. Second, we will calculate the probability that a system in the Canonical ensemble is in quantum state j with energy Ej(N,V). Finally, we will provide a physical interpretation of the Canonical partition function. Statistical mechanics studies macroscopic systems from a microscopic, or molecular, viewpoint. The goal of statistical mechanics is both to understand and to predict macroscopic behavior given the properties of the individual molecules comprising the system, including their interactions. As we showed in Parts I and II, thermodynamics provides mathematical relations between experimental properties of macroscopic systems at equilibrium. However, it provides no information about the magnitude of any of these properties. Furthermore, thermodynamics does not seek to connect the relations that it describes to molecular models. As such, thermodynamics is limited by its inability to calculate, at the molecular level, physical properties of the type discussed in Parts I and II, including U, S, H, A, G, Cv, μ, etc., or to provide molecular interpretations of its governing equations. When our goal is to formulate a molecular theory that can accomplish the above, we enter the realm of statistical mechanics, which assumes the existence of molecules, to both calculate and interpret thermodynamic behavior from a molecular perspective. As the title of the book indicates, in Part III, we will present an introductory, albeit rigorous, exposure to the fundamentals of statistical mechanics. We hope that this introductory exposure will sufficiently spark the interest of the readers to pursue a broader immersion into this increasingly relevant subject for chemical and mechanical engineers and for chemists, physicists, and materials scientists.

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_38

411

412

38

Statistical Mechanics, Canonical Ensemble, Probability and the Boltzmann. . .

In Part III, when possible, connections will be established with many of the concepts and methodologies that we discussed in Parts I and II. Specifically, we will present (i) a statistical mechanical interpretation of the First, Second, and Third Laws of Thermodynamics; (ii) a statistical mechanical derivation of the ideal gas EOS, the virial EOS, and the van der Waals EOS; (iii) a statistical mechanical description of ideal binary liquid mixtures using a lattice theory approach; and (iv) a statistical mechanical formulation of chemical reaction equilibria.

38.2

Canonical Ensemble and the Boltzmann Factor

Consider a macroscopic system, for example, a liter of oxygen, a gallon of water, or a kilogram of nickel. From a mechanical viewpoint, such macroscopic systems may be described by specifying the number of molecules, N; the volume, V; and the interactions operating between the molecules. Although N is of the order of Avogadro’s number, from a quantum mechanical perspective, each system may be described in terms of its Hamiltonian operator and associated wave functions, which depend on the coordinates of all the molecules comprising the system. In Part III, we will not derive the fundamental equations of quantum mechanics. Instead, we refer the interested reader to an introductory textbook on this fascinating subject. However, we will provide the most important quantum mechanical tools needed to obtain the statistical mechanical results presented in Part III. We begin with the celebrated Schrödinger equation for an N-body (molecule) system, given by: HN ψj ¼ Ej ψj , j ¼ 1, 2, 3, . . .

ð38:1Þ

where HN is the Hamiltonian operator of the N-body (molecule) system, ψj is the quantum mechanical wave function associated with quantum state j, and Ej is the quantum mechanical energy (eigenvalue) associated with quantum state j. It is noteworthy that the discrete energy, Ej, of quantum state j (1, 2, 3,. . .) depends on N and V, that is: Ej ¼ Ej ðN, VÞ

ð38:2Þ

For the special case of a non-interacting system (e.g., an ideal gas), because the molecules are independent, the total energy, Ej(N, V), can be expressed as a sum of the individual energies of each of the N molecules comprising the system, that is, Ej ðN, VÞ ¼ ɛ1 þ ɛ2 þ . . . þ ɛN where εi is the energy of molecule i (i ¼ 1, 2, . . ., N).

ð38:3Þ

38.2

Canonical Ensemble and the Boltzmann Factor

413

Next, we would like to determine the probability, pj, that a system will be in quantum state j having energy, Ej(N, V). As we will show in Lecture 39, knowledge of this probability is essential because it will allow us to calculate average thermodynamic properties of a macroscopic system, including those discussed in Parts I and II. To calculate the desired probability, we consider a very large collection of identical systems in thermal contact with each other, as well as with an infinite heat reservoir maintained at a constant temperature T. We should recognize that each system in the ensemble has the same values of N, V, and T but is likely to be in a

Fig. 38.1

different quantum state, consistent with the values of N and V. Such a collection of systems is referred to as an ensemble and when N, V, and T are specified is referred to as the Canonical ensemble, illustrated in Fig. 38.1. In Fig. 38.1, the number of systems (s) in quantum state j with energy Ej(N,V) is denoted as sj, and the total number of systems in the ensemble is denoted as S. Because the ensemble is a conceptual construction, S can be chosen to be as large as desired. This will become important when we define statistical probabilities (see below). Clearly, the following relation applies: X

sj ¼ S

ð38:4Þ

j

Next, we would like to calculate the relative number of systems in the Canonical ensemble in each quantum state j. As an illustration, we consider quantum states 1 and 2, having energies E1(N,V) and E2(N,V), respectively. Clearly, the relative number of systems in quantum states 1 and 2 must depend on E1 and E2, so that: s2 s1 ¼ f ðE 1 , E 2 Þ where the function f will be determined below.

ð38:5Þ

414

38

Statistical Mechanics, Canonical Ensemble, Probability and the Boltzmann. . .

Because energy is always defined relative to a zero of energy, the dependence of the function, f, on E1 and E2 is expected to have the following form: f ðE1 , E2 Þ ¼ f ðE1  E2 Þ

ð38:6Þ

Using Eq. (38.6) in Eq. (38.5), we obtain: s2 s1 ¼ f ðE1  E2 Þ

ð38:7Þ

Because Eq. (38.7) also applies to quantum states 3 and 2 and 3 and 1, it follows that: s3 s2 ¼ f ðE2  E3 Þ

ð38:8Þ

s3 s1 ¼ f ðE1  E3 Þ

ð38:9Þ

and



Because s3/s1 ¼ (s2/s1) (s3/s2), Eqs. (38.7), (38.8), and (38.9) indicate that f should satisfy: f ðE1  E3 Þ ¼ f ðE1  E2 Þ f ðE2  E3 Þ Recalling that the exponential function satisfies ex suggests that: f ðEÞ ¼ eβE

+ y

ð38:10Þ



¼ ex ey, Eq. (38.10) ð38:11Þ

where β is an arbitrary constant to be determined later. For any quantum states n and m, using Eq. (38.11), with E ! Em  En , in Eq. (38.7), with 2 ! n and 1 ! m, yields: sn ¼ eβðEm En Þ sm

ð38:12Þ

The form of Eq. (38.12) implies that: sj ¼ CeβEj

ð38:13Þ

where j is either quantum state n or m and C is a constant to be determined below.

38.3

38.3

Probability That a System in the Canonical Ensemble Is in Quantum. . .

415

Probability That a System in the Canonical Ensemble Is in Quantum State j with Energy Ej(N, V)

Equation (38.13) has two unknown quantities, C and β, that we need to determine. Determining C is quite simple.PIndeed, summing both sides of Eq. (38.13) with respect to j and recalling that sj is equal to the total number of systems in the j

Canonical ensemble, S, we obtain: X X sj ¼ S ¼ C eβEj j

ð38:14Þ

j

or C¼P

S eβEj

ð38:15Þ

j

Using Eq. (38.15) in Eq. (38.13), including rearranging, yields: sj eβEj ¼ P βE S e j

ð38:16Þ

j

In Eq. (38.16), the ratio sj/S is the fraction of systems in the Canonical ensemble that are in quantum state j with energy Ej. In the limit of large S, which we can certainly take because our ensemble may be chosen to be as large as we would like it to be, the ratio sj/S becomes the statistical probability that a randomly chosen system in the Canonical ensemble will be in quantum state j with energy Ej(N, V). Denoting this probability as pj, it follows that: eβEj pj ¼ P βE e j

ð38:17Þ

j

Equation (38.17) is a central result of statistical mechanics because, as we will show in the following lectures, it will allow us to calculate average thermodynamic properties of macroscopic systems. It is customary to denote the denominator in Eq. (38.17) as Q, where: Q ðN, V, βÞ ¼

X j

eβEj ðN,VÞ

ð38:18Þ

416

38

Statistical Mechanics, Canonical Ensemble, Probability and the Boltzmann. . .

In a coming lecture, we will show that: β¼

1 kB T

ð38:19Þ

where kB is the Boltzmann constant (1.381  1023J/K) and T is the Kelvin temperature. Using Eqs. (38.19) and (38.18) in Eq. (38.17) yields: pj ¼

eEjðN,VÞ=kB T QðN, V, TÞ

ð38:20Þ

Equation (38.20) can also be expressed in terms of β, as follows: pj ¼

eβ EjðN,VÞ QðN, V, βÞ

ð38:21Þ

The function Q(N, V, β), or Q(N, V, T), is known as the partition function of the system in the Canonical ensemble representation or, in short, as the Canonical partition function. In the coming lectures, we will show how to calculate all the thermodynamic properties of a macroscopic system if we know Q(N, V, β) or Q(N, V, T). Specifically, we will learn how to calculate the Canonical partition function for a number of interesting systems.

38.4

Physical Interpretation of the Canonical Partition Function

According to Eq. (38.18), Q is a sum of Boltzmann factors, eEj =kB T , that determine how molecules are partitioned throughout the accessible quantum states of the system. For simplicity, if we assume that the energy of the ground state (1) is zero, that is, if E1 ¼ 0, then: Q¼

t X

eEj =kB T ¼ 1 þ eE2 =kB T þ eE3 =kB T þ     þ eEt =kB T

ð38:22Þ

j¼1

where t is the number of accessible quantum states, which is determined by the underlying physics of the system. The Canonical partition function, Q, accounts for the number of quantum states that are effectively accessible to the system, out of the t accessible quantum states. To better understand what this means, it is instructive to consider two limiting cases:

38.4

Physical Interpretation of the Canonical Partition Function

417

(i) The energies are small, or equivalently, the temperatures are high, that is, Ej =kB T ! 0, for all js > 1. Accordingly, t Times

zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{ Q ¼1 þ 1 þ 1 þ    þ 1 ¼ t

ð38:23Þ

and pj ¼

eEj =kB T 1 ¼ t Q

ð38:24Þ

Equation (38.24) indicates that, in this limit, all the t states become accessible with equal probability, 1/t. (ii) The energies are large, or equivalently, the temperatures are low, that is, Ej =kB T ! 1, for all js > 1. Accordingly, ðt1Þ Times

zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{ Q ¼ 1þ 0 þ 0 þ 0 þ    þ 0¼ 1

ð38:25Þ

and 1 ¼1 1 0 ¼ ¼0 1

p1 ¼ pj>1

ð38:26Þ

Equation (38.26) indicates that, in this limit, only the ground state is accessible. In conclusion, the magnitude of Ej relative to kBT, namely, Ej =kB T, determines whether or not quantum state j is effectively accessible. Indeed, quantum states that possess energies that are higher than kBT are relatively inaccessible and unpopulated at temperature T. On the other hand, quantum states that possess energies that are lower than kBT are accessible and well populated at temperature, T. This criterion will be utilized extensively in the coming lectures.

Lecture 39

Calculation of Average Thermodynamic Properties Using the Canonical Partition Function and Treatment of Distinguishable and Indistinguishable Molecules

39.1

Introduction

The material presented in this lecture is adapted from Chapter 3 in M&S. First, we will utilize material presented in Lecture 38 to calculate various Canonical ensemble-averaged thermodynamic properties of the system, including the average energy, the average heat capacity at constant volume, and the average pressure. Second, we will derive an expression for the Canonical partition function of a system of independent and distinguishable molecules. Third, we will derive an expression for the Canonical partition function of a system of independent and indistinguishable molecules, including stressing the role of the Pauli exclusion principle when carrying out the summation over the indistinguishable quantum states. Fourth, we will decompose the molecular Canonical partition function into contributions from the translational, vibrational, rotational, and electronic degrees of freedom. Finally, we will contrast energy states and energy levels, including discussing the degeneracy of an energy level.

39.2

Calculation of the Average Energy of a Macroscopic System

In Lecture 38, we saw that in the Canonical ensemble, the probability that a system is in quantum state j is given by: pjðN, V, βÞ ¼


 exp βEjðN, VÞ QðN, V, βÞ

ð39:1Þ

where Q is the Canonical partition function, given by: © Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_39

419

420

39

Calculation of Average Thermodynamic Properties Using the Canonical Partition. . .

QðN, V, βÞ ¼

X


 exp βEjðN, VÞ

ð39:2Þ

j

Using the well-known statistical definition of an average property, the Canonical ensemble-averaged energy, hEi, which is equal to the experimentally observed energy, U, is given by: U ¼ hEi ¼

X

pj E j ¼

X Ej ðN, VÞeβEjðN,VÞ

j

QðN, V, βÞ

j

ð39:3Þ

where Eq. (39.1) was used. Although the quantum mechanical energies, Ej , do not depend on temperature, the Canonical ensemble-averaged energy, hEi ¼ U, depends on T or equivalently on β, as well as on N and V. It is possible to express hEi ¼ U in Eq. (39.3) entirely in terms of Q(N,V, β). Specifically, differentiating lnQðN, V, βÞ with respect to β, at constant N and V, yields: 

∂lnQðN, V, βÞ ∂β

 ¼ V,N

X Ej ðN, VÞeβEjðN,VÞ j

QðN, V, βÞ

ð39:4Þ

A comparison of Eqs. (39.3) and (39.4) shows that: U ¼ hEi ¼ 

  ∂lnQðN, V, βÞ ∂β V,N

ð39:5Þ

We can also express hEi ¼ U as a temperature derivative, rather than as a β derivative. Using the fact that: ∂ ∂ ¼ kBT2 ∂β ∂T

ð39:6Þ

and then using Eq. (39.6) in Eq. (39.5), we obtain:   ∂lnQðN, V, TÞ U ¼ hEi ¼ kB T2 ∂T V,N

ð39:7Þ

39.4

39.3

Calculation of the Average Pressure of a Macroscopic System

421

Calculation of the Average Heat Capacity at Constant Volume of a Macroscopic System

Recall that the heat capacity at constant volume, Cv, was introduced in Part I as follows: Cv ¼

    ∂U 1 ∂U ¼ ∂T V N ∂T V,N

ð39:8Þ

Using the fact that the internal energy of a macroscopic system, U ¼ hEi, it follows that: Cv ¼

    ∂hEi 1 ∂hEi ¼ ∂T V N ∂T V,N

ð39:9Þ

We can use Eq. (39.7), along with Eq. (39.2), to calculate Cv using Eq. (39.9).

39.4

Calculation of the Average Pressure of a Macroscopic System

In one of the coming lectures, we will show that the pressure of a macroscopic system in quantum state j is given by:   ∂Ej Pj ðN, VÞ ¼  ∂V N

ð39:10Þ

Equation (39.10) is analogous to the definition of the pressure presented in Part I, where: P ¼ ð∂U=∂VÞS,N

ð39:11Þ

Accepting Eq. (39.10) for now, the Canonical ensemble-averaged pressure, 〈P〉, which is equal to the experimentally observed pressure, P, is given by: P ¼ hPi ¼

X j

pj ðN, V, βÞPj ðN, VÞ

ð39:12Þ

422

39

Calculation of Average Thermodynamic Properties Using the Canonical Partition. . .

Using Eq. (39.1) for pj and Eq. (39.10) for Pj, in Eq. (39.12), yields: P ¼ hPi ¼

X ∂Ej  eβEj ðN,VÞ  ∂V N QðN, V, βÞ j

ð39:13Þ

We can express P ¼ hPi solely in terms of Q as follows. Starting from the definition of Q in Eq. (39.2), and differentiating Q with respect to V, keeping N and β constant, yields: 

∂Q ∂V

 ¼ β N,β

X∂Ej  j

∂V

eβEj ðN,VÞ

ð39:14Þ

N

A comparison of Eqs. (39.13) and (39.14) shows that: P ¼ hPi ¼

  kB T ∂Q QðN, V, βÞ ∂V β,N

ð39:15Þ

Equation (39.15) can also be expressed as follows:   ∂lnQðN, V, βÞ P ¼ hPi ¼ kB T ∂V β,N

39.5

ð39:16Þ

Canonical Partition Function of a System of Independent and Distinguishable Molecules

The general results that we have presented so far in Part III are valid for arbitrary systems. In order to actually use these results, we need to calculate the Canonical partitionfunction,Q. For this purpose, we need to know the set of energy eigenvalues, Ej ðN, VÞ , for the N-body (molecule) Schrödinger equation. In general, because of the complexity of this interacting multi-molecule system, this turns out to be an intractable problem. However, for many important systems, the total energy of the N-molecule system can be expressed as a sum of individual energies associated with each molecule comprising the system. This approximation, when applicable, leads to a great simplification in the evaluation of the Canonical partition function, Q. For a system of independent, distinguishable molecules, we will denote the n o energies of the individual molecules as εaj , where a denotes the molecule in question (they are distinguishable) and j denotes the quantum state of the molecule in question.

39.5

Canonical Partition Function of a System of Independent and Distinguishable. . .

423

For such a system, the total energy, Eijk... ðN, VÞ, can be written as follows: Eijk... ðN, VÞ ¼ εai ðVÞ þ εbj ðVÞ þ εck ðVÞ þ    |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

ð39:17Þ

N Terms

In principle, because the εis in Eq. (39.17) depend on V, they should carry an underbar. However, to simplify the notation, we have not included the underbars. In addition, as stressed above, the temperature does not appear at the quantum mechanical level. Using Eq. (39.17), the Canonical partition function of the system can be written as follows: QðN, V, TÞ ¼

X X β εa þεb þεc þ eβEi,j,k,... ¼ e ði j k Þ

ð39:18Þ

i, j, k, ...

ijk...

Recall that in Eq. (39.18), the molecules are distinguishable and independent. Therefore, we can sum over i, j, k, etc. independently. As a result, Q(N, V, T) in Eq. (39.18) can be written as a product of partition functions associated with molecules a, b, c, etc., that is, QðN, V, TÞ ¼

X a eβεi

!

X βεb e j

i

!

j

X

! e

βεck

...

ð39:19Þ

k

or QðN, V, TÞ ¼ qa ðV, TÞ qb ðV, TÞ qc ðV, TÞ . . .

ð39:20Þ

where each of the molecular Canonical partition functions, q(V,T), is given by: q ðV, TÞ ¼

X X eβεi ¼ eεi =kB T i

ð39:21Þ

i

and the number of molecular Canonical partition functions in Eq. (39.20) is equal to N. Because the calculation of q(V,T) in Eq. (39.20) only requires knowledge of the discrete quantum mechanical energy states of a single molecule, its calculation is often feasible, as we will show shortly. If the quantum mechanical energy states of all the molecules are the same, then, all the molecular Canonical partition functions are equal, and Eq. (39.20) reduces to: QðN, V, TÞ ¼ ½qðV, TÞN where q(V, T) is given by Eq. (39.21).

ð39:22Þ

424

39.6

39

Calculation of Average Thermodynamic Properties Using the Canonical Partition. . .

Canonical Partition Function of a System of Independent and Indistinguishable Molecules

In the case of indistinguishable molecules, the total energy is given by: Eijk ¼ εi þ εj þ εk þ    |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}

ð39:23Þ

N Terms

where, because the N molecules are indistinguishable, no superscripts appear in the energies {εi}. Each molecular energy in Eq. (39.23) depends on the system volume, V, and therefore should carry an underbar, which we have not included for notational simplicity. Using Eq. (39.23) in the expression for the Canonical partition function of the system, we obtain: QðN, V, TÞ ¼

X

eβðεi þεj þεk þÞ

ð39:24Þ

i, j, k, 

Because the molecules are indistinguishable, the quantum mechanical Pauli exclusion principle applies and states that: “No two electrons in a molecule can be in the same quantum state.” As a result, it is clear that the indices i, j, k,    in Eq. (39.24) are not independent. Therefore, the summations in Eq. (39.24) cannot be carried out independently as we did in the case of molecules which are distinguishable (see Eq. (39.19)). Nevertheless, if the number of terms in Eq. (39.24) in which two or more indices are the same, and hence, are not independent, is small, we could carry out the summations in an unrestricted manner and obtain ½qðV, TÞN , as we did in the distinguishable molecule case. Subsequently, we would need to divide this result by N! in order to account for the overcounting because the molecules are indistinguishable in this case. This, in turn, would result in ½qðV, TÞN =N! It is reasonable to assume that if the number of quantum states available to any molecule is much larger than the number of molecules, it would be very unlikely for any two molecules to be in the same quantum state. The following quantum mechanical criterion can be used to ensure that this condition is satisfied:  3=2 N h2 1 V 8mkB T

ð39:25Þ

The inequality in Eq. (39.25) is favored by a large molecular mass (m), a high temperature (T), and a low number density ðρ ¼ N=VÞ. Table 39.1 lists the values of
3=2 ðN=VÞ h2 =8mkB T , at a pressure of one bar and various temperatures, for several systems.

39.7

Decomposition of a Molecular Canonical Partition Function into Canonical. . .

425

Table 39.1 System Liquid helium Gaseous helium Gaseous helium Gaseous helium Liquid hydrogen Gaseous hydrogen Gaseous hydrogen Liquid neon Gaseous neon Liquid krypton Electrons in metals

T/K 4 4 20 100 20 20 100 27 27 127 300

N V



h2 8mkB T

3=2

1.5 0.11 1.8  103 3.3  105 0.29 5.1  103 9.4  105 1.0  102 7.8  105 5.1  105 1400


3=2 As Table 39.1 shows, systems for which ðN=VÞ h2 =8mkB T is not  1, therefore violating the criterion in Eq. (39.25), include liquid helium and liquid hydrogen (due to their small masses and low temperatures) and electrons in metals (due to their small mass). When Eq. (39.25) is satisfied, the summations in Eq. (39.24) can be carried out independently, which after dividing by N!, yields: QðN, V, TÞ ¼

½qðV, TÞN N!

ð39:26Þ

where q(V, T) is given in Eq. (39.22). When Eqs. (39.25) and (39.26) are satisfied, the molecules obey Boltzmann statistics. As Eq. (39.25) shows, Boltzmann statistics becomes increasingly applicable as the temperature increases.

39.7

Decomposition of a Molecular Canonical Partition Function into Canonical Partition Functions for Each Degree of Freedom

We would like to express the average energy of a macroscopic system, hEi, in terms of the molecular Canonical partition function, qðV, TÞ. We begin with Eq. (39.7), where for indistinguishable molecules taking the natural logarithm of Eq. (39.26) yields:  N q lnQ ¼ ln ¼ Nlnq  lnN! N!

ð39:27Þ

426

39

Calculation of Average Thermodynamic Properties Using the Canonical Partition. . .

Using Eq. (39.27) in Eq. (39.7), while keeping N and V constant, yields: hE i ¼

  NkB T2 ∂q q ∂T V

ð39:28Þ

Because (see Eqs. (39.22) and (39.6)) q¼

X

eβεj and

j

dβ 1 ¼ dT kB T 2

ð39:29Þ

it follows that: 

 ∂q 1 X βεj ¼ εj e ∂T V kB T2 j

ð39:30Þ

Using Eq. (39.30) in Eq. (39.28) yields: hEi ¼ Nhεi

ð39:31Þ

where the average molecular energy, hεi, is given by: h εi ¼

X eεj =kB T εj qðV, TÞ j

ð39:32Þ

Equation (39.32) shows that the probability that a molecule is in quantum state j, denoted by πj ðV, TÞ, is given by: πj ðV, TÞ ¼

eεj =kB T eεj =kB T ¼ P ε =k T qðV, TÞ e j B

ð39:33Þ

j

Equations (39.31), (39.32), and (39.33) can be simplified even further if we assume that the energy of a molecule can be decomposed into several contributions associated with the various molecular degrees of freedom: translational (trans), rotational (rot), vibrational (vib), and electronic (elec). Specifically: vib elec ε ¼ εtrans þ εrot i j þ εk þ εl

ð39:34Þ

39.8

Energy States and Energy Levels

427

Because the various molecular degrees of freedom are distinguishable, it follows that: qðV, TÞ ¼

X

! e

εtrans =kB T i

i

X

! e

εrot j =kB T

j

X

! e

εvib k =kB T

k

X

! e

εelec l =kB T

l

|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} qtrans

qvib

qrot

qelec

or
 q V, T ¼ qtrans qrot qvib qelec

39.8

ð39:35Þ

Energy States and Energy Levels

So far, we have expressed partition functions as summations over energy states. Each energy state is represented by a quantum mechanical wave function with an associated energy, that is, qðV, TÞ ¼

X

eεj =kB T

ð39:36Þ

j ðStatesÞ

Alternatively, if we refer to sets of states that have the same energy as levels, we can write qðV, TÞ as a summation over energy levels by including the degeneracy, gj, of each energy level j, that is, qðV, TÞ ¼

X

gj eεj =kB T

ð39:37Þ

j ðLevelsÞ

In Eq. (39.36), the terms representing a degenerate energy state j are repeated gj times. On the other hand, in Eq. (39.37), degenerate energy state j is written only once and then multiplied by gj. Including degeneracies explicitly, as we did in Eq. (39.37), is usually more convenient from a computational point of view, as we will show in the coming lectures.

Lecture 40

Translational, Vibrational, Rotational, and Electronic Contributions to the Partition Function of Monoatomic and Diatomic Ideal Gases and Sample Problem

40.1

Introduction

The material presented in this lecture is adapted from Chapter 4 in M&S. First, we will derive an expression for the Canonical partition function of a monoatomic ideal gas, including calculating the translational contribution to the partition function and its average translation energy. Second, we will discuss the Energy Equipartition Theorem. Third, we will derive an expression for the electronic contribution to the atomic partition function. Fourth, we will solve Sample Problem 40.1 to calculate the fraction of helium atoms in the first excited state at 300 K, given information about the degeneracies and energies of the excited states involved. Fifth, we will calculate the average energy, average heat capacity at constant volume, and average pressure of a monoatomic ideal gas. Finally, we will begin discussing a diatomic ideal gas, where in addition to the translational and electronic degrees of freedom encountered in the case of a monoatomic ideal gas, we will need to incorporate the vibrational and rotational degrees of freedom, which we will discuss in Lecture 41. Here, we will calculate the translational and electronic contributions to the molecular partition function of a diatomic ideal gas.

40.2

Partition Functions of Ideal Gases

We already saw that if the number of available quantum states is much larger than the number of molecules (or atoms) in the system, then, we can write the Canonical partition function of the entire system in terms of the individual Canonical molecular partition functions. Specifically,

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_40

429

430

40

Translational, Vibrational, Rotational, and Electronic Contributions. . .

QðN, V, TÞ ¼

½qðV, TÞN N!

ð40:1Þ

Equation (40.1) for Q is particularly valid for ideal gases, because the molecules are independent in that case and the densities (N/V) of gases that behave ideally are
3=2 sufficiently low that the criterion, ðN=VÞ h2 =8mkB T  1, required for Eq. (40.1) to be valid, is satisfied. We will discuss a monoatomic ideal gas first, followed by a diatomic ideal gas. At this introductory level, we will not discuss a polyatomic ideal gas.

40.3

Translational Partition Function of a Monoatomic Ideal Gas

The energy of an atom in a monoatomic ideal gas is the sum of its translational energy and electronic energy, that is, εatomic ¼ εtrans þ εelec

ð40:2Þ

In Eq. (40.2), nuclear degrees of freedom are not included because they are not accessible at the temperatures of interest. Using Eq. (40.2) in the general expression for q involving these two degrees of freedom (see Lecture 39), we obtain: qðV, TÞ ¼ qtrans ðV, TÞqelec ðTÞ

ð40:3Þ

To compute qtrans in Eq. (40.3), we need to make use of some well-known quantum mechanics results. Specifically, the solution of the Schrödinger equation for the translational energy states of an atom in a cubic container of side length, d, is given by: εnx ny nz ¼

  h2 2 2 2 n þ n þ n y z , for nx , ny , nz ¼ 1, 2, . . . 8md2 x

ð40:4Þ

where the translational energy states are quantized (discrete) and are labeled by the quantum numbers (nx, ny, and nz) corresponding to the three spatial dimensions (x, y, and z). Substituting Eq. (40.4) in the expression for qtrans derived in Lecture 39 yields: qtrans ¼

X βεtrans e j j

ð40:5Þ

40.3

Translational Partition Function of a Monoatomic Ideal Gas

431

with j ¼ nx, ny, nz ¼ 1, 2, etc. Carrying out the summations in Eq. (40.5), we obtain: 1 X

qtrans ¼

eβεnx ny nz

nx , ny , nz ¼1

" ¼

1 X nx ¼1

"





βh2 n2x exp  8md2

1 X nz ¼1

#" X 1

exp 

ny ¼1

 # βh2 n2z exp 8md2

βh2 n2y

!#

8md2 ð40:6Þ

Because each summation in Eq. (40.6) is identical, we can rewrite it as follows: qtrans ¼

" 1 X n¼1

 #3 βh2 n2 exp 8md2

ð40:7Þ

Recall that in Eqs. (40.4), (40.6), and (40.7), m is the mass of the atom, β ¼ 1/ kBT, and h is Planck’s constant ¼ 6.626  1034Js. To a very good approximation, the summation in Eq. (40.7) can be replaced by an integral over n. Specifically, qtrans ðV, TÞ ¼

ð 1

eβh n =8md dn 2 2

2

3 ð40:8Þ

0

Recall that the integral in Eq. (40.8) is a Gaussian integral. If we denote βh2/8md2 by z, then, this integral is given by: 1 ð

ezn dn ¼ 2

 1=2 π 4z

ð40:9Þ

0

Using Eq. (40.9) in Eq. (40.8) yields:  qtrans ðV, TÞ ¼

3=2 2πmkB T d3 h2

ð40:10Þ

where d3 ¼ V. In Eq. (40.10), the factor (h2/2πmkBT)1/2 has units of length and is known as the thermal de Broglie wavelength of the atom. Denoting this length by Λ, Eq. (40.10) can be written as follows:

432

Translational, Vibrational, Rotational, and Electronic Contributions. . .

40

qtrans ¼

V Λ3

ð40:11Þ

Recall that the criterion for the applicability of Boltzmann statistics presented in
Lecture 39, ðN=VÞ h2=8mkB T  1, can be expressed in terms of Ʌ as follows:  3=2 3 π Λ 1 4 d

ð40:12Þ

Equation (40.12) shows that for the number of available quantum states to be much larger than the number of atoms in the system, so that Boltzmann statistics is valid, we require that Λd, that is, that the spatial spread of the quantum mechanical wave function be much smaller than the interatomic distance. This, in turn, guarantees that the wave functions do not overlap. Clearly, quantum mechanical effects become less important as Λ  d, which can be attained when m is large, when T is high, or when N/V is small. Next, we will calculate the average translational energy, hεtransi, of an atom using Eq. (40.10) in the expression for hεtransi derived in Lecture 39. Specifically,   ∂ lnqtrans hεtrans i ¼ kB T ∂T V

ð40:13Þ

3 hεtrnas i ¼ kBT 2

ð40:14Þ

2



  ∂ 3 lnT þ ðTerms that do not depend on TÞ hεtrans i ¼ kB T ∂T 2 V 2

Equation (40.14) indicates that each of the three atomic translational degrees of freedom contributes 12 kBT to the average translational energy of the atom. This result is an example of the Energy Equipartition Theorem to be discussed below.

40.4

Electronic Contribution to the Atomic Partition Function

It is convenient to write the electronic partition function of an atom as a sum over energy levels, with known degeneracies. Specifically, qelec ¼

X gei eβ εei

ð40:15Þ

i

where gei is the degeneracy of quantum level i and εei is the electronic energy of quantum level i. If we measure all the electronic energies relative to the ground

40.5

Sample Problem 40.1

433

electronic state (denoted by i ¼ 1), or equivalently, if we choose εe1 ¼ 0, we can express Eq. (40.15) as follows: qelec ðTÞ ¼ ge1 þ ge2 eβεe2 þ   

ð40:16Þ

Because qelec is an atomic property, it does not depend on V, and depends solely on T. Quantum mechanics indicates that, typically, βεelec ¼

εelec 104 K  T kB T

ð40:17Þ

which is equal to 10, even for a relatively high temperature like T ¼ 1000 K. Clearly, as T decreases, βεelec becomes even larger. In view of Eq. (40.17), eβεe2 in Eq. (40.16) has a typical value of around 105 for most atoms at ordinary temperatures. As a result, in Eq. (40.16), only the first term, ge1 – the degeneracy of the ground electronic level – is 6¼ 0.

40.5

Sample Problem 40.1

Calculate the fraction of helium atoms in the first excited state at T ¼ 300 K. It is known that ge1 ¼ 1, ge2 ¼ 3, and βεe2 ¼ 767 at 300 K. It is also known that ge3 ¼ 1 and βεe3 ¼ 797 at 300 K.

40.5.1 Solution The fraction of helium atoms in the first excited state (j ¼ 2) is given by the probability of finding the helium atoms in that state. In other words, f 2 ¼ p2 ¼

ge2 eβεe2 qelec ðTÞ

where qelec ðTÞ ¼ ge1 þ ge2 eβεe2 þ ge3 eβεe3 þ   

434

40

Translational, Vibrational, Rotational, and Electronic Contributions. . .

Using the information provided in the Problem Statement, it follows that: f2 ¼

1þ

3e767 ’ 10334 þ 1e797

3e767

which clearly shows that the first excited state is not populated. In fact, for most atoms, including the first two terms of the electronic partition function is usually sufficient, that is, qelec ðTÞ ¼ ge1 þ ge2 eβεe2

ð40:18Þ

In summary, for a monoatomic ideal gas: N qN ðqtrans qelec Þ ¼ N! N!  3=2 2πmkB T ¼ V h2

QðN, V, TÞ ¼ qtrans

ð40:19Þ

qelec ¼ ge1 þ ge2 eβεe2

40.6

Average Energy of a Monoatomic Ideal Gas

Denoting the average energy by U ¼ hEi and using the result obtained in Lecture 39, it follows that:     ∂lnQ 2 ∂lnq U ¼ kB T ¼ NkB T ∂T V,N ∂T V 2

ð40:20Þ

Using the expression for q ¼ qtrans  qelec, along with Eq. (40.19), in Eq. (40.20) yields: Average Electronic Energy

3 U ¼ NkB Tþ 2|fflfflffl{zfflfflffl}

zfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflffl{ Nge2 εe2 eβεe2 qelec

Average Translational ðKineticÞ Energy

In most cases, the electronic contribution in Eq. (40.21) is negligible.

ð40:21Þ

40.9

40.7

Diatomic Ideal Gas

435

Average Heat Capacity at Constant Volume of a Monoatomic Ideal Gas

Neglecting the electronic contribution to U, the average heat capacity at constant volume, CV, is given by (see Part I): Per Molecule

z}|{     ∂U 1 ∂U 3 3 ¼ ¼ kB ¼ R CV ¼ 2 2 ∂T V N ∂T V,N |{z}

ð40:22Þ

Per Mole

40.8

Average Pressure of a Monoatomic Ideal Gas

As shown in Lecture 39:     ∂lnQ ∂lnq ¼ NkB T P ¼ kB T ∂V T,N ∂V T

ð40:23Þ

Using the expression for q in Eq. (40.19) in Eq. (40.23), we obtain:  ∂ ðlnV þ Terms that do not depend on VÞ P ¼ NkB T ∂V T or P¼

NkB T V

ð40:24Þ

Equation (40.24) is, of course, the ideal gas equation of state, expressed on a per molecule basis (see Part I), that we have now derived molecularly using statistical mechanics. Further, Eq. (40.24) was obtained because qðV, TÞ is of the form f ðTÞV. As a result, only the translational degrees of freedom of the atoms depend on V and contribute to P. Below, we will show that this is also the case for a diatomic ideal gas.

40.9

Diatomic Ideal Gas

In addition to the translational and electronic degrees of freedom, diatomic molecules possess vibrational and rotational degrees of freedom. As we did to model a monoatomic ideal gas, for a diatomic ideal gas, we will use solutions of the

436

40

Translational, Vibrational, Rotational, and Electronic Contributions. . .

Schrödinger equation to obtain expressions for the allowable energy levels and their degeneracies. Specifically, to describe/model the rotations and vibrations, we will utilize the rigid rotator-harmonic oscillator approximation to solve the Schrödinger equation. In this approximation, the rotational energy levels are quantized (discrete) and are given by: εJ ¼

ℏ2 JðJ þ 1Þ, for J ¼ 0, 1, 2, . . . 2I

ð40:25Þ

where ℏ ¼ h=2π I ¼ Moment of Inertia with each energy level, J, having a degeneracy, gJ, given by: gJ ¼ 2J þ 1

ð40:26Þ

The vibrational energies of the harmonic oscillator are also quantized (discrete) and are given by:   1 εn ¼ hν n þ , for n ¼ 0, 1, 2, . . . 2

ð40:27Þ

where ν ¼ Vibrational frequency ¼ c=λ ¼ ce ν where c – Speed of light λ – Wavelength e ν – Wave number In Eq. (40.27), each energy level, n, is non-degenerate, that is, it has: gn ¼ 1

ð40:28Þ

In the rigid rotator-harmonic oscillator approximation, we can write the total energy of the diatomic molecule as a sum of its translational, rotational, vibrational, and electronic energies, that is, ε ¼ εtrans þ εrot þ εvib þ εelec

ð40:29Þ

40.9

Diatomic Ideal Gas

437

As in the case of a monoatomic ideal gas, the criterion for the applicability of
Boltzmann’s statistics, ðN=VÞ h2 =8mkB T  1, is readily satisfied at normal temperatures. For a diatomic molecule comprising atoms 1 and 2, m ¼ m1 + m2, which is the diatomic mass. As before, we can write: QðN, V, TÞ ¼

½qðV, TÞN N!

ð40:30Þ

In addition, the energy decomposition in Eq. (40.29) allows us to write:
 q V, T ¼ qtrans qrot qvib qelec

ð40:31Þ

Using Eq. (40.31) in Eq. (40.30) yields: QðN, V, TÞ ¼

ðqtrans qrot qvib qelec ÞN N!

ð40:32Þ

As discussed above, the translational partition function of a diatomic molecule is similar to that of an atom, except that the atomic mass, m, is replaced by the diatomic mass, m1 + m2, that is:  3=2 2πðm1 þ m2 ÞkB T qtrans ðV, TÞ ¼ V h2

ð40:33Þ

Next, in order to calculate qrot, qvib, and qelec, we first need to select the zero of each energy contribution. Specifically, (i) The zero of the rotational energy corresponds to the J ¼ 0 state, where εJ ¼ 0 ¼ 0 (see Eq. (40.25)). (ii) The zero of the vibrational energy corresponds to the bottom of the internuclear potential well of the lowest electronic state, so that the energy of the ground vibrational state is hν/2 (see Eq. (40.27)). (iii) The zero of the electronic energy corresponds to that of the separated atoms comprising the diatomic molecule at rest in their ground electronic states. If we denote the depth of the ground electronic energy state by De(>0), the energy of the ground electronic state corresponds to εe1 ¼  De with a degeneracy ge1. The energy of the first excited electronic state corresponds to εe2 with a degeneracy ge2. Figure 40.1 illustrates all the energies involved as a function of the internuclear separation, R.

438

40

Translational, Vibrational, Rotational, and Electronic Contributions. . .

Fig. 40.1

Note that, in Fig. 40.1, De ¼ Do + hν/2, where Do corresponds to the energy difference between the lowest vibrational state and the dissociated molecule and can be measured spectroscopically. Using ge1, εe1 ¼  De, ge2 and εe2, the electronic partition function can be written as follows: qelec ¼ ge1 eDe =kB T þ ge2 eεe2 =kB T

ð40:34Þ

Lecture 41

Thermodynamic Properties of Ideal Gases of Diatomic Molecules Calculated Using Partition Functions and Sample Problems

41.1

Introduction

The material presented in this lecture is adapted from Chapter 4 in M&S. In Lecture 40, we presented expressions for the translational and electronic contributions to the partition function of a diatomic molecule. In this lecture, first, we will discuss the vibrational contribution to the partition function of a diatomic molecule, including deriving expressions for the average vibrational energy and vibrational contribution to the heat capacity at constant volume of an ideal gas of diatomic molecules. Second, we will introduce the characteristic vibrational temperature, whose value relative to the system temperature will determine the probability that a given vibrational energy level is populated. Further, we will solve Sample Problem 41.1 to calculate the probability that a vibrational energy level is populated in the case of nitrogen molecules, including discussing the effect of temperature. Third, we will discuss the rotational contribution to the partition function of a diatomic molecule, including deriving expressions for the average rotational energy and rotational contribution to the heat capacity at constant volume of an ideal gas of diatomic molecules. In addition, we will introduce the characteristic rotational temperature, whose value relative to the system temperature will determine the probability that a given rotational energy level is populated. Fourth, we will emphasize that the rotational partition function of a diatomic molecule contains a symmetry number, which is equal to one when the two atoms comprising the diatomic molecule are different (referred to as the heteronuclear case) and is equal to two when the two atoms are identical (referred to as the homonuclear case). Fifth, we will present an expression for the total partition function of a diatomic molecule, which includes translational, vibrational, rotational, and electronic contributions. Finally, we will solve Sample Problem 41.2 to calculate the average molar internal energy and molar heat capacity at constant volume of an ideal gas of diatomic molecules, including identifying the various contributions to each thermodynamic property.

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7_41

439

440

41.2

41

Thermodynamic Properties of Ideal Gases of Diatomic Molecules Calculated. . .

Vibrational Partition Function of a Diatomic Molecule

If we measure the vibrational levels relative to the bottom of the internuclear potential well, then, as discussed in Lecture 40, the vibrational energies are given by:
 1 ɛn ¼ n þ hν, for n ¼ 0, 1, 2, . . . 2

ð41:1Þ

where ν(Vibrational frequency) ¼ (K/μ)1/2/2π μ (Reduced mass) ¼ (m1m2)/(m1 + m2) K – Hooke spring constant and where, ν ¼ ce v, where c is the speed of light and e v is the wave number. Using Eq. (41.1), the vibrational partition function of a diatomic molecule is given by: qvib ðTÞ ¼

1 X X 1 eβɛn ¼ eβðnþ2Þhν n

ð41:2Þ

n¼0

or qvib ðTÞ ¼ eβhν=2

1 X

eβhνn

ð41:3Þ

n¼0

The sum in Eq. (41.3) corresponds to a geometric series. Denoting eβhν ¼ x < 1, it follows that: 1 X

1 1x

ð41:4Þ

eβhν=2 1  eβhν

ð41:5Þ

xn ¼

n¼0

Using Eq. (41.4) in Eq. (41.3) yields: qvib ðTÞ ¼

It is convenient to introduce a characteristic vibrational temperature, θvib, defined as follows: θvib ¼

hν hce v ¼ kB kB

ð41:6Þ

41.2

Vibrational Partition Function of a Diatomic Molecule

441

Using Eq. (41.6) in Eq. (41.5) yields: qvib ðTÞ ¼

eθvib =2T 1  eθvib =T

ð41:7Þ

As we will see, θvib/T is a central quantity whose value will determine the occupancy of the various vibrational energy levels. Next, we will calculate the average vibrational energy of an ideal gas of N diatomic molecules using qvib (T). Specifically,   dlnqvib hEvib i ¼ Nhɛvib i ¼ NkB T2 dT

ð41:8Þ

Using Eq. (41.7) for qvib in Eq. (41.8), we obtain: 
 d θvib θvib =T   ln 1  e hEvib i ¼ NkB T dT 2T 2

which, after taking the temperature derivative, yields:   θ θ hEvib i ¼ NkB vib þ θ =Tvib 2 e vib  1

ð41:9Þ

Next, we will calculate the average vibrational contribution to the molar heat capacity at constant volume of an ideal gas of diatomic molecules. To this end, we take the temperature derivative of Eq. (41.9), with N ¼ NA (Avogadro’s number) and kBNA ¼ R (the Gas constant). This yields: Cv,vib ¼

dhEvib i dT

ð41:10Þ

or Cv,vib

  d θvib ¼R dT eθvib =T  1

Taking the temperature derivative in the last equation and rearranging, we obtain: Cv,vib

 2 θ eθvib =T ¼ R vib 2 T ð1  eθvib =T Þ

ð41:11Þ

Equation (41.11) shows that the high-temperature limit of Cv,vib (that is, θvib 0  0.

41.4

Rotational Partition Function of a Diatomic Molecule

41.4

443

Rotational Partition Function of a Diatomic Molecule

As we saw in Lecture 40, the discrete energy levels of a rigid rotator are given by: ɛJ ¼

ħ2 JðJ þ 1Þ , J ¼ 0, 1, 2, . . . 2I

ð41:14Þ

where I is the moment of inertia, given by: I ¼ μd2 , with

μ ðReduced massÞ ¼

m1 m2 ðm1 þ m2 Þ

ð41:15Þ

In Eq. (41.15), d is the mass-to-mass separation. In Lecture 40, we saw that each rotational energy level, J, has a degeneracy, gJ, given by: gJ ¼ 2J þ 1

ð41:16Þ

Using Eqs. (41.14) and (41.16), we can write the partition function of a rigid rotator as follows: qrot ðTÞ ¼

1 X J¼0

ð2J þ 1Þeβ ħ |fflfflfflffl{zfflfflfflffl}

2

JðJþ1Þ=2I

ð41:17Þ

gJ

For convenience, as we did in the harmonic oscillator case, we introduce the rotational temperature, θrot, as follows: θrot ¼

ħ2 hB ¼ 2IkB kB

ð41:18Þ

where B¼

h 8π2 I

Using Eq. (41.18) for θrot in Eq. (41.17) yields: qrot ðTÞ ¼

1 X

ð2J þ 1Þeθrot

JðJþ1Þ=T

ð41:19Þ

J¼0

The summation in Eq. (41.19) cannot be expressed in closed form. Nevertheless, at ordinary temperatures, we find that the value of (θrot/T) is quite small for diatomic molecules that do not contain hydrogen atoms. For example, for CO(g), θrot ¼ 2.77 K, and, therefore, θrot/T  102 at room temperature. In that case, we can approximate the summation in Eq. (41.19) by an integral, because (θrot/T) is small for most

444

41

Thermodynamic Properties of Ideal Gases of Diatomic Molecules Calculated. . .

diatomic molecules at ordinary temperatures. It is therefore a very good approximation to write qrot(T) in Eq. (41.19) as follows: 1 ð

qrot ðTÞ ¼

ð2J þ 1Þeθrot

JðJþ1Þ=T

ð41:20Þ

dJ

0

The integral in Eq. (41.20) can be evaluated if we let x ¼ J(J + 1), for which dx ¼ (2J + 1)dJ, so that Eq. (41.20) can be expressed as follows: 1 ð

qrot ðTÞ ¼ 0

eθrot x=T dx ¼

1 ð

0

y y 01 1 zfflfflffl}|fflfflffl{
zffl}|ffl{     ð θ x T T rot θrot x=T y @ A e d e dy ¼ T θrot θrot 0 |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} ¼1

or qrot ðTÞ ¼

T 8π2 IkB T ¼ , for θrot ¼ P ¼ kBT ∂V N,β

ð50:8Þ

Average entropy SðN, V, TÞ: S ¼ kB

X

pj ln pj

ð50:9Þ

j

  ∂lnQðN, V, TÞ S ¼ kBT þ kB lnQðN, V, TÞ ∂T N,V

ð50:10Þ

50.4

Calculation of the Canonical Partition Function

533

Average Helmholtz free energy AðN, V, TÞ: A ¼ kBTlnQ

ð50:11Þ

Average chemical potential μðN, V, TÞ:   ∂lnQðN, V, TÞ μ ¼ kBT ∂N T,V

50.4

ð50:12Þ

Calculation of the Canonical Partition Function

In general, use the definition: QðN, V, TÞ ¼

X

  exp βEj ðN, VÞ

ð50:13Þ

j

For independent and distinguishable molecules (or atoms) with identical energy states: QðN, V, TÞ ¼ ½qðV, TÞN

ð50:14Þ

where qðV, TÞ ¼

X

  exp βEj

ð50:15Þ

j

where Ej is the energy of state j. For independent and indistinguishable molecules (or atoms) with identical energy states: QðN, V, TÞ ¼

½qðV, TÞN N!

ð50:16Þ

  exp βEj

ð50:17Þ

where qðV, TÞ ¼

X j

Equation (50.16) is valid only when the condition for the applicability of Boltzmann’s statistics is satisfied, namely, when:

534

50

Review of Part III and Sample Problem

 2 3=2 N h 1 V 8mkBT

ð50:18Þ

  q V, T ¼ qtrans :qrot :qvib :qelec

ð50:19Þ



exp βEtrans j

ð50:20Þ



exp βErot j

ð50:21Þ



exp βEvib j

ð50:22Þ



exp βEelec j

ð50:23Þ

Determining qðV, TÞ:

where qtrans ðV, TÞ ¼

X j

qrot ðTÞ ¼

X j

qvib ðTÞ ¼

X j

qelec ðTÞ ¼

X j

and j refers to a quantum state.

50.5

Molecular Partition Functions of Ideal Gases

Monoatomic ideal gas: Translation:  qtrans ðV, TÞ ¼

3=2 2πmkBT V h2

ð50:24Þ

Electronic: qelec ðTÞ ¼ ge1 þ ge2 eβEe2

ð50:25Þ

50.5

Molecular Partition Functions of Ideal Gases

535

Diatomic ideal gas: Translation:

qtrans ðV, TÞ ¼

 3=2 2πðm1 þ m2 ÞkBT V h2

ð50:26Þ

Electronic: qelec ðTÞ ¼ ge1 eþβDe þ ge2 eβEe2

ð50:27Þ

In Eq. (50.27), De ¼ D0 + hν/2. Vibration:   vib exp  θ2T   qvib ðTÞ ¼ 1  exp  θTvib

ð50:28Þ

where θvib ¼

hν hce ν ¼ kB kB

ð50:29Þ

Rotation: qrot ðTÞ ¼

T σθrot

for

θrot  T

ð50:30Þ

where θrot ¼

h2 hB ¼ 8π2 IkB kB

and

B¼

h 8π2 I

ð50:31Þ

where I¼

m1 m2 2 d m1 þ m2

σ ¼ 1, for a heteronuclear diatomic molecule σ ¼ 2, for a homonuclear diatomic molecule

ð50:32Þ

536

50.6

50

Review of Part III and Sample Problem

Summary of Thermodynamic Functions of Ideal Gases

Below, we present expressions for some thermodynamic functions of monoatomic and diatomic ideal gases. Monoatomic ideal gas: " # 3=2 A 2πmkBT V eg ¼  ln kBT N e1 h2

ð50:33Þ

U 3 ¼ kBT 2

ð50:34Þ

CV 3 ¼ kB 2

ð50:35Þ

" # 3=2 V 5=2 S 2πmkBT e ge1 ¼ ln N kB h2

ð50:36Þ

  θ =T U 3 2 θ D ¼ þ þ vib þ θ vib  e kBT 2 2 2T e vib =T  1 kBT

ð50:37Þ

Diatomic ideal gas:

CV 3 2 ¼ þ þ kB 2 2

50.7

"

θvib T

2

#

eθvib =T ðeθvib =T  1Þ

2

ð50:38Þ

Grand-Canonical Ensemble

Grand-Canonical partition function: ΞðV, T, μÞ ¼

XX ENj þ μN e kBT e kBT N

j

ð50:39Þ

50.8

Average Properties in the Grand-Canonical Ensemble

537

Probability of finding the system with N molecules in state j: 

ENj

þ μN

e kBT e kBT pNj ðV, T, μÞ ¼ ΞðV, T, μÞ

ð50:40Þ

Relation between Ξ and Q: ΞðV, T, μÞ ¼

X þ μN QðN, V, TÞe kBT

ð50:41Þ

N

Equation (50.41) shows that we can use the Canonical partition function Q given in Eq. (50.1) to calculate the Grand-Canonical partition function Ξ.

50.8

Average Properties in the Grand-Canonical Ensemble

Average energy:   ∂lnΞ < E >¼ U ¼  ∂β V,μ

ð50:42Þ

  ∂lnΞ < P >¼ PðV, T, μÞ ¼ kBT ∂V T,μ

ð50:43Þ

Average pressure:

Average number of molecules:   ∂lnΞ < N >¼ N ¼ kBT ∂μ V,T

ð50:44Þ

Relative deviations from the average number of molecules : σN 1 ¼ pffiffiffiffi

N

rffiffiffiffiffiffiffiffiffiffiffiffi kBTκT V

ð50:45Þ

where κT is the isothermal compressibility:   1 ∂V κT ¼  V ∂P N,T

ð50:46Þ

538

50

50.9

Review of Part III and Sample Problem

Micro-Canonical Ensemble

Probability of finding a system in state j: pj ¼

1 W

ð50:47Þ

where WðN, V, EÞ is the degeneracy of the system characterized by ðN, V, EÞ, that is, A! W¼Q j aj !

50.10

Average Entropy in the Micro-Canonical Ensemble

S ¼ kB

50.11

ð50:48Þ

W

X X 1 1 ln ¼ kB lnWðN, V, EÞ pjlnpj ¼ kB W W j j¼1

ð50:49Þ

Classical Statistical Mechanics

Definition of the Hamiltonian: ! !

!

!

Hðq , p Þ ¼ Kð p Þ þ Φðq Þ !

ð50:50Þ

!

where Kðp Þ is the molecular kinetic energy and Φðq Þ is the intermolecular potential energy. Classical molecular partition function: 1 qclass ðV, TÞ ¼ s h

ð

ð

. . . eβH

s Y

dpj dqj

j¼1

where s denotes the number of degrees of freedom of the molecule.

ð50:51Þ

50.12

Calculation of Virial Coefficients

539

Classical Canonical partition function for N independent and indistinguishable molecules (or atoms): Qclass ¼

1 N!hsN

ð

ð !! ! ! . . . eβHclass ð q , p Þ d p d q

ð50:52Þ

! !

where Hclass ð q , p Þ is the classical N-body Hamiltonian for interacting particles and is given by: ! !

Hclass ð q , p Þ ¼

N 1 X 2 ðp þ p2jy þ p2jz Þ þ Φðx1 , y1 , z1 , . . . , xN , yN , zN Þ 2m j¼1 jx

ð50:53Þ

Using Eq. (50.53) in Eq. (50.52) and integrating over the momenta coordinates, we obtain: Qclass ¼

 3N=2 1 2πmkBT ZN N! h2

ð50:54Þ

where the classical configurational integral, ZN, in Eq. (50.54) is given by: ð

eβΦðx1 ,y1 ,z1 ,...,xN ,yN ,zN Þ dx1 dy1 dz1 . . . dzN

ZN ¼

ð50:55Þ

V

Canonical ensemble partition function for interacting molecules (or atoms): Q ¼ Qclass QQM

50.12

ð50:56Þ

Calculation of Virial Coefficients

Virial equation of state (EOS): P ¼ ρ þ B2 ðTÞρ2 þ B3 ðTÞρ3 þ . . . kBT

ð50:57Þ

540

50.13

50

Review of Part III and Sample Problem

Statistical Mechanical Treatment of Chemical Reaction Equilibria

Chemical potential μj:   qj ðT, VÞ μj ¼ kBTln Nj

ð50:58Þ

Equilibrium constant K(T): For a general homogeneous gas-phase chemical reaction: jνAjAþjνBjB ⇄ jνCjCþjνDjD jν j jν j

ρC C ρD D

jν j jν j ρAA ρB B

50.14

¼ KðTÞ ¼

ðqC =VÞjνC j ðqD =VÞjνD j

ðqA =VÞjνA j ðqB =VÞjνB j

ð50:59Þ

ð50:60Þ

Statistical Mechanical Treatment of Binary Liquid Mixtures

Refer to Lecture 49 for details.

50.15

Useful Constants in Statistical Mechanics

Plank’s constant: h ¼ 6.626  1034 Js Boltzmann’s constant: kB ¼ 1.381  1023 J K1

50.16

Useful Relations in Statistical Mechanics

Useful conversions: ∂ ∂ ¼ kBT2 ∂β ∂T

ð50:61Þ

50.17

Sample Problem 50.1

541

∂ ∂ ¼ kBT ∂γ ∂μ

ð50:62Þ

Stirling’s approximation: For large N, lnN! ¼ NlnN  N

ð50:63Þ

The Gaussian integral: ð1 e

α2 x2

1

pffiffiffi π dx ¼ α

ð50:64Þ

Infinite sum for e: ex ¼

1 X xn n! n¼0

ð50:65Þ

Statistical fluctuations: σ2x ¼< ðx < x >Þ2 >¼< x2 >  < x>2

50.17

ð50:66Þ

Sample Problem 50.1

Consider a lattice consisting of 10,000 lattice cells which are occupied by 3,000 indistinguishable particles at temperature, T. It is known that only one particle can reside in each lattice cell. It is also known that the particles do not interact. The initial state is such that the 3,000 particles occupy 30% of the lattice, which is separated by an impermeable partition from the rest of the lattice. Calculate the change in the system Gibbs free energy after the partition is removed and the system reaches equilibrium.

50.17.1

Solution

To compute the change in the system Gibbs free energy for the given process, we recall that: G ¼ H  TS

ð50:67Þ

542

50

Review of Part III and Sample Problem

Because the particles do not interact (e.g., there is no bond breakage or formation in the lattice representation), the enthalpy change associated with this process is zero! Accordingly, for the constant temperature process considered here, Eq. (50.67) indicates that: ΔGi!f ¼ Gf  Gi ¼ ðHf  Hi Þ  T ðSf  Si Þ ΔGi!f ¼ 0  T ðSf  Si Þ ¼ TΔSi!f

ð50:68Þ

Equation (50.68) shows that for the process considered here, the evaluation of ΔGi!f is equivalent to the evaluation of TΔSi!f ! Because the energy of all the states is the same, we can use the entropy definition in the Micro-Canonical ensemble (E, V, N). Note that for indistinguishable particles where only one particle can reside at each lattice cell, it follows that: S ¼ kB lnΩ

ð50:69Þ

where Ω is the number of distinct available configurations, or the degeneracy, of the system. Note that Ω in Eq. (50.69) is the same as W in Eq. (50.48). We can therefore write that: ΔSi!f ¼ Sf  Si ¼ kB ln Ωf  kB ln Ωi

ð50:70Þ

or that: ΔSi!f

  Ω ¼ kB ln f Ωi

ð50:71Þ

We are told that the lattice consists of M ¼ 10,000 lattice cells, which are occupied by N ¼ 3,000 indistinguishable particles. We are also told that in the initial state, i, only 30% of the lattice cells are occupied. In other words, initially, only 3,000 lattice cells are occupied. This implies that in the initial state, there are only (3,000!/3,000!) ¼ 1 distinct configurations available to arrange the 3,000 indistinguishable particles in the 3,000 lattice cells. That is: Ωi ¼ 1 ) Si ¼ kBlnΩi ¼ 0

ð50:72Þ

In the final state, f, after the partition is removed, the entire lattice comprising M ¼ 10,000 lattice cells becomes available to the N ¼ 3,000 indistinguishable particles. Therefore, we need to compute the number of distinct configurations where N particles and (M-N) empty lattice cells are arranged in a lattice comprising M lattice cells. The answer is:

50.17

Sample Problem 50.1

543

Ωf ¼

M! N!ðM  NÞ!

ð50:73Þ

Using Eq. (50.73) in Eq. (50.69), it follows that:  Sf ¼ kBln Ωf ¼ kBln

M! N!ðM  NÞ!

 ð50:74Þ

Expanding the natural logarithm in Eq. (50.74) yields: Sf ¼ kB flnM!  lnN!  lnðM  NÞ!g

ð50:75Þ

Using Stirling’s approximation for each ln(factorial) in Eq. (50.75) yields: Sf ¼ kB fðMlnM  MÞ  ðNlnN  NÞ  ½ðM  NÞ lnðM  NÞ  ðM  NÞg ð50:76Þ Manipulation of Eq. (50.76) yields: Sf ¼ kB fMlnM  NlnN  ðM  NÞ þ ðM  NÞ  ðM  NÞlnðM  NÞg ð50:77Þ or Sf ¼ kB fMlnM  NlnN  ðM  NÞlnðM  NÞg

ð50:78Þ

Using M ¼ 10,000 and N ¼ 3,000 in Eq. (50.78) yields: Sf ¼ 6,109 kB

ð50:79Þ

Using Si ¼ 0, Sf ¼ 6,109 kB , in Eq. (50.68) for ΔGi!f , yields: ΔGi!f ¼ T ðSf  Si Þ ¼ T Sf

ð50:80Þ

ΔGi!f ¼ 6,109 kBT

ð50:81Þ

or

The negative sign in Eq. (50.81) indicates that the process considered is favored and, therefore, occurs spontaneously.

Solved Problems for Part I

© Springer Nature Switzerland AG 2020 D. Blankschtein, Lectures in Classical Thermodynamics with an Introduction to Statistical Mechanics, https://doi.org/10.1007/978-3-030-49198-7

545

546

Solved Problems for Part I

Problem 1 Problem 3.4 in Tester and Modell Two cylinders are attached as shown in the following figure. Both cylinders and pistons are adiabatic and have walls of negligible heat capacity. The connecting rod is nonconducting.

The initial conditions and pertinent dimensions are as follows: Initial pressure (bar) Initial temperature (K) Initial volume (m3) Piston area (m2)

Cylinder A 10 300 6.28
103 3.14
103

Cylinder B 1 300 1.96
103 1.96
103

The pistons are, initially, prevented from moving by a stop on the outer face of piston A. When the stop is removed, the pistons move and finally reach an end state characterized by a balance of forces on the connecting rod. There is some friction in both piston and cylinders during this process. Gases A and B are ideal and have constant values of Cv ¼ 20.9 J/mol K. What are the final pressures in both A and B? Consider two cases, one where the ambient pressure is 0 bar and another where it is 1 bar.

Solution to Problem 1 Solution Strategy To solve this problem, we will use the four-step strategy discussed in Part I, where we: 1. 2. 3. 4.

Draw the pertinent problem configuration Summarize the given information Identify critical issues Make physically reasonable approximations

Solved Problems for Part I

547

1. Draw the pertinent problem configuration

Fig. 1

2. Summarize the given information Two cylinders, each containing a known quantity of an ideal gas, are connected to each other via a double piston connected by a single rod (see Fig. 1). The entire assembly of cylinders, pistons, and rod is adiabatic and nonconducting. At time zero, a stopper that prevents the piston assembly from moving is removed. We would like to determine the pressure in each of the cylinders as a function of the atmospheric pressure for the cases, patm ¼ 0 bar and patm ¼ 1 bar, when the system reaches its final equilibrium state. In addition to the information provided above, we know the initial pressure, temperature, and volume in each cylinder, as well as the areas of the pistons in each cylinder. We are also told that the heat capacities at constant volume of gases A and B are constant and equal to 20.9 J/mol K. Finally, we are told that at equilibrium, all the forces acting on the pistons are balanced. • Data provided • Initial conditions of gas A: – pA1 ¼ 10 bar – VA1 ¼ 6.28
103 m3 – TA1 ¼ 300 K • Initial conditions of gas B: – pB1 ¼ 1 bar – VB1 ¼ 1.96
103 m3 – TB1 ¼ 300 K

548

Solved Problems for Part I

• Physical dimensions of the pistons: – AA ¼ 3.14
103 m2 – AB ¼ 1.96
103 m2 • Other: – CV ¼ 20.9 J/mol K • Given assumptions • Nonconducting walls and pistons are adiabatic with negligible heat capacities • Gases A and B are ideal 3. Identify critical issues • The geometry In this problem, the two cylinders do not have equal areas. This leads to a net displacement of the atmospheric gas as the pistons move. Therefore, care must be taken in writing down the force balance between the two pistons when the atmospheric pressure is nonzero. • Friction There is some friction associated with the motion of the pistons. However, we do not have enough information to uniquely specify the friction. Therefore, we anticipate that the system of equations may be underspecified. Consequently, we may have to make additional assumptions in order to obtain a reasonable range for the solution. • System boundary Many system boundaries are possible and if solved correctly will lead to the same solution. Because the pistons and the walls have negligible heat capacities and the pistons have negligible masses, it is advantageous to include the pistons and the walls within the system boundary. This will ensure that the pressure at the system boundary is the well-defined atmospheric pressure, rather than the gas pressures which vary during the process. Alternatively, one could choose a system which includes the atmosphere (but not the pistons or the chambers) and then relate the work done on the atmosphere to the work done on the gas/pistons system. The least desirable boundaries would involve the gas in the cylinders, but not the pistons or the walls. In this case, the system is simple, but the assumption of an adiabatic system is not valid. In Fig. 2, we show a schematic depicting the boundary used in the solution presented here, which contains the cylinder walls, the pistons, and the gas in the chambers, but none of the atmosphere (see the dashed black line).

Solved Problems for Part I

549

Fig. 2

4. Make physically reasonable assumptions • Quasi-static process If sufficient friction is present, we can assume a slow quasi-static process • Negligible gravity Due to the horizontal configuration and the low density of the gases, we can assume that gravity has a negligible effect on the system

Solving the Problem There are a number of steps which, taken together, provide a good method to solve problems involving the First Law and Second Law of Thermodynamics. Below, we first outline these steps and then use them to develop our solution. 1. Define the system, keeping in mind the need to implement the First Law of Thermodynamics to the system. A convenient choice for the system is the entirety of the piston/cylinder assembly including the gases within. We will not include any of the gas at atmospheric pressure outside of the system (see Fig. 2). 2. Determine the properties of the system boundaries (i.e., adiabatic vs. diathermal, permeable vs. impermeable, movable vs. rigid). The Problem Statement indicates that the walls are nonconducting, and based on our choice of system boundary, we can further conclude that the boundary is impermeable (no gas enters or leaves). Finally, because the pistons can move, the boundary is movable. It is important at this point to consider the complexity of the system in question. Is it simple or composite? The presence of an adiabatic and impermeable piston between the two chambers represents an internal boundary. Systems with internal boundaries are, by definition, composite. Because the system is

550

Solved Problems for Part I

composite, Postulate I does not apply to this system. Therefore, we will need more than two independently variable properties to define the system completely. 3. Carry out a First Law of Thermodynamics analysis of the system using your physical understanding of the problem to deduce heat and work interactions associated with the system. We can begin by using the First Law of Thermodynamics for closed systems modeling a differential change in state. Specifically: dU ¼ δQ þ δW

ð1Þ

Note: We have replaced E with U because the pistons are horizontally aligned (no potential energy effect) and stationary at the initial and final positions (no kinetic energy effect). Remember that we have defined the system to include both gases, A and B, and therefore, recognizing that the walls and the pistons have negligible heat capacities, it follows that: dU ¼ dU A þ dU B

ð2Þ

All the walls in the system are adiabatic and nonconducting. Therefore, we can set δQ in Eq. (1) to zero. Furthermore, we can define the work done on the system by adding up the works done by the environment on both pistons. Following the definition, we can define a pressure/volume work interaction as follows (recall that PdV is the differential work done by the system and –PdV is the differential work done on the system): δW ¼ PdV

ð3Þ

Applying Eq. (3) to our specific system and inserting it into Eq. (1), we obtain: dU ¼ Patm dV A  Patm dV B

ð4Þ

4. Replace the internal energy term using the ideal gas law. We can also expand the dU term by expressing the internal energy of an ideal gas in terms of the temperature and the gas heat capacity at constant volume. Specifically: dU ¼ dU A þ dU B ¼ N A CV dT A þ N B CV dT B

ð5Þ

Combining Eqs. (4) and (5), we obtain: N A CV dT A þ N B C V dT B ¼ Patm dV A  Patm dV B

ð6Þ

Solved Problems for Part I

551

We can then integrate Eq. (6) from the initial state (state 1) to the final state (state 2): TðA2

TðB2

dT A þ N B C V

N A CV T A1

VðA2

dT B ¼ Patm T B1

VðB2

dV A  Patm V A1

dV B

ð7Þ

V B1

Because Patm, NA, NB, and CV are all constant, all the integrals in Eq. (7) are simple to carry out and yields: N A C V ðT A2  T A1 Þ þ N B C V ðT B2  T B1 Þ ¼ Patm ðV A2  V A1 Þ  Patm ðV B2  V B1 Þ

ð7aÞ

Next, we can use the ideal gas law to express each of the four temperatures, TA1, TA2, TB1, and TB2, in Eq. (7a) in terms of pressures, volumes, and number of moles. For example: PA1 V A1 ¼ T A1 NAR

ð7bÞ

Substituting the four temperatures in Eq. (7a), using expressions similar to that in Eq. (7b), we obtain: CV ½ðPA2 V A2  PA1 V A1 Þ þ ðPB2 V B2  PB1 V B1 Þ R ¼ Patm ½ðV A2  V A1 Þ þ ðV B2  V B1 Þ

ð8Þ

Note that there are four unknowns in Eq. (8): PA2, PB2, VA2, and VB2. Therefore, we recognize that three additional equations are required to obtain a unique solution. 5. Use your physical understanding of the problem to define the internal constraints imposed on the system. A second equation can be derived by carrying out a force balance on the pistons, because according to the Problem Statement, the pistons are balanced, and therefore: X

F ¼ 0 ¼ ðPA2  Patm ÞAA  ðPB2  Patm ÞAB

ð9Þ

A third equation can be derived based on the geometry of the pistons to relate the changes in volume in cylinders A and B. Specifically: V A2  V A1 A ¼ A V B2  V B1 AB

ð10Þ

552

Solved Problems for Part I

At this point, we have used all the provided information but still are one equation short of fully defining the problem. This is because we are told that there is some friction, but are not given a friction coefficient. As we will see, the problem is still solvable when Patm equals zero. However, when Patm equals 1 bar, we will need to make further assumptions to obtain reasonable ranges rather than exact solutions. In this respect, a good engineering approximation can be made by using the solutions when (1) there is no friction in cylinder A and (2) there is no friction in cylinder B. This will give us the upper/lower bounds of the final pressures and temperatures in the two cylinders. 6. Do the math. To simplify the math, we can begin by rearranging Eqs. (9) and (10) to eliminate the dependence on the area in one of the equations. Specifically: V A2  V A1 P  Patm ¼  B2 V B2  V B1 PA2  Patm

ð11Þ

Equation (11) will be one of our three linearly independent equations (replacing Eq. (10)). For Patm = 0 Combining Eq. (8) and Eq. (11) when Patm equals zero leads to the cancelation of both VA2 and VB2 from the system of equations, that is: V A1 ðPA2  PA1 Þ þ V B1 ðPB2  PB1 Þ ¼ 0

ð12Þ

Note that combining only two equations led to the cancelation of two variables rather than just one. We can now combine Eq. (12) with Eq. (9) to solve explicitly for the values of PA2 and PB2. Specifically, we find that: PA1 V A1 þ PB1 V B1 ¼ 6:87 bar A V A1 þ V B1 A AB PA1 V A1 þ PB1 V B1 ¼ ¼ 11:01 bar A V A1 B þ V B1 AA

PA2 ¼

PB2

ð13Þ

For Patm = 1 Unfortunately, in this case, the favorable cancelation does not occur, and our system is still incompletely defined. To proceed, we must make additional reasonable approximations to study the limiting behaviors of the system. These limiting cases will yields a range for the possible system pressures rather than yielding a single solution. First, we will assume that friction is only present in cylinder A and, subsequently, that it is only present in cylinder B. These two cases represent the

Solved Problems for Part I

553

extremes of the possible processes and, hence, should give us the full range of feasible solutions. If no friction is present in cylinder A, we can assume an adiabatic expansion of the gas in that compartment modeled by:    CV þR V A1 CV PA2 ¼ PA1 V A2

ð14Þ

Combining this result with Eqs. (8), (9), and (11), we have four equations and four unknowns. Using a numerical solving technique (e.g., MATLAB), it is possible to solve for the system pressures. Doing this, we find: PA2 ¼ 6:89 bar PB2 ¼ 10:44 bar

ð15Þ

This procedure can be repeated to solve for the case of no friction present in cylinder B. Doing this, we find:    CV þR V B1 CV PB2 ¼ PB1 V B2

ð16Þ

By solving Eqs. (8), (9), (11), and (16), we find: PA2 ¼ 6:86 bar PB2 ¼ 10:38 bar

ð17Þ

Equations (15) and (17) show that the final pressure in cylinder A must be between 6.86 bar and 6.89 bar and that the final pressure in cylinder B must be between 10.38 bar and 10.44 bar. We can compare the calculated final pressures in chambers A and B for the two different Patm scenarios. Interestingly, the values are very close. We find that PB2 is slightly lower and that PA2 is slightly higher when the atmospheric pressure is 1 bar as opposed to 0 bar. This makes physical sense because when the atmospheric pressure is nonzero, the atmosphere will exert a larger force on the larger piston in chamber A than on the smaller piston in chamber B. This will provide extra resistance against the movement of gas A, preventing the pistons from moving as far to the right (toward chamber B) than in the case with vacuum. The relatively small difference between the pressures calculated with and without atmosphere shows that the atmosphere yields only a small contribution to the change in the energy of the system when compared to the gases in the cylinders.

554

Solved Problems for Part I

Other Possible Solution Strategies There are, of course, many other solution strategies to arrive at the same answers. Initially, as stated earlier, several different boundary conditions are possible and should all eventually lead to an expression analogous to Eq. (8). In addition, while unnecessary, it is valid to solve the Patm ¼ 0 bar case by making the same friction assumptions used in the Patm ¼ 1 case. We encourage the interested reader to try other solution strategies to further penetrate this interesting and challenging problem.

Solved Problems for Part I

555

Problem 2 Problem 3.8 in Tester and Modell Bottles of compressed gases are commonly found in chemistry and chemical engineering laboratories. They present a serious safety hazard unless they are properly handled and stored. Oxygen cylinders are particularly dangerous. Pressure regulators for oxygen must be kept scrupulously clean, and no oil or grease should ever be applied to any threads or on moving parts within the regulator. The rationale for this rule comes from the fact that if oil were present – and if it were to ignite in the oxygen atmosphere – this “hot” spot could lead to ignition of the metal tubing and regulator and cause a disastrous fire and failure of the pressure container. Yet it is hard to see how a trace of heavy oil or grease could become ignited even in pure, compressed oxygen since ignition points probably are over 800 K if “nonflammable” synthetic greases are employed. Let us model the simple act of opening an oxygen cylinder that is connected to a closed regulator (see the following figure). Assume that the sum of the volumes of the connecting line and the interior of the regulator is VR. VR is negligible compared to the bottle volume. Opening valve A pressurizes VR from some initial pressure to full bottle pressure. Presumably, the temperature in VR also changes. The question we would like to raise is: Can the temperature in VR ever rise to a sufficiently high value to ignite any traces of oil or grease in the line or regulator?

Data: The oxygen cylinder is at 15.17 MPa and 311 K. The connecting line to the regulator and the regulator interior (VR) are initially at 0.101 MPa, 311 K, and contain pure oxygen. Assume no heat transfer to the metal tubing or regulator during the operation. Oxygen is essentially an ideal gas. CP ¼ 29.3 J/mol K, CV ¼ 20.9 J/mol K, and both are independent of pressure or temperature.

556

Solved Problems for Part I

(a) If gas entering VR mixes completely with the initial gas, what is the final temperature in VR? (b) An alternative model assumes that there is no mixing between the gas originally in VR and that which enters from the bottle. In this case, after the pressures are equalized, we would have two identifiable gas slugs which presumably are at different temperatures. Assuming no axial heat transfer between the gas slugs, what is the final temperature of each? (c) Comment on your assessment of the hazard of this simple operation of bottle opening. Which of the models in (a) and (b) is more realistic? Can you suggest other improved models?

Solution to Problem 2 Solution Strategy To solve this problem, we will use the following information: 1. Properties of an ideal gas: PV ¼ NRT

ð1Þ

CP  CV ¼ R

ð2Þ

2. The First Law of Thermodynamics for: (i) Closed systems: dE ¼ δQ þ δW

ð3Þ

dE ¼ δQ þ δW þ H in δN in  H out δN out

ð4Þ

(ii) Open systems:

Part (a) We are told that upon opening an oxygen cylinder connected to a closed regulator, the entering oxygen from the cylinder mixes with the already existing gas, thereby raising its pressure. Let the gas in the connecting tubes and interior of the regulator be our system (see the colored region in Fig. 1). Based on the Problem Statement, we recognize that the initial state of this system is characterized by: Pi ¼ 0:101 MPa

Solved Problems for Part I

557

Fig. 1

Vi ¼ VR T i ¼ 311 K Ni ¼ ? The final state of this system is characterized by: P f ¼ 15:17 MPa V f ¼ VR Tf ¼? Nf ¼? Next, we proceed to determine the characteristics of the system boundary: 1. Permeable or Impermeable? Oxygen from the tank enters our system. Therefore, the system boundary is permeable, indicating that the chosen system is open. The conditions of the inlet (in) stream are given by: Pin ¼ 15:17 MPa ¼ 15:17
106 Pa T in ¼ 311 K

558

Solved Problems for Part I

2. Rigid or Movable? There are no movable parts in the system boundary. Therefore, it is rigid. Due to the rigidity of the boundary, there is no PV work interaction between the system and its environment. In addition, no other work is done on the chosen system. As a result: δW ¼ 0 3. Adiabatic or Diathermal? The Problem Statement tells us to neglect any heat transfer to the metal tubing. In addition, we assume that there is no heat transfer across the boundary separating our system and the oxygen cylinder. As a result, the operation is adiabatic, that is: δQ ¼ 0 Because the system does not have any internal boundaries, it is a simple system. We can use the information above to write down the First Law of Thermodynamics for our open system. Specifically: dE ¼ dU ¼ δQ þ δW þ H in δN in  H out δN out

ð5Þ

From this point on, we can either follow the steps presented in Lecture 6 or directly integrate Eq. (5) as is. Either way, we will rewrite the internal energy and enthalpy in terms of a reference state and be able to cancel out several terms. Below, we directly integrate Eq. (5) between the initial (i) and final ( f ) states. This yields: U f  U i ¼ H in N in

ð6Þ

Using the definitions of U and H for ideal gases presented in Part I (see Eq. (7a)), Eq. (6) can be simplified to yields Eq. (7b):   U j ¼ N j Cv T j  T o þ N j U o , for j ¼ f , i H in ¼ Cp ðT in  T o Þ þ H o

ð7aÞ

  N f C V T f  T 0 þ N f U 0  ðN i C V ðT i  T 0 Þ þ N i U 0 Þ ¼ N in C P ðT in  T 0 Þ þ N in H 0     ) N f C V T f  N i C V T i ¼ N in C P T in þ N in H 0  N f  N i U 0  N in C P T 0   þ N f  N i CV T 0     ) N f C V T f  N i C V T i ¼ N in C P T in þ N in ðU 0 þ RT 0 Þ  N f  N i U 0   N in ðC V þ RÞT 0 þ N f  N i C V T 0

ð7bÞ

Solved Problems for Part I

559

Using a mole balance, we can eliminate Nin in favor of Nf and Ni to obtain: N f  N i ¼ N in

ð8Þ

Substituting Eq. (8) on the right-hand side of the equality in the last line of Eq. (7b) and cancelling the equal terms, we obtain:      N f CV T f  N i C V T i ¼ ðNf  NiÞCP T in þ N f  N i ðU 0 þ RT 0 Þ  N f  N i U 0      N f  N i ðCV þ RÞT 0 þ N f  N i CV T 0     ) N f CV T f  N i CV T i ¼ ðNf  NiÞCP T in þ N f  N i RT 0  N f  N i ðC V þ R  CV ÞT 0     ) N f CV T f  N i CV T i ¼ ðNf  NiÞCP T in þ N f  N i RT 0  N f  N i RT 0   ) N f CV T f  N i CV T i ¼ N f  N i CP T in

ð9Þ Using the ideal gas law in the last line of Eq. (9), we can express Nf and Ni in terms of Pf, V f , Tf and Pi, V i , and Ti, respectively. After cancelling terms and rearranging, we obtain Eq. (10) below: 0 @

1 P f Vf RT f

0

AC V T f 

1 P V i iA @ RT i

00 C V T i ¼ @@

1 Pf Vf RT f

0

A

11 P V i i AA @

RT i !!

CP T in

! P f VR Pi V R ) P f V R C V  Pi V R C V ¼  C P T in Tf Ti     P f C V  Pi C V P f Pi Pi P f ) P f C V  Pi C V ¼  þ ¼ CP T in ) C P T in T f Ti Ti T f ) Tf ¼ 

Pf CV ðP f Pi Þ CP T in

¼ þ TPii

1þ

Pi Pf

κT  in  , where κ ¼ Cp =Cv κTTini  1

ð10Þ

Note that Eq. (10) is identical to Eq. (6.14) presented in Lecture 6 where P ¼ Pf. Substituting numerical values of the different variables in Eq. (10) yields: ð29:3=20:9Þð311KÞ 29:3 311  ¼ 434:8K T f ¼¼  1 þ 0:101 15:17 20:9 311  1

ð11Þ

Part (b) In this case, the volume VR houses two subsystems (see colored regions A and B in Fig. 2). Subsystem A contains the gas that occupied the volume VR before the valve was opened and that gets compressed as the gas enters from the oxygen tank. Subsystem B contains the gas that enters from the oxygen tank. Let us first consider subsystem A. The initial and final conditions for subsystem A are given by:

560

Solved Problems for Part I

Fig. 2

Initial state: Pi ¼ 0:101 MPa Vi ¼ VR T i ¼ 311 K Ni ¼ ? Final state: P f ¼ 15:17 MPa Vf ¼? Tf ¼? N f ¼ Ni Next, we proceed to determine the characteristics of the boundary of this system: 1. Permeable or Impermeable? There is no mixing between the incoming gas and the gas already present in the connecting pipe. As a result, the boundary of the system is impermeable and the system is closed.

Solved Problems for Part I

561

2. Rigid or Movable? The bottom boundary of subsystem A is moved by the gas which enters into system B. Therefore, the bottom boundary of subsystem A is movable and undergoes PV work (δW ¼ PdV). 3. Adiabatic or Diathermal? The Problem Statement asks us to neglect any heat transfer to the metal tubing or between the two gas slugs A and B. As a result, the operation is adiabatic (δQ ¼ 0). Next, we can use the information above to write down the First Law of Thermodynamics for our closed system. This yields: dE ¼ dU ¼ δQ þ δW

ð12Þ

Equation (12) can be solved by writing dU and δW in terms of P, V , T, and N, including rearranging as needed. This yields: NC V dT ¼ PdV ¼  ) NC V dT ¼ 

NRT dV V

NRT dV V

CV R dT ¼  dV T V Vf CV T f ln ) ¼  ln R Ti Vi

ð13Þ

)

Using the ideal gas law, we can write the two Vs in Eq. (13) in terms of P, N, and T. Rearranging as needed and finally integrating from the initial (i) to the final ( f ) condition, we obtain: 2 13 !0 N f RT f CV T f P i @ A5 ln ¼  ln 4 R Ti Pf N i RT i

   CV T f =N i T f Pi ln ¼  ln ) R Ti Pf NiT i =

   CV T f Tf Pi ln ) ¼  ln R Ti Ti Pf     CV Tf Pi ) þ 1 ln ¼  ln R Ti Pf  ðR=ðCv þRÞÞ P ) T f ¼ T i Pif

ð14Þ

Substituting numerical values of the different variables in the last expression in Eq. (14) yields:

562

Solved Problems for Part I

  8:314 15:17 ð20:9þ8:314Þ T f ¼ 311T f ¼ 1294:8 K 0:101

ð15Þ

To determine the temperature of the gas in subsystem B, we consider the entire volume VR, consisting of subsystems A and B, as our new system (see the shaded region in Fig. 1). Because the system is a composite system, to carry out any analysis, we need to determine the initial and final conditions for both subsystem A and subsystem B. We already know the conditions for subsystem A. Those for subsystem B are given by: Pi,B ¼ n=a V i,B ¼ 0 T i ¼ n=a N i,B ¼ 0 P

f ,B

V

¼ 15:17 MPa

f ,B

¼ VR  V f

T

f ,B

¼?

N

f ,B

¼?

Again, we proceed to determine the characteristics of the boundary of our composite system (A + B): 1. Permeable or Impermeable? Oxygen from the tank enters our system. As a result, the boundary is permeable, and the system is open. The conditions for the inlet stream are given by: Pin ¼ 15:17 MPa T in ¼ 311 K 2. Rigid or Movable? Similar to Part (a), there are no movable parts in the system boundary. As a result, the boundary is rigid, and no PV work is incurred, that is: δW ¼ 0 Note that the movable boundary between subsystems A and B is internal to our chosen system. Therefore, any PV work done across this boundary does not appear in our analysis.

Solved Problems for Part I

563

3. Adiabatic or Diathermal? Based on an argument identical to that made in Part (a), we can conclude that the system boundary is adiabatic. Therefore, δQ ¼ 0 Next, we can use the information above to write down the First Law of Thermodynamics for our open system. Although the system is a composite system with an adiabatic internal boundary, it is comprised of two simple subsystems (A and B). Therefore, the sum of the internal energies of these two simple subsystems is equal to the energy of the composite system, that is: dE ¼ dU A þ dU B ¼ δQ þ δW þ H in δN in  H out δN out

ð16Þ

Integrating Eq. (16) between the initial (i) and final ( f ) conditions yields: U

f ,A

 U i,A þ U

f ,B

 U i,B ¼ H in N in

ð17Þ

Because our system is comprised of two subsystems, the internal energy change of the composite system can be written as the sum of the internal energy changes of the two subsystems comprising the composite system. Using a derivation similar to the derivation in Part (a), Eq. (17) can be further simplified to yields (note that Nf,A is the same as Nf, because we continue to use the same notation as in the first section of Part (b)):  ) N

f ,B C V



T

f ,B

  T0 þ N

f ,B U 0

   þ N f CV T f  T 0 þ N f U 0

ðN i,B C V ðT i,B  T 0 Þ þ N i,B U 0 þ N i C V ðT i  T 0 Þ þ N i U 0 Þ ¼ N in C P ðT in  T 0 Þ þ N in H 0   ) N f ,B C V T f ,B þ N i C V T f  ðN i C V T i Þ ¼ N in C P T in   þ N f ,B C V T 0 þ N f C V T 0  N i,B C V T 0  N i C V T 0  N in C P T 0   þ N f ,B U 0  N f U 0 þ N i,B U 0 þ N i U 0 þ N in H 0   ) N f ,B C V T f ,B þ N i C V T f  ðN i C V T i Þ ¼ N in C P T in   þ N f ,B C V T 0 þ N i C V T 0  N i C V T 0  N in ðC V þ RÞT 0   þ N f ,B U 0  N i U 0 þ N i U 0 þ N in H 0

ð18Þ Using a mole balance, we conclude that: N

f ,B

¼ N in

Substituting Eq. (19) into Eq. (18) as needed, including rearranging, yields:

ð19Þ

564

Solved Problems for Part I



 N f ,B C V T f ,B þ N i C V T f  ðN i C V T i Þ ¼ N f ,B CP T in     þ N f ,B CV T 0  N f ,B ðC V þ RÞT 0 þ N f ,B U 0 þ N f ,B H 0   ) N f ,B CV T f ,B þ N i C V T f  ðN i CV T i Þ ¼ N f ,B CP T in

ð20Þ

Using the ideal gas law in the last expression in Eq. (20), we can eliminate Nf, B and Ni in terms of Pf, B, V f ,B , Tf, B, and Pi, V i , and Ti, respectively. This yields: 00 @@

1 P f , BV f , B =RT f , B

0

AC V T f , B þ @

1 Pf Vf

1

0

1

AC V T f A  @P i V i AC V = Ti ¼ =RT f =RT i



P f , BV f , B RT f , B =

 C P T in

       T in ) P f ,B V R  V f CV þ P f V f C V  Pi V R C V ¼ P f ,B V R  V f C P T f ,B         Vf Vf Vf T in ) P f ,B 1  C C þ Pf  Pi CV ¼ P f ,B 1  C VR V VR V V R P T f ,B 

ð21Þ Note that in the derivation above, we used the fact that Ni ¼ Nf. To further simplify the last expression in Eq. (21), we can use the ideal gas law to express V f =V R in terms of Pi, Ti, Pf, and Tf. This yields: 

      P   N f =RT f Pi i P f ,B 1  CV þ P f N f =RT f P f C V  Pi C V P N R = T N R = Ti f i i i       N f =RT f Pi T in ¼ P f ,B 1  CP P N R = T T f i i f,B 0 0 11 0 1 1 !0   P NiT f P @ i AAC V þ N i T f @ i ACV A  Pi C V ) @P f , B @1  Pf NiT i NiT i 0 11 !0   NiT f @ Pi AACP T in ¼ P f , B @1  Pf T f,B NiT i         Tf Pi Pi ) P f ,B 1  CV þ T f CV  Pi C V P T T   f  i   i Tf Pi T in ¼ P f ,B 1  CP Pf Ti T f ,B

) T f,B

    Tf Pi T in P f , B CP 1  Ti Pf        ¼ Tf Pi Tf P f,B 1  CV þ Pi C V  Pi C V Ti Pf Ti ð22Þ

Solved Problems for Part I

565

Substituting numerical values of the different variables in the last expression in Eq. (22) yields:

T

f ,B

    1294:84 0:101
106 311
15:17
106
29:3 1  6 311 15:17
10    ¼     1294:84 0:101
106 6 1294:84 20:9  0:101
106
20:9 20:9 þ 0:101
10 15:17
106 1  6 311 311 15:17
10 ¼ 426:75 K

ð23Þ Part (c) From Parts (a) and (b), we conclude that there is a possibility for the temperature in VR to reach a value higher than 800 K if the system behaves similar to that in Part (b), i.e., the incoming gas and the already existing gas do not mix with each other. A comparison of the solutions to Parts (a) and (b) suggests that, in Part (b), a very small fraction of the gas in VR is at a temperature of 1294.84 K. This follows because the solution to Part (a) should be the average temperature of the gas in Part (b), because the state of the gas in Part (a) can be attained by going through the state of the gas in Part (b) and then going through an additional step of mixing the two gas slugs. The final temperature obtained in Part (a) is very close to Tf, B obtained in Part (b), which suggests that most of the gas is at a temperature of Tf, B. Quantitatively, it can be shown that the gas at a higher temperature in Part (b) occupies only 2.8% of the total volume and constitutes only 0.93% of the total mass. Therefore, it is highly unlikely that such a small amount of gas will not get mixed with the gas coming in from the valve. In addition, it is also highly unlikely that a gas slug at a temperature of over 1000 K will not transfer heat to the metal tubing and the adjacent gas which are at a significantly lower temperature. Therefore, the probability of the gas catching fire is very low. However, such precautions are warranted when handling potential safety hazards. A more realistic stratified model for the system would be one where heat transfer between the two gas slugs is allowed, i.e., where the boundary between systems A and B in Fig. 2 is not adiabatic.

566

Solved Problems for Part I

Problem 3 Problem 4.11 in Tester and Modell In several parts of the world, there exist ocean currents of different temperatures that come into contact. An example of this takes place off the coast of Southern Africa, where the warm Agulhas and the cold Benguela currents meet at Cape Point, near Cape Town (see the figure below). It has been proposed that work may be obtained from these ocean currents by operating a heat engine between the warm current as a source and the cold current as a sink (shown schematically in the illustration). This renewable form of energy production has been called ocean thermal, and the Department of Energy needs your help in evaluating the proposal.

Assume that the system may be simplified into two channels of water in contact and flowing cocurrently. Furthermore, assume that no mixing occurs between the streams and that heat transfer between the stratified streams under natural conditions is negligible. Finally, assume that the two streams have equal specific heat capacities at constant pressure equal to 4.186 J/gK and mass densities of 1000 kg/m3. (a) Derive a general expression for the maximum amount of power that could be obtained from the system. Express your result in terms of temperatures, flow rates, and physical properties of the streams. (b) What is the pinch temperature, and where does it occur? The pinch is defined as the limiting condition where the stream temperatures approach each other. (c) Repeat part (a) if the two streams flow countercurrent rather than cocurrent. (d) It has been estimated that the Benguela current is 16
106 m3/s and its initial temperature is 278 K. The Agulhas current is 20
106 m3/s and its initial temperature is 300 K. Calculate and compare the power obtainable from these two currents assuming that they flow (1) cocurrently and (2) countercurrently.

Solved Problems for Part I

567

Solution to Problem 3 Solution Strategy To solve this problem, we will use the following four-step strategy discussed in Part I: 1. 2. 3. 4.

Draw the pertinent problem configuration Summarize the given information Identify critical issues Make physically reasonable approximations

1. Draw the pertinent problem configuration (see Fig. 1)

Fig. 1

2. Summarize the given information Two ocean streams at different temperatures are used to power a series of differential Carnot engines. The streams each have given mass flow rates and inlet temperatures. Because the heat capacities of these streams are finite, their temperatures will change throughout the heat interaction. The goal is to extract the maximum amount of work from the two streams. To extract the work, the streams can be contacted either cocurrently or countercurrently. It is necessary to obtain an algebraic solution for the maximum work attainable in both cases in terms of the temperatures, the flow rates, and the stream properties. Using the given information about the two streams in question, a numerical solution is also required. Because no information is given about the specific heat capacities and densities of the streams, we will assume these to be equal for both streams.

568

Solved Problems for Part I

Given Data: • Cold Stream (Benguela Current): • TC,IN ¼ 278 K • FC ¼ 16
106 m3/s • Hot Stream (Agulhas Current): • TH,IN ¼ 300 K • FH ¼ 20
106 m3/s • Assumptions: • No heat transfer between the stratified layers • No mixing between the streams • The specific heat capacities at constant pressure and mass densities of the streams are equal (CP ¼ 4.186 J/g K, ρ ¼ 1000 kg/m3) • The heat engine operates at an ideal (Carnot) efficiency 3. Identify critical issues Constantly Varying Temperatures The most difficult aspect of this problem is that the temperatures of both streams vary throughout the process. Finding the Maximum Work When solving this problem, we are asked to find the maximum work that can be extracted. Although it is possible to make assumptions of what the end conditions should be, it is more rigorous to define the amount of work in terms of the outlet temperatures and then to optimize the result using calculus. Defining the System Another challenging part of this problem is defining an appropriate system to analyze. This problem does not immediately suggest a logical system. While many systems may be possible, for the analysis presented here, we will consider a composite system of differential slices of each stream which are connected by a differential Carnot engine. Figure 2 shows a diagram where the areas surrounded by the dashed lines represent our system. Defining the Boundaries The boundaries of this system are open because the hot and cold streams are fluxing in and out of the differential volume. The boundaries are adiabatic because the Problem Statement specifically indicates that the stratified layers do not interact. Finally, the boundary is rigid. It should be noted, however, that although the boundary does not move, we will be integrating over the entirety of both streams.

Solved Problems for Part I

569

Fig. 2

Therefore, we will expand our knowledge of this one generic differential element to the full process in order to calculate the total work. 4. Make physically reasonable assumptions In general, if possible, it is best to keep the number of assumptions to a minimum. As will be stressed below, it is easy to make intuitive, but incorrect, assumptions for this problem that will lead to erroneous solutions. Cocurrent Streams Reaching Thermal Equilibrium One assumption that can be made (although is not required) is that the streams, when in a cocurrent configuration, will reach a final equilibrated temperature where heat transfer will no longer occur. This assumption does simplify the analysis and understanding of the problem. System Steady State Because we are dealing with an open system where the supply of the two streams should be constant, we can assume that there is no accumulation (steady state) in the differential system that we have defined.

Solving the Problem Parts (a and b): Cocurrent Flow Apply the First Law of Thermodynamics to the System To apply the First Law of Thermodynamics to the system, we will first consider each differential element separately. Specifically, we will use the First Law of Thermodynamics for the hot (H) and cold (C) streams, each an open system. Specifically:

570

Solved Problems for Part I

X X dEH ¼ dU H ¼ δQH þ δW H þ H H,in δnH,in  H H,out δnH,out X X dEC ¼ dU C ¼ δQC þ δW C þ H C,in δnC,in  H C,out δnC,out

ð1Þ

In Eq. (1), the negative sign in δQH indicates that δQH is removed from system “H” (see also Fig. 1). Assuming no mass accumulation in the system, the number of moles entering the system must equal to the number of moles leaving the system. Because there are no work interactions in the boundaries that we have defined, the two equations in Eq. (1) reduce to: dU H ¼ δQH þ δnH ðH H,in  H H,out Þ dU C ¼ δQC þ δnC ðH C,in  H C,out Þ

ð2Þ

Next, we can express the two equations in Eq. (2) during a differential change in time. We recognize that the change in internal energy per time is zero because the dU dU system is at steady state. In other words, dtH ¼ 0 and dtC ¼ 0. Instead of heats or molar quantities, it is convenient to use heat flow rates (δQ_ i ) and volumetric flow rates (Fi) (molar or mass flow rates would also be acceptable). We also use the mass density, ρ, to match our heat capacity definition which is on a per mass basis. Specifically, the two equations in Eq. (2) can be expressed as follows: dU H ¼ 0 ¼ δQ_ H þ F H ρðH H,in  H H,out Þ dt dU C ¼ 0 ¼ δQ_ C þ F C ρðH C,in  H C,out Þ dt

ð3Þ

Next, we can rewrite the difference in enthalpies as simply the difference in temperatures multiplied by the specific heat capacity (CP). Doing that and then using the results in Eq. (3) yields the last two expressions in Eq. (4), that is:   h   i H H , in  H H , out ¼ C p T H , in  T0 þ =Cp =H 0  Cp T H , out  T 0 þ CpH 0 ¼ C P dT H   h   i H C, in  H C, out ¼ C p T C, in  T 0 þ Cp H 0  Cp T C, out  T 0 þ C p H 0 ¼ C P dT C 0 ¼ δQ_ H  F H ρC P dT H 0 ¼ δQ_ C  F C ρC P dT C

ð4Þ Note that we changed the sign of the second terms in the last two expressions in Eq. (4) to be consistent (the derivative refers to out minus in rather than to in minus out). Our First Law of Thermodynamics analysis is nearly complete, and next we will use the Second Law of Thermodynamics in order to derive another equation relating the differential heat flows.

Solved Problems for Part I

571

Apply the Second Law of Thermodynamics to the System of the Carnot Engine As indicated above, application of the Second Law of Thermodynamics will provide us another equation to relate the differential heat flows. This relation will be based on the efficiency of the Carnot engine, which is a function of the temperatures of the two streams in our system. Using the Second Law of Thermodynamics and the fact that the Carnot engine undergoes a reversible (δSgen, the generated entropy is zero), cyclic process (dS ¼ 0), it follows that: P δQi þ δ=Sgen Ti δQ_ H δQ_ C 0¼ þ TH TC δQ_ H T H ) ¼ δQ_ C T C dS ¼

ð5Þ

Combining the First Law and the Second Law of Thermodynamics By combining our results from the First Law of Thermodynamics in Eq. (4) and our results from the Second Law of Thermodynamics in Eq. (5), we obtain: δQ_ H F H dT H T H ¼ ¼ F C dT C TC δQ_ C

ð6Þ

Rearranging Eq. (6), we obtain: dT H F dT C ¼ C TH FH T C

ð7Þ

Through integration of Eq. (7) from the inlet to the outlet (noting that the flow rates are constant), it follows that: T H,OUT ð

T H,IN

dT H F ¼ C TH FH

T C,OUT ð

dT C TC

ð7aÞ

T C,IN

    T H,OUT T C,OUT F ln ¼  C ln T H,IN FH T C,IN  α T C,IN T H,OUT ¼ T H,IN T C,OUT

ð7bÞ ð8Þ

where α ¼ FC=FH . If we define the pinch (P) temperature, TP, to be the equilibrated final temperatures of the hot and cold streams, it follows that:

572

Solved Problems for Part I

 T P ¼ T H,IN

T C,IN TP

α

ð8aÞ

 1 T P ¼ T H,OUT ¼ T C,OUT ¼ T H,IN T αC,IN 1þα

ð9Þ

The pinch temperature will be found at the outlet of the process after the streams have had ample time to equilibrate via the heat engines. Through our understanding of thermodynamics, we can assume that when the outlets reach the pinch temperatures, the maximum possible quantity of work would have been extracted. As will be shown in the countercurrent section, this assumption is true, but not necessary to solve the problem. In making use of this assumption, we can define the total fluxes of heat transfer and work generation as follows: Q_ H ¼ F H ρCP ðT H,IN  T P Þ Q_ C ¼ F C ρC P ðT C,IN  T P Þ _ ¼ ðQH þ QC Þ ¼ F H ρCP ðT H,IN  T P Þ þ F C ρC P ðT C,IN  T P Þ W h h  1i  1i _ ¼ F H ρC P T H,IN  T H,IN T αC,IN 1þα þ F C ρC P T C,IN  T H,IN T αC,IN 1þα W ð10Þ Part (c): Countercurrent Flow The solution strategy for the countercurrent flow is very similar to that for the cocurrent flow. What we find is that the change in the flow direction of one of the streams leads to a change in the last expression in Eq. (4), resulting in (this can be explained by switching the order of the enthalpies in Eq. (3) to out minus in): 0 ¼ δQ_ H  F H ρCP dT H 0 ¼ δQ_ C þ F C ρC P dT C

ð11Þ

(Note: It would have been equally valid to change the sign in the hot stream term in Eq. (11), but either way, the end result is identical). Propagating the change caused by the sign change through to Eq. (7), we find that (note the sign change relative to that in Eq. (7) in the cocurrent case): dT H F C dT C ¼ TH FH T C

ð12Þ

As we consider Eq. (12), we recognize that we must be very careful about the integration limits that we use. While in Part (a) we simply integrated both sides from the inlet to the outlet temperatures, for the countercurrent case, one of the integrals must be flipped. Therefore, we obtain:

Solved Problems for Part I T H,OUT ð

T H,IN

573

dT H F C ¼ TH FH

T C,IN ð

dT C F ¼ C TC FH

T C,OUT

T C,OUT ð

dT C TC

ð13Þ

T C,IN

Integrating Eq. (13) and rearranging leads to the following result:  T H,OUT ¼ T H,IN

T C,IN T C,OUT

α

ð14Þ

Note that Eq. (14) is identical to Eq. (8)! In other words, even with the changing of the flow direction, the basic equations relating the inlet and outlet temperatures are identical. However, it is unclear if the total amount of work will be equal, because the final temperatures of the process could be different. It is tempting to assume that the outlet temperature of the hot stream should be equal to the inlet temperature of the cold stream (which may be the case in a pure heat exchanger). It can be shown, however, that if that were doable, it would lead to a pure conversion of heat into work in violation of the Second Law of Thermodynamics (see the comment section at the end of the solution). To solve for the optimal outlet temperatures, we can setup a simple optimization to maximize the work with respect to the outlet cold temperature. To this end, we will set the derivative of the work with respect to the outlet cold temperature equal to zero and then solve to find a maximum. This yields: _ ðT C,OUT Þ ¼ ðQH þ QC Þ W

 α  T C,IN ¼ F H ρC P T H,IN  T H,IN T C,OUT þ F C ρC P ½T C,IN  T C,OUT  _ F H ρCP T H,IN T αC,IN dW ¼ ðαÞ  F C ρCP ¼ 0 ðαþ1Þ dT C,OUT T

ð15Þ ð16Þ

C,OUT

Solving for the outlet cold stream temperature in Eq. (16), we find that:  1 T C,OUT ¼ T H,IN T αC,IN 1þα

ð17Þ

Note that Eq. (17) is identical to the expression for the pinch temperature in Eq. (9)! Using Eq. (14), we can show that the outlet temperature of the hot stream must be equal to the outlet temperature of the cold stream, which equals the pinch temperature of the cocurrent process. We can check to make sure that this is indeed a maximum by checking the sign of the second derivative:

574

Solved Problems for Part I α

_ F H ρC P T H,IN T C,IN d2 W ¼ ðαÞðα þ 1Þ 2 ðαþ2Þ dT C,OUT T C,OUT

ð18Þ

Equation (18) clearly shows that the second derivative is always negative, ensuring that our solution is indeed a maximum. We can define the total work done using Eq. (10), and obtain the exact same solution for the countercurrent and cocurrent cases: h h  1i  1i _ counter ¼ W _ co ¼ F H ρCP T H,IN  T H,IN T αC,IN 1þα þ F C ρCP T C,IN  T H,IN T αC,IN 1þα W

ð19Þ Part (d): Plugging in the Numbers To calculate the total amounts of work, we simply use Eq. (19) with the values given in the Problem Statement. Specifically: F H ¼ 20
106 m3 =s F C ¼ 16
106 m3 =s C P ¼ 4:186 J=gK ρ ¼ 1000 kg=m3 α ¼ F C =F H ¼ 0:8 T H,IN ¼ 300 K T C,IN ¼ 278 K Results T p ¼ 290:01 K _ co ¼ W _ counter ¼ 3:12
1013 J=s W

Additional Comments Why Is the TH,OUT = TC,IN Assumption Invalid for the Countercurrent Flow? Several logical arguments exist which show why the final condition of TH,OUT ¼ TC,IN will not lead to more work production than using the final condition presented above. Below is just one example: Consider the system as a black box (see Fig. 3).

Solved Problems for Part I

575

Fig. 3

Now, if the flow rates of the hot and cold streams are equal, then, the energy of the exiting hot stream is equal to that of the entering cold stream. We can then redirect the hot outlet stream to the cold inlet stream and redraw the diagram (see Fig. 4):

Fig. 4

Examining Fig. 4, it is clear that heat (being taken from the hot inlet stream) is being converted directly into work in violation of the Second Law of Thermodynamics (unless TC,OUT ¼ TH,IN, which would be predicted by Eqs. (8) and (14), thus confirming their validity)!

576

Solved Problems for Part I

Problem 4 Problem 4.29 in Tester and Modell We are given a cylindrical vessel with an initial volume of 2 m3 that is filled with helium at 1 bar, 300 K. We plan to pressurize this vessel to 10 bar using a large external source of helium gas maintained at 300 bar, 300 K. Our model assumes that the initial helium present in the vessel does not mix with the entering helium and that the initial gas is layered (layer A) and compressed by the entering gas (layer B) from 1 bar to 10 bar. The final system then contains two “layers” of helium, both at 10 bar, but presumably at different temperatures. You can assume that (1) there is no heat transfer between the gas layers, (2) there is no heat transfer to the vessel during the operation, and (3) helium behaves as an ideal gas with a constant value of Cp ¼ 20.9 J/mol K. (a) Calculate the final temperature in layer B. (b) Calculate the entropy change of the universe after pressurization. (c) If the system consisting of layers A and B was thermally isolated from the environment and no additional gas allowed to enter, what is the maximum work that one could obtain for the reversible mixing of A and B (in the same vessel) to some final homogeneous temperature Tf? What would be the final temperature and pressure?

Solution to Problem 4 Solution Strategy This problem deals with pressurizing a vessel, initially filled with helium, using high-pressure helium from a cylinder. The process is assumed to occur such that the entering helium does not mix with the helium already present in the vessel. In addition, it is assumed that there are no heat interactions between the two gas layers or between the gas layers and the vessel. Part (a) The figure below shows a schematic of the elements of this problem, where system A denotes the gas already present in the vessel and system B represents the gas that enters the vessel from the helium cylinder. Note that determining the final state of the two systems is identical to what we did in the Solution to Problem 2, Part (b). Therefore, the relevant equations are directly reproduced here without a derivation. For a detailed derivation, please refer to the Solution to Problem 2. To solve the rest of the problem below, we list the initial and final states of the two systems.

Solved Problems for Part I

577

NA NA

NB

NB NT>>NB

Final State

Inial State

Initial state of system A: Pi,A ¼ 1 bar ¼ 1
105 Pa V i,A ¼ 2 m3 T i,A ¼ 300 K N i,A ¼ ? ¼

Pi,A V i,A 1
105
2 ¼ ¼ 80:19 moles 8:314
300 RT i,A

Final state of system A: Pi,A ¼ 10 bar ¼ 10
105 Pa Nf ,A ¼ N i,A ¼ 80:19 moles Tf ,A ¼?



  R=Cp Pf ,A R=ðCv þRÞ Pf ,A ¼T i,A ¼ T i,A Pi,A Pi,A  8:314=20:9 10 ¼300 ¼ 749:76 K 1

See Eq. (14) in the Solution to Problem 2, Part (b). Vf ,A ¼ ? ¼

Nf ,A RTf ,A 80:19
8:314
749:76 ¼ ¼ 0:5 m3 Pf ,A 10
105

578

Solved Problems for Part I

Note that to be consistent with this problem, in the equations above, we added the subscript “A” to the gas system that was originally in the cylinder. Inlet conditions of the helium stream: Pin ¼ 300 bar ¼ 300
105 Pa T in ¼ 300 K Final state of system B: Pf ¼ 10 bar ¼ 10
105 Pa Tf ,B ¼?

     Tf ,A Pi,A T in Pf ,B CP 1  T i,A Pf ,A        ¼  Tf ,A Tf ,A Pi,A C V þ Pi,A C  Pi,A CV Pf ,B 1  T i,A Pf ,A T i,A V ðsee Eq:ð22Þ in the Solution to Problem 2, Part ðbÞÞ     749:76 1
105 300
10
105
20:9 1  300 10
105         ¼ 5 749:76 1
10 5 749:76 5 þ 1
10 ð20:9  8:314Þ 10
105 1   1
10 300 300 10
105

¼415:19 K

Note that to be consistent with this problem, in the equation above, we again added the subscript “A” to the gas system that was originally in the cylinder: Vf ,B ¼ ? ¼ V i,A  Vf ,A ¼ 2  0:5 ¼ 1:5 m3 Nf ,B ¼ N B ¼

Pf ,B V f ,B 10
105
1:5 ¼ ¼ 434:54 moles 8:314
415:19 RTf ,B

Part (b) To find the change in the entropy of the universe, let us first identify the important components of the universe that are undergoing a change in state. The universe consists of the vessel, the helium cylinder, and the ambient in which these two are kept. There are no interactions between the vessel and the ambient or between the helium cylinder and the ambient. Therefore, the entropy change of the universe is purely due to the entropy change of the vessel (consisting of systems A and B) and the helium cylinder. Specifically:

Solved Problems for Part I

579

ΔSuniverse ¼ΔSvessel þ ΔScylinder ¼ΔSA þ ΔSB þ ΔScylinder

ð1Þ

We begin the calculation of ΔSuniverse by first computing the entropy change of system A, ΔSA . The gas in system A is undergoing an adiabatic, reversible compression, and therefore, the entropy change of system A is 0, that is: ΔSA ¼



δQ T

 reversible

¼0

ð2Þ

To calculate ΔSB, let us consider the NB moles of helium that were initially in the helium cylinder at a temperature and pressure of 300 K and 300 bar, respectively, and finally occupy the vessel at a temperature and pressure of 415.19 K and 10 bar, respectively. Although we know nothing about the path taken by the NB moles of gas during this process, we know that entropy is a state function, and therefore, we can construct a hypothetical reversible path between the initial and final states to calculate the entropy change undergone by the NB moles of gas. For a simple system following a reversible path, we know that: dU ¼ δQ þ δW ¼ TdS  PdV dU P þ dV ) dS ¼ T T

ð3Þ

Because the working fluid is an ideal gas, we can simplify Eq. (3), including integrating it from the initial state i to the final state f. This yields:   NC V dT NR C dT R þ dV ¼ N V þ dV T T V V    ðf ðf  Vf Tf dT dV ) dS ¼ ΔS ¼ N CV þR ¼ N C V ln þ R ln T Vi V Ti i i   Tf Tf Pi ¼ N C V ln þ R ln Ti T i Pf   Tf P ) ΔS ¼ N ðCV þ RÞ ln þ R ln i Ti Pf   Tf P ) ΔS ¼ N Cp ln þ R ln i Ti Pf dS ¼

ð4Þ

For system B, the last expression in Eq. (4) can be written as follows:   Tf ,B P ΔSB ¼ Nf Cp ln þ R ln in T in Pf ,B

ð5Þ

580

Solved Problems for Part I

Substituting values of Tf,B, Tin, Pf,B, and Pin in Eq. (5) yields:   415:19 300 þ 8:314 ln ¼ 15238:92 J=K ΔSB ¼ 434:54 ð20:9Þ ln 300 10

ð6Þ

The last contribution that we need to evaluate is the entropy change of the moles of helium that never exited the cylinder. Here, we assume that the helium cylinder behaves like an infinite reservoir, such that even after removing NB moles of gas from the cylinder, the temperature and pressure of the cylinder do not change appreciably. As a result, the entropy change of the cylinder is negligible compared to that of system B, that is: ΔScylinder ¼ 0

ð7Þ

Combining Eqs. (1), (2), (6), and (7), we obtain: ΔSuniverse ¼ ΔSA þ ΔSB þ ΔScylinder ¼ 0 þ 15238:92 þ 0 ¼ 15238:92 J=K

ð8Þ

As expected, ΔSuniverse > 0 because the overall process is irreversible. Indeed, the overall process is irreversible because it involves the rapid expansion followed by compression of gas as it transfers from the gas reservoir to the cylinder. There is no way to return the system and the environment back to their original state. Part (c) To solve this part of the problem, let us first understand why the system is considered to be unmixed. A solution is considered well mixed when the pressure, temperature, and composition of the solution are uniform throughout. Because the vessel consists of two identical gas layers (see the figure below), the only reason that it is considered unmixed is because the two helium layers in the vessel are at different temperatures. Therefore, the process of mixing can be envisioned as one that would ultimately result in the vessel having a uniform pressure and temperature. Furthermore, we seek to carry out this process reversibly to extract useful work out of it. Therefore, let us run a Carnot engine such that the hot helium layer A acts as the hot source and the cold helium layer B acts as the cold sink. At any point during the process, the Carnot engine absorbs heat equal to δQH from layer A, rejects heat equal to δQC to layer B, and converts the rest to work, δW. To calculate δQH, let us carry out a First Law of Thermodynamics analysis of layer A. Specifically:

Solved Problems for Part I

581

NA

δQH

NB

δW δQC

dE A ¼ dU A ¼ N A C V dT A ¼ δQA þ δW A ¼ jδQH j  PA V A N RT ) jδQH j ¼ N A CV dT A  A A dV A VA

ð9Þ

Similarly, a First Law of Thermodynamics analysis of layer B yields: dEB ¼ dU B ¼ N B CV dT B ¼ δQB þ δW B ¼ jδQC j  PB V B N RT ) jδQC j ¼ N B C V dT B þ B B dV B VB

ð10Þ

Based on the efficiency of the Carnot engine, we know that: δQC T C δQH ¼ T H

ð11Þ

Here, we recognize that TC is equal to TB and TH is equal to TA. Substituting Eqs. (9) and (10) in Eq. (11) yields: RT B dV B VB T ¼ B N A RT A TA N A C V dT A  dV A VA dV dT N B CV B þ N B R B TB VB ) ¼1 dV dT A N A C V þ N AR A TA VA dV B dV dT B dT þ NBR ¼ N A C V A þ N A R A ) N B CV TB VB TA VA N B C V dT B þ N

ð12Þ

As stated earlier, the gas layers are considered mixed when the temperature and pressure of both gas layers are equal. Let us denote this  temperature and pressure by  T and P, respectively. Integrating Eq. (12) between T B,f , T A,f , PB,f , PA,f , V B,f , V A,f   and T, T, P, P, V B,m , V A,m yields:

582

Solved Problems for Part I

    V B,f V A,f T B,f T A,f N B CV ln þ R ln þ R ln ¼ N A C V ln T V B,m T V A,m       T B,f T B,f P T A,f T A,f P ) N B CV ln þ R ln þ R ln ¼ N A CV ln T T PB,f T T PA,f   T B,f T B,f P ) N B CV ln þ R ln þ R ln PB,f T T   T A,f T A,f P ¼ N A C V ln þ R ln þ N A R ln PA,f T T   ) N B ðC V þ RÞ ln T B,f  ðCV þ RÞ ln T þ R ln P  R ln PB,f   ¼ N A ðC V þ RÞ ln T A,f  ðCV þ RÞ ln T þ R ln P  R ln PA,f     ) C P N B ln T B,f þ N A ln T A,f  R N B ln PB,f þ N A ln PA,f ¼ ðN A þ N B ÞðCP ln T  R ln PÞ     ) C P N B ln T B,f þ N A ln T A,f  R N B ln PB,f þ N A ln PA,f !! ðN A þ N B ÞRT  ¼ ðN A þ N B Þ C P ln T  R ln  V A,m þ V B,m     ) C P N B ln T B,f þ N A ln T A,f  R N B ln PB,f þ N A ln PA,f ! ðN A þ N B ÞR  ¼ ðN A þ N B ÞððC P  RÞ ln T Þ ¼ ðN A þ N B ÞR ln  V A,m þ V B,m !CR C C V αB C P αA C P αB CR αA CR ð N þ N ÞR A B  ) T ¼ T B,f V T A,f V PB,f V PA,f V  V A,m þ V B,m ð13Þ where αi ¼ Ni/(NA + NB) for i¼ A or B. Note that V A,m and V B,m represent the volumes occupied by layers A and B at the end of the mixing process, such that V A þ V B ¼ 2 m3 . Substituting the values of the known variables yields:  ð 80:19 Þð 8:314 Þ 434:54 20:9 80:19 20:9 T ¼ ð415:19Þð80:19þ434:54Þð20:98:314Þ ð749:76Þð80:19þ434:54Þð20:98:314Þ 10
105 80:19þ434:54 20:98:314   8:314  ð 434:54 Þð 8:314 Þ ð80:19 þ 434:54Þ8:314 ð20:98:314Þ 10
105 80:19þ434:54 20:98:314 2 ¼ 447:40 K

ð14Þ Using the ideal gas law, we can calculate the final pressure: P¼

ðN A þ N B ÞRT ð80:19 þ 434:54Þ
8:314
447:40 ¼ 957316:37 Pa ¼ 2 VA þ VB

¼ 9:57 bar

ð15Þ

Solved Problems for Part I

583

Finally, we can calculate the maximum available work using a First Law of Thermodynamics analysis of the Carnot engine. Specifically: ð ð ð W ¼ δW ¼ jδQH j  jδQC j

ð16Þ

Substituting the expressions for δQH and δQC from Eqs. (9) and (10), respectively, in Eq. (16) yields: ð ð W ¼ ðN A CV dT A  PA dV A Þ  ðN B CV dT B  PB dV B Þ

ð17Þ

To maintain mechanical equilibrium at the interface separating the two helium layers, it is necessary that the pressures on either side of the interface be equal, i.e., PA ¼ PB. In addition, we know that the volume of the vessel is fixed. Therefore, dV A þ dV B ¼ 0 ) dV A ¼ dV B . It may seem odd to maintain the identity of V A and V B in this mixing process as one could expect that, upon mixing, both gas layers would occupy the entire vessel and not some part of the vessel. However, we can assume that we carry out the process of equilibrating the temperature and pressure of the vessel by modeling the interface between the two gas layers as impermeable, and at the end of the process, we make the interface permeable to allow “mixing” between the two identical gases. However, once the temperatures and pressures are equilibrated, we realize that the states of the two gas layers are identical to each other. Therefore, even after we remove the interface (or make it permeable), it is not going to change the state of the system. Consequently, we cannot obtain any more useful work. Therefore, the thought process presented here to obtain useful work indeed gives the maximum possible work that can be done by the two gas layers in the isolated vessel. Note that if the two gas layers consisted of different gases (say helium and argon instead of helium and helium), then, due to the presence of the impermeable interface, the two gas layers would be at a different composition (but at the same temperature and pressure). Therefore, upon making the interface permeable, one would need to carry out an additional step of mixing to obtain the final fully mixed state (uniform temperature, pressure, and composition). Returning to calculating the work, we substitute the relations between PA, PB, V A, and V B in Eq. (17) to obtain: ð ð W ¼ ðN A C V dT A  PA dV A Þ  ðN B C V dT B  PA dV A Þ ð     ð18Þ ) ðN A CV dT A  N B C V dT B Þ ¼ N A C V T  T A,f  N B C V T  T B,f Substituting values for the different variables in Eq. (18) yields:

584

Solved Problems for Part I

W ¼ ð20:9  8:314Þ
ð80:19
ð447:40  749:76ÞÞ þ 434:54
ð447:40  415:19Þ

ð19Þ

¼ 129002:91 J ¼ 129:00 kJ

Other Possible Solution Strategies Alternate Way of Solving for the Final Temperature in Part (c) From the Second Law of Thermodynamics, we know that for any reversible process, the entropy change of the universe is 0. Our system (the vessel) is an isolated system. Therefore, the mixing process is going to be reversible only when the entropy change of the system is equal to 0, that is: ΔSA þ ΔSB ¼ 0

ð20Þ

Like in Part (b), we know the initial and final states of the constituents of the vessel (both helium layers attain a temperature and pressure of T and P, respectively). Therefore, we can make use of Eq. (4) to evaluate the entropy change of systems A and B. Specifically:  T ΔSA ¼ N A C p ln þ R ln Tf ,A  T ΔSB ¼ N B C p ln þ R ln Tf ,B

Pf ,A P Pf ,B P

 ð21Þ  ð22Þ

Substituting Eqs. (21) and (22) in Eq. (20) then yields:     Pf ,A Pf ,B T T þ R ln þ R ln þ N B C P ln ¼0 N A C P ln Tf ,A Tf ,B P P   ) ðN A þ N B ÞðC P ln T  R ln PÞ ¼ C P N B ln T B,f þ N A ln T A,f   R N B ln PB,f þ N A ln PA,f      ðN A þ N B ÞRT ) ðN A þ N B Þ CP ln T  R ln ¼ CP N B ln T B,f þ N A ln T A,f ðV A þ V B Þ   R N B ln PB,f þ N A ln PA,f   ) ðN A þ N B ÞððC P  RÞ ln T Þ ¼ C P N B ln T B,f þ N A ln T A,f     ðN A þ N B ÞR R N B ln PB,f þ N A ln PA,f þ ðN A þ N B ÞR ln ðV A þ V B Þ R   C C αB C P αA C P αB CR αA CR ðN A þ N B ÞR CV ) T ¼ T B,f V T A,f V PB,f V PA,f V ðV A þ V B Þ ð23Þ

Solved Problems for Part I

585

Note that, as expected, the expression for T in Eq. (23) is identical to the expression for T in Eq. (13)! Alternate Way of Solving for the Total Work in Part (c) Instead of carrying out a First Law of Thermodynamics analysis on the Carnot engine, we can select a system that includes gas A, gas B, and the Carnot engine. This yields: ΔE ¼ W on sys þ Qinto sys

ð24Þ

However, during the final work extraction process, there is no heat interaction between the composite system and the environment, and therefore, Qinto sys ¼ 0. Furthermore, the work done on the system is related to the total work as follows: W ¼  Won sys. Finally, the total energy of the system can be decomposed into the sum of the energies of the three subsystems, that is: ΔE ¼ ΔEA þ ΔEB þ ΔE engine

ð25Þ

However, because the engine operates in a cyclic manner, its energy is constant over the process. Further, because the two gaseous subsystems are simple ideal gases, their energies are simply their internal energies, that is:     ΔE ¼ ΔEA þ ΔE B ¼ ΔU A þ ΔU B ¼ N A C V T  T A,f þ N B C V T  T B,f ð26Þ

Combining Eqs. (24), (25), and (26), we obtain:     W ¼ ΔE ¼ N A C V T  T A,f  N B C V T  T B,f

ð27Þ

Note that, as expected, the expression for W in Eq. (27) is identical to the expression for W in Eq. (18)!

586

Solved Problems for Part I

Problem 5 Problems 5.17 + 5.27 in Tester and Modell 1. Problem 5.17 Show that: κ¼

  Cp ∂V ∂P ¼ Cv ∂P T ∂V S

and check to see if the relation holds for an ideal gas. 2. Problem 5.27 A non-ideal gas of constant heat capacity Cv ¼ 12.56 J/mol K undergoes a reversible adiabatic expansion. The gas is described by the van der Waals equation of state: 

 P þ a=V 2 ðV  bÞ ¼ RT

where a ¼ 0.1362 Jm3/mol2 and b ¼ 3.215
105 m3/mol. Derive an expression for the temperature variation of the gas internal energy, and calculate its value when the gas volume of 400 moles is 0.1 m3 and its temperature is 294 K.

Solution to Problem 5 Solution to Problem 5.17 Solution Strategy 1. Summarize what we know From Part I, we already know the definitions of both the constant pressure and the constant volume heat capacities. Specifically:   ∂H CP  ∂T P

ð1Þ

  ∂U ∂T V

ð2Þ

CV 

Solved Problems for Part I

587

Therefore, to solve the problem, we must show that: ∂H      CP ∂V ∂P ∂T P  ¼ ¼ ∂U CV ∂P T ∂V S

ð3Þ

∂T V

2. Select Possible Solution Strategies 1. Use the differential forms of the fundamental equations which include heat capacities. 2. Standard calculus strategies: (i) (ii) (iii) (iv)

Derivative inversion Triple product rule Chain rule expansion Maxwell reciprocity relationships

3. Advanced calculus strategies: (i) Jacobian transformations 3. Logically analyze the problem What we should recognize from Eq. (3) is that the variables that appear in the heat capacity definitions (H, U, T, P, V) are somewhat different than the variables in the desired proof (V, P, T, S). Therefore, we anticipate that it will be important through our calculus transitions to remove the U and H dependencies and to somehow add S terms. The simplest way to accomplish these goals is to replace the standard heat capacity definitions in Eqs. (1) and (2) with their other, well-known, entropy-based definitions. Specifically, as shown in Part I:     ∂H ∂S CP  ¼T ∂T P ∂T P  CV 

∂U ∂T



 ¼T

V

∂S ∂T

ð4Þ

 ð5Þ V

For completeness, we will first show that Eqs. (4) and (5) are valid representations. The simplest way to do is to consider the molar enthalpy and molar internal energy fundamental equations, that is: dH ¼ TdS  VdP

ð6Þ

588

Solved Problems for Part I

dU ¼ TdS  PdV

ð7Þ

By taking the derivative with respect to temperature of both sides of Eqs. (6) and (7), with the appropriate variables held constant (P for Eq. (6) and V for Eq. (7)), the results in Eqs. (4) and (5) are readily obtained. The use of the chain rule expansion is also possible: 

      ∂H ∂S ∂S ¼T ∂S P ∂T P ∂T P

ð8Þ

        ∂U ∂U ∂S ∂S ¼ ¼T ∂T V ∂S V ∂T V ∂T V

ð9Þ

∂H ∂T

 ¼ P

Using Eqs. (4) and (5) in our Problem Statement yields: T

 ∂S 

 ∂T P ∂S T ∂T V

¼

 ∂S 

 ∂T P ∂S ∂T V

 ¼

∂V ∂P

  T

∂P ∂V

 ð10Þ S

We must prove Eq. (10) in order to solve the problem. At this point, the problem may still seem challenging, because the variables that are held constant in our derivation (P, V) differ from the variables that are held constant in the proposed solution (T, S). Many of our strategies, including derivative inversion and chain rule expansion, do not change the variables that are held constant. As a result, they will not be useful to us. We could try to use the Maxwell reciprocity relationships, but eventually, this would lead us to a dead end (as would be found by trial and error). Our remaining strategies would be to use the fundamental equation (unfortunately, leading again to a dead end) or the triple product rule. Interestingly, the triple product rule may turn out to be an effective strategy. Indeed, we know that using the triple product rule on both the top and the bottom of the left-hand side of Eq. (10) will result in partial derivatives where T and S are held constant, similar to the proposed solution that we are trying to prove (the right-hand side of Eq. (10)). Utilizing the triple product rule on the numerator and the denominator of the lefthand side of Eq. (10), we obtain: 

∂S T ∂T

 P

T ¼ ∂T  ∂P

ð11Þ

T ¼ ∂T  ∂V 

ð12Þ

∂P S ∂S T

and 

∂S T ∂T

 V

∂V S ∂S T

Solved Problems for Part I

589

Dividing Eq. (11) by Eq. (12), including cancelling the two Ts and rearranging, we obtain:  ∂S 

 ∂T P ∂S ∂T V

∂T  ∂V       ∂V ∂P ∂V ∂S T  ¼ ¼ ∂T S ∂P ∂V S ∂P T  ∂P S ∂S T

Equation (13) shows that κ ¼ CCVP ¼

∂V   ∂P 

∂P T ∂V S

ð13Þ

is correct!

4. Validate the proof for an ideal gas Recall that, for an ideal gas, the equation of state is given by: PV ¼ RT

ð14Þ

and that U is only a function of T, where: dU ¼ CV dT

ð15Þ

Accordingly, the molar internal energy form of the fundamental equation can be written as follows: C V dT ¼ TdS  PdV To obtain

∂V 

∂P T

, we use the ideal gas law. This yields:  RT    ∂P ∂V RT V ¼ ¼ 2 ¼ P ∂P T ∂P T P

To obtain

 ∂P 

∂V S

ð16Þ

ð17Þ

, we use Eq. (16) at constant entropy, which yields: C V dT ¼ PdV

ð18Þ

d ðPV Þ ¼ RdT

ð19Þ

VdP þ PdV ¼ RdT

ð20Þ

V P dP þ dV R R

ð21Þ

From the ideal gas law:

dT ¼

Substituting Eq. (21) in Eq. (18) and simplifying yields:

590

Solved Problems for Part I

CV

V P dP þ CV dV ¼ PdV R R

ð22Þ

CV

  V P dP ¼  P þ C V dV R R

ð23Þ

CV VdP ¼ ðR þ CV ÞPdV

ð24Þ

Recall that CV þ R ¼ CP, and therefore, Eq. (24) can be expressed as follows: CV VdP ¼ CP PdV

ð25Þ

Rearranging Eq. (25) yields Eq. (26), which upon integration from an initial state i to a final state f yields Eq. (27): dP C dV ¼ P P CV V

ln

    Pf C Vi ¼ P ln Pi CV Vf

ð26Þ

ð27Þ

Recall that κ ¼ CCVP , and therefore, Eq. (27) can be simplified as follows:    κ Pf Vi ¼ Pi Vf

ð28Þ

Therefore, at constant entropy, Eq. (28) yields: P f V κf ¼ Pi V κi ¼ const:

ð29Þ

Note: Equation (29) is the same expression as that for the reversible adiabatic expansion, or compression, of an ideal gas ) constant entropy! Accordingly, for an ideal gas: 

∂P ∂V

 ¼ S

∂α=V κ ∂V

! ¼ S

κα , where α ¼ PV κ ða constantÞ V κþ1

Combining Eqs. (17) and (30), we obtain:

ð30Þ

Solved Problems for Part I



∂P ∂V

 S

591

∂V ∂P



 ¼ T

κα V κþ1



 V κα κα ¼ ¼κ ¼ P PV κ α

ð31Þ

Equation (31) shows that, as expected, the general relationship also holds for an ideal gas! 5. Alternative solution using Jacobian transformations We begin with Eq. (3), which we repeat below for completeness: CP ∂U ¼ ð∂H ∂T ÞP =ð ∂T Þ CV V

ð32Þ

We then express the two partial derivatives in Eq. (32) using the appropriate Jacobians. This yields: ∂ðH, PÞ

CP ∂ðT, PÞ ∂U ¼ ð∂H ∂T Þp=ð ∂T Þ ¼ ∂ðU, V Þ CV V

ð33Þ

∂ðT, V Þ

Next, we carry out a chain rule expansion on the top and bottom derivatives in Eq. (33) such that we end up with the derivative for entropy, that is: ∂ðH, PÞ

CP ∂ðS, PÞ ¼ CV ∂ðU, V Þ ∂ðS, V Þ

∂H 

CP ∂S P ¼ C V ∂U  ∂S V

CP T ¼ CV T

∂ðS, PÞ ∂ðT, PÞ ∂ðS, V Þ ∂ðT, V Þ

ð34Þ

∂ðS, PÞ ∂ðT, PÞ ∂ðS, V Þ ∂ðT, V Þ

ð35Þ

∂ðS, PÞ ∂ðT, PÞ ∂ðS, V Þ ∂ðT, V Þ

C P ∂ðS, PÞ ∂ðT, V Þ ¼ C V ∂ðT, PÞ ∂ðS, V Þ Finally, we can rearrange Eq. (37) to obtain the desired result, that is:

ð36Þ

ð37Þ

592

Solved Problems for Part I

C P ∂ðT, V Þ ∂ðS, PÞ ¼ CV ∂ðT, PÞ ∂ðS, V Þ

ð38Þ

    ∂V ∂P ∂P T ∂V S

ð39Þ

CP ¼ CV

Solution to Problem 5.27 We first rephrase the problem in a clearer manner. Basically, we are asked to derive an expression for ð∂U=∂T ÞS and then to evaluate it at the conditions given below: • • • • • •

V ¼ 0.1 m3/400 mol ¼ 0.25
103m3/mol T ¼ 294 K a ¼ 0.1362 Jm3/mol b ¼ 3.25
105m3/mol CV ¼ 12.56 J/molK R ¼ 8.314 J/molK

We are told that the volumetric behavior of the gas is described by the van der Waals equation of state (EOS), given by:  Pþ

 a ðV  bÞ ¼ RT V2

We are also told that the process under consideration involves the reversible expansion of a non-ideal gas.

Solution Strategy 1. Summarize what we know In considering the problem, we should recognize several key points: (a) The system is closed, and the process is adiabatic and reversible (i.e., dS ¼ 0). (b) The heat capacity at constant volume is given, although at the moment there is no apparent use for it. (c) The cubic van der Waals EOS is given, which has explicit solutions for T(P,V) and P(T,V), but not for V(T,P). 2. Use the internal energy fundamental equation While the initial solution strategy is not entirely clear based on the Problem Statement, beginning with fundamental equations is always a good option. Because the internal energy appears in the partial derivative that we need to calculate, we begin with the molar internal energy fundamental equation, given by:

Solved Problems for Part I

593

dU ¼ TdS  PdV

ð1Þ

Because the entropy term in Eq. (1) is equal to zero (recall that the process is adiabatic and reversible), it follows that: dU ¼ PdV

ð2Þ

Because our desired solution is ð∂U=∂T ÞS , we can take the partial derivative of Eq. (2) with respect to temperature, at constant entropy, to obtain: 

∂U ∂T



  ∂V ¼ P ∂T S

S

ð3Þ

The derivative on the right-hand side of Eq. (3) cannot yet be evaluated using the given information because the van der Waals EOS is pressure explicit, that is, of the form P(T,V), with no information given about S. Again, we need to use our calculus techniques to find a relationship that we can evaluate with the given information. As usual, it is not immediately obvious which strategy will lead to a useful solution. If we consider the different strategies, we find that: • Using derivative inversion would not get us very far (try it out). • Using the triple product rule could be useful (our technique of choice; see below). • Using a chain rule expansion would not remove the difficult constant entropy constraint (try it out). • Using Maxwell’s reciprocity theorem would only move the entropic term inside the derivative, and would not really help (try it out and you will find that the variable kept constant will always end up inside). Therefore, the best option among the standard techniques appears to be using the triple product rule:  ∂S      ∂U ∂V V ¼ P ¼ P ∂T ∂S ∂T S ∂T S ∂V

ð4Þ

T

where Eq. (3) was used. The partial derivative in the numerator of Eq. (4), as shown in the solution of the previous problem, is equal to CV/T. The partial derivative in the denominator of Eq. (4), again, cannot be immediately evaluated. However, using Maxwell’s reciprocity relationships will be useful to replace the entropy term in the partial derivative. Specifically: 

∂S ∂V

 ¼ T

∂

∂A ! ∂T V

dV

T

¼

∂

 ∂A  ! ∂V T

dT

V

 ¼

∂P ∂T

 ð5Þ V

594

Solved Problems for Part I

Equation (5) is an expression that we can readily evaluate based on the pressureexplicit EOS given in the Problem Statement. Combining all our results, we obtain: 

∂V ∂T

 S

  C CV ∂T ¼ ∂PV ¼ T ∂P V T ∂T V

ð6Þ

Using Eq. (6) in Eq. (3), we obtain: 

∂U ∂T



  CV ∂T ¼P T ∂P V

S

ð7Þ

Using the van der Waals EOS, we can expand both the pressure and the partial derivative with respect to P terms in Eq. (7) as follows:       að V  bÞ ∂U RT a C V ðV  bÞ ¼ CV 1   2 ¼ R V b V T ∂T S RV 2 T

ð8Þ

Using the variable values given in the Problem Statement in Eq. (8) yields: 

∂U ∂T

 S

 ! 0:1362 Jm3 =mol 0:25
103 m3 =mol  3:25
105 m3 =mol ¼ 12:56 J=molK 1   2 ð8:314 J=molKÞ 0:25
103 m3 =mol ð294 KÞ

ð9Þ This leads us to our final result of:   ∂U ¼ 10:12 J=molK ∂T S 3. Alternative solution using Jacobian transformations An alternative method of getting from Eq. (3) to Eq. (7) above is to utilize Jacobian transformations. Specifically, we again begin with Eq. (3), which we repeat below for completeness: 

∂U ∂T

 S

  ∂V ¼ P ∂T S

Then, we use a Maxwell relationship to obtain:

ð10Þ

Solved Problems for Part I

595

      ∂U ∂V ∂S ¼ P ¼P ∂T S ∂T S ∂P V

ð11Þ

Expanding Eq. (11) using Jacobians yields: 

∂U ∂T

 ¼P S

∂ðS, V Þ ∂ðP, V Þ

ð12Þ

Subsequently, carrying out a chain rule expansion, including adding in d(T,V) to obtain one derivative in terms of T, V, and P, yields:   ∂ðS, V Þ ∂ðT, V Þ ∂U ¼P ∂ðT, V Þ ∂ðP, V Þ ∂T S

ð13Þ

Carrying out another chain rule expansion, including adding in d(U,V) to obtain an expression for 1/T, yields:   ∂ðS, V Þ ∂ðU, V Þ ∂ðT, V Þ ∂U ¼P ∂ðU, V Þ ∂ðT, V Þ ∂ðP, V Þ ∂T S 

∂U ∂T

 S

      ∂S ∂U ∂T ¼P ∂U V ∂T V ∂P V

    ∂U 1 ∂T ¼ P CV T ∂T S ∂P V Equation (16) is identical to Eq. (7)!

ð14Þ

ð15Þ

ð16Þ

596

Solved Problems for Part I

Problem 6 Problem 5.28 in Tester and Modell The basic thermodynamic relationships for an axially stressed bar can be written as follows: dQrev ¼ TdS,

dW rev ¼ τdε,

ε ¼ Nε

where τ is the stress and ε is the strain. Derive the fundamental equation for a one-component bar and show that: 

∂μ ∂T





τ,N

∂S ¼ ∂N

 T,τ

Solution to Problem 6 Solution Strategy Obtaining the Fundamental Equation Let us assume that the axially stressed bar forms a closed, simple system. Carrying out a First Law of Thermodynamics analysis of the bar yields: dE ¼ dU ¼ δQrev þ δW rev ¼ TdS  τdε ) dU ¼ TdS  τdε

ð1Þ

From Postulate I, we know that the state of a closed, simple system can be characterized by two independently variable properties in addition to the masses of the components comprising the system. From Eq. (1), we see that, in this system, the two independently variable properties are the entropy of the system, S, and the strain, ε. If the mass of the bar is denoted by N, we can conclude that: U ¼ U ðS, ε, N Þ       ∂U ∂U ∂U ) dU ¼ dS þ dε þ dN ∂S ε,N ∂ε N,S ∂N S,ε where   ∂U ¼ T, ∂S ε,N



∂U ∂ε

 ¼ τ, N,S

and

  ∂U ¼μ ∂N S,ε

ð2Þ

Solved Problems for Part I

597

Therefore, the differential form of the internal energy fundamental equation can be written as follows: dU ¼ TdS  τdε þ μdN

ð3Þ

Euler integrating Eq. (3), we obtain: U ¼ TS  τε þ μN

ð4Þ

Partial Derivative of the Chemical Potential In the second half of the problem, we want to establish the following relationship: 

∂μ ∂T



 ¼

τ,N

∂S ∂N

 ð4aÞ T,τ

Because the relationship in Eq. (4a) contains many variables (μ, T, N, S, τ), let us first try to create the two partial derivatives separately and then device a way to prove their equality. To create the partial derivative on the left-hand side of Eq. (4a), we begin with Eq. (3) and manipulate it such that we can obtain an equation for dμ. To do that, let us separate the intensive variables from the extensive ones as follows: dðNU Þ ¼ Td ðNSÞ  τdðNεÞ þ μdN ) NdU þ UdN ¼ NTdS þ TSdN  Nτdε  τεdN þ μdN

ð5Þ

) N ðdU  TdS þ τdεÞ þ ðU  TS þ τε  μÞdN ¼ 0 For any simple system, we can choose the size of the system and vary it arbitrarily while maintaining all other intensive properties at a fixed value. Consequently, for Eq. (5) to be true, it is necessary that the coefficients of N and dN be both zero, that is: U ¼ TS  τε þ μ

ð6Þ

dU ¼ TdS  τdε

ð7Þ

Note that Eq. (6) is simply the integrated version of the fundamental equation, which is already given in Eq. (4). Equation (7) shows that U is a function of the two variables S and ε. Equivalently, we can also choose U to be a function of T and τ (the conjugate variables of S and ε). Similarly, from the Corollary to Postulate I, we can conclude that S is a function of T and τ, that is: S ¼ SðT, τÞ We will use Eq. (8) later in the proof.

ð8Þ

598

Solved Problems for Part I

Returning back to the problem of finding an expression for dμ, we next proceed to derive an equation which is analogous to the Gibbs-Duhem equation, by taking the total differential of Eq. (6). This yields: dU ¼ TdS þ SdT  τdε  εdτ þ dμ ) dU  TdS þ τdε ¼ SdT  εdτ þ dμ ) 0 ¼ SdT  εdτ þ dμ ) dμ ¼ εdτ  SdT

ð9Þ

To derive Eq. (9), we made use of Eq. (7). We can obtain the left-hand side of Eq. (4a) by differentiating Eq. (9) with respect to T at constant τ and N. This yields: 

∂μ ∂T

 τ,N

  ∂T ¼0S ¼ S ∂T τ,N

ð10Þ

Because Eq. (9) already provides a relationship between the left-hand side of Eq. (4a) in terms of the entropy of the system, next, let us consider the right-hand side of Eq. (4a) and try to simplify it. Specifically:     ∂ðNSÞ ∂S ¼  ∂N T,τ ∂N T,τ     ∂N ∂S ¼S N ∂N T,τ ∂N T,τ   ∂S ¼SN ∂N T,τ

ð11Þ

From Eq. (8), we know that S is a function of T and τ. Accordingly, (∂S/∂N)T,τ ¼ 0. Substituting this result in Eq. (11) yields: 

∂S  ∂N

 ¼ S

ð12Þ

T,τ

Combining Eqs. (12) and (10) yields: 

   ∂μ ∂S ¼ S ¼  ∂T τ,N ∂N T,τ     ∂μ ∂S ) ¼ ∂T τ,N ∂N T,τ Equation (13) is precisely what we set out to prove!

ð13Þ

Solved Problems for Part I

599

Other Possible Solution Strategies   Alternate proof for

∂μ ∂T τ,N

¼

 

∂S ∂N T,τ

Method 1 We begin with the internal energy fundamental equation given in Eq. (3), which we repeat below for completeness: dU ¼ TdS  τdε þ μdN

ð14Þ

By inspection, we recognize that direct differentiation of Eq. (14) will yields the desired derivative that contains S. Therefore, we differentiate Eq. (14) with respect to N, at constant T and τ, which will give us μ in terms of the other derivatives, including ð∂S=∂N ÞT,τ . Specifically:         ∂U ∂S ∂ε ∂N ¼T τ þμ ∂N T,τ ∂N T,τ ∂N T,τ ∂N T,τ       ∂U ∂S ∂ε ) ¼T τ þμ ∂N T,τ ∂N T,τ ∂N T,τ       ∂U ∂S ∂ε T þτ )μ¼ ∂N T,τ ∂N T,τ ∂N T,τ

ð15Þ

Next, we need to obtain the partial derivative of μ with respect to T, at constant N and τ. Therefore, we differentiate the last expression in Eq. (15) with respect to T, at constant N and τ. This yields: "   # " "    !#   !# ∂μ ∂ ∂U ∂ ∂S ∂ ∂ε ¼  þ T τ ∂T N,τ ∂T ∂N T,τ ∂T ∂N T,τ ∂T ∂N T,τ N,τ N,τ N,τ "   # "   #     ∂μ ∂ ∂U ∂S ∂ ∂S ) ¼  T ∂T N,τ ∂T ∂N T,τ ∂N T,τ ∂T ∂N T,τ N,τ N,τ "   # ∂ ∂ε þτ ∂T ∂N T,τ N,τ "   # "   #     ∂μ ∂S ∂ ∂U ∂ ∂S ) ¼ þ T ∂T N,τ ∂N T,τ ∂T ∂N T,τ ∂T ∂N T,τ N,τ N,τ "   # ∂ ∂ε þτ ∂T ∂N T,τ



N,τ

ð16Þ

600

Solved Problems for Part I

To complete the proof, we need to show that the sum of the last three terms on the right-hand side of Eq. (16) is 0, that is: "

  # ∂ ∂U ∂T ∂N T,τ

"

N,τ

  # ∂ ∂S T ∂T ∂N T,τ

"

N,τ

  # ∂ ∂ε þτ ∂T ∂N T,τ

¼0

ð17Þ

N,τ

Assuming that all the physical observables behave smoothly, the order of differ   ∂ ∂U entiation of every term in Eq. (17) can be switched. For example, ∂T ∂N T,τ N,τ

   ∂ ∂U can be expressed as ∂N ∂T . Next, after switching the order of differentiaN,τ T,τ

tion in every term in Eq. (17), we would like to show that: "

" "   #   #   # ∂ ∂U ∂ ∂S ∂ ∂ε 0¼ T þτ ∂N ∂T N,τ ∂N ∂T N,τ ∂N ∂T N,τ T,τ T,τ T,τ "       !# ∂ ∂U ∂S ∂ε ¼ T þτ ∂N ∂T N,τ ∂T N,τ ∂T N,τ

ð18Þ

T,τ

Because T and τ are held constant in the partial derivatives with respect to N, moving T and τ terms into the derivative is allowed. However, it can be readily observed that the terms within the outmost derivative sum up to zero, because: dU ¼ TdS  τdε þ μdN       ∂U ∂S ∂ε ) ¼T τ þ0 ∂T N,τ ∂T N,τ ∂T N,τ       ∂U ∂S ∂ε T þτ ¼0 ) ∂T N,τ ∂T N,τ ∂T N,τ "   # "   # "   # ∂ ∂U ∂ ∂S ∂ ∂ε T þτ ∂T ∂N T,τ ∂T ∂N T,τ ∂T ∂N T,τ N,τ

N,τ

ð18aÞ

¼0

ð19Þ

N,τ

Therefore, we have shown that Eq. (16) is equivalent to the relationship that we set out to prove, that is: 

∂μ ∂T

Method 2 A simple way to prove that





τ,N

∂S ¼ ∂N

∂μ ∂T τ,N

  ∂S ¼  ∂N

 

 ð20Þ T,τ

T,τ

is to use a suitable Maxwell

relation. Because we need to calculate the derivatives of μ and S with respect to

Solved Problems for Part I

601

T and N, respectively, under fixed τ, we recognize that we are searching for a Maxwell relation established by a thermodynamic function which is naturally described by T, τ, and N. It turns out that this corresponds to the Gibbs free energy, G . To see this, we begin by writing the differential form of the internal energy fundamental equation given by: dU ¼ TdS  τdε þ μdN

ð21Þ

However, Eq. (21) shows that U is a function of S, ε, and N. To change the dependence to T, τ, and N, we carry out the following variable transformation: dU  dðTSÞ þ dðτεÞ ¼ TdS  τdε þ μdN  TdS  SdT þ τdε þ εdτ ) d ðU  TS þ τεÞ ¼ SdT þ εdτ þ μdN

ð22Þ

From the analogy between PV and τε, we recognize that the Gibbs free energy has the form G ¼ U  TS þ τε in this case, with dG given by: dG ¼ SdT þ εdτ þ μdN

ð23Þ

Because the order of differentiation does not matter, the following second derivatives associated with Eq. (23) are identical:   ! ∂ ∂G ∂N ∂T τ,N

¼ τ,T

  ! ∂ ∂G ∂T ∂N τ,T

ð24Þ τ,N

In addition, as shown in Part I, we have:   ∂G ¼ S, ∂T τ,N



∂G ∂N

 τ,T

¼μ

ð25Þ

Substituting Eq. (25) in Eq. (24) yields: 

∂S  ∂N



 ¼

τ,T

which is precisely what we set out to prove.

∂μ ∂T

 ð26Þ τ,N

602

Solved Problems for Part I

Problem 7 Problem 8.2 in Tester and Modell Our research laboratory has synthesized a new material, and the properties of this material are being studied under conditions where it is always a vapor. Two sets of experiments have been carried out, and they are described below. Given the results of these experiments, the relationship PV ¼ NCT is proposed to describe the PVT properties of the material (C is constant). If you agree with this proposal, show a rigorous proof. If you do not agree, either prove that the relationship cannot be applicable or describe clearly what additional experiments you would recommend, and demonstrate how you would use these data to show whether or not PV ¼ NCT. Experiment A: A rigid and well-insulated container consists of compartment I separated from compartment II by an impermeable partition. Compartment I is initially filled with gas, and compartment II is initially evacuated. When the experiment begins, the partition between compartments I and II is broken, and gas fills both compartments I and II. Over a wide range of initial temperatures, pressures, and volumes of compartment I, the final temperature, after expansion, equals the initial temperature. Experiment B: Gas flows in an insulated pipe and through an insulated throttling valve wherein the pressure is reduced. Over a wide range of upstream temperatures, pressures, as well as downstream pressures, the temperature of the gas does not change when the gas flows through the valve.

Solution to Problem 7 Solution Strategy We need to determine the PVTN equation of state (EOS) of a new pure material and then compare it to the given PV ¼ NCT EOS to determine if they are equivalent. The only sources of information that we have are the two experiments A and B. Because both experiments provide information about the temperature and its change under different scenarios, it is convenient to express the EOS (i.e., the relation between P, V, N, and T ) as follows: T ¼ f ðP, V, N Þ

ð1Þ

Taking the differential of Eq. (1), we obtain:  dT ¼

∂T ∂P





∂T dP þ ∂V V,N





∂T dV þ ∂N P,N

 dN P,V

ð2Þ

Solved Problems for Part I

603

In order to determine the function f in Eq. (1) and therefore the equation of state, we first need to determine the three partial derivatives in Eq. (2), that is: ∂T   ∂T  ∂T , ∂V , and ∂N utilizing the information provided in the Problem ∂P V,N P,V P,N

Statement. 1. Experiment A Experiment A describes the adiabatic expansion of the gas from an initial PVT state to a final PVT state. We are told that the final temperature is always equal to the initial one. Let us consider the system to be the gas occupying the total volume of the container (consisting of compartments I and II). The system is simple, closed, rigid, and adiabatic. We can therefore use the First Law of Thermodynamics as follows: dU ¼ δQ þ δW

ð3Þ

Because δQ ¼ 0 (adiabatic) and δW ¼ 0 (rigid), Eq. (3) yields: dU ¼ 0 or U ¼ constant

ð4Þ

Note: We would obtain the same result if we considered the gas in compartment I as our system. Assuming a quasi-static expansion, the system is closed and adiabatic but with a movable boundary. However, the gas in compartment I is expanding against vacuum, and therefore, δW ¼ 0. Using the First Law of Thermodynamics in this case, we would obtain the same result: the internal energy of the gas remains constant! Therefore, experiment A provides us with information about the change in the temperature of the gas with respect to the volume, when the total number of moles, N, and the internal energy of the system, U, both remain constant. We learn that, in this case, there is no change in the temperature of the gas. Mathematically, then: 

∂T ∂V

 ¼0

ð5Þ

U,N

However, note that the partial derivative in Eq. (5), while similar, is not equal to the partial derivative multiplying dV in Eq. (2). Nevertheless, as shown next, it can be transformed into something useful. Indeed, using the triple product rule to move U into the partial derivative, we obtain: 

∂T ∂V

or



 U,N

∂V ∂U



 T,N

∂U ∂T

 ¼ 1 V,N

ð6Þ

604

Solved Problems for Part I



∂T ∂V

 

 ¼ U,N

∂U ∂V   T,N ∂U ∂T V,N

ð7Þ

According to Eqs. (5) and (7), it follows that:   ∂U ¼0 ∂V T,N

ð8Þ

We can then use Eq. (8) to obtain one of the partial derivatives that we need in Eq. (2). To do so, we would like to express the derivative in Eq. (8) as a function of P, V, T, and N. For this purpose, we write the differential form of U ¼ f ðS, V, N Þ as follows: dU ¼ TdS  PdV þ μdN

ð9Þ

Taking the derivative of Eq. (9) with respect to V, at constant T and N, yields:         ∂U ∂S ∂V ∂N ¼T P þμ ∂V T,N ∂V T,N ∂V T,N ∂V T,N

ð10Þ

In Eq. (10), the third term is equal to -P, and the last term is equal to 0, which leads to:     ∂U ∂S ¼T P ∂V T,N ∂V T,N

ð11Þ

Using a Maxwell relation, it follows that:   ! ∂ ∂A  ∂V ∂T V,N

 ¼ T,N

∂S ∂V



 ¼ T,N

∂P ∂T

 V,N

  ! ∂ ∂A ¼ ∂T ∂V T,N

Combining Eqs. (8), (11), and (12), we obtain:  T

∂P ∂T

 P¼0 V,N

or 

∂P ∂T

 ¼ V,N

P T

ð12Þ V,N

Solved Problems for Part I

605

or   ∂T T ¼ ∂P V,N P

ð13Þ

Equation (13) is the first partial derivative on the right-hand side of Eq. (1). 2. Experiment B Experiment B describes the flow of gas in an insulated pipe and through an insulated throttling valve. Considering the gas going through the valve as our system, we can use the First Law of Thermodynamics and describe the process as we did with Experiment A. In this case, the system is open, adiabatic, and simple. Using the First Law of Thermodynamics on the open system, we obtain: dU ¼ δQ þ δW þ H in δnin  H out δnout

ð14Þ

We know that: dU ¼ 0

ðSteady stateÞ

ð15Þ

ðAdiabaticÞ

ð16Þ

ðRigidÞ

ð17Þ

δQ ¼ 0 δW ¼ 0

δnin ¼ δnout ¼ dN

ðSteady stateÞ

ð18Þ

Using Eqs. (15), (16), (17), and (18) in Eq. (14) yields: ðH in  H out ÞdN ¼ 0 or H in ¼ H out , or dH ¼ 0

ð19Þ

Therefore, Experiment B describes the change in the temperature of the gas with respect to pressure when the molar enthalpy remains constant. This change can be described mathematically as the derivative of T with respect to P when H remains constant. From the Problem Statement, we know that there is no change in the temperature. Therefore, it follows that: 

∂T ∂P

 ¼0 H

ð20Þ

606

Solved Problems for Part I

Because H ¼ NH, Eq. (20) is equivalent to: 

∂T ∂P

 ¼0

ð21Þ

H,N

Note: When we deal with a throttling valve, the molar enthalpy remains constant. Next, we follow the same procedure as the one in Experiment A, in order to express the partial derivative in Eq. (21) in terms of measurable quantities. Using the triple product rule (to move H into the derivative), we obtain:       ∂T ∂P ∂H ¼ 1 ∂P H,N ∂H T,N ∂T P,N or 

∂T ∂P

 

 ¼ H,N

∂H ∂P   T,N ∂H ∂T P,N

ð22Þ

Combining Eqs. (21) and (22), we obtain:   ∂H ¼0 ∂P T,N

ð23Þ

Again, we would like to make use of Eq. (23) by converting the partial derivative to one of the partial derivatives in Eq. (2). To this end, we write the differential form of H as follows: dH ¼ TdS þ VdP þ μdN

ðFor a one component systemÞ

ð24Þ

Taking the partial derivative of Eq. (24) with respect to P, at constant T and N, yields:         ∂H ∂S ∂P ∂N ¼T þV þμ ∂P T,N ∂P T,N ∂P T,N ∂P T,N In the last equation, the third term is equal to V and the last term is equal to 0. As a result, we obtain:     ∂H ∂S ¼T þV ∂P T,N ∂P T,N

ð25Þ

Solved Problems for Part I

607

Using one of the Maxwell relations, it follows that:   ! ∂ ∂G  ∂P ∂T P,N

 ¼ T,N

     ! ∂S ∂V ∂ ∂G ¼ ¼ ∂P T,N ∂T P,N ∂T ∂P T,N

ð26Þ P,N

Combining Eqs. (25) and (26), we obtain: 

∂H ∂P





∂V ¼ T ∂T

T,N

 þV

ð27Þ

P,N

Using Eq. (23) in Eq. (27), we obtain:   ∂V V ¼ ∂T P,N T

ð28Þ

Equation (28) can be rewritten as follows: 

∂T ∂V

 ¼ P,N

T V

ð29Þ

Equation (29) is the second partial derivative on the right-hand side of Eq. (1). To summarize, below, we repeat the main results derived so far:   ∂T T ¼ ∂P V,N P 

∂T ∂V

Finally, we need to determine

 ∂T 



∂N P,V

¼ P,N

T V

ð13Þ

ð29Þ

in Eq. (1) in order to construct the complete

differential of dT. Recall that for a one-component (n ¼ 1) system, the number of independent first-order partial derivatives is n + 1 ¼ 2. Accordingly, because we have already determined two independent ones (see Eqs. (13) and (29)), any other one can be expressed as a function of those two. We can actually obtain the third partial derivative by first using the triple product rule as follows: 

∂T ∂N

    ∂V ∂N ¼ 1 ∂T V,P N,P ∂V T,P



608

Solved Problems for Part I

or 

 



∂V ∂N T,P   ∂V ∂T N,P

ð30Þ

    ∂NV ∂V ¼V þN ¼V ∂N T , P ∂N T , P

ð31Þ

∂T ∂N

¼ V,P

or 

∂V ∂N

 ¼ T, P

Combining Eqs. (29), (30), and (31), we obtain: 

∂T ∂N

 V,P

V V T ¼  V ¼  NV ¼  N T

ð32Þ

T

Euler integrating Eq. (2), recognizing that T and P are intensive variables, it follows that: 

∂T 0¼0þ ∂V





∂T Vþ ∂N P,N

 N P,V

or 

∂T ∂N



 P,V

∂T ¼ ∂V



V N P,N

Using Eq. (29) in the last equation yields the third required partial derivative in Eq. (1), that is: 

∂T ∂N

 ¼ P,V

T V T ¼ N V N

ð33Þ

Substituting Eqs. (13), (29), and (33) in Eq. (2), we finally obtain: dT ¼

T T T dP þ dV  dN P N V

or dT dP dV dN ¼ þ  N T P V

ð34Þ

Solved Problems for Part I

609

Integrating Eq. (34) yields: lnT ¼ lnP þ lnV  lnN þ C 0

ð35Þ

where C0 is the constant of integration. We can rewrite Eq. (35) as follows: lnT þ lnN ¼ lnP þ lnV þ lnC1

ð36Þ

where lnC1 ¼ C0. Combining the logarithmic terms in Eq. (36) yields: ln ðNT Þ ¼ ln ðPVC 1 Þ or CNT ¼ PV

ð37Þ

where C ¼ C11 is a constant. We therefore conclude that the EOS in Eq. (37) does describe the volumetric behavior of the new synthesized material.

610

Solved Problems for Part I

Problem 8 Problem 8.4 in Tester and Modell We have a constant volume, closed vessel filled with dichlorodifluoromethane gas. We plan to heat the gas and would like to know how the molar entropy of the gas varies with pressure. Derive a general relation to calculate the desired derivative, (∂S/∂P)V, assuming that we know the total volume of the vessel, the moles of gas, CP as a function of T and P, and have access to a pressure-explicit equation of state. Illustrate your result at the start of the heating process, where T ¼ 365.8 K, P ¼ 16.5 bar, and the total volume ¼ 1.51
103 m3. Assume that at this condition, CP ¼ 94.9 J/mol K and the pressure-explicit Redlich-Kwong equation of state is applicable, with a ¼ 20.839 J m3 K1/2/mol2 and b ¼ 6.725
105 m3/mol.

Solution to Problem 8 Solution Strategy Summarize What We Know We are asked to find a general relation to calculate how the molar entropy of dichlorodifluoromethane gas varies with pressure at constant molar volume using solely: • The heat capacity at constant pressure • A general, pressure-explicit equation of state • The volume and number of moles (or, equivalently, the intensive volume) Following that, we are asked to choose the pressure-explicit Redlich-Kwong (RK) equation of state, along with all the information provided in the Problem Statement, to obtain a numerical value for the desired partial derivative. Select Possible Solution Strategies As discussed in Part I, because we are given a general pressure-explicit EOS, where the variables T, P, and V appear, as well as heat capacity data, we should be able to calculate any thermodynamic property of a one-component (n ¼ 1) system, including the partial derivative of the molar entropy with respect to pressure at constant molar volume. Recall that the heat capacities at constant pressure and volume can be expressed as follows:     ∂H ∂S CP  ¼T ∂T P ∂T P

ð1Þ

Solved Problems for Part I

611

 CV 

∂U ∂T

 V



∂S ¼T ∂T

 ð2Þ V

Below, we will make use of Eqs. (1) and (2), along with the general pressureexplicit EOS. Strategy I: Using Eq. (1) to Obtain the Desired Result Choosing T and P as the two independent intensive variables, according to the Corollary to Postulate I, the molar entropy can be expressed as S ¼ S(T,P). The differential of S is then given by:  dS ¼

∂S ∂T



 dT þ

P

∂S ∂P

 dP

ð3Þ

T

Using Eq. (1) to simplify the first partial derivative in Eq. (3) and using a Maxwell relationship to transform the second partial derivative in Eq. (3), we obtain:    CP ∂V dS ¼ dP dT  T ∂T P 

ð4Þ

Next, if we take the derivative of Eq. (4) with respect to pressure, at constant molar volume, we obtain: 

             ∂S CP ∂T ∂V ∂P C P ∂T ∂V ¼  ¼  T T ∂P V ∂P V ∂T P ∂P V ∂P V ∂T P

ð5Þ

An examination of Eq. (5) indicates that we are quite close to the desired solution, because everything that appears on the right-hand side of Eq. (5) is expressed in terms of T, P, and V, as well as the heat capacity at constant pressure. However, in order to use the general pressure-explicit equation of state, we need to modify our derivatives slightly. Specifically, using a derivative inversion of the (∂T/∂P)V term and the triple-product rule on the (∂V/∂T)P term, we obtain: 

   1    1 ∂S C P ∂P ∂P ∂P ¼ þ T ∂P V ∂T V ∂T V ∂V T

ð6Þ

We can now evaluate Eq. (6) using the given pressure-explicit EOS and heat capacity data. We will do so after we consider Strategy II. Strategy II: Using Eq. (2) to Obtain the Desired Result

612

Solved Problems for Part I

Considering Eq. (2), we recognize that we can use the heat capacity definition by a quick chain rule expansion: 

        ∂S ∂S ∂T CV ∂T ¼ ¼ T ∂P V ∂T V ∂P V ∂P V

ð7Þ

Unfortunately, neither of the last two terms in Eq. (7) can be evaluated directly using the information provided. However, we can use transformations to obtain useful results. The constant volume heat capacity can be expanded as shown in Part I, that is: 

∂V CV ¼ CP  T ∂T

  P

∂P ∂T

 ð8Þ V

Using Eq. (8) in Eq. (7), we obtain: 

∂S ∂P

 ¼ V

             CP ∂V ∂P ∂T CP ∂T ∂V  ¼  ð9Þ T T ∂T P ∂T V ∂P V ∂P V ∂T P

Equation (9) is identical to Eq. (5)! Evaluate the Desired Partial Derivative To solve for the desired partial derivative, we will evaluate all the terms in Eq. (6) independently and then combine them all to obtain the desired result. The first term, CP/T, requires no further simplification because it is given. The other terms can be readily calculated using the RK EOS as follows:



 RT a  1=2 P¼ ð V  bÞ T V ð V þ bÞ    h i

∂P R a ¼ þ V b ∂T V 2T 3=2 V ðV þ bÞ " # " #   að2V þ bÞ ∂P RT ¼ þ 1=2 2 ∂V T ð V  bÞ 2 T V ð V þ bÞ 2

ð10Þ ð11Þ ð12Þ

Using the variable values provided in the Problem Statement, it is possible via a numerical solver to evaluate the volume at the specified temperature and pressure. Doing so, we obtain: V ¼ 1:51
103 m3 =mol

ð13Þ

Solved Problems for Part I

613

(Note: If we use this result to find the number of moles in the system, we find N ¼ 1 mole). Next, we can evaluate the two derivatives based on Eqs. (11) and (12): 

∂P ∂T



∂P ∂V

 ¼ 6:387
103 V

 ¼ 8:674
108 T

N m2  K

ð14Þ

N  mol m5

ð15Þ

Using the results in Eqs. (13), (14), and (15), as well as the CP and T information given in the Problem Statement, we obtain our final result: 

 ∂S J ¼ 3:33
105 mol  K  Pa ∂P V

ð16Þ

614

Solved Problems for Part I

Problem 9 Problem 8.6 in Tester and Modell Sulfur dioxide gas at 520 K and 100 bar fills one-half of a rigid, adiabatic cylinder. The other half is evacuated, and the two halves are separated by a metal diaphragm. If this should rupture, what would be the final temperature and pressure? Assume that the gas is well mixed and that expansion is sufficiently rapid so that negligible heat transfer occurs between the walls and the SO2 gas. For SO2, Tc ¼ 430.8 K, Pc ¼ 78.8 bar, Vc ¼ 1.22
104 m3/mol, ω ¼ 0.251, and 0 C P (J/mol K) is given by: C 0P ¼ 23:852 þ 6:699
102 T  4:961
105 T 2 þ 1:328
108 T 3 with T in degrees Kelvin.

Solution to Problem 9 Solution Strategy 1. Summarize what we know The problem asks us to find the final temperature and pressure of a gas undergoing an expansion against vacuum. To find the final state of the system (the gas), we can either use the First Law of Thermodynamics to find the change in the internal energy of the gas, and then relate it to the final state of the gas or we can use the Second Law of Thermodynamics to calculate the entropy change of the gas and then relate it to the final state of the gas. The most convenient choice here is to use the First Law of Thermodynamics to model the expansion of the gas. However, the gas expansion is not reversible (the gas is initially at a pressure of 100 bar and expands against a pressure of 0 bar). Consequently, although the gas expands adiabatically, we cannot claim that the entropy is conserved, and as a result, do not have enough information to calculate the entropy change directly. All that we know is that the change in entropy will be positive because the expansion of the gas is irreversible. However, because of the rigid, adiabatic nature of the cylinder, the gas is an isolated system, and because it expands against vacuum, no mechanical work is done by the gas as it expands. As a result, a First Law of Thermodynamics analysis of the system (the gas) yields: dU ¼ δQ þ δW ¼ 0 þ 0 ) dU ¼ 0

ð1Þ

Solved Problems for Part I

615

Because the number of moles of gas does not change throughout the expansion, Eq. (1) can be rewritten in terms of the molar (intensive) internal energy as follows: dU ¼ 0 ) U f ¼ Ui

ð2Þ

Therefore, in this problem, we need to find the final state of the gas following its expansion at constant molar internal energy. We will proceed using solely the following information provided in the Problem Statement: • Heat capacity • Critical parameters • Initial T and P of the gas 2. Make reasonable approximations If we assume that the SO2 gas behaves like an ideal gas, Eq. (2) implies that the final temperature of the gas is equal to its initial temperature. This is because the molar internal energy of an ideal gas is only a function of temperature. In that case, the pressure of the ideal gas would decrease by a factor of two due to the doubling of the volume occupied by the ideal gas at the end of its expansion. However, at pressures as high as 100 bar, it is evident that the gas would not behave ideally. As a result, its molar internal energy would not necessarily be a function of temperature only. From the Corollary to Postulate I introduced in Part I, we know that the molar internal energy for any pure (one-component, n ¼ 1) fluid can be written in terms of (n + 1) ¼ 2 independent intensive variables, for example, T and V, T and P, or P and V. One of the key steps in solving this problem is to select the correct set of variables that would make the analysis simple. Because we have been given heat capacity data, it is natural to select T as one of the two variables. In order to decide between P and V, we note that we know the final value of V (Vfinal ¼ 2Vinitial). Moreover, if we use a pressure-explicit equation of state, it would be easier for us to eliminate P in favor of T and V. Therefore, we choose T and V as our two independent intensive variables. From Part I, we know that the differential of U(T, V ) is given by:

   ∂P dU ¼ CV dT þ T  P dV ∂T V

ð3Þ

Combining Eqs. (2) and (3), including integrating from the initial (i) to the final ( f ) states, yields:

616

Solved Problems for Part I Uð f

Tðf

dU ¼ U f  U i ¼ 0 ¼ Ui

Vðf

C V dT þ Ti

Tðf

)

Vðf

CV dT þ Ti

Vi

Vi

   ∂P T  P dV ∂T V

   ∂P T  P dV ¼ 0 ∂T V

ð4Þ

We know that the internal energy is a state function, and therefore, to use Eq. (4), it would be beneficial to construct a convenient path that allows us to calculate the change in internal energy using the limited information available to us. The Problem Statement provides us with heat capacity data for the SO2 gas when it behaves ideally. Therefore, we want to carry out the temperature integration in Eq. (4) when the gas behaves ideally (i.e., at a very large V value in this case). Accordingly, for the purpose of integration, the gas first traverses an isothermal path at Ti from a volume Vi to a volume V ! 1. Next, the gas traverses an isochoric path at a volume V from a temperature Ti to a temperature Tf. Finally, to reach the final state, the gas again traverses an isothermal path at a temperature Tf from a volume V to a volume Vf. This three-step integration method is what we referred to in Part I as the attenuated-state approximation. Mathematically, this can be written as follows: Vðf

Tðf

C V dT þ Ti

Vi

Vð 

     ∂P ∂P T  P dV ¼ T P dV ∂T V ∂T V T¼T i Vi

Tðf

þ

Vðf

C V jV¼V  dT þ Ti

V

   ∂P T P dV ∂T V T¼T f

ð5Þ

3. Choose a suitable EOS and solve So far, our derivation is general. However, to proceed and evaluate the first and third integrals on the right-hand side of Eq. (5), we need to choose a suitable equation of state. Because we have chosen T and V as our two independent intensive variables, it is advantageous to use a pressure-explicit equation of state for which P is a function of T and V. We need to use a pressure-explicit equation of state that utilizes all the information provided in the Problem Statement. With some reflection, this gives us a choice between the Peng-Robinson (PR) equation of state and the Redlich-Kwong-Soave (RKS) equation of state. From the Problem Statement, we know that the critical compressibility factor ZC is PCVC/RTC ¼ 0.268, which is close to the ZC for the Peng-Robinson equation of state discussed in Part I. Therefore, we assume that the volumetric behavior of the SO2 gas can be accurately modeled using the PengRobinson equation of state. For the Peng-Robinson equation of state, the integrands

Solved Problems for Part I

617

of the first and third integrals on the right-hand side of Eq. (5) can be evaluated as follows: aðω, T r Þ RT P¼  ðV  bÞ V ðV þ bÞ þ bðV  bÞ   da=dT ∂P R )  ¼ ∂T V ðV  bÞ V ðV þ bÞ þ bðV  bÞ   T ðda=dT Þ ∂P RT RT P¼  )T  ðV  bÞ V ðV þ bÞ þ bðV  bÞ ðV  bÞ ∂T V aðω, T r Þ þ V ðV þ bÞ þ bðV  bÞ   aðω, T r Þ  T ðda=dT Þ ∂P P¼ )T V ð V þ bÞ þ bð V  bÞ ∂T V

ð6Þ

where  pffiffiffiffiffi2 0:45724R2 T 2C  1 þ κ 1  Tr PC

ð7Þ

κ ¼ 0:37464 þ 1:54226ω  0:26992ω2

ð8Þ

a¼

b¼

0:07780RT C PC

ð9Þ

Substituting values of TC, PC, VC, and ω in Eqs. (7), (8), and (9) yields: b¼

0:07780
8:314
430:8 ¼ 3:54
105 m3 =mol 78:8
105

κ ¼ 0:37464 þ 1:54226
0:251  0:26992
0:2512 ¼ 0:7447 rffiffiffiffiffiffiffiffiffiffiffi!!2 0:45724
8:3142
430:82 T a¼ 1 þ 0:7447 1  430:8 78:8
105 rffiffiffiffiffiffiffiffiffiffiffi!!2 T ¼0:7444
1 þ 0:7447 1  J m3 =mol2 430:8

ð10Þ ð11Þ

ð12Þ

618

Solved Problems for Part I

rffiffiffiffiffiffiffiffiffiffiffi!!   da T 1
0:7447
ð1Þ
¼ 0:7444
2
1 þ 0:7447 1  dT 430:8 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1
430:8
T 0:0466 ¼ 9:5798
104  pffiffiffiffi Jm3 =K mol2 T V Ð2 V1

ð13Þ

In view of Eq. (6), to evaluate Eq. (5), we need to evaluate the integral, aðω, T r ÞT ðda=dT Þ V ðVþbÞþbðVbÞ dV

, with the appropriate limits of integration. Because the

numerator in the integrand is independent of V, we can take it out of the integral, leaving the following integral for evaluation: Vð2

1 dV ¼ V ð V þ bÞ þ bð V  bÞ

V1

Vð2

V1 Vð2

¼ V1 Vð2

¼ V1

1 dV V þ 2Vb  b2 2

1 dV V 2 þ 2Vb þ b2  2b2 1 dV ðV þ bÞ2  2b2

1 ¼ pffiffiffi 2 2b 1 ¼ pffiffiffi 2 2b

Vð2



pffiffiffi   pffiffiffi  V þ b þ 2b  V þ b  2b pffiffiffi  pffiffiffi  dV  V þ b  2b V þ b þ 2b



1 1 pffiffiffi    pffiffiffi  dV V þ b  2b V þ b þ 2b

V1 Vð2

V1

pffiffiffi V þ b  2b V 2 1 pffiffiffi ¼ pffiffiffi ln 2 2b V þ b þ 2b V 1 Using Eq. (14), we can simplify Eq. (5) as follows:

ð14Þ

Solved Problems for Part I Vð

U f  Ui ¼ 0 ¼ Vi

619

Tðf Vðf  

    ∂P ∂P T P dV þ C V jV¼V  dT þ T P dV ∂T V ∂T V T¼T i T¼T f Ti

V

Tðf pffiffiffi  V þ b  2b V !1 pffiffiffi ln þ C V jV¼V  !1 dTþ V þ b þ 2b V i

aðω, T i,r Þ  T i ðda=dT ÞT¼T i pffiffiffi 2 2b Ti   pffiffiffi V f ¼2V i a ω, T f ,r  T f ðda=dT ÞT¼T f V þ b  2b pffiffiffi pffiffiffi ln 2 2b V þ b þ 2b V  !1 0 1 pffiffiffi   1 þ b  2b pffiffiffi C B  V þ b  2b C aðω, T i,r Þ  T i ðda=dT ÞT¼T i B V pffiffiffi pffiffiffi   ln i pffiffiffi Cþ ¼ ln  B lim  @ V !1 V 2 2b b þ 2 b þ b þ 2b A i 1 þ V ¼

Tðf

C V jV¼V  !1 dT Ti

pffiffiffi  1  b  2b pffiffiffi C 1 þ B V f ,r  T f ðda=dT ÞT¼T f B 2V i þ b  2b C pffiffiffi pffiffiffi  lim p ffiffi ffi ln   þ C B ln A @ 2V i þ b þ 2b V  !1 2 2b 1 þ b þ  2b V pffiffiffi   V i þ b  2b aðω, T i,r Þ  T i ðda=dT ÞT¼T i pffiffiffi pffiffiffi ¼ ln 1  ln 2 2b V i þ b þ 2b   Tðf pffiffiffi   a ω, T f ,r  T f ðda=dT ÞT¼T f 2V þ b  2b pffiffiffi pffiffiffi  ln 1 ln i þ C V jV¼V  !1 dT þ 2V i þ b þ 2b 2 2b Ti   pffiffiffi   a ω, T f ,r  T f ðda=dT ÞT¼T f 2V þ b  2b pffiffiffi pffiffiffi  ¼ ln i 2 2b 2V i þ b þ 2b pffiffiffi  Tðf  V i þ b  2b aðω, T i,r Þ  T i ðda=dT ÞT¼T i pffiffiffi pffiffiffi þ CV jV¼V  !1 dT ln V i þ b þ 2b 2 2b  a ω, T



0

Ti

ð15Þ Using the given values of the parameters in Eq. (15), we obtain an equation that we can solve for Vi. Specifically: qffiffiffiffiffiffiffiffi2   520 0:7444
1 þ 0:7447 1  430:8 8:314
520      100
105 ¼   V i  3:54
105 V i V i þ 3:54
105 þ 3:54
105
V i  3:54
105 4323:28 0:639  2 ) 107 ¼  V i  3:54
105 V i þ V i
7:08
105  1:253
109 ) V i ¼ 3:2125
104 m3 =mol

ð16Þ

620

Solved Problems for Part I

Note that in carrying out the integration in Eq. (5), shown in Eq. (15), we did not encounter any singularity, and therefore, did not have to add or subtract any term to remove the singularity, as we discussed in Part I. The last piece in calculating the change in the molar internal energy of the system involves evaluating the temperature integral. Because the temperature integral is carried out in the attenuated state (V ! 1), we can use the ideal gas heat capacity to compute the integral as shown below: Tðf

C V jV¼V  !1 dT Ti Tðf

Tðf

¼

C 0V dT Ti Tðf

¼

¼



 C 0P  R dT

Ti



 23:852 þ 6:699
102 T  4:961
105 T 2 þ 1:328
108 T 3  8:314 dT

Ti



¼ 15:538
T þ 6:699
102



T2 2



 4:961
105

 3  4 T f T T þ 1:328
108 3 4 Ti

ð17Þ Combining Eqs. (15) and (17), and substituting values of a, da/dT, b, Ti, and Vi, yields an equation that can be solved for Tf:

Solved Problems for Part I

621

!  qffiffiffiffiffiffiffiffi2 0:0466 Tf 4 0:7444
1 þ 0:7447 1  430:8  T f 9:5798
10  pffiffiffiffiffiffi Tf pffiffiffi U f  Ui ¼ 0 ¼ 2 2
3:54
105 ! 2
3:2125
104 þ 3:54
105 1  pffiffi2ffi p ffiffi ffi ln   2
3:2125
104 þ 3:54
105 1 þ 2 

  qffiffiffiffiffiffiffiffi2   0:0466 520 0:7444
1 þ 0:7447 1  430:8  520 9:5798
104  pffiffiffiffiffiffiffiffi 520 pffiffiffi  2 2
3:54
105 ! 3:2125
104 þ 3:54
105 1  pffiffi2ffi pffiffiffi  ln 3:2125
104 þ 3:54
105 1 þ 2 T 2f 2

þ 15:538
T f þ 6:699
102  15:538
520  6:699
10 þ 4:961
105



0 ¼ @7434:61


5203 3



2



!  4:961
105

5202 2

T 3f 3

! þ 1:328
108

T 4f 4

!



 1:328
108



5204 4



1 rffiffiffiffiffiffiffiffiffiffiffi!!2 pffiffiffiffiffiffi Tf 1 þ 0:7447 1   9:5677
T f þ 465:41
T f A 430:8


ð0:1480Þ  12020:07
ð0:2826Þ þ 15:538
T f þ 0:0335
T 2f  1:654
105
T 3f þ 3:32
109
T 4f  15054:37 ¼ 1100:32


1 þ 0:7447 1 

rffiffiffiffiffiffiffiffiffiffiffi!!2 pffiffiffiffiffiffi Tf  68:88
T f þ 16:954
T f 430:8

þ 0:0335
T 2f  1:654
105
T 3f þ 3:32
109
T 4f  11657:5

ð18Þ Equation (18) can be solved for Tf using a nonlinear equation solver. Solving it using the goal seek function in MS Excel yields Tf ¼ 479.35 K. Using this value of the final temperature and final volume ¼ 2 (initial volume), we can obtain the final pressure from the equation of state as follows:

622

Solved Problems for Part I

8:314
479:3478  P¼ 2
3:2125
104  3:54
105

qffiffiffiffiffiffiffiffiffiffiffiffiffi2   0:7444
1 þ 0:7447 1  479:3478 430:8      2
3:2125
104 2
3:2125
104 þ 3:54
105 þ 3:54
105
2
3:2125
104  3:54
105 ¼ 50:66 bar

ð19Þ Therefore, the final temperature and pressure of the gas are 479.35 K and 50.66 bar, respectively. Note that the final temperature of the gas modeled using the Peng-Robinson equation of state is lower than that for an ideal gas. This is because, in this case, the molar internal energy of the gas increased due to an increase in the volume. However, because the system (the gas) is isolated, the molar internal energy of the system cannot increase. Therefore, to maintain constant molar internal energy, the gas temperature had to decrease so that it offsets the increase in internal energy due to the increase in volume. 4. Steps to avoid 1. Setting ΔS ¼ 0 – Although the system is an isolated system undergoing an adiabatic expansion, the expansion is not reversible in this case, and therefore, the process is not isentropic. 2. Using T and P as the two independent intensive variables when using a pressure-explicit equation of state – Although there is nothing fundamentally incorrect in using T and P as the two independent intensive variables, it is not advisable to use them in this case because this will lead to a significantly more complicated mathematical derivation. 5. Other solution strategies We could also use the departure-function approach presented in Part I. In addition, other suitable equations of state may be used.

Solved Problems for Part I

623

Problem 10 Adapted from Problem 8.15 in Tester and Modell Oxygen gas at 150 K and 30 bar is to be pressurized to 100 bar in a compressor with an efficiency of 80% (based on an isentropic process). If the flow rate is 10 kg/s, what would be the power required? Carry out your calculation using the Peng-Robinson equation of state, and use both the attenuated state approach and the departure function approach. The following information is available: Tc ¼ 154.6K, Pc ¼ 50.46 bar, and ω ¼ 0.021. Also assume that CP is 29.3 J/mol K, independent of temperature.

Solution to Problem 10 Solution Strategy To solve this problem, we begin with: δW ¼

δW isen 0:80

where δW is the real, differential work done by the compressor on the gas, which includes pressurizing the δn moles and the flow work associated with the δn moles, δWisen is the differential work done by the compressor on the gas in an isentropic process, and 0.80 is the efficiency of the actual compressor relative to the isentropic process. Recall that this is not the Carnot efficiency introduced in Part I! We recognize that δW is also the differential amount of work that must be supplied to the compressor so that it can compress δn moles of gas in this open system. Therefore, dividing both sides of the equation above by dt yields: δW 1 δW isen ¼ dt 0:80 dt _ ¼ 1 W _ W 0:80 isen

ð1Þ

Based on Eq. (1), it is clear that more power needs to be supplied to the compressor in the real process than in the isentropic process. This makes physical sense because the isentropic process is reversible. Therefore, we can first calculate the power required in the isentropic process and then divide this value by 0.80 to evaluate the actual power required.

624

Solved Problems for Part I

O2

O2

10 kg/s 150 K 30 bar

10 kg/s 100 bar

δWisen Fig. 1

In Fig. 1, δQ ¼ 0, because an isentropic process is equivalent to a reversible, adiabatic process (see below). Carrying out an entropy balance on the gas in the compressor yields: dS ¼ 0 ¼

δQ þ ðSin  Sout Þδn þ dσ ðSteady stateÞ T

ð2Þ

To derive Eq. (2), we used the fact that at steady state, δnin ¼ δnout ¼ δn. If the process is adiabatic (δQ ¼ 0) and reversible (dσ ¼ 0), Eq. (2) reduces to: ðSin  Sout Þδn ¼ 0 Because δn 6¼ 0, the equation above shows that: Sin ¼ Sout ðFor an isentropic processÞ Next, let us carry out a First Law of Thermodynamics analysis of the gas in the compressor: dU ¼ δQ þ δW þ H in δnin  H out δnout

ð3Þ

dU ¼ 0 ðSteady stateÞ δQ ¼ 0 ðAdiabaticÞ δnin ¼ δnout ¼ δn ðSteady stateÞ Therefore, δW isen þ ðH in  H out Þδn ¼ 0 where we have highlighted the fact that the differential work corresponds to that in an isentropic process! Dividing the last equation by dt and rearranging yields:

Solved Problems for Part I

625

_ isen þ ðH in  H out Þ n_ 0¼W _ W isen ¼ ðH out  H in Þ n_ ¼ ðH 2  H 1 Þ n_

ð3aÞ

Derivations We are given an ideal gas heat capacity, and therefore, we need to carry out the temperature change when the gas is ideal. We are also given an equation of state that describes the gas. With this given information, as discussed in Part I, we can use the “attenuated state approach” or the “departure function approach” to determine the change in any derived intensive property of the gas. Essentially, as shown in Eq. (3a), we need to calculate (H2  H1), so that we can _ isen and then W _ using Eq. (1). In order to calculate (H2  H1), we must evaluate W also use the fact that (S2  S1) ¼ 0. The “attenuated state approach” or the “departure function approach” can be used to derive expressions for (S2  S1) and (H2  H1). We begin with the attenuated state approach.

Attenuated-State Approach Because we are given a pressure-explicit EOS, we choose T and V as the two independent intensive variables. In the (V-T) phase diagram shown in Fig. 2, we see that:

V∞Æ∞

V

V1

Actual

V2 T1

T Fig. 2

T2

626

Solved Problems for Part I

(i) V2 < V1, because the gas is compressed, and (ii) T2 > T1, because work is done on the gas without heat transfer, and as a result, the internal energy of the gas increases. Let us first derive an expression for (S2  S1), where S (T,V ). The differential of S can be written as follows:  dS ¼

∂S ∂T



 dT þ

V

∂S ∂V

 dV T

In Part I, we saw that the fist partial derivative in the last equation can be related to the heat capacity at constant volume and the second partial derivative can be related to a partial derivative that can be calculated using a pressure-explicit EOS. Specifically:   CV ∂P dS ¼ dV dT þ T ∂T V

ð4Þ

Working with CV, although we are given CP, is no problem when we deal with an ideal gas. Indeed, as shown in Fig. 2, we will be changing the temperature when the gas behaves ideally, for which the heat capacities are related by the following equation derived in Part I: CV ¼ CP  R

ð5Þ

As shown in Fig. 2, (S2  S1) can be calculated using an isothermal step followed by an isochoric step and then completed with a second isothermal step. Specifically: S2  S1 ¼

ðV 1  V1

∂P ∂T

constant T

 ðV 2   ðT2 0 CV ∂P dV þ dT þ T ∂T V T1

T1

constant V

V1

dV

V T2

ð6Þ

constant T

As discussed in Part I, it is generally good practice to add and subtract Ð V 1 ∂Pideal  Ð V 1 ∂Pideal  dV as well as V 2 dV to handle the two ln(V1 ! 1) V1 ∂T ∂T V T1

V T2

singularities. Nevertheless, in this solution, we will work directly with Eq. (6), without adding and subtracting integrals. As discussed in Part I, the Peng-Robinson EOS is given by: P¼

aðω, T r Þ RT  V  b V ð V þ b Þ þ bð V  bÞ

ð7Þ

where aðω, T r Þ ¼ ac αðω, T r Þ

ð8Þ

Solved Problems for Part I

627

ac ¼

∂P ∂T

 ¼ V

ð9Þ

  pffiffiffiffiffi 2 αðω, T r Þ ¼ 1 þ κ 1  T r

ð10Þ

κ ¼ 0:37464 þ 1:54226ω  0:26992ω2

ð11Þ

b¼ 

0:45724R2 T 2c Pc

0:07780RT c Pc

ð12Þ

dað, T r Þ |fflfflfflffldT {zfflfflfflffl}

R 1  V  b V ð V þ bÞ þ bð V  bÞ

a0 ðT Þ is a full derivative, because }a} is not a function of P or V

where   pffiffiffiffiffi 2  da dα d  ¼ a0 ð T Þ ¼ ac ¼ ac 1 þ κ 1  Tr dT dT dT It then follows that: "

 1= # 1= !#" T 2 1 1 1 2 a ð T Þ ¼ ac 2 1 þ κ 1  κ TC 2 TC T

    1=2 ac κ 1 þ κ 1  TTC pffiffiffiffiffiffiffiffiffi ¼ TT C Ð V ∂P Let us next derive a general expression for V i f ∂T dV, which we will need to V 0



T0

use twice in Eq. (6):

628

Solved Problems for Part I

ðV f  Vi

∂P ∂T

 ðV f ðV f dV dV 0 dV ¼ R  a ðT Þ 2 V  b V þ 2bV  b2 Vi Vi V T0 |fflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 



Vf b a0 ðT Þ  pffiffiffi ¼ R ln Vi  b 2b 2

(

Using Maple to find the integral

pffiffiffi! pffiffiffi!)   Vf þb 1 2 Vi þ b 1  2 pffiffiffi  ln pffiffiffi ln   Vf þb 1þ 2 Vi þ b 1 þ 2

ð13Þ In addition, we know that: ðT2 T1

  C 0V T2 0 dT ¼ CV ln T T1

ð14Þ

Using Eqs. (13) and (14) in Eq. (6) yields:  ðV2   ðT2 0 CV ∂P dT þ S2  S1 ¼ dV þ dV T ∂T V1 T1 V1 V T1 V T2 ( p ffiffi ffi !    V1 þ b 1  2 a0 ð T 1 Þ V1  b pffiffiffi  ln  ¼ R ln  pffiffiffi ln V1  b V1 þ b 1 þ 2 2b 2   T2 þC 0V ln T1 ( pffiffiffi!    V2 þ b 1  2 a0 ðT 2 Þ V2  b pffiffiffi  ln  þR ln  pffiffiffi ln V1  b V2 þ b 1 þ 2 2b 2 ðV1 

∂P ∂T

pffiffiffi!)  V1 þ b 1  2 pffiffiffi  V1 þ b 1 þ 2 pffiffiffi!)  V1 þ b 1  2 pffiffiffi  V1 þ b 1 þ 2

ð15Þ Rearranging Eq. (15) and recognizing that S2  S1 ¼ 0, we obtain:

Recall that a'(T ) refers to the derivative of a(ω, Tr) with respect to T. Because V1 ! 1, we can further simplify the last equation as follows:

Solved Problems for Part I

629

or pffiffiffi!   V1 þ b 1  2 a0 ð T 1 Þ V2  b pffiffiffi   pffiffiffi ln þ R ln V1  b 2b 2 V1 þ b 1 þ 2 pffiffiffi!  V2 þ b 1  2 a0 ðT 2 Þ pffiffiffi  pffiffiffi ln  2b 2 V2 þ b 1 þ 2 

T2 0 ¼ C0V ln T1





ð16Þ

We know that T1 ¼ 150 K, and we can calculate V1 by numerically solving Eq. (7) with T1 ¼ 150 K and P1 ¼ 30
105Pa. Specifically, we solved Eq. (7) using Maple, and V1 was found to be 2.936
104m3/mol. Therefore, in Eq. (16), T2 and V2 are the only two unknowns. Because we know that P2 ¼ 100
105Pa, T2 and V2 are also the only two unknowns in Eq. (7), the Peng-Robinson equation of state. Consequently, T2 and V2 can be calculated numerically by simultaneously solving Eq. (7) and Eq. (16) using Maple. This yields: T 2 ¼ 217:7 K V 2 ¼ 1:361
104 m3 =mol Now that we have (T1, V1) and (T2, V2), we can calculate (H2  H1) by expanding dH in terms of dT and dV. Recall that (H2  H1) is the key quantity that we are after. Like we did for the entropy, in order to utilize the pressure-explicit PengRobinson EOS, let us expand dH in terms of dT and dV. Specifically: dH ¼ In Part I, we saw that:

    ∂H ∂H dT þ dV ∂T V ∂V T

630

Solved Problems for Part I

    ∂H ∂P ¼ CV þ V ∂T V ∂T V       ∂H ∂P ∂P ¼T þV ∂V T ∂T V ∂V T Using the last two results in the expression for dH above, we obtain:

        ∂P ∂P ∂P þV dH ¼ C V þ V dT þ T dV ∂T V ∂T V ∂V T We again proceed with the integration of the last equation along the isothermal (at T1) ! isochoric (at V1) ! isothermal (at T2) path indicated in the (V-T) phase diagram shown in Fig. 2. This yields: Tð2

   Ideal      ∂P ∂P ∂P 0 dV þ dT H2  H1 ¼ T þV C þ V V ∂T ∂T V ∂V T T 1 V V1 V1 T1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Vð1

dT¼0

dV¼0

     ∂P ∂P dV þ T þV ∂T V ∂V T T 2 V1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Vð2

dT¼0

or Vð1



H2  H1 ¼ T 1 V1 Vð2



þT 2 V1

∂P ∂T

Vð1    dV þ V ∂P ∂V V T1

Tð2    dV þ C 0 ðT 2  T 1 Þ þ V ∂ RT V ∂T V

T T1

V1

T1



dT

V V1

Vð2    ∂P ∂P dV þ V dV ∂T V T 2 ∂V T T 2 V1

Carrying out the temperature integration in the last equation yields: Vð1



H2  H1 ¼ T 1 V1 Vð2

þT 2 V1



∂P ∂T

∂P ∂T

Vð1    dV þ V ∂P ∂V V T1

V1

dV þ C 0 ðT 2  T 1 Þ þ RðT 2  T 1 Þ V

T T1

Vð2    dV þ V ∂P dV ∂V V T2 T T2 V1

ð17Þ

Solved Problems for Part I

631

Using Eq. (13), the general expression for

Ð V f ∂P V i ∂T V dV , in the two pertinent T0

integrals in Eq. (17), we obtain: Vð2    ∂P T1 dV dV þ T 2 ∂T V T 2 V T1 V1 V1 8 9 > > > > > pffiffiffi! pffiffiffi!>   > >   < 0 V1 þ b 1  2 V1 þ b 1  2 = a ðT 1 Þ V1  b pffiffiffi  ln pffiffiffi   ¼ T 1 R ln  T 1 pffiffiffi ln V1  b 2b 2 > V1 þ b 1 þ 2 V1 þ b 1 þ 2 > > > > > > > |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl {zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl } : ; 0, as V 1 goes to infinity 8 9 > > > > > ! ! pffiffiffi pffiffiffi >   > >   < 0 V2 þ b 1  2 V1 þ b 1  2 = a ðT 2 Þ V2  b pffiffiffi  ln pffiffiffi    T 2 pffiffiffi ln þT 2 R ln V1  b > 2b 2 > V2 þ b 1 þ 2 V1 þ b 1 þ 2 > > > > > |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}> : ; Vð1



∂P ∂T

pffiffiffi! V1 þ b 1  2 a ðT 1 Þ V1  b pffiffiffi  þ T 1 pffiffiffi ln ¼ T 1 R ln V1  b V1 þ b 1 þ 2 2b 2 pffiffiffi!    V2 þ b 1  2 a0 ðT 2 Þ V2  b pffiffiffi þT 2 R ln þ T 2 pffiffiffi ln  V1  b 2b 2 V2 þ b 1 þ 2 





0

0, as V 1 goes to infinity

ð18Þ

Before we can proceed further, we need to find the general expression for: ðV f   ∂P V dV ∂V T T 0 Vi which we will then use with the appropriate integration limits in Eq. (17). Based on Eq. (7), we can directly take the partial derivative of P with respect to V at constant temperature. This yields: 

∂P ∂V

 ¼ T

aðT Þð2V þ 2bÞ RT þ 2 ð V  bÞ 2 V 2 þ 2bV  b2

Then, ðV f   ðV f ðV f V ð2V þ 2bÞ ∂P V V dV þ aðT 0 Þ  2 2 dV dV ¼ RT 0 2 ∂V Vi V i ð V  bÞ V i V þ 2bV  b2 T T0

632

Solved Problems for Part I

Using Maple, we can simplify the last result as follows:

   ðV f   Vf b ∂P b b V dV ¼ RT 0   RT 0 ln V f  b Vi  b Vi  b ∂V T T 0 Vi " # Vf Vi aðT 0 Þ  ð19Þ V f 2 þ 2bV f  b2 V i 2 þ 2bV i  b2 ( pffiffiffi! pffiffiffi!)   Vf þb 1 2 Vi þ b 1 þ 2 a0 ð T 0 Þ pffiffiffi pffiffiffi þ pffiffiffi ln   2b 2 Vf þb 1þ 2 Vi þ b 1  2 We can then use Eq. (19) for the two relevant dV integrals in Eq. (17). This yields: Vð2    ∂P ∂P V dV þ V dV ∂V T T 1 ∂V T T 2 V1 V1

   b b V1  b ¼ RT 1   RT 1 ln V1  b V1  b V1  b

 V1 V1 aðT 1 Þ  V 1 2 þ 2bV 1  b2 V 1 2 þ 2bV 1  b2 ( pffiffiffi! pffiffiffi!)   V1 þ b 1  2 V1 þ b 1 þ 2 a0 ð T 1 Þ pffiffiffi pffiffiffi   þ pffiffiffi ln 2b 2 V1 þ b 1 þ 2 V1 þ b 1  2

   b b V2  b þRT 2   RT 2 ln V2  b V1  b V1  b

 V2 V1 aðT 2 Þ  V 2 2 þ 2bV 2  b2 V 1 2 þ 2bV 1  b2 ( pffiffiffi! pffiffiffi!)   V2 þ b 1  2 V1 þ b 1 þ 2 a0 ð T 2 Þ pffiffiffi pffiffiffi   þ pffiffiffi ln 2b 2 V2 þ b 1 þ 2 V1 þ b 1  2 Vð1



Substituting Eqs. (19) and (20) in Eq. (17), we obtain:

ð20Þ

Solved Problems for Part I

633



   b b V1  b   RT 1 ln V1  b V1  b V1  b

 V1 V1 aðT 1 Þ  V 1 2 þ 2bV 1  b2 V 1 2 þ 2bV 1  b2 ( pffiffiffi! pffiffiffi!)   V1 þ b 1  2 V1 þ b 1 þ 2 a0 ð T 1 Þ pffiffiffi pffiffiffi þ pffiffiffi ln   2b 2 V1 þ b 1 þ 2 V1 þ b 1  2

   b b V2  b   RT 2 ln þRT 2 V2  b V1  b V1  b

 V2 V1 aðT 2 Þ  V 2 2 þ 2bV 2  b2 V 1 2 þ 2bV 1  b2 ( pffiffiffi! pffiffiffi!)   V2 þ b 1  2 V1 þ b 1 þ 2 a0 ð T 2 Þ pffiffiffi pffiffiffi þ pffiffiffi ln   2b 2 V2 þ b 1 þ 2 V1 þ b 1  2

H 2  H 1 ¼ RT 1

þC 0V ðT 2  T 1 Þ þ RðT 2  T 1 Þ pffiffiffi!    V1 þ b 1  2 a0 ð T 1 Þ V1  b pffiffiffi þ T 1 pffiffiffi ln þT 1 R ln  V1  b 2b 2 V1 þ b 1 þ 2 pffiffiffi!    V2 þ b 1  2 a0 ð T 2 Þ V2  b pffiffiffi þT 2 R ln þ T 2 pffiffiffi ln  V1  b 2b 2 V2 þ b 1 þ 2 Combining terms in the last equation and letting V1 go to infinity, we obtain:

 b b V1  RT 1 þ að T 1 Þ V2  b V1  b V 1 2 þ 2bV 1  b2

 V2 aðT 2 Þ V 2 2 þ 2bV 2  b2 pffiffiffi!  V1 þ b 1  2 a 0 ð T 1 Þ  T 1 a0 ð T 1 Þ pffiffiffi pffiffiffi   ln 2b 2 V1 þ b 1 þ 2 pffiffiffi!  V2 þ b 1  2 aðT 2 Þ  T 2 a0 ðT 2 Þ pffiffiffi pffiffiffi  þ ln 2b 2 V2 þ b 1 þ 2

H 2  H 1 ¼ RT 2

ð21Þ

þC 0P ðT 2  T 1 Þ Equation (21) is the final expression for H2  H1. Substituting T1, V1, T2, and V2, we find that: ΔH ¼ H 2  H 1 ¼ 1336 J=mol

634

Solved Problems for Part I

_ isen ¼ ðH 2  H 1 Þn_ ¼ ð1336 J=molÞð10 kg=s  1000 g=kg  mol=32 gÞ W ¼ 417500 J=s _ _ ¼ W isen ¼ 417500 J=s ¼ 522 kW W 0:80 0:80 Next, we will pursue the departure function approach.

Departure Function Approach As discussed in Part I, it is convenient to first analyze the Helmholtz free energy departure function, DA, because we chose T and V as the two independent intensive variables. Again, recall that this choice was motivated because we have access to the Peng-Robinson pressure-explicit EOS. The (V-T) phase diagram in Fig. 3 shows the three steps that we need to follow to implement the departure function approach.

V10 V20

V

V1

Actual

V2 T1

T

T2

Fig. 3

Following the derivation in Part I, it follows that:   AðT, V Þ  A0 T, V 0 ¼ 

 0 ðV h i RT V dV þ RT ln P V V V1

ð22Þ

Solved Problems for Part I

635

In addition, in Part I, we showed that: 0

2

  B∂ 6 SðT, V Þ  S0 T, V 0 ¼ @ 4 ∂T

ðV h

31  0 i RT V 7C dV 5A  R ln P V V

V1

ð23Þ

V

        U ðT, V Þ  U 0 T, V 0 ¼ AðT, V Þ  A0 T, V 0 þ T SðT, V Þ  S0 T, V 0 ð24Þ      H ðT, V Þ  H 0 T, V 0 ¼ U ðT, V Þ  U 0 T, V 0 þ ðPV  RT Þ

ð25Þ

Carrying out the integrals in Eqs. (23) and (24), one can show that: 

0



0

SðT, V Þ  S T, V 0

" pffiffiffi#  V þb 1 2 a0 ð T Þ V0 pffiffiffi  pffiffiffi ln ¼ R ln  V  b 2 2b V þb 1þ 2



ð26Þ

and H ðT, V Þ  H T, V 0



" pffiffiffi#  V þb 1 2 að T Þ pffiffiffi ¼ pffiffiffi ln  2 2b V þb 1þ 2 " pffiffiffi#  V þb 1 2 Ta0 ðT Þ pffiffiffi þ PV  RT   pffiffiffi ln 2 2b V þb 1þ 2

ð27Þ

Using Eq. (26), let us next calculate S2  S1 ¼ S(T2, V2)  S(T1, V1). Specifically:    SðT 2 , V 2 Þ  SðT 1 , V 1 Þ ¼ SðT 2 , V 2 Þ  S0 T 2 , V 02 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ð1Þ

     þ S0 T 2 , V 02  S0 T 1 , V 01 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ð2Þ

ð28Þ

   þ S0 T 1 , V 01  SðT 1 , V 1 Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ð3Þ

The first term (1) and the third term (3) are departure functions at T2 and T1, respectively. The second term involves a temperature and volume change when the gas behaves ideally. Using Eq. (26) as needed, we obtain: " pffiffiffi#  V2 þ b 1  2 V 02 a0 ð T 2 Þ pffiffiffi SðT 2 , V 2 Þ  S T 2 , V 2 ¼ R ln  pffiffiffi ln  V 2  b 2 2b |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl} V2 þ b 1 þ 2 

 0

ð1Þ

 0

636

Solved Problems for Part I

and " pffiffiffi#   0  V1 þ b 1  2 V 01 a0 ð T 1 Þ 0 pffiffiffi  pffiffiffi ln S T 1 , V 1  SðT 1 , V 1 Þ ¼ R ln  V 1  b 2 2b |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} V1 þ b 1 þ 2 ð3Þ

For the ideal gas contribution (2), we obtain:    0  0    V2 T2 S T 2 , V 02  S0 T 1 , V 01 ¼ C 0V ln þ R ln T1 V 01 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ð2Þ

Substituting the three last equations in Eq. (28), we obtain:   T2 V b SðT 2 , V 2 Þ  SðT 1 , V 1 Þ ¼ C 0V ln þ R ln 2 V1  b T1 " " pffiffiffi# pffiffiffi#   V1 þ b 1  2 V2 þ b 1  2 a0 ð T 1 Þ a0 ðT 2 Þ pffiffiffi  pffiffiffi ln pffiffiffi þ pffiffiffi ln   2 2b 2 2b V1 þ b 1 þ 2 V2 þ b 1 þ 2 ð29Þ

Because S2  S1 ¼ S(T2, V2)  S(T1, V1) ¼ 0, Eq. (29) yields: " pffiffiffi#    0 V þ b 1  2 a ð T Þ T V  b 1 1 2 pffiffiffi  þ pffiffiffi ln C 0V ln þ R ln 2 V 1  b 2 2b T1 V1 þ b 1 þ 2 " pffiffiffi#  V2 þ b 1  2 a0 ð T Þ pffiffiffi  pffiffiffi2 ln  2 2b V2 þ b 1 þ 2 ¼0 Note that the last equation is identical to Eq. (16) that we derived above using the attenuated state approach! Next, let us calculate H2  H1 ¼ H(T2, V2)  H(T1, V1). Specifically:    H ðT 2 , V 2 Þ  H ðT 1 , V 1 Þ ¼ H ðT 2 , V 2 Þ  H 0 T 2 , V 02 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 



ð1Þ

    H 0 T 1 , V 01 þ H |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 0

T 2 , V 02

ð2Þ

   þ H 0 T 1 , V 01  H ðT 1 , V 1 Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ð3Þ

ð30Þ

Solved Problems for Part I

637

Like in the entropy case, the first term (1) and the third term (3) are departure functions at T2 and T1, respectively. The second term involves a temperature and volume change when the gas behaves ideally. The three contributions in Eq. (30) are given by: " pffiffiffi#  V2 þ b 1  2 aðT 2 Þ pffiffiffi ¼ pffiffiffi ln H ðT 2 , V 2 Þ  H  |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 2 2b V2 þ b 1 þ 2 

0



T 2 , V 02



ð1Þ

" pffiffiffi#  V2 þ b 1  2 T 2 a0 ð T 2 Þ pffiffiffi  pffiffiffi ln  2 2b V2 þ b 1 þ 2 þP2 V 2  RT 2 " pffiffiffi#   0 aðT 1 Þ  V1 þ b 1  2 0 pffiffiffi  H T 1 , V 1  H ðT 1 , V 1 Þ ¼ pffiffiffi ln |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 2 2b V1 þ b 1 þ 2 ð3Þ

" pffiffiffi#  V1 þ b 1  2 T 1 a0 ð T 1 Þ pffiffiffi  pffiffiffi ln  2 2b V1 þ b 1 þ 2 þP1 V 1  RT 1  0    H T 2 , V 02  H 0 T 1 , V 01 ¼ C 0P ðT 2  T 1 Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ð2Þ

Substituting the last three equations in Eq. (30) yields:

H ðT 2 , V 2 Þ  H ðT 1 , V 1 Þ ¼

C 0P ðT 2

 aðT 2 Þ RT 2  T 1Þ þ V  RT 2  V 2  b V 2 2 þ 2bV 2  b2 2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}





P2

að T 1 Þ RT 1  V þ RT 1 V 1  b V 1 2 þ 2bV 1  b2 2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} P1

pffiffiffi!  V1 þ b 1  2 a ðT 1 Þ  T 1 a ðT 1 Þ pffiffiffi pffiffiffi   ln 2b 2 V1 þ b 1 þ 2 pffiffiffi!  V2 þ b 1  2 aðT 2 Þ  T 2 a0 ðT 2 Þ pffiffiffi pffiffiffi  ln þ 2b 2 V2 þ b 1 þ 2 0

Collecting terms, we obtain:

0

638

Solved Problems for Part I

 b b V1  RT 1 þ aðT 1 Þ V2  b V1  b V 1 2 þ 2bV 1  b2 pffiffiffi! 

 V1 þ b 1  2 a0 ð T 1 Þ  T 1 a0 ð T 1 Þ V2 pffiffiffi pffiffiffi  ln  aðT 2 Þ V 2 2 þ 2bV 2  b2 V1 þ b 1 þ 2 2b 2 pffiffiffi!  V2 þ b 1  2 aðT 2 Þ  T 2 a0 ðT 2 Þ pffiffiffi þ C 0P ðT 2  T 1 Þ pffiffiffi  þ ln V2 þ b 1 þ 2 2b 2

H ðT 2 , V 2 Þ  H ðT 1 , V 1 Þ ¼ RT 2

Note that the last equation is identical to Eq. (21) that we derived above using the attenuated state approach!

Solved Problems for Part II

640

Solved Problems for Part II

Problem 11 Problem 9.24 in Tester and Modell The enthalpy of mixing of a ternary solution containing components 1, 2, and 3 is given by: ΔH 123 ¼ ½100x1 x2 þ 50ðx1 x3 þ x2 x3 Þ=½x1 þ 2x2 þ 0:5x3  where ΔH123 is measured in J/(mole of solution) and xi is the mole fraction of component i for i ¼ 1, 2, and 3. (a) Determine the partial molar enthalpy of mixing of component 2 at x1 ¼ 0.2 and x2 ¼ 0.5. Express your result in J/(mole of component 2). (b) A blender is used to mix two binary solutions to make an equimolar ternary solution of composition x1 ¼ x2 ¼ x3. The compositions of the two binary solutions are (x1 ¼ 0.333, x2 ¼ 0.667) and (x1 ¼ 0.333, x3 ¼ 0.667), respectively. If the mixing is carried out isothermally and isobarically as a continuous process operating at steady state, determine the heating load required. Express your result in J/(mole of ternary product).

Solution to Problem 11 Solution strategy Two general methods can be used to solve this problem: 1. Working with the extensive form of the enthalpy of mixing 2. Working with the intensive form of the enthalpy of mixing Of course, the two methods are equally valid, but use different sets of variables to calculate the same quantity, which is the partial molar enthalpy of mixing with respect to component 2. Part (a) Method 1 This method involves correctly converting the mole fractions to extensive moles. To implement this method, we first multiply the given intensive enthalpy of mixing expression by N to obtain the extensive form and then take the partial derivative with respect to N2. Specifically, we first multiply ΔH123given  in the Problem Statement ∂ by N and then we apply the partial molar operator ∂N 2 , that is: T,P,N 1 ,N 3

Solved Problems for Part II

ΔH 123 ¼

641

½100x1 x2 þ 50ðx1 x3 þ x2 x3 Þ ½x1 þ 2x2 þ 0:5x3 

ð1Þ

N ½100x1 x2 þ 50ðx1 x3 þ x2 x3 Þ ΔH 123 ¼ NΔH 123 ¼ ½x1 þ 2x2 þ 0:5x3 

 2  N x x N x1 x 3 þ N 2 x2 x3 N 100 12 2 þ 50 N N2

 ¼ Nx1 þ 2Nx2 þ 0:5Nx3 N 100N 1 N 2 þ 50N 1 N 3 þ 50N 2 N 3 ¼ N 1 þ 2N 2 þ 0:5N 3 

 ∂ΔH 123 ∂N 2 T,P,N 1 ,N 3    ∂ 100N 1 N 2 þ 50N 1 N 3 þ 50N 2 N 3 ¼ N 1 þ 2N 2 þ 0:5N 3 ∂N 2 T,P,N 1 ,N 3

ΔH 123,2 ¼

Simplifying the last equation, we obtain: ΔH 123,2 ¼

100x1 2 þ 25ð1  x1  x2 Þ2 ½x1 þ 2x2 þ 0:5ð1  x1  x2 Þ2

At x1 ¼ 0.2 and x2 ¼ 0.5, we obtain: ΔH 123,2 ¼

100
0:22 þ 25ð1  0:2  0:5Þ2 ¼ 3:42 J=ðmol of component 2Þ ½0:2 þ 2
0:5 þ 0:5ð1  0:2  0:5Þ2

Therefore, the partial molar enthalpy of mixing of component 2 at x1 ¼ 0.2 and x2 ¼ 0.5 is 3.42 J/(mol of component 2). Method 2 In this method, we use an expression for ΔH 123,2 discussed in Part I, that is: ΔH 123,2 ¼ ΔH 123  x1

    ∂ΔH 123 ∂ΔH 123  x3 ∂x1 ∂x1 T,P,x3 T,P,x1

ð2Þ

642

Solved Problems for Part II

To calculate the partial derivatives in Eq. (2), it is first convenient to replace x2 by (1  x1  x3) in Eq. (1). Specifically: ½100x1 ð1  x1  x3 Þ þ 50ðx1 x3 þ ð1  x1  x3 Þx3 Þ ½x1 þ 2ð1  x1  x3 Þ þ 0:5x3  100x1  100x1 2  100x1 x3 þ 50x3  50x3 2 ¼ 2  x1  1:5x3

ΔH 123 ¼

ð3Þ

We then need to calculate the two partial derivatives in Eq. (2) using Eq. (3). Specifically:   ∂ΔH 123 200  300x3 þ 100x1 2 þ 300x1 x3  400x1 þ 100x3 2 ¼ ∂x1 ð2  x1  1:5x3 Þ2 T,P,x3

ð4Þ

and   ∂ΔH 123 100  50x1 2 þ 100x1 x3  100x1  200x3 þ 75x3 2 ¼ ∂x3 ð2  x1  1:5x3 Þ2 T,P,x1

ð5Þ

Substituting Eqs. (4) and (5) in Eq. (2) yields: 200  300x3 þ 100x1 2 þ 300x1 x3  400x1 þ 100x3 2 ð2  x1  1:5x3 Þ2 ð6Þ 100  50x1 2 þ 100x1 x3  100x1  200x3 þ 75x3 2 x3 ð2  x1  1:5x3 Þ2

ΔH 123,2 ¼ ΔH 123  x1

At x1 ¼ 0.2, x2 ¼ 0.5, and x3 ¼ 1  x1  x2 ¼ 0.3, Eq. (6) yields: ΔH 123,2 ¼ 3:42 J=ðmol of component 2Þ As expected, this is the same answer that we obtained using Method 1. Part (b) To solve this part of the problem, we first recall our discussions in Part I to evaluate the thermodynamic behavior of a given system (the blender, in this case). A diagram of the system in question is shown in Fig. 1.

Solved Problems for Part II

n& A x1,A = 0.333;

643

Boundary: open, diathermal, and rigid

x2,A = 0.667; x3,A = 0

n&C x1,C = 0.333 x2,C = 0.333

Blender

x3,C = 0.333

n&B x1,B = 0.333

Q

x2,B = 0 x3,B = 0.667

Fig. 1

We first write the First Law of Thermodynamics for the blender (chosen as the system), which is surrounded by a boundary that is open, diathermal, and rigid (see the dashed boundary in Fig. 1). Specifically: dU ¼ δW þ δQ þ

X

H in δnin 

X

H out δnout

ð7Þ

Because the boundary is rigid (δW ¼ 0) and the blender operates at steady state dU ( dt ¼ 0), Eq. (7) reduces to: dN dN dN 0 ¼ Q_ þ H A A þ H B B þ H C C dt dt dt

ð8Þ

Although Eqs. (7) and (8) look very similar to those that we encountered many times in Part I, it is imperative to recognize that each stream that appears in these equations involves a mixture. Given the steady-state operation of the blender, we know that there is no mole accumulation in the blender. Carrying out a mass balance for the overall system, it follows that: dN A dN B dN þ ¼ C dt dt dt Similarly, carrying out a mole balance for each of the components in the system (i.e., components 1, 2, and 3), it follows that:

644

Solved Problems for Part II

2

dN A dN dN ¼2 B¼ C dt dt dt

ð9Þ

Using Eq. (9) in Eq. (8), we obtain:     1 dN C 1 dN C dN _ 0 ¼ Q þ HA  þ HB  þ HC C 2 dt 2 dt dt     1 dN C 1 dN C dN Q_ ¼ H A þ HB  HC C 2 dt 2 dt dt   dN H þ HB Q_ ¼  C H C  A dt 2

ð10Þ

Next, we can calculate the enthalpies of the three streams (each a mixture) in terms of their pure component enthalpies and the enthalpy of mixing. Specifically: H A ¼ ΔH 123 ðxA,1 , xA,2 Þ þ

X xi,A H i ¼ ΔH 123,A þ xA,1 H 1 þ xA,2 H 2 þ xA,3 H 3 i

X H B ¼ ΔH 123 ðxB,1 , xB,2 Þ þ xi,B H i ¼ ΔH 123,B þ xB,1 H 1 þ xB,2 H 2 þ xB,3 H 3 i

X H C ¼ ΔH 123 ðxC,1 , xC,2 Þ þ xi,C H i ¼ ΔH 123,C þ xC,1 H 1 þ xC,2 H 2 þ xC,3 H 3 i

ð11Þ Combining Eqs. (10) and (11), we obtain: 0

ΔH 123,C þ xC,1 H 1 þ xC,2 H 2 þ xC,3 H 3

1

B C dN B  ΔH 123,A  xA,1 H 1  xA,2 H 2  xA,3 H 3 C Q_ ¼  C B 2 2 2 2 C C dt B @ A ΔH 123,B xB,1 H 1 xB,2 H 2 xB,3 H 3     2 2 2 2 0 ΔH 123,A ΔH 123,B x H x H 1  þ xC,1 H 1  A,1 1  B,1 1 ΔH 123,C  _Q 2 2 2 2 C B ¼@ A dN C =dt xA,2 H 2 xB,2 H 2 xA,3 H 3 xB,3 H 3 þxC,2 H 2   þ xC,3 H 3   2 2 2 2   x þ xB,1  x þ xB,2  Q_ H 1 þ xC,2  A,2 H2 ¼ ΔH 123,C þ xC,1  A,1 2 2 ðdN C =dt Þ ð12Þ   xA,3 þ xB,3 ΔH 123,A þ ΔH 123,B H3  þ xC,3  2 2

Solved Problems for Part II

645

Using Eq. (1), we can calculate the enthalpy of mixing for each of the three species: ΔH 123,A ðxA,1 , xA,2 Þ ¼ 19:0476 J=mol ΔH 123,B ðxB,1 , xB,2 Þ ¼ 13:3333 J=mol

ð13Þ

ΔH 123,C ðxC,1 , xC,2 Þ ¼ 16:6667 J=mol Combining the results in Eq. (13) with Eq. (12), we obtain the heat duty required per mole of ternary product: Heat=mole of ternary product ¼

Q_ ¼ 4:05 J=ðmol of CÞ ðdN C =dt Þ

ð14Þ

646

Solved Problems for Part II

Problem 12 Problem 9.2 in Tester and Modell We are faced with the problem of diluting a 90 wt % H2SO4 solution with water in the following manner. A tank contains 500 kg of pure water at 298 K; it is equipped with a cooling device to remove any heat of mixing. This cooling device operates with a boiling refrigerant reflux condenser system to maintain the temperature at 298 K. Because of the peculiarities of the system, the rate of heat transfer (W/m2) must be constant. We wish to add 1500 kg of acid solution (at a variable rate) in 1 h. The acid is initially at 298 K. Enthalpy data are provided in the table below. (a) Plot the heat of solution (kJ/kg solution) versus weight fraction H2SO4 with the reference states as pure water and pure H2SO4, liquid, at 298 K. (b) What is the total heat transferred in the dilution process described? (c) Derive a differential equation to express the mass flow of 90 wt % acid, kg/min, as a function of the acid concentration in the solution. (d) Using the result from Part (c), determine the mass flow of 90 wt % acid when the overall tank liquid is 64.5 wt % acid. 





xH2 SO4

H  H0 (J/Mol)

0.00 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

0 183 1228 2428 4187 6071 7997 10,340 12,810 16,250 20,310 23,990 26,380 28,010 29,350 30,480 31,360 32,240 32,950 33,700 34,440

H2 O

H  H0 (J/Mol)

0 17,290 32,360 38,980 46,850 53,090 58,490 63,350 67,660 72,180 76,660 79,720 81,770 82,650 83,360 83,850 84,150 84,300 84,380 84,460 84,570

 H2 SO4

Solved Problems for Part II

647

Solution to Problem 12 Solution Strategy For this solution, the subscript “w” denotes water and the subscript “a” denotes sulfuric acid. Part (a) To begin solving this problem, we would like to derive an expression for the heat (or enthalpy) of mixing of the solution on a mass basis as a function of the weight fraction of sulfuric acid. In general, as discussed in Part II, the enthalpy of mixing of a solution is given by: ΔH mix ¼ H 

X

þ

ð1Þ

xi H i

i

In Eq. (1), H is the weighted sum of the partial molar enthalpies H i , that is: H¼

X

xi H i:

i þ

xi is the mole fraction of component i, and H i is the partial molar enthalpy of component i in the reference state (+). Because the solution consists of sulfuric acid and water, Eq. (1) combined with the last equation yields:     þ þ ΔH mix ¼ xa H a þ xw H w  xa H a þ xw H w     þ þ ¼ xa H a  H a þ xw H w  H w

ð2Þ

þ

In Eq. (2), the reference state of water is pure water, H w ¼ H0w, and the reference state of sulfuric acid is pure sulfuric acid. Comparing the reference state for sulfuric acid in Eq. (2) with the data provided in the table in the Problem Statement, we recognize that the table entries for sulfuric acid are defined with respect to an infiniteþ dilution reference state, that is, H a 6¼ H0a! Therefore, we cannot simply use the table for this column. Instead, we can define both sulfuric acid enthalpies in terms of this reference state:       þ 0 þ 0 xa H a  H a ¼ xa H a  H a  H a  H a Using Eq. (3) in Eq. (2) yields:

ð3Þ

648

Solved Problems for Part II

ΔH mix ¼ xa



     0 þ 0 þ H a  H a  H a  H a þ xw H w  H w

ð4Þ

The mole fractions are related by: xw ¼ 1  xa , Using the last equation in Eq. (4), we obtain: ΔH mix ¼ xa

      0 þ 0 þ H a  H a  H a  H a þ ð1  xa Þ H w  H w

ð5Þ

Equation (5) is written on a molar basis. However, we need to express everything on a mass basis. The simplest way to do this is to recognize that the molecular weight of 1 mole of solution is given by:     kg kg þ ð1  xa Þ 0:018 M sol ¼ xa M a þ xw M w ¼ xa 0:098 mol mol

ð6Þ

and to divide Eq. (5) by Msol given in Eq. (6). This yields: e mix ¼ ΔH

h      i 1 0 þ 0 þ xa H a  H a  H a  H a þ ð 1  xa Þ H w  H w M sol

ð7Þ

e mix is the enthalpy of mixing on a mass basis. Finally, we can determine where ΔH þ 0 þ H a  H a from the table in the Problem Statement. We recognize that H a is H when xa ¼ 1.00, and therefore, the table shows that: þ

0

H a  H a ¼ 8:457
104 J=mol Using the last result in Eq. (7) yields: e mix ¼ ΔH

1 M sol h     i  0 þ
xa H a  H a  8:457
104 J=mol þ ð1  xa Þ H w  H w ð8Þ

To create the required plot, we also need to determine the weight fraction corresponding to each value of xa. Using Eq. (6), this can be readily done as follows: wa ¼

0:098xa 0:098xa ¼ 0:098xa þ 0:018ð1  xa Þ 0:08xa þ 0:018

ð9Þ

Solved Problems for Part II

649

and ww ¼ 1  wa

ð10Þ

Alternatively, we can work on a term-by-term basis and divide each term by its appropriate molar weight and then multiply the appropriate weight fractions, which are given in Eqs. (9) and (10). After some simple algebra, we obtain:  e mix ¼ ΔH

wa

   0 þ 0 Ha  Ha  Ha  Ha 0:098 kg=mol

þ

  þ ð1  w a Þ H w  H w 0:018 kg=mol

ð11Þ

where the tilde indicates the mass basis of the enthalpy of mixing. Analyzing the data in the table using Eq. (11) and plotting them in a spreadsheet, we obtain the following data table and plot (Table 1 and Fig. 1 below). Note that the numerical answers obtained using either Eq. (8) or Eq. (11) will be identical.

Table 1 Data to calculate the heat of mixing in Part (a)

 xa 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  H0

J/Mol 0 183 1228 2428 4187 6071 7997 10,340 12,810 16,250 20,310 23,990 26,380 28,010 29,350 30,480 31,360 32,240 32,950 33,700 34,440

 w

  0 HH J/Mol 0 17,290 32,360 38,980 46,850 53,090 58,490 63,350 67,660 72,180 76,660 79,720 81,770 82,650 83,360 83,850 84,150 84,300 84,380 84,460 84,570

a

wa 0.000 0.223 0.377 0.490 0.576 0.645 0.700 0.746 0.784 0.817 0.845 0.869 0.891 0.910 0.927 0.942 0.956 0.969 0.980 0.990 1.000

e mix ΔH kJ/kg 0.00 160.81 243.32 296.74 320.40 326.93 319.57 307.57 289.00 268.76 243.28 217.15 185.33 157.88 130.43 104.62 80.59 58.90 38.51 19.04 0.00

650

Solved Problems for Part II

Heat of Mixing, ∆Hmix (kJ/kg)

0 -50

0.0

0.2

0.4

0.6

0.8

1.0

-100 -150 -200 -250 -300 -350

Weight Fraction of Sulfuric Acid, wa

Fig. 1

Part (b) Next, we examine the process of adding sulfuric acid to the water, considering the complete solution to be our system. This system is clearly open, but it is also diathermal and movable, because the volume of the system is changing, although the pressure is (effectively) constant. Because we want to calculate heat transfer in the system, we begin with the First Law of Thermodynamics for an open system given by: dU ¼ δQ þ δW þ

X X H in δnin  H out δnout in

ð12Þ

out

Because there is only one inlet stream and no outlet streams, it follows that δnin ¼ dN, and δnout ¼ 0. The work term δW in Eq. (12) can be written as PdV . However, because the pressure is constant, it follows that: δW ¼ PdV ¼ dðPV Þ Combining these results, we can express Eq. (12) as follows: dU þ d ðPV Þ ¼ δQ þ H in dN

ð13Þ

The left-hand side of Eq. (13) is equal to the differential enthalpy, and therefore, Eq. (13) can be expressed as follows:

Solved Problems for Part II

651

dH ¼ δQ þ H in dN

ð14Þ

Integrating Eq. (14) from the initial (i) to the final ( f ) states yields:   H f  H i ¼ Q þ H in N f  N i

ð15Þ

Expressing the left-hand side of Eq. (15) on a molar basis we obtain:   N f H f  N i H i ¼ Q þ H in N f  N i

ð16Þ

Examining the terms in Eq. (16), we recognize that Ni ¼ Nwi is the number of moles of water initially charged and Hi ¼ Hw is the enthalpy of pure water. Similarly, Nf ¼ Nin + Nwi is the final number of moles of solution. The expression for Hf is related to the enthalpy of mixing in the final state. Specifically:   þ þ ΔH mix,f ¼ H f  xwf H w þ xaf H a

ð17aÞ

which we can rewrite as follows:   þ þ H f ¼ ΔH mix,f þ xwf H w þ xaf H a

ð17bÞ

We can also describe the enthalpy of the incoming solution using an expression similar to Eq. (17b), that is:   þ þ H in ¼ ΔH mix,in þ xw,in H w þ xa,in H a

ð18Þ

In Eq. (17b) and Eq. (18), we have defined Hf and Hin relative to a pair of þ þ reference states H w and H a . For convenience, we choose these reference states to be the pure component reference states at temperature T and pressure P of the þ þ solution. As a result, we have H w ¼ H 0w and H a ¼ H 0a, where H 0w  pure component 0 enthalpy of water and H a  pure component enthalpy of the acid. This allows us to rewrite Eqs. (17b) and (18) as follows:   H f ¼ ΔH mix,f þ xwf H 0w þ xaf H 0a   H in ¼ ΔH mix,in þ xw,in H 0w þ xa,in H 0a Using Eqs. (19) and (20) in Eq. (16), we obtain:    N f ΔH mix,f þ xwf H 0w þ xaf H 0a  N wi H 0w    ¼ Q þ N in ΔH mix,in þ xw,in H 0w þ xa,in H 0a

ð19Þ ð20Þ

652

Solved Problems for Part II

which we can rearrange as an expression for Q, that is:    Q ¼ N f ΔH mix,f þ xwf H 0w þ xaf H 0a  N wi H 0w     N in ΔH mix,in þ xw,in H 0w þ xa,in H 0a     ¼ N f ΔH mix,f þ N f xwf  N wi  N in xw,in H 0w þ N f xaf  N in xa,in H 0a  N in ΔH mix,in

ð21Þ

Equation (21) indicates that the coefficients of H 0w and H 0a are both equal to zero. This important result follows because we selected the pure component reference states for all our calculations. Indeed: 

 N f xwf  N wi  N in xw,in ¼ N wf  N wi  N w,in ¼ 0   N f xaf  N in xa,in ¼ N af  N a,in ¼ 0

and therefore, Eq. (21) reduces to: Q ¼ N f ΔH mix,f  N in ΔH mix,in

ð22Þ

Before we can determine the heat transfer using Eq. (22) and Table 1, we recognize that we again need to convert to a mass basis. Doing so yields: Q ¼ N f M sol,f

ΔH mix,f ΔH mix,in e mix,f  min ΔH e mix,in  N in M sol,in ¼ m f ΔH M sol,f M sol,in

ð23Þ

Using the data provided in the Problem Statement, we see that in Eq. (23), mw ¼ e mix,in, we read 500 kg, min ¼ 1500 kg, and mf ¼ mw + min ¼ 2000 kg. To determine ΔH off the plot in Fig. 1 at wa ¼ 0.9. Specifically, we can fit the data with a fourth-order e mix,in ðwa ¼ 0:9Þ. This polynomial and then use the resulting equation to calculate ΔH yields: e mix,in  175 kJ=kg ΔH e mix,f , we have to determine the weight fraction of acid in In order to calculate ΔH the final solution, which is given by: wa ¼

0:90ð1500 kgÞ ¼ 0:675 1500 kg þ 500 kg

from which we can find from Fig. 1 that: e mix,f  325 kJ=kg ΔH

Solved Problems for Part II

653

Alternatively, we can obtain estimates via linear interpolation of the relevant values in Table 1, finding that: 0:9  0:891 0:910  0:891
ð157:88 kJ=kg þ 185:33 kJ=kgÞ ¼ 172:33 kJ=kg

ð24Þ

0:675  0:645 0:700  0:645
ð319:57 kJ=kg þ 326:93 kJ=kgÞ ¼ 322:92 kJ=kg

ð25Þ

e mix,in ¼ 185:33 kJ=kg þ ΔH

e mix,f ¼ 326:93 kJ=kg þ ΔH

Substitution of the results in Eqs. (24) and (25) in Eq. (23) yields:     kJ kJ  ð1500 kgÞ 175 ¼ 388 MJ Q ¼ ð2000 kgÞ 325 kg kg     kJ kJ Q ¼ ð2000 kgÞ 322:92  ð1500 kgÞ 172:33 ¼ 387 MJ kg kg Part (c) The most logical way to begin our derivation of the differential equation describing the mass flow as a function of acid concentration is to let m total mass of the liquid in the tank and use the differential relationship in Eq. (14) given by: dH ¼ δQ þ H in dN

ð26Þ

Because the Problem Statement requests a mass flow rate, rather than a molar flow rate, it is necessary to describe the respective amounts of water and acid in the tank in terms of mass (m) and mass fractions (w), rather than in terms of moles (N ) and mole fractions (x). With this need in mind, Eq. (26) can be expressed as follows: e in dm dH ¼ δQ þ H

ð27Þ

e in represents the enthalpy per kg of solution added to the tank. According to where H Postulate I, H for this binary (n ¼ 2) system containing water and acid depends on n + 2 ¼ 2 + 2 ¼ 4 independent variables. Because we need to work in mass units, we will choose H to be a function of T, P, mw, and ma. In terms of these four variables, it follows that:

654

Solved Problems for Part II

    ∂H ∂H e e  w dmw þ H  a dma dH ¼ dT þ dP þ H ∂T P,ma ,mw ∂P T,ma ,mw

ð28Þ

e  w represents the change in the enthalpy of the In Eq. (28), we recognize that H solution resulting from the addition of 1 kg of water at constant T and P. Note that e  w is not equivalent to H w  partial molar enthalpy of water in the solution. Setting H dT ¼ dP ¼ 0 in Eq. (28), we obtain: e e  w dmw þ H  a dma dH ¼ H

ð29Þ

Equation (20) taken on a mass basis becomes:   e in ¼ ΔH e 0w þ wa,in H e 0a e mix,in þ ww,in H H

ð30Þ

We can now equate Eq. (27) to Eq. (29) to obtain: e e  w dmw þ H  a dma e in dm ¼ H δQ þ H

ð31Þ

Substituting Eq. (30) in Eq. (31) yields:     e e e mix,in dm þ ww,in H  a dma  w dmw þ H e 0w þ wa,in H e 0a dm ¼ H δQ þ ΔH

ð32Þ

Writing the species balance separately for the water and the acid, we obtain: wa,in dm ¼ dma

ð33Þ

ww,in dm ¼ dmw

ð34Þ

and

Equations (33) and (34) allow us to express Eq. (32) as follows: e e  w ww,in dm þ H  a wa,in dm ¼ δQ þ ΔH e mix,in dm þ H e 0a wa,in dm þ H e 0w ww,in dm H

ð35Þ

We can rearrange Eq. (35) to obtain:     δQ e e a  H w  H e 0w ww,in þ H e 0a wa,in  ΔH e mix,in ¼ H dm

ð36Þ

We can rearrange Eq. (36) and take the derivative of both sides with respect to time to obtain:

Solved Problems for Part II

dm ¼ dt



655

   1 δQ e e w  H a  H e 0w ww,in þ H e 0a wa,in  ΔH e mix,in H dt

ð37Þ

Part (d) After deriving the differential equation in Eq. (37), we would like to evaluate it numerically for a specific solution composition. To this end, we only need to evaluate the individual terms in Eq. (37). The simplest term is δQ/dt, and because it is constant, we obtain: δQ ΔQ 387 MJ 387000 kJ ¼ ¼ ¼ ¼ 107:5 kW dt Δt 1h ð60 min Þð60 sÞ Based on our knowledge of the incoming solution, we know that ww, in ¼ 0.1 and e mix,in ¼ 172 kJ=kg. Next, wa, in ¼ 0.9. In addition, from Eq. (24), we know that ΔH 0 e e w  H a  H e w and H e 0a , which should be we need to evaluate the enthalpy changes H determined at a weight fraction of wa ¼ 0.645. If we convert mass fractions to mole fractions, the two required enthalpy changes can be calculated from the table provided in the Problem Statement and from the results in Part (a), respectively. We can utilize the table by using the relation:   0 e w  H e 0w ¼ H w  H w H 18 g=mol

ð38Þ

followed by converting wa to xa as follows:  xa ¼ 

wa 98 g=mol

wa 98 g=mol



þ





1wa 18 g=mol



ð39Þ

Substituting wa ¼ 0.645 in Eq. (39) yields xa ¼ 0.25. From the table in the Problem Statement, it follows that: 

H  H0

 wxa ¼0:25

¼ 6:07 kJ=mol

  e w  H e 0w ¼ 337 kJ=kg . To Substituting the last result in Eq. (38) yields H   e a  H e 0a , we need to revisit Part (a), including using the data provided in evaluate H the Problem Statement. We can begin with: 

 0 e a  H e 0a ¼ H a  H a H 98 g=mol

  0 and then determine H a  H a . To this end, examining Eq. (3), we obtain:

ð40Þ

656

Solved Problems for Part II



þ

Ha  Ha



    0 þ 0 ¼ Ha  Ha  Ha  Ha

  0 We recognize that the  left-handside of Eq. (3) is equivalent to H a  H a : From þ

0

Part (a), we know that H a  H a ¼ 84:57 kJ=mol. For xa ¼ 0.25, we can use the   0 table in the Problem Statement to find that H a  H a ¼ 53:09 kJ=mol. Substitut  ing the above values in Eq. (3) and equating to H a  H 0a yields:   H a  H 0a ¼ 53:09 kJ=mol  84:57 kJ=mol ¼ 31:48 kJ=mol Next, we can convert the last result to a mass basis by using Eq. (40). This yields: 

 e a  H e 0a ¼ 31:48 kJ=mol ¼ 321 kJ=mol H 98 g=mol

Finally, we can substitute the mass basis enthalpies above in Eq. (37) to obtain: ð107:5 kJ=sÞ dm ¼ ¼ 0:714 kg=s dt ð337 kJ=kgÞð0:1Þ þ ð321 kJ=kgÞð0:9Þ—172 kJ=kg ¼ 42:8 kg= min

Solved Problems for Part II

657

Problem 13 Problem 9.23 in Tester and Modell A stream composed of a binary mixture of benzene and cyclohexane that contains 0.5 mole fraction benzene is separated into two product streams. One stream contains a 0.98 mole fraction of benzene and the second contains a 0.90 mole fraction of cyclohexane. The separation process is performed at a constant temperature of 300 K and a constant pressure of 1 bar under steady-state conditions. You can assume that the activity coefficient for benzene, γB, is given by: h i γB ¼ exp 0:56ð1  xB Þ2 where xB is the mole fraction of benzene in the mixture. What is the minimum work required to perform the separation?

Solution to Problem 13 Solution Strategy This problem deals with separating a mixture stream of a given composition into two mixed streams of different compositions. The only information given to us about the process is that it is carried out under isothermal and isobaric conditions. Using this information, we are asked to calculate the minimum work required to perform this separation process. Because we have been asked to calculate work, it is clear that we will have to carry out a First Law of Thermodynamics analysis of the system. Further, because we have been asked to calculate the minimum work, it is very likely that we will also have to use the Second Law of Thermodynamics to relate the heat interaction in terms of the properties of the inlet and the outlet streams. In addition, we are dealing with mixtures, and therefore, we anticipate that mixing free energies will be useful. In summary, the key concepts to be used in the solution of this problem include: 1. First Law of Thermodynamics analysis 2. Second Law of Thermodynamics analysis 3. Mixing energies for a binary mixture using activity coefficient models

Selection of System and Boundaries As discussed in Part I, we begin solving the problem by defining our system and its boundaries. As shown in Fig. 1, our system consists of a black box in which the

658

Solved Problems for Part II

separation process is carried out. Next, we proceed to determine the characteristics of the boundary of the system (see Fig. 1).

xB = 0.98 nout,1

xB,in = 0.50 nin Black Box

xB = 0.10 nout,2 Fig. 1

1. Permeable or impermeable? The mixture stream to be separated enters our system (denoted as Black Box in Fig. 1). Therefore, the system boundary is permeable, and the system is open. The conditions of the inlet stream are given by: Pin ¼ 1 bar ¼ 105 Pa T in ¼ 300 K :

nin ¼ ? The conditions of the two outlet streams are given by: Pin ¼ 1 bar ¼ 105 Pa T in ¼ 300 K :

nout,1 ¼ ? xB,out,1 ¼ 0:98 :

nout,2 ¼ ? xB,out,2 ¼ 0:10 2. Rigid or Movable? There are no movable parts in the system boundary, and therefore, it is rigid. Due to the rigidity of the boundary, there is no P-V work interaction between the system and its environment. However, we have been told that external work is done to perform this operation. Therefore: :

δ W 6¼ 0 ¼ ?

Solved Problems for Part II

659

3. Adiabatic or Diathermal? The Problem Statement provides no information about this property of the boundary, and therefore, we can assume that there is some heat interaction between the system and the surroundings. Moreover, we are told that the process is carried out isothermally, which provides an additional reason to assume that heat is supplied to, taken away from, taken away from the system in order to maintain the process isothermal. Therefore, :

δ Q6¼ 0 ¼ ? Next, we can use the information above to write the First Law of Thermodynamics for our open system. Specifically: : : : : : dE dU ¼ ¼ δ Q þδ W þH in δnin  H out,1 δnout,1  H out,2 δnout,2 dt dt

ð1Þ

In Eq. (1), we assumed that kinetic energy effects and potential energy effects are negligible, such that E ¼ U. Further, assuming that the process is carried out at steady state, the first and second terms in Eq. (1) are both zero. We can then integrate Eq. (1) to obtain: :

:

:

:

:

Q þ W þH in nin  H out,1 nout,1  H out,2 nout,2 ¼ 0

ð2Þ

where in the equations above, the dots in Q, W, nin, nout,1, and nout,2 denote time derivatives. We recognize that the flow rates of the different inlet and outlet streams are not independent of each other, because they have to satisfy the mass balance for each of the two species comprising the system. Below, we seek to relate the different flow rates. The species mass balance for benzene can be written as follows: :

:

:

xB,in nin ¼ xB,out,1 nout,1 þ xB,out,2 nout,2 : : : ) 0:50nin ¼ 0:98nout,1 þ 0:10nout,2 :

:

:

ð3Þ

) nin ¼ 1:96nout,1 þ 0:20nout,2 Similarly, the species mass balance for cyclohexane yields: :

:

:

xC,in nin ¼ xC,out,1 nout,1 þ xC,out,2 nout,2 :

:

:

) 0:50nin ¼ 0:02nout,1 þ 0:90nout,2 : : : ) nin ¼ 0:04nout,1 þ 1:80nout,2 Substituting Eq. (3) in Eq. (4) yields:

ð4Þ

660

Solved Problems for Part II :

:

:

:

1:96nout,1 þ 0:20nout,2 ¼ 0:04nout,1 þ 1:80nout,2 :

:

ð5Þ

) 1:92nout,1 ¼ 1:60nout,2 :

:

) nout,1 ¼ 5=6nout,2 Substituting Eq. (5) in Eq. (3) yields: :

:

:

:

nin ¼ 1:96ð5=6Þnout,2 þ 0:20nout,2 ¼ ð11=6Þnout,2

ð6Þ

Using Eqs. (3), (5), and (6), we can relate the two outlet flow rates to the inlet flow rate as follows: :

:

:

:

nout,2 ¼ ð6=11Þnin

ð7Þ

nout,1 ¼ ð5=11Þnin

ð8Þ

Substituting Eqs. (7) and (8) in Eq. (2) yields: :

:

:

:

:

Q þ W þH in nin  H out,1 ð5=11Þnin  H out,2 ð6=11Þnin ¼ 0 :

:

:

)W ¼  Q þnin ðH out,1 ð5=11Þ þ H out,2 ð6=11Þ  H in Þ

ð9Þ

Our next task is to calculate the heat interaction between the system and its : environment that would result in the smallest value of W . The work requirement would be minimum when the operation is carried out reversibly such that the cumulative entropy change of the system and the surroundings would amount to 0. Therefore, we write an entropy balance on the system when the process is carried out reversibly so that we can relate the heat interaction to the entropy changes of the streams. Specifically: dS δQ_ þ Sgeneration ¼ Sin δn_ in  Sout,1 δn_ out,1  Sout,2 δn_ out,2 þ T dt Q_ ) Sin n_ in  Sout,1 ð5=11Þn_ in  Sout,2 ð6=11Þn_ in þ ¼ 0 T ) Q_ ¼ T n_ in ðð5=11ÞSout,1 þ ð6=11ÞSout,2  Sin Þ

ð10Þ

where again we have assumed that the process is carried out reversibly and is at steady state. In addition, we have used Eqs. (7) and (8) to eliminate the outlet flow rates in terms of the inlet flow rate. Substituting Eq. (10) in Eq. (9) yields:

Solved Problems for Part II :

661

:

:

W ¼ Tnin ðð5=11ÞSout,1 þ ð6=11ÞSout,2  Sin Þ þ nin ðH out,1 ð5=11Þ þ H out,2 ð6=11Þ  H in Þ :

:

:

:

)W ¼ nin ðð5=11ÞðH out,1  TSout,1 Þ þ ð6=11ÞðH out,2  TSout,2 Þ  ðH in  TSin ÞÞ )W ¼ nin ðð5=11ÞGout,1 þ ð6=11ÞGout,2  Gin Þ ð11Þ

Equation (11) shows that the minimum work required to perform the separation process is given by the change in the Gibbs free energy of the streams. Accordingly, the last step required to solve this problem involves calculating the Gibbs free energies of the different streams. This can be done using the information about the activity coefficient model provided in the Problem Statement. In Part II, we showed that the excess Gibbs free energy can be related to the activity coefficient model using the following relationship: GEX ¼ RT

X N i ln γ i i

) dGEX ¼ RT

X ðN i d ln γ i þ ln γ i dN i Þ

ð12Þ

i

From Postulate 1, we know that any extensive property of a simple system can be written in terms of the masses (moles) of the different species and two other independent variables. If we take T and P as the two independent variables (recall that the Problem Statement indicates that T and P are held constant), we can write GEX as follows: GEX ¼ GEX ðT, P, N B , N C Þ  EX   EX  X∂GEX  ∂G ∂G ) dGEX ¼ dT þ dP þ ∂T P,N B ,N C ∂P T,N B ,N C ∂N i T,P,N i  EX   EX  X ∂G ∂G ) dGEX ¼ dT þ dP þ RT ln γ i dN i ∂T P,N B ,N C ∂P T,N B ,N C i

dN i j6¼i

ð13Þ Equating Eqs. (13) and (12), at constant temperature and pressure, yields: dGEX ¼ RT

X

ln γ i dN i ¼ RT

i

) RT

X N i d ln γ i ¼ 0 i

) RTN

X i

xi d ln γ i ¼ 0

X xi d ln γ i ¼ 0 ) i

X

ðN i d ln γ i þ ln γ i dN i Þ

i

ð14Þ

662

Solved Problems for Part II

Substituting the activity model given in the Problem Statement in Eq. (14) yields: xB d ln γ B þ xC d ln γ C ¼ 0  h i ) xB d ln exp 0:56ð1  xB Þ2 þ ð1  xB Þd ln γ C ¼ 0 ) xB  0:56  2ð1  xB Þð1ÞdxB þ ð1  xB Þd ln γ C ¼ 0 ) d ln γ C ¼ 1:12xB dxB lnðγ C xðB ) d ln γ C ¼ 1:12xB dxB

ð15Þ

xB ¼0

ln γ C ¼0

  ) ln γ C  0 ¼ 0:56 x2B  0 h i ) γ C ¼ exp 0:56ð1  xC Þ2 Substituting Eq. (15) in Eq. (12) yields:   GEX ¼RT ðN B ln γ B þ N C ln γ C Þ ¼ NRT xB 0:56ð1  xB Þ2 þ xC 0:56ð1  xC Þ2   ¼0:56NRT x2C xB þ x2B xC ¼0:56NRTxB xC ðxC þ xB Þ ¼0:56NRTxB xC ð16Þ Using the definition of GEX , we can find the Gibbs free energy of any stream as follows: GEX ¼ ΔGEX ¼ ΔG  ΔGID ¼ G  G0  ΔGID X  G ¼G0 þ ΔGID þ GEX ¼ N i G0i þ N i RT ln xi þ 0:56NRTxB xC i

      ¼NRT xB G0B =RT þ xC G0C =RT þ xB ln xB þ xC ln xC þ 0:56xB xC

ð17Þ

where G0B and G0C are the molar Gibbs free energies of pure benzene and cyclohexane at T ¼ 300 K and P ¼ 1 bar, respectively. Substituting Eq. (17) in Eq. (11) yields:

Solved Problems for Part II

663

! 1 _ xB,out,1 ðG0B =RT ÞþxC,out,1 ðG0C =RT ÞþxB,out,1 ln xB,out,1 þxC,out,1 ln xC,out,1 ð 5=11 Þ B C B C þ0:56xB,out ,1 xC ,out,1 B C ! B C B C : 0 0 G G ln þ ln x =RT þ x =RT þ x x x x ð Þ ð Þ B,out ,2 B C ,out ,2 C B,out,2 B,out ,2 C ,out,2 C ,out,2 C þ ð 6=11 Þ W ¼ n_ in RT B B C þ0:56xB,out,2 xC ,out,2 B C B C ! B C @ A G0B =RT ÞþxC , in ðG0C =RT ÞþxB, in ln xB, in þxC , in ln xC , in x ð B , in  þ0:56xB, in xC , in 0

0

ð5=11Þð0:98 ln 0:98 þ 0:02 ln 0:02 þ 0:56  0:98  0:02Þ

1

B C ¼ n_ in RT @ þð6=11Þð0:10 ln 0:10 þ 0:90 ln 0:90 þ 0:56  0:10  0:90Þ A ð0:50 ln 0:50 þ 0:50 ln 0:50 þ 0:56  0:50  0:50Þ ¼ n_ in RT ðð5=11Þð0:087Þ þ ð6=11Þð0:275Þ  ð0:553ÞÞ ¼ 0:364n_ in RT ¼ 907n_ in ð18Þ Note that in Eq. (18), all the standard-state contributions cancelled out due to the mass balance constraints on the system (see Eqs. (3)–(8)). From Eq. (18), we conclude that the minimum work required to separate the stream is 907 J/mole flow rate of the inlet stream.

664

Solved Problems for Part II

Problem 14 Problem 15.4 in Tester and Modell In a binary solution of two components, the eutectic point is the lowest freezing point of the mixture. It is less than the freezing point of either pure component. Assume that the liquid phase forms an ideal solution and that all solid phases are pure components (i.e., no mixed crystals form). The vapor phase forms an ideal gas mixture. Some data that may be of use are given below. Neglect any pressure effects. Assume that ΔH values do not vary with temperature.

Freezing point (K) ΔHvaporization (J/mol) ΔHsublimation (J/mol) ΔHfusion (J/mol)

Nitrogen

Oxygen

63.3 6000 6720 721

54.4 7490 7940 447

Estimate the eutectic point (i.e., composition and temperature) for a liquid-air mixture (O2 and N2).

Solution to Problem 14 Solution Strategy To solve this problem, we will utilize the general strategy presented in Part II to solve phase equilibria problems: 1. Draw the system and describe the boundaries 2. Use the Gibbs Phase Rule to determine how much information is required to find a unique solution based on the given boundaries 3. Draw a phase diagram to better understand the problem 4. Decide whether the integral approach or the differential approach is more appropriate to calculate the chemical potential equalities 5. Use the approach selected in 4 above to solve the problem

Draw the System and Describe the Boundaries As per the Problem Statement, the system boundaries are diathermal, movable, and permeable. The only constraint mentioned in the Problem Statement is that both solid phases consist of pure component solids (rather than of a solid mixture). However, this does not prevent the pure component solid phases from coexisting with each other. Using this information, we can draw the diagram shown in Fig. 1.

Solved Problems for Part II

665

Liquid N2/O2 Mixture

Solid O2

Solid N2 Fig. 1

The existence of all three phases, as we show below using the Gibbs Phase Rule, can only occur at one temperature and composition for a given pressure, which is referred to as the Eutectic Point.

Use the Gibbs Phase Rule At the Eutectic Point, there are two components in three phases (pure solid O2, pure solid N2, and liquid mixture containing both O2 and N2). Because two of the solid phases are pure components – indicating that there is a barrier preventing O2 from entering the solid N2 phase and a barrier preventing N2 from entering the solid O2 phase – the system is not simple, and therefore, not all the conditions of thermodynamic equilibria apply. It then follows that we must use the Gibbs Phase Rule analysis for composite systems presented in Part II. We begin by determining the number of independent intensive variables needed to describe each phase (a simple system). From the Corollary to Postulate I, it follows that we need (n + 1) independent intensive variables to define each phase: Liquid Phase (α): n ¼ 2 (O2 and N2), n + 1 ¼ 2 + 1 ¼ 3 variables Solid O2 Phase (β): n ¼ 1 (O2 only), n + 1 ¼ 1 + 1 ¼ 2 variables Solid N2 Phase (γ): n ¼ 1 (N2 only), n + 1 ¼ 1 + 1 ¼ 2 variables Total number of independent intensive variables needed to characterize each phase ¼ 3 + 2 + 2 ¼ 7 variables. After we determine the total number of intensive variables needed to describe the three phases, we need to determine the number of constraints relating these intensive variables by identifying the conditions of thermodynamic equilibria that must be satisfied. We know that there are no adiabatic or rigid internal barriers, and therefore, thermal and mechanical equilibria are satisfied between the three phases. Specifically: Thermal Equilibrium (T.E.): Tα ¼ Tβ ¼ Tγ (π  1 ¼ 3  1 ¼ 2 constraints) Mechanical Equilibrium (M.E.): Pα ¼ Pβ ¼ Pγ (π  1 ¼ 3  1 ¼ 2 constraints)

666

Solved Problems for Part II

The composite system has internal barriers that are impermeable to certain species. Therefore, the diffusional equilibria conditions are different than those in a simple three-phase system. Specifically: Diffusional Equilibrium (D.E.) for O2: μαO2 ¼ μβO2 (1 constraint) Diffusional Equilibrium (D.E.) for N2: μαN 2 ¼ μγN 2 (1 constraint) Total number of constraints that apply ¼ 2 + 2 + 1 + 1 ¼ 6. The variance for a composite system is equal to the total number of independent intensive variables needed to characterize all the phases (seven in this case) minus the total number of constraints that apply (six in this case). Therefore, for the system under consideration, it follows that: L¼76¼1

ðaÞ

Therefore, at the eutectic point (three phases), the system is monovariant, indicating that only one variable needs to be specified to find a unique solution. If we specify the pressure, there exists only one temperature and liquid phase composition at which three phases are simultaneously present. At other liquid-solid equilibrium conditions, only one of the solid phases will be present. Using the same procedure as above, we find that five variables are needed to characterize each phase independently (three for the liquid phase and two for the solid phase). We also recognize that there are three constraints on the system (one for T.E., one for M.E., and one for D.E.). Therefore, the variance is given by: L¼53¼2

ðbÞ

Therefore, the Gibbs Phase Rule indicates that the system is divariant, indicating that if we specify the pressure, another variable would have to be specified to find a unique solution.

Draw a Schematic Phase Diagram To better understand the problem, we can draw a sketch of the phase diagram that represents our problem (see Fig. 2). We know that there are two liquid-solid equilibrium lines, whose temperatures vary with composition (see the left and the right curves in Fig. 2). At the pure component mole fractions (0 and 1), the liquidsolid equilibrium temperatures are equal to the pure component temperatures (63.3 K and 54.4 K). The temperatures decrease along the left and the right lines as the system evolves from the pure components to the eutectic composition due to the interactions between the two components.

Solved Problems for Part II

667

Fig. 2

Decide Whether the Integral Approach or the Differential Approach Is More Appropriate To decide which approach to use, it is useful to examine the information given in the Problem Statement. In order to use the integral approach, we would need the vapor pressures of the different phases in order to calculate the values of the fugacities. Instead, we are given several different enthalpies of phase change, which can be used in the differential approach. We may be concerned that no information is provided about the volume changes associated with the phase changes. However, these changes are related to pressure effects which the Problem Statement asks us to neglect. Accordingly, we will utilize the differential approach.

Solve the Problem To solve this problem, we need to find the point at which the liquid mixture is in equilibrium with both pure solid phases. To find this special point (the Eutectic), we will first calculate the two curves that describe each individual component in the liquid mixture being in equilibrium with that component in the corresponding solid phase (these correspond to the left and right curves shown in Fig. 2). Subsequently, we will calculate the point at which the two curves intersect, that is, the Eutectic point. At that intersection, both liquid components will be in equilibrium with their solid phases. Note that because the equilibrium curves are not necessarily continuous and differentiable, setting a derivate equal to zero and solving could lead to erroneous results!

668

Solved Problems for Part II

N2 Liquid/Solid Equilibrium Figure 3 shows the relevant phases in thermodynamic equilibrium:

Fig. 3

At equilibrium: TL ¼ TS ¼ T f

ð1Þ

P ¼P ¼P     μLN 2 T L , PL , xLN 2 ¼ μSN 2 T S , PS

ð2Þ

L

S

    L ¼> bf N 2 T f , P, xLN 2 ¼ f SN 2 T S , P

ð3Þ

    L d ln bf N 2 T f , P, xLN 2 ¼ d ln f SN 2 T S , P

ð4Þ

Therefore:

Expanding each fugacity term in Eq. (4) yields: 0

1 L ∂ ln bf N 2 @ A ∂T ¼

∂ ln ∂T

P,xLN

f SN 2

1 L ∂ ln bf N 2 A dT þ @ ∂P

0

1 L ∂ ln bf N 2 A dP þ @ ∂xN 2

0

!2 dT þ P

∂ ln ∂P

f SN 2

!

T,xLN

2

dP T

dxN 2 T,P

ð5Þ

Solved Problems for Part II

669

At constant pressure, Eq. (5) becomes: 0

1 L ∂ ln bf N 2 @ A ∂T

0

P,xLN 2

1 L ∂ ln bf N 2 A dT þ @ ∂xN 2

∂ ln f SN 2 ∂T

dxN 2 ¼

!

T,P

ð6Þ

dT P

Because 0

1 L ∂ ln bf N 2 @ A ∂T

"

L

¼

H N 2  H 0N 2 ðT Þ

"

#

#

RT 2

P,xLN 2

and ∂ ln f SN 2 ∂T

! ¼

H SN 2  H 0N 2 ðT Þ RT 2

P

it follows that: " 

# L H N 2  H 0N 2 ðT Þ RT 2

0

1 L ∂ ln bf N 2 A dT þ @ ∂xN 2

" dxN 2 ¼  T,P

H SN 2  H 0N 2 ðT Þ

#

RT 2

dT

ð7Þ

We are told that the liquid phase forms an ideal solution, and therefore: H N 2 ¼ H LN 2 ðT, PÞ

L

ð8Þ

bf L ¼ xN f L 2 N2 N2

ð9Þ

L

¼> ln bf N 2 ¼ ln xN 2 þ ln f LN 2 L

¼> d ln bf N 2 ¼ d ln xN 2 þ d ln f LN 2 1 1 0  0  L L   ∂ ln bf N 2 ∂ ln f N2 A ¼ ∂ð ln xN 2 Þ A ¼> @ þ@ ∂xN 2 ∂x ∂xN 2 N2 T,P T,P

Substituting Eqs. (8) and (10) in Eq. (7) yields:

¼ T,P

1 xN 2

ð10Þ

670

Solved Problems for Part II



" L # H N 2  H 0N 2 ðT Þ RT 2

" # H SN 2  H 0N 2 ðT Þ 1 dx ¼  dT þ dT xN 2 N 2 RT 2

ð11Þ

Rearranging Eq. (11) yields: " L # H N 2  H SN 2 1 dx ¼ dT ¼> xN 2 N 2 RT 2 ¼>

ΔH Nf 2 1 dxN 2 ¼ dT xN 2 RT 2

¼> d ln xN 2 ¼ 

ð12Þ

ð13Þ

ΔH Nf 2  1  d T R

ð14Þ

Because ΔH Nf 2 is not a function of temperature, we can integrate Eq. (14) from the freezing point of pure N2 (xN 2 ¼ 1) to the freezing temperature of a liquid mixture with N2 composition (xN 2 ). This yields: ð

xN 2

d ln xN 2

ΔH Nf 2 ¼ R

1

Tðf

d T Nf

¼> ln xN 2 ¼  ¼>

ΔH Nf 2 R

  1 T

2

!

1 1  T f T Nf 2

1 R 1 ¼ ln xN 2 þ f f Tf ΔH N 2 T N2

ð15Þ

Because ln xN 2 < 0, we have T1f > T1f or T Nf 2 > T f . N2

O2 Liquid/Solid Equilibrium Following a similar approach to model this solid/liquid equilibrium, we obtain: 1 R 1 ¼ ln xO2 þ f f Tf ΔH O2 T O2 As before, we have T Of 2 > T f .

ð16Þ

Solved Problems for Part II

671

At the eutectic point, the two equilibrium curves intersect, and we have: T f ¼ T E , xN 2 ¼ xEN 2 , xO2 ¼ xEO2 ¼ 1  xEN 2

ð17Þ

Using Eqs. (15) and (16), it follows that: 

¼> 

R 1 R 1 ln xN 2 þ f ¼  ln xO2 þ f f f ΔH N 2 T N2 ΔH O2 T O2

R 1 R 1 ln xN 2 þ f ¼  ln ð1  xN 2 Þ þ f ΔH Nf 2 T N2 ΔH Of 2 T O2 !

¼>

1 1  T Nf 2 T Of 2

"

ln xN 2 ln ð1  xN 2 Þ  ¼R ΔH Of 2 ΔH Nf 2

# ð18Þ

Using the values given in the Problem Statement in Eq. (18), we obtain:  3 E E ln 1  x ln x N 2 1 1 N2 5  ¼ 8:3144  447 63:3 54:4 721





2

ð19Þ

Solving Eq. (19), we obtain: xEN 2 ¼ 0:373 and xEO2 ¼ 0:627 Using these two mole fraction values in Eqs. (15) or (16), we can calculate TE: 1 1 8:314 1 ln 0:373 þ ¼ ¼ TE T f 721 63:3 or T E ¼ 36:8K

672

Solved Problems for Part II

Problem 15 Problem 15.13 in Tester and Modell The International Critical Tables, Vol. III, p. 313, lists the boiling points for mixtures of acetaldehyde and ethyl alcohol at various pressures. The data are summarized below for mixtures containing 80 mole % ethyl alcohol in the liquid. From these data, determine the molar enthalpy of vaporization of ethyl alcohol at 320.7 K, from   a liquid mixture containing 80% ethyl alcohol. [That is, what is V L H  H alcohol?] T (K)

P (N/m2)

Mole fraction ethyl alcohol in vapor

331.3 9.319
10 0.318 320.7 5.306
104 0.385 299.9 1.027
104 0.330 4

Note: The vapor phase may be considered to be an ideal gas mixture, but the liquid phase is a non-ideal solution.

Solution to Problem 15 Solution Strategy This problem asks us to evaluate partial molar enthalpies using phase equilibria data. From our knowledge of phase equilibria, we know that at thermodynamic equilibrium, the chemical potentials (and therefore the fugacities) of each of the components in each of the phases are equal. Furthermore, we know that the temperature derivatives of ln of the fugacities can be expressed in terms of the partial molar enthalpies. In addition, we know that the pressure derivatives of ln of the fugacities can be expressed in terms of the partial molar volumes. Therefore, it is convenient to solve this problem by writing the phase equilibria criteria in terms of the fugacities and then use the differential approach to phase equilibria to calculate the difference between the partial molar enthalpies of ethyl alcohol in the vapor phase and in the liquid phase. Solution Using the Differential Approach to Phase Equilibria For the binary mixture in liquid/vapor equilibrium, the conditions of thermodynamic equilibrium are given by: Thermal equilibrium:

Solved Problems for Part II

673

TV ¼ TL ¼ T

ð1Þ

PV ¼ PL ¼ P

ð2Þ

Mechanical equilibrium:

Diffusional equilibrium: μVE ¼ μLE

) bf E ¼ bf E

μVA ¼ μLA

) bf A ¼ bf A

V

L

ð3Þ

V

L

ð4Þ

where in Eqs. (1), (2), (3), and (4), the subscripts A and E denote acetaldehyde and ethyl alcohol, respectively, and the superscripts V and L denote the vapor and liquid phases, respectively. Because the problem asks us to evaluate the difference between the partial molar enthalpies in the vapor and the liquid phases for ethyl alcohol only, we only need to use Eqs. (1), (2), and (3). In order to relate Eq. (3) to the partial molar enthalpy, we take its total differential and simplify it using the information given in the Problem Statement. Specifically, we will eliminate any dependence on xE because the Problem Statement tells us that the liquid mole fraction is held constant at 0.8. It then follows that: ^f V ¼ ^f L ) ln ^f V ¼ ln ^f L ) d ln ^f V ¼ d ln ^f L E E E  E  E  E    ∂ ^V ∂ ^V ∂ ∂ ^L ln f E ln f E ln f E ) dT þ dP þ ln ^f EV dyE ¼ dT ∂T ∂P ∂yE ∂T P, y y ,T T, P P, x    ∂ ^L ∂ ln f E dP þ ln ^f EL dxE þ ∂P ∂x E x, T T , P  ΔH EV VV ∂ ΔH EL VL ) dT þ E dP þ ln ðyE PÞ dyE ¼ 1dT þ E dP 2 2 ∂yE RT RT RT RT T, P Ideal Gas Mixture

  V    ΔH EV  ΔH EL V E  V EL =P dP þ dy ¼ 0 ) dT þ yE = RT P E RT 2

However, because the same reference state is used for the two phases, it follows that: V

L

V

L

ΔH E  ΔH E ¼ H E  H E : V

L

In addition, because the pressure is not high, it is safe to assume that V E V E V L V and V E  V E  V E . Accordingly, the last equation becomes:

674

Solved Problems for Part II

 V   E  H EL  H V EV dy dP þ E ¼ 0 ) dT þ Ideal Gas Mixture 2 RT yE RT  V  E  H  EL RT=P  H dy dP þ E ¼ 0 dT þ Ideal Gas Mixture ) 2 yE RT RT    EL  H EV  H dP dyE þ dT þ ¼0 ) 2 P yE RT  V   EL  1  H E  H ) d þ d ln P þ d ln yE ¼ 0 T R  V   V    H EL  EL  1  H  H H d ln ðyE PÞ ¼ E d ) E þ d ln ðyE PÞ ¼ 0 ) T dð1=T Þ R R

ð5Þ

Using the data given in the Problem Statement, we can obtain a polynomial fit for ln(yEP) in terms of (1/T )  and then evaluate its derivative at the required condition  V L (T ¼ 320.7 K) to obtain H E  H E . From the data provided, the highest-order polynomial fit that we can obtain is a quadratic fit (we have only three data points). Although it may seem inappropriate to fit a quadratic equation when we have only three data points, it should be noted that the three points are quite close to each other (299.9–331.3 K), and therefore, fitting a quadratic equation is a reasonable approximation. Note that this is widely done in numerical schemes. For example, in finite differences, this approximation is used to obtain the three-point finite difference formula for the second derivative. Similarly, in numerical integration, this approximation is the basis for Simpson’s rule. The main difference between these algorithms and the approximation that we use in this problem is that our points are not equidistant. Performing a quadratic fit between ln(yEP) in terms of (1/T) (see Fig. 1), we obtain:

11

ln(yP)

10

9

y = -1.45E+07x 2 + 8.52E+04x - 1.15E+02 R² = 1.00E+00

y = -7093.7x + 31.845 R² = 0.9768

8 0.003

0.0031

0.0032

1/T (1/K)

Fig. 1

0.0033

0.0034

Solved Problems for Part II

ln ðyE PÞ ¼ 1:45
107
ð1=T Þ2 þ 8:52
104
ð1=T Þ  1:15
102

675

ð6Þ

Differentiating Eq. (6) with respect to 1/Tand evaluating it at T ¼ 320.7 K yields: ∂ ln ðyE PÞ ¼ 1:45
107
2
ð1=T Þ þ 8:52
104 ∂ð1=T Þ ∂ ln ðyE PÞ ) ¼ 2:90
107
2
ð1=320:7Þ þ 8:52
104 ¼ 5227:19 ∂ð1=T Þ T¼320:7

ð7Þ Substituting Eq. (7) in Eq. (5) yields:   V L HE  HE d ln ðyE PÞ ¼ 5227:19 ¼ R d ð1=T Þ

  V L ) H E  H E ¼ 5227:19
8:314 ¼ 43458:68 J=mol ¼ 43:46 kJ=mol

ð8Þ

  V L Instead of fitting a quadratic, if we had assumed that H E  H E was a constant and that ln(yEP) was  a linear function of (1/T), we would have obtained the V L following value for H E  H E : ln ðyE PÞ ¼ 7093:7
ð1=T Þ þ 31:845 d ln ðyE PÞ ) ¼ 7093:7 dð1=T Þ   V L HE  HE ¼ 7093:7 ) R   V L ) H E  H E ¼ 7093:7
8:314 ¼ 58977:02 J=mol   V L ) H E  H E ¼ 58:98 kJ=mol

ð9Þ

which is more than 30% higher than the value obtained using Eq. (8)! However, in the absence of error bars on the data points and additional data points, it is difficult to evaluate the statistical significance of this difference.

676

Solved Problems for Part II

Problem 16 Problem 16.7 in Tester and Modell Michael K. Jones, a close relative of Rocky and Rochelle and an avid inventor, claims to be able to produce diamonds from β-graphite at room temperature by a process involving the application of 37 kbar pressure. In view of the data shown below, are his claims to be taken seriously? Specific gravity, β  graphite ¼ 2:26 Specific gravity, diamond ¼ 3:51 C β‐graphite ¼ C diamond ΔG0298 ¼ 2870 J=g‐atom Both solids are incompressible and no solid solutions are formed.

Solution to Problem 16 Solution Strategy The strategy to solve this problem involves the following two steps: 1. Derive a thermodynamic criterion to determine whether or not the process in question is physically possible 2. Use the given information to determine if the actual process satisfies the criterion in 1 To carry out Step 1, we will use the techniques discussed in Part I. We recall that two thermodynamic laws (the First and the Second) govern all open systems: dU ¼ δW þ δQ þ

X

H in δnin 

X

H out δnout

ð1Þ

and dSuniverse 0

ð2Þ

For the process under consideration, we can draw the diagram shown in Fig. 1 to represent the inlets and outlets of the system. Using this diagram, we can rewrite Eq. (1) as follows:

Solved Problems for Part II

677 Diamond Produced

ND

Nβ0

Black Box

Nβ

β-Graphite Charged

Unreacted β-Graphite

W

Q

Fig. 1

in out out out out dU ¼ δW þ δQ þ H in β δnβ  H D δnD  H β δnβ ¼ 0

ð3Þ

where the differential change in the internal energy in Eq. (1) was set to zero because the system is at steady state. By integrating and cancelling out the enthalpies of the unreacted graphite, we obtain: W þ Q þ H β N 0β  H D N D  H β N β ¼ 0   ) W þ Q þ H β N 0β  H D N D  H β N 0β  N D ¼ 0 ðFrom a stoichiometric balanceÞ ) W þ Q  HDND þ HβND ¼ 0   ) Q ¼ W þ H β  H D N D ð4Þ After obtaining the last expression in Eq. (4) based on the First Law of Thermodynamics, we can begin working through the Second Law of Thermodynamics. To this end, we can expand Eq. (2) by splitting the entropy into two terms as follows: ΔSuniverse ¼ ΔSprocess þ ΔSenv

ð5Þ

We can then separately evaluate the two entropy terms in Eq. (5). For the entropy term associated with the process, we obtain: ΔSprocess ¼ Sout  Sin ¼ SD N D þ Sβ N β  Sβ N 0β   ¼ SD N D þ Sβ N 0β  N D  Sβ N 0β ðFrom a stoichiometric balanceÞ   ¼ N D SD  Sβ

ð6Þ

For the entropy term associated with the environment, we can show that: ΔSenv ¼ 

Q T env

ð7Þ

where Tenv is the temperature of the environment. Combining Eqs. (5), (6), and (7), we obtain:

678

Solved Problems for Part II

  Q ΔSuniverse ¼ N D SD  Sβ 

0 T env

ð8Þ

By rearranging Eq. (8), we obtain:   Q  T env N D SD  Sβ

ð9Þ

Combining Eqs. (4) and (9), we obtain:     W H D  H β N D  T D SD  Sβ N D

ð10Þ

If we assume that there is no shaft work in the system and use the definition for the Gibbs free energy, we can express Eq. (9) as follows:   0 GD  Gβ N D

ð11Þ

) 0 ΔGðT, PÞ

where T is the system temperature and P ¼ 37 kbar is the system pressure. Equation (11) provides us with a criterion that must be satisfied for the process to be possible. Next, we must take the information provided in the Problem Statement and evaluate the reaction free energy at the system temperature and pressure. This can be done as follows: 

 ∂Gi ¼ Vi ∂P T,N   ∂ΔG ) ¼ ΔV ∂P T,N 37 ð kbar

ΔG0 ðT,ð37 kbarÞ

dΔG ¼

) ΔG ðT, 1 barÞ

ΔVdP 1 bar

0

37 ð kbar

) ΔGðT, 37 kbarÞ  ΔGðT, 1 barÞ ¼



 V D  V β dP

1 bar 37 ð kbar

) ΔGðT, 37 kbarÞ  ΔG0298 ¼



 V D  V β dP

1 bar 37 ð kbar

) ΔGðT, 37 kbarÞ ¼ 1 bar



 V D  V β dP þ 2870 J=mol

ð12Þ

Solved Problems for Part II

679

Because the volumes can be approximated as being independent of pressure, the integral in Eq. (12) can be readily evaluated. Specifically:   ΔGðT, 37 kbarÞ ¼ V D  V β ð37 kbar  1 barÞ þ 2870 J=mol    12
106 12
106  ¼  37
108  1
105 þ 2870 1
3:51 1
2:26

ð13Þ

¼ 4126:J=mol < 0 Because the free energy of the reaction is negative, Eq. (11) is satisfied, demonstrating that the process is possible!

Other Possible Solution Strategies It is possible that considering an equilibrium reaction, one would decide to solve the problem by evaluating the equilibrium constant and showing that the conversion of the reaction is nonzero. If we proceed along those lines, we will realize that there is some sort of discrepancy between the equilibrium constant calculated using the standard Gibbs free energy of reaction and calculated using the fugacity of the reacting species. If we define our reference states as pure solids at the system temperature and 1 bar pressure, we can write the equilibrium constant as follows: 

ΔGðT, 1 barÞ K ¼ exp  RT



 ¼ exp 

 2870 ¼ 0:314 8:314
298

ð14Þ

However, if we calculate the equilibrium constant based on its definition in terms of the fugacities of the reacting species, we obtain: !1   f D ðT; 37 kbarÞ f D ðT; 1 barÞ 1 11 0 37 ð kbar 37 ð kbar   Bf ðT; 1 barÞexp Bf β ðT; 1 barÞexp ðV D =RT ÞdPC V β =RT dPC C C B D B C C B B 1 bar 1 bar C C B ¼B C C B B f β ðT; 1 barÞ C C B B f D ðT; 1 barÞ A A @ @

f β ðT; 37 kbarÞ K¼ f β ðT; 1 barÞ 0

680

Solved Problems for Part II

  V D  V β ð37 kbar  1 barÞ=RT     12
106 12
106  8 5  37
10  1
10 =ð8:314
298Þ ¼ exp 1
3:51 1
2:26     12
106 12
106  8 5 ¼ exp  37
10  1
10 =ð8:314
298Þ 1
3:51 1
2:26

K ¼ exp



¼0:059 ð15Þ which is different from the value obtained in Eq. (14). To explain this difference, let us invoke the Gibbs Phase Rule to see if we indeed have two phases at the given T and P condition. In the last analysis, we have assumed that we have two purecomponent phases (n + 1 ¼ 2 per phase, total four independent intensive variables), two conditions of equilibrium (thermal and mechanical), and one chemical reaction. This results in a variance which is equal to 4–2–1 ¼ 1. Therefore, once the temperature is specified, equilibrium exists at a unique pressure. In other words, we are not allowed to independently specify both temperature and pressure. This is the reason behind the discrepancy in the values obtained for the equilibrium constant in Eqs. (13) and (14) using two seemingly valid approaches to calculate it. In light of the fact that equilibrium need not exist at a pressure of 37 kbar, it is clear that we cannot use Eq. (14) to calculate the equilibrium constant because it inherently assumes that there exists equilibrium at 37 kbar. We can calculate the system pressure using the following equation:       f β ðT, PÞ 1 ΔGðT, 1 barÞ f D ðT, PÞ K ¼ exp  ¼ 0:314 ¼ RT f D ðT, 1 barÞ f β ðT, 1 barÞ    ) 0:314 ¼ exp V D  V β ðP  1 barÞ=RT ð16Þ    12
106 12
106  5  P  1
10 =ð8:314
298Þ ) ln 0:314 ¼ 1
3:51 1
2:26 ) P ¼ 15:18 kbar This pressure corresponds to a standard molar Gibbs free energy of reaction which is equal to 0. At any pressure lower or higher than this pressure, the reaction will proceed to completion, and there would exist only one component and one phase. In other words, there would be no chemical equilibrium between the two components, and application of the Gibbs Phase Rule will yields a variance of 2 (if n is equal to 2, π is also equal to 2, while if n is equal to 1, π is also equal to 1). Note that, in both cases, r is equal to 0.

Solved Problems for Part II

681

Problem 17 Problem 16.11 in Tester and Modell We do not seem to be able to locate any data for the standard Gibbs free energy of reaction, nor for the standard enthalpy of reaction, ΔH0, for the following chemical reaction: DðgÞ ¼ AðgÞ þ BðgÞ However, one of our research students has conducted a few experiments as described below. A constant volume pressure vessel was immersed in a constanttemperature bath. The vessel was first evacuated and then filled with pure gaseous D at 298 K and 1 bar. After heating the closed vessel to 473 K, reaction occurred to form A and B. When no further pressure change was noted, the gauge read 3 bar. Further heating to 523 K caused a pressure increase to 3.30 bar. With these data, can one calculate ΔG0523 K and ΔH0? If so, provide numerical values. If not, describe what additional data are required. You may assume that the vapor phase forms an ideal gas mixture and that ΔH0 is not a function of temperature.

Solution to Problem 17 Solution Strategy This problem provides information about the equilibrium temperatures and pressures of a gas-phase chemical reaction and asks us to evaluate the standard molar Gibbs free energy of reaction using the data provided. Let us begin by summarizing all the experiments which were carried out in a constant volume vessel having volume V: Experiment 1 P1 ¼ 1 bar T 1 ¼ 298 K V1 ¼ V N 1,A ¼ 0 N 1,B ¼ 0 N 1,D ¼ ? ¼

P1 V RT 1

682

Solved Problems for Part II

Experiment 2 P2 ¼ 3 bar T 2 ¼ 473 K V2 ¼ V N 2,A ¼ ? N 2,B ¼ ? N 2,D ¼ ? In Experiment 2, the moles of the three gaseous species are related to each other based on the stoichiometry of the chemical reaction. The relationship is reported in the Stoichiometric Table below.

Initial moles Final moles Final mole fraction

D N1,D N2,D ¼ N1,D  N2,A (N1,D  N2,A)/(N1,D + N2,A)

A 0 N2,A N2,A/(N1,D + N2,A)

B 0 N2,B ¼ N2,A N2,A/(N1,D + N2,A)

From the Stoichiometric table, we can deduce that the total number of moles of gas in the reactor vessel at the end of Experiment 2 is given by N2,A + N1,D, which can be related to P2 and T2 using the ideal gas law as follows: N 2,A þ N 1,D ¼

P2 V RT 2

ð1Þ

Finally, we can relate the standard molar Gibbs free energy of reaction to the measurable experimental properties using the equilibrium constant for the chemical reaction as follows:

Solved Problems for Part II

683

! ^f B f 0B 1 !1 0 ! ^ 2, A P2 ^ 2, B P2 ^ 2, D P2 y2, A ϕ ϕ y y2 , D ϕ 2 , B @ A ¼ P0 P0 P0 ! y2, A y2, B P2 ^i ¼ 1 ¼ ideal gas mixture; ϕ y2, D P0 0 1 B N 2, A N 2, B ðN 1, D þ N 2, A ÞP2 C ¼@ A 2 0 ðN 1, D þ N 2, A Þ ðN 1, D  N 2, A ÞP ! N 22, A P2 ¼ ðN 1, D þ N 2, A ÞðN 1, D  N 2, A ÞP0    ΔG0 T 2 ; P0 ¼ exp  RT 2

^ 1 ^f A K ¼ ff 0D D f 0A

!

ð2Þ

Substituting the values of N1,D and N2,A in terms of V using Eq. (1) yields: N 22, A P2

!

  ΔG0 T 2 ; P0 ¼ exp  RT 2 

ðN 1, D þ N 2, A ÞðN 1, D  N 2, A ÞP0 0 1  2 P2V = P1V =    0 0  =P2 B C RT = 2  RT = 1 C ¼ exp  ΔG T 2 ; P   )B @=P2 =V P1 =V P2 =V 0 A RT 2 2  P =RT 2 =RT 1 =RT 2 !    ΔG0 T 2 ; P0 T 2 ðP2 =T 2  P1 =T 1 Þ2 ) ¼ exp  RT 2 ð2P1 =T 1  P2 =T 2 ÞP0 !    ΔG0 T 2 ; P0 473ð3=473  1=298Þ2 ) ¼ exp  RT 2 ð2=298  3=473Þ1     ΔG0 T 2 ; P0 ) 11:4739 ¼ exp  8:314
473   0 0 ) ΔG T 2 ; P ¼ 9583:2748 J=mol

ð3Þ

684

Solved Problems for Part II

Experiment 3 P3 ¼ 3:30 bar T 2 ¼ 523 K V3 ¼ V N 3,A ¼ ? N 3,B ¼ ? N 3,D ¼ ? An analysis identical to that done for Experiment 2 would lead to the following Stoichiometric Table: D N1,D N3,D ¼ N1,D  N3,A (N1,D  N3,A)/(N1,D + N3,A)

Initial moles Final moles Final mole fraction

A 0 N3,A N3,A/(N1,D + N3,A)

B 0 N3, B ¼ N3,A N3, A/(N1,D + N3,A)

Similar to Experiment 2, the total number of moles of gas in the reactor vessel at the end of Experiment 3 is given by N3, A + N1, D, which can be related to P3 and T3 as follows: N 3,A þ N 1,D ¼

P3 V RT 3

ð4Þ

In addition, we can relate the standard molar Gibbs free energy of reaction for Experiment 3 to the measurable experimental properties using the equilibrium constant as follows: 0 K ¼@

^ P3 y3, D ϕ 3, D P0

11 0 A @

^ P3 y3 , A ϕ 3, A P0

1 A

^ P3 y3, B ϕ 3, B

!

P0

! ! N 23, A P3 y3, A y3, B P3 ¼ ¼ y3, D P0 ðN 1, D þ N 3, A ÞðN 1, D  N 3, A ÞP0    0 0  ΔG T 3 ; P ¼ exp  RT 3 Substituting the values of N1,D and N3,A in terms of V yields:

ð5Þ

Solved Problems for Part II

685

N 23, A P3

!

   ΔG0 T 3 ; P0 ¼ exp  RT 3

ðN 1, D þ N 3, A ÞðN 1, D  N 3, A ÞP0 0 1  2 P3V = P1V =    0 0  =P2 B C ΔG T ; P RT = 3  RT = 1 3 C )B @=P3 =V  P1 =V P3 =V  A ¼ exp  RT 3 0 2  P =RT 3 =RT 1 =RT 3 !    ΔG0 T 3 ; P0 T 3 ðP3 =T 3  P1 =T 1 Þ2 ) ¼ exp  RT 3 ð2P1 =T 1  P3 =T 3 ÞP0 !    ΔG0 T 3 ; P0 523ð3:3=523  1=298Þ2 ) ¼ exp  RT 3 ð2=298  3:3=523Þ1     ΔG0 T 3 ; P0 ) 11:3627 ¼ exp  8:314
523   0 0 ) ΔG 523 K; P ¼ 10567:6256 J=mol

ð6Þ

We can then calculate ΔH0 from the standard molar Gibbs free energy of reaction using the Gibbs-Helmholtz equation. Specifically:   0  ∂ ΔG =T ΔH 0 ¼ 2 ∂T T P   0  ∂ ΔG =T ) ¼ ΔH 0 ∂ð1=T Þ P

ð7Þ

The Problem Statement indicates that the standard molar enthalpy of reaction is independent of temperature. Therefore, we can calculate the partial derivative in Eq. (7) using the two data points for ΔG0 as follows:   0  ∂ ΔG =T ¼ ΔH 0 ∂ð1=T Þ P     0  ΔG T 3 , P0 =T 3  ΔG0 T 2 , P0 =T 2 ) ¼ ΔH 0 ð1=T 3 Þ  ð1=T 2 Þ   ð10567:6256=523Þ  ð9583:2748=473Þ ) ¼ ΔH 0 ð1=523Þ  ð1=473Þ ) ΔH 0 ¼ 271:3162 J=mol

ð8Þ

686

Solved Problems for Part II

Problem 18 Problem 9.3 in Tester and Modell In an experiment, a mixture of helium and ammonia was prepared as follows. As shown in the figure below, separate supply manifolds are available for the helium and the ammonia gases. The aluminum mixing tank is first evacuated to a very low pressure. Helium gas is then admitted very rapidly until the tank pressure is at 2 bar. The helium supply valve is then closed. Ten minutes later, the ammonia supply valve is opened to allow ammonia gas to flow rapidly into the tank. The valve is closed when the tank pressure reaches 3 bar. Data: 1. Assume that the gases behave ideally. The heat capacities of helium and ammonia may be considered to be constants with the following values: C P ðHeÞ ¼ 20:9 J=mol K C P ðNH3 Þ ¼ 35:6 J=mol K 2. Assume that the He-NH3 gas mixture is ideal. 3. The mixing tank dimensions are 0.3 m (inside diameter), 0.3 m tall. 4. The initial wall temperature of the mixing tank is 310 K and may be assumed to remain constant. 5. It may be assumed that the heat transfer from the gas to the mixing tank walls occurs by natural convection. For purposes of computation, assume that the heat transfer coefficient has a constant value of 15 W/m2K. 6. The helium gas in its manifold is at 310 K and 10 bar, and the ammonia gas in its manifold is at 310 K and 5 bar. With only this description and the given data, use your engineering reasoning to answer the following questions: (a) What is the amount of He gas in the mixing tank and its temperature when the tank pressure reaches 2 bar? (b) What is the temperature and the pressure of the He gas in the mixing tank 10 min later? (c) What is the temperature and composition of the He-NH3 gas mixture in the mixing tank when the tank pressure reaches 3 bar?

Solved Problems for Part II

687

Solution to Problem 18 Solution Strategy To setup this problem, before solving it, we will use the four-step strategy discussed in Part I when we utilized the First Law of Thermodynamics for closed and open systems. In this problem, we will also need to do this for a one component gas (in Parts (a) and (b)), as well as for an ideal binary gas mixture in Part (c). The four steps include: 1. 2. 3. 4.

Draw the pertinent problem configuration Summarize the given information Identify critical issues Make physically reasonable approximations

1. Draw the pertinent problem configuration A figure depicting the tank and the gas supply tubes is provided in the Problem Statement. To solve the three parts of this problem, we will treat the contents of the tank at any given time as the system to be studied (see 3 below). 2. Summarize the given information This problem consists of three parts, with a different process occurring in each part. For Part (a), the helium valve is opened, and helium flows into the tank until the tank pressure reaches 2 bars. Once this pressure is reached, the helium valve is closed. We are also told that this helium filling process occurs very rapidly. For Part (b), we are asked to examine how the temperature of helium in the tank changes in 10 min. During this 10-min period, no additional gases are added to the tank. Finally, for Part (c), the ammonia valve is opened, and ammonia is allowed to flow into the

688

Solved Problems for Part II

helium-containing tank until the tank pressure reaches 3 bars. Subsequently, the ammonia valve is closed. We are also told that this filling process occurs rapidly. 3. Identify critical issues Defining the System As stated above, we choose as our system the contents of the tank at any time. This allows us to neglect any PdV-type work (i.e., δW ¼ 0), which will simplify our First Law of Thermodynamics analysis and equations. Defining the Boundaries For all three parts, the boundaries are rigid (solid, unmovable tank walls). However, depending on the process that we are considering, some of the system boundaries will change. In Part (a), the system has open boundaries because helium flows into the tank. The boundaries are also considered adiabatic because we are told that this filling process occurs very rapidly, and therefore, we can assume that there is no time for heat transfer to occur between helium and the tank walls. In Part (b), the system has closed boundaries. Unlike in Part (a), however, we are specifically told that there is heat transfer between helium and the tank walls. Finally, in Part (c), the boundaries are open because ammonia flows into the tank. The boundaries are similar to those in Part (a), including being adiabatic. Mixing of Gases In this problem, helium and ammonia are considered to be ideal gases. We will assume that helium and ammonia will be well-mixed within the tank. As a result, each gas will occupy the entire tank volume and will have uniform temperatures and pressures. 4. Make physically reasonable approximations Ideal Gas Behavior The problem states that helium and ammonia behave ideally. Therefore, helium and ammonia will be assumed to have constant CV and CP values. Nevertheless, we note that the two gases have different CV (and therefore CP) values! Adiabatic process for Parts (a) and (c) As stated above, we take our system to be the contents of the tank at any time. For Parts (a) and (c), we are told that the filling occurs rapidly. Consequently, we will assume that there is not enough time for heat transfer between the gas and the tank walls to occur (i.e., adiabatic filling process). Heat Transfer by Convection for Part (b) For Part (b), we are told that heat transfer from helium in the tank to the tank walls occurs by natural convection. Because the tank is kept at a constant temperature of 310 K, we will also assume that any heat transferred from helium to the tank walls

Solved Problems for Part II

689

will be immediately transferred to the surroundings, so that the tank walls are kept at a constant temperature. Part (a) To solve this part of the problem, we can make the following assumptions: • Helium is an ideal gas of constant CV and CP ¼ CV + R. • Helium is well-mixed and occupies the entire aluminum tank. As a result, dV ¼ 0, and δW ¼ 0 (no PdV-type work). • This is a fast filling process, implying that there is no heat transfer. That is, δQ ¼ 0 (adiabatic operation). • The initial (i) tank pressure is assumed to be zero (Pi ¼ 0). That is, Ni ¼ 0 for helium (The aluminum tank is initially empty). We begin by carrying out a First Law of Thermodynamics analysis of the tank contents at any time. This system is simple, open, rigid, and adiabatic. As a result, dU ¼ δQ þ δW þ H in δnin  H out δnout δQ ¼ 0 δW ¼ 0

ð1Þ

δnout ¼ 0 δnin ¼ dN H in ¼ H in ðT in Þ ¼ H in ð310K Þ ¼ constant Therefore, the top expression in Eq. (1) can be rewritten as follows: dU ¼ H in dN

ð2Þ

We can integrate Eq. (2) directly from (Ni, Ui) to (Nf, Uf). Note that Ni ¼ 0, and therefore, Ui ¼ 0. Upon integration, we obtain: U f ¼ H in N f U f N f ¼ H in N f U f ¼ H in dH Because dU dT ¼ C V and dT ¼ C P , upon integration, we can write:

  U f ¼ U 0 þ CV T f  T 0 H in ¼ H 0 þ C P ðT in  T 0 Þ Using the expressions above for Uf and Hin in Eq. (3) yields:

ð3Þ

690

Solved Problems for Part II

U f ¼ H in U 0 þ C V T f  CV T 0 ¼ H 0 þ C P T in  C P T 0 CV T f ¼ CP T in þ ðH 0  ðC P  CV ÞT 0  U 0 Þ

ð4Þ

CV T f ¼ CP T in þ ðH 0  H 0 Þ CV T f ¼ CP T in Equation (4) shows that: Tf ¼

CP T in CV

ð5Þ

Denoting Tf  T1 and using the data provided in the Problem Statement, we obtain: T f ¼ T 1 ¼ ð310 KÞ

h

i 20:9 ¼ 514:8 K 20:9  8:314

T 1 ¼ 514:8 K

ð6Þ

Because helium is treated as an ideal gas, we know that it is described by the ideal gas EOS. Therefore: N1 ¼

P1 V RT 1

ð7Þ

Using the required values in Eq. (7), we find that: P1 ¼ 2 bar ¼ 2
105 Pa T 1 ¼ 514:8 K R ¼ 8:314 J=mol K 2 πD2 L 3:14ð0:3Þ 0:3 ¼ 2:12
102 E 3 ¼ 4 4    2
105 Pa 2:12
102 m3 N1 ¼ ¼ 0:991 mol ð8:314Þð514:8 KÞ

V¼

N 1 ¼ 0:991 mol

ð8Þ

Part (b) After a 10-min waiting period, during which helium in the aluminum tank exchanges heat with the tank walls via a natural convection process, helium cools down due to

Solved Problems for Part II

691

the heat transfer to the vessel walls. We will assume that the wall temperature, Tw, does not change. Then, if we consider the helium gas as a well-mixed, closed, simple system that exchanges heat, a First Law of Thermodynamics analysis yields: dU ¼ δQ þ δW ¼ δQ

ð9Þ

where δQ refers to the heat absorbed by the gas, and δW ¼ 0 because of the rigid tank walls. We also know that: dU ¼ N 1 CV dT

ð10Þ

According to the Problem Statement, we know that heat transfer between helium and the tank walls occurs via convection, that is: δQ ¼ hQ AðT  T w Þdt

ð11Þ

In Eq. (11), hQ denotes the heat transfer coefficient given in the Problem Statement, and A denotes the surface area of the tank. A negative sign is necessary to account for the fact that helium loses heat (it is at a temperature of 515 K, which is higher than the tank wall temperature of 310 K). Using Eqs. (10) and (11) in Eq. (9) yields: N 1 C V dT ¼ hQ AðT  T w Þdt

ð12Þ

Because Tw and hQ are constant, Eq. (12) can be integrated as follows:   hQ A dT dt ¼ N 1 CV ðT  T w Þ Tð2   t¼600 ð hQ A dT ¼ N 1 CV ðT  T w Þ T1 0     hQ A T2  Tw ln ¼ t N 1 CV T1  Tw

sec

dt

or

  hQ A t T 2 ¼ T w þ ðT 1  T w Þ exp  N 1 CV Using the information given below:

ð13Þ

692

Solved Problems for Part II

 2 2ð3:14Þð0:3 mÞ2 πD A ¼ πDL þ 2 ¼ ð3:14Þð0:3 mÞð0:3 mÞ þ ¼ 0:424m2 4 4 N 1 ¼ 0:991 mol CV ¼ ð20:9  8:314Þ J=mol K ¼ 12:586 J=mol K hQ ¼ 15 W=m2 K t ¼ 10 min ¼ 600 sec T 1 ¼ 514:8 K ðFrom Part ðaÞÞ in Eq. (13), we obtain:

T 2 ¼ 310 K þ ð514:8  310ÞK exp

ð1Þð15Þð0:424Þð600Þ ð0:991Þð12:586Þ



T 2 ¼ 310 K þ ð204:8Þ exp ð305Þ or T 2 ffi 310 K

ð14Þ

Equation (14) shows that in 10 min, helium cools down to the wall temperature. Further, the helium pressure changes to P2 ¼ N 1VRT 2 . However, P1 ¼ N 1VRT 1 , so that: P1 P2 ¼ T1 T2   P T 310 P2 ¼ 1 2 ¼ ð2 barÞ 514:8 T1 or P2 ¼ 1:2bar

ð15Þ

Part (c) This part of the problem deals with the addition of ammonia to the tank already filled with helium. We can make the same assumptions that we made in Part (a): • • • •

Helium and ammonia behave ideally There is rapid addition of ammonia, so that δQ ¼ 0 δnout ¼ 0, δnin ¼ dN We will analyze the well-mixed case. This implies that δW ¼ 0

We first note that inside the aluminum tank, we have a binary mixture of ideal gases (helium and ammonia), and we are told that the gas mixture itself is also ideal. As in Part (a), we carry out a First Law of Thermodynamics analysis of the binary

Solved Problems for Part II

693

gas mixture at any time. The system is simple, open, adiabatic, and rigid. Accordingly: dU ¼ δQ þ δW þ H in δnin  H out δnout

ð16Þ

In Eq. (16), we have: H in ¼ H in ðT in Þ ¼ H in ð310K Þ ¼ constant

ð17Þ

Like in Part (a), Eq. (16) reduces to: dU ¼ H in δnin

ð18Þ

As we did in Part (a), we can integrate Eq. (18) directly from state i to state f, which yields: U f  U i ¼ H in N in

ð19Þ

where Nin ¼ NNH3 (the total number of moles of ammonia that entered into the aluminum tank). We can next relate U to H, P, and V as follows: U f ¼ H f  PfV f U i ¼ H i  Pi V i

ð20Þ

V f ¼ Vi ¼ V Using the three equations in Eq. (20), we can write that:     U f  U i ¼ H f  H i  P f  Pi V

ð21Þ

Because the final ( f ) state corresponds to an ideal binary mixture of helium and ammonia in the tank and the initial (i) state corresponds to pure helium in the tank, the following enthalpy expressions can be written down: f f f f H f ¼ N He H He þ N NH H NH 3 3

H i ¼ pure helium ¼ N iHe H iHe H in ¼ H iNH 3 Recall that NHef ¼ NHei ¼ NHe and N NH 3 f  N NH 3 . Hence, using the three expressions above in Eq. (21), rearranging, and using Eq. (19) where in denotes ammonia, we obtain:

694

Solved Problems for Part II

    U f  U i ¼ N He H He f þ N NH 3 H NH 3 f  P f V  NHe H He i  Pi V ¼ H NH 3 N NH 3

ð22Þ

Rearranging Eq. (22), we obtain:       N He H He f  H He i þ N NH 3 H NH 3 f  H NH 3 i  V Pf  Pi ¼ 0

ð23Þ

Using the fact that helium and ammonia are considered here as ideal gases, we can express the molar enthalpy changes of each gas in Eq. (23) as follows:     H He f  H He i ¼ C P He Tf  T 0  C P He ðT 2  T 0 Þ ¼ C P He Tf  T 2

ð24Þ

    H NH 3 f  H NH 3 i ¼ C P NH 3 Tf  T 0  C P NH 3 ðT in  T 0 Þ ¼ C P NH 3 Tf  T in ð25Þ Using Eqs. (24) and (25) in Eq. (23), we obtain:         N He C P He T f  T 2 þ N NH 3 CP NH 3 T f  T in  V Pf  Pi ¼ 0

ð26Þ

where NHe ¼ N1 ¼ 0.991 mol, T2 ¼ 310 K, Tin ¼ 310 K, V ¼ 2.12
102 m3, Pf ¼ 3 bar, and P2 ¼ 1.2 bar. Equation (26) relates the two unknowns, NNH3 and Tf. In order to solve for them, we need a second equation relating them that we can then solve simultaneously with Eq. (26). Because ammonia is treated here as an ideal gas, the second equation is the ideal gas EOS relating Tf and NNH3. Specifically: P f V ¼ N total RT f    3
105 2:12
102 PfV N total ¼ ¼ RT f ð8:314ÞT f   765 N NH 3 ¼ N total  N 1 ¼  0:991 Tf   765 N NH 3 ¼  0:991 Tf Using Eq. (27) in Eq. (26), we obtain:

ð27Þ

Solved Problems for Part II

695

      765 ð0:991Þð20:9Þ T f  310 þ  0:991 ð35:6Þ T f  310 Tf    5 2  2:12
10 ð3  1:2Þ 10 ¼ 0 14:57T f 2 þ 27935T f  8442540 ¼ 0 or T f 2  1917T f þ 579447 ¼ 0

ð28Þ

Solving the quadratic equation above, we find that that there are two roots, that is:   1917 1165 K 2 ¼ 376 K

Tf ¼ T f

T f þ ¼ 1541 K To determine which of the two Tf values obtained above is the correct one, we use these values in Eq. (27). This yields: 765  0:991 ¼ 1:044 mol 376 765  0:991 < 0! ¼ 1541

N NH 3  ¼ N NH 3 þ

Clearly, using Tf ¼ 1541 K results in a negative value for NNH3, which is not physical. Therefore, the correct final temperature of the helium-ammonia mixture is 376 K. This implies that 1.044 mol of NH3 were added to the tank during the final operation and, therefore, that ammonia makes up 51.3% of the final gas mixture in the tank. To summarize, we obtain: T f ¼ 376 K, N NH 3 ¼ 1:044 mol Note that the final helium-ammonia mixture temperature is higher than its initial temperature (Tf > T2). This reflects ammonia entering the tank, which originally contained helium only. Because of the well-mixing assumption made, there is no compression of the existing helium gas (i.e., the stratified layer model discussed in Part I does not apply). Instead, the increase in temperature of the helium-ammonia mixture results from the enthalpy of ammonia flowing in. An examination of Eq. (19) reveals that the internal energy of the helium-ammonia mixture (and therefore, its temperature) increases as NH3 enters the system. Therefore, as expected, Tf > T2.

696

Solved Problems for Part II

Problem 19 A creative Chemical Engineering graduate student is trying his luck in the business world to supplement his monthly stipend. He plans to produce freshwater for the crew of a nuclear submarine by inserting a semipermeable membrane, which allows solely passage of freshwater, in the hull of the submarine, and uses the pressure difference between the outside (the sea) and the inside of the submarine to provide a driving force to produce freshwater by reverse osmosis. The student believes that the US Navy will be interested in his idea. As proof of concept, the engineer would like to calculate what the minimum depth of the submarine, hmin, has to be for the first drop of freshwater to pass through the membrane. You are asked to utilize your expertise in classical thermodynamics to answer the following question: Derive an explicit expression for hmin, and then use it to obtain a numerical value for the minimum submarine depth. For this purpose, you can make the following assumptions: 1. Seawater can be considered as a binary water-NaCl solution. 2. NaCl in seawater is not dissociated. 3. The NaCl concentration in seawater is uniform and has a value of 5 wt%. The mass density of seawater, ρSW, and the partial specific volume of water, V w , are independent of pressure, and they have constant values of: ρSW ¼ 1:145 g=cm3 V w ¼ 0:994 cm3 =g 4. The seawater temperature is uniform and has a value of 25  C. 5. The temperature in the hull of the submarine is maintained at a constant value of 25  C. 6. The pressure in the hull of the submarine is maintained at a constant value of 1 atm. 7. The graph of water chemical potential versus weight fraction of NaCl, given below, can be utilized.

Solved Problems for Part II

697

Solution to Problem 19 Solution Strategy This is an interesting problem which explores the possibility of obtaining freshwater in the hull of a nuclear submarine from seawater through reverse osmosis. Because as the submarine descends there is a pressure difference (higher pressure outside the submarine, P, relative to that inside the submarine, P0 ¼ 1 atm), the equality of the chemical potentials (or fugacities) at the given T0 will be the required condition. Before we begin to solve this problem, we would like to take a moment to think about what is actually going on. We know that when pure water is separated from an aqueous solution of a solute by a semipermeable membrane (one that is impermeable to the solute), water flows from the pure water side to the solution side. This is known as osmosis, which is the result of the higher water chemical potential of pure water. However, in this problem, this would mean that freshwater is flowing from the hull of the submarine into the sea, which is not very helpful. Fortunately, because of the presence of hydrostatic pressure, the pressure outside the submarine is much larger than that in the hull of the submarine. When this pressure difference is large enough to overcompensate for the concentration difference, we can have water flowing in the reverse direction, that is, from the sea into the hull of the submarine. This process is known as reverse osmosis. Therefore, if we

698

Solved Problems for Part II

need to calculate the submarine depth at which the first drop of freshwater flows from the sea into the hull of the submarine, we simply need to calculate the depth at which the pressure difference just balances out the concentration difference to allow reverse osmosis to occur.

Calculation of Phase Equilibria With the above discussion in mind, we can write the required phase equilibria condition as follows: μw ðT 0 , P0 Þ ¼ μw ðT 0 , P, xs Þ

ð1Þ

where T0 ¼ 25  C, P > P0 ¼ 1 atm, and xs refers to the sea salt concentration (or equivalently, xw ¼ 1 – xs). Because the Problem Statement provides graphical information about the water chemical potential, it is convenient to continue using Eq. (1) without switching to water fugacities, as reflected in the integral or the differential approaches to phase equilibria that we discussed in Part II. In order to utilize the given graph of chemical potential (which is given at T0 ¼ 25  C and P ¼ 1 atm) at T0 ¼ 25  C and P > 1 atm, we need to express μw(T0, P, xs) in terms of μw(T0, P0, xs). This can be readily done because, as discussed in Part II:   ∂μi ¼ Vi ∂P T,x

ð2Þ

Integrating Eq. (2) for i ¼ w, at T ¼ T0 and x ¼ xs, from P0 to P, including rearranging, we obtain:  

∂μw ∂P ∂μw ∂P

 ¼ Vw 

T 0 ,xs

dP ¼ V w dP T 0 ,xs

 ðP  ðP ∂μw dP ¼ μw ðT 0 , P, xs Þ  μw ðT 0 , P0 , xs Þ ¼ V w dP ∂P T 0 ,xs

P0

P0

or ðP μw ðT 0 , P, xs Þ ¼ μw ðT 0 , P0 , xs Þ þ

V w dP P0

Using Eq. (3) in Eq. (1), we find that:

ð3Þ

Solved Problems for Part II

699

ðP μw ðT 0 , P0 , xs Þ þ

V w dP ¼ μw ðT 0 , P0 Þ P0

or ðP μw ðT 0 , P0 , xs Þ  μw ðT 0 , P0 Þ ¼ 

V w dP

ð4Þ

P0

Note that the left-hand side of Eq. (4) is given by the chemical potential graph (see Fig. 1), because T0 ¼ 25  C and P0 ¼ 1 atm for any salt concentration. In particular, for xs ¼ 5 wt% ¼ 0.05, we read:

Fig. 1

700

Solved Problems for Part II

      Δμw ¼ μw 25 C, 1 atm, 0:05  μw 25 C, 1 atm ¼ 4:2 Joule=gH2 O Because V w is constant, Eq. (4) shows that: Δμw ¼ V w ðP  P0 Þ or that: P  P0 ¼ 

Δμw Vw

ð5Þ

Because of the hydrostatic pressure, in Eq. (5), P  P0 ¼ ρswgh, which yields: h¼

w  Δμ V w

ρsw g

ð6Þ

Using all the information given in the Problem Statement, that is, Δμw ¼ 4 J=gH2 O ¼ 4 Nm=gH2 O, V w ¼ 0:994cm3 =gH2 O, ρsw ¼ 1:145 gH2 O=cm3 , and g ffi 9.8 m/sec2, in Eq. (6), we find that:   4N=m==g 40 N 105 2 6 32 = 0:994
10 m ==g 0:994 m h¼  ¼ 1:145
103 kg  m  4N 1:145
10 3 10 6 32 = m sec 2 10 m   40
105 0:994 h¼ m 1:145
104 or h ffi 351 m

ð7Þ

Solved Problems for Part II

701

Problem 20 As discussed in Part II, the azeotropic point in a binary mixture of components A and B coexisting in liquid and vapor phases results in special properties. For example, at the azeotropic point, the relative volatility is unity because the liquid and the vapor compositions are identical. In addition, a plot of the logarithm of the azeotropic pressure, Paz, versus 1/T is typically linear for many real systems. Starting with the phase equilibria criteria for this liquid-vapor equilibrium azeotropic system, develop an explicit expression to relate Paz ¼ f(T ). Describe under what conditions you would expect a plot of ln(Paz) to be linear in 1/T.

Solution to Problem 20 Solution Strategy First, we recall that at the azeotrope, the Gibbs Phase Rule requires that: L¼nþ2πrs¼1 where n ¼ 2, π ¼ 2, r ¼ 0, and s ¼ 1 (recall that at the azeotrope, the liquid composition is equal to the vapor composition, which is a constraint, such that s ¼ 1). This indicates that as discussed in Part II, the system is indeed monovariant. Because L ¼ 1, we know that if we choose T as the intensive variable, at the azeotrope, it should be possible to express Paz as a unique function of T. That is, Paz ¼ f(T )!

Conditions of Phase Equilibria Next, we utilize the three conditions of phase equilibria, that is: ði Þ T V ¼ T L  T ðiiÞ PV ¼ PL  P V L V L ðiiiÞ bf A ¼ bf A and bf B ¼ bf B

ðThermalÞ ðMechanicalÞ ðDiffusionalÞ

Choosing the Differential Approach to Phase Equilibria Because we need to calculate how Paz ¼ f(T ), it is convenient to utilize the differential approach to phase equilibria. As discussed in Part II, we write:

702

Solved Problems for Part II V

HA  HA  RT 2

L

! dT þ

V

VA  VA RT

L

! dP þ

V ∂ ln bf A ∂yA

! dyA  T,P

L ∂ ln bf A ∂xA

! dxA ¼ 0 T,P

ð1Þ and V

HB  HB  RT 2

L

! dT þ

V

VB  VB RT

L

! dP þ

∂ ln bf B ∂yA

V

! dyA  T,P

∂ ln bf B ∂xA

L

! dxA ¼ 0 T,P

ð2Þ Next, we multiply Eq. (1) by xA and Eq. (2) by xB, which yields:  xA  xA

! ! ! V V L V L ∂ ln bf A HA  HA VA  VA dyA dP þ xA dT þ xA RT ∂yA RT 2 T,P ! L ∂ ln bf A dxA ¼ 0 ∂xA

ð3Þ

! ! ! V V L V L ∂ ln bf B HB  HB VB  VB dyA dP þ xB dT þ xB RT ∂yA RT 2 T,P ! L ∂ ln bf B dxA ¼ 0 ∂xA

ð4Þ

T,P

and  xB  xB

T,P

Adding up Eqs. (3) and (4) yields:   1 0  V L V L xA H A  H A þ xB H B  H B AdT @ RT 2   1 0  V L V L xA V A  V A þ xB V B  V B AdP þ@ RT " ! ! # V V ∂ ln bf A ∂ ln bf B þ xA þ xB dyA ∂yA ∂yA T,P T,P " ! ! # L L ∂ ln bf A ∂ ln bf B  xA þ xB dxA ¼ 0 ∂xA ∂xA T,P

T,P

ð5Þ

Solved Problems for Part II

703

Because L ¼ 1, we recognize that we should be able to get rid of the dyA and dxA terms in Eq. (5) to obtain a unique relation between dT and dP. To do that, we use the Gibbs-Duhem equation in each phase. Solving Equation (5) • In the liquid phase, at constant T and P, the Gibbs-Duhem Equation applies. Specifically:   L xA d ln bf A

T,P

  L þ xB d ln bf B

T,P

¼0

or

xA

L ∂ ln bf A ∂xA

! þ xB T,P

L ∂ ln bf B ∂xA

! ¼0

ð6Þ

T,P

• In the gas phase, at constant T and P, the Gibbs-Duhem Equation applies. Specifically:     V V yA d ln bf A þ yB d ln bf B ¼0 T,P T,P ! ! V V ∂ ln bf A ∂ ln bf B þ yB ¼0 yA ∂yA ∂yA T,P

V ∂ ln bf B ∂yA

T,P

! T,P

y ¼ A yB

V ∂ ln bf A ∂yA

! T,P

or V ∂ ln bf B ∂yA

!



T,P

yA ¼ 1  yA



V ∂ ln bf A ∂yA

! ð7Þ T,P

Equation (6) shows that the term in the square brackets multiplying dxA in Eq. (5) is zero! Equation (7) shows that the term in the square brackets multiplying dyA in Eq. (5) can be expressed as follows:

704

Solved Problems for Part II

! ! V V ∂ ln bf A ∂ ln bf B xA þ xB ∂yA ∂yA T,P T,P ! !   V b yA ∂ ln f A þxB  1  yA ∂yA T,P ! ! V V ∂ ln bf A ∂ ln bf B xA þ xB ∂yA ∂yA T,P T,P ! ! V V ∂ ln bf A ∂ ln bf B xA þ xB ∂yA ∂yA T,P T,P ! ! V V ∂ ln bf A ∂ ln bf B xA þ xB ∂yA ∂yA T,P

T,P

¼ xA

V ∂ ln bf A ∂yA

! T,P

! V ∂ ln bf A ∂yA T,P !

 V ð1  xA ÞyA ∂ ln bf A ¼ xA  1  yA ∂yA T,P !

 V y ð1  xA Þ ∂ ln bf A ¼ xA 1  A xA ð 1  yA Þ ∂yA

x y ¼ xA  B A 1  yA



T,P

However, at the azeotrope, xA ¼ yA. Therefore, the last term in the square brackets in the last equation is equal to zero! Accordingly, as expected because L ¼ 1, at the azeotrope, Eq. (5) reduces to a relation between dP and dT! Next, in Eq. (5), we can simplify the numerators of the terms multiplying dT and dP. The vaporization of NA moles of component A and NB moles of component B involves the following changes in enthalpy and volume: L

L

V

V

L

H L ¼ N A H A þ N B H B ) H L ¼ xA H A þ xB H B

L

V

H V ¼ N A H A þ N B H B ) H V ¼ xA H A þ xB H B

V

Therefore, we can express the mixture molar enthalpy of vaporization as follows:     V L V L ðH V  H L Þ ¼ xA H A  H A þ xB H B  H B ¼ ΔH vap mix

ð8Þ

Similarly, we can express the mixture molar volume of vaporization as follows: L

L

V

V

L

V L ¼ N A V A þ N B V B ) V L ¼ xA V A þ xB V B V

L

V V ¼ N A V A þ N B V B ) V V ¼ xA V A þ xB V B

V

    V L V L ðV V  V L Þ ¼ xA V A  V A þ xB V B  V B ¼ ΔV vap mix

ð9Þ

Using Eqs. (8) and (9) in Eq. (5), with the dyA and dxA terms set equal to zero, yields:

Solved Problems for Part II

705

    ΔH vap mix ΔV vap mix  dP ¼ 0 dT þ RT RT 2 or 

dP dT



 ¼ L=V,azeotrope

ΔH vap mix TΔV vap mix

 ð10Þ

Equation (10) is the Clapeyron equation for the [L/V] equilibrium at the azeotrope. If we make the following assumptions: ð 1Þ

ΔH vap mix  constant

ð 2Þ

V V >> V L V

ΔV vap mix  V V ¼ xA V A þ xB V B V

V A  V A V ¼ RT=P V

V B  V B ¼ RT=P V

V

ðIdealÞ ðIdealÞ

V V ¼ ðxA þ xB ÞðRT=PÞ ¼ RT=P Using the last result in Eq. (10), including rearranging, we obtain: 

 ΔH vap mix dP ¼ dT L=V,azeotrope T ðRT=PÞ   ΔH vap mix dT dP ¼ P L=V,azeotrope R T2



d ln P d ð1=T Þ

 ¼ L=V,azeotrope

ΔH vap mix R

ð11Þ

Equation (11) is the Clausius-Clapeyron equation at the azeotrope. Because ΔHvapmix  constant, Eq. (11) shows that Paz is indeed a linear function of 1/T!

Solved Problems for Part III

Solved Problems for Part III

707

Problem 21 Problem 21.1 A molecule has the following three energy levels and degeneracies: Level 1 2 3

Energy 0 ε 2ε

Degeneracy 1 1 γ

(a) Write an expression for the molecular partition function, q, as a function of ε, γ, and the temperature, T. (b) Write an expression for the average molecular energy, , as a function of ε, γ, and T. (c) For ε/kBT ¼ 1, and γ ¼ 1, compute the populations or probabilities of occupancy, p1, p2, and p3, of the three levels. (d) Find the temperature, T*, at which p1 ¼ p3. Express your result in terms of ε and γ. (e) Compute p1, p2, and p3 at the condition given in (d). Express your result in terms of γ. (f) Explain what happens to the results in (d) and (e) when γ ¼ 1.

Problem 21.2 Consider a system of independent, distinguishable particles that have only two quantum states with energies 0 and ε. Calculate the molecular heat capacity of such a system, and show that CV/kB plotted against βε passes through a maximum at βε ¼ 2.40, which corresponds to the solution of the equation, βε/2 ¼ coth(βε/2).

Problem 21.3 The Canonical partition function of a monoatomic van der Waals gas is given by the following expression:  3N=2   1 2πmk B T QðN, V, T Þ ¼ ðV  NbÞN exp aN 2 =Vk B T N! h2 where a and b are the van der Waals constants.

708

Solved Problems for Part III

(a) Derive an expression for the energy of a monoatomic van der Waals gas. Compare your result with that for a monoatomic ideal gas. (b) Calculate the heat capacity, CV, of a monoatomic van der Waals gas. Compare your result with that of a monoatomic ideal gas. (c) Derive an expression for the pressure of a monoatomic van der Waals gas.

Solution to Problem 21 Solution to Problem 21.1 Part (a) The molecular partition function is given by: q¼

  Ej g j exp  kB T j¼1

3 X

ð1Þ

In Eq. (1), j represents energy levels and not energy states, because we have already considered the existence of distinct states having the same energy by including the gj term which indicates degeneracy. A table specifying different energy levels and their respective degeneracies is provided in the Problem Statement. Using the information provided in the Table in Eq. (1) yields: E

q ¼ 1e0 þ 1e =kB T þ γe E

q ¼ 1 þ 1e =kB T þ γe

2E=k T B

2E=k T B

ð2Þ

Part (b) The average molecular energy is given by: < E >¼

3 X

p jE j

ð3Þ

j¼1

where pj, the probability of occurrence of energy level, Ej, is given by: pj ¼

  E g j exp  kB jT q

ð4Þ

Solved Problems for Part III

709

In Eq. (4), the molecular partition function, q, is given by Eq. (2). Using Eqs. (4) and (2), it follows that: 1 exp ð0Þ 1 ¼ q q     E 1 exp kE exp kB T BT p2 ¼ ¼ q q   γ exp 2E kB T p3 ¼ q p1 ¼

ð5Þ

ð6Þ

ð7Þ

Substituting p1, p2, and p3 from Eqs. (5), (6), and (7), respectively, as well as q from Eq. (2), in Eq. (3) then yields: < E >¼ p1 0 þ p2 E þ p3 ð2EÞ

    1 E E þ 2Eγ exp < E >¼ E exp q kB T kB T     E exp kE þ 2Eγ exp kE BT BT < E >¼ E=k B T 2E=kB T 1 þ 1e þ γe

ð8Þ

Equation (8) can also be obtained from the relation: < E >¼ kB T 2

  ∂ ln q ∂T V

ð9Þ

Indeed, starting from Eq. (2) and recognizing that q does not depend on V in this case, it follows that: 

    d ln q d E 2E ¼ ln 1 þ 1e =kB T þ γe =kB T dT dT V           E E=k T þ γ 2E 2 exp 2E=k T 0 þ exp 2 B B kB T kB T ∂ ln q     ¼ ∂T V 1 þ 1 exp E=kB T þ γ exp 2E=kB T ∂ ln q ∂T





¼

ð10Þ

Substituting the expression in Eq. (10) in Eq. (9) yields:     E exp E=kB T þ 2γE exp 2E=kB T     < E >¼ 1 þ 1 exp E=kB T þ γ exp 2E=kB T which is identical to the result derived in Eq. (8).

ð11Þ

710

Solved Problems for Part III

Part (c) When E/kBT ¼ 1 and γ ¼ 1, the molecular partition function q can be calculated using Eq. (2) as follows: q ¼ 1 þ e1 þ e2 ¼ 1 þ 0:3678 þ 0:1354 ¼ 1:5032

ð12Þ

Using Eq. (12) in Eqs. (5), (6), and (7), it follows that: p1 ¼

1 ¼ 0:6652 q

p2 ¼

e1 ¼ 0:2447 q

p3 ¼

e2 ¼ 0:0981 q

ð13Þ

Part (d) We are asked to find the temperature, T, at which p1 ¼ p3. Using Eqs. (5) and (7), we obtain:

p1 ¼ p3

! 

1 ¼ q

  γ exp  kB2ET  

q

2E 1 ¼ kB T  γ   2E 1  ¼ ln ¼  ln γ kB T  γ   2 E  T ¼ ln γ kB exp 

ð14Þ

Clearly, to obtain a finite value of T, γ has to be greater than 1. Part (e) At the conditions given in Part (d), we have: kB T  ¼

2E ln γ

ð15Þ

From Eq. (2), it follows that: E

qðT  Þ ¼ 1 þ 1e =kB T  þ γe

2E=k T  B

Solved Problems for Part III

711

ln γ 1 γ qðT  Þ ¼ 1 þ e 2 þ γelnγ ¼ 1 þ pffiffiffi þ γ γ

1 qðT  Þ ¼ 2 þ pffiffiffi γ

ð16Þ

With q(T) given by Eq. (16), we can use Eqs. (5), (6), and (7) to determine p1, p2, and p3. Specifically: 1 1 ¼ pffiffiffi q 2 þ 1= γ pffiffiffi γ  pffiffiffi p1 ¼ 1þ2 γ 0

p1 ¼

 p2 ¼

exp

E kB T 

A

E  E kB

q

  p1ffiffi exp  lnγ γ 2 ¼ q 2 þ p1ffiffiγ p2 ¼

γ exp



2E kB T 

1 pffiffiffi 1þ2 γ 0

ð18Þ 1

γ exp @



q ¼

1

k B ln2 γ

¼

q

¼

p3 ¼

exp @



ð17Þ

¼

2E  kB ln2 γ

q

E kB

A ð19Þ

γ exp ð ln γ Þ γ=γ ¼ q 2 þ p1ffiffiγ p3 ¼

pffiffiffi γ pffiffiffi 1þ2 γ

ð20Þ

As expected, at the conditions in Part (d), p1 ¼ p3. Part (f) When γ ¼ 1, Eq. (14) reveals that:   2 E T ¼ ¼1 ln 1 kB 

ð21Þ

712

Solved Problems for Part III

In this case, the three levels become equally probable, with a probability of 1/3 each. This can be verified from Eqs. (17), (18), and (19) when γ ¼ 1

Solution to Problem 21.2 The particle partition function q is given by: q¼

X

  exp βE j ¼ exp ð0Þ þ exp ðβEÞ ¼ 1 þ exp ðβEÞ

ð22Þ

j

The average energy of a particle is given by:   X E j exp βE j E exp ðβEÞ ¼0þ < E >¼ q q j

ð23Þ

Because the particles are independent, it follows that: < E >¼ N < E >¼

NE exp ðβEÞ 1 þ exp ðβEÞ

ð24Þ

We know that:   1 ∂< E > CV ¼ N ∂T N,V

ð25Þ

Next, substituting Eq. (24) for E in Eq. (25), we obtain: 1 CV ¼ N

 !

 E exp ðβEÞ ∂ NE exp ðβEÞ ∂ ¼ ∂T 1 þ exp ðβEÞ N,V ∂T 1 þ exp ðβEÞ

d ¼  k 1T 2 Recalling that β ¼ 1/(kBT ) and that dT B expressed as follows:

CV ¼ 

d dβ ,

ð26Þ

it follows that Eq. (26) can be

 E exp ðβEÞ 1 d kB T 2 dβ 1 þ exp ðβEÞ

ð27Þ

Solved Problems for Part III

713

1 kB T 2 " # ð1 þ exp ðβEÞÞ exp ðβEÞðEÞ  exp ðβEÞð exp ðβEÞÞðEÞ
ð28Þ ð1 þ exp ðβEÞÞ2 " # 1 E exp ðβEÞ  E exp ð2βEÞ þ E exp ð2βEÞ CV ¼  kB T 2 ð1 þ exp ðβEÞÞ2

CV ¼ 

¼

E2 exp ðβEÞ k B T 2 ð1 þ exp ðβEÞÞ2

ð29Þ

In order to express CV in terms of β, we write T in Eq. (29) as T ¼ kB1β . This yields: CV ¼

kB ðβEÞ2 exp ðβEÞ ð1 þ exp ðβEÞÞ2

ð30Þ

To find the maximum value of CkBV , let βE ¼ x in Eq. (30). This yields: x2 exp ðxÞ CV ¼ kB ð1 þ exp ðxÞÞ2

ð31Þ

At the maximum value of CkBV , we know that: d ðCV =kB Þ ¼0 dx

ð32Þ

Therefore, differentiating Eq. (31) with respect to x yields: d ðCV =kB Þ 2x exp ðxÞ x2 exp ðxÞ ¼  dx ð1 þ exp ðxÞÞ2 ð1 þ exp ðxÞÞ2 ðx2 exp ðxÞÞð2 exp ðxÞÞ ð1 þ exp ðxÞÞ3

 dðCV =kB Þ exp ðxÞx 2x exp ðxÞ ¼ 2xþ dx 1 þ exp ðxÞ ð1 þ exp ðxÞÞ2 þ

ð33Þ ð34Þ

Substituting Eq. (34) in Eq. (32) and equating to zero yields: 2xþ

2x exp ðxÞ ¼0 1 þ exp ðxÞ

!

x

2x exp ðxÞ ¼2 1 þ exp ðxÞ

ð35Þ

714

Solved Problems for Part III

The solution of Eq. (35) yields the value of x when CkBV attains its maximum value. Equation (35) can be written in a more compact form after some manipulation. Specifically: x

1 þ exp ðxÞ2 exp ðxÞ ¼2 1 þ exp ðxÞ

1  exp ðxÞ x 1 þ exp ðxÞ ¼ ¼2 ! 2 1  exp ðxÞ 1 þ exp ðxÞ   exp ðx=2Þ exp ðx=2Þ þ exp ðx=2Þ x ¼ 2 exp ðx=2Þ exp ðx=2Þ  exp ðx=2Þ

x

ð36Þ ð37Þ ð38Þ

Cancelling the equal terms in Eq. (38), and recognizing that the remaining term on the right-hand side of the equal sign is equal to coth(x/2), we obtain: x=2 ¼ cothðx=2Þ

ð39Þ

The solution of Eq. (39) is x ¼ βE ¼ 2.40. Remark A rigorous analysis to obtain the value of x at which CV/kB attains its maximum value should include checking the second-order derivative condition, in addition to the first-order derivative condition. An analytical differentiation of Eq. (34) to obtain the second differential may be quite messy. In such cases, one way out is to plot the function CV/kB numerically. Figure 1 shows the plot of CV/kB as a function of x. One can clearly see that x ¼ 2.40 corresponds to the maximum of the function.

Fig. 1

Solved Problems for Part III

715

Solution to Problem 21.3 Part (a) We know that:  < E >¼ kB T 2

 ∂ ln Q ∂T N,V

The calculation of < E > involves finding



ð40Þ



∂ ln Q ∂T N,V

given the expression for

Q in the Problem Statement. It is first useful to isolate the T-dependent terms in Q. Specifically: QðN, V, T Þ ¼

   3N=2 ðV  NbÞN 2πmk B aN 2 T 3N=2 exp N! VkB T h2

ð41Þ

Taking the logarithm of Eq. (41) yields: ln ðQÞ ¼

3N aN 2 ln T þ þ Terms which do not contain T 2 VkB T

ð42Þ

Taking the derivative of lnQ with respect to T, at constant N and V, yields:   ∂ ln Q 3N aN 2 ¼  ∂T N,V 2T Vk B T 2

ð43Þ

Substituting Eq. (43) in Eq. (40) yields the desired result:   3N aN 2  < E >¼ k B T 2T VkB T 2

ð44Þ

3 aN 2 < E >¼ Nk B T  2 V

ð45Þ

2

In Eq. (45), the van der Waals parameter a represents attractive interactions between the atoms and has a positive value. For a > 0, we see that the energy of a monoatomic van der Walls gas is lower than that of an ideal gas (32 NkB T ) by an 2 amount, aNV , which reflects the pairwise attractions between the atoms in the van der Waals gas, which are absent in the monatomic ideal gas.

716

Solved Problems for Part III

Part (b) After calculating < E >, we can calculate CV by recalling that:  CV ¼

∂U ∂T

 V

U ¼< E >     1 ∂U 1 ∂ ¼ ¼ N ∂T V,N N ∂T V,N

ð46Þ ð47Þ

Substituting Eq. (45) in Eq. (47) yields: CV ¼



 1 ∂ 3 aN 2 Nk B T  N ∂T 2 V V,N 3 CV ¼ kB 2

ð48Þ ð49Þ

Equation (49) shows that CV is the same for the monoatomic van der Waals gas and for a monoatomic ideal gas. This reflects the fact that the van der Waals parameter, a, is not a function of temperature! Part (c) The gas pressure can be calculated from the partition function as follows: P ¼ kB T

  ∂ ln Q ∂V N,T

ð50Þ

This time, we will isolate the terms in Q that depend on V, then take the logarithm, and finally differentiate with respect to V at constant N and T. Specifically:  3N=2   1 2πmk B T ðV  NbÞN exp aN 2 =Vk B T 2 N! h  3N=2   1 2πmk B T QðN, V, T Þ ¼ ðV  NbÞN exp aN 2 =Vk B T 2 N! h  3N=2   1 2πmk B T QðN, V, T Þ ¼ ðV  NbÞN exp aN 2 =Vk B T 2 N! h

QðN, V, T Þ ¼

aN 2 þ Terms which do not contain V VkB T   ∂ ln Q N aN 2 ¼  2 ∂V N,T V  Nb V kB T

ln Q ¼ N ln ðV  NbÞ þ

ð51Þ ð52Þ ð53Þ

Solved Problems for Part III

717

Substituting Eq. (53) in Eq. (50) then yields the desired expression for P. Specifically: P¼

NkB T aN 2  2 V  Nb V

ð54Þ

Equation (54) is, of course, the celebrated van der Waals equation of state first introduced in Part I, which we have now derived molecularly using statistical mechanics!

718

Solved Problems for Part III

Problem 22 Problem 22.1 In analogy with the characteristic vibrational and rotational temperatures, θvib and θrot, respectively, one can define a characteristic electronic temperature by: θelec,j ¼ εej=k B where εej is the energy of the j th excited electronic state relative to the ground state. (a) If one defines the ground electronic state to be the zero of energy, derive an expression for the electronic partition function, qelec, expressed in terms of θelec, j. (b) The first ( j ¼ 1) and the second ( j ¼ 2) excited electronic states of O(g) lie 158.2 cm1 and 226.5 cm1 above the ground electronic state ( j ¼ 0). Given the degeneracies, ge0 ¼ 5, ge1 ¼ 3, and ge2 ¼ 1, calculate the values of θelec,1, θelec,2, and qelec (ignoring any higher excited electronic states) for O(g) at 5000 K. (c) Calculate the fraction of O(g) atoms in the ground, first, and second electronic states at 5000 K. What is the fraction of O(g) atoms in all the remaining excited electronic states? Note: When energies are given in cm1 (which is typical in the case of electronic states, it is convenient to use kB ¼ 0.69509 cm1 K1).

Problem 22.2 Molecular nitrogen is heated in an electronic arc. The spectroscopically determined relative populations of the excited vibrational states are listed below: n

0

1

2

3

4

...

fn /f0 1.000 0.200 0.040 0.008 0.002 . . . You are asked to test if the spectroscopic data provided above corresponds to molecular nitrogen being in thermodynamic equilibrium with respect to vibrational energy, as well as to determine the actual temperature at which this equilibrium would be established.

Solved Problems for Part III

719

Problem 22.3 For O2(g), the following spectroscopic data is available: Mass ¼ 53:15
1027 kg Bond Length ¼ 1:21
1010 m v ¼ 1567 cm1 D0 ¼ 118 kcal mol1 (a) What is the fraction of O2(g) in the ground translational state when T ¼ 298 K and V ¼ 1000 cm3? (b) What is the fraction of O2(g) in the ground rotational state when T ¼ 298 K? (c) What is the fraction of O2(g) in the ground vibrational state when T ¼ 298 K? (d) Compute the enthalpy per molecule of O2(g) at 298 K and 1 atm.

Solution to Problem 22 Solution to Problem 22.1 Part (a) As discussed in Part III, if we define the ground electronic state ( j ¼ 0) to be the zero of energy (such that Ee0¼0), the electronic partition function can be written as follows: qelec

1 X



Eej ¼ gej exp kB T j¼0



  ϴelec,j ¼ gej exp T j¼0 1 X

ð1Þ

Note that j in Eq. (1) represents energy levels. Applying Eq. (1) to the given case: qelec ¼ ge0 exp ð0Þ þ

1 X

gej exp

j¼1

qelec

1 X



  ϴelec,j T

θelec,j ¼ ge0 þ gej exp T j¼1

ð2Þ



Part (b) Using the definition of ϴelec,j, we can compute ϴelec,1 and ϴelec,2 as follows:

ð3Þ

720

Solved Problems for Part III

ϴelec,1 ¼

Ee1 158:2 cm1 ¼ ¼ 227:6 K kB 0:69509 cm1 K 1

ϴelec,2 ¼

Ee2 226:5 cm1 ¼ ¼ 325:8 K kB 0:69509 cm1 K 1

Ignoring excited electronic states with energy levels j > 2, we can calculate qelec using Eq. (3) and the data given in the Problem Statement, as follows:  qelec ¼ ge0 þ ge1 exp

ϴelec,1 T

 þ ge2 exp

  ϴelec,2 T



qelec

   227:6 325:8 ¼ 5 þ 3 exp þ 1 exp 5000 5000 qelec ¼ 8:8034

ð4Þ ð5Þ

Part (c) The various fractions can be calculated using the following expression:

fj¼

  E gej exp  kB jT qelec

ð6Þ

Recall that j in Eq. (6) denotes energy levels. Because each energy level can be gej degenerate, the gej factor appears in the expression for fj. Using Eq. (6) for j ¼ 0, 1, and 2, we obtain: f0 ¼

f1 ¼

f2 ¼

ge0 5 ¼ 0:5679 ¼ qelec 8:8034

  ge1 exp  kEB1T qelec   ge2 exp  kEB2T qelec

ð7Þ

  3 exp  227:6 5000 ¼ 0:3256 ¼ 8:8034

ð8Þ

  1 exp  325:8 5000 ¼ 0:1064 ¼ 8:8034

ð9Þ

Because higher electronic states were ignored in the calculation of qelec, the fraction of O(g) in the higher states will be 0. However, if the contributions of the

Solved Problems for Part III

721

higher-energy states are included in the calculation of qelec, the fraction will turn out to be a very small positive quantity.

Solution to Problem 22.2 At thermal equilibrium, the population of state n is given by:    1  exp β n þ 12 hν fn ¼ qvib

ð10Þ

The term “1” in the numerator of Eq. (10) is due to the fact that the vibrational energy levels are nondegenerate. For n ¼ 0, Eq. (10) yields the population of the ground vibrational state:   exp β hν 2 f0 ¼ qvib

ð11Þ

Dividing Eq. (10) by Eq. (11) yields: fn ¼ exp ðβnhνÞ f0

ð12Þ

According to Eq. (12), if nitrogen isin thermodynamic equilibrium with respect to vibrational energy, then a plot of ln ff n versus n will be a straight line passing 0

through the origin with a slope of (βnhν). Let us examine  the  spectroscopic data given in the Problem Statement, and plot the values of ln ff n versus n. The table below summarizes the results: 0

n

0

1

2

3

4

fn/f0 1.000 0.200 0.040 0.008 0.002 Ln( fn/f0) 0.000 1.609 3.218 4.828 6.221 The best fit to the data is a straight line passing through the origin with a slope  1.5804 (see Fig. 1). From Eq. (12), the slope is given by:

722

Solved Problems for Part III

0 -1

n 0

1

2

3

4

ln(fn/fo)

-2 -3 -4

y = -1.5804x R² = 0.9991

-5 -6 -7

Fig. 1

βnhν ¼ 

hν hcν ¼ ¼ 1:5804 kB T kB T

T¼

hcev 1:5804kB

ð13Þ

Substituting h ¼ 6:626
1034 Js, c ¼ 2:998
1010 cms1 kB ¼ 1:381
1023 J K 1 ev ¼ 2330 cm1 in Eq. (13), we obtain: T ¼ 2121 K Solution to Problem 22.3 The fraction of O2(g) in a specific state j is equivalent to the probability of finding O2(g) in that state and is given by:

Solved Problems for Part III

723

  exp βE j f j ¼ pj ¼ q

ð14Þ

Equation (14) shows that in order to calculate pj, one has to determine Ej and q. Contrary to Eq. (6), gj does not appear in the numerator of Eq. (14). This is because fj in Eq. (14) represents the fraction in a specific energy state, while fj in Eq. (6) represents the fraction in a specific energy level. Part (a) As discussed in Part III, the ground-state energy for translational motion can be obtained from the expression:

Enx ,ny ,nz ¼

  h2 n2x þ n2y þ n2z 8mV 2=3

,

with

nx ¼ ny ¼ n z ¼ 1

ð15Þ

That is: E111 ¼

3h2 8mV 2=3

ð16Þ

The translational partition function is given by: qtrans

 3=2 2πmk B T ¼ V h2

ð17Þ

Using the given values of m and V in Eq. (16) yields: E111

 2 3 6:626
1034 Js ¼   2 8 53:15
1027 kg ð1000 cm3 Þ3 E111 ¼ 3:097
1040 J

Next, we calculate qtrans using the values of h ¼ 6.626
1034 Js, m ¼ 53.15
1027 kg, kB ¼ 1.381
1023 J K1, T ¼ 298 K, and V ¼ 1000 cm3 in Eq. (17). This yields: qtrans ¼ 1:75
1029 From Eq. (14), the fraction of molecular oxygen in the ground translational state is given by:

724

Solved Problems for Part III

f trans 111

  exp βE j ¼ qtrans

ð18Þ

Substituting the values of E111 ¼ 3.097
1040 J and qtrans ¼ 1.75
1029 in Eq. (18) yields: 30 f trans 111 ¼ 5:7143
10

Because kBT ¼ 4.115
1025 E111, most translational states are accessible. As a result, the probability of occupancy of any one of the available translational states is the same and is given by q 1 . trans

Part (b) Similar to our approach in Part (a), we will first determine the ground-state rotational energy and the rotational partition function and then use Eq. (14) to calculate the fraction of O2(g) in the ground-rotational state. The energy levels of a rotational state are given by: EJ ¼

h2 J ð J þ 1Þ 2I

ð19Þ

The ground-state rotational energy corresponds to J ¼ 0 in Eq. (19), such that: E0 ¼ 0

ð20Þ

The rotational partition function of O2(g) is given by: qrot ¼

T σϴrot

ð21Þ

where in the case of O2(g), the symmetry number, σ ¼ 2, and where: ϴrot ¼

h 8π 2 Ik B

ð22Þ

In Eq. (22), the moment of inertia, I, is given by: I ¼ μd2 where μ is the reduced mass of the molecule and is given by:

ð23Þ

Solved Problems for Part III

725

μ¼

mðOÞmðOÞ mðOÞ mðO2 Þ ¼ ¼ 2 4 mðOÞ þ mðOÞ

ð24Þ

Using the value of m(O2) ¼ 53.15
1027 kg in Eq. (24) yields: μ¼

mðO2 Þ ¼ 13:28
1027 kg 4

ð25Þ

Using the given value of the bond length, d, and μ ¼ 13.28
1027 kg in Eq. (23) yields: I ¼ 1:945
1046 kgm2 Using this value of I in Eq. (22), ϴrot is given by: ϴrot ¼ 2:07 K

ð26Þ

Recall that Eq. (22) was derived under the assumption that ϴrot  T. Indeed, at T ¼ 298 K, we see that ϴrot  T. We can now use ϴrot in Eq. (26), and σ ¼ 2, to calculate qrot in Eq. (21). Specifically, qrot ¼

298 ¼ 71:98 2ð2:07Þ

We can next calculate the fraction of O2(g) in the ground rotational state using Eq. (14), which yields: f rot 0 ¼

exp ðβE0 Þ qrot

Substituting the values of E0¼ 0 from Eq. (20), and qrot ¼ 71.98 from Eq. (26), in the last result yields: f rot 0 ¼

exp ðβð0ÞÞ 1 ¼ 71:98 71:98 f rot 0 ¼ 0:0139

Part (c) We will follow the same approach as in Parts (a) and (b) to determine the fraction of O2(g) in the ground vibrational state. That is, we will first calculate the ground-state energy, the vibrational partition function, and then use Eq. (14) to obtain the desired quantity.

726

Solved Problems for Part III

The vibrational energies are given by:     1 1 ν En ¼ n þ hν ¼ n þ hce 2 2

ð27Þ

The ground-state vibrational energy corresponds to n ¼ 0, and hence: 1 ν E0 ¼ hce 2

ð28Þ

The vibrational partition function is given by: qvib ¼

exp ðϴvib =2T Þ 1  exp ðϴvib =T Þ

ð29Þ

hce ν kB

ð30Þ

where ϴvib ¼

Now that we derived expressions for E0 and qvib, let us compute the values of these quantities. Using the values of h ¼ 6.626
1034 Js, c ¼ 2.998
1010 cm s1, kB ¼ 1.381
1023 J, and e ν ¼ 1567 cm1 in Eqs. (28) and (30), E0 and ϴvib can be calculated as follows: E0 ¼ 1:5564
1020 J

ð31Þ

ϴvib ¼ 2254 K

ð32Þ

Because ϴvib T, we anticipate that few vibrational states will be accessible besides the ground vibrational state. That is, we can expect to observe a high fraction of O2(g) in the ground vibrational state. Let us see if our calculation reflects that. Using ϴvib from Eq. (32) in Eq. (29), we find: qvib ¼ 0:2279

ð33Þ

We can now calculate the fraction of O2(g) in the ground vibrational state by substituting the values of E0 ¼ 1.5564
1020J from Eq. (31) and qvib ¼ 0.2279 from Eq. (33) in Eq. (14). This yields: f vib 0 ¼

exp ðβE0 Þ ¼ 0:99956 qvib

ð34Þ

As expected, Eq. (34) clearly shows that O2(g) is essentially in the ground vibrational state.

Solved Problems for Part III

727

Part (d) Calculating the molecular enthalpy for O2(g) requires that we calculate the molecular internal energy of O2(g) using the expression derived in Part III for diatomic molecules and then use the thermodynamic relation: H ¼ U þ PV

ð35Þ

" #) ϴvib 3 2 ϴvib T  U ¼ kB T þ þ þ  De 2 2 2T exp ϴTvib  1

ð36Þ

where (

In Eq. (35), because T ¼ 298 K and P ¼ 1 atm, oxygen can be modeled as an ideal gas, for which: PV ¼ kB T

ð37Þ

Using Eqs. (36) and (37) in Eq. (35), we obtain: (

" #) θvib 3 2 θvib T   þ þ1þ H ¼ kB T þ  De 2 2 2T exp θTvib  1

ð38Þ

Next, we will calculate the values of each of the terms in Eq. (38) separately and then will substitute them in Eq. (38) to determine H. From the calculation carried out in Part (c), we obtain: ϴvib ¼ 2254 K 1 1 ν De ¼ D0 þ hν ¼ D0 þ hce 2 2   118
4:184 Jmol1 1  De ¼ þ 6:626
1034 Js 2:998
1010 cm s1 1 23 2 6:023
10 mol   1
1567 cm

ð39Þ

ð40Þ

ð41Þ

In Eq. (41), we note that we converted the basis of the given value of D0 from moles to molecules by dividing by Avogadro’s number. Use of Eq. (41) yields: De ¼ 8:5097
1019 J

ð42Þ

728

Solved Problems for Part III

We are now ready to compute H in Eq. (39) using ϴvib from Eq. (39) and De from Eq. (42). Substituting the values of ϴvib ¼ 2254K, De ¼ 8.5097
1019J, and T ¼ 298 K in Eq. (38), we obtain: H ¼ 8:2099
1019 J Note: There is no need to be concerned about the negative value of H. This is because of the negative contribution by De to H and the fact that De has a negative contribution because of the choice of reference state for the electronic energy levels.

Solved Problems for Part III

729

Problem 23 Problem 23.1 Consider a system of N identical, but distinguishable, particles, each having two energy levels with energies 0 and ε > 0, respectively. The upper level is g-fold degenerate and the lower level is nondegenerate. The total energy of the system is E. (a) Use the Micro-Canonical ensemble to calculate the entropy of the system. Express your result in terms of g, N, and the occupation numbers of the upper and lower levels, n+ and n0, respectively, where n+ + n0 ¼ N and E ¼ n+ε. Note that your result corresponds to the entropy fundamental equation, S ¼ S (E, N), where in the present case, there is no explicit dependence on the system volume, V. (b) Use the result in Part (a) to derive an expression for the temperature of the system. (c) Use the result in Part (b) to derive expressions for the occupation numbers n0 and n+. Show that the same expressions can be derived much more readily in the context of the Canonical ensemble.

Problem 23.2 Consider a vessel containing a gas consisting of N independent and indistinguishable atoms at pressure, P, and temperature, T. The walls of the vessel have n adsorbing sites, each of which can only adsorb one atom. Let – ε be the energy of an adsorbed atom. (a) Use the Grand-Canonical ensemble to derive an expression for the fugacity, λ ¼ exp (βμ). Express your result in terms of T, P, and the necessary atomic constants. (b) Derive an expression for the average number of atoms, , that adsorb from the gas onto the walls of the vessel. Express your result in terms of n, ε, T, P, and the necessary atomic constants. Discuss the low- and high-pressure and temperature behaviors of , and provide a physical interpretation of these behaviors.

Problem 23.3 (a) Calculate the vibrational partition function of a diatomic molecule treating the vibrational mode classically. (b) Calculate the average vibrational energy of a diatomic molecule treating the vibrational mode classically.

730

Solved Problems for Part III

(c) Compare your results in Parts (a) and (b) with the corresponding quantum mechanical results derived in Part III, and determine the temperature conditions at which the classical and the quantum mechanical results become identical.

Solution to Problem 23 Solution to Problem 23.1 In this problem, we are asked to find the entropy, S, of a system of N identical, but distinguishable, particles using the Micro-Canonical ensemble, where the independent variables are E, V, and N. Knowing S as a function of E, V, and N, T can be evaluated using concepts from classical thermodynamics discussed in Part I. Lastly, we are asked to compare the results obtained using the Micro-Canonical ensemble to the results obtained using the Canonical ensemble, where the independent variables are T, V, and N. Part (a): Calculating W and S We begin with Boltzmann’s celebrated expression for the entropy in the context of the Micro-Canonical ensemble. Specifically: S ¼ kB lnðWÞ

ð1Þ

In Eq. (1), kB is the Boltzmann constant, and W is the number of distinct states of the system for a given E, V, and N, that is, the degeneracy corresponding to E, V, and N. Note that W is not the degeneracy, g, of an energy level. Recall that according to the Problem Statement, we have: E ¼ nþ ε

ð2Þ

Therefore, we can also regard W as the number of distinct ways in which we can arrange N particles such that the total energy is always E. An examination of Eq. (2) shows that only the number of particles in the upper energy level, n+, contributes to the total energy. Calculation of W, therefore, involves counting the distinct number of ways in which n+ distinguishable particles out of the available N particles can occupy the upper energy level, ε, including recognizing that the n+ particles can be assigned to any one of the g available degenerate states. Accordingly, W can be expressed as follows:  W¼

 N! ðgnþ Þ nþ !ðN  nþ Þ!

ð3Þ

In Eq. (3), the first quantity in parentheses represents the number of distinct ways in which n+ distinguishable particles can occupy the upper energy level, and the

Solved Problems for Part III

731

second quantity in parentheses represents the number of distinct ways of assigning the n+ particles to g degenerate states. Using Eq. (3) in Eq. (1), we obtain:



  N! ð gn þ Þ nþ !ðN  nþ Þ!

S ¼ kB ln

ð4Þ

Rearranging Eq. (4) yields: S ¼ kB fln N!  ln nþ !  ln ðN  nþ Þ! þ nþ ln gg

ð5Þ

Because N >>1, we can use Stirling’s approximation, that is: ln y! ¼ y ln y  y, for y >> 1

ð6Þ

Using Eq. (6) in Eq. (5), we obtain: S ¼ k B fN ln N  N  nþ ln nþ þ nþ  ðN  nþ Þ ln ðN  nþ Þ þ N  nþ þ nþ ln gg Rearranging and simplifying the last expression yields:  S ¼ kB

ð N  nþ Þ n  nþ ln g nþ ln þ þ ðN  nþ Þ ln N N

 ð7Þ

For convenience, let us define the fraction of molecules in the upper level as follows: x¼

nþ E=ε E ¼ ¼ εN N N

ð8Þ

Recall that E¼n+ε (see the Problem Statement). Substituting Eq. (8) in Eq. (7) yields: S ¼ kB N fx ln x þ ð1  xÞ ln ð1  xÞ  x ln gg

ð9Þ

Equation (9) can also be expressed as follows. n S ¼ k B N ln

1 1x þ x ln þ x ln g 1x x

o

ð10Þ

Part (b): Obtaining T Note that because x¼E/Nε, Eq. (9), or Eq. (10), expresses S as a function of E and N (the volume V does not appear explicitly in Eq. (9), although it could affect the value of ε which is assumed to be a constant). Therefore, Eq. (9), or Eq. (10), is

732

Solved Problems for Part III

essentially the entropy fundamental equation corresponding to this system. As discussed in Part I, we can calculate the temperature recalling that S and E are related as follows: 

∂S ∂E

 ¼ N

1 T

ð11Þ

Because E ¼ Nεx, Eq. (11) can be expressed as follows: 

∂S ∂E

 N

  1 ∂S ¼ Nε ∂x N

ð12Þ

Differentiating Eq. (9) (or Eq. (10)) with respect to x, at constant N, yields:   n o ∂S x 1x  ln g ¼ kN ln x þ  ln ð1  xÞ  x 1x ∂x N   gð1  xÞ ∂S ¼ kN f ln ð1  xÞ þ ln g  ln xg ¼ kN ln x ∂x N

ð13Þ

Using Eq. (3) in Eq. (12) yields: 

∂S ∂E

 ¼ N

gð1  xÞ k gð1  xÞ 1 ¼ ln kN ln x x Nε ε

ð14Þ

Using Eqs. (11) and (14), we can now calculate the temperature which is given by: T¼

1 gð 1  x Þ ε n E ln ,x ¼ þ ¼ x k Nε N

ð15Þ

Part (c): Determining n+ and no For large values of N, nNþ represents the probablility of finding a particle in the upper energy level (p+). In that case, rearranging Eq. (15) for x yields:   gexp ε gexp ðεβÞ nþ kT  x¼ ¼ pþ ¼ ¼ ε N 1 þ gexp ðεβÞ 1 þ gexp kT

ð16Þ

Equation (16) can be readily derived using the molecular partition function in the context of the Canonical ensemble, where q is given by:

Solved Problems for Part III

733

q ¼ exp ðβεo Þ þ g exp ðβεÞ ¼ 1 þ g exp ðβεÞ recalling that εo ¼ 0. The fraction of particles in the lower-energy level, or po, is given by: po ¼

1 1 þ gexp ðβεÞ

ð17Þ

pþ ¼

gexp ðεβÞ 1 þ gexp ðεβÞ

ð18Þ

Similarly, p+ is given by:

Upon comparison, Eq. (18) (obtained using the Canonical ensemble) is identical to Eq. (16) (obtained using the Micro-Canonical ensemble). We can next calculate no and n+ using Eqs. (17) and (18), respectively. Specifically: no ¼

N 1 þ gexp ðβεÞ

ð19Þ

nþ ¼

Ngexp ðεβÞ 1 þ gexp ðεβÞ

ð20Þ

Solution to Problem 23.2 This problem asks us to use the Grand-Canonical ensemble to study this system. The Grand-Canonical ensemble is typically used to study systems in equilibrium. The system described in this problem contains two simple systems: the gas in the threedimensional region and the two-dimensional adsorption sites. We will focus on each of these systems separately. Then, we can relate the two systems by imposing the condition of equality of the chemical potentials. In Part (a), we are asked to derive an expression for λ, or the fugacity, in terms of P and T, and in Part (b), we are asked to derive an expression for , the average number of atoms adsorbed onto a wall. Since this is an equilibrium situation, where atoms are in equilibrium between the gas phase and the adsorbed phase, the use of the Grand-Canonical ensemble is most appropriate. Recall that the Grand-Canonical ensemble was discussed in detail in Part III.

734

Solved Problems for Part III

Part (a): Calculate the Fugacity, λ, Using the Grand-Canonical Ensemble Because the gas atoms can adsorb onto the walls of the vessel, the number of gas atoms in the three-dimensional gas region does not remain constant. As discussed in Part III, the Grand-Canonical partition function is given by: ΞðV, T, μÞ ¼

1 X

QðN, V, TÞλN , where λ ¼ exp ðβμÞ

ð21Þ

N¼0

For a collection of N independent, indistinguishable atoms, we know that: ½qðV, T ÞN N!

ð22Þ

1 X ½qðV, T ÞλN N! N¼0

ð23Þ

QðN, V, T Þ ¼ where q is the atomic partition function. Substituting Eq. (22) in Eq. (21) yields: ΞðV, T, μÞ ¼

where

1 q V, T λ N P ½ð Þ N¼0

N!

¼ eqλ : Using the last result in Eq. (23), we obtain: ΞðV, T, μÞ ¼ exp ðqλÞ

ð24Þ

In Part III, we showed that: PV ¼ k B T ln Ξ

ð25Þ

Combining Eq. (25) with Eq. (24) yields: λ¼

PV kB Tq

ð26Þ

For a monoatomic ideal gas, we know that:  3=2 2πmk B T q¼ Vge1 h2 Using Eq. (27) in Eq. (26), we obtain:

ð27Þ

Solved Problems for Part III

735

λ ¼ Pðk B T Þ5=2



3=2 h2 1 ge1 2πm

ð28Þ

where λ is not a function of V. Part (b): Calculate We are asked to calculate the average number of atoms in the adsorbed state. Naturally, the focus is the adsorbing wall, which we regard as being in equilibrium with the gas. Accordingly, μ and T are the same for both the gas and the adsorbed atoms. In fact, each adsorbing site acts independently, is separately in equilibrium with the gas, and has a Grand-Canonical partition function derived in Part III. Specifically: Ξsite ¼

1 X

QðN, V, T ÞλN

ð29Þ

N¼0

and QðN, V, T Þ ¼

X eβE j j

where the summation is over all possible states corresponding to N and V. In our case, there exists only one energy level for a given N and V, that is, -Nε. Therefore, Eq. (29) becomes: Ξsite ¼

1 X

exp ðβNεÞλN

ð30Þ

N¼0

In other words: Ξsite ¼ e0 λ0 þ eβε λ ¼ 1 þ λeβε

ð31Þ

The average number of adsorbed atoms per site was derived in Part III. Specifically: < N>site ¼ kT  < N>site ¼ kT

     ∂ln Ξ site ∂ ln 1 þ eβε eβμ ¼ kT ∂μ ∂μ T,V T,V

βeβε eβμ 1 þ eβε eβμ

 ¼

eβε eβμ 1 1 ¼ ¼ 1 þ eβε eβμ 1 þ eβε eβμ 1 þ λ1 eβε

ð32Þ

ð33Þ

736

Solved Problems for Part III

It then follows that for n adsorbing sites, the average number of adsorbing atoms is given by: < N >¼< N>site n ¼

n 1 þ λ1 eβε

ð34Þ

At this stage, we impose the condition of thermodynamic equilibrium between the gas and the adsorbing wall. Accordingly, using the expression for λ (see Eq. (28)) in Eq. (34), we obtain the desired expression for the average number of adsorbed atoms as a function of T and P. Specifically: < N >¼ 1þ



2πm h2

3=2

n 

ð35Þ

ge1 P1 ðkT Þ5=2 eβε

At fixed temperature, < N > varies with pressure in a manner that is expected intuitively. At high pressures, P 1, P1 n. Physically, at high pressures, the gas density is high and the atoms frequently approach and adsorb onto the adsorbing sites, so that < N >  n. At low pressures, P  1, P1 >> 1, and < N > 0. The effect of temperature is also apparent. At high temperatures, T !1, eβε  1, and (kT)5/2 1, so that < N > 0, because the adsorbed atoms are easily released from the adsorbing sites. At low temperatures, T !0, eβε  0, and (kT)5/2 ! 0, so that < N > n. Clearly, at high temperatures, a high pressure is required to keep the adsorbing sites filled, while only a modest pressure is required at low temperatures. With respect to the temperature dependence, the atoms can be thought of as possessing kinetic energy in the three-dimensional phase and zero kinetic energy in the two-dimensional phase. When adsorbed onto the two-dimensional wall, an atom has no kinetic energy but gains -ε of energy through its interaction with the wall. When the temperature is very low, the kinetic energy of an atom is so small relative to ε, that the atom prefers to gain ε of energy by adsorbing onto the wall. On the other hand, when the temperature is high, the atom has far more kinetic energy than ε and, therefore, prefers to access the three-dimensional gas phase.

Solution to Problem 23.3 In this problem, the main goal is to compare the classical and quantum mechanical representations of q and . At some temperature T, the classical and the quantum mechanical values approach each other. Although we are asked to calculate the vibrational contribution, similar calculations could be carried out for the translational or the rotational contributions. In Part III, we discussed the vibrational partition function in detail and provided an introductory exposition to the classical partition function.

Solved Problems for Part III

737

Part(a): Calculate the Classical Vibrational Partition Function We have already calculated the quantum mechanical vibrational contribution to the partition function for a diatomic molecule. Let us recall the results: (i) The energy levels of the harmonic oscillator are quantized according to:   1 εn ¼ n þ hν, n ¼ 0, 1, 2 . . . 2

ð36Þ

(ii) Results presented in Part III indicate that: qQM vib ¼

exp ðθvib =2T Þ , θ ¼ hν=k 1  exp ðθvib =T Þ vib (

< εvib >

QM

θvib  ) θvib θvib T exp T  þ ¼ kT 2T 1  exp θvib

ð37Þ

ð38Þ

T

where QM stands for quantum mechanics. In this problem, we are asked to calculate qvib and treating the vibrational modes classically. We therefore consider the energy of each possible vibrational state as a continuous variable. We consider the classical Hamiltonian description of ! ! the energy of a diatomic molecule in terms of its position, q , and momentum, p : Here, we envision a diatomic molecule as made up of two atoms connected by a Hookean spring. In this case, we have two degrees of freedom: the displacement, x, and the momentum associated with that displacement, p. We can write the classical Hamiltonian describing the motion of the two atoms and their interactions (through the stretch of the spring) as follows (see Part III): H¼

p2 1 2 þ kx 2m 2 s

where m is the reduced mass of the diatomic molecule and ks is the spring constant. As shown in Part III, the spring constant is given by k s ¼ 4π 2 ν2 m

ð39Þ

where ν is the characteristic vibrational frequency of the system. We can also use a result derived in Part III to calculate qCvib , where C stands for classical. Specifically:

738

Solved Problems for Part III

1 ¼ h

qcvib

1 ð

1 ð

eβH ðp,xÞ dpdx

ð40Þ

1 1

Using the expression for H above in Eq. (40) and separating the integrals over p and x, we obtain: 0 1 qcvib ¼ @ h

10

1 ð

e

2 βp 2m

dpA@

1

1

1 ð

e

β2k s x2

dxA

ð41Þ

1

where the two integrals in Eq. (41) are Gaussian integrals. In Part III, we showed that: 1 ð

ea y dy ¼ 2 2

1

pffiffiffi π a

ð42Þ

Using Eq. (42) in Eq. (41), we obtain: qcvib

 1=2  1=2 1 2πm 2π 2πm1=2 ¼ ¼ h β βks hβks1=2

ð43Þ

or qcvib ¼

! 2πm1=2 kB T hk 1=2 s

ð44Þ

Part (b): Calculate the Average Vibrational Energy To calculate the average vibrational energy, we use the following equation derived in Part III:  < εvib >c ¼ kB T 2

∂ ln qcvib ∂T

 ð45Þ V

Using Eq. (44) in Eq. (45) yields:  < εvib >c ¼ k B T 2

∂ ð ln T þ terms that do not contain T Þ ∂T

< εvib >c ¼ kB T

 ð46Þ V

ð47Þ

Solved Problems for Part III

739

Part (c): Compare the Classical and the Quantum Mechanical Limits An examination of Eq. (37) shows that if T >> θvib, we obtain the following result (see Part III): lim T >> θvib , qQM vib ¼

1  θvib =2T T T k T ¼ ¼ B ¼ hν 1  ð1 þ ðθvib =T ÞÞ θvib hν k

ð48Þ

B

According to Eq. (44): qcvib

! 2πm1=2 kB T hk 1=2 s

¼

Using the expression for ks, as well as Eq. (56), in Eq. (61) yields: qcvib ¼

kB T hν

ð49Þ

A comparison of Eq. (66) and Eq. (65) shows that when T θvib, qcvib ¼ qQM vib . Next, let us consider the average energy in the quantum mechanical representation. Recall that for the QM case: (

< εvib >QM

θ  ) θvib vib exp θvib T θ  þ T ¼ kT vib 2T 1  exp

ð50Þ

T

In the limit T θvib, Eq. (67) reduces to: (

< εvib >QM

 ) θvib θvib 1  θvib T ¼ kT þ T ¼ kB T 2T 1  1 þ θTvib

ð51Þ

A comparison of Eq. (51) and Eq. (47) shows that in the limit T θvib, C ¼ < εvib>QM.

740

Solved Problems for Part III

Problem 24 Adapted from Molecular Driving Forces - Statistical Thermodynamics in Chemistry and Biology by Ken A. Dill and Sarina Bromberg, Garland Science, Taylor & Francis Group, New York and London (2003). Hereafter, we will abbreviate the names of the authors as D&B.

Problem 24.1 (a) One mole of a molecular system can occupy any one of the four energy states below: ________ E3 ¼ 11 kCal/mol E1 ¼ 8 kCal/mol ________

________ E2 ¼ 8 kCal/mol

________ E0 ¼ 3 kCal/mol 1. Calculate U at T ¼ 300 K. 2. Calculate the probability that a given snapshot of the system will have an energy of 3 kCal/mol at 300 K. 3. If the energy of each state is increased by 2 kCal/mol, what is the probability that a given snapshot of the system will have an energy of 3 kCal/mol at 300 K? 4. Calculate U as T gets very large. 5. Calculate U as T gets very small. (b) A four-bead chain can adopt several conformations that may be grouped as shown in the energy ladder below:

Solved Problems for Part III

741

The distance between the chain ends is 1 latticepunit ffiffiffi in the compact conformation, 3 lattice units in the extended conformation, and 5 lattice units in each of the other three chain conformations. 1. Calculate the average end-to-end distance of the chain (in lattice units) as a function of temperature. 2. Calculate the maximum value of the average end-to-end distance of the chain (in lattice units). At which temperature will this be realized? 3. Calculate the temperature at which the average end-to-end distance of the chain is equal to one half of its maximum value.

Problem 24.2 (a) The protein below has four distinguishable binding sites (α, β, γ, and δ) for the ligand L. Find the protein equilibrium binding population for a case where the ligand-protein association and dissociation constants are equal. Specifically, calculate the number of distinct arrangements, W, and the entropy (in units of kB) when:

742

Solved Problems for Part III

1. 2. 3. 4. 5. 6. 7. 8.

No ligands are bound (NL ¼ 0). One ligand is bound (NL ¼ 1). Two ligands are bound (NL ¼ 2). Three ligands are bound (NL ¼ 3). Four ligands are bound (NL ¼ 4). Which states have the highest entropy? Which states have the lowest entropy? If the binding constants are not equal, namely, if ligand L has a higher probability of being bound, say: Pbound ¼ 75% and Punbound ¼ 25%

Calculate the probability distribution for all the available states. (b) Consider a zipper that has N links. Each link can be in two states, where state 1 means that the zipper is closed and has energy of zero (the zipper ground state) and state 2 means that the zipper is open and has energy ε (the zipper excited state). The zipper can only unzip from the left end, and the ith link cannot open unless all the links to its left (1, 2, 3. . . i1) are already open. HINT:

k P n¼0

kþ1

ar n ¼ a 1r 1r , when r < 1.

1. Think carefully about what microstates are allowed, and derive an explicit expression for the Canonical partition function, Q, for the zipper. 2. If the average energy of the zipper is , find the average number of open links, , in the low-temperature limit, ε/kBT >> 1. (Do not leave in your final expression, but instead, calculate it.)

Problem 24.3 (a) Given the intermolecular potential, φ2(r), as a function of intermolecular separation, r, shown below, derive an expression for the second virial coefficient, B2(T ). Express your result solely in terms of ε, σ, Rσ, and kBT.

Solved Problems for Part III

743

(b) Consider an ideal gas of molecules which possess permanent dipole moments, ! ! μ , in an external electric field, ε . It is known that the potential energy of a single dipolar molecule in an external electric field is u ¼ μεcosθ, where θ is ! ! the angle between the vectors μ and ε . The Hamiltonian for a single molecule possessing a permanent dipole moment and interacting with an external electric field can be expressed as the sum of translational, rotational, and potential energy contributions. The contribution to the Hamiltonian related to rotational energy can be modeled using a rigid rotor. 1. Write the Hamiltonian expression for a single molecule possessing a permanent dipole moment in an external electric field. 2. Derive the classical partition function for the single gas molecule in Part (a). 3. Calculate the additional contribution to the ideal gas internal energy resulting from the dipole-electric field interactions.

Solution to Problem 24 Solution to Problem 24.1 (a1) Because the number of moles, the temperature, and the system volume are constant, we can use the Canonical ensemble to solve this problem. Working on a per mole basis, underbars will not be carried in the solution of this problem. It is most convenient to compute the average energy, hEi, or internal energy, U, as follows:

744

Solved Problems for Part III 3 P

U ¼ hE i ¼

Ei eEi =RT

i¼0

Q

ð1Þ

where the Canonical partition function, Q, is given by: Q¼

3 X

eEi =RT

ð2Þ

i¼0

In Eqs. (1) and (2), we consider four distinguishable energy states, E0, E1, E2, and E3 (see the energy ladder in the Problem Statement). At T ¼ 300 K, it follows that:   RT T¼300 K ¼ 1:98717
103 kCal=ðmolKÞ ð300 KÞ ¼ 0:5962 kCal=mol

ð3Þ

Using Eqs. (2) and (3), along with E0 ¼ 3 kCal/mol, E1 ¼ E2 ¼ 8 kCal/mol, and E3 ¼ 11 kCal/mol, we obtain: Q ¼ eð3Þ=ð0:5962Þ þ 2eð8 Þ=ð0:5962Þ þ eð11Þ=ð0:5962 Þ ¼ 0:006527

ð4Þ

In addition, the numerator in Eq. (1) is given by: 3 X

h i E i eEi =RT ¼ ð3Þeð3Þ=ð0:5962Þ þ 2ð8Þeð8 Þ=ð0:5962Þ þ ð11Þeð11Þ=ð0:5962 Þ kCal=mol

i¼0 3 X

Ei eEi =RT ¼ 0:019596 kCal=mol

ð5Þ

i¼0

Using Eqs. (5) and (4) in Eq. (1) yields the desired result: U¼

0:019532 kCal=mol ¼ 3:00229 kCal=mol 0:006506

ð6Þ

(a2) The probability that the system will be in the ground state (0) having an energy E0 ¼ 3 kCal/mol is given by: p0 ¼

eE0 =RT Q

ð7Þ

Using E0 ¼ 3 kCal/mol, along with Eqs. (3) and (4), in Eq. (7) yields: p0 ¼

e3=0:5962 ¼ 0:99954, or 99:954%, a very high probability! 0:006506

ð8Þ

(a3) If each energy is increased by 2 kCal/mol, then, the possible energy states will

Solved Problems for Part III

745

be E0 ¼ 5kCal/mol, E1 ¼ E2 ¼ 10kCal/mol, and E3 ¼ 13kCal/mol. Therefore, the lowest possible energy state that the system can attain is now 5 kCal/mol. As a result, the system can never have E ¼ 3 kCal/mol, and hence P (E ¼ 3 kCal/ mol) ¼ 0! (a4) As T ! 1, the term eEi =RT tends to 1. It then follows that: pi ¼

eEi =RT 1 1 ¼ ¼ Q 4 Q

ð9Þ

Equation (9) indicates that the four available energy states are likely to be equally populated. As a result, the ensemble-averaged energy, U, is equal to the average of E0, E1, E2, and E3. That is: U ¼ hE i ¼

½3 þ ð2Þð8Þ þ 11 kCal=mol 30 ¼ kCal=mol ¼ 7:5 kCal=mol 4 4 ð10Þ

(a5) As T ! 0, the term eEi =RT tends to zero. However, it tends to zero more rapidly for the higher-energy states than for the lower-energy states. As a result, the only energy state that is likely to be populated is the ground state (0) having energy E0 ¼ 3 kCal/mol. In other words, as T ! 0, the ensemble-averaged energy is equal to the energy of the ground state. In mathematical terms, one obtains:

U jT!0

3 P E i eEi =RT E eE0 =RT i¼0 ¼ hEi T!0 ¼ lim 3 ¼ lim 0E =RT ¼ E 0 ¼ 3 kCal=mol T!0 P T!0 e 0 eEi =RT i¼0

ð11Þ (b1) As can be seen from the energy ladder in the Problem Statement, there are five conformations of the four-bead chain: the conformation with the bead-bead contact is taken to have energy ε ¼ 0 (the ground state). The other four conformations have no bead-bead contacts, and all have energy ε ¼ ε0, where ε0 is a constant. The average end-to-end distance of the four-bead chain is given by:

hd i ¼

5 X

d i pi

i¼1

where according to the Problem Statement (in lattice units):

ð12Þ

746

Solved Problems for Part III

d1 ¼ 1 pffiffiffi d2 ¼ d3 ¼ d4 ¼ 5 d5 ¼ 3

ð13Þ

ε1 ¼ 0 ε2 ¼ ε3 ¼ ε4 ¼ ε5 ¼ ε0 In addition, the various probabilities are given by: X eε1 =kB T 1 eε0 =kB T eεi =kB T ¼ , p2 ¼ p3 ¼ p4 ¼ p5 ¼ , and q ¼ q q q i¼1 5

p1 ¼

¼ 1 þ 4eε0 =kB T

ð14Þ

Using Eqs. (13) and (14) in Eq. (12) yields: pffiffiffi   3 5 eε0 =kB T 3eε0 =kB T 1 hd i ¼ 1 þ þ q q q

ð15Þ

Using q in Eq. (14) in Eq. (15) yields: hd i ¼

1 þ 9:71eε0 =kB T 1 þ 4eε0 =kB T

ð16Þ

(b2) The maximum value of the end-to-end distance of the chain is obtained when T ! 1 and eε0 =kB T ! 1. In that case (in lattice units): lim hdi ¼ hdimax ¼

T!1

1 þ 9:71 10:71 ¼ ffi 2:14 1þ4 5

ð17Þ

(b3) We are asked to calculate at what temperature the end-to-end distance of the chain is equal to half of hdimax given in Eq. (17). We therefore require that hdi in hd i Eq. (16) be equal to 2max . Specifically: 1 þ 9:71eε0 =kB T hdimax ¼ ¼ 1:07 2 1 þ 4eε0 =kB T   1 þ 9:71eε0 =kB T ¼ 1:07 1 þ 4eε0 =kB T

ð18Þ

Solved Problems for Part III

747

ð9:71  1:07ð4ÞÞeε0 =kB T ¼ 5:43eε0 =kB T ¼ ð1:07  1Þ ¼ 0:07 eε0 =kB T ¼ 

0:07 ¼ 0:01305 5:43

ε0 ¼ ln ð0:01305Þ ¼ 4:339 kB T

  ε T ¼ 0:2305 0 ¼ 0:2305T 0 kB

ð19Þ

Solution to Problem 24.2 4! (a1) N L ¼ 0 ¼> W ¼ 0!4! ¼ 1, only one arrangement is possible.

S ¼ kB ln W ¼ k B ln ð1Þ ¼ 0 4! (a2) N L ¼ 1 ¼> W ¼ 1!3! ¼ 4, four ways to arrange on α, β, γ, or δ.

S ¼ kB ln W ¼ kB ln ð4Þ

4! (a3) N L ¼ 2 ¼> W ¼ 2!2! ¼ 6, six ways to arrange: αβ, αγ, αδ, βγ, βδ, or γδ.

S ¼ kB ln W ¼ kB ln ð6Þ 4! (a4) N L ¼ 3 ¼> W ¼ 3!1! ¼ 4, four ways to arrange: the vacant spot is on α, β, γ, or δ.

S ¼ kB ln W ¼ kB ln ð4Þ 4! (a5) N L ¼ 4 ¼> W ¼ 4!0! ¼ 1, only one arrangement is possible.

S ¼ kB ln W ¼ k B ln ð1Þ ¼ 0 (a6) The states of highest entropy have NL ¼ 2. (a7) The states of lowest entropy have NL ¼ 0 or NL ¼ 4.

748

Solved Problems for Part III

(a8) The probability distribution associated with binding NL ligands with probability p and not binding (MNL) ligands with probability (1p) onto M sites on the protein is given by:

pð N L , M Þ ¼

M! pN L ð1  pÞMN L N L !ðM  N L Þ!

Using Eq. (20), it follows that: (i) (ii) (iii) (iv) (v)

4! ð0:75Þ0 ð0:25Þ4 For NL ¼ 0, pð0, 4Þ ¼ 0!4! 4! For NL ¼ 1, pð1, 4Þ ¼ 1!3! ð0:75Þ1 ð0:25Þ3 4! For NL ¼ 2, pð2, 4Þ ¼ 2!2! ð0:75Þ2 ð0:25Þ2 4! For NL ¼ 3, pð3, 4Þ ¼ 3!1! ð0:75Þ3 ð0:25Þ1 4! For NL ¼ 4, pð4, 4Þ ¼ 4!0! ð0:75Þ4 ð0:25Þ0

¼ 3:91
103 . ¼ 4:69
102 . ¼ 0:211. ¼ 0:422. ¼ 0:316.

The resulting probability distribution is plotted in Fig. 1.

Fig. 1

ð20Þ

Solved Problems for Part III

749

(b1) The possible states of the zipper are determined by the open link number i. Each state of the zipper has an energy of iε, and the Canonical partition function can be obtained by summing eEi =kB T for every possible state. Specifically:

Q¼

N X

eiε=kB T ¼ e0 þ eε=kB T þ e2ε=kB T þ . . . þ eNε=kB T

ð21Þ

i¼0

Equation (21) is a finite geometric series (recall that eε=kB T < 1 ) and can be k P kþ1 summed according to the HINT given in the Problem Statement ( ar n ¼ a 1r 1r , n¼0

for r < 1). Specifically: Q¼

1  eðNþ1Þε=kB T 1  eε=kB T

ð22Þ

(b2) The average energy of the zipper can be obtained by differentiating lnQ with respect to β ¼ 1/kBT. Specifically, from Eq. (22), it follows that:     ln Q ¼ ln 1  eðNþ1Þβε  ln 1  eβε   ∂ ln Q hE i ¼  ∂β 

ð23Þ ð24Þ

    ∂ 1 ∂  ðNþ1Þβε βε 1e 1e  hE i ¼  1  eβε ∂β 1  eðNþ1Þβε ∂β

 ðN þ 1ÞεeðNþ1Þβε εeβε ¼   ð Nþ1 Þβε 1  eβε 1e 1

For eβε > > < ε, φð r Þ ¼ > ε, > > : 0,

for r < σ

ðI Þ

for Rσ > r σ for 2Rσ > r Rσ

ðII Þ ðIII Þ

for r 2Rσ

ðIV Þ

ð27Þ

As discussed in Part III, the second virial coefficient is related to the intermolecular potential as follows: B2 ðT Þ ¼ 2π

ð1h

i eβφ2 ðrÞ  1 r 2 dr

ð28Þ

0 IV Using B2 ðT Þ ¼ BI2 ðT Þ þ BII2 ðT Þ þ BIII 2 ðT Þ þ B2 ðT Þ, we find:

ðσ ðσ 2 BI2 ðT Þ ¼ 2π ½e1  1r 2 dr ¼ 2π r 2 dr ¼ πσ 3 3 0 0

BII2 ðT Þ

¼ 2π

ð Rσ h σ

i   2  eβðεÞ  1 r 2 dr ¼ π 1  eβε ðRσ Þ3  σ 3 3

ð 2Rσ

  2 eβε  1 r 2 dr ¼ π ðRσ Þ3 1  eβε ð8  1Þ 3 Rσ   14 ¼ π ðRσ Þ3 1  eβε 3

BIII 2 ðT Þ ¼ 2π

ð29Þ

ð30Þ



BIV 2 ðT Þ ¼ 2π

ð1



e0  1 r 2 dr ¼ 2π

2Rσ

ð1

0r 2 dr ¼ 0

ð31Þ

ð32Þ

2Rσ

Adding up Eqs. (29)–(32) yields the desired result:       2 B2 ðT Þ ¼ πσ 3 1 þ 1  eβε Rσ 3  1 þ 7Rσ 3 1  eβε 3

ð33Þ

Solved Problems for Part III

751

(b) The solution to this problem involves (i) deriving an expression for the Hamiltonian of a dipolar molecule interacting with an external electric field, (ii) calculating the partition function of the dipolar molecule, and (iii) calculating the additional contribution to the ideal gas internal energy resulting from the dipole-electric field interactions. In this problem, the kinetic energy is not affected by the presence of the dipoles in the molecules. In addition, while the dipolar molecules do not interact with each ! other by assumption (ideal gas), they interact with the external electric field, ɛ , according to the interaction given in the Problem Statement (u ¼  μɛ cos ϴ) (see Fig. 2).

Fig. 2

Figure 1 shows only the in-plane projection, where the out-of-plane angle, ϕ, is not shown. The five degrees of freedom of one dipolar molecule include: • Translational: (x, y, z, px, py, pz) – 3 translations. The limits for x, y, and z are (1 to +1). • Rotational: ϴ, ϕ, Pϴ, Pϕ – 2 rotations. The limits for ϴ and ϕ are (0 to π) and (0 to 2π), respectively. • Because the dipole is fixed, no vibrations are possible. The Classical Hamiltonian for the Rigid Rotor is given by:

752

Solved Problems for Part III

H¼

p2x þ p2y þ p2z 2m

! 1 þ 2I

p2ϴ þ

!

p2ϕ

 μɛcosϴ

sin 2 ϴ

ð34Þ

where the first two terms in Eq. (34) were discussed in Part III and the third term is new and corresponds to the potential energy of the dipole in an external electric field. The classical partition function of the dipolar molecule possessing five degrees of freedom (see above) is then given by: q¼

q¼

V h5

ð1

ð1

1

V 2π Λ 3 h2

dpx

ð1 1

ð1

1

dpy

ð1

1

dpz

ePϴ =2IkB T dpϴ 2

1

ð1

ð1 dpϴ

1

ð 2π dpϕ

0

ePϕ =2IkB T sin ϴ dpϕ 2

1

ðπ dϕ dϴeH=kB T

2

ð35Þ

0

ðπ

eμɛcosϴ=kB T dϴ

ð36Þ

0

The integrals over pϴ and pϕ in Eq. (36) are both Gaussian integrals that can be evaluated using the expression presented in Part III, which is repeated below for completeness: ð1 e

a2 x2

1

pffiffiffi π dx ¼ a

ð37Þ

Carrying out the two Gaussian integrals in Eq. (36) using Eq. (37) yields:  q¼

V Λ3



8π2 IkB T h2

ð π

1 μɛcosϴ=kB T e sin ϴdϴ 02

ð38Þ

As shown in Part III, in Eq. (38), the first two terms in parentheses correspond to the translational and the rotational contributions to the molecular partition function of a rigid rotor, respectively. In other words, we can rewrite Eq. (38) as follows: q ¼ qtran qrot

ðπ 0

1 μɛcosϴ=kB T e sin ϴdϴ 2

ð39Þ

An examination of Eq. (39) reveals that the additional contribution of the μ-ɛ interaction to the molecular partition function, that is, of qdipoleɛ in Eq. (39), corresponds to: qdipoleɛ ¼

ðπ 0

1 μɛcosϴ=kB T e sin ϴdϴ 2

ð40Þ

To calculate the integral in Eq. (40), it is convenient to change variables as follows:

Solved Problems for Part III

753

x ¼ cos ϴ, dx ¼  sin ϴdϴ

ð41Þ

Using Eq. (41) in Eq. (40) yields: qdipoleɛ ¼

ð 1 1

1 kμɛx 1 e B T ðdxÞ ¼  2 2

qdipoleɛ ¼

qdipoleɛ

1 2

ð1

μɛx

ð42Þ

ekB T dx 1

μɛx

1

ð43Þ

ekB T dx

 1 1 kB T kμɛx k T ¼ ¼ B e BT 2 μɛ μɛ 1

qdipoleɛ

ð 1

μɛ

μɛ

ek B T  e k B T 2

! ð44Þ

  kB T μɛ ¼ sinh μɛ kB T

ð45Þ

Accordingly, the expression for the total partition function of a dipolar molecule interacting with an external electric field is given by:  q¼

V Λ3

   2  8π IkB T kB T μɛ sinh μɛ kB T h2

ð46Þ

(c) The additional contribution (i.e., excluding the translations and the rotations) to the total internal energy of an ideal gas of N dipolar molecules, each interacting with an external electric field, is given by:

 ∂lnqdipoleɛ U dipoleɛ ðN, V, T, ɛÞ ¼ NkB T ∂T N,V,ɛ

ð47Þ

2

Using Eq. (46) in Eq. (47), including taking the temperature partial derivative, we obtain the desired result:    3 2  BT ∂ ln kμɛ ∂ln sinh kμɛ BT 5 þ U dipoleɛ ðN, V, T, ɛÞ ¼ NkB T 2 4 ∂T ∂T

ð48Þ N,V,ɛ

754

Solved Problems for Part III



2 kB=μɛ

U dipoleɛ ðN, V, T, ɛÞ ¼ Nk B T 2 4k

=μɛ

BT



cosh

μɛ kB T

3

μɛ  5 kB T 2 sinh μɛ kB T

  με με U dipoleε ðN, V, T, εÞ ¼ Nk B T 1  cosh kB T kB T

ð49Þ



ð50Þ

Solved Problems for Part III

755

Problem 25 Problem 25.1 (a) The energies and degeneracies of the two lowest electronic levels of atomic iodine are listed below:

Level 1 2

Energy (cm1) 0.0 7603.2

Degeneracy 4 2

Calculate the temperature at which 2% of the iodine atoms will be in the excited electronic state. NOTE: 1 cm1 ¼ 1.986
1023 J, and kB ¼ 0.69509 cm1 K1. (b) Two monoatomic ideal gases consisting of N1 atoms and N2 atoms, respectively, are mixed in a vessel of volume, V, at temperature, T. It is known that the masses of the atoms in each gas are m1 and m2, respectively. (i) Calculate the Canonical partition function of the binary gas mixture. (ii) Calculate the energy and pressure of the binary gas mixture.

Problem 25.2 (Adapted from D&B) Statistical mechanics can be used to predict the dependence of protein folding on temperature. Consider a polypeptide chain consisting of six beads having the energy ladder shown below, where the energy increment of the unfolded states, ε0, is positive. You are asked to answer the following questions:

756

Solved Problems for Part III

(a) What is the degeneracy of each energy state? (b) What is the Canonical partition function of the polypeptide chain? (c) What are the probabilities of observing the polypeptide chain in each of the three states as a function of temperature? Discuss the low-temperature and the hightemperature limits.

Problem 25.3 (a) Use the Grand-Canonical ensemble to calculate the standard-state chemical potential, μ0(T), of Ar (g) at 298 K. It is known that (i) the standard-state pressure, P0, equals 1 bar, (ii) the mass of argon is 0.03995 kg/mol, and (iii) the first electronic state of argon is nondegenerate. How does your result compare with the experimentally measured value of 39.97 kJ/mol. (b) It has been suggested that the triangular potential: 8 1, > > > >   < r  σ1 Φð r Þ ¼  E , σ0  σ1 > > > > : 0,

for r  σ 0 for σ 0 < r  σ 1 for r > σ 1

may provide adequate second viral coefficients, B2(T). Use the model for Φ(r) above to derive an expression for B2(T) in the high-temperature limit (to linear order in Eβ) in terms of the model parameters σ 0, σ 1, and E.

Solved Problems for Part III

757

Solution to Problem 25 Solution to Problem 25.1 (a) The probability that the iodine atoms will be in excited electronic (e) state i is given by:

pei ¼

  gei exp  kBEiT

ð1Þ

qe

where qe, the electronic partition function, is given by: qe ¼

X i

  E gei exp  i kB T

ð2Þ

and gei is the degeneracy of the electronic energy level i. Assuming that only the ground (i ¼ 1) and the first (i ¼ 2) excited electronic states contribute significantly to qe, that is, assuming that the energy of the second (i ¼ 3) and the higher (i > 3) excited electronic states are extremely high, qe in Eq. (2) is given by:     E E qe ¼ ge1 exp  1 þ ge2 exp  2 kB T kB T

ð3Þ

Using the values of ge1, ge2, E1, and E2 given in the Table in the Problem Statement, Eq. (3) yields: 

7603:2 qe ¼ 4 þ 2x, where x ¼ exp  0:69509T

 ð4Þ

where in Eq. (4), T has units of Kelvin. We are asked to find the temperature at which 2% of the iodine atoms will be in the excited (i ¼ 2) electronic state. From Eqs. (1) and (4), it follows that:

pe2 ¼

  ge2 exp  kEB2T qe

¼

2x ¼ 0:02 4 þ 2x

ð5Þ

Clearly, x can be determined from Eq. (5). Once x is known, T can be calculated using Eq. (4). Rearranging Eq. (5) yields:

758

Solved Problems for Part III

0:08 þ 0:04x ¼ 2x ! 0:08 ¼ 1:96x

ð6Þ

x ¼ 0:040816

ð7Þ

Using Eq. (4), T can be calculated from x as follows: T¼

10938:44 10938:44 ¼ ln ðxÞ ln ð0:040816Þ T ¼ 3419:67 K

ð8Þ ð9Þ

(bi) Because the atoms in a monoatomic ideal gas are independent and atoms of type 1 and 2 are distinguishable, the Canonical partition function of a mixture of monoatomic ideal gases can be written as the product of the Canonical partition functions of each gas, that is: QðN 1 , N 2 , V, T Þ ¼ Q1 ðN 1 , V, T ÞQ2 ðN 2 , V, T Þ

ð10Þ

Because in each gas the atoms are independent and indistinguishable, it follows that: Q1 ðN 1 , V, T Þ ¼

½q1 ðV, T ÞN 1 N1!

ð11Þ

Q2 ðN 2 , V, T Þ ¼

½q2 ðV, T ÞN 2 N2!

ð12Þ

and

where q1 and q2 are the atomic partition functions of gases 1 and 2, respectively. For a monoatomic ideal gas, these are given by: q1 ðV, T Þ ¼

 3=2 2πm1 k BT Vge1,1 h2

ð13Þ

where ge1,1 is the degeneracy of the ground electronic state for gas 1. Similarly: q2 ðV, T Þ ¼

 3=2 2πm2 k BT Vge1,2 h2

where ge1,2 is the degeneracy of the ground electronic state for gas 2. Substituting Eqs. (11) through (14) in Eq. (10) yields the desired result:

ð14Þ

Solved Problems for Part III

759

1 QðN 1 , N 2 , V, T Þ ¼ N 1 !N 2 !

" #N 1 " #N 2 3=2 3=2 2πm1 k B T 2πm2 kB T Vge1,1 Vge1,2 h2 h2 ð15Þ

(bii) To calculate the energy of the gas mixture, we use the expression presented in Part III, that is:

< E >¼ U ¼ kB T 2

  ∂ ln Q ∂T N 1 ,N 2 ,V

ð16Þ

It is convenient to first write down the expression for lnQ where we isolate the terms that depend explicitly on T. From Eq. (15), it follows that: ln Q ¼

3N 1 3N ln T þ 2 ln T þ Terms that do not depend on T 2 2

ð17Þ

Differentiating Eq. (17) with respect to T, at constant N1, N2, and V, yields:   ∂ ln Q 3N 1 3N 2 þ ¼ 2T 2T ∂T N 1 ,N 2 ,V

ð18Þ

Using Eq. (18) in Eq. (16) yields the desired result: 3 < E >¼ U ¼ ðN 1 þ N 2 Þk B T 2

ð19Þ

To calculate the pressure of the gas mixture, we use the expression presented in Part III, that is: P ¼ kB T

  ∂ ln Q ∂V N 1 ,N 2 ,T

ð20Þ

Again, it is convenient to isolate the terms that depend explicitly on V in the expression of lnQ before differentiation with respect to V. Specifically: ln Q ¼ N1 ln V þ N2 ln V þ Terms that do not depend on V Differentiating Eq. (21) with respect to V, at constant N1, N2, and T, yields:

ð21Þ

760

Solved Problems for Part III

  ∂ ln Q N N ¼ 1þ 2 V V ∂V N 1 ,N 2 ,T

ð22Þ

Using Eq. (22) in Eq. (20) yields the desired result: P¼

k B T ðN 1 þ N 2 Þ V

ð23Þ

Equations (19) and (23) indicate that a gas mixture of the two monoatomic ideal gases behaves like an ideal gas consisting of (N1 + N2) atoms.

Problem Solution 25.2 (a) The degeneracy corresponds to the number of microstates in each energy level. Examination of Fig. 1 in the Problem Statement shows that: E 0 ¼ 0,

g0 ¼ 4

ð24Þ

E 1 ¼ E0 ,

g1 ¼ 11

ð25Þ

E2 ¼ 2E0 , g2 ¼ 21

ð26Þ

(b) The partition function of the polypeptide chain, corresponding to the energy ladder in Fig. 1 in the Problem Statement, is given by: q¼

X

gi exp ðβEi Þ

ð27Þ

i

q ¼ g0 exp ðβE 0 Þ þ g1 exp ðβE 1 Þ þ g2 exp ðβE 2 Þ

ð28Þ

Using the information from Fig. 1 and the Problem Statement, as well as using the degeneracies found in Part (a), yields: q ¼ 4 exp ðβð0ÞÞ þ 11 exp ðβðE0 ÞÞ þ 21 exp ðβð2E0 ÞÞ

ð29Þ

q ¼ 4 þ 11 exp ðβE0 Þ þ 21 exp ð2βE0 Þ

ð30Þ

q ¼ 4 þ 11 exp ðE0 =kB T Þ þ 21 exp ð2E0 =k B T Þ

ð31Þ

or

Solved Problems for Part III

761

(c) The probabilities of finding the polypeptide chain in the folded state (0) and in the two unfolded states (1 and 2) are given by: p0 ðfolded stateÞ ¼

ð32Þ

g1 exp ðE0 =k B T Þ ¼ 11 exp ðE0 =k B T Þ=q q

ð33Þ

g2 exp ð2E0 =k B T Þ ¼ 21 exp ð2E0 =kB T Þ=q q

ð34Þ

p1 ðpartially unfolded stateÞ ¼ p2 ðfully unfolded stateÞ ¼

g0 exp ð0=kB T Þ ¼ 4=q q

Because each energy level has a different degeneracy (g0 6¼ g1 6¼ g2), the energy level with the largest degeneracy is more likely to be observed at higher temperatures. At high temperature (T ! 1), lim q ¼ 4 þ 11 þ 21 ¼ 36

ð35Þ

lim p0 ¼ 4=36  0:1111

ð36Þ

lim p1 ¼ 11=36  0:3056

ð37Þ

lim p2 ¼ 21=36  0:5833

ð38Þ

T!1

T!1 T!1 T!1

As expected, p2 > p1 > p0. We can see that as T increases, the unfolded states (denatured polypeptide) are more likely to be observed, as seen in nature. At low temperatures, the lowest-energy levels are more likely, because at low temperatures (T ! 0), it follows that: lim q ¼ 4

ð39Þ

lim p0 ¼ 4=4 ¼ 1

ð40Þ

lim p1 ¼ 0=4 ¼ 0

ð41Þ

lim p2 ¼ 4=4 ¼ 0

ð42Þ

T!0 T!0 T!0 T!0

As expected, p0 ¼ 1 and p1 ¼ p2 ¼ 0 as T ! 0.

762

Solved Problems for Part III

Solution to Problem 25.3 (a) We are asked to calculate the chemical potential, μ, using the Grand-Canonical ensemble. The Grand-Canonical ensemble partition function is given by:

Ξ¼

1 X

QðN, V, T Þ exp ðNμ=k B T Þ

ð43Þ

N¼0

where QðN, V, T Þ is the Canonical partition function. We anticipate that at P0 ¼ 1 bar and T ¼ 298 K, argon behaves like an ideal gas. In that case, the Canonical partition function is given by: QðN, V, T Þ ¼

½qðV, T ÞN N!

ð44Þ

where the atomic partition function, q, is given by:  qðV, T Þ ¼

2πmk B T h2

3=2 Vge1

ð45Þ

We are asked to calculate μ when P0 ¼ 1 bar and T ¼ 298 K. Therefore, if we could express Ξ in terms of P0, we could use Eq. (43) to calculate μ as a function of P0. Recall that Ξ and P are related by: Ξ ¼ exp ðPV=k B T Þ

ð46Þ

Using Eq. (46) in Eq. (43) and substituting Eqs. (44) and (45) yields: exp ðPV=kB T Þ ¼

1 X ½qðV, T ÞN N λ , where N! N¼0

λ ¼ exp ðμ=k B T Þ

ð47Þ

Recall that: ex ¼

1 X xN N! N¼0

ð48Þ

In view of Eq. (48), Eq. (47) can be written as follows: exp ðPV=k B T Þ ¼ exp ðqðV, T ÞλÞ

ð49Þ

Substituting Eq. (45) for qðV, T Þ in Eq. (49), and subsequently taking the natural logarithm of both sides of the resulting equation, yields:

Solved Problems for Part III

763

 3=2 2πmk B T Vge1 λ h2

ð50Þ

 2 3=2 P h ðkB T Þ5=2 ge1 2πm

ð51Þ

PV ¼ kB T Rearranging Eq. (50) yields: λ¼

Recalling that λ ¼ exp (μ/kBT ), we obtain the desired expression: "

 2 3=2 # P h 5=2 ðk T Þ μ ¼ k B T ln ge1 B 2πm

ð52Þ

Note that the expression for μ in Eq. (52) is the same as the expression obtained in Part III using the Canonical ensemble approach. Equation (52) enables calculation of the chemical potential at a given P and T. Note that the standard-state chemical potential is simply the chemical potential at the reference pressure P ¼ P0, that is: u(r)

0

∞

σ0

σ1 r

Fig. 1

"

 2 3=2 # P0 h 5=2 μ0 ðT, P0 Þ ¼ kB T ln ðk T Þ ge1 B 2πm

ð53Þ

Next, let us calculate the chemical potential of argon at P ¼ P0 ¼ 1 bar and T ¼ 298 K. The values of the various terms that appear in Eq. (53) are as follows:

764

Solved Problems for Part III

ge1 ¼ 1 ðnondegenerate first electronic ground stateÞ h ¼ 6:626
1034 Js m ¼ 0:03995kg=mol ¼ 6:633
1026 kg=molecule kB ¼ 1:381
1023 J=K T ¼ 298 K P0 ¼ 1 bar ¼ 1:0
105 N=m2

ð54Þ

Substituting the values in Eq. (54) in Eq. (53) yields: μ0 ð298 K, 1 barÞ ¼ 6:636
1023 kJ=molecule

ð55Þ

We next need to convert the value in Eq. (55) from a per molecule basis to a per mole basis and then compare the resulting predicted value with the experimental value given in the Problem Statement. To this end, we simply multiply the result in Eq. (55) by Avogadro’s number. This yields: μ0 ð298 K, 1 barÞ ¼ 39:97 kJ=mol

ð56Þ

It turns out that μexperiment ð1 bar, 298 KÞ ¼ 39:97 kJ=mol. It then follows that 0 the statistical mechanical prediction is in remarkable agreement with the experimental result. It is important to note that the calculated μ(1 bar, 298 K) is based on the assumption that the ground electronic state of the argon atom is zero. If this is not the case, additional contributions may appear in the expression for μ0(1 bar, 298 K). (b) This problem involves calculating the second virial coefficient from a given interaction potential. We begin with the expression for B2(T) in terms of u(r) presented in Part III: ð1h i B2 ðT Þ ¼ 2π eβuðrÞ  1 r 2 dr

ð57Þ

0

The given interaction potential, u(r), is plotted in Fig. 1 below: Because u(r) shows distinct behaviors in three regions (see Fig. 1), we break the integration range in Eq. (57) into three regions as follows: B2 ðT Þ ¼ 2π

ð σ 0 h e 0

βuðrÞ

 ð σ1 h ð1h i i βuðrÞ 2 βuðr Þ 2  1 r dr þ e  1 r dr þ e  1 r dr i

2

σ0

B2 ðT Þ ¼ 2π ½I 1 þ I 2 þ I 3  We will next calculate I1, I2, and I3 as follows:

σ1

ð58Þ

Solved Problems for Part III

I1 ¼

765

ð σ0

½e

1

 1r dr ¼

0

I2 ¼ I3 ¼

ð1h σ1

2

½1r 2 dr ¼ 

0

ð σ1 h σ0

ð σ0

e

rσ 1 0 σ 1

βɛσ

σ 30 3

i  1 r 2 dr

ð σ0 i eβð0Þ  1 r 2 dr ¼ ½0r 2 dr ¼ 0

ð59Þ ð60Þ ð61Þ

0

Using Eqs. (59), (60), and (61) in Eq. (58) yields: B 2 ðT Þ ¼

2πσ 30  2π 3

ð σ1 h rσ i βɛ 1 e σ0 σ1  1 r 2 dr σ0

ð62Þ

At high temperatures, βɛ  1, and we can expand the exponential term in Eq. (62) as follows: e

rσ 1 0 σ 1

βɛσ

 1 ¼ 1 þ βɛ

  r  σ1 βɛr βɛσ 1  1 þ O ðr  σ 0 Þ2   σ0  σ1 σ0  σ1 σ0  σ1

ð63Þ

Using Eq. (63) in Eq. (60) yields: ð σ1 βɛr 2 βɛσ 1 2 r dr  r dr σ0 σ 0  σ1 σ0 σ 0  σ 1  4   3  βɛ σ1  σ04 βɛσ 1 σ1  σ03 I2 ¼  σ0  σ1 4 σ0  σ1 3  2  ðσ 1  σ 0 2 Þðσ 1 2 þ σ 0 2 Þ βɛ βɛσ 1  I2 ¼ 4 σ0  σ1 σ0  σ1  2  ðσ 1 þ σ 0 σ 1 þ σ 0 2 Þðσ 1  σ 0 Þ
3 I2 ¼

ð σ1

ð64Þ ð65Þ

ð66Þ

or    2  ðσ 1 þ σ 0 Þðσ 1 2 þ σ 0 2 Þ ðσ 1 þ σ 0 σ 1 þ σ 0 2 Þ I 2 ¼ βɛ þ βɛσ 1 4 3

ð67Þ

Combining the two terms in Eq. (67) yields:  3  ðσ 1 þ σ 0 2 σ 1 þ σ 1 2 σ 0  3σ 0 3 Þ I 2 ¼ βɛ 12 Equation (68) can be further simplified as follows:

ð68Þ

766

Solved Problems for Part III

 2  ðσ 1 þ 2σ 0 σ 1 þ 3σ 0 2 Þðσ 1  σ 0 Þ I 2 ¼ βɛ 12

ð69Þ

Using Eqs. (59), (61), and (69) in Eq. (58) yields the desired result: B2 ðT Þ ¼

 2  2πσ 30 ðσ 1 þ 2σ 0 σ 1 þ 3σ 0 2 Þðσ 1  σ 0 Þ  πβɛ 6 3

ð70Þ

The first term in Eq. (70) is due to the hard-sphere repulsive part of the interaction potential, u(r), and, therefore, has a positive contribution to B2. As Fig. 1 shows, ɛ > 0 corresponds to an attractive interaction and, consequently, has a negative contribution to B2. On the other hand, ɛ < 0 corresponds to a repulsive interaction and, consequently, has a negative contribution to B2.