Heat and Thermodynamics [1 ed.] 9781783320592, 9781842659021

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Heat and Thermodynamics

Heat and Thermodynamics

A.K. Saxena C.M. Tiwari

a Alpha Science International Ltd. Oxford, U.K.

Heat and Thermodynamics 328 pgs. | 145 figs. | 09 tbls.

A.K. Saxena C.M. Tiwari Department of Physics A.P.S. University Rewa (M.P.) Copyright © 2014 ALPHA SCIENCE INTERNATIONAL LTD. 7200 The Quorum, Oxford Business Park North Garsington Road, Oxford OX4 2JZ, U.K.

www.alphasci.com ISBN 978-1-84265-902-1 E-ISBN 978-1-78332-059-2 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without prior written permission of the publisher.

Preface

This book has been written to cover the syllabus of B.Sc. Physics (and chemistry) of all Colleges and Universities. The subject matter includes Basic Ideas (Temperature and heat, measurement of temperature, gas laws and gas equation, intensive and extensive properties, thermodynamic system and coordinator, PvT surfaces, heat capacity, mechanical equivalent of heat and thermal expansion), Kinetic theory (Brownian motion, kinetic model of gases and deductions of gas laws, degrees of freedom, principle of equipartition of energy, Maxwell-Boltzmann distribution law and its experimental verification, mean free path, transport phenomena and thermal conductivity), Real gases and their liquefaction and production and measurement of very low temperatures (van der Waals equation, Joules expansion, Joule-Thomson cooling, porous plug experiment, adiabatic demagnetization, liquefaction of gases and approach to absolute zero), The first law of thermodynamics and thermodynamic properties, application of the first law and concept of a perfect gas, reversible and irreversible processes and the second and third laws of thermodynamics, Carnots’ cycle, thermodynamic temperature scale, entropy, real heat engines, negative temperatures, thermodynamic potentials and Maxwell’s relations and change of phase and, in the last thermal radiation (Emission and absorption, Prevost theory and laws of radiation). The authors are thankful to Mr. N.K. Mehra and Mr. Shashikant (Narosa Publishing House) for bringing out the book in a short time. We hope that the book would be helpful to undergraduate students and aid to their understanding of the subject. A.K. Saxena C.M. Tiwari

Contents

Preface

v

1. Basic Ideas 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22

Thermodynamics Zeroeth Law of Thermodynamics and Temperature Temperature and Heat Gas Thermometers Absolute Zero Triple-Point Cell Temperature Measurement with a Constant-Volume Gas Thermometer International Practical Temperature Scale Resistance Thermometer Thermocouple Ideal Gases and Gas Laws Avogadro’s Number Universal Gas Constant, Boltzmann Constant and Ideal Gas Equation The “System” Intensive and Extensive Properties Thermodynamic System and Thermodynamic Coordinates P-v-T Surface for an Ideal Gas and Boyle’s Law Behaviour of Real Gases (Equation of State and P-v-T Surfaces) Heat Capacity Heat: Its Nature and Units Mechanical Equivalent of Heat Thermal Expansion Questions Objective Type Questions

1.1 1.1 1.1 1.1 1.3 1.5 1.6 1.8 1.9 1.10 1.11 1.12 1.15 1.16 1.17 1.18 1.19 1.20 1.22 1.26 1.27 1.28 1.30 1.36 1.37

viii Contents 2. Kinetic Theory 2.1 2.2 2.3

2.1

Brownian Motion Theories of Brownian Motion Vertical Distribution of Brownian Particles and Determination of Avogadro’s Number Ideal Gas (Macroscopic Description) Kinetic Model of Gases Some Deductions from Kinetic Theory The Internal Energy and Degrees of Freedom Principle of Equipartition of Energy Maxwell-Boltzmann Law of Distribution of Velocities in an Ideal Gas Maxwell Boltzmann’s Energy Distribution Law Experimental Verification of Maxwell’s Velocity Distribution Law Mean Free Path Transport Phenomena Gas Viscosity Thermal Conductivity Gas Self Diffusion Questions Objective Type Questions

2.36 2.38 2.39 2.42 2.43 2.52 2.53

3. Real Gases, Liquefaction of Gases and Production and Measurement of Very Low Temperatures

3.1

2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16

3.1 3.2 3.3 3.4 3.5 3.6 3.7

Perfect Gas and Real Gas Experiments on the Behaviour of Real Gases Critical Point Continuity of State Form of Equation of State van der Waals Equation and the Critical Point Comparison between Experimental and Theoretical P-V Curves 3.8 van der Waals Equation and the Boyle Temperature 3.9 Corresponding States 3.10 Joule’s Expansion of an Ideal Gas and van der Waal’s Gas 3.11 Joule’s Coefficient for Real Gas

2.1 2.2 2.6 2.7 2.8 2.12 2.16 2.18 2.20 2.29 2.33

3.1 3.2 3.6 3.7 3.7 3.8 3.11 3.12 3.14 3.15 3.16

Contents ix

3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20

Joule-Thomson Cooling Temperature of Inversion Principle of Joule Thomson’s Porous Plug Experiment Change in Temperature in an Adiabatic Change Distinction between Joule’s Expansion, Joule-Thomson’s Expansion and Adiabatic Expansion Adiabatic Demagnetisation Liquefaction of Gases and Approach to Absolute Zero Production of Very Low Temperatures Measurement of Very Low Temperatures Questions Objective Type Questions

4. The First Law of Thermodynamics 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18

Introduction Work Depends on the Path (Indicator Diagram) Heat Depends on the Path First Law of Thermodynamics and Internal Energy Work Done by a Thermodynamic System in Expansion Against External Pressure Proof for the First Law of Thermodynamics The Thermodynamic Property ‘Enthalpy’ Heat Capacity Specific Heat Capacity Relationship between Cp and Cv Applications of the First Law of Thermodynamics to an Ideal Gas Impossibility of a Perpetual-Motion Machine Surfaces Work in Changing the Area of a Surface Film Paramagnetic Rod Magnetic Work Partial Derivatives Concept of a Perfect Gas Questions Objective Type Questions

3.17 3.21 3.22 3.27 3.29 3.29 3.35 3.41 3.44 3.47 3.48

4.1 4.1 4.1 4.4 4.5 4.8 4.13 4.14 4.15 4.16 4.17 4.23 4.26 4.27 4.28 4.29 4.30 4.32 4.35 4.37 4.38

x Contents 5. The Second Law and Third Law of Thermodynamics 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21 5.22 5.23 5.24 5.25 5.26 5.27 5.28 5.29 5.30

Reversible and Irreversible Processes Significance of Reversible Processes Need for Formulations of Second Law of Thermodynamics The Second Law of Thermodynamics Heat Engine and its Efficiency Carnot’s Engine: Carnot’s Cycle and its Efficiency Working of a Heat Engine/Refrigerator Operating on a Carnot Cycle Demonstration of the Fact that the Carnot Cycle is the Most Efficient Cycle Operating between Two Fixed-Temperature Reservoirs Thermodynamic Temperature Scale Relation between Thermodynamic Scale, Celsius Scale, Rankine Scale and the Fahrenheit Scale of Temperatures The Clausius Theorem The Clausius Inequality Entropy Physical Significance of Entropy Thermodynamic Definition of Temperature T-S Diagram Entropy Change in an Irreversible Process Entropy Change of a Perfect Gas in a Reversible Process Real Heat Engines Principle of Degradation of Energy Theorem of Maximum Work Negative Temperatures Thermodynamic Potentials Maxwell’s Relations How to Remember the Maxwell’s Relations Thermodynamic Variables in Terms of Thermodynamic Potentials TdS Equations Expressions for Cp – CV Change of Phase: Equilibrium between a Liquid and its Vapour Clausius-Clapeyron Equation

5.1 5.1 5.2 5.3 5.4 5.6 5.7 5.13 5.15

5.17 5.21 5.23 5.24 5.26 5.27 5.28 5.30 5.31 5.33 5.34 5.41 5.42 5.43 5.44 5.48 5.52 5.55 5.56 5.57 5.59 5.60

Contents xi

5.31 First Order Phase Transition and Clausius-Clapeyron Equation 5.32 Second Order Phase Transition 5.33 The Third Law of Thermodynamics 5.34 Apparent Violations of the Third Law Questions Objective Type Questions

6. Radiation 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.7 6.8 6.9 6.10 6.11 6.12 6.13 6.14 6.15 6.16 6.17 6.18 6.19 6.20 6.21 6.22

References Index

Introduction Heat Waves Detection and Measurement of Thermal Radiation Emission and Absorption of Radiation Prevost’s Theory of Exchanges Emissive Power Radiation in an Enclosure Kirchhoff’s Law Blackbody Intensity and Energy Density Pressure due to Radiation Stefan-Boltzmann Law Verification of Stefan’s Law Determination of Stefan’s Constant in the Laboratory Newton’s Law of Cooling Adiabatic Expansion of Blackbody Radiation Effect of Adiabatic Expansion on Blackbody Radiation Wien’s Law Wien’s Formula Rayleigh-Jeans Formula Planck’s Formula Determination of the Solar Constant Temperature of the Sun Questions Objective Type Questions

5.61 5.64 5.68 5.70 5.82 5.84

6.1 6.1 6.1 6.2 6.4 6.6 6.6 6.7 6.8 6.9 6.10 6.13 6.14 6.17 6.18 6.20 6.20 6.21 6.22 6.25 6.26 6.30 6.33 6.34 6.36 6.37 R.1 I.1

Chapter

1

1.1

THERMODYNAMICS

Basic Ideas

Thermodynamics is the branch of physics in which we study the relationship between the heat energy and mechanical energy or in general, any other form of energy. In thermodynamics, we neither consider the internal structure of the substance, nor we consider the motion of its particles, rather we consider only the macroscopic properties of the substance.

1.2 ZEROETH LAW OF THERMODYNAMICS AND TEMPERATURE According to the Zeroeth Law, if two systems are separately in thermal equilibrium with a third system then they both are also in thermal equilibrium with each other. According to Zeroeth law of thermodynamics, temperature is the quantity which determines the direction of flow of heat when the two systems are kept in contact. If there is no exchange of heat energy between the two systems kept in contact, the systems are said to be in thermal equilibrium (i.e., at the same temperature). But if there is a transfer of heat energy from one system to the other system, the system imparting heat energy is said to be at a higher temperature and the system which receives the heat energy is said to be at a lower temperature. Temperature is thus the thermodynamic property which controls the state of thermal equilibrium of the systems.

1.3 TEMPERATURE AND HEAT Except for judging by the feel of our skin, which is not a very accurate quantitative guide, the only way we can measure temperature is to measure the effects temperature changes have on the physical properties of materials e.g., thermal expansion of certain liquids such as mercury, change in electrical resistance of a wire, the variation with temperature of electric current generated by dissimilar metals joined together and so on. The most commonly employed effect to measure temperature is that of expansion. Thermometers are usually constructed in such a way that a small thermal expansion results in a large displacement of an indicator.

1.2 Heat and Thermodynamics In thermometers based on the expansion of a Hot liquid, a comparatively large amount of liquid (usually mercury or alcohol) is confined in a bulb, and its expansion causes the excess liquid to rise in a narrow capillary tube (Fig. 1.1) shows a mercury thermometer. Whatever kind of thermometer we choose, the scale on which the indicator moves must be calibrated in some definite easily reproducible manner. The two thermometric scales in general use are calibrated at the freezing point and boiling Cold point of water (at a pressure of one standard Hg atmosphere). On the Centigrade or Celsius scale, which is used generally all over the world, Fig. 1.1 A mercury thermometer the freezing point is 0°C and the boiling point is 100°C. On the Fahrenheit scale, 32°F is the freezing point and the boiling point is 212°F (Fig. 1.2). On the Fahrenheit scale, the interval between freezing and boiling point is divided into 180 degrees (21232), and on the Celsius scale, there are 100 degrees in this same interval. The Fahrenheit and Celsius scales are thus related as TC TF – 32 ___ = _______ 9 5 9 e.g., 80°C corresponds to 80 × __ + 32 = 176°F. 5 °F °F

°C

°F

°C

°C 212

100

y

80

100°

212°

0°

32°

Crushed ice and water A

Boiling water B

68

x

32

0

C

Fig. 1.2 Calibration of Fahrenheit and Centigrade (or Celsius) thermometers

Basic Ideas 1.3

1.3.1

Pressure

The pressure P is defined as the magnitude of the force per unit area and the unit of pressure in the MKS system is 1 newton per square meter (1 Nm–2). A pressure of exactly 105 Nm–2 (= 106 dyne.cm–2) is called 1 bar, and a pressure of 10–1 Nm–2 (= 1 dyne cm–2) is 1 microbar (1 mbar). A pressure of 1 standard atmosphere (atm) is defined as the pressure produced by a vertical column of mercury exactly 76 cm in height, of density r = 13.5951 g cm–3 at a point where g has its standard value of 980.665 cm s–2. From the equation P = rgh, we find 1 standard atmosphere = 1.01325 × 106 dyne cm–2 = 1.01325 × 105 Nm–2 Hence, 1 standard atmosphere is very nearly equal to 1 bar, and 1 m bar is very nearly 10–6 atm. A unit of pressure commonly used in experimental work at low pressures is 1 Torr (named after Torricelli (1608-1647)) and defined as the pressure produced by a mercury column exactly 1 mm. in height, under the above conditions. Therefore 1 Torr = 133.3 Nm –2

1.4

GAS THERMOMETERS

Thermometers using liquid materials (e.g., mercury, alcohol or water) will disagree among themselves in smaller details due to the fact that different materials react somewhat differently to an increase of temperature. Thus, we need some other solution for an exact and universal definition of temperature scale. The solution is provided by gases since it has been observed that all gases subjected to heating expand in almost exactly the same way. Thus, we can accept as a standard the temperature scale provided by a gas thermometer (Fig. 1.3), regardless of what gas is used to fill it.

1.4.1

Empirical Scale of Temperature

The temperature is defined as a quantity, which prescribes a thermal equilibrium between two bodies in contact. Among a group of bodies, a particular body A may be used as a thermometer. The temperature of other body may be compared by bringing the body A in contact with it. The body used as a thermometer should posses an easily observable thermometric property, such as the length of a mercury column in a capillary tube, the pressure of a gas in a bulb, or the resistance of a platinum wire. The common temperature scale is the Celsius or Centigrade scale. On this scale, the melting point of ice and boiling point of water at normal pressure are 0°C and 100°C respectively.

1.4 Heat and Thermodynamics

Open top

Bulb containing known volume of any gas

Mercury

Rubber tube

Fig. 1.3 A gas thermometer. The changing volume of the trapped gas is read from the height of the mercury column in front of the scale. (In order to keep the mercury level the same on both sides, the right hand tube is lowered as the gas heats; the gas in the bulb will therefore remain at atmospheric pressure at all temperature)

Let the temperature t be the linear function of the thermodynamic property X. If X0, X100 and Xt be the values of X at 0°C, 100°C and t°C respectively, then X t – X0 (1) t°C = _______ × 100 X100 – X0 Although the choice of a temperature scale at this point is arbitrary, its absolute choice will be established with the introduction of the second law of thermodynamics. The absolute temperature will be represented by T. We use gas thermometer for an empirical temperature scale. An arbitrary temperature scale q can the converted into T K by (PV)qs T __ = lim _____ T0 p Æ 0 (PV)q

0

(2)

Basic Ideas 1.5

Thus, T is established uniquely, since T0 is known. If the interval between steam and ice-points is given a value 100, then (PV)qs T __s = lim _____ = 1.366 T0 P0 Æ 0 (PV)q

(3)

0

Ts – T0 = 100

(4)

Equations (3) and (4) define the Kelvin or Absolute Scale. The solution of these equations gives T0 = 273.16 K Ts = 373.16 K The absolute and centigrade temperatures are therefore related by T K = 273.16 + t°C

1.5

(5)

ABSOLUTE ZERO

Let us take a gas thermometer (Fig. 1.3) and carefully measure the gas volume, first at the boiling point of water and then again when the thermometer is in ice (In defining the Celsius temperature scale, we called these temperatures 100°C and 0°C). We can now plot them as points B and F on a graph against our measured volumes (Fig. 1.4). A straight line through these two points can be drawn and extended to left and right to define a scale for measurement over a wide range of temperature.

Fig. 1.4 The behaviour of a gas at low temperature

1.6 Heat and Thermodynamics There is something peculiar that should be noted about the left (low-temperature) side of the graph. We cannot extend the line indefinitely as we can on the other (high-temperature) side, because it soon runs into the axis of the graph indicating a zero volume. Then it would be ridiculous to extend the line any farther since, as far as we know, a gas with a negative volume is an idea that has no meaning. So this point, where our graph seems to be heading, is called the absolute zero of temperature. The apparent intention of the gas to shrink to zero volume at absolute zero is naturally never fulfilled. All gases liquefy before this point is reached; in fact even before it begins to liquefy, a gas commences to deviate considerably from its more regular behaviour at higher temperatures. Ingenious experimental procedures and theoretical corrections, however, have revealed that the temperature of absolute zero is – 273.16°C. A temperature scale beginning with 0 at absolute zero is an absolute temperature scale. The most common absolute scale uses the same size degree as the Celsius scale and is called the Kelvin scale (K), named after Lord Kelvin (England). Since the Kelvin degree is the same size as the Celsius degree, and begins counting 273° lower on the scale, it follows that in order to convert a temperature given in °C into K, we need only to add 273. This concept of an absolute zero is an important one in physics, since the triple-point of water is – 273.16°C and internationally chosen as the standard fixed point for the absolute scale.

1.6

TRIPLE‐POINT CELL

Before 1954, the international metric temperature scale was the Celsius scale, which was based on the temperature interval between two fixed points: 1. the temperature at which pure ice coexisted in equilibrium with airsaturated water at standard atmospheric pressure (the ice-point), and 2. the temperature of equilibrium between pure water and pure steam at standard atmospheric pressure (the steam point). The temperature interval between these two fixed points was assigned 100 “degrees” (of hotness) i.e., 100°C. Hundreds of attempts were made all over the world to measure the temperature of the ice point with great accuracy-without much success. The main difficulty was achieving equilibrium between airsaturated water and pure ice. When ice melts, it surrounds itself with pure water that prevents intimate contact between ice and air-saturated water. Attempts to measure the steam point also presented problem, because the temperature of the steam point is very sensitive to pressure. In 1954, a single fixed point was chosen as the basis for a new international temperature scale, the Kelvin scale. The state in which ice, liquid water and water

Basic Ideas 1.7

vapour coexist in equilibrium (a state known as the triple point of water), provides the standard reference temperature. The temperature of the triple point of water which can be very accurately and reproducibly measured was assigned the value 273.16 Kelvin, corresponding to 0.01°C, in order to maintain the magnitude of a unit of temperature. Note that the word “degree” has been dropped from the Kelvin scale, so the triple point temperature is abbreviated as 273.16 K. This temperature is the standard fixed point of thermometry. To achieve the triple point, water of the highest purity which has substantially the isotopic composition of ocean water is distilled into a vessel like that shown schematically in Fig. 1.5.

Fig. 1.5 Triple-point cell with a thermometer in the well, which melts a thin layer of ice nearby

When all air has been removed, the vessel is sealed off. With the aid of a freezing mixture in the inner well, a layer of ice is formed around the well. When the freezing mixture is removed and replaced with a thermometer, a thin layer of ice is melted nearby. So long as the solid, liquid and vapour coexist in equilibrium, the system is at the triple point. Celsius temperature t (i.e., centigrade temperature) is defined by the equation t = T – Ti where T is the absolute or thermodynamic temperature and Ti is the thermodynamic temperature of the ice point (= 273.15 K). At the ice point, T = Ti, t = 0°C and at the triple point of water, T = 273.16 K t = 0.01°C and at the steam point t = 100°C. Some Kelvin and Celsius temperatures are compared in Fig. 1.6.

1.8 Heat and Thermodynamics K

C 100°C

Steam point 373 K 100 Kelvins 100 deg C Ice point 273 K

0°C –78°C

NSP CO2 195 K

Fig. 1.6

1.7

NBP oxygen

90 K

–183°C

Absolute zero

0

–273°C

Comparison of some Kelvin and Celsius temperatures. Temperatures have been rounded off to the nearest degree

TEMPERATURE MEASUREMENT WITH A CONSTANT‐ VOLUME GAS THERMOMETER

The standard thermometer, against which all other thermometers are calibrated, is based on the pressure of a gas in a fixed volume. Figure 1.7 shows such a constant volume gas thermometer. It consists of a gas-filled bulb connected by a tube to a mercury manometer. By raising or lowering reservoir R, the mercury level on the left can always be brought to the zero of the scale to keep the gas volume constant. The temperature of anybody in thermal contact with the bulb is then defined to be T = Cp

(1)

Fig. 1.7

A constant volume gas thermom-

where p is the pressure within the gas eter. Its bulb is immersed in a liquid whose temperature T is to be measured and C is a constant. The pressure p = p0 – r hg

(2)

p0 is the atmospheric pressure, r is the density of mercury in the manometer, h is the measured difference between the mercury levels in the two arms of the manometer tube and g is acceleration due to gravity (1 atm = 1.01 × 105 Pa = 760 torr).

Basic Ideas 1.9

If we next put the bulb in a triple point cell, the temperature now being measured is T3 = Cp3

(3)

where p3 is the gas pressure now. Eliminating C between Eqns. (1) and (3), we get temperature p p __ T = T3 __ p3 = (273.16 K) p3

( )

( )

(4)

We face a problem with this thermometer: if we use it to measure (say) the boiling point of water, we find that different gases in the bulb give slightly different results. However, as we use smaller and smaller amounts of gas to fill the bulb, the readings converge to a single temperature, no matter what gas we use. Thus, temperature T = (273.16 K)

(

p lim __ gas Æ 0 p3

)

(5)

and unknown temperature T can be calculated, knowing p and p3.

1.8

INTERNATIONAL PRACTICAL TEMPERATURE SCALE

The accurate measurement of temperature by a standard gas thermometer is a difficult and tedious process. To help in the calibration and correction of other thermometers, several basic fixed points have been measured on the Celsius scale with great accuracy using helium gas constant volume thermometer (Table 1.1). Between these points, temperatures are easily interpolated by means of resistance thermometers (– 190°C to 630.5°C), thermocouples (630.5°C to 1063°C) and radiation thermometers (beyond 1063°C). To facilitate measurements, secondary fixed points have been added to the International Practical Temperature Scale based on substances easily available in pure form. Table 1.1 Basic fixed points on the International Practical Temperature Scale of 1948. Substance

Designation

Oxygen

normal boiling point

– 182.97

90.18

Water

normal freezing point

0.00

273.15

standard triple point

0.01

273.16

normal boiling point

100.00

373.15

Sulphur

normal boiling point

444.60

717.75

Antimony

normal melting point

630.5

903.6

960.8

1233.9

Silver

normal melting point

Gold

normal melting point

°C

1063

°K

1336

1.10 Heat and Thermodynamics 1.9

RESISTANCE THERMOMETER

Pure platinum (Pt) is a suitable metal for the resistance thermometer due to its resistance to chemical attack and its high melting point (1770°C). A long fine platinum wire is wound around a thin frame in such a way that excessive strain is avoided when the wire contracts on cooling. The resistance of the Pt wire, Rt, in a resistance thermometer with compensating leads can be measured by a Wheatstone bridge (Fig. 1.8), with RP and RQ as known fixed resistances and R as an adjustable measuring resistance. We have Rt = (RQ /RP)R.

Fig. 1.8

Resistance thermometer with compensating leads

The resistance Rt can also be measured by passing a known constant current in the thermometer and measuring the potential difference across it with the help of a potentiometer (Fig. 1.9). The current is kept constant by adjusting a rheostat so that the potential fall across a standard resistance in series with the thermometer, as checked with a monitoring potentiometer, remains constant.

Basic Ideas 1.11

Fig. 1.9

Potentiometer method measuring Rt

From the oxygen point (– 182.97°C) to the triple point of water (0.01°C), the temperature t is given by Rt = R0 [1 + At + Bt2 + C (t – 100) t3] where Rt is the resistance of the platinum wire at t°C and R0 at 0°C. The constants R0, A, B and C are determined by measurements at the oxygen point, the triple point of water, the steam point and the sulphur point. From the triple point of water to the antimony point (630.5°C), the temperature t is given by Rt = R0 [1 + At + Bt2] The constants are determined by the triple point of water, the steam point and the sulphur point.

1.10

THERMOCOUPLE

When two wires of different metals are joined to form a closed circuit, an electric current flows round the circuit so long as the two junctions are kept at different temperatures. The thermal emf E produced is measured with a potentiometer (Fig. 1.10). The thermocouple is calibrated by measuring the emf at various known temperatures, the reference junction being kept at 0°C. The temperature t at the test junction is given by E = a + bt + ct2 for the range 650.5°C to the gold point (1063°C).

1.12 Heat and Thermodynamics

Fig. 1.10 Thermocouple

1.11

IDEAL GASES AND GAS LAWS

The gas which perfectly obeys the Boyle’s and Charle’s laws is called an ideal or perfect gas.

Assumptions for an Ideal Gas (i) This gas obeys perfectly the Boyle’s, Charle’s and pressure laws at each pressure and temperature. (ii) The coefficient of volume expansion and the pressure coefficient of this gas are equal = 1/273 °C–1 (iii) The molecules are of negligible size in comparison to the volume of the gas. (iv) There are no forces of attraction amongst the molecules of the gas. Thus an ideal gas cannot be obtained in the liquid or solid state. Indeed, no gas is a perfectly an ideal gas, but in practice we can assume oxygen, nitrogen, hydrogen and helium to be nearly the ideal gases at ordinary temperature and ordinary pressure because it is difficult to liquefy them under these conditions. At a constant temperature, the volume of a given mass of gas is inversely proportional to its pressure:

Boyle’s Law

1 V μ __ P or or

PV = constant (at T = constant) P1V1 = P2V2

(1)

Basic Ideas 1.13

If a graph is plotted by taking the pressure P on the Y axis and volume V on the x-axis, a rectangular hyperbola is obtained (Fig. 1.11). This curve is called an isothermal curve. At a constant pressure, the volume of a given mass of a gas increases (decreases) by (1/273) of its volume at 0°C on increasing (decreasing) the temperature by 1°C. V Let at a constant pressure, the volume of given mass Fig. 1.11 P-V graph of a gas at 0°C be V0 and at t°C be Vt. Since increase in volume of gas on increasing its temperature by t°C will be V0t/273, therefore, the volume of a gas at t°C is given by P

Charle’s Law

V0t Vt = V0 + ____ 273 or

t Vt = V0 1 + ____ 273

[

]

(2)

Volume V

If a graph is plotted, taking volume V on Y-axis and temperature t on X-axis, we get a straight line (Fig. 1.12) which meets the temperature axis at – 273°C.

Vt V0

– 273

0 t Temperature t (in °C)

Fig. 1.12 V-t graph

Charle’s Law in Terms of Absolute Temperature Let, at a constant pressure, the volume of given mass of a gas at 0°C, t1°C and t2°C be respectively V0, V1 and V2, then by Charle’s law. t1 V1 = V0 1 + ____ ; 273

( (

t2 V2 = V0 1 + ____ 273

) )

1.14 Heat and Thermodynamics V1 [1 + (t1/273)] __ = ___________ V2 [1 + (t2/273)] 273 + t1 T1 = _______ = __ 273 + t2 T2 or

V __ = constant T

i.e.,

V μT

(3)

Thus, at a constant pressure, the volume of given mass of a gas is directly proportional to its absolute temperature.

Pressure Law Keeping the volume constant, if given mass of gas is heated, its pressure increases. Experimentally, it is found that at a constant volume, the pressure of a given mass of a gas increases (decreases) by 1/273 of its pressure at 0°C or increasing (decreasing) its temperature by 1°C. Let at a constant volume, the pressure of given mass of a gas at 0°C be P0 and at t°C be Pt. Then by pressure law P0t Pt = P0 + ____ 273 or

t Pt = P0 1 + ____ 273

[

]

(4)

Pressure P

If a graph is plotted by taking the pressure P on Y-axis and temperature t on X-axis, a straight line is obtained (Fig. 1.13) which meets the temperature axis at – 273°C.

Pt P0

– 273

0 t Temperature t (in °C)

Fig. 1.13

Basic Ideas 1.15

Pressure Law in Terms of Absolute Temperature Let at a constant volume, the pressure of given mass of a gas at 0°C, t1°C and t2°C be respectively P0, P1 and P2, then by pressure law t1 P1 = P0 1 + ____ ; 273

( (

t2 P2 = P0 1 + ____ 273 \

) )

P1 1 + (t1/273) __ = __________ P2 1 + (t2/273) 273 + t1 T1 = _______ = __ 273 + t2 T2

or

P __ = constant T

i.e.,

PμT

(5)

Thus, at a constant volume, the pressure of a given mass of a gas is directly proportional to its absolute temperature.

1.12

AVOGADRO’S NUMBER

While thinking of molecular aspect of gases, one has to measure the sizes of gas – samples in moles. If we do so, then we can be certain that we are comparing samples that contain the same number of atoms or molecules. The mole is one of the seven SI base units and is defined as follows: One mole is the number of atoms in a 12g sample of Carbon-12. Then one may enquire “How many atoms or molecules are there in a mole”? The answer is determined experimentally and is NA = 6.02 × 1023 mol–1

(1)

(This is known as Avogadro’s number). Here mol–1 represents “per mole”. This is based on Avogadro’s assertion that all gases contain the same number of atoms or molecules when they occupy the same volume under the same conditions of temperature and pressure. The number of moles n contained in a sample of any substance having N molecules is given by N n = ___ NA

(2)

1.16 Heat and Thermodynamics 1.13

UNIVERSAL GAS CONSTANT, BOLTZMANN CONSTANT AND IDEAL GAS EQUATION

Experiments have confirmed that if we confine 1 mole samples of various gases in boxes of identical volume and hold the gases at the same temperature, then their measured pressures are nearly (though not exactly) the same. If the measurements are at lower gas densities, then the small differences in the measured pressures tend to disappear. Further, at low enough densities, all real gases tend to obey the relation. pV = nRT (ideal gas law)

(1)

where p is the absolute (not gauge) pressure, n is the number of moles of gas present and T is the absolute temperature (in Kelvin). The symbol R is a constant called the Universal Gas Constant (that has the same value for all gases). R = 8.31 J/mol.K

(2)

Equation (1) is called the ideal gas law. Provided the gas density is low, this law holds for any single gas or any mixture of different gases (For a mixture, n is the total number of moles in the mixture). We can rewrite Eqn. (1) in an alternative form in terms of a constant called the Boltzmann constant, which is defined as 8.31 J/mol.K R k = ___ = _______________ NA 6.02 × 1023 mol–1 = 1.38 × 10–23 J/K \

(3)

R = kNA But

N n = ___ NA

N or NA = __ n

(4)

\

kN R = ___ n

(5)

or

nR = Nk

(6)

Substituting this into Eqn. (1) gives a second expression for the ideal gas law: pV = NkT

(ideal gas law)

(7)

(Note the difference between the Eqns. (1) and (7), for the ideal gas law: Eqn. (1) involves the number of moles n, whereas Eqn. (7) involves the number of molecules N). The data for CO2 are plotted in Fig. 1.14 for three different temperatures. The remarkable feature of these curves is (a) that they all converge to exactly the same point on the vertical axis, whatever the temperature and (b) that the curves for all

Basic Ideas 1.17

other gases converge to exactly the same point. This common limit of the ratio Pv/T as P approaches zero is, the universal gas constant (R). The unit of Pv/T is 1 (Nm–2) (m3 kilomole–1) (K–1) = 1 (Nm) (Kilomole–1 K–1) = 1 J kilomole–1 K–1 and the value of R in this system is R = 8.3143 × 103 J kilomole–1 K–1 At sufficiently low pressures, we can write for all gases Pv/T = R

or Pv = RT

Pv ___ = R (ideal gas) T

or Since \

v = V/n PV = nRT

For an ideal gas, the curves in Fig. 1.14 coalesce into a single horizontal straight line at a height R above the pressure axis.

–1

Pv/(J kilomole K

–1

)

10 3 R = 8.3143 × 10 Ideal gas

8

T3

6

T2 4

T3 > T2 > T1

T1

2 0

2

4 6 –2 Pressure (Nm )

8

× 10

7

Fig. 1.14 The limiting value of Pv/T is independent of T for all gases. For an ideal gas, Pv/T is constant

1.14

THE “SYSTEM”

In order to understand the world around us, it is just not possible to make observations on the whole world at a time. For our convenience, we can extract relevant portions of the world (depending on the view to use it to our advantage) and put them under observations. These relevant parts of the physical world separated from the rest of the world by means of well-defined boundaries are called systems. Whatever is left out is called environment or surroundings.

1.18 Heat and Thermodynamics If ‘mass’ is held constant the system is called a closed system. The surrounding walls of a closed system must not permit any material transport across them. Although mass of a closed system remains constant, yet the energy may vary. This implies that surrounding walls though impermeable to matter should be permeable to energy (This implies that the walls should be made of good conductors of heat or electricity). If the enclosing walls are impermeable to both matter and energy, the system is said to be an isolated system. Thus, the enclosing walls of an isolated system should be impervious, rigid and made of non-conducting materials (One of the nearest approaches to an isolated system is a Dewar flask). The mass and energy of an isolated system remains constant because it does not interact with its environment in terms of both mass transfer and energy transfer. Thus, the isolated system though highly idealized and practically unattainable is a useful concept in performing energy analysis because system plus environment can always be taken as an isolated system. If a system communicates with its environment in terms of both mass transfer and energy transfer, it is called an open system. Apart from the terms ‘closed’, ‘open’ and ‘isolated’, there are also other adjectives to qualify systems further. These are, for example, isothermal (meaning same temperature), isobaric (meaning same pressure), adiabatic (meaning no exchange of heat with the environment), isochoric (meaning no change of volume) etc. These adjectives can be used to qualify a process e.g., an isothermal process (i.e., a process at constant temperature) isochoric process (a process in which volume remains constant). Further, the minimum number of independent variables necessary for complete description of the system are called state variables. These variables define the state of the system. State variables in fact are mostly the experimentally determinable properties of a system (Property is an attribute which belongs to a system and which, in principle, can be quantitatively evaluated). It must be borne in mind that property should be relevant to the context. For example, while talking about thermodynamics, the relevant properties are energy, volume, temperature, pressure, etc. Note that heat and work are not properties as they do not belong to the system. These are operations which are performed on the system to alter its energy. In thermodynamics, the central theme is energy and energy transfer. The relevant properties are therefore, those that are involved with energy in some way or the other.

1.15

INTENSIVE AND EXTENSIVE PROPERTIES

Imagine a system in equilibrium to be divided into two equal parts, each with equal mass. Those properties of each half of the system that remain the same to be

Basic Ideas 1.19

intensive and those that are halved are called extensive. The intensive coordinates of a system, such as temperature and pressure, are independent of the mass; the extensive properties are proportional to the mass. Extensive properties are thus additive i.e., their value for the whole system is equal to the sum of the values for the individual parts. Intensive properties are not additive. An extensive variable X, when divided by the mass m or the number of moles n of system, becomes an intensive variable x X x = __ m

X or x = __ n

nM = m where M is the molecular mass (or molecular weight). The ratio of an extensive variable to the mass of the system is called the specific value of that variables e.g., the volume V and specific volume (or volume per unit mass) v are related by V v = __ m The ratio of an extensive variable to the number of moles of a system is called the molar (or molal) specific value of that variables e.g., molar specific volume v, is V v = __ n No confusion arises from the use of the same letter to denote both the extensive variable per unit mass and the extensive variable per mole. In almost every equation in such a variable occurs, there comes some other quantity that indicates which variable, X/m or X/n, is meant. If in some rare case there happens to be no such indication, the equation holds equally well for either case.

1.16

THERMODYNAMIC SYSTEM AND THERMODYNAMIC COORDINATES

A system whose state depends on the thermodynamic quantities is called thermodynamic system. For example a gas enclosed in cylinder is a thermodynamic system because state of the gas depends on its pressure, volume and temperature. A thermodynamic system means a macroscopic body in equilibrium which is made up of large number of microparticles e.g., molecules, atoms or ions etc. The quantities required for the complete knowledge of thermodynamic state of a system are called the thermodynamic coordinates e.g., pressure P, volume V and temperature T are the thermodynamic coordinates of the gaseous systems. These quantities are related by an equation called the equation of state of the

1.20 Heat and Thermodynamics system (e.g., PV = RT is the equation of state for a perfect gas). It is clear from the equation of state that out of the thermodynamic coordinates P, V and T, only two coordinates are independent variables, which can be chosen arbitrarily, the third coordinate is the dependent variable and can be determined with the help of equation of state.

1.17

P‐v‐T SURFACE FOR AN IDEAL GAS AND BOYLE’S LAW

Pressure

The equation of state of a PvT system defines a surface in a rectangular coordinate system in which P, v, and T are plotted along the three axes. A portion of this surface for an ideal gas is shown in Fig. 1.15. Every possible equilibrium state of an ideal gas is represented by a point on its P-v-T surface, and every point on the surface represents a possible equilibrium state. A quasistatic process, i.e., a succession of equilibrium states, is represented by a line on the surface. The full lines in Fig. 1.15 represent processes at constant temperature, or isothermal processes. The dotted lines represent isochoric processes, and the dashed lines, isobaric processes.

Vo lu

me

Fig. 1.15 P-v-T surface for an ideal gas

e

tur

era

p Tem

Basic Ideas 1.21

Pressure

Pressure

Figures 1.16(a) and 1.16(b) are projections of the lines in Fig. 1.15 onto the P-v and P-T planes.

Temperature

Volume (a)

(b)

Fig. 1.16 Projections of the ideal gas P-v-T surface onto (a) the P-v plane, and (b) the P-T plane

In an isothermal process, for a fixed mass of an ideal gas, Pv = RT = constant

(1)

Robert Boyle, in 1660, discovered experimentally that the product of the pressure and volume is very nearly constant for a fixed mass of a real gas at constant temperature. This fact is known as Boyle’s law. It is, of course, exactly true for an ideal gas, by definition. The curves in Fig. 1.16(a) are graphs of Eqn. (1) for different temperatures and hence for different values of the constant. They are equilateral hyperbolas. In a process at constant volume, for a fixed mass of an ideal gas,

( )

nR P = ___ T = constant × T V

(2)

That is, the pressure is a linear function of the temperature T. The dotted lines in Fig. 1.16(b) are graphs of Eqn. (2) for different volumes and hence different values of the constant. If the pressure of a fixed mass of an ideal gas is constant,

( )

nR V = ___ T = constant × T P and the volume is a linear function of the temperature at constant pressure.

(3)

1.22 Heat and Thermodynamics 1.18

BEHAVIOUR OF REAL GASES EQUATION OF STATE AND P‐v‐T SURFACES

The assumption of non-interacting molecules is satisfactory upto a point. After all, the elastic collision between two molecules implies a repulsive force at sufficiently small distances. Figure 1.17(a) is a sketch of potentials of such a force where ro represents the radius of the hard-sphere model molecule. A mutual interaction between molecules of a real gas comes into existence due to rearrangement of the charge distribution in each of the molecules.

van der Waal’s Interaction The interaction is explained in terms of van der Waals forces, as follows: Consider two nearby atoms. It is true that on an average, the centre of negative charge in an atom coincides with the centre of positive charge (i.e., with the nucleus). But the electrons are moving around the nucleus and at any instant it may happen that the two centres fail to coincide and an instantaneous electric dipole moment of magnitude m1 develops which produces an electric field E = 2m1/r3 at the centre of the second atom distant r from the first atom. This field induces an instantaneous dipole moment m2 = aE = 2am1/r3 on the second atom where a is the electronic polarizability. The interaction potential energy of the dipole moments is Ve(r) = – m2E = – aE2 4am12 c = – _____ ∫ – __6 (1) r6 r This is called van der Waals interaction. The negative sign shows that the interaction is attractive (Ve ~ – 2 × 10–14 erg). As the atoms (or molecules) approach each other more closely, a repulsive term arises from the action of electric charges of one atom on the other (This repulsive force must exist otherwise two atoms would not repel one another after impact). Experimental data on the inert gases can be fitted well by assuming that the repulsive potential is of the form B/r12, where B is a positive constant, when used with a long range attractive potential of the form (Eqn. 1). Then, we can write the total potential energy of two interacting atoms at separation r as V(r) = 4 e

[ ( __sr ) – ( __sr ) ] 12

6

(2)

where e and s are the parameters with 4 e s6 = C and

4 e s12 = B

(3)

Basic Ideas 1.23

Repulsion Separation

Potential Energy V

Potential Energy V (r)

The potential (2) is called the Lennard-Jones potential and is plotted in Fig. 1.17(a). The minimum occurs at ro = s 21/6.

Separation

r s

e

e

d r0 Attraction (a)

Fig. 1.17

(b)

(a) Lennard-Jones potential (b) Idealized interatomic potential

The value of V(r) at the minimum is – e and V(r) = 0 at r = s. The constants B and C (or e and s) are empirical parameters determines from measurements made on gases (The data used include the virial coefficients and the viscosity). The point of minimum V(r), ro is the equilibrium separation at 0°K and corresponds to the density of the solid. Gases always have r > ro. van der Waals idealized the situation by approximating the repulsive part by an infinite hard sphere repulsion so that the potential energy looks like that shown in Fig. 1.17(b). Thus, each atom is imagined to be a hard sphere of radius ro surrounded by an attractive force field. The effects of the repulsive and attractive parts are then taken into account separately. Some typical values of ro and e are given in Table 1.2. Table 1.2 Values of ro and e. Molecule

2ro (Å)

e (ergs)

He

2.2

1 × 10–15

H2

2.7

4 × 10–15

Ar

3.2

5 × 10–15

CO2

4.5

40 × 10–15

The van der Waals force (F = – ∂V/∂r) is repulsive for distances less than ~ 2.5 × 10–10 m which is of the order of diameter of the molecule. The attractive force becomes vanishingly small for distances ~ four times the diameter. Within

1.24 Heat and Thermodynamics this range the attractive force is about 5 × 10–13 Newtons. Though this force is small, considering the fact that molecular masses are of the order of a 10–26 kg, this attractive force plays a significant role in molecular motion.

Modification of the Ideal Gas Equation The above fact suggests that the ideal gas equation PV = nRT

(4)

needs to be modified for the case of a real gas. The need for such a modification is more so if the above equation is to be employed in the measurement of temperatures or calibration of standard thermometers employing real gases. The modifications as applicable to real gases on the following lines were first suggested by van der Waals. As a consequence of the finite size of a molecule, the centre of the molecule cannot approach the wall of the container closes than ro. The minimum distance between two colliding molecules is 2ro. Hence the actual volume available for the molecule is less than the volume of the container. The precise decrement depends on the number of molecules. Let us write the gas equation for one mole of gas i.e., in terms of the molar volume v (= V/n). If v is the molar volume with a real gas, to achieve the same pressure with point-mass molecules of an ideal gas we would need a volume less than v by an amount b called the covolume. Therefore, P (v – b) = RT

(5)

The effect of short range forces amongst the molecules is to render the molecular density somewhat more compact on the macroscopic scale. The internal pressure in the gas is due to the mutual interaction between the molecules. If there are No molecules per unit volume, the number of mutually interacting pairs is No (No – 1)/2, which for large N is proportional to No2. Thus, in addition to the external pressure, there is an additional term which is proportional to N2 or equivalently to the inverse square of molar volume (i.e., 1/v2). The equation of state, therefore, is (P + a/v2) (v – b) = RT

(6)

where a is a constant. This is known as the van der Waals equation of state for a real gas. The term (a/v2) is called internal pressure. It is clear that at low pressure v >> b and also a/v2 qc, only one real root exists for all values of p. The maxima and minima of the isotherms for q > qc are not usually observed. However, by using very pure gases and liquids, the portions xA (supersaturated vapour) and By (supercooled liquid) can be experimentally realized. They are metastble states. We have drawn the locus of the maxima and minima of the isotherms as a dotted curve in Fig. 3.8. Its equation is found by the condition that the tangent to dp the isotherm is horizontal, ___ = 0 where q = constant i.e., from Eqn. (7) dv q

( )

( )

q

( )

d 2p ∫ ___2 q dv

dp ___ dv

Rq 2a = – _______2 + ___3 = 0 (8) (v – b) v The maximum of the dotted curve determines the critical point X. It is obtained by the condition that the tangent to the dotted curve described by Eqn. (8) is horizontal, dp d ___ ___ dv dv

( )

=0

(9)

q

2Rq 6a _______ – ___4 = 0 3 (v – b) v

(10)

At the critical point X (pc, vc, qc), the three Eqns. (7), (8) and (10) must be simultaneously satisfied i.e., Rqc a pc = _______ – ___2 (vc – b) vc

(11)

– Rqc 2a _______ + ___3 = 0 2 (vc – b) vc

(12)

2Rqc 6a _______ – ___4 = 0 3 (vc – b) vc

(13)

From Eqns. (12) and (13), we get

or

vc _____ vc – b __ = 3 2 vc = 3b

(14)

Substituting this value in Eqn. (12), we get 8a qc = _____ 27bR

(15)

3.10 Heat and Thermodynamics Equation (11) gives a pc = ____2 27b

(16)

Thus, the constants a and b can be determined from the critical point data. It may be verified that Rqc __ 8 ____ pc vc ∫ 3 = 2.67 (for all gases)

(17)

The observed values of Rqc/(pc vc) fall in the neighbourhood of 3.5. Thus van der Waals equation is not closely obeyed by any gas in the neighbourhood of its critical point. However, it is still a very convenient equation over a wide range.

Values of a and b in Terms of Critical Constants From Eqns. (15) and (16), 2 q2c [8a/(27Rb)] 64a ___ ___________ = ____ = 2 pc a/(27b ) 27R2

Therefore, 2 27R2 q c a = ____ ___ 64 pc

(18)

and qc _________ 8a/(27Rb) ___ 8a __ pc = a/(27b2) = R \

R qc b = __ __ 8 pc

(19)

Thus, knowing qc and pc, the values of van der Waals constants a and b can be calculated.

Gas and Vapour In general conversation, when the substances which are solid ir liquid at normal temperature and pressure come in gaseous state are known as vapour whereas the substances which are in gaseous state at normal temperature and pressure are said to be gases. In scientific language, the gaseous state above the critical temperature of a substance is known as gas and the gaseous state below the critical temperature of the substance is known as vapour.

Real Gases, Liquefaction of Gases and Production and... 3.11

3.7

COMPARISON BETWEEN EXPERIMENTAL AND THEORETICAL P‐V CURVES

On comparing the experimental P-V curve (Fig. 3.9(a)] and theoretical P-V curve obtained from van der Waals Equation of state (Fig. 3.9b), we see that Bb and cC parts obtained in the theoretical curve (Fig. 3.9b) which respectively represent the supersaturated vapour state and supersaturated liquid state, are not obtained in the experimental curve (Fig. 3.9a). On the other hand, only a horizontal line BC is obtained in the experimental curve which shows that the gas always remains in steady state. Indeed, the supersaturated vapour state (in which volume decreases with the decrease in pressure) and supersaturated liquid state are the unsteady states which are impossible to be obtained in practice.

Fig. 3.9

P-V isotherms for van der Waals equation of state (a) obtained from Andrew’s experiment for CO2; (b) theoretical curves for CO2

Show that the van der Waals gas law departs from perfect gas law by 62.5% at critical temperature. Solution The van der Waals equation is Example 3.1

( p + ___Va ) (V – b) = RT 2

or

a ab pV – bp + __ – ___2 = RT V V The perfect gas equation is pV = RT The deviation is obtained by subtracting Eqn. (ii) from Eqn. (i) at Tc: a ab – bp + __ – ___2 = R (Tc – T) V V

(i)

(ii)

3.12 Heat and Thermodynamics or

a ab RDT = – bp + __ – ___2 V V 8a a Now, Tc = _____, pc = ____2 , Vc = 3b 27Rb 27b So, at the critical point

a ab a RDT = – b ____2 + ___ – ___2 3b 9b 27b Using Tc value i.e., 27RTc a ______ __ = , 8 b we have 1 27RTc 1 27RTc 1 27RTc RDT = – ___ × ______ + __ × ______ – __ × ______ 3 9 27 8 8 8 \ or

3.8

5 DT __ ___ = Tc 8 5 DT __ ___ = × 100% Tc 8 = 62.5%

VAN DER WAALS EQUATION AND THE BOYLE TEMPERATURE

In general, instead of the Boyle law, pv = constant a fixed Q, the real (or imperfect) gases obey an empirical power equation of the form (1) pv = A + Bp + Cp2 + Dp3 + … where A, B, C, D, … are the virial coefficients. Usually C, D, … are very small. Therefore, apart from A, the most important coefficient is B. The variation of pv and p is shown in Fig. 3.10.

Fig. 3.10 Variation of pv with p for a real gas at different temperatures

Real Gases, Liquefaction of Gases and Production and... 3.13

If we find the slope of the curves near p = 0, we have

( ) ∂(pv) _____ ∂p

= (B + 2Cp + 3Dp2 + …)p = 0 p=0

=B

(2)

We define a temperature QB at which B = 0. From Fig. 3.10, we see that B is negative at Q < QB and positive at Q > QB. For Q = QB and p Æ 0 we have B = 0 so that the expansion (1) reduces to pv = A that is, Boyle law holds approximately again. For this reason QB is called the Boyle temperature. For a van der Waals gas we can find QB by calculating ∂(pv)/∂p and equating it RQ a to zero at p = 0. Thus van der Waals equation pv = _____ – __2 gives v–b v RQv a pv = _____ – __ v–b v ∂(pv) v a 1 _____ = RQ _____ – RQ ______2 + __2 v–b (v – b) v ∂p

[

[

RQb a = – _______2 + __2 (v – b) v

]( ) ∂v ___ ∂p

]( ) ∂v ___ ∂p

Q

(3)

Q

If we require it to be zero at Q = QB and p = 0, it is equivalent to considering v as approaching infinity so that v – b can be replaced by v. Therefore, from Eq. (3), we have – RQB b + a = 0 or

a 27 QB = ___ = ___ Qc Rb 8 The van der Waals equation can be written as

(

b pv = RQ 1 – __ v

)

–1

(4)

a – __ v

b b__2 a __ = RQ 1 + __ v + v2 + … – v b – (a/RQ) __ b2 = RQ 1 + _________ + 2+… v v

(

[

)

]

(5)

At QB, we can put Q = QB = a/Rb. Then for p small, or large v, Eqn. (5) reduces to

3.14 Heat and Thermodynamics pv = RQB

(6)

for a real van der Waals gas. It is useful to plot (pv/RQ) vs. p as shown in Fig. 3.11 for CO2. Then all the curves converge exactly to the same point on the vertical axis, whatever the temperature, because in the limit of low pressure lim (pv) = A = RQ. The curve at QB is parallel to the p-axis for small p values.

Fig. 3.11 (pv/RQ) vs. p graph for CO2 [After B.K. Agarwal, Thermal Physics, Lokbharti Publications, Allahabad, India (1988)]

3.9

CORRESPONDING STATES

The van der Waals equation can be put in a form that is applicable to any substance by introducing the reduced quantities pr, vr, Qr as follows p pr = __ p, c

v vr = __ v, c

Q Qr = ___ Qc

(7)

Then, a p = ____2 pr, 27b 8a Q = _____ Qr 27Rb

v = 3b vr, (8)

and the van der Waals equation becomes

(

)

3 pr + ___2 (3vr – 1) = 8Qr vr

(9)

Real Gases, Liquefaction of Gases and Production and... 3.15

This is a remarkable equation because it does not involve the constants a and b characteristic of a particular gas. It holds for any van der Waals gas. The statement that it is a universal equation valid for all gases is called the law of corresponding states to the extent that real gases obey the van der Waals equation. Different gases with the same pr, vr and Qr are said to be in corresponding states.

3.10 JOULE’S EXPANSION OF AN IDEAL GAS AND VAN DER WAAL’S GAS When a definite mass of a gas is expanded such that the external work done by the gas or on the gas is zero, neither the gas absorbs nor rejects any heat, then the expansion of gas is known as free expansion or Joule’s expansion. Scientist Joule in his experiments, studied the expansion of ideal gas and real (or van der Waal) gas and concluded that there exists no force between the molecules of an ideal gas whereas there exist attractive forces between the molecules of van der Waal gas which are known as van der Waal forces. Figure 3.12 shows the experimental arrangement of Joule’s experiment. It consists of two copper cylindrical vessels A and B which are mutually connected by a tube C. There is a stop cock S in the tube C. Vessel A is filled with the experimental gas at a high pressure while the vessel B is kept completely evacuated. Both the cylinders are immersed in a vessel filled with water. Fig. 3.12 Experimental arrangement of Joule’s experiment Temperature of water is measured with the help of a sensitive thermometer T. It is clear that on opening the stop cock S, free expansion of experimental gas from cylinder A to cylinder B takes place since it involves neither any external work done nor any exchange of heat. Experimental

(a)

Arrangement

Joule’s Expansion of an Ideal Gas

Assuming air to be an ideal gas when scientist Joule filled air in the cylinder A and then allowed the air to expand freely by opening the stop cock S, he found that there is no change in temperature of water. It is thus concluded that there are no intermolecular attractive forces acting between the molecules of an ideal gas because if there will be any attractive force between the molecules, then the temperature of water should fall during this expansion because there should be some internal work done by the gas against the attractive forces in separating the molecules apart from each other.

3.16 Heat and Thermodynamics Thus, on bringing any change in pressure and volume of an ideal gas, its temperature remains constant i.e., its internal energy remains unchanged. This is known as Joule’s law. In other words, we can say that according to Joule’s law, the internal energy U of an ideal gas only depends on its temperature T and does not depend on its volume V and pressure P. Mathematically, for an ideal gas U = f(T)

and

( ) ( ) ( ) ( ) ∂U ___ ∂V

T

∂U ___ ∂T

P

∂U ___ ∂P

= 0,

∂U = ___ ∂T

=0 T

(1)

V

Thus, internal energy of an ideal gas remains unchanged with pressure and volume at a constant temperature and change in internal energy per degree change in temperature remains same whether pressure remains constant or volume remains constant. Remember that Joule’s law is true only for an ideal gas.

(b)

Joule’s Expansion of van der Waal’s or Real Gas

When scientist Joule used a real gas at a high pressure in cylinder A in his experiment and opened the stop cock to allow the free expansion of gas, then he found that there is decrease in temperature of water. From this, it is concluded that there exist inter-molecular attractive forces in the molecules of a real gas. Hence, in expansion of gas, there is some internal work done by the molecules against the attractive forces in going away from each other. This increases the potential energy of gas. Since during this expansion, neither any external heat is taken up by the gas nor the gas rejects any heat, therefore, as a result of increase in potential energy, the internal kinetic energy of the gas decreases. As a result, the temperature of the gas falls. It is clear that inter-molecular forces in a real gas are of attractive nature. Remember that if inter-molecular forces are of repulsive nature, then in free expansion of gas, the potential energy of gas should decrease i.e., kinetic energy should increase, as a result the temperature of gas should increase. But experimentally, it is found that there is always cooling in Joule’s expansion of a real gas, therefore, the van der Waal’s forces are of attractive nature only.

3.11

JOULE’S COEFFICIENT FOR REAL GAS

We have read that cooling is produced in Joule’s (or free) expansion of a real gas (i.e., the temperature of gas falls). The temperature change in free expansion of gas can be represented in terms of Joule’s coefficient h. If in free expansion, there is a temperature change dT with change in volume dV, then

Real Gases, Liquefaction of Gases and Production and... 3.17

Joule’s coefficient

( )

∂T h = ___ ∂V

(2) U

Since the internal pressure caused due to inter-molecular attractive forces in a van der Waal gas is p = a/V2 where a is van der Waal’s constant of gas and V is the initial volume of gas, hence, the internal work done by gas against this internal pressure in Joule’s expansion of gas is a dW = pdV = ___2 dV V Clearly, this internal work done by gas will be equal to the increase in internal potential energy of gas or decrease in internal kinetic energy of gas. Hence, a (3) Decrease in internal kinetic energy of gas = – ___2 dV V Now, if the molar specific heat of gas at constant volume is CV and decrease in temperature of gas is dT, then for 1 mole of gas Decrease in internal kinetic energy = CV dT

(4)

From Eqns. (3) and (4) a CV dT = – ___2 dV V dT 1 a ___ = – ___ ___2 CV V dV

or

Hence, Joule’s coefficient

( )

∂T h = ___ ∂V

U

1 a = – ___ ___2 CV V

(5)

From above Eqn. (5), since a, CV and V are the positive quantities, hence Joule’s coefficient for a real gas is negative. It means that cooling effect is produced on Joule’s expansion of a real gas which is in arrangement to the experimental fact.

3.12

JOULE‐THOMSON COOLING

To establish completely the existence of inter-molecular forces in real gases, scientist Joule along with Thomson performed an experiment in 1952 which is known as porous plug experiment. In their experiment, they found that when a gas is passed through an insulated porous plug from a to a constant low pressure region, then there is a change in temperature of gas. This phenomenon is known as Joule-Thomson’s effect and this process is known as throttling process.

3.18 Heat and Thermodynamics The apparatus used in the experiment is shown in Fig. 3.13. In this experiment, experimental gas is taken in a cylinder A, compressed at a high pressure by means of a piston P and is passed through a spiral tube S. The spiral tube S is kept immersed in a water bath W maintained at a constant temperature so that the temperature of the compressed gas remain constant. The compressed gas at a constant temperature coming from the tube S enters a wide tube B, the upper end of which is open to atmosphere. Inside the tube B, there is a porous plug G made of cotton or silk and placed in between the two perforated discs D, D which are surrounded by a container C filled with asbestos so that it remains insulated. There are two platinum resistance thermometers T1 and T2 inside the tube B which measure respectively the temperature of gas before and after passing through the porous plug. The pressure of gas before entering the porous plug is equal to the pressure of gas entering the tube B which is measured by a manometer (not shown in the figure), while the pressure of gas coming out of the porous plug is equal to the atmospheric pressure. Porous Plug Experiment

Fig. 3.13

Arrangement of porous plug experiment

This experiment was performed for different gases and varying the initial pressure and temperature of the gas before entering the porous plug, different observations were obtained. From the various observations, following conclusions were drawn.

Results Obtained from the Experiment In each real gas, there are intermolecular forces of the attractive nature because on passing through the porous plug, there is a large decrease in the pressure of the gas due to which its molecules become far separated and so, the gas has to do the internal work against the intermolecular attractive forces. As a result, the internal kinetic energy of the gas decreases and the gas shows the cooling effect.

Real Gases, Liquefaction of Gases and Production and... 3.19

For a given pressure difference across the plug, there is a definite temperature for each gas, called the temperature of inversion Ti. If the initial temperature of the gas is lower than Ti, the gas shows the cooling effect after passing through the plug and if the initial temperature of the gas is higher than Ti, the gas shows the heating effect after passing through the plug.

3.12.1 How Temperature Changes in Joule-Thompson Expansion of a van der Waal’s Gas Let 1 mole of a real gas suffer Joule-Thompson expansion at a constant pressure P1 through a porous plug to a constant pressure P2. The initial volume and temperature be V1 and T1 and the final volume and temperature of gas after passing through the porous plug be V2 and T2 respectively. Refer to Fig. 3.15. External work done on the gas by the piston 1 on left side = – P1 V1. External work done on piston 2 by the gas on right side = P2 V2. \ Not external work done by the gas = P2 V2 – P1 V1.

(1)

Now, there are also attractive forces amongst the gas molecules and so, in this expansion, some internal work is also done by the gas against the intermolecular a attractive forces. The internal pressure due to attractive forces is ___2 where a is V van der Waals constant of the gas. Therefore, the internal work done by 1 mole gas in expansion of volume from V1 to V2 is V2

a –a = Ú ___2 dV = ___ V V1 V

[ ]

V2 V1

a a = ___ – ___ V1 V2

(2)

From Eqns. (1) and (2), total work done by the gas a a W = (P2 V2 – P1 V1) + ___ – ___ V1 V2

(

)

(3)

Now, from van der Waal’s gas equation

( P + ___Va ) (V – b) = RT 2

or

a ab PV = RT + Pb – __ + ___2 V V a = RT + Pb – __ V Substituting the above value of PV in Eqn. (3)

(

ab ___ is negligible V2

)

3.20 Heat and Thermodynamics a a a a W = RT2 + P2b – ___ – RT1 + P1b – ___ + ___ – ___ V2 V1 V1 V2

(

) (

)

2a 2a = R(T2 – T1) – b(P1 – P2) + ___ – ___ V1 V2 Now, if from the ideal gas equation PV = RT, the approximate values of V1 and RT1 RT2 V2 are assumed to be V1 = ____ and V2 = ____ respectively, then P1 P2 2aP1 2aP2 W = R(T2 – T1) – b(P1 – P2) + _____ – _____ RT1 RT2 Now, since T1 and T2 are nearly equal, hence assuming T1 – T2 = dT and T1 ª T2 = T (initial temperature), we get 2a W = – RdT – b(P1 – P2) + ___ (P1 – P2) RT

[

]

2a = (P1 – P2) ___ – b – RdT RT

(4)

Since the porous plug is insulated hence neither the gas absorbs heat nor it rejects heat. As a result, the total work done by the gas is obtained from the kinetic energy of its molecules due to which the temperature of gas decreases by dT. If molar specific heat of the gas at constant volume is CV, the decrease in internal kinetic energy of gas due to fall in temperature by dT is dU = CV (T2 – T1) = CV dT

(5)

Now, from the law of conservation of energy W = dU

(

(

or Since or \ or

)

2a (P1 – P2) ___ – b – RdT = CV dT RT

or

2a (CV + R) dT = (P1 – P2) ___ – b RT CP – CV = R

)

CV + R = CP

(

2a CP dT = (P1 – P2) ___ – b RT

(

)

)

1 2a dT = ___ ___ – b (P1 – P2) CP RT

(6)

Real Gases, Liquefaction of Gases and Production and... 3.21

The above equation gives the expression for cooling of a real gas in Joule Thompson effect. In this expression, since P2 < P1 therefore P1 – P2 is always positive. Obviously, in Joule-Thompson effect, the change in temperature of a real gas is directly proportional to the pressure difference P1 – P2 across the porous plug. Now, from Eqn. (6), it is clear that

(

)

(

)

2a 2a 2a (i) If ___ > b or T < ___, the quantity ___ – b is positive, hence dT will be RT RT Rb positive i.e., the temperature decreases with decrease in pressure. In other words, the gas on passing through the porous plug gets cooled. 2a 2a 2a (ii) If ___ < b or T > ___, the quantity ___ – b is negative, hence dT will be RT RT Rb negtive i.e., the temperature increases with decrease in pressure. In other words, the gas after passing through the porous plug gets heated.

(

)

2a 2a 2a (iii) If ___ = b or T = ___, the quantity ___ – b = 0, hence dT = 0, i.e., there is RT RT Rb no change in temperature of gas on passing through the porous plug. This temperature is called the temperature of inversion 2a Ti = ___ Rb

(7)

The temperature of inversion of hydrogen and helium are below the ordinary temperature. Therefore, these gases (H2 and He) show heating effect in adiabatic throttling process. If these gases (H2 and He) are first cooled below their temperature of inversion and then passed through the porous plug, they also show the cooling effect.

3.13

TEMPERATURE OF INVERSION

The temperature of inversion of a gas is the temperature below which a gas gets cooled in Joule-Thomson expansion and above which the gas gets heated, i.e., at this temperature, the sign of Joule-Thomson effect is reversed. In other words, the gas must be initially cooled below its temperature of inversion to produce cooling by Joule-Thomson effect. The expression for temperature of inversion given by Eqn. (7) is an approximate expression. The actual expression is obtained by equating the quantity a [T(∂V/∂T)P – V] to zero.* From van der Waal’s gas equation P + ___2 (V – b) V = RT, at T = Ti i.e., 2aV (V – b)2 – RTV 3b = 0, at T = Ti.

(

*

Refer to Section 3.14.

)

3.22 Heat and Thermodynamics or or

2aV (V – b)2 = RTiV 3b 2a (V – b)2 Ti = _________ RV 2b

(8)

Temperature (in °C)

Robeck and Oesterberg experimentally observed the Joule-Thomson effect at 400 different initial pressures. They found that there are two temperatures of inversion 300 225°C of a gas for a given initial pressure and 200 the values are different at different initial 100 pressures. A graph plotted between the Heating Cooling initial pressure and the temperature of 0 inversion is a parabola (Fig. 3.14). It is – 100 –120°C called the inversion curve. At a temperature and pressure corresponding to the left – 200 0 100 200 300 400 of this curve, cooling is produced while Pressure (in atmosphere) corresponding to the right of this curve, heating is produced. Figure 3.14 shows Fig. 3.14 Temperature of inversion curve for N2 the inversion curve for nitrogen. It is clear that if the initial pressure of nitrogen is 200 atmosphere, then cooling is produced in Joule-Thomson expansion when the initial temperature is in the range from – 120°C to 225°C. But if initial temperature is below – 120°C or above 225°C, heating is produced. Thus, the temperatures of inversion of nitrogen at 200 atmospheric pressure are – 120°C and 225°C.

3.14 PRINCIPLE OF JOULE THOMSON’S POROUS PLUG EXPERIMENT Joule’s experiment was not capable of measuring precisely the extremely small temperature changes in a free expansion of a gas. However, Joule and Thomson (later Joule and Kelvin) devised another better experiment during 1850-1860. They devised their experiment in such a way that the temperature change due to expansion of a gas would not be masked by the comparatively large heat capacity of the surroundings. The result of this experiment provided information about intermolecular forces. Further, temperature drop produced in the experiment could be used in the liquefaction of gases such as hydrogen and helium. The schematic representation of the experimental set up is shown in Fig. 3.15. A cylindrical tube, insulated to prevent any transfer of heat to the surroundings is fitted with two pistons and a porous plug which is capable of allowing gas to flow slowly through it. The tube A is initially filled with a certain amount of gas at temperature T1, volume V1 and pressure P1; tube B is empty. The gas is then

Real Gases, Liquefaction of Gases and Production and... 3.23

P1

V1

P2

A

B

V2 T2

T1

Porous Plug

Fig. 3.15

Schematic arrangement of porous plug experiment

allowed to flow slowly through the plug in such a way that its pressure in A is kept constant at P1 by the movement of the piston towards the plug. At the same time piston B is adjusted in such a way that the low pressure P2(< P1) is kept constant. Let the final volume in B, after all the gas has streamed through the porous plug, be V2 and its temperature T2. The significant datum obtained in this experiment is the change in temperature due to flow of the gas through the porous plug (by measuring temperatures T1 and T2). The whole system is insulated, so there is not heat transfer to the surroundings and therefore Q = 0. Therefore, the change in internal energy is equal to the work done by the system. The total work done is W = P1V1 – P2V2 Then

(1)

DU = U2 – U1 = P1V1 – P2V2

or

(U2 + P2V2) – (U1 + P1V1) = 0

or

H2 – H1 = 0

or

DH = 0

(2)

This shows that the Joule-Thomson experiment is carried out under constant enthalpy conditions. When the gas involved is perfect, H is a function of T only and therefore DH = 0 implies that DT = 0 or T2 = T1 (no temperature change for a perfect gas). For an imperfect gas it generally depends on whether T2 > T1 or T2 < T1> If T2 < T1, the gas will be cooled and if T2 > T1, the gas will be heated. The crucial temperature is called the Joule-Thomson inversion temperature. Above this temperature, there will be heating and below this temperature there will be cooling upon JouleThomson expansion. For most gases, the Joule-Thomson inversion temperature lies above the room temperature so, no precooling is necessary before they are allowed to expand

3.24 Heat and Thermodynamics through a porous plug. Helium and hydrogen are exemptions; these are heated when Joule-Thomson expansion occurs at room temperature (They require precooling for liquefaction). The Joule-Thomson coefficient m is defined as the change in temperature per unit change in pressure when the enthalpy is constant i.e.,

( )

∂T m = ___ ∂P

(3) H

From this equation it follows that if the gas cools in the process, the JouleThomson coefficient m is positive, because the pressure always decreases in the experiment. Conversely, a negative Joule-Thomson coefficient implies an increase in temperature. At a definite temperature for each gas, called the inversion point, the Joule-Thomson effect must therefore change sign. If we consider now H as a function of temperature and pressure, the total differential of H is

( )

∂H dH = ___ ∂T

( )

∂H But ___ ∂T

P

P

( )

∂H dT + ___ ∂P

dP

(4)

T

= CP, therefore

( )

∂H dH = CP dT + ___ ∂P

dP

(5)

T

Since in Joule-Thomson experiment, there is no change in enthalpy, i.e., dH = 0, so

( ) ( ) ( ) ( )

∂H 0 = CP (dT)H + ___ ∂P ∂T 0 = CP ___ ∂P

or

( ) ∂H ___ ∂P

or

∂H + ___ ∂P H

∂T = – CP ___ ∂P T

H

T

(dP)H

T

= – CP m

(6)

Since dH = 0, it follows from Eqn. (4) that

( ) ( )( ) ∂T ___ ∂P

∂H = – ___ ∂P H

T

∂H ___ ∂T

–1

(7)

P

It can be shown* that

( ) ∂H ___ ∂P

*

( )

∂V = V – T ___ ∂T T

(8) P

H = U + PV fi dH = dU + PdV + VdP = TdS + VdP

( )

∂H fi ___ ∂P

T

( )

∂S = V + T ___ ∂P

T

( )

∂V = V – T ___ ∂T

P

( ) ( )

∂S using Maxwell’s relation ___ ∂P

T

∂V = – ___ ∂T

P

Real Gases, Liquefaction of Gases and Production and... 3.25

and we already know that (∂H/∂T)P = CP, therefore Eqn. (7) gives

( ) ∂T ___ ∂P

T(∂V/∂T)P – V = ____________ CP H =m

(Joule-Thomson coefficient)

(9i)

If CP is assumed to be constant over a small temperature range, then Eqn. (9i) can be written as T(∂V/∂T)P – V DT = ____________ DP CP

(9ii)

This is the equation for the differential Joule-Thomson effect, and is in agreement with experiments. At the Joule-Thomson inversion temperature, m = 0. Below this temperature m is negative and above, it is positive. Most gases at room temperature have a positive Joule-Thomson coefficient. They do not require precooling for liquefaction by Joule-Thomson expansion, since (∂T/∂P)H > 0 fi gas temperature falls with pressure. On the other hand, (∂T/∂P)H < 0 fi when the gas goes from a higher to a lower pressure in a Joule-Thomson process, its temperature rises. Case (i) Perfect Gas The equation of state is PV = RT

( ) ( ) ( )

fi or or

∂V ___ ∂T

P

R = __ P

∂V T ___ ∂T

P

RT = ___ = V P

∂V T ___ ∂T

–V=0

P

\

m=0

(From Eqn. (9i).

i.e., Joule-Thomson coefficient for a perfect gas is zero. Also DT = 0 from Eqn. (9ii). Case (ii) Real Gas

( P + ___Va ) (V – b) = RT 2

(10)

Differentiating it with respect to T taking P constant, we have

( P + ___Va ) ( ___∂V∂T ) – ___2aV ( ___∂V∂T ) (V – b) = R 2

P

3

P

(11)

3.26 Heat and Thermodynamics or

( ) ∂V ___ ∂T

R = ________________ a 2a P P + ___2 – ___3 (V – b) V V R = _______________ 2a RT _____ – ___ (V – b) V – b V3 R(V – b) = ______________ 2a RT – ___3 (V – b)2 V RT (V – b) = ______________ 2a P RT – ___3 (V – b)2 V

( )

∂V T ___ ∂T

RT (V – b) = _________ ( b b or T < ___ then m (= (∂T/∂P)H) is positive. But since ∂P is RT Rb negative (pressure on emergent side of the porous plug is lower) so, ∂T will be negative and the gas will cool on passing through the porous plug. If 2a/RT < b or T > 2a/Rb, the gas will be heated up. At T = 2a/Rb, there will be no change in temperature. Therefore, T = 2a/Rb = Ti (inversion temperature). For most of the gases, the ordinary working temperatures are below Ti , hence they show a cooling effect. Ti for H2 and He are much below the ordinary temperatures, hence, they show a heating effect. If, however, these gases are

Real Gases, Liquefaction of Gases and Production and... 3.27

precooled below Ti before undergoing Joule-Thomson expansion, they will also show a cooling effect. It may be mentioned here that the Boyle temperature at a which the gas obeys Boyle’s law is TB = ___. Therefore, Ti = 2TB i.e., temperature Rb of inversion = 2 × Boyle’s temperature.

3.15

CHANGE IN TEMPERATURE IN AN ADIABATIC CHANGE

From Maxwell’s thermodynamic relations

( ) ( ) ( ) ( )

or

∂T ___ ∂V

∂P = – ___ ∂S S

∂T ___ ∂V

S

(see Chapter 5).

V

T ∂P = – __ ___ T ∂S

V

( )

∂P = – T ___ ∂Q

(1) V

( T∂S = ∂Q) since the pressure of the system increases at a constant volume hence the ∂P ∂T quantity ___ is positive and the quantity ___ is negative. It means that at ∂Q V ∂V S a constant entropy (i.e., in an adiabatic process), the temperature decreases on increasing volume, i.e., cooling is produced in adiabatic expansion.

( )

Since

( )

(∂Q)V = CV dT

\ From Eqn. (1)

( ) ∂T ___ ∂V

( )

T ∂P = – ___ ___ CV ∂T S

(2)

V

Now, if pressure coefficient of gas is b, then increase in pressure b = _________________________________ initial pressure × increase in temperature

( )

1 ∂P = __ ___ P ∂T or

( ) ∂P ___ ∂T

V

= Pb V

Hence, from Eqn. (2)

( )

TPb dT = – ____ dV CV

(3)

3.28 Heat and Thermodynamics It is thus clear that in adiabatic expansion, temperature decreases with increase in volume and the above Eqn. (3) represents the decrease in temperature in adiabatic expansion. Similarly, from Maxwell’s thermodynamic relation

( ) ( ) ( ) ( ) ( ) ∂T ___ ∂P

S

∂T ___ ∂P

or

∂V = ___ ∂S

P

T ∂V = __ ___ T ∂S

P

∂V = T ___ ∂Q S

P

(4)

Since, on imparting heat to the system at a constant pressure, its volume ∂V ∂T increases i.e., the quantity ___ is positive, hence the quantity ___ will also ∂Q P ∂P S

( )

( )

be positive which means that at a constant entropy, (i.e., in adiabatic process), temperature increase with the increase in pressure or temperature decreases with decrease in pressure i.e., heating is produces in adiabatic compression and cooling is produced in adiabatic expansion. Since

(∂Q)P = CP dT

\ From Eqn. (4)

( ) ∂T ___ ∂P

S

( )

T ∂V = ___ ___ CP ∂T

(5)

P

Now, if the coefficient of volume expansion of gas is a, then increase in volume a = _________________________________ initial volume × increase in temperature 1 ∂V = __ ___ V ∂T P

( )

or

( ) ∂V ___ ∂T

= Va P

Now, from Eqn. (5)

( )

TVa (6) dT = ____ dP CP It is clear that in adiabatic expansion, temperature decreases with the decrease in pressure and the above Eqn. (6) represents the decrease in temperature in adiabatic expansion. The Joule-Thomson (also known as Joule-Kelvin) effect discussed in the previous section, is an adiabatic throttling process in which cooling or heating

.

Real Gases, Liquefaction of Gases and Production and... 3.29

is produced depending on whether the initial temperature of the gas is below or above the temperature of inversion.

3.16 DISTINCTION BETWEEN JOULE’S EXPANSION, JOULE‐ THOMSON’S EXPANSION AND ADIABATIC EXPANSION In all the three Joule’s expansion, Joule-Thomson’s expansion and adiabatic expansion, the system is thermally isolated from its surroundings. Consequently, in all the three processes, there is no exchange of heat with the surroundings. Even then, these processes differ in the following respects: (i) In Joule’s expansion, the system is mechanically isolated from the surroundings, hence the total external work done on the system or by the system is zero (i.e., W = 0). In Joule-Thomson’s expansion, the system is not mechanically isolated from its surroundings and pressure of the gas in it is kept constant before and after the expansion as P1 and P2 respectively, hence in this expansion external work is done by the system and on the system (net external work done by the system is W = P2V2 – P1V1). In adiabatic expansion, the system is neither isolated from its surroundings, nor its pressure is kept constant, hence in this expression, only the external work is done against the atmospheric pressure by the system (W = Ú PdV). (ii) In Joule’s expansion, there is no change in internal energy (dU = 0). In Joule-Thomson expansion, internal energy decreases (dU = P1V1 – P2V2). In adiabatic expansion also, the internal energy decreases (dU = – Ú PdV). (iii) In Joule’s expansion, there is no change in temperature of an ideal gas, whereas there is always cooling in a real gas because the internal kinetic energy decreases because of internal work done by the gas. In JouleThomson’s expansion, although there is no change in temperature of an ideal gas, but a real gas shows cooling when P2V2 > P1V1 or heating when P2V2 < P1V1. In adiabatic expansion, the external work is obtained from its own internal energy due to which there is always a decrease in temperature of the system.

3.17

ADIABATIC DEMAGNETISATION

The production of low temperatures i.e., temperatures far below the freezing point of ice was a subject of great scientific interest. The quest for reaching temperatures close to the absolute zero (0°K) was the chief interest of the physicists. The adiabatic expansion of a real gas was found to give rise to cooling of a gas. When a gas passes through an orifice, or a set of orifices from a region of constant high pressure to a region of constant low pressure adiabatically, it is said to undergo throttling or Joule-Thomson expansion. A number of gases were

3.30 Heat and Thermodynamics cooled to their boiling points. Of course, the gases, initially, should be at their inversion temperatures. The gases which could be cooled to their boiling points were hydrogen (33 K), oxygen 82K), Nitrogen (77 K) and finally helium (4.2 K). Helium remains as liquid even below 1 K. The production of temperatures below 1K is not possible by Joule-Thomson process. Adiabatic demagnetization is found to be highly useful in the production of temperatures below 1 K.

3.17.1

Principle

A paramagnetic substance (or salt) can be treated as a thermodynamic system containing atoms which behave like small magnets, all lying in a disorderly manner. When it is magnetized, these groups of dipoles set themselves parallel to the lines of force, an external work is done on the system. If this already magnetized substance is suddenly demagnetized, the axes of atomic magnets (due to thermal agitation) tend to resume their natural disorderly state. Now work will be done by the substance drawing energy from the substance itself and consequently cooling will be produced. This is known as magnetocaloric effect. The entropy S of a paramagnetic system is a function of B/T and is given by m 2 mB dS ___ (1) = – Nk ___ B sec h2 ___ dB kT kT Since sec h2 (mB/kT) is always positive, the entropy S decreases with increasing B for a given T. Figure 3.16 shows entropy S of a paramagnetic system for two fields B1 and B2 (B1 < B2) as a function of temperature. We first increase the magnetic field

( )

( )

B1 a

B2 S

c

b

Tf

Ti

B1< B2

T

Fig. 3.16 Principle of adiabatic demagnetization

Real Gases, Liquefaction of Gases and Production and... 3.31

from B1 to B2, under isothermal condition, at a initial temperature Ti (from state a to state b). During this process, there is a heat transfer from the paramagnetic specimen to the heat bath. The specimen is then thermally isolated and the magnetic field is reduced very slowly (adiabatic demagnetization) under adiabatic conditions from B2 to B1. In this process, the entropy S of the specimen remains constant (This is illustrated in the change from state b to state c), leading to the lower temperature Tf (cooling by adiabatic demagnetization).

3.17.2

Theory

When a paramagnetic salt is placed in a magnetic field of strength H, the intensity of magnetization per gm-mole I of the system change by dI. The work done by the field on the substance = HdI. or, work done by the substance = – HdI. From first law, heat supplied dQ = dU + dW

(2)

dW = PdV – HdI \ dQ = dV + PdV – HdI

(3)

But here

Since change in volume of a solid for small changes in temperature is negligible i.e., dV = 0, thus dQ = dU – HdI

(4)

dQ = TdS

(5)

From second law where dS is change in entropy \ or

TdS = dU – HdI dU = TdS – HdI

(6)

Now, entropy S, intensity of magnetization I and internal energy U are functions of absolute temperature T and magnetic field strength H. Consequently, we can write

( ) ( ) ( )

( ) ( ) ( )

∂S dS = ___ ∂T

∂S dT + ___ ∂H H

∂I dI = ___ ∂T

H

∂I dT + ___ ∂H

H

∂U dU = ___ ∂T

H

∂U dT + ___ ∂H

H

dH

T

dH dH

(7)

3.32 Heat and Thermodynamics Substituting the values of dS, dI and dU in Eqn. (6), we get

( ) ∂U ___ ∂T

( )

∂U dT + ___ ∂H H

dH = T

H

{( ) ( ) } { ( ) ( ) } ∂S ___ ∂T

∂S dT + ___ ∂H H

∂I + H ___ ∂T

dH

T

∂I dT + ___ ∂H H

dH

(8)

H

Comparing coefficients of dT on both sides,

( ) ( ) ( ) ∂U ___ ∂T

H

∂S = T ___ ∂T

H

∂I + H ___ ∂T

(9) H

Comparing coefficients of dH on both sides of Eqn. (8)

( ) ( ) ( ) ∂U ___ ∂T

∂S = T ___ ∂H H

∂I + H ___ ∂H T

(10) T

As U is a function of state of the system, dU must be a perfect differential i.e.,

[ ( )] [ ( )] [ { ( ) ( ) }] [ { ( ) ( ) }] ∂ ___ ∂U ___ ∂H ∂T

or

∂ ∂S ___ T ___ ∂H ∂T

H

∂I + H ___ ∂T

H T

H

T

∂ ∂U = ___ ___ ∂T ∂H

T H

∂ ∂S = ___ T ___ ∂T ∂H

T

∂I + H ___ ∂H

T

H

This yields

( ) ( ) ∂I ___ ∂T

H

∂S = ___ ∂H

(11) T

Substituting this in Eqn. (10) gives

( ) ( ) ( ) ∂U ___ ∂H

∂I = T ___ ∂T T

∂I + H ___ ∂H H

(12) T

Using the above equation, we can calculate the internal energy (or the entropy) in an arbitrary field from the internal energy (or entropy) in zero field at the same temperature when the magnetization curves at different temperatures are known. From Eqns. (4) and (7), we have dQ =

=

{( ) [( )

( ) } {( ) ( ) } { ( ) ( )} ] {( ) ( ) }

∂U ___ ∂T

H

∂U dT + ___ ∂H

∂U ___ ∂T

∂I dT + T ___ ∂T H

dH – H

T

–H

∂I + H ___ ∂H H ∂I ___ ∂T

H

∂I ___ ∂T

H

∂I dT + ___ ∂H

dH

T

dH

T

∂I dT + ___ ∂H

dH

T

(using Eqn. 12)

Real Gases, Liquefaction of Gases and Production and... 3.33

\

dQ =

{( ) ( ) } ∂U ___ ∂T

H

∂I – H ___ ∂T

H

( )

∂I dT + T ___ ∂T

dH

(13)

H

fi Thermal capacity at constant field

( ) ( ) ( )

(14)

( )

(15)

∂Q CH = ___ ∂T

∂U = ___ ∂T T

∂I – H ___ ∂T H

∂I dQ = CH dT + T ___ ∂T

or

H

dH H

During adiabatic demagnetization, no heat enters or leaves the system (dQ = 0) so Eqn. (15) gives

( )

∂I CH dT + T ___ ∂T

dH = 0

(16)

H

Now, according to Curie’s law, molar susceptibility C c = __ T

(17)

C is Curie’s constant. M IV c = __ = ___ H H where V is the volume occupied by 1 gm mole of the substance and M is molar magnetization. Therefore, Hc C H (using Eqn. 17) I = ___ = __ . __ V T V KH (18) = ___ T C where K = __ is the Curie constant per unit volume. V Differentiating Eqn. (18) w.r.t. T at constant field H, But

( ) ∂I ___ ∂T

K dH = – __2 H H T Substituting it in Eqn. (16), we get

(19)

( )

K CH dT + T – __2 HdH = 0 T KH or TdT = ___ dH CH Integrating Tf

0

Ti

H

KH dH Ú TdT = Ú ___ C H

(20)

3.34 Heat and Thermodynamics or or or

[ ] T__2 2

[ ]

Tf

K H2 = ___ ___ Ti CH 2

0 H

K Tf2 – Ti2 = – ___ H2 CH

[

( )]

K H Tf = Ti 1 – ___ __ CH Ti

2 1/2

(21)

It may be noted that Tf < Ti i.e., the temperature of the salt falls during adiabatic demagnetization. The fall in temperature is greater for larger values of the magnetizing field H and low values of initial temperature Ti.

3.17.3

Experimental Method

Figure 3.17 gives the apparatus used in the experiment. The paramagnetic salt (gadolinium sulphate) is suspended in vessel surrounded by liquid helium cooled to ~1°K contained in a Dewar flask D1 which in turn is surrounded by a Dewar flask D2 containing liquid H2. The whole arrangement is placed between the poles of a strong electromagnet, providing a magnetic field of the order of 10 K-gauss. The following procedure is now adopted: Fig. 3.17 (i) The magnetic field is switched on so that the specimen is magnetized. (ii) The heat produced during magnetization is conducted away by the gaseous helium in A to the liquid He in D1. Thus, the specimen in A is left both cold and highly magnetized. It corresponds to isothermal magnetization. (iii) The He-gas from the cylinder A is now pumped off with a high vacuum pump. The specimen is now thermally isolated from D1 and D2. (iv) The magnetic field is now switched off. Thus, instantaneous adiabatic demagnetization of the salt takes place and its temperature falls to ~ 0.25 K. The temperature of the specimen is determined by fitting a coaxial solenoid coil round the tube A and measuring the self inductance and hence susceptibility of the salt with the help of an a.c. bridge at the beginning and the end of the experiment. If c1 is the susceptibility at temperature T1 of the He-bath and c2, the

Real Gases, Liquefaction of Gases and Production and... 3.35

susceptibility after adiabatic demagnetization at temperature T2,then according to Curie’s law c1 T2 __ = __ c2 T1 or

c1 T2 = __ T1 c2

whence T2 can be calculated, which is called the Curie temperature or magnetic temperature. From it, the Kelvin temperature can be determined.

3.18 LIQUEFACTION OF GASES AND APPROACH TO ABSOLUTE ZERO The science of the production and use of low temperatures is known as cryogenics. The methods employed for the liquefaction of gases by cooling can be classified as follows: 1. The rapid evaporation of volatile liquids. 2. The adiabatic expansion of cold compressed gases. 3. The Joule-Thomson effect. 4. The adiabatic demagnetization. We briefly describe these methods below.

3.18.1

The Rapid Evaporation of Volatile Liquids

When a liquid evaporates, latest heat of vaporization has to be supplied to it. If the liquid is thermally isolated, this heat comes from the liquid itself. As a result the liquid is cooled. We know that a gas can be liquefied only if it has been cooled below its critical temperature Tc, Table 3.1. By rapidly evaporating one liquefied gas it is possible to obtain a temperature low enough to liquefy another gas. This is called the cascade process. It was first used by Cailletet and by Pictet in 1877 to liquefy oxygen (Tc = – 119 C). Kamerlingh-Onnes, at Leyden, set up a cascade system using methyl chloride (Tc = 143°C > room temperature, liquefied by compression, cooled to – 90°C by evaporation at low pressure) and ethylene (Tc = 10°C, cooled to – 140°C, by evaporation at low pressure) and was able to produce oxygen in quantity. His apparatus is shown in Fig. 3.18.

3.18.2

The Adiabatic Expansion of Cold Compressed Gases

In all gases an approximately reversible adiabatic expansion (dS = 0) of a gas against a piston produces cooling. We can write

3.36 Heat and Thermodynamics

Fig. 3.18 Cascade process

( ) ( )( ) ∂T ___ ∂V

S

∂T = – ___ ∂S

V

∂S ___ ∂V

T

( )

T ∂p = – ___ ___ CV ∂T

V

where all the terms on the right are positive. Therefore, if V increases, T decreases. The physical reason for cooling is simply that in adiabatic expansion the work is done on the moving piston at the expense of the kinetic energy of the molecules, resulting in temperature decrease. The amount of cooling can be easily calculated. For the adiabatic process Tf = Ti (pi/pf)(1 – g)/g If pi = 51 atm, pf = 1 atm, Ti = 27°C and g = 1.4, we have Tf = 300 × (51)– 0.4 / 1.4 = 97.4°K Ti – Tf = 300 – 97.4 = 202.6°K = 202.6°C The apparatus of Claude is shown in Fig. 3.19. Purified air is compressed to 40 atm and passed through the tube A into the piston-cylinder arrangement B where it expands doing external work. The cooled and expanded air moves through the liquefier and finally passes along the outer tube of the heat exchanger A, cooling the incoming gas. This goes on until liquid air is formed.

Real Gases, Liquefaction of Gases and Production and... 3.37

Adiabatic expansion

A

B Piston

Heat exchanger

Liquefier

Fig. 3.19

3.18.3

Claude’s apparatus

The Joule-Thomson Effect

For cooling to result in the Joule-Thomson expansion we must have m > 0 or T < Ti. The amount of cooling produced is given by (using Eqns. 9(i) and (14) of 3.14). pi – pf ∂v ___ Ti – Tf = _____ –v cp T ∂T p pi – pf 2a ___ – b = _____ cp RTi

[( ) ] [ ]

For example, for oxygen using pi – pf = 150 atm @ 150 × 106 dynes per cm2 cp = 7 cal/mole. °K, Ti = 27°C = 300°K, a = 1.32 litre2 atm per mole2, and b = 3.12 × 10– 2 litre per mole, we have 2 × 1.32 × 1012 2a ___ – b = ________________ – 31.2 RTi 2 × 4.2 × 107 × 300 = 73.6 cm3 per mole 150 × 106 × 73.6 = 37.5°K Ti – Tf = ______________ 7 × 4.2 × 107 = 37.5°C. If Ti is taken to be 75°K instead of 300°K, we get Ti – Tf = 198°K = 198°C. (a) Hampson’s Regenerative Cooling Method and Linde’s Method Joule-Thomson cooling is small at room temperatures but increases as the initial temperature is lowered. This fact has been used in the regenerative cooling

3.38 Heat and Thermodynamics method given by Hampson of England. The principle is shown in Fig. 3.20. The compressed air after passing through a water bath, which removes the compression heat, passes to the heat exchanger. Air at a pressure of 150 atm is allowed to undergo Joule-Thomson expansion through a porous plug or throttle valve. The expanded air returns to the compressor through the outer jacket of the heat exchanger (counter-current heat exchanger), cooling the incoming air (regenerative cooling). As the temperature of the incoming air is lowered, the Joule-Thomson cooling increases rapidly. After a few cycles the air gets liquefied under Joule-Thomson effect alone. Water bath

20°C 150 atm

Heat exchange

Compressor

Porousplug or throttle valve

1 atm

– 180°C

Fig. 3.20 Regenerative cooling method of Hampson [After: B.K. Agarwal, Thermal Physics, Lokbharti Publications, Allahabad (India) (1988)]

Linde independently developed a similar method. The difference is that the Joule-Thomson expansion takes place from 150 atm to 40 atm and the gas needs to be compressed back from 40 atm to 150 atm, instead of 150 atm. If we assume isothermal compression the work involved is proportional to ln (pf /pi). Since cooling is proportional to (pi – pf), we have ln (pf /pi)LINDE ln (150/40) _____________ = __________ ln (pf /pi)HAMPSON ln (150/1) = 0.26

Real Gases, Liquefaction of Gases and Production and... 3.39

(Ti – Tf)LINDE 150 – 40 ____________ = ________ (Ti – Tf)HAMPSON 150 – 1 = 0.74 Thus, the Linde process gives about 3/4 the cooling for 1/4 the work, that is, it is more efficient. Figure 3.21 gives a flow diagram of the Linde process. The pump P1 compresses the air from 1 atm to 40 atm. The circulating pump P2 compresses the air from 40 atm to 150 atm. The compressed air gets cooled in ammonia bath E1 passes through the heat exchanger E2, and expands through the throttle value V1, becoming liquid at 40 atm and – 183°C. The unliquefied gas rises up through E2 back to P2, cooling the incoming air on the way. (b)

Claude’s Process and Kamerlingh-Onnes Method

A combination of methods 2 and 3 is more successful. The adiabatic reversible expansion is used to achieve a temperature below Ti and then the Joule-Thomson effect completes the liquefaction. Claude process shown in Fig. 3.22 is based on this principle. Air at 40 atm is cooled in exchanger E1 and divided into two streams: 80% goes into the adiabatic expansion cylinderpiston arrangement, gets cooled and moves up the heat exchanger E2 to cool the other 20%. This cooled 20% stream undergoes Joule-Thomson expansion at the

Fig. 3.21 Linde process

Fig. 3.22

Claude process

3.40 Heat and Thermodynamics throttle valve V, a part of it gets liquefied and the remainder rises up the exchanger E2 to cool the incoming air. It is useful to know the points of distinction between adiabatic and JouleThomson expansion, Table 3.2. Table 3.2 Points of distinction between adiabatic and Joule-Thomson expansions. Adiabatic expansion

Joule-Thomson expansion

1. Net external work done by the gas.

1. No net external work done by the gas.

2. Entropy remains constant.

2. Enthalpy remains constant.

3. Cooling produced for perfect gas.

3. Cooling not produced for perfect gas.

4. Cooling produced for real gas at all initial temperatures.

4. For real gas heating above Ti, no effect at Ti, and cooling below Ti.

5. Cooling decreases as initial temperature decreases.

5. Cooling increases as initial temperature decreases.

6. Reversible process.

6. Irreversible process.

Liquefaction of hydrogen is difficult because its critical temperature is 33°K, which is much below the liquid air temperature 78°K. Its inversion temperature is 194°K. So, it can be precooled to below 194°K with the help of liquid air and then liquefied by Joule-Thomson effect. Kamerlingh-Onnes designed a simple apparatus on this principle, Fig. 3.23. Purified hydrogen is passed through a bath

Fig. 3.23

Kamerlingh-Onnes method for hydrogen liquefaction by Joule-Thomson effect

Real Gases, Liquefaction of Gases and Production and... 3.41

of liquid air boiling under reduced pressure at – 208°C. The hydrogen at 150 atm and – 208°C undergoes Joule-Thomson expansion to a pressure of 1 atm and gets liquefied. The inversion temperature for helium is about – 240°C or 33°K. KamerlinghOnnes liquefied it in the same way by precooling to – 258°C by the aid of a bath of hydrogen boiling under reduced pressure.

3.18.4

The Adiabatic Demagnetization

This method was first perfected by Giauque in 1933 to produce extremely low temperatures, within a very small fraction of a degree of absolute zero. Figure 3.24 shows the variation of entropy with temperature for a paramagnetic salt for two values of external magnetic field, 0 and Hi. For H = 0, the sudden fall in entropy at the Curie temperature TC corresponds to the onset of spontaneous ordering. A rare earth paramagnetic salt is first magnetized by applying a field Hi at an initial temperature Ti (~1°K). By evaporating liquid He4 under reduced pressure, a temperature of about 1°K can be easily obtained. The heat evolved during magnetization is absorbed by the helium bath. The entropy falls and the salt goes from state a to state i (Fig. 3.24). Now the salt is thermally isolated and demagnetized. For slow demagnetization the process is reversible, entropy remains constant, and the temperature falls. The salt goes from state i to state f with temperature Tf. The lowest temperature attainable Fig. 3.24 The entropy vs. temperature graph for a paramagnetic salt at H = 0 and H = Hi by demagnetization is effectively the Curie temperature. The experimental arrangement for adiabatic demagnetization is given in Fig. 3.25. The salt A is placed in the chamber R which is immersed in the helium bath. For isothermal magnetization R is filled with helium gas to provide thermal contact with the liquid helium bath. After magnetization in a field of about 10,000 oersted, R is evacuated, thermally isolating the salt. The field is now reduced to zero. There is an accompanying drop in the temperature of the salt.

3.19

PRODUCTION OF VERY LOW TEMPERATURES

For the liquefaction of gases, it is required to produce temperatures below 0°C. Apart from this, production of low temperature is important because: (i) many

3.42 Heat and Thermodynamics

Fig. 3.25 Adiabatic demagnization

physical properties of a substance show a specific behaviour at low temperatures; (ii) the phenomenon of superconductivity is only possible at low temperatures. Following are the major methods of production of very low temperatures: 1. By adding salt in ice, 2. By vaporization of liquids at low pressure, 3. By adiabatic expansion of gases, 4. By Joule-Thomson effect, 5. By regenerative cooling, 6. By adiabatic demagnetization. When a soluble salt (such as common salt, NaCl) is mixed in ice, the ice takes some heat from it and starts melting. The latest heat required for melting of ice is obtained from the mixture. Moreover, the salt dissolves in water which is formed by melting of ice, for which the heat of solution required, is also taken from the mixture. As a result, the temperature of mixture falls. As the amount of iced water in mixture increases, the temperature of mixture decreases and a stage is reached when

1. By adding salt in ice

Real Gases, Liquefaction of Gases and Production and... 3.43

no more salt dissolves and the mixture attains a minimum temperature. The minimum temperature of mixture is different for different salts. For example, on adding ammonium chloride, it reaches – 55°C and on adding potassium hydroxide, it reaches – 65°C. When a liquid vaporizes in changing from liquid state to the gaseous state, it take the required latent heat of vaporization from the liquid and its surroundings due to which the temperature of liquid falls. More the latent heat of vaporization of liquid, more is the cooling produced. Moreover, if the pressure at the liquid surface is decreased, the boiling point of liquid will decrease due to which its vaporization will be rapid and the temperature of liquid will reduce more. Thus, the amount of cooling produced depends on the nature of liquid and its rate of vaporization. This principle is used in a refrigerator in which a low temperature is produced by the evaporation of liquid ammonia, liquid methyl chloride, sulphur-dioxide, freon etc. under the reduced pressure.

2. By vaporization of liquids at low pressure

We have read that when a gas expands adiabatically, the gas does work against the external pressure due to which the internal energy of the gas decreases and cooling is produced. By this method, we can reach to a low temperature at which the gas freezes to a solid. For example, if the tap of a cylinder containing CO2 gas at a high pressure is suddenly opened, we get the gas in form of white snow on a piece of cloth placed at the jet of gas.

3. By adiabatic expansion of gases

When a compressed gas at an initial temperature below its temperature of inversion is passed through a narrow nozzle from a constant high pressure to a constant low pressure, the temperature of gas decreases. The fall in temperature of a gas is directly proportional to the pressure difference across the nozzle. Lower the initial temperature of gas below its temperature of inversion, higher is the cooling of gas produced. This is called the Joule-Thomson expansion. 5. By regenerative cooling For liquefaction of gases, the gas which has suffered Joule-Thomson expansion is used to cool the coming gas before its suffers the Joule-Thomson expansion. As a result, this after suffering JouleThomson expansion gets further cooled. This process is repeated again and again, hence, the process is called the regenerative cooling and by this method, a very low temperature can be attained. 4. By Joule-Thomson expansion

When a paramagnetic substance is magnetized at a low temperature, the molecular magnets of the substance align along the magnetizing field in which work done by the magnetic field is stored in the substance in form of its internal energy and the temperature of substance increases. Now the substance is first cooled to the initial

6. By adiabatic demagnetization

3.44 Heat and Thermodynamics temperature and then it is adiabatically demagnetized due to which the temperature of substance decreases to a great extent. This method is used to attain a temperature very near to absolute zero.

3.20

MEASUREMENT OF VERY LOW TEMPERATURES

To measure the temperatures below 0°C, following thermometers are used: (i) Liquid thermometer, (ii) Gas thermometer, (iii) Resistance thermometer, (iv) Thermocouple thermometer, (v) Vapour pressure thermometer, (vi) Magnetic thermometer. The freezing point of mercury is – 39°C, hence, a mercury thermometer can be used to measure a temperature upto – 39°C. To measure a temperature below it, mercury can be replaced by toluene (upto – 95°C), alcohol (upto – 112°C), pentane (upto – 132°C) and isopentane (upto – 160°C).

(i) Liquid thermometer

A constant volume hydrogen gas thermometer is used to measure a temperature upto – 253°C and a constant volume helium gas thermometer is used upto – 268.7°C.

(ii) Gas thermometer

A platinum resistance thermometer can measure a temperature upto – 190°C to a great accuracy but the essential requirement is that the platinum used must be in the pure state. To measure temperature in the range of –190°C to – 258°C, lead resistance thermometer is used instead of platinum resistance thermometer. In the temperature range below it (upto – 272°C) an alloy resistance thermometer such as that of constantan or phosphor bronze is used. (iv) Thermocouple thermometer A copper-constantan or iron-constantan thermocouple can measure a temperature upto – 200°C easily, but to measure temperature below – 200°C (upto – 258°C), gold-silver thermocouple is used.

(iii) Resistance thermometer

The vapour pressure thermometers are much sensitive for the measurement of low temperatures. The choice of vapour used in a thermometer depends on the range of temperature to be measured by it. Oxygen is used in the temperature range – 150°C to – 210°C, neon in the temperature range – 246°C to – 249°C, hydrogen in temperature range – 253°C to – 262°C and helium is used in the temperature range – 268°C to – 272.29°C.

(v) Vapour pressure thermometer

Real Gases, Liquefaction of Gases and Production and... 3.45

This thermometer is used to measure a temperature below – 272°C. Its principle is based on Curie law, according to which the magnetic susceptibility of a paramagnetic substance is inversely proportional to its absolute temperature.

(vi) Magnetic thermometer

Example 3.1 The volume of 1 g mole of gas at 0°C is 0.55 litre. Calculate its pressure if the gas obeys (i) Ideal gas equation (ii) van der Waals equation. Given a = 0.37 Nm4 mole–2, b = 43 cm3 mole–1 and R = 8.31 J mole–1 K–1. Solution

Given

V = 0.55 litre = 0.55 × 10–3 m3 R = 8.31 J/(mole K). T = 0°C = 273 K a = 0.37 Nm4 mole–2, b = 43 cm3/mole = 43 × 10–6 m3/mole

(i) From the ideal gas equation PV = RT for 1 mole of gas RT P = ___ V 8.31 × 273 P = __________ 0.55 × 10–3

or \

= 4.125 × 106 N/m2 (ii) From van der Waals equation

( P + ___Va ) (V – b) = RT for 1 mole of gas 2

\

a RT P = _____ – ___2 V–b V

or

8.31 × 273 0.37 P = _____________________ – ____________ –3 –6 (0.55 × 10 – 43 × 10 ) (0.55 × 10–3)2 8.31 × 273 ____________ 0.37 = _________ – –6 (0.55 × 10–3)2 507 × 10 = 4.475 × 106 – 1.223 × 106 = 3.252 × 106 N/m2

The van der Waals constants of CO2 gas are a = 172 atm cm6 mole–2 and b = 0.002 cm3 mole–1. If R = 8.3 J mol–1 K–1, calculate its critical temperature, Boyle temperature and temperature of inversion. Example 3.2

3.46 Heat and Thermodynamics Solution

Given

a = 172 atm cm6 mole–2 = 172 × 10–7 Nm4 mole–2 b = 0.002 cm3 mole–1 = 2 × 10–9 cm3 mole–1 R = 8.3 Joule mol–1 K–1

Critical temperature 8a 8 × 172 × 10–7 TC = _____ = ________________ 27Rb 27 × 8.3 × 2 × 10–9 = 307 K (= 34°C) Boyle temperature 172 × 10–7 a TB = ___ = ____________ Rb 8.3 × 2 × 10–9 = 1036.14 K (= 763.14°C) Temperature of inversion 2a 2 × 172 × 10–7 Ti = ___ = ____________ Rb 8.3 × 2 × 10–9 = 2072.3K (= 1799.3°C) The van der Waals constants for a gas are a = 0.37 Nm4/mole2 and b = 43 cm3/mole. Find its critical constants. Example 3.3 Solution

Given

a = 0.37 Nm4/mole2, b = 43 × 10–6 m3/mole

\ Critical volume VC = 3b = 3 × 43 × 10–6 = 1.29 × 10–4 m3/mole Critical pressure 0.37 a PC = ____2 = ______________ 27b 27 × (43 × 10–6)2 = 7.4 × 106 N/m2 Critical temperature 8a 8 × 0.37 TC = _____ = ___________________ 27Rb 27 × 8.31 × (43 × 10–6) = 305.6 K or 32.6°C

Real Gases, Liquefaction of Gases and Production and... 3.47

QUESTIONS 1. Describe in brief the experiment performed by Andrews on CO2 and discuss the results of the experiment. 2. Write the van der Waals equation of state. Compare the theoretical isothermal curves obtained from this equation with the Andrews experimental curves. 3. Write the van der Waals equation of state and derive expressions for critical constants in terms of van der Waals constants a and b. 4. Obtain expressions for van der Waals constants a and b of a gas in terms of critical constants. 5. What do you mean by Boyle temperature? Use van der Waals equation of state to obtain an expression for Boyle-temperature of a gas. 6. Define critical temperature, Boyle temperature and temperature of inversion of a gas and deduce relationship among them. 7. What is meant by temperature of inversion of a real gas? Obtain a expression for it. 8. Show that Ti = 2TB and Ti = 6.75 TC where Ti, TB and TC are respectively the temperature of inversion, Boyle’s temperature and critical temperature of a gas. 9. Explain the principle of regenerative cooling of a gas. 10. What is cascade process of cooling? Explain the liquefaction of O2 by this method with the help of a neat diagram. 11. What is Joule’s law for an ideal gas? Describe Joule’s experiment to establish the existence of inter-molecular forces in a gas. What conclusions were drawn from this experiment? 12. What is Joule-Thomson effect? Describe porous plug experiment and discuss the results obtained from it. 13. What is Joule-Thomson effect? Obtain expression for Joule-Thomson cooling produced in van der Waals gas and explain why at ordinary temperatures hydrogen and helium gases show the heating effect while the other gases show cooling effect. 14. Show that in adiabatic throttling process, the enthalpy of a system remains constant. Obtain an expression for the change in temperature of a van der Waals gas due to Joule-Thomson effect. 15. Describe the Linde’s method of liquefaction of air. 16. Describe Kammerlingh Onne’s method for liquefaction of hydrogen and helium.

3.48 Heat and Thermodynamics 17. Describe with diagram the method of liquefaction of hydrogen and explain the principle on which this method is based. 18. Write short notes on (a) Boyle temperature (b) Critical point (c) Temperature of inversion (d) Liquefaction of gases

OBJECTIVE TYPE QUESTIONS 1. A graph plotted for PV against P for an ideal gas is (a) straight line parallel to P axis. (b) a straight line parallel to PV axis. (c) an inclined line passing through origin and making an angle of 45° with P-axis. (d) a hyperbola. 2. The isotherms of CO2 in Andrew’s experiment show that (a) CO2 can be liquefied by increasing the pressure at any temperature. (b) states are discontinuous. (c) CO2 is a perfect gas. (d) states are continuous. 3. The isothermals of a real gas are hyperbolic (a) below the critical temperature. (b) at the critical temperature. (c) at temperature much higher than the critical temperature. (d) at all the temperatures. 4. In Joule’s expansion of an ideal gas, the temperature of gas (a) increases (b) decreases (c) remains unchanged (d) first decreases and then increases 5. van der Waals forces are a result of interaction of (a) only dipole-dipole (b) only dispersion (c) only dipole-induced dipole (d) all of these

Real Gases, Liquefaction of Gases and Production and... 3.49

6. The critical temperature of a gas is 8a (b) _____ 27Rb 8a a (d) ____2 (c) ____ 27b 27b The highest temperature at which a gas can be liquefied is called the (a) Critical temperature (b) Boyle temperature (c) Melting point (d) Boiling point The critical temperature of water vapour is (a) 0°C (b) below 0°C (c) 100°C (d) above 100°C Hydrogen gas cannot be liquefied at the room temperature because (a) its density is very low (b) it is a diatomic gas (c) its thermal conductivity is quite high (d) its critical temperature is below the room temperature In Joule’s expansion of a van der Waal’s gas, the temperature of gas (a) sometimes decreases and sometimes increases (b) increases (c) remains unchanged (d) decreases In Joule-Thomson expansion, the temperature of inversion of a gas for a given pressure difference is 260°C. It can be cooled by keeping its initial temperature at (a) only at 260°C (b) 15 K (c) 0°C (d) 3 K The Boyle temperature TB and temperature of inversion Ti are related as (b) Ti = 0.5TB (a) Ti = TB (c) Ti = 2TB (d) Ti = 6.75TB At Boyle’s temperature, a van der Waal gas (a) behaves like an ideal gas (b) cannot be liquefied (c) causes heating effect on Joule-expansion (d) has maximum deviation from the behaviour of an ideal gas Liquefaction of helium is possible (a) at ordinary temperature (b) below – 268°C (c) at – 240°C (d) below – 83°C (a) 3b

7.

8.

9.

10.

11.

12.

13.

14.

3.50 Heat and Thermodynamics 15. Minimum temperature can be produced by (a) adiabatic demagnetization (b) vaporization of liquid at low pressure (c) adiabatic expansion (d) none of these Answers 1. (a); 8. (d); 15. (a).

2. (d); 9. (d);

3. (c); 10. (d);

4. (b); 11. (d);

5. (d); 12. (c);

6. (b); 13. (a);

7. (a); 14. (b);

Chapter 4.1

4

The First Law of Thermodynamics

INTRODUCTION

As pointed out in the previous chapter, the first law of thermodynamics is an extension of the principle of conservation of mechanical energy. Such an extension became reasonable after it was shown that expenditure of work could cause production of heat. The first quantitative experiments on inter conversion of work and heat were carried out by Benjamin Thompson. But in 1840, the law of conservation of energy was accepted in purely mechanical systems and the inter-conversion of heat and work was well established (the conversion factor was named the mechanical equivalent of heat, denoted by J). The value of J calculated by Mayor was 3.56 J cal–1. The accepted modern value of J is 4.484 J cal–1. Mayor was able to state the principle of conservation of energy, the first law of thermodynamics (in general terms) and to give one (rather rough) numerical example of its application. The exact evaluation of J and the proof that it is a constant independent of the method of measurement was accomplished by Joule. Although Mayer was the father of the philosophy of the first law of thermodynamics, the precise experiments carried out by Joule firmly established the law on an experimental and inductive foundation.

4.2

WORK DEPENDS ON THE PATH INDICATOR DIAGRAM

A large mass heat source which can supply any amount of heat to a system of finite mass without changing its own prescribed temperature is called a heat reservoir (Fig. 4.1). To raise the temperature of the system by a given amount, we can either add heat Q without performing any work, or perform work W on the system without Fig. 4.1 A system in thermal adding any heat such that Q = – W. (The Q is contact with a reservoir considered positive when heat enters the system and W is positive when work is done by the system). Let us calculate the quantities Q and W for a simple process. Consider a gas in a cylinder with a movable piston at one end and a heat reservoir at the other

4.2 Heat and Thermodynamics

Fig. 4.2 Energy added to the system (gas) as a result of thermal contact with the reservoir is called heat; energy added by the displacement of piston is called work (After B.K. Agarwal “Thermal Physics”, Lokbharti Publications, Allahabad, India (1988))

end (Fig. 4.2). The walls of the cylinder and the piston are made of insulating (adiabatic) material. Let gas be the system. The piston and the heat reservoir form the environment with which the system can interact. The temperature of the reservoir can be changed in steps as desired, or we can think of a series of reservoirs with temperatures q, q + dq, q + 2dq… etc. The heat Q can flow into the system or out of it due to thermal contact with the reservoir and work W can be performed on or by the system by compressing or expanding the gas with the piston. When any of the properties (for example, volume, pressure temperature, etc.) of a system change, the state of the system changes and the system is said to undergo a process. For example, when one of the weights on the piston (Fig. 4.2) is removed, the piston rises and a change in state occurs. The process is said to be quasi-static (i.e., almost static) when it takes place in such a way (usually very slowly) that at every instant the system differs only

The First Law of Thermodynamics 4.3

infinitesimally from an equilibrium state. A quasistatic process can be regarded as a succession of equilibrium states. If the weights on the piston in Fig. 4.2 are small and are taken off one by one, the process could be considered quasistatic. (Many actual processes closely approach a quasistatic process). Consider a quasistatic process in which the gas expands by moving the piston of area A against a pressure p through an infinitesimal distance dx (Fig. 4.2). The work performed by the system is dW = p (Adx) = pdV

(1)

where we have assumed the piston to be friction-less. By our convention, the work done by the system is positive and is consistent with the fact that the volume change dV = + Adx in an expansion process is positive. If a finite change in volume is carried out in such a way that the external pressure p is known at each successive state of expansion (or compressor), we can plot the process on a p-v diagram. Such a plot is called an indicator diagram (Fig. 4.3). The work done by the system in the expansion process i Æ f, is Vf

Wi, f = Ú pdV

(2)

Vi

and is equal to the area under the curve. To evaluate the integral mathematically, we have to know how pressure p varies with V (for a perfect gas pV/T a constant).

Fig. 4.3

(a) Indicator diagram for calculating work done by a gas during expansion; (b) The work done by a system depends upon the path

4.4 Heat and Thermodynamics It is possible to take the system from the initial state i, to the final state f in many different ways. For example, the pressure can be kept constant from i to a and then the volume kept constant from a to f (Fig. 4.3). Then the work done by the gas is given by the area under the line ai. Another possible path is ibf, in which case the work done by the gas is different and is given by the area under the line bf. To carry out processes iaf and ibf, the temperature of the heat reservoir must be changed during the process. The continuous path from i to f is another possible path in which the work done by the gas is still different from the previous cases. We conclude, therefore, that the work done by the system depends not only on the initial and final states but also on the intermediate states i.e., on the path taken by the process. Clearly, it is meaningless to speak of ‘the work in (or of) a system’ in the same way that we can speak of the pressure or temperature of a system. Work is not a property of the system. At this stage, we should consider point and path functions or to use another term, exact (or complete) and inexact (or incomplete) differentials. Thermodynamic properties like p, V or q (empirical temperature) are point functions. This means that for a given point on the diagram (e.g., Fig. 4.3), the state is fixed, and thus there is a definite or exact value of each property corresponding to this point. The differentials dV of a point function V are exact differentials and can be integrated along any path 2

Ú dV = V2 – V1 1

We can speak of volume V2 in state 2 and the volume V1 in state 1. The change in volume V2 – V1 is independent of the path and depends only on the initial and final states. Work on the other hand, is a path function. The differentials dW of path functions are inexact differentials. We generally use the symbol d to denote inexact differentials instead of d (used for exact differentials). The integration of dW depends on the path and so for work we write 2

Ú dW = W1, 2 1

as we cannot speak of work W2 in state 2 and W1 in state 1 and so cannot write W2 – W1.

4.3

HEAT DEPENDS ON THE PATH

A result similar to that for work is obtained if we compute the flow of heat Q into the system during the quasistatic process when the temperature of the reservoir

The First Law of Thermodynamics 4.5

(Fig. 4.2) is slowly changed in infinitesimal steps. The initial state i is characterized by a temperature qi and the final state by qf . We can heat the system (gas) at a constant pressure pi, for example, until we reach the temperature qf , and then change the pressure to the final value pf keeping the temperature constant. Or we can first decrease the pressure to pf and then heat it to the final temperature qf keeping the pressure constant. Or we can follow many other different possible paths. Each path involves a different quantity of heat flowing into the system. This is an experimental fact. We can, therefore, say that the heat lost or gained by a system depends not only on the initial and final states but also on the intermediate states, that is, on the path taken by the process. Thus, heat, like work, is a path function or an inexact differential dQ. Heat is not a property of the system. On integrating we write 2

Ú dQ = Q1, 2 1

In words, Q1, 2 is the heat transferred during the given process between state 1 and state 2.

4.4

FIRST LAW OF THERMODYNAMICS AND INTERNAL ENERGY

Thermodynamics can be defined as the science that deals with heat and work and those properties of matter that bear a relation to heat and work. Like all other sciences, the basis of thermodynamics is experimental observation. When a system in a given state goes through a series of different processes and finally returns to its initial state, the system is said to have undergone a cycle. At the conclusion of a cycle all properties have the same value that had at the beginning. Consider a system (gas) that undergoes a cycle, changing from state i to state f along path iaf (called process a), and returning from state f to state i along path fbi (called process b). Such a cycle is shown, for example, in Fig. 4.3b. An alternative possible return path is fci (called process c). The first law of thermodynamics states that for a closed system undergoing a cycle, the cyclic integral of the work is equal to the cyclic integral of the heat, dW = J dQ

(1)

The symbol designates the cyclic integral. If both W and Q are expressed in the same unit, the first law is dW = J dQ

(first law for a cycle)

(2)

4.6 Heat and Thermodynamics Every experiment that has been conducted thus far, for a wide variety of systems and for various amounts of work and heat, has verified the first law. The basis of this law is, therefore, experimental evidence. It has never been disproved. Often we are interested in a process rather than in a cycle. We can formulate the first law for a closed system that undergoes a change of state. This can be done by introducing a new property, called the internal energy U. From the first law of thermodynamics, Eqn. (2), we have for two separate processes a and b, Fig. 4.3(b), (a)f

Ú

dW +

(a)i

(b)i

(a)f

Ú

dW = Ú dQ +

Ú

(b)f

(a)i

(b)f

(b)i

dQ

If the system returns to the state i by process c, we can write (a)f

Ú

dW +

(a)i

(c)i

(a)f

(c)i

Ú

dW = Ú dQ +

Ú

(c)f

(a)i

(c)f

dQ

Subtracting the second of these equations from the first, we have (b)i Ú (b)f dW

–

(c)i

(b)i

(c)i

Ú

dW = Ú dQ –

Ú

(c)f

(b)f

(c)f

dQ

or, by rearranging, (b)i

Ú (b)f

(c)i

(dQ – dW) = Ú (dQ – dW) (c)f

Since b and c represent arbitrary processes between states i and f, we conclude that the quantity (dQ – dW) is the same for all processes between state i and state f. Therefore, (dQ – dW) depends only on the initial and final states and not on the path followed between the two states. It is a point function. Physicists are always happy to discover such quantities and like to give them a name. Since both Q and W are different forms of energy, we conclude that there exists an energy, called the internal energy U, such that the first law reads dU = dQ – dW

(first law for a process)

(3)

Since U is a point function, a property, its derivative is written dU. When Eqn. (3) is integrated from an initial state i to the final state f, we have Uf – Ui = Qi, f – Wi, f

(4)

where Qi, f is the heat transferred to the system during the process i Æ f, Wi, f is the work done by the system during the process i Æ f, and Ui and Uf are the initial and final values of the property U of the system. U is an extensive property. The first law Eqn. (4) is consistent with the fact that if we want we can add any amount of energy to a system in the form of work (turning of paddle in Joule’s arrangement) and take away any amount in the form of heat (by connecting the

The First Law of Thermodynamics 4.7

water to a cooler system). Therefore, neither heat Q nor work W can be conserved alone. However, the difference Q – W = DU is conserved. The essential content of the first law is that there exists a useful quantity called internal energy U. Internal energy U is the energy possessed by the body which can be drawn off either as heat or as work, depending upon the circumstances. The first law as stated in Eqn. (4) implies the familiar principle of energy conservation. The change in the energy of a system during a thermodynamic process is exactly balanced by the energy exchanged with the surroundings in the form of heat transferred to the system and in the form of work done by the system. This justifies the three sign conventions regarding heat Q, work W and internal energy U adopted in thermodynamics, shown in Fig. 4.4.

Fig. 4.4 Sign conventions with Q, W and U regarded as positive quantities or positive numbers

Note that in thermodynamics, the sign convention for work is opposite to that in mechanics. The convention here is designed to fit the behaviour of heat engines where there is an inflow of heat and an output of work. If a system receives both heat energy and work, Eqn. (3) would read dU = dQ – (– dW)

(7)

The law of energy conservation will include the changes in the kinetic energy (system may gain velocity) and the changes in the potential energy (system may be elevated). Then total energy E = internal energy + kinetic energy + potential energy = U + KE + PE dE = dU + d(KE) + d(PE)

(6) (7)

Then, for a closed system, we can write dE = dQ – dW

(8)

4.8 Heat and Thermodynamics 4.5

WORK DONE BY A THERMODYNAMIC SYSTEM IN EXPANSION AGAINST EXTERNAL PRESSURE

Consider a conducting cylinder fitted with a movable piston with a gas enclosed in it. Since material of cylinder is conducting, there can be an exchange of heat from its surroundings. Some weights can be placed on the piston. If weights are withdrawn, gas expands and the gas does some work in expansion against external pressure. On the other hands, if weights are increased, gas is compressed and work is done on the gas in compression by external pressure. Force

F = Pressure × Area =P×A

B

(1)

B DX

A

A Gas pressure P Cylinder

Fig. 4.5 Expansion of gas

If weights are withdrawn, gas expands. Let the piston moves from A to B, and work done by the gas in expansion DW = force × displacement = P × A × Dx = PDV

(2)

where DV = ADx (increase in volume of the gas) If the volume increases from V1 to V2, at a pressure P, total work done by the gas. V2

W = Ú PdV

(3)

V1

(where dV is a small increase in volume) If the volume decreases by DV, at a pressure P, work is done on the gas: DW = – PDV

(4)

If the volume decreases from V1 to V2, at a pressure P, total work done by the gas

The First Law of Thermodynamics 4.9

P

1(i)

2(f)

W

Vi

Fig. 4.6

V

Vf

PV diagram

V2

W = – Ú PdV

(5)

V1

(where dV is a small increase in volume) Equation (5) is analogous to the general result for the work done on a system by a variable force. If we plot F against x, the work done by F is simply area under the curve between xi and xf. Similarly, a graph can be obtained by plotting P on the vertical axis and V on the horizontal axis. This is called a PV diagram. The magnitude of work done on the gas is equal to the area under the curve representing the process on a PV diagram. The sign of W is determined according to whether Vf > Vi (in which case W is negative) or Vf < Vi (in which case W is positive), if W represents work done on the gas.

4.5.1

The Pressure-Force is Non-Conservative

This has been demonstrated in Fig. 4.7. We can find W1, the work done on the gas along path 1 by considering the work done along the two segments AB and BD. W1 = WAB + WBD Because the volume is constant along AB, it follows that WAB = 0. Along BD, pressure is constant at pf, it comes out of the integral and the result is

4.10 Heat and Thermodynamics

B

Pf

1

2

1

P Pi

D

2

A

C

F

E Vi

Vf

V

Fig. 4.7 A gas is taken from the pressure and volume at point A to the pressure and volume at point D along two different paths ABD and ACD. Along path 1 (ABD), the work is equal to the area of rectangle BDFE, whereas along path 2 (ACD) the work is equal to the area of rectangle ACFE

W1 = WAB + WBD = 0 – Ú pdV Vf

= – pf Ú dV Vi

= – pf (Vf – Vi)

(6)

For path 2 (ACD) W2 = WAC + WCD = – Ú pdv + 0 Vf

= – pi Ú dV Vi

= – pi (Vf – Vi)

(7)

Clearly, W1 π W2 and the work depends on the path. So, pressure force is nonconservative.

4.5.2

Work Done at Constant Volume

The work is zero for any process in which the volume remains constant (as in segments AB and CD in Fig. 4.7. W = 0 (constant V)

The First Law of Thermodynamics 4.11

We deduce directly from Eqn. (5) that W = 0 if dV = 0 (or V = constant). Note that it is not sufficient that the process start and end with the same volume; the volume must be constant throughout the process for the work to vanish. For example, consider process ACDB in Fig. 4.7. The volume starts and ends at Vi, but the work is certainly not zero. The work is zero only for vertical paths such as AB, representing a process at constant volume.

4.5.3

Work Done at Constant Pressure

Here, we can easily apply Eqn. (5), because the constant P comes out of the integral, W = – p Ú dv = – P(Vf – Vi)

(8)

Examples are the segments AC and BD in Fig. 4.7. Note that the work done on the gas is negative for both of these segments, because the volume increases in both processes.

4.5.4

Work Done at Constant Temperature

If the gas expands or contracts at constant temperature, the relationship between P and V, given by ideal gas law (PV = nRT)is PV = constant On a PV diagram, the plot of the equation PV = constant is exactly like a plot of the equation xy = constant: it is a hyperbola, as shown in Fig. 4.6. A process done at constant temperature is called an isothermal process and the corresponding hyperbolic curve on the PV diagram is called an isotherm. To find the work done on a gas during an isothermal process, we use Eqn. (5), but we must find a way of carrying out the integral when p varies. To do this, we write p = nRT/V, then Vf

W = – Ú pdv Vi Vf

nRT = – Ú ____ dV Vi V Vf

dV = – nRT Ú ___ Vi V (because T is constant)

(9)

4.12 Heat and Thermodynamics Carrying out the integral, we find,

()

Vf W = – nRT ln __ Vi

(T = constant)

(10)

Note that this is negative whenever Vf > Vi (ln x is positive for x > 1) and positive whenever Vf < Vi.

4.5.5

Work Done in Thermal Isolation

Suppose, a gas cylinder which was initially in contact with a thermal reservoir (below it) is now placed on a slab of insulating material (after removing the thermal reservoir). The gas will then be in complete thermal isolation from its surroundings, if we do work on it, its temperature will change, in contrast to its behavior when it was in contact with the thermal reservoir. A process carried out in thermal isolation is called an adiabatic process. If we allow the gas to change its volume with no other constraints, then, the path it will follow is represented on a pV diagram by the parabola-like curve pV g = constant as shown in Fig. 4.8. where the dimensionless parameter g is the ratio of the specific heats Cp and Cv (determined by experiments for any particular gas). Its values are typically in the range 1.1 to 1.8. Because g is greater than 1, the curve pV g = constant at any

Fig. 4.8 An adiabatic process is represented on a pV diagram by the hyperbola-like curve pV g = constant. The work done in changing the volume is equal to the area under the curve between Vi and Vf.

The First Law of Thermodynamics 4.13

point at which they intersect (see example 1.4). As Fig. 4.8 shows, this means that the work done by the gas in expanding adiabatically from Vi and Vf will be somewhat smaller in magnitude than the work done in expanding isothermally between these same two volumes. We can find the “constant” in Eqn. (10), if we know g and also the pressure and volume at any particular point on the curve. If we choose the initial point pi, Vi in Fig. 4.8, the “constant” has the value piVig and we can write Eqn. (10) as pV g = piVig

(11)

g

piVi p = ____ Vg

or

(12)

We can now find the adiabatic work Vf

W = – Ú pdV Vi Vf

Vf

piVig dV dV = – piVig Ú ___g = – Ú ____ g Vi V Vi V piVig = – ____ V1i – g – V1f – g g–1

(

)

By bringing a factor of V gi – 1 inside the bracket, we can write piVi W = ____ g–1

[( ) ] Vi __ Vf

g–1

–1

(13)

If the gas expands then Vi /Vf < 1 and since a number less than one raised to any positive power remains less than one, the work W is negative. By further using piVig = pfVfg

(14)

we can write the adiabatic work in the equivalent form as 1 W = ____ (pfVf – piVi)(adiabatic) g–1

4.6

(15)

PROOF FOR THE FIRST LAW OF THERMODYNAMICS

The first law of thermodynamics is essentially a balance sheet of energy transactions and an extension of the principle of conservation of energy. To illustrate this point, let us consider a closed system whose energy we want to alter. This can be achieved through heat operation or

W

Q DE (increase)

Fig. 4.9 Operations as being done on a closed system to alter its energy

4.14 Heat and Thermodynamics work operation or both. If heat Q is supplied to the system and a work W is work done on the system the consequence will be an increase in energy E of the system. The situation is as depicted pictorially in Fig. 4.9. The energy-balance can now be written as Q + W = DE (increase) = Efinal – Einitial

(1)

The positive signs with Q, W and DE indicate a gain be the system. If heat is supplied to the system and a work is done by the system, as a consequence, Q – W = DE

(2)

In the differential form, Eqn. (1) can be written as dQ + dW = dE

(3)

If internal energy U is the only energy the system has Eqn. (3) gives dQ + dW = dU

(4)

Equations (1) to (4) are quantitative statements of the first law of thermodynamics, and may be states as follows “The change DU in internal energy must be equal to the energy absorbed in the process from the surroundings in the form of heat minus the energy lost to the surroundings in the form of external work done by the system” i.e., DU = DQ – DW

4.7

THE THERMODYNAMIC PROPERTY ‘ENTHALPY’

In analyzing certain processes, we come across, certain combinations of thermodynamic properties which are as a result, also properties of the substance undergoing the change of state. For example, let us consider a control-mass* undergoing a quasi-equilibrium constant-pressure process as shown in Fig. 4.10. Assume that there are no changes in kinetic or potential energy and that the only work done during the process is that associated with

2

1

Q

Gas

Fig. 4.10 The constant-pressure quasi-equilibrium process * In case of a system with a control surface that is closed to mass flow, so that no mass can escape or enter the control volume, it is called a control mass (containing the same amount of matter at all times).

The First Law of Thermodynamics 4.15

the boundary movement. Taking the gas as our control mass and applying the first law, we have (in terms of Q) 1Q2

= U2 – U1 + 1W2 2

The work done can be calculated as 1W2 = Ú PdV since the pressure is constant 1 2 1W2 = P Ú dV = P(V2 – V1)

\

1

Therefore, 1Q2

= U2 – U1 + P2V2 – P1V1

(in general)

= (U2 + P2V2) – (U1 + P1V1) Thus, we find that in this very restricted case, the heat transfer during the process is given by change in the quantity U + PV between the initial and final states. Because all these quantities are thermodynamic properties i.e., functions only of the state of the system, their combination must also have these same characteristics. Therefore, we find it convenient to define a new extensive property, the enthalpy: H ∫ U + PV or per unit mass h = u + Pv As for internal energy, we could speak of specific enthalpy h, and total enthalpy H. However, we refer to both as enthalpy, and the context will make it clear which is being used.

4.8

HEAT CAPACITY

We define heat capacity as the amount of heat required to raise the temperature of 1 mole of a system by 1°C. It is denoted by C. In order to define heat capacity, two notations are used as follows: 1. Specific heat capacity: Heat absorbed by 1 g of material that undergoes a rise in temperature of 1°C. 2. Molar heat capacity: Heat absorbed by 1 g mol of material that undergoes a rise in temperature of 1°C. Thus, molar heat capacity is the specific heat capacity multiplied by the gram molecular weight.

4.16 Heat and Thermodynamics The heat effect (Q) in a process for a change in temperature from T1 to T2 can be calculated from the relation Q = CDT = C(T2 – T1)

(1)

assuming that heat capacity C (which generally varies with temperature), remains essentially constant between T1 and T2. Then heat absorbed T2

Q = Ú (C as a function of T) dT

(2)

T1

To obtain a quantity which is characteristic of a substance and which does not depend on how much of the substance one considers, we define the specific heat capacity c of a substance as the ratio of the true heat capacity C of a system composed of that substance to the mass m of the system i.e., C c = __ m

(3i)

then, the true molar specific heat capacity would be defined as C c = __ n

(3ii)

where n is the number of moles.

4.9

SPECIFIC HEAT CAPACITY

In the preceding section, we have talked about C as though it were a well defined quantity but, of course, the heat required to pass from one specific state of a system to another will vary with the path taken. Then, heat capacity will be well defined for a specific path. In fact, we find useful two species of heat capacities, corresponding to two simply specified paths, namely that at constant volume and that at constant pressure. (i) At constant volume, over a temperature range in which the heat capacity is constant, Eqn. (1) becomes Qv = Cv DT

(4)

but Qv = DU definition, hence DU = Cv DT

(5)

If over the temperature range concerned, Cv, is not constant, then from Eqn. (2) T2

DU = Qv = Ú (Cv as a function of T) dT T1

(6)

The First Law of Thermodynamics 4.17

For n mol, T2

DU = n Ú Cv dT T1

= nCv (T2 – T1) = nCv DT

(7)

if Cv is constant. (ii) At constant pressure, we have similar to Eqn. (4) Qp = Cp DT

(8)

but Qp is by definition, the heat content or enthalpy, then DH = Cp DT

(9)

If Cp varies over the temperature range concerned, then, we have T2

DH = Qp = Ú (Cp as a function of T) dT

(10i)

T1

For n mole T2

DH = n Ú Cp dT T1

= nCp (T2 – T1) = nCp DT

(10ii)

if Cp is constant

4.10

RELATIONSHIP BETWEEN Cp AND Cv

When an ideal gas is heated at constant volume, no work is done, and all the heat goes into increasing the temperature of the gas. If an ideal gas is heated at constant pressure, although the same amount of heat is still required to raise the temperature of the gas, an additional investment of heat is required for the PDV work done by the heated gas, as at constant pressure, it expands against the atmosphere. Then, if Cv is the heat required to increase the temperature of 1 mol of an ideal gas by 1°C at constant volume and Cp is the corresponding quantity at constant pressure, we have Cp = Cv + PDV

(11)

However, from the ideal gas law, we have PV = nRT. Then for 1 mol of gas, i.e., for n = 1, PV = RT. This can also be written (for small changes in volume and in temperature) as PDV = RDT. By definition of “heat capacity”, the quantity of material is 1 mol and the temperature increase is 1°C. Hence,

4.18 Heat and Thermodynamics PDV = RDT = R(1) = R

(12)

Combining Eqns. (11) and (12), we get Cp = Cv + R

(13)

This equation has been abundantly confirmed for real gases at reasonably low pressures. The ratio of the heat capacities is given by Cp Cv + R g = ___ = ______ (14) Cv Cv

4.10.1

Heat Capacities of Gases

We have seen in Chapter 2 (article 2.8) that on the basis of the Boltzmann statistics, if the energy associated with any degree of freedom is a quadratic function of the variable specifying the degree of freedom, the mean value of corresponding energy-equals kT/2. For example, the kinetic energy associated with the velocity component vx is 1/2 mvx2, and so, its mean value is kT/2. Similarly for rotation, (kinetic energy = 1/2 Iw2), the mean rotational kinetic energy is kT/2, and for a harmonic oscillator, where the potential energy is Kx2/2 (K = force constant), the mean potential energy is kT/2. In general, from the principle of equipartition of energy, the mean total energy of a molecule with f degrees of freedom, is therefore, f _ __ e = kT 2 and the total energy of N molecules is f f _ Ne = __ NkT = __ nRT 2 2 where n is the number of moles and R, the universal gas constant. Classical Theory In thermodynamics, the change in internal energy U of a system, between two equilibrium states, is defined by the equation Ua – Ub = Wad where Wad is the work in any adiabatic process between the states (only changes in internal energy are defined). Starting with a molecular model of a system (e.g., monatomic, diatomic etc.), we can identify the internal energy with the sum of the energies of the individual molecules. Just above, we have desired a theoretical expression for the total energy associated with the f degrees of freedom of each of the N molecules of a gas. We thus set this equal to the internal energy U: f f U = __ NkT = __ nRT 2 2

The First Law of Thermodynamics 4.19

The specific internal energy per mole is U __f u = __ n = 2 RT How can we test the validity of the atmosphere made in the foregoing derivation? The most direct way is from measurements of specific heat capacities. The molal specific heat capacity at constant volume is u Cv = ___ ∂T v Hence, if the hypothesis above is correct, we should have

( )

( )

f d f Cv = __ __ RT = __ R 2 dt 2 But for an ideal gas (from thermodynamic reasoning) Cp = Cv + R (from Eqn. 13, Section 4.10) f f+2 Cp = __ R + R = ____ R 2 2 f____ +2 Cp f+2 2 and g = ___ = _____ = ____ Cv f__ f 2 Thus, while the principles of thermodynamics give us only an expression for the difference between the specific heat capacities Cp and Cv, the molecular theory together with the equipartition principle, predicts the actual magnitudes of the specific heat capacities and their ratio g, in terms of the number of degrees of freedom f and the universal gas constant R. Note that according to the theory, Cp, Cv and g are all constants independent of the temperature. Consider first a gas whose molecules are monatomic and the energy is wholly kinetic or translational. Since, here, there are three translational degrees of freedom, f = 3, and we expect f 3 Cv = __ R = __ R = 1.5 R 2 2 \

f+2 5 Cp = ____ R = __ R = 2.5 R 2 2 and

Cp 5 g = ___ = __ = 1.67 Cv 3

This is in good agreement with the values of cv and g for the monatomic gases listed in Table 4.1.

4.20 Heat and Thermodynamics Table 4.1 Molal specific heat capacities of a number of gases, at temperatures near room temperature. The quantities measured experimentally are Cp and g. The former is determined by use of a continuous flow calorimeter and the latter is obtained from measurements of the velocity of sound in the gas. Cp /R

Cv /R

Cp – Cv _____ R

Gas

g

He

1.66

2.50

1.506

.991

Ne

1.64

2.50

1.52

.975

Ar

1.67

2.51

1.507

1.005

Kr

1.69

2.49

1.48

1.01

Xe

1.67

2.50

1.50

1.00

H2

1.40

3.47

2.47

1.00

O2

1.40

3.53

2.52

1.01

N2

1.40

3.50

2.51

1.00

CO

1.42

3.50

2.50

1.00

NO

1.43

3.59

2.52

1.07

Cl2

1.36

4.07

3.00

1.07

CO2

1.29

4.47

3.47

1.00

NH3

1.33

4.41

3.32

1.10

CH4

1.30

4.30

3.30

1.00

Air

1.40

3.50

2.50

1.00

y Furthermore, the specific heat capacities of these gases are found to be practically independent of temperature, in agreement with the theory. Consider next a diatomic molecule having the dumbbell structure shown in Fig. 4.11. Its moment of inertia about the x and y axes is very much greater than that about the z-axis, and if x the latter can be neglected, the molecule has two rotational degrees of freedom. Also since the atomic bond is not perfectly rigid, the atoms z can vibrate along the line joining them. This Fig. 4.11 A diatomic molecule consistintroduces two vibrational degrees of freedom ing of two atoms (considered to be point (since the vibrational energy is partly kinetic particles) with its axis along the z-axis of the coordinate system and partly potential and is specified by the separation of the atoms:

The First Law of Thermodynamics 4.21

(

1 1 1 U = __ mvx2 + __ mvy2 + __ mvz2 2 2 2

(

)

) (

)

1 1 1 1 + __ Ixwx2 + __ Iywy2 + __ kz2 + __ kz2 2 2 2 2 We might, therefore, expect seven degrees of freedom for a diatomic molecule (3 for translation, 2 for rotation and 2 for vibration). For f = 7, the theory predicts 7 Cv = __ R = 3.5 R, 2 9 g = __ = 1.29 7 These values are not in good agreement with those observed for the diatomic gases listed in table. However, letting f = 5, we find 5 Cv = __ R = 2.5 R, 2 7 g = __ = 1.40 5 These are almost exactly equal to the average values of Cv and g for the diatomic molecules in the second part of the table (except Cl2). Thus near room temperature, these molecules behave as if either the rotational or vibrational degrees of freedom (but not both), are shared equally with the translational degrees of freedom in the total molecular energy. As the number of atoms in a molecule increases, the number of degrees of freedom can be expected to increase; and the theory predicts a decreasing ratio of specific heat capacities, in general agreement with experiment: (Theoretical Values): 3 Cv = __ R = 12.5 J/mol K (monatomic gas) 2 5 Cv = __ R = 20.8 J/mol K (diatomic gas) 2 Cv = 3 R = 24.9 J/mol K (polyatomic gas) Cp = Cv + R or

5 Cp = __ R = 20.8 J/mol K (monatomic gas) 2 7 Cp = __ R = 29.1 J/mol K (diatomic gas) 2 6+2 Cp = _____ R = 4 R = 33.3 J/mol K (polyatomic gas) 2

4.22 Heat and Thermodynamics 5 g = __ = 1.67 (monatomic gas) 3 7 g = __ = 1.40 (diatomic gas) 5 4 g = __ = 1.33 (polyatomic gas) 3

4.10.2

Variation of Specific Heat of Gases with Temperature

The experimental values of cp, cv and g for monatomic gases agree with the values obtained from the law of equipartition of energy. These values are independent of temperature. But the values of cp, cv and g for diatomic and polyatomic gases are found to be different at different temperatures, whereas theoretically these values do not depend on the temperature. The variation of molar specific heat at constant volume with temperature for hydrogen gas is shown in Fig. 4.12. It is clear from the curve that for Hydrogen (diatomic gas), the value of cv is equal to its theoretical value in the temperature range 250 K to 750 K. Below 250 K, the experimental value of cv falls upto 3/2 R, while above 750 K, it rises upto 7/2 R. This variation of specific heat with temperature could be explained by quantum theory from which following conclusions are drawn:

3.5R Vibrational

3R 2.5R 2R

Rotational

Cv 1.5R R Translational

0

200 K

750 K

T

Fig. 4.12 Variation of specific heat of hydrogen gas with temperature

(i) At low temperatures, a molecule of diatomic gas can have only the translational motion (i.e., it behaves like a molecule of monatomic gas). Therefore, each molecule has only 3 degrees of freedom. The mean energy of a molecule is 3/2 kT and the mean energy per mole of gas is 3/2 RT. Hence, the molar specific heat at constant volume is CV = 3/2 R.

The First Law of Thermodynamics 4.23

(ii) At intermediate (or moderate) temperatures, a molecule can have translational motion with 3 degrees of freedom and rotational motion with 2 degrees of freedom. Therefore, the mean energy of molecule is 5/2 kT and mean energy per mole of gas is 5/2 RT. Hence the molar specific heat at constant volume is CV = 5/2 R. (iii) At high temperatures, a molecule can have translational motion with 3 degrees of freedom, rotational motion with 2 degrees of freedom and vibrational motion with 1 degree of freedom (for which mean energy is kT). Therefore, the mean energy of a molecule is 7/2 kT, mean energy per mole of gas is 7/2 RT and molar specific heat at constant volume is CV = 7/2 R. The value of CV changes from 3/2 R to 5/2 R or from 5/2 R to 7/2 R at different temperatures for different diatomic gases. Example 4.1 A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 0.105 MPa to a final state of 0.15 m3 and 0.105 MPa, the pressure remaining constant during the process. There is a transfer of 37.6 kJ of heat from the gas during the process. How much does the internal energy of the gas change? Solution

or

Q = DU + W Q1 – 2 = U2 – U1 + W1 – 2

(i)

V2

Here

W1 – 2 = Ú pdV = p(V2 – V1) V1

= 0.105 (0.15 – 0.30) MJ = – 15.75 kJ Q1 – 2 = – 37.6 kJ Substituting in Eqn. (i), we get – 37.6 kJ = U2 – U1 – 15.75 kJ or

U2 – U1 = – 21.85 kJ

i.e., internal energy of the gas decreases by 21.85 kJ in the process.

4.11 4.11.1

APPLICATIONS OF THE FIRST LAW OF THERMODYNAMICS TO AN IDEAL GAS Adiabatic Process

In an adiabatic process, the system is well is insulated so that no heat enters or leaves, in which case Q = 0. The first law becomes in this case DU or

4.24 Heat and Thermodynamics DEint = W

(adiabatic process)

(1)

Let us derive the relationship between p and V for an adiabatic process carried out on an ideal gas. We assume the process to be carried out slowly so that the pressure is always well defined. For an ideal gas, we can write dEint = nCv dT

(2)

(where n = number of moles and Cv is molar heat capacity at constant volume) Thus,

pdV = – dW = – dEint = – nCv dT

(3)

The equation of state of the gas can be written in differential form as d(pV) = d(nRT) pdV + Vdp = nRdT or

Vdp = nCv dT + nRdT = nCv dT (using Cp – Cv = R)

(4)

We now take the ratio between Eqns. (3) and (4) nCpdT Vdp ________ ____ = pdV – nCv dT Cp = – ___ ∫ – g Cv where g denotes the ratio of molar heat capacities. Rewriting, dp dV ___ ___ p =–g V Integrating between initial state i and final state f, we get pf

Vf

i

i

dp dV ___ Ú ___ p -= – g VÚ V p

(5)

(6)

(7)

()

pf V __f = – g ln ln __ pi Vi which can be written as or

()

piVig = pfVfg or

pVg = constant (adiabatic process)

(8) (9)

We can rewrite this result in terms of temperature using the ideal gas equation of state:

The First Law of Thermodynamics 4.25

(pV)V g – 1 = constant

(10)

or

TV g – 1 = constant

(11)

or

TiVig – 1 = TfVfg – 1

(12)

()

(13)

or

Vi Tf = Ti __ Vf

g–1

Suppose we compress a gas in a adiabatic process. Then Vi > Vf. This requires that Tf > Ti. The temperature of the gas rises as it is compressed. Conversely, the temperature falls when a gas expands (which is often used as a means to achieve low temperatures in the laboratory).

4.11.2

Isothermal Processes

In an isothermal process, the temperature remains constant. If the system is an ideal gas, then the internal energy must therefore also remain constant. With DEint = 0 the first law gives Q + W = 0 (isothermal process, ideal gas)

(14)

If an amount of (positive) work W is done on the gas, an equivalent amount of heat Q = – W is released by the gas to the environment. None of the work done on the gas remains with the gas as stored internal energy.

4.11.3

Constant-Volume Processed (Isochoric Process)

If the volume of a gas remains constant, it can do no work. Thus, W = 0 and the first law gives DEint = Q (constant volume process)

(15)

In this case all the heat that enters the gas (Q > 0) is stored as internal energy DEint > 0.

4.11.4

Cyclic Processes

In a cyclic process, we carry out a sequence of operations that eventually restores the system to its initial state as for example, a three-step process illustrated in Fig. 4.13. Because the process starts and finishes at point A, the change in internal energy for the cycle is zero. Thus, according to first law Q + W = 0 (cyclic process)

(16)

where Q and W represent the totals for the cycle. In the figure, the total work is positive because there is more positive area under the curve 3 than there is negative area under the line 2.

4.26 Heat and Thermodynamics

Fig. 4.13

A gas undergoing a cyclical process starting at A and consisting of (1) a constant volume process (no work), (2) a constant pressure process and (3) an isothermal process

In fact, for any cycle that is done in a counter clockwise direction, we must have W > 0 (and therefore, Q < 0), whereas for cycles performed in clockwise direction W < 0 (and therefore Q > 0).

4.12

IMPOSSIBILITY OF A PERPETUAL‐MOTION MACHINE

Let us imagine that a closed system undergoes a process by which it passes from state A to state B as shown in Fig. 4.14.

System A UA

Fig. 4.14

The Process +Q

–W

System B UB

A closed system undergoing a process to pass from state A to state B

If the only interaction with its surroundings is in the form of transfer of heat Q to the system, or performance of work W by the system, the change in internal energy U will be DU = UB – UA = Q – W or

dU = dQ – dW

This first law is often stated in terms of the universal human experience i.e., it is impossible to construct a “perpetual-motion machine” viz., a machine that produces useful work by a cyclic process with no change in its surroundings. To see how this is embodied in the first law, consider a cyclic process from state A to state B and back to A again. If perpetual motion were ever possible, it would be sometimes possible to obtain net energy increase DU > 0 by such a cycle. That this is impossible can be ascertained from the equation:

The First Law of Thermodynamics 4.27

DU = (UB – UA) + (UA – UB) = 0,

(i.e., DU >| 0)

or in general dU = 0 for any cyclic process.

4.13

SURFACES

The study of a surface, which can be considered to be a two-dimensional system is an interesting branch pertaining to thermodynamics. The surfaces of liquids and solids have properties distinctly different from the bulk properties of the underlying material. These are some important examples of surfaces: 1. The interface of a liquid in equilibrium with a vapor. 2. A soap bubble stretched across a wire framework, consisting of two surfaces trapping in between a small amount of liquid. 3. A thin film of oil on the surface of water. A surface such as a soap bubble on a wire-framework, is somewhat like a stretched membrane. The surface on one side of any imaginary line pulls this line perpendicular to this line with a force equal and opposite to that exerted by the surface on the other side of the line. The force acting perpendicular to such a line of unit length is called the surface tension t (Greek letter ‘tau’). An adequate thermodynamic description of a surface is given by specifying the following three coordinates: 1. The surface tension t in newtons per meter (N/m) 2. The area of the film A (m2). 3. The absolute temperature T (K). In dealing with a surface, which is a thin layer on a substrate, the underlying substrate must always be considered as part of the system. This may be done, however, without introducing the pressure and volume of the composite system, because (as a rule), the pressure remains constant and volume changes are negligible. The surface of a pure liquid in equilibrium with its vapour has a particularly simple equation of state. Experiment shows that the surface tension of such a surface does not depend on the area but is a function of the temperature only. For most pure liquids in equilibrium with their vapour phase, the surface tension has an equation of state of the form originated by van der Waals, but further developed by Guggenheim:

(

T t = t0 1 – __ Tc

)

n

(1)

where t0 is the surface tension at a standard temperature (usually 20°C), Tc is the critical temperature (the temperature above which no amount of pressure will

4.28 Heat and Thermodynamics condense vapour into liquid). It is clear from this equation that the surface tension decreases as T increases, becoming zero when T = Tc. The equation of state of a thin film of insoluble oil on water is particularly interesting. If tw denotes the surface tension of a clear water surface and t the surface tension of the water covered by the oil, then, within a restricted range of values of area A (t – tw) A = aT

(2)

where a is a constant. The difference (t – tw) is sometimes called the surface pressure or “two-dimensional” pressure. Equation (2) is the surface analog of the ideal-gas law. Oil films can be compressed and expanded and when deposited on glass, produce the interesting optical effect of Newton’s rings.

4.14

WORK IN CHANGING THE AREA OF A SURFACE FILM

Consider a soap film, which consists of two surfaces enclosing water, stretched across a wire framework. The right side of the wire framework is movable (Fig. 4.15). If the movable wire has a length L and the surface tension in one surface is t, then the external force F exerted on both surfaces is equal in magnitude to 2tL. For an infinitesimal displacement dx, the work is

F

L

dx

Fig. 4.15 A soap film stretched across a rigid wire framework which has a movable wire on the right. An external force F displaces the movable wire an infinitesimal distance dx

d W = 2tL dx

(3)

but for two films 2Ldx = dA where A is the total syrface area. Hence d W = t dA

(4)

When surface tension t is expressed in newtons for meter and area A in square meters, the work W is expressed in joules. Although Eqn. (4) has been derived for a film with two surfaces, it is also valid for a single surface, such as the boundary between the bulk of a liquid and the gaseous environment. For a finite change from an initial area Ai to a final area Af,the expression for work is Af

W = Ú t dA Ai

The First Law of Thermodynamics 4.29

A quasi-static process may be approximated by maintaining the external force at all times only slightly different from the force exerted by the film.

4.15

PARAMAGNETIC ROD

A paramagnetic, unlike a ferromagnet such as iron, has no permanent magnetism and no hysteresis. Upon being introduced into an external magnetic field, a paramagnet becomes slightly magnetized in the direction of the field. Within the paramagnetic material, microscopic currents are induced by the magnetic field. The macroscopic effect of the induced currents is called the magnetic dipole moment __›

__›

per unit volume or the magnetization M. The magnetization M is analogous to the _›

__›

polarization P. Similarly, analogous to the electric displacement D, we define the __›

magnetic field strength H. __›

_›

B __› H = ___ – M m0

(1)

where m0 is the permeability of vacuum a scalar quantity for isotropic paramagnets _›

_› __›

__›

and B is the magnetic induction. Further, the vectors B, M and H have the same direction. We now introduce the total magnetization M such that electrodynamic __› magnetization M equals M/V multiplied by the systems volume (This is done to make explicit the thermodynamic coordinate volume). Then, Eqn. (1) becomes B M H = ___ – ___ m0 V

(2)

Experiments on paramagnetic materials are not usually performed on samples of arbitrary shape, rather, on samples in the form of a long cylinder, ellipsoid or sphere. The restricted geometry is necessary in order to solve Ampere’s law in closed form. We here limit ourselves to a long thin cylinder (i.e., a rod) inside a current carrying solenoid. Most experiments on magnetic rode are performed at constant atmospheric pressure and involve only negligible volume changes. Consequently, we may ignore the pressure and the volume and describe a paramagnetic solid with the aid of only three thermodynamic coordinates, which are scalar quantities in simple thermodynamic systems: 1. The magnetic field H, measured in amperes/m (A/m). 2. The total magnetization M measured in A m2. 3. The temperature T (K) The states of thermodynamic equilibrium of a paramagnetic solid can be represented by an equation of state among these coordinates. Experiment shows that the magnetization of many paramagnetic solids is a function of the ratio

4.30 Heat and Thermodynamics of the magnetic intensity to the temperature. For small values of this ratio, the function reduces to a very simple form, namely Cc H (3) M = ____ T which is known as Curie’s law. Cc is called the Curie constant. This SI unit for the Curie constant in Eqn. (3) is expressed as A.m2 K Unit of Cc = _______ = m3.K A/m Since the Curie constant depends upon the amount of material, its unit may be taken to be any one of the four listed in Table 4.2. Table 4.2 Units of the Curie constant Total m3.K

Per mole

Per kilogram

3

Per cubic meter

3

m .K ____ mol

m .K ____ kg

K

Materials exhibiting either electronic paramagnetism or nuclear paramagnetism are particularly important in thermodynamics. They are used to obtain extremely low temperatures.

4.16

MAGNETIC WORK

Consider a sample of paramagnetic material in the form of a ring of crosssectional area A and of mean circumference L. Suppose that an insulated wire is wound around the sample, forming a toroidal winding of N closely spaced turns (Fig. 4.16). A constant current may be maintained in the winding by a battery and changed by means of the variable resistor.

Fig. 4.16

The First Law of Thermodynamics 4.31

The direct current in the windings sets up a magnetic induction B. If the dimensions are as shown in Fig. 4.16, the B will be nearly uniform over the crosssection of the toroid. Suppose that the current is changed and that in time dt, the magnetic induction changes by an amount dB. Then by Faraday’s principle of electromagnetic induction, there is induced in the winding a back emf e, where dB e = – NA ___ (1) dt During the time interval dt, a quantity of charge dz moves in the circuit and the work done by the battery to maintain the current is given by dW = – e dz

(2)

dB = NA ___ dz dt dz = NA __ dB dt = NA I dB

(3)

where the current I equals dz/dt. The magnetic field H due to a current I inside a toroidal winding is given by NI NAI NAI H = ___ = ____ = ____ L V AL

(4)

where V is the volume of the paramagnetic material. Therefore and

NAI = V H

(5)

dW = V H dB

(6)

If M is the total magnetization of the paramagnetic (assumed to be isotropic), then we have the relation M B = m0 H + m0 ___ V

(7)

Therefore, dW = V H d(m0 H) + m0 H dM

(8)

If no material were present within the solenoid winding then M would be zero, B would equal m0 H and dW = V H d(m0 H)

(vacuum only)

(9)

This is the work necessary to increase the magnetic field in a volume V of empty space by an amount d(m0H). The second term m0 H dM, is the work done in increasing the total magnetization of the material by an amount dM. We are generally concerned with changes of temperature, energy etc. of the paramagnet only, (brought about by work done on or by the material), consequently

4.32 Heat and Thermodynamics dW = m0 H dM

(10)

If m0 H is measured in newtons per ampere-meter and the total magnetization M in ampere-m2 then the work W will be expressed in joules. If the total magnetization is caused to change a finite amount from Mi to Mf , the work will be Mf

W = m0 Ú H dM

(11)

Mi

In any actual case, a change of total magnetization is accomplished very nearly quasi-statically and therefore, an equation of state may be used in the integration of the expression denoting the work.

4.17

PARTIAL DERIVATIVES

In the equation of state there is a relationship between at least three variables (for example, p, v, Q). We can choose any pair them to be independent variables; the remaining one is then a dependent variable. We are often required to calculate the rate of change of the dependent variable with respect to one of the independent variables, keeping the other constant. This rate is called a partial derivative. For example, in the case of the equation of state Q = Q (p, v)

(1)

we shall be interested in the partial derivatives

( ) ∂Q ___ ∂p

and v

( ) ∂Q ___ ∂v

p

(2)

The variable which is kept constant in the process of differentiation is placed as a subscript. This cumbersome notation is essential in thermodynamics, though it is found superfluous in mathematics. The advantage is that the subscript v in (∂Q/∂p)v denotes the variable v held constant and also indicates that Q is to be expressed explicitly as a function of independent variables p and v, Q = Q(p, v). Both of these information are important in thermodynamics. Partial derivatives like (∂v/∂Q)p, (∂v/∂p)Q, and (∂Q/∂p)v can be interpreted geometrically as the slopes of lines on a p-v-Q surface. For example, if AB is the curve of intersection of a portion of an arbitrary p-v-Q surface (Fig. 4.17) with the plane v = constant, the partial derivative (∂Q/∂p)v represents the gradient of path AB. Its numerical value is that of the tangent of the angle between AB and the p-v plane. The distance BB¢ is the difference in altitude which is covered by moving along AB by dp. The principal value of this difference in altitude is given by the product

( )

∂Q ___ dp ∂p v

(3)

The First Law of Thermodynamics 4.33

Fig. 4.17

Geometrical meaning of (∂Q∂/∂p)v

The principal value of the total difference dQ in altitude, which is covered when changes in dp and dv are considered simultaneously, or in succession in any order, is given by the total differential

( ) ( )

∂Q ∂Q dQ = ___ dp + ___ ∂p v ∂v

p

dv

(4)

of the function Q (p, v) By interchanging the roles played by the variables p, v, Q, we get two more total differentials

( ) ( )

( )

∂p dp = ___ ∂v

∂p dv + ___ dQ ∂Q v Q

∂v dv = ___ ∂p

∂v dp + ___ ∂Q Q

( )

p

dQ

(5) (6)

In general, suppose we have any three variables x, y, z, satisfying the relation f (x, y, z) = 0

(7)

We can solve it for x and y, to get x = x(y, z), y = y(x, z)

(8)

Differentiating by parts, we have

( ) ( ) ( ) ( )

∂x ∂x dx = ___ dy + ___ dz ∂y z ∂z y

(9)

∂y ∂y dy = ___ dx + ___ dz ∂x z ∂z x

(10)

4.34 Heat and Thermodynamics Substituting Eqn. (10) in (9), we get

[ ( ) ( ) ] [( ) ( ) ( ) ] ∂x 1 – ___ ∂y

z

∂y ___ ∂x

dx =

z

∂x ___ ∂y

z

∂y ∂x ___ + ___ ∂z x ∂z

dz

(11)

y

This result must always be true whichever pair of variables we choose to be independent. In particular, we can choose x and z as independent. Then, we can give any value to dx and any other value to dz. Suppose we put dx π 0 and dz = 0. Then (11) gives

( )( )

∂x 1 – ___ ∂y

z

∂y ___ =0 ∂x z

(12)

or

( )

∂x 1 ___ = _______ ∂y z (∂y/∂x)z

(reciprocal theorem)

(13)

This is the reciprocal theorem. It allows us to replace any partial derivative by the reciprocal of the inverted derivative with the same variable held constant. On the other hand, if we put dx = 0 and dz π 0, we obtain

( )( ) ( ) ∂x ___ ∂y

z

∂y ∂x ___ + ___ ∂z x ∂z

=0

(14)

y

or

( ) ∂y ___ ∂z

(∂x/∂z)y = – _______ (∂x/∂y)z x

(15)

We can combine Eqns. (13) and (15) to put the latter in the form

( )( ) ( ) ∂x ___ ∂y

z

∂y ∂z ___ + ___ ∂z x ∂x

y

= – 1 (reciprocity theorem)

(16)

This is the reciprocity theorem. It is used to split up a derivative into a product of more convenient derivatives. Point functions like p, v and Q are functions of the state of the system alone. Such functions are called state function or state variables. State variables, like p, v, Q have specific values determined only by the equilibrium state of the system and not by the previous history of the system. They return to the same values whenever the system returns to the same equilibrium state. Work w and heat q, being path functions, cannot be expressed as functions of the state of a system. It follows that derivatives such as (dw/∂Q)p or (dq/∂p)Q have no meaning, since w or q cannot be expressed as functions of p and Q. However, unlike w and q, the internal energy u is a function of the state of the system and therefore its partial derivatives will have meaning.

The First Law of Thermodynamics 4.35

For the variables p, v, Q the chain relation (16) becomes

( )( )( ) ∂p ___ ∂v

Q

∂v ___ ∂Q

p

∂Q ___ ∂p

=–1

(17)

v

The partial derivatives constitute new physical quantities. They are all properties and definite functions of the independent variables. Their values are fixed when the values of the independent variables are fixed. Some of them are important enough to be given a name. When a system is heated at constant pressure (dp = 0), we get from Eqn. (6)

( )

∂v dv = ___ ∂Q

p

dQ

(18)

If we consider the relative change in volume, we can define the coefficient of volume expansion as

( )

∂v 1 ___ b = __ v ∂Q

(19)

p

When heating is performed at constant volume (dv = 0), we can use Eqn. (5) to define the coefficient of pressure

( )

∂p 1 ___ p = __ p ∂Q

(20) v

Finally, if the pressure is increased at constant temperature (dQ = 0), we can define from Eqn. (6) the isothermal coefficient of compressibility

( )

∂v 1 ___ k = – __ v ∂p

(21) Q

The negative sign is added because for all known substances the pressure and volume changes are of opposite signs. When these coefficients ate substituted in (17), we get the interesting result b=pkp

4.18

(22)

CONCEPT OF A PERFECT GAS

In 1843 Joule performed the following experiment. Two copper vessels, A and B connected by a tube with stopcock C, were immersed in a thermally isolated bath of water, Fig. 4.18. Initially A contained air and B was evacuated. When thermal equilibrium was attained the valve C was opened, allowing the pressure in A and B to equalize. No change in temperature of the bath was observed during or after this process of expansion of gas into a vacuum or free expansion of gas.

4.36 Heat and Thermodynamics

C A

B

Water

Fig. 4.18 Apparatus for Joule’s experiment

As there was no change in the temperature of the bath, Joule concluded that there had been no heat transferred to the air (system), Q = 0. In a free expansion no work is done by the gas on its surroundings, W = 0. It is true that during the expansion the air remaining in A is doing work on the air already in B. But this work is done by one part of the system (air) on another part and is not done by the system as a whole on its surroundings. The work done in free expansion cannot be represented by an arrow piercing the system boundary. Since Q = 0 and W = 0, Joule concluded from the first law, DU = Q – W, that there was no change in the internal energy of the air DU = 0 or

U = constant

As the pressure and volume changed during this process, one concludes that the internal energy of a perfect gas is independent of p and V and is a function of temperature Q only, U = U(Q). This is called Joule’s law. Clearly the isotherms of a perfect gas are also curves of constant internal energy. When very careful experiments are made with real gases a very small change in temperature is detected. Therefore, Joule’s law like Boyle’s law, holds strictly only for a perfect gas. In fact, the perfect gas is defined as one which obeys both Boyle’s law and Joule’s law. In general, U is a function of any two of the variables p, V and Q. Considering U as a function of V and Q, U = U(V, Q),

( )

∂U dU = ___ ∂V

Q

( )

∂U dV + ___ dQ ∂Q v

(1)

or, for specific internal energy, u = u (v, Q),

( )

∂u du = ___ ∂v

Q

( )

∂u dv + ___ dQ ∂Q v

(2)

The First Law of Thermodynamics 4.37

In a free expansion of perfect gas (du = 0), there is no temperature change (dQ = 0). Therefore

( ) ∂u ___ ∂v

=0

(perfect gas)

(3)

Q

which means u does not depend upon v. Considering u as a function of p and Q, u = (p, Q),

( )

∂u du = ___ ∂p

( )

∂u dp + ___ ∂Q Q

p

dQ

(4)

For free expansion du = 0 and dQ = 0. Therefore,

( ) ∂u ___ ∂p

=0

(perfect gas)

(5)

Q

which means that u does not depend upon p. It follows that for a perfect gas u is a function of Q only u = u(Q)

(perfect gas)

(6)

We can now define a perfect gas (or an ideal gas) in thermodynamic terms as follows pv = RQ

( ) ( ) ∂u ___ ∂v

Q

∂u = ___ ∂p

(7) (definition of perfect gas)

=0

(8)

Q

A real gas at a pressure below about 2 atm can be regarded as a perfect gas with only a little error in calculations. In principle, the limit of zero pressure must be taken. We will come across with partial derivatives again, in the next chapter.

QUESTIONS 1. Differentiate between the path function and point function. 2. State first law of thermodynamic and explain it. 3. State the first law of thermodynamics explaining the meanings of the symbol used, hence explain (i) isothermal expansion, (ii) adiabatic expansion and (iii) cyclic process. 4. Explain internal energy of a system on the basis of first law of thermodynamics.

4.38 Heat and Thermodynamics 5. What is meant by P-V diagram or indicator diagram? How is the work done on the system or the work done by the system during a change of state obtained from this graph? Show that the work done on the system or the work done by the system is the path function. 6. Demonstrate that the pressure-force is non-conservative. 7. Deduce expression for the work done when a perfect gas undergoes expansion (i) adiabatically (ii) isothermally. 8. Explain “Heat depends on the path”. 9. Define the term Enthalpy. State its units and explain its physical significance. 10. Prove that the ratio of specific heat at constant pressure to that at constant volume for an ideal gas with f degrees of freedom per molecule is g = Cp /Cv = 1 + 2/f. Hence show that the value of g is 1.66 for monatomic gas, 1.4 for diatomic gas and 1.33 for polyatomic gas. How does the experimental values agree with the theoretical values? 11. What do you mean by perpetual motion machine? Explain its physical meaning. 12. What are surfaces. Give some examples. Derive an expression for work required in changing the surface area of a soap film from an initial value Ai to a final value Af. 13. Explain the concept of magnetic work by considering a sample of paramagnetic material in the form of a ring.

OBJECTIVE TYPE QUESTIONS 1. In the perfect gas equation, PV = RT, V is the volume of (a) 1 g gas (b) 1 mole of gas (c) whole amount of gas (d) 1 litre gas 2. The potential energy of molecules of a perfect gas is (a) equal to its kinetic energy (b) zero (c) equal to its internal energy (b) infinity 3. Internal energy of a system changes in (a) isothermal change (b) adiabatic change (c) free expansion (d) cyclic process 4. The first law of thermodynamics is the law of conservation of (a) momentum (b) energy (c) both (a) and (b) (d) temperature

The First Law of Thermodynamics 4.39

5. In a cyclic process, the change in internal energy is (a) infinite (b) zero (c) equal to area of cycle (d) none of these Answers 1. (b);

2. (b);

3. (b);

4. (b);

5. (b).

Chapter 5.1

5

The Second Law and Third Law of Thermodynamics

REVERSIBLE AND IRREVERSIBLE PROCESSES

Consider a gas enclosed in the frictionless piston-cylinder arrangement, Fig. 5.1a. We can go from the initial equilibrium state i to the final equilibrium state f in the following three ways.

1.

Irreversible Process

The piston is suddenly drawn up. Gas would rush in to fill the space, pressure differences would be created throughout the gas volume, shock waves or sound waves would be set up. The state of the gas is not of equilibrium and so variables p and Q of the gas are not well defined. In fact, many variables would be required corresponding to various values of p and Q at different points in the gas. If we wait long enough, the thermal equilibrium with the reservoir is re-established. This gives the final equilibrium state f, Fig. 5.1b.

Fig. 5.1 (a) Piston-cylinder arrangement; (b) irreversible process; (c) reversible process; and (d) quasistatic irreversible process (discrete sequence of states of equilibrium)

The system passes from equilibrium state i (pi, vi) to equilibrium state f (pf , vf) through a series of nonequilibrium states. These intermediate states, being not in equilibrium, cannot be plotted on a p-V diagram. This is an example of a typical irreversible process. It is not reversible because, for example, we cannot extract the sound waves or shock waves by moving the piston down again.

5.2 Heat and Thermodynamics 2.

Reversible Process

The frictionless piston is drawn very slowly, in the limit infinite slowly (quasistatically). Then at every stage the variables p, V, Q have well defined values because during the entire process the system remained in equilibrium. Such a process consists of a continuous sequence of states of equilibrium and is said to be reversible. Each successive intermediate equilibrium state can be represented by a point in the p-V diagram. The line joining such points is called a reversible path, Fig. 5.1(c). The process is called reversible because it can be made to retrace its path without producing any change in the surroundings. Reversible processes are ideal abstractions. They are never realized in actuality. They must be carried out (i) quasistatically, and (ii) without dissipation of energy (for example, without friction or turbulence). These conditions are never fully satisfied.

3.

Quasistatic Irreversible Process

If the process remains quasistatic but the condition of ‘no dissipation of energy’ is relaxed (for example, by slow motion of piston with friction) we no longer have a reversible process but a quasistatic irreversible process. It follows that all reversible processes are quasistatic, but that the converse is not true. In a quasistatic irreversible process each successive intermediate state is a near equilibrium state and can be represented by a discrete sequence of points, Fig. 5.1(d).

5.2

SIGNIFICANCE OF REVERSIBLE PROCESSES

All natural process are irreversible because the two conditions of reversible process can never be fully satisfied. If a process is irreversible it does not mean that the system cannot be restored to its initial state. It only means that the initial state cannot be restored ‘without producing any change in the surroundings’. Although all natural processes are irreversible, the reversible processes constitute convenient idealizations as they can be studied by analytical methods. For example, consider frictionless quasistatic (i.e., reversible) expansion of a gas. The work done by the gas is pdV. For a frictionless reversible expansion pdV is uniquely defined for the increment dV. If work dW is done by the system and heat dQ is added to the system, the first law gives dU = (unique)

dQ (not unique)

–

dW (not unique)

(first law)

(1)

The Second Law and Third Law of Thermodynamics 5.3

dU is determined uniquely by the initial and final states while dQ and dW are path functions. However, if we specify that the work dW is a frictionless reversible expansion, we have dU = dQ – pdV (unique)

(unique)

It follows that dQ must also uniquely defined for this process. Therefore, dU = dQ – pdV

(2)

where each of the quantities is now in the form of a differential. If the relevant paths are known, they can all be integrated exactly. This shows why frictionless reversible expansions (or compressions) are so important in thermodynamics. Some Examples of Reversible and Irreversible Processes 1. Suppose we place a hot block of metal in contact with a cool block. Then heat is transferred from the hotter block to the cooler one. This is an irreversible process – we cannot reverse any step in the procedure that would cause the heat to flow in reverse direction and restore the blocks to their original temperatures. 2. A cup of hot coffee left on a table gradually cools down. It never gets hotter all by itself. (irreversible-process). In fact (as explained above) all naturally occurring processes proceed in one direction only and such spontaneous, one way processes are irreversible. 3. Consider a piece of metal on a hot plate at a temperature T. If we increase the temperature of the hot plate by a small step dT, a small amount of heat dQ is transferred from the hot plate to the block. If we then decrease the temperature of the hot plate by dT, an equal amount of heat dQ is transferred from the block to the hot plate. The block and the hot plate are restored to their original conditions; the heat transferred in this way is by a reversible process. It is to be noted here that irreversibility does not imply that it is impossible to force a process in the reverse direction but only indicates that such a reversal cannot be achieved simply by changing parameters by infinitesimal amounts.

5.3

NEED FOR FORMULATIONS OF SECOND LAW OF THERMODYNAMICS

Joule’s experiment for the determination of J showed that if the mechanical work is converted into heat, there is exact equivalence. _

W =› Q

5.4 Heat and Thermodynamics Such a possible process a shown in Fig. 5.2(a). If, however, we wish to convert heat into work (Fig. 5.2(b)), this is not true _›

QπW although admissible by the first law of thermodynamics e.g., a piece of stone resting on the floor has never been seen to cool itself spontaneously and jump upto the ceiling (thereby converting heat into equivalent amount of potential energy). In all actual heat Æ work processes, we find W < Q. The purpose of the second law is to incorporate such experimental facts into thermodynamics.

Fig. 5.2 (a) A possible process; (b) a hypothetical process

A cyclic device which would abstract heat from a single reservoir and convert the heat completely to work is called a perpetual motion machine of the second kind. Such a machine would not violate the first law, since it would not create energy but would enable us (for example) to propel ships simply by extracting heat from the oceans. The second law rules out the possibility of constructing a perpetual motion machine of the second kind. Thereby, it also distinguishes between the two directions of the energy transfer viz. W Æ Q and Q Æ W.

5.4

THE SECOND LAW OF THERMODYNAMICS

From our common experience, we know that there are processes that satisfy the law of conservation of energy yet never occur. For instance, a piece of stone resting on the floor is never seen to cool itself spontaneously and jump upto the ceiling. The purpose of the second law (as explained above) is to incorporate reasons for such facts into thermodynamics. There are two equivalent statements of the second law of thermodynamics. There exists no thermodynamic transformation whose sole effect is to extract a quantity of heat from a given heat reservoir and to convert it entirely into work.

Kelvin Statement

The Second Law and Third Law of Thermodynamics 5.5

There exists no thermodynamic transformation whose sole effect is to extract a quantity of heat from a colder reservoir and deliver it to a hotter reservoir. The meanings of the Clausius statement (C) and the Kelvin statement (K) can be understood by referring to the impossible processes shown in Fig. 5.3. In both statements the key word is ‘sole’. We give an example to illustrate the point. Suppose a perfect gas expands reversibly and isothermally. Since dU = 0 in this process, the work done by the gas is equal to the heat absorbed by the gas during the expansion. Hence, a certain amount of heat is converted entirely into work. However, this is not the sole effect because at the end of the process the gas occupies a larger volume. This process is allowed by the second law.

Clausius Statement

Fig. 5.3

Schematic diagrams of the impossible processes according to (a) the Clausius Statement, and (b) the Kelvin Statement

The two statements, C and K, are equivalent. To prove this we show that if K is false, and vice versa. Proof that K false fi C false Suppose K is false. Then we can extract heat Q from a reservoir at temperature Q2 and convert it entirely into work W(= Q), with no other effect. Next we can convert this work into heat Q by means of friction and deliver it to a reservoir at temperature Q1 > Q2, Fig. 5.4 with no other effect. The net result of this two step process is the transfer

Heat Reservoir Q1 > Q2

Q=W W=Q

K False

(Friction)

Q Q2

Heat Reservoir

Fig. 5.4

Equivalence of Kelvin and Clausius statements

5.6 Heat and Thermodynamics of an amount of heat Q from a colder reservoir Q2 to a hotter reservoir Q1, with no other effect. Therefore C is false. Proof that C false fi K false First Heat Reservoir we define an engine to be a Q1 > Q 2 thermodynamic system that can (Friction) undergo a cyclic process in which Q + Q¢ the system does the following things, Q and only the following things: Engine W = Q¢ (i) absorbs an amount of heat C False Q1 > 0 from a hot reservoir Q Q1; Q2 (ii) performs an amount of work Heat Reservoir W; (iii) rejects an amount of heat Fig. 5.5 Equivalence of Clausius and Q2 > 0 to a cold reservoir Q2, Kelvin statements with Q2 < Q1. Suppose C is false. Then we can extract Q from cold reservoir Q2 and deliver it to hot reservoir Q1, with Q1 > Q2. Operate an engine between Q1 and Q2 for one cycle, and arrange the engine so that it absorbs Q + Q¢ from Q1, converts Q¢ to work W, and rejects Q to Q2, Fig. 5.5. The net result is that an amount of heat Q¢ is extracted from the single reservoir Q1 and entirely converted into work, with no other effect. Hence K is false.

5.5

HEAT ENGINE AND ITS EFFICIENCY

Heat engine is a cyclic process in which heat is continuously converted into mechanical work. A heat engine has the following three main parts: It is kept at a constant high temperature T1 so that heat can be extracted from it, without the change in its temperature.

(i) Source of heat

It is kept at a constant high temperature T2 so that heat can be supplied to it without the change in its temperature. The atmosphere acts as a proper sink in practice because it has an infinite capacity to absorb heat without change in its temperature.

(ii) Sink of heat

It absorbs heat from the source, converts a part of it into mechanical work and rejects the remaining heat at the sink. This is called a cycle. Since the working substance returns back to its initial state after a complete cycle, its internal energy* remains unchanged. Thus work is obtained continuously due to repetition of the cycle.

(iii) Working substance

* Under the same conditions of pressure, volume and temperature, the internal energy remains the same.

The Second Law and Third Law of Thermodynamics 5.7

Figure 5.6 shows the operation of heat engine in one cycle. Let in a cycle, the working substance absorbs heat Q1 from the source and rejects heat Q2 at the sink, converting the remaining energy (Q1 – Q2) into the mechanical work. Then from the first law of thermodynamics, Q1 – Q2 = W (if there is no loss in energy due to friction etc.). The ratio of useful work W obtained by the engine in one cycle to the heat absorbed Q1 from the source, is called the efficiency of engine. It is written by the Greek letter h. Therefore, efficiency of engine.

Source of Heat T1 Q1 Cycle

W = Q1 – Q 2

Efficiency of engine

Work output W h = _____________ Heat input Q1

Q2 Sink of Heat T2

Fig. 5.6

Heat engine

(1)

(When W and Q1 are in same units) or

Q2 Q1 – Q2 h = _______ = 1 – ___ Q1 Q1

(2)

Obviously, the efficiency of heat engine is 1 (or 100%) only if Q2 = 0 (i.e., no heat is rejected at the sink in a cycle). In other words, if whole amount of heat extracted from the source is converted into the mechanical work, the engine is said to be 100% efficient. But experimentally, we find that such an engine is not realized in practice, i.e., no engine is 100% efficient.

5.6

CARNOT’S ENGINE: CARNOT’S CYCLE AND ITS EFFICIENCY

To study the efficiency of an engine, Sadi Carnot in 1824, considered a theoretical ideal engine in which there is no loss of heat due to friction etc., and the working substance is the perfect gas. Such an engine can not be realized in practice, but is helpful for the theoretical study as the mathematical calculations become easier. The engine is reversible and its efficiency depends only on the temperatures of source and sink between which it works. No engine in practice can have efficiency more than it. Main parts — The mainparts of the Ideal Carnot engine (Fig. 5.7) are: (i) Source of heat, (ii) Sink, (iii) Perfect insulating plate, (iv) Piston-cylinder assembly.

5.8 Heat and Thermodynamics

Fig. 5.7 Main parts of Carnot’s engine

It is a source of heat of infinite capacity kept at a constant high temperature T1 K, such that its temperature is unaffected on extracting any amount of heat from it.

(i) Source of heat

It is kept at a constant low temperature T2 K. Its capacity is also infinite, i.e., its temperature is unaffected even if any amount of heat is rejected to it.

(ii) Sink

It is a perfect insulating plate which does not allow heat to pass through it.

(iii) Perfect insulating plate

The walls of the cylinder are perfectly insulating and its base is perfectly conducting. Piston is also of non-conducting material and the piston can slide inside the cylinder without any friction. Cylinder contains the working substance (i.e., perfect gas). The cylinder can be kept in contact with either of the heat source, insulating plate and sink. No work is needed in taking it from one to the other.

(iv) Piston-cylinder assembly

A(T1)

Isoth e expa rmal nsion

D (T2)

P

B(T1)

Isoth comp ermal ressio n

batic Adia sion n expa

tic aba n Adi ressio p com

Carnot Cycle According to Carnot, to obtain work continuously from the Ideal engine, it has to be operated in a definite cycle such that the working substance returns back to its initial state after each cycle. A cycle is completed by the four operations in order. These operations are represented on the P-V diagram in Fig. 5.8. The cycle is called the Carnot cycle. Let P1, V1 and T1 be the initial pressure, volume and temperature of the gas. The four operations are: (i) Isothermal expansion, (ii) Adiabatic expansion, (iii) Isothermal compression, (iv) Adiabatic compression

C(T2)

a

d

v

b

Fig. 5.8 Carnot cycle

c

The Second Law and Third Law of Thermodynamics 5.9

The base of the cylinder is kept in contact with the heat source at temperature T1 K and the piston is slowly withdrawn out of the cylinder so that the gas expands isothermally. During this process work is done by the gas, but its internal energy remains constant (since temperature remains constant = T1), therefore, the gas absorbs heat from the source. The state of the gas changes from A (P1, V1, T1) to B (P2, V2, T1) as shown by the curve AB on the P-V diagram. In the process, work done by the gas W1 is equal to the heat absorbed Q1 from the source (since there is no change in internal energy of gas).

(i) First operation: isothermal expansion

Heat absorbed from the source Q1 = work done by the gas V2

W1 = Ú PdV V1 V2

or

RT1 Q1 = W1 = Ú ____ dV V1 V V2 = RT1 loge __ V1

(since PV = RT1)

= area ABbaA enclosed by the curve AB on the P-V diagram with the V-axis

(3)

Now the cylinder is taken away from the source and its base is kept in contact with the insulating plate. Piston continues to move outwards due to inertia, i.e., the gas expands adiabatically (since the gas is enclosed from all sides inside the insulator). In this process, work is done by the gas due to which internal energy of gas and hence the temperature of gas decreases from T1 to T2 (= temperature of sink). The state of the gas changes from B(P2, V2, T1) to C(P3, V3, T2) as shown by the curve BC on the P-V diagram.

(ii) Second operation: Adiabatic expansion

Decrease in internal energy of gas = work done by the gas V3

W2 = Ú PdV V2 V3

or

K W2 = Ú __g dV V2 V

(

1 1 K – _____ = ______ _____ (g – 1) Vg – 1 Vg – 1 2 3 or

)

1 W2 = ______ (P2 V2 – P3 V3) (g – 1)

(since P2 V2g = P3 V3g = K)

5.10 Heat and Thermodynamics R = ______ (T1 – T2) (g – 1)

(since P2 V2 = RT1 and P3 V3 = RT2)

= area BCcbB enclosed by the curve BC on the P-V diagram with V-axis

(4)

Now the cylinder is withdrawn from the insulating plate and is kept in contact with the sink. The piston is moved slowly inwards the cylinder so that the gas is compressed isothermally at temperature T2. The work is done on the gas, but the internal energy of gas remains constant (since temperature of gas does not change), therefore, heat is rejected at the sink. The state of the gas changes from C(P3, V3, T2) to D(P4, V4, T2) which is shown by the curve CD on the P-V diagram.

(iii) Third operation: isothermal compression

Heat rejected at the sink Q2 = work done by the gas V4

W3 = – Ú PdV V3 V

or

4 RT2 Q2 = W3 = Ú ____ dV V3 V

V3 = RT2 loge ___ V4

(since PV = RT2)

= area CDdcC enclosed by the curve CD on the P-V diagram with the volume axis

(5)

The cylinder is now removed from the sink and is again kept in contact with the insulating plate. This piston is moved inwards the cylinder to compress the gas till the gas come back to its initial state (P1, V1, T1). In this operation, the compression of gas is adiabatic. The work is done on the gas due to which the internal energy of gas increases and hence the temperature of gas rises from T2 to T1. The state of gas changes from D(P4, V4, T2) to A (P1, V1, T1) which is represented by the curve DA on the P-V diagram.

(iv) Fourth operation: adiabatic compression

Increase in internal energy of gas = work done by the gas V1

W4 = – Ú PdV V4 V1

or

K W4 = – Ú __g dV V V4

(

1 1 K – _____ = ______ _____ (g – 1) Vg – 1 Vg – 1 1 4

)

(since P1 V1g = P4 V4g = K)

The Second Law and Third Law of Thermodynamics 5.11

1 = ______ (P1 V1 – P4 V4) (g – 1) R (since P1 V1 = RT1 and P4 V4 = RT2) = ______ (T1 – T2) (g – 1) = area DAadD enclosed by the curve DA on the P-V diagram with V-axis (6) The cycle is repeated to get the work continuously. From Eqns. (4) and (6), it is clear that W2 = W4, i.e., the work done by the gas in adiabatic expansion = the work done on the gas in adiabatic compression. Hence the area BCcbB and area DAadD on the P-V diagram are equal. The net work done by the working substance in the complete cycle of Carnot engine is

Efficiency of Carnot Engine

W = W1 + W2 – W3 – W4 = W1 – W 3

(since W2 = W4)

= Q1 – Q2 = area ABCDA on the P-V diagram. Putting the values of W1 and W3 from Eqns. (3) and (5) V3 V2 W = RT1 loge ___ – RT2 loge ___ V1 V4

(7)

Since the points B and C are on the adiabatic curve BC, therefore, from TVg – 1 = constant, we get T1 Vg2 – 1 = T2 Vg3 – 1 or

(T1/T2) = (V3/V2)g – 1

(8)

Similarly, the points A and D are on an adiabatic curve AD, therefore, T1 Vg1 – 1 = T2 Vg4 – 1 or

(T1/T2) = (V4/V1)g – 1 Form Eqns. (8) and (9), we get V3 ___ V4 ___ = =r V2 V1

where r is the adiabatic expansion ratio. or

V3 V2 ___ ___ = =r V1 V4

where r is the isothermal expansion ratio.

(9)

5.12 Heat and Thermodynamics \ From Eqn. (7), W = RT1 loge r – RT2 loge r = R(T1 – T2) loge r

(10)

Heat absorbed by the working substance in one cycle V2 Q1 = RT1 loge ___ V1 = RT1 loge r Net work output \ Efficiency of engine h = ______________________ Heat absorbed from source Q2 W Q1 – Q2 = ___ = _______ = 1 – ___ Q1 Q1 Q1

[from Eqn. (3)] (11) (12)

R(T1 – T2) loge r h = ______________ RT1 loge r

or

T2 T1 – T2 = ______ = 1 – __ T1 T1

(13)

It is clear from the above expression that (i) The efficiency of Carnot engine depends only on the temperatures of the source and sink. The efficiency of all the engines working between the same two temperatures is the same. (ii) The efficiency of an engine can be 1.0 (or 100%) only if the temperature of sink is absolute zero. But it is impossible to acquire the temperature of sink equal to absolute zero, hence no engine can be 100% efficient. (iii) The efficiency of engine is more if (T1 – T2) is more. The efficiency of the Carnot engine can be increased in two ways: (a) by keeping the temperature of source T1 constant and decreasing the temperature of sink T2, (b) by keeping the temperature of sink T2 constant and increasing the temperature of source T1. But out of these, the first way is more effective because T2 < T1, therefore, the decrease in T2 is more effective than the increase in T1 by the same amount. Ways of Increasing the Efficiency

Note

1. The efficiency of Carnot engine can also be expressed in terms of adiabatic expansion ratio r. From Eqns. (8) and (9), T1 __ = rg – 1 T2

The Second Law and Third Law of Thermodynamics 5.13

\ Efficiency of engine T2 1 h = 1 – __ = 1 – __ r T1

()

g–1

(14)

2. Comparing Eqns. (12) and (13), for an ideal engine T2 Q2 __ ___ = Q1 T1

(15)

Thus, the ratio of heat rejected at the sink to the heat absorbed from the source is equal to the ratio of temperature of sink to the temperature of source.

5.7

WORKING OF A HEAT ENGINE/REFRIGERATOR OPERATING ON A CARNOT CYCLE

Let us assume that we have a heat engine, operating between the given high temperature and low temperature reservoirs does so in a cycle in which every process is reversible. (It is also assumed that the high temperatures and the low temperature of the two reservoirs remain constant regardless of the amount of heat transferred). If every process is reversible, the cycle is also reversible, and if the cycle is reversed, the heat engine becomes a refrigerator. We now consider the heat engine (refrigerator) that operates on a Carnot cycle (i.e., the most efficient cycle we can have) as depicted in Fig. 5.9. The figure shows a power plant that is similar in many ways to a simple steam power plant. High-temperature reservoir QH

QH

Boiler (condenser) W

Pump (turbine)

Turbine (pump) W

Condenser (evaporator) QL

QL

Low-temperature reservoir

Fig. 5.9 A heat engine that operates on a Carnot cycle (The refrigerator is shown by dotted lines and parentheses)

5.14 Heat and Thermodynamics Consider the working fluid to be a pure substance, such as steam. Heat is transferred from the high temperature reservoir to the water (steam) in the boiler. For this process to be a reversible heat transfer, the temperature of the water (steam) must be only infinitesimally lower than the temperature of the reservoir. This result also implies (since the temperature of the reservoir remains constant) that the temperature of the water must remain constant. Therefore, the first process in the Carnot cycle is a reversible isothermal process in which heat is transferred from the high temperature reservoir to the working fluid. The next process occurs in the turbine without heat transfer and is therefore, adiabatic-process during which the temperature of the working fluid decreases from the temperature of the high temperature reservoir to the temperature of the low temperature reservoir. In the next process, heat is rejected from the working fluid to the low temperature reservoir. This must be a reversible isothermal process in which the temperature of the working fluid is infinitesimally higher than that of the low temperature reservoir. During this isothermal process some of the steam is condensed. The final process, which completes the cycle, is a reversible adiabatic process in which the temperature of the working fluid increases from the low temperature to the high temperature.

5.7.1

Efficiency of Refrigerators

In case of a refrigerator, we supply work in order to extract an amount Q2 of heat from a cold body (e.g., something to be freezed in a refrigerator) and then throw it out into the surrounding (atmosphere). The heat rejected is not just Q2 but Q1 = Q2 + W (Fig. 5.10). In this case, one does not define efficiency but a coefficient of performance: Q2 K = ___ W T2 = ______ T1 – T2

Hot T1 Q1 = (W + Q2)

W

Q2

T2 Cold

Fig. 5.10 Schematic figure of a refrigerator. Work is done to remove heat from a cold source and then dump it into a hot reservoir

Thus, it is obvious that K can be > 1. For an ideal refrigerator, the variation of K as a function of the ratio (T2/T1) is shown in Fig. 5.11. For a domestic refrigerator K ~ 9. Figure 5.11 also reveals that K Æ 0 as T2 Æ 0. In other words,

The Second Law and Third Law of Thermodynamics 5.15

Fig. 5.11 Temperature variation of K, the coefficient of performance of an ideal refrigerator, i.e., a Carnot engine reversed. When T2 = T1, the cold and the hot reservoirs have the same temperature. Usually, T1 is the room temperature (~ 300 K) while T2 denotes the temperature to be attained by cooling. Down to (T2/T1) = 0.5, more heat is absorbed than the work supplied but below that the work required to extract a given quantity of heat increases sharply (W = Q/K). After G. Venkataraman, “A Hot Story”, Universities Press (India) 100. (1993)]

refrigerators meant to cool objects down to T ~ 0 K are very efficient. However, if going very low in temperature is our objective, then there is no choice.

5.8

DEMONSTRATION OF THE FACT THAT THE CARNOT CYCLE IS THE MOST EFFICIENT CYCLE OPERATING BETWEEN TWO FIXED‐TEMPERATURE RESERVOIRS

There are two propositions regarding the efficiency of a Carnot cycle: First Proposition “It is impossible to construct an engine that operates between two given reservoirs and is more efficient than a reversible engine operating between the same two reservoirs”. The proof of this statement is accomplished through a “thought experiment”: We make an initial assumption, then show that this leads to impossible conclusion. Then, the only possible conclusion is that our initial assumption was incorrect. This is as follows. Let us assume that there is an irreversible engine operating between two given reservoirs that has a greater efficiency than a reversible engine operating between the same two reservoirs.

5.16 Heat and Thermodynamics Let the heat transfer to the irreversible engine be QH, the heat rejected be Q¢L and the work performed be WIE (= QH – Q¢L) as shown in Fig. 5.12. System boundary High-temperature reservoir QH

Wnet = QH – Q ¢L

WIE = QH – Q ¢L

Irreversible engine

QH

Reversible engine

WRE = QH – Q ¢L Q ¢L

QL

Low-temperature reservoir

Fig. 5.12 [After: R.E. Sonntag, C. Borgnakke and G.J.V. Boyle Fundamentals of Thermodynamics, John Wiley (2002)]

Let the reversible engine operate as a refrigerator (this is possible since it is reversible). Finally, let the heat transfer with the low temperature reservoir be QL, the heat transfer with the high temperature reservoir be QH and the work required be WRE (= QH – QL). Since the initial assumption was that the irreversible engine is more efficient, it follows (because QH is the same for both engines) that Q¢L < QL and WIE > WRE. Now, the irreversible engine can drive the reversible engine and still deliver the net work Wnet = WIE – WRE = QL – Q¢L. If we consider the two engines and the high temperature reservoir as a system (as indicated in Fig. 5.12), we have a system that operates in a cycle, exchanges heat with a single reservoir and does a certain amount of work. However, this would constitute a violation of the second law and we conclude that our initial assumption (that the irreversible engine is more efficient than a reversible engine) is wrong. Therefore, we cannot have an irreversible engine that is more efficient than a reversible engine operating between the same two reservoirs. Second Proposition “All engines that operate on the Carnot cycle between two given constanttemperature reservoirs have the same efficiency”. The proof of this proposition is similar to the proof just outlined (above), which assumes that there is one Carnot cycle that is more efficient than the another Carnot cycle operating between the same reservoirs. Let the Carnot cycle

The Second Law and Third Law of Thermodynamics 5.17

with the higher efficiency replace the irreversible cycle of the previous argument and let the Carnot cycle with the lower efficiency operate as a refrigerator. The proof proceeds with the same line of reasoning as above and details are left as an exercise for the reader.

5.9

THERMODYNAMIC TEMPERATURE SCALE

In the first chapter, we pointed out that the zeroth law of thermodynamics provides a basis for temperature measurement, but that a temperature scale must be defined in terms of a particular thermometer substance and device. A temperature scale that is independent of any particular substance, which might be called an absolute temperature scale, would be most desirable. In the previous section, we have noted that the efficiency of a Carnot cycle is independent of the working substance and depends only on the temperature. This fact provides the basis for such an absolute temperature scale which we call the thermodynamic scale. The concept of this temperature scale may be developed with the help of Fig. 5.13, which shows three reservoirs and three engines that operate on the Carnot cycle. T1 is the highest temperature, T3 is the lowest temperature and T2 is an intermediate temperature, and the engines operate between the various reservoirs as indicated. Q1 is the same for both A and C and since, we are dealing with reversible cycles, Q3 is the same for both B and C. T1 Q1 WA

Q1

A WC

C

Q2 T2 Q2 WB

Q3 B Q3 T3

Fig. 5.13

Since the efficiency of a Carnot cycle is a function only of the temperature, we can write QL hthermal = 1 – ___ = 1 – y (TL, TH) QH

(1)

5.18 Heat and Thermodynamics where y designates a functional relation. Let us apply this functional relation to the three Carnot cycles of Fig. 5.13. Q1 ___ = y(T1, T2) Q2

(2)

Q2 ___ = y(T2, T3) Q3

(3)

Q1 ___ = y(T1, T3) Q3

(4)

Since Q1 Q2 Q1 _____ ___ = , Q3 Q2 Q3 it follows that y(T1, T3) = y(T1, T2) × y(T2, T3)

(5)

Note that the LHS is a function of T1 and T3 (and not of T2) and therefore, the RHS must also be a function of T1 and T3 (and not of T2). From this fact we conclude that the form of the function y must be such that f(T1) y(T1, T2) = ____ (6) f(T2) f(T2) y(T2, T3) = ____ f(T3)

(7)

so that f(T2) will cancel from the product of y(T1, T2) and y(T2, T3). Therefore, we conclude that Q1 f(T1) ___ = y(T1, T3) = ____ Q3 f(T3)

(8)

f(TH) QH _____ ___ = QL f(TL)

(9)

In general terms

Now, there are several functional relations that will satisfy this equation. For the thermodynamic scale of temperature, which was originally proposed by Lord Kelvin, the selected relation is QH ____ (tH) ___ = QL (tL)

(10)

(where tH and tL are the Kelvin temperature itself corresponding to f(TH) and f(TL) respectively).

The Second Law and Third Law of Thermodynamics 5.19

With absolute temperature t so defined, the efficiency of a Carnot cycle may be expressed in terms of the absolute temperatures: QL tL hthermal = 1 – ___ = 1 – ___ tH QH

(11)

This means that if the thermal efficiency of a Carnot cycle operating between two given constant-temperature reservoirs is known, the ratio of the two absolute temperatures is also known.

5.9.1

Zero of the Thermodynamic Scale QL tL h = 1 – ___ = 1 – __ tH QH

(12)

If tL = 0 then QL = 0 and h = 1. Thus, zero of the thermodynamic scale is the temperature of sink at which the reversible engine is 100% efficient (i.e., the engine converts whole of the heat into work without rejecting any heat at the sink). This temperature is called absolute zero. It may be mentioned here that it is not possible to define absolute zero by any empirical gas scale because each gas becomes solid before reaching this temperature and the thermometer does not work. Now, we can see that zero on the thermodynamic scale is the lowest possible temperature on the absolute scale and there is no negative temperature on this scale. For this, consider that a reversible engine rejects heat QL at the sink temperature – tL, then, its efficiency tL tL QL h = 1 – ___ = 1 – – ___ = 1 + ___ tH tH QH

( )

i.e., h > 1 or efficiency obtained from the reversible engine is more than 100% which is impossible. Hence, negative temperature is not possible on the thermodynamic scale.

5.9.2

Size of Degree on the Thermodynamic Scale

To fix the size of degree, like the Celsius scale, the interval between the steam point ts and the ice point ti is divided into 100 equal parts. Each part is called a degree. Thus, ts – ti = 100 Kelvin degrees. Now, if a reversible engine absorbs heat Qs at the steam point ts and rejects heat Q at a temperature t, then Qs __ ts ___ = Q t

(13)

5.20 Heat and Thermodynamics Similarly, if another reversible engine absorbs heat Q at a temperature t and rejects heat Qi at the ice point ti (where ti < t), then Q __ t ___ = Qi ti Qi __ ti ___ = Q t

or

(14)

From the above two equations ts – ti Qs – Qi _____ ______ = t Q

(15)

But ts – ti = 100, therefore Qs – Qi ____ 100 ______ = t Q

(

Q t = 100 × ______ Qs – Qi

or

)

(16)

Thus, an unknown temperature t can be calculated.

5.9.3 Equivalence of the Thermodynamic Scale with the Perfect Gas Scale Let the temperatures of source and sink on the perfect gas scale be T1 and T2 respectively (T1 > T2). A reversible engine working between these temperatures absorbs heat Q1 from the source and rejects heat Q2 to the sink. Then, the efficiency of Carnot engine Q2 t2 h = 1 – ___ = 1 – __ t1 Q1 Q2 __ T2 ___ = Q1 T1

or

(17)

Now, if the temperatures of these source and sink on the thermodynamic (i.e., absolute) scale are T1 and T2 respectively, then Q2 __ t2 ___ = t Q1 1

(18)

Thus, T2 __ t2 __ = T1 t 1

(i.e., the ratios are equal)

(19)

The Second Law and Third Law of Thermodynamics 5.21

Now, if the engine works between the steam point and the ice point, then ti Ti __ __ = t Ts s or

Ti ti 1 – __ = 1 – __ ts Ts

or

Ts – Ti _____ ts – ti ______ = ts Ts

or

100 ____ 100 ____ = ts Ts

i.e.,

Ts = ts

(20)

(21)

Similarly, Ts __ ts __ = Ti ti fi

ts Ts __ – 1 = __ – 1 ti Ti

or

ts – ti Ts – Ti _____ ______ = ti Ti

or

100 ____ 100 ____ = ti Ti

or

Ti = ti

(22)

i.e., the steam point and the ice point are same on the two scales. Now, if a reversible engine operates between the steam point and any other temperature (T or t), then t T __ __ = Ts t s \

T=t

(\ Ts = ts),

i.e., both the scales are perfectly equivalent.

5.10 RELATION BETWEEN THERMODYNAMIC SCALE, CELSIUS SCALE, RANKINE SCALE AND THE FAHRENHEIT SCALE OF TEMPERATURES Suppose we have a heat engine operating on the Carnot cycle that receives heat at the temperature of steam point and rejects heat at the temperature of ice point.

5.22 Heat and Thermodynamics To gain an idea about the magnitude of a degree on the thermodynamic scale, we cannot construct an engine on the Carnot cycle and perform the experiment for its efficiency. However, we can follow the reasoning as the following “thought experiment”. If the efficiency of such an engine could be measured, we would find it to be 26.8%. then TL h = 1 – ___ TH Tice = 1 – _____ = 0.2680 Tsteam Tice _____ = 1 – 0.2680 = 0.7320 Tsteam

\

(1)

This is one equation concerning the two unknowns TH and TL. The second equation comes from deciding the magnitude of the degree on the thermodynamic (i.e., absolute) scale to be equal to the magnitude of the degree on the Celsius scale. Then Tsteam – Tice = 100

(2)

Solving Eqns. (1) and (2) simultaneously, we find Tsteam = 373.15 K, Tice = 273.15 K

(3)

It follows that T(°C) + 273.15 = T(K)

(4)

The absolute scale related to the Fahrenheit scale is the Rankine scale, designated by R. On both these scales, there are 180 degrees between the ice point and the steam point. Therefore, for a Carnot cycle heat engine operating between the steam point and the ice point, we would have the two relations: Tsteam – Tice = 180

(5)

Tice _____ = 0.7320 Tsteam solving these two equations simultaneously, we find Tsteam = 671.67 R, Tice = 491.67 R

(6)

The Second Law and Third Law of Thermodynamics 5.23

It follows that since 0° R = 32°F, we have the following relation between Fahrenheit and Rankine scales 491.67 – 32 = T (R) – T (F) or

T (F) + 459.67 = T (R)

(7)

As already pointed out, the measurement of efficiencies of Carnot cycles is not a practical way to approach the problem of temperature measurements on the thermodynamic scale. Rather, the actual approach is based on ideal gas thermometer and an assigned value for the triple point of water (as described in the first chapter). At the tenth conference on Weights and Measures (held in 1954), the temperature of the triple point of water was assigned the value 273.16 K (the triple point of water is approximately 0.01°C above the ice point). The ice point is defined, as the temperature of a mixture of ice and water at a pressure of 1 atmosphere (101.13 kPa) of air that is saturated with water vapour.

5.11

THE CLAUSIUS THEOREM

Consider a smooth closed curve representing a reversible cycle CR (Fig. 5.14). Let the curve be divided into a large number of strips by means of reversible adiabatics. Each strip may be closed at the top and bottom by reversible isotherms. The original closed cycle is thus replaced by a zig-zag closed path consisting of alternate adiabatic and isothermal processes such that the heat transferred during all the isothermal processes is equal to the heat transferred in the original cycle (If the adiabatics are close to one another and the number of Carnot cycles is large, the zig-zag closed path will coincide with the original cycle). P Reversible adiabatics dQ3 T3 T1 a

e b

f

Reversible isotherms

dQ1 T2

c d

T4

g

Original reversible cycle

dQ2 dQ 4 V

Fig. 5.14 A reversible cycle divided into a large number of Carnot cycles

5.24 Heat and Thermodynamics For the element cycle abcd, dQ1 heat is absorbed reversibly at T1 and dQ2 heat is rejected reversibly at T2.* So, dQ2 dQ1 ____ = – ____ (1) T1 T2 (taking heat supplied as positive and heat rejected as negative) dQ2 dQ1 ____ ____ + =0 T1 T2

or

(2)

Similarly, for the elemental cycle efgh dQ3 ____ dQ4 ____ + =0 T3 T4

(3)

If similar equations are written for all the elemental Carnot cycles, then for the whole original cycle CR dQ3 dQ1 ____ dQ2 ____ ____ + + +…=0 T1 T2 T3 dQ ___ =0 T for any closed reversible cycle CR. This result is known as Clausius theorem. or

5.12

CR

(4)

THE CLAUSIUS INEQUALITY

Let us consider a cycle ABCD (Fig. 5.15). Let AB be a general process, either reversible or irreversible, while the other processes in the cycle are reversible. Let the cycle be divided from a number of elementary cycles as shown. For one of these elementary cycles: dQ2 h = 1 – ____ (5) dQ where dQ is the heat supplied at T and dQ2 is the heat rejected at T2. Now, the efficiency of a general cycle will be equal to or less than the efficiency of a reversible cycle, i.e.,

( or

)( ) ( )

dQ2 dQ2 1 – ____ £ 1 – ____ dQ dQ dQ2 dQ2 ____ ≥ ____ dQ dQ

(6) rev

rev

* Note that an imperfect differential has been represented by dQ or dQ.

The Second Law and Third Law of Thermodynamics 5.25 T=C

B A T

dQ

Reversible adiabatics

C

D

T2

dQ2

V

Fig. 5.15

Pertaining to Clausius inequality

( )

dQ dQ ____ £ ____ dQ2 dQ2

or

Since

( ) dQ ____ dQ2

(7) rev

T = __ T 2 rev

(8)

Therefore, dQ __ T ____ £ dQ2 T2 dQ2 dQ ____ ___ £ T T2 (reversible or irreversible) For a reversible process

or

for any process AB

dQrev dQ2 ds = _____ = ____ T T2

(9)

(10)

Hence, for any process AB dQ ___ £ ds T

(11)

dQ ___ £ ds T

(12)

and for any cycle

5.26 Heat and Thermodynamics Since

ds = 0

therefore dQ ___ £0 (13) T This is known as Clausius inequality. It provides the criterion of the reversibility of a cycle. dQ ___ £ 0, the cycle is reversible (14) If T dQ ___ < 0, the cycle is irreversible and possible T

(15)

dQ ___ > 0, the cycle is impossible, since it violates the second law. (16) T

5.13

B

ENTROPY

Consider an arbitrary reversible cycle represented by the closed curve in Fig. 5.16. A and B areany two points on the curve. From Clausius theorem, we have B A dQ(rev) dQ(rev) dQ(rev) ______ = Ú ______ + Ú ______ T T T B A (path I)

Path I p

Path II

A V

(path II)

=0

(1)

Fig. 5.16

An arbitrary reversible cycle

The cycle being reversible, we can traverse the path II from B to A in the opposite direction. Then for path II A

Ú B (path II)

dQ(rev) ______ =– T

B

Ú A (path II)

dQ(rev) ______ T

(2)

So, Eqn. (1) becomes B

Ú A (path I)

dQ(rev) ______ = T

B

Ú A (path II)

dQ(rev) ______ T

(3)

Since paths I and II were chosen at random, and represent any two reversible paths, we have the result that B

dQ(rev)

= independent of path between A and B. Ú ______ T

A

The Second Law and Third Law of Thermodynamics 5.27

In mathematical terms, it means that the integrand dQ(rev)/T can be expressed as the perfect differential of a point function or a state variable. Denoting this variable by S, we can write dQ(rev) dS = ______ (4) T The point function S was introduced by Clausius in 1850, and is called the entropy. B B dQ(rev) (5) \ SB – SA = Ú dS = Ú ______ T A A Equation (4) indicates that when the inexact differential dQ(rev) is multiplied by 1/T, it becomes an exact differential; 1/T is the integrating factor. The integral B

B

dQ(rev)

is independent of the path. Ú dQ(rev) is dependent on the path whereas Ú ______ T

A

A

Entropy is an extensive property. Common units for entropy are Joule/K or cal/K. The Clausius theorem now reads dS = 0

(6)

Since S is a state variable, this result also follows from Eqn. (5). Entropy like other state variables, is associated with a state and not with a process. Given two equilibrium states A and B, it is possible to determine the entropy difference SB – SA, regardless of the actual process (reversible or irreversible), which the system may have undergone. To calculate this difference, we can choose any reversible path joining them without reference to the process being analyzed (In practice, a reversible path yielding the simplest form of the integral is chosen). To summarize, we have the following consequences of the second law of thermodynamics: 1. For all substances, there exists a thermodynamic temperature scale T. 2. The thermodynamic temperature T constitutes the integrating denominator for dQ. 3. There exists a useful state variable S, called entropy.

5.14

PHYSICAL SIGNIFICANCE OF ENTROPY

Entropy is a mathematical function, which is a definite single valued function of the thermodynamic coordinates, defining the state of a system. The entropy of a system is a real physical quantity given by A

dQ SA = SO + Ú ___ O T

5.28 Heat and Thermodynamics where SO denotes the entropy of the system in state O. Here SA is a function of A only since the state O is fixed. The entropy thus defined, requires the arbitrary choice of the standard state O. Thus, SA can be calculated except for an additive constant. We also know that the entropy of a system increases in an irreversible process and remains constant in the limit of reversible process. Thus, the flow of heat always takes place in a direction in which the entropy increases. It is rather very difficult to visualize the concept of entropy because there is nothing physical to measure it. We cannot feel it like pressure or temperature. However, the entropy can be considered as a measure of the disorder of the molecular distribution of the system. When heat is added to a system, thermal agitation of molecules increases and the entropy of the system increases. Thus, the increase of entropy implies a transition from ordered to disordered state of the system.

5.15

THERMODYNAMIC DEFINITION OF TEMPERATURE

While dealing with Carnot’s cycle, we had attempted to give absolute or Kelvin scale of temperature in terms of efficiencies of Carnot’s engine. However, the main question remains that what we should understand by temperature in thermodynamic terms. To do this, we follow a simple approach. We take two closed systems, each in thermodynamic equilibrium and put them in contact so that they interact with each other only in terms of energy transfer as heat and then look for the property which the two systems should have in common when the energy transfer as heat ceases. Let us consider a total system C which consists of two subsystems A and B separated from each other by a rigid wall which is thermally conducting but does not allow mass transfer across it. The total system C is an isolated system. The two subsystems A and B though individually in thermodynamic equilibrium are not in equilibrium with each other. The situation is depicted in Fig. 5.17 which also indicates the direction of heat flow. Rigid wall

Isolated system C A

B dQ

Fig. 5.17

An isolated system consisting of two closed systems separated from each other by a thermally conducting rigid wall

The Second Law and Third Law of Thermodynamics 5.29

For the system C, we can write the entropy of the total system in terms of entropies of subsystems A and B i.e., SC = SA + SB

(1)

Since entropy is a state function \

S = S (U, V)

and we can write Eqn. (1) as SC = SA (UA, VA) + SB (UB, VB)

(2)

Since the total system C is isolated, its entropy cannot decrease and since, the wall between the subsystems A and B is rigid, interaction between A and B cannot be in terms of energy transfer as work; it can only be in terms of energy transfer as heat. As a result, the total internal energy UC = UA + UB = constant

(3)

which although remains constant for the isolated system C is shared by the two subsystems A and B. If r represents the fraction of the total internal energy shared by A, we can write UA = rUC; UB = (1 – r) UC

(4)

Since the total system C is an isolated system and the wall separating the subsystems A and B is a rigid wall, therefore, UC, VA and VB would remain constant. The only parameter which may vary during the course of interaction between A and B is only r. When the systems A and B attain thermal equilibrium with each other, the value of SC should attain its maximum. For this dSC ____ =0 dr

(5)

From Eqns. (2) and (4), we have

( ) ( )

dSC ∂SA ____ = ____ dr ∂UA ∂SA = ____ ∂UA

( ) ( )

dUA ∂SB ____ + ____ ∂UB VA dr ∂SB UC + ____ ∂UB VA

dUB ____ VB dr

VB

(– UC)

(6)

From Eqns. (5) and (6), it follows that at thermal equilibrium between A and B

( ) ( ) ∂SA ____ ∂UA

∂SB = ____ ∂UB VA

VB

(7)

5.30 Heat and Thermodynamics Thus (∂S/∂U)V is the property which the two systems A and B should have in common when thermal equilibrium is attained. From dimensional analysis, ∂S/∂U has the dimensions of temperature–1. (S = energy/temperature and U = energy). Therefore, 1 T ∫ ________ (∂S/∂U)V

(8)

This equation gives us thermodynamic definition of temperature. We can also establish the direction of heat flow using Eqn. (6) from which it follows that

(

1 1 dSC = UC dr ___ – ___ TA TB

)

(9)

Since TB π TA initially and dSC > 0 for the total isolated system, it follows that dr > 0 and 1/TA > 1/TB (i.e., TB > TA). UA increases and UB decreases, when TB is greater than TA i.e., the heat flows from B to A. Let us take this analysis a little further to learn about the relationship between dSC T and U. Since ____ = 0 represents the condition of a maximum, the second dr 2 derivative i.e., d SC/dr2 should be negative. Hence from Eqn. (6), we can write

( )

d2SC ∂2SA ____ _____ = dr2 ∂UA2

( )

∂2SB UC2 + _____2 ∂UB VA

UC2 < 0

(10)

VB

If A and B are identical systems, we get from Eqn. (10)

or

∂2S ____ ∂U2

( )

V

( )

V

∂T ___ ∂U

0

(11)

i.e., the temperature must be a monotonically increasing function of internal energy at a fixed volume.

5.16

T‐S DIAGRAM

For the heat transferred in a reversible process, we have dQ(rev) = T dS

(1)

f

Q(rev) i, f = Ú T dS i

(2)

The Second Law and Third Law of Thermodynamics 5.31

The integral represents the amount of Area = TdS . dQ heat Q(rev) added to the system and can B be interpreted graphically as the area A under a curve on a T-S diagram, in which T T is plotted along the y-axis and S along the x-axis, Fig. 5.18. An isothermal process (dT = 0) appears as a horizontal line on a T-S SA SB S diagram. For a reversible adiabatic process, we have Fig. 5.18 Area under a curve in a T-S diagram represents heat absorbed dQ(rev) = 0 dQ(rev) dS = _____ = 0 T S = constant

(adiabatic process) (3)

Therefore, entropy remains constant during a reversible adiabatic process and it is called an isentropic process. Such a process will be represented by a vertical line on a T-S diagram. A Carnot cycle consists of two isothermal and two adiabatic processes. Therefore, on a T-S diagram it will look like a rectangle, Fig. 5.19.

5.17 ENTROPY CHANGE IN AN IRREVERSIBLE PROCESS

T1

T

T2 S

Fig. 5.19

Carnot cycle on a T-S diagram

For any process undergone by a system dQ ___ £ ds T dQ ds ≥ ___ T

or

(1) (2)

This is further clarified if we consider the cycle as shown in Fig. 5.20, where A and B are reversible processes and C is an irreversible process. For the reversible cycle consisting of A and B, 2

1

dQ dQ dQ = Ú ___ + Ú ___ = 0 Ú ___ R T 1A T 2B T 2

or

dQ

1

dQ

= – Ú ___ Ú ___ T T

1A

2B

(3)

5.32 Heat and Thermodynamics

A Rev.

2

T Rev.B

C

1 Irrev.

S

Fig. 5.20

Pertaining to entropy change in an irreversible process

For the irreversible cycle consisting of A and C, by the inequality of Clausius 2

1

dQ dQ dQ ___ = Ú ___ + Ú ___ < 0 T A1 T C 2 T

(4)

From Eqns. (3) and (4) 1

1

dQ dQ – Ú ___ + Ú ___ < 0 T C2 T B2 1

or

1

dQ dQ > Ú ___ Ú ___ T T B2 C2

(5)

Since the path B is reversible 1

\

dQ

1

= Ú dS Ú ___ T

B2

(6)

B2

Since entropy is a state property, so entropy changes for the paths B and C would be the same i.e., 1

1

Ú dS = Ú dS B2

C2

From Eqns. (5) and (7) 1

1

dQ

Ú dS > Ú ___ T

C2

C2

\ For any irreversible process dQ dS > ___ T whereas for a reversible process

(7)

The Second Law and Third Law of Thermodynamics 5.33

dQ(rev) dS = ______ T For the general case dQ dS ≥ ___ T 2

dQ S2 – S1 ≥ Ú ___ 1 T

or

(8)

5.18 ENTROPY CHANGE OF A PERFECT GAS IN A REVERSIBLE PROCESS Consider 1 mole of a perfect gas at temperature T, pressure P and volume V. If dW is the work done by the gas in change of state from i to f, dU is the increase in internal energy and dQ is the heat absorbed by the system, then from the first law of thermodynamics dQ = dU + dW But if CV is the molar specific heat at constant volume and the temperature of the gas increases by dT, volume increases by dV in the change of state, then increase in internal energy of gas dU = CV dT and the work done by the gas dW = PdV \

dQ = CV dT + PdV

(1)

Hence, change in entropy f

f

dQ 1 Sf – Si = Ú ___ = Ú __ (CV dT + PdV) T T i i f

f

f

f

dT P = Ú CV ___ + Ú __ dV T i iT dT R = Ú CV ___ + Ú __ dV T V i i or

()

()

Tf Vf (Sf – Si) = CV loge __ + R loge __ Ti Vi

(2)

In Terms of P, T From perfect gas equation PV = RT, we have V T Pi __f = __f × __ Vi Ti Pf

(3)

5.34 Heat and Thermodynamics Hence, from Eqn. (1)

( )

()

Tf Tf Pi Sf – Si = CV loge __ + R loge __ × __ Ti Ti Pf

()

Tf Pi = (CV + R) loge __ + R loge __ Ti Pf

()

()

()

Tf Pf (Sf – Si) = CP loge __ – R loge __ Ti Pi

or

(4)

In Terms of P, V From perfect gas equation PV = RT, we have T P i Vf __f = __ × __ Ti Pf Vi

(5)

Hence, from Eqn. (1), we get

( ) ()

() ()

Pf Vf Vf Sf – Si = CV loge ____ + R loge __ Pi Vi Vi or

5.19

Pf Vf (Sf – Si) = CV loge __ + CP loge __ Pi Vi

(6)

REAL HEAT ENGINES

Reversible Carnot engine with perfect gas (or air) as working substance is not a practical engine because isothermal processes have to be performed very slowly and the source and sink are required to have very large heat capacity for heat so as not to change in temperature during the process. Moreover, the piston should move without friction. There should also be no loss of heat by conducing from the system (gas) to surroundings. All T1 this makes the air-standard Carnot T2 engine very heavy, slow and difficult to construct. It is only an ideal engine Q1 with no practical value. B A p1 In spite of the above difficulties, T1 one may think of constructing a p reversible Carnot steam engine p2 T2 C C¢ D¢ D Q2 operated in the liquid-vapour region as shown in Fig. 5.21. V The heat Q1 is absorbed from Fig. 5.21 Carnot steam cycle a boiler (source) and Q2 rejected

The Second Law and Third Law of Thermodynamics 5.35

to a condenser (sink) at constant temperatures and pressures T1, p1 and T2, p2, respectively. An efficient turbine can nearly achieve the reversible adiabatic expansion, since steam expands rapidly and with negligible friction. However, it is not possible to achieve the reversible adiabatic compression of a two-phase system. If the compression is performed slowly the conduction losses occur (dQ π 0). If it is done rapidly, the temperature of the liquid phase is raised slightly while the vapour phase gets superheated. Therefore, it is not possible to maintain equilibrium between the two phases, that is, reversibility cannot be achieved. To overcome the last difficulty the steam cycle has been modified into what is known as the Rankine cycle. Other possible practical cycles belong to internal combustion engines: (a) Petrol Otto cycle (heat absorbed at constant volume), and (b) Diesel cycle (heat absorbed at constant pressure). We shall now discuss their idealized versions where effects of friction, conduction, etc. are neglected.

5.19.1

Rankine Cycle

The basis of ideal steam engine (Fig. 5.22(a)) is provided by the reversible Rankine cycle (Fig. 5.22(b)). The cycle is: (i) a Æ b Starting at point a representing the state of saturated liquid water at the temperature and pressure T1, p1 of boiler, reversible isobaric isothermal vaporization of water into saturated steam (point b). (ii) b Æ c Reversible adiabatic expansion of steam to T2, p2 (point c). This process corresponds to the passage of steam through the steam engine or turbine. (iii) c Æ d Reversible isobaric isothermal condensation of steam into saturated water (point d). (iv) d Æ e Reversible adiabatic compression of water to the pressure of the boiler p1 (point e, only a small change of temperature occurs). This process is performed by the pump shown in Fig. 5.22(a). (v) e Æ a Reversible isobaric heating of water to the boiler temperature T1. This heating takes place after the liquid has been pumped into the boiler (Fig. 5.22(a)). However, for the cycle to be reversible this heat must be supplied by a series of heat reservoirs. The Rankine cycle differs from the Carnot cycle in the step e Æ a. The mean temperature for this reversible heating being less than the source temperature T1, the efficiency of the Rankine cycle will be lower than that of a Carnot cycle, which absorbs all heat Q1 at the single temperature T1 and rejects Q2 at T2.

5.36 Heat and Thermodynamics

p

Cylinder Boiler T1,p1

Con Denser Valve T2,p2

Valve

T2

T1 e

b

a

p1 T1

Feed pump

d

p2

c T2

Valve

V

Valve (a)

(b)

Fig. 5.22 (a) Theoretical steam engine; (b) Rankine cycle

We can easily compare the efficiencies of Carnot cycle and Rankine cycle by drawing the T-s diagrams. From Fig. 5.23(a), Work engine hCarnot = ____________ Heat absorbed Area ABC¢D¢ = ___________ Area ABEF (T1 – T2) (s1 – s2) = ______________ T1 (s1 – s2) T 1 – T2 = ______ T1 T

T

T1

(1)

A

B

a

T1

b

F T2

D

T2

C F

e c

d

E s

s (a)

(b)

Fig. 5.23 (a) Carnot steam cycle in a T-s diagram; (b) Rankine cycle in a T-s diagram

The Second Law and Third Law of Thermodynamics 5.37

For the Rankine cycle (Fig. 5.23(b)), we can define Tm1 = Q1/Ds1 and Tm2 = Q2/Ds2 as the mean effective temperatures at which heat is absorbed or rejected, respectively. Then Tm1 (sb – se) – Tm2 (sc – sd) hRankine = ______________________ Tm1 (sb – se) Tm1 – Tm2 = ________ Tm1

(2)

since de and bc are isentropic processes, sb – sc = sc – sd. Clearly, T2 = Tm2 but Tm1 < T1. Therefore, we have h (Rankine) < h (Carnot steam)

5.19.2

Otto Cycle

It is an internal combustion engine, that is, heat is generated inside the cylinder itself by spark-ignition of petrol vapour and air mixture. The Otto engine is shown in Fig. 5.24. The engine consists of a cylinder-piston arrangement with two valves. The cycle (Fig. 5.25) consists of six parts. Four of these involve motion of the piston and are called strokes. Intake valve

Exhaust valve

Intake

Exhaust

Compression

Power

Fig. 5.24

Otto engine

Exhaust

A mixture of petrol (2%) and air (98%) is drawn into the cylinder at atmospheric pressure through the intake valve by the outward stroke of the piston. (ii) Compression stroke a Æ b Both the valves are closed and the piston moves up the cylinder compressing the mixture rapidly. The compression is nearly adiabatic and temperature rises considerably. (i) Intake stroke x Æ a

5.38 Heat and Thermodynamics At b the fuel is caused to explode by spark-ignition. The piston is hardly able to move during the explosion so that volume remains constant, but a very high temperature and pressure are reached.

(iii) Explosion b Æ c

The adiabatic expansion c-d sets in and the piston is driven out. There is considerable drop in pressure and temperature.

(iv) Power stroke c Æ d

At the end of the power stroke the exhaust valve opens and the combustion products flow out rapidly into the atmosphere. The pressure drops at once to atmospheric.

(v) Valve exhaust d Æ a

The piston moves up the cylinder throwing out the remaining gases. The exhaust valve then closes and the intake valve opens for the start of the same cycle. The petrol engine (Otto) cycle is clearly irreversible. Figure 5.25 shows the idealized air standard Otto cycle with the working substance as air (petrol vapour merely helps in ignition) which behaves like a perfect gas. All processes are assumed to be reversible (quasistatic and frictionless). The various steps of the cycle for one mole of air are then represented as follows: (vi) Exhaust stroke a Æ x

A quasistatic isobaric intake stroke of air at pa to a specific volume v1.

(i) x Æ a

Fig. 5.25

Otto cycle

A quasistatic, adiabatic compression stroke from v1 to v2 during which the temperature rises from Ta to Tb. We have

(ii) a Æ b

Ta vg1 – 1 = Tb vg2 – 1

(1)

where g is the ratio of the specific heats. A quasistatic isomeric rise of temperature and pressure brought about by the absorption of heat q1 from a series of reservoirs between Tb and Tc. We have

(iii) b Æ c

q1 = cv (Tc – Tb)

(2)

A quasistatic, adiabatic expansion (power stroke) producing a drop in temperature such that

(iv) c Æ d

Tc vg2 – 1 = Td vg1 – 1

(3)

The Second Law and Third Law of Thermodynamics 5.39

A quasistatic isomeric drop of temperature to Ta and of pressure to pa bought about by rejection of heat q2 with a series of sinks b between Td and Ta. We have

(v) d Æ a

Q2 = cv (Td – Ta)

(4)

A quasistatic exhaust stroke. From Eqns. (1) to (4), we can write g–1 Td – Ta Td Ta __ = __ = v2 ______ = v1 T T T –T

(vi) a Æ x

c

b

( )

c

b

Td – Ta q1 – q2 ______ h = ______ q1 = 1 – Tc – Tb v2 =1– v 1

( )

g–1

1 = 1 – ____ g–1 r

(5)

where r = v1/v2 is called the compression (or expansion) ratio. Note that h (Otto) < h (Carnot) because when r is kept same for both, it is found that the highest temperature of the Otto cycle exceeds and/or the lowest temperatures of the Otto cycle is lower than the corresponding source and sink temperatures for the Carnot cycle. In an actual __ engine r @ 9 and if we put g = 1.5 (for air g = 1.4), we find h (Otto) = 1 – ( 1/÷9 ) = 0.67 = 67 per cent under ideal conditions. In practice efficiency probably only reaches half this value. A cyclic heat engine operates between a source temperature of 800°C and a sink temperature of 30°C. What is the least rate of heat rejection per kW net output of heat engine?

Example 5.1

Solution

T2 hmax = hrev = 1 – __ T1 30 + 273 = 1 – _________ 800 + 273 = 1 – 0.282 = 0.718

Now, \

Wnet ____ = hmax = 0.718 Q1 1 Q1 = _____ = 1.392 KW 0.718

Fig. 5.26

5.40 Heat and Thermodynamics Q2 = Q1 – Wnet = 1.392 – 1 = 0.392 kW This is the least rate of heat rejection. Example 5.2 Which is the more efficient way to increase the efficiency of a Carnot engine: to increase T1, keeping T2 constant; or to decrease T2, keeping T1 constant? Solution The efficiency of a Carnot engine is given by

T2 h = 1 – __ T1 If T2 is constant

( ) ∂h ___ ∂T1

T2 = ___2 T2 T1

( )

∂h As T1 increases, h increases and the slope ___ ∂T1

decreases (Fig. 5.27a). T2

Fig. 5.27

If T1 is constant

( ) ∂h ___ ∂T2

1 = – __ T T1 1

( )

∂h As T2 decreases, h increases, but the slope ___ ∂T2 Also,

( ) ∂h ___ ∂T1

T2 = ___2 T2 T1

remains constant. T1

The Second Law and Third Law of Thermodynamics 5.41

( ) ∂h ___ ∂T2

T1 = – ___2 T1 T1 Since T1 > T2, therefore

and

( ) ( ) ∂h ___ ∂T2

∂h > ___ ∂T1 T1

T2

So, the more effective way to increase the efficiency is to decrease T2.

Perpetual Motion Machine of Second Kind (PMM2) Q2 Wnet The efficiency of a heat engine h = 1 – ___ = ____. If Q2 = 0 (Wnet = Q1, or h = 1.00), Q1 Q1 the heat engine will produce net work in a complete cycle by exchanging heat with only one reservoir, thus violating the Kelvin-Planck statement of the 2nd law. Such a heat engine is called a perpetual motion machine of the second kind. Such a PMM2 is impossible.

5.20

PRINCIPLE OF DEGRADATION OF ENERGY

In universe, due to occurrence of natural processes that are irreversible, entropy is increasing and ultimately, it may attain the state of maximum entropy when all the temperatures will be equalized and no work, then, would be possible. It can be shown that, when entropy is increasing, available energy for doing work, will be decreasing. Let us consider the example of conduction. Conduction is irreversible process. Let a quantity dQ of heat be conducted from the body at a higher temperature T1 to a body at lower temperature T2. Suppose T0 is the lowest available temperature in the system. Then, the available T0 T0 energy to start with is dQ 1 – __ . After the transfer, it becomes dQ 1 – __ . T1 T2

( )

( )

Hence, the loss of available energy in the process of conduction is

( ) ( ) ( ) ( )

T0 T0 T0 T0 dQ 1 – __ – dQ 1 – __ = dQ __ – __ T1 T2 T2 T1

dQ dQ = ___ – ___ T0 T2 T1

Since T1 < T2 and right hand side is positive, so there is a net loss of available energy. Further, dQ ___ = (S)T 2 T2

5.42 Heat and Thermodynamics dQ ___ = (S)T 1 T1

and so that

(S)T > (S)T 2

1

i.e., entropy is increasing while the available energy is decreasing. Since irreversible processes are continually going on in nature, the available energy of the universe is continually decreasing. The principle of increase of entropy is equivalent to the principle of decrease of available energy or degradation of energy. There is yet no violation of conservation of energy or of first law of thermodynamics. The reason being that the energy which becomes unavailable is not lost but is transformed to unavailable form. This principle can be stated now as: “Whenever an irreversible process takes place, certain amount of energy which could have been utilized for doing useful work changes to a form in which it becomes unavailable”.

5.21

THEOREM OF MAXIMUM WORK

For a thermally isolated system consisting of several bodies not in thermal equilibrium with one another, when equilibrium is established, the system may do work on some external objects. The transition may, however, occur in different ways and the final equilibrium state will also be different i.e., energy and entropy of the final equilibrium states will be different. The total work done on external objects will depend on the manner in which equilibrium is established. Thus, there arises a question: How the final equilibrium state must be reached in order that the system does the maximum possible amount of work? Now, let the original energy of the system be U0 and the energy in the equilibrium state, as a function of entropy in the equilibrium state be U(S). As the system is thermally isolated, the work done by the system would be |W| = U0 – U(S) Here, we have written |W| (instead of W) because W < 0 according to the convention, as the work is done by the system. Differentiating |W| with respect to entropy S of the final state (at constant volume),

( )

∂|W| ∂U ____ = – ___ ∂S ∂S

(

=–T V

( )

∂U because dU = TdS – pdV and U = U (S, V) fi dU = ___ ∂S

( )

∂U consequently ___ ∂S

V

( )

∂U = T and ___ ∂V

=–p S

)

V

( )

∂U dS + ___ ∂V

dV S

The Second Law and Third Law of Thermodynamics 5.43

Here T is the temperature of the final state. Thus, the derivative ∂|W|/∂S is negative i.e., |W| decreases with increasing S. As the system is thermally isolated, its entropy cannot decrease, the greater possibility is that W would be greatest when S is constant throughout the process. Hence, we conclude that the system does maximum work when its entropy remains constant i.e., when the process of reaching equilibrium is reversible. This is called the theorem of maximum work.

5.22

NEGATIVE TEMPERATURES

It can be seen that the Boltzmann equation

( )

n E __ ___ n0 = exp – kT

(1)

formally permits the temperature to take on not only positive but also negative values. From Eqn. (1), we have n E ___ ln __ n0 = – kT Hence, E T = – ______ n k ln __ n0 It can be seen from this expression that if n < n0 then T > 0. If, however, an atomic system happened to exist in which n could be greater than n0, then T will be negative. The more, the energy supplied to the system, the greater will be the number of particles at the highest energy levels. In the limit, we can imagine a state in which all the particles will gather at the highest levels. Such a state is obviously a highly ordered one. We can, therefore, designate the temperature at which the ordered state sets in by (– 0), in contrast to the conventional absolute zero (+ 0). The difference between these two ‘zeros’ is that we approach the first from negative side and the second from positive side. Thus, the temperatures of a system are not limited to 0 to • but extend from + 0 to + •, – • to – 0, + • to – •. The energy S dependence of entropy is shown in Fig. 5.28. T>0 T n0 correspond to all of them whereas for equilibrium we must have the reverse relationship of these quantities.

5.23

THERMODYNAMIC POTENTIALS

We know that the state of a fixed mass of a thermodynamic system can be expressed in terms of its pressure P, volume V, temperature T and entropy S. These are called the thermodynamic variables. These are related as dQ = dU + PdV (1st law of thermodynamics) (1)

and for a reversible process dQ = TdS

nd

(2 law of thermodynamics)

In the four thermodynamic variables P, V, T and S, only two variables are independent and remaining two variables are functions of the first two. These variables are the extensive properties of the system (their values depend on the mass of the system). From Eqn. (1), combining both the laws of thermodynamics TdS = dU + PdV or

dU = TdS – PdV

(2)

This eqn. represents internal energy of the system in terms of all the four thermodynamic variables. But, for the complete knowledge of the behaviour of the system, we require some other relations (in addition to these variables) which are known as thermodynamic potentials (or energy functions). Following are the four main thermodynamic potentials: (i) Internal or intrinsic energy U; (ii) Helmholtz free energy F; (iii) Enthalpy or total heat function H; (iv) Gibb’s free energy G.

5.23.1

Internal Energy U

The internal energy U of a system is a thermodynamic variable. It is the energy due to molecular constitution and molecular motion of the system. The total energy of a system is the sum of internal kinetic energy due to its molecular motion and internal potential energy due to mutual interaction amongst the molecules. It is an extensive property of the system. It is a unique function of the state of the system

The Second Law and Third Law of Thermodynamics 5.45

i.e., its value depends only on the initial and final states of the system and does not depend on the path or the process by which the state of the system is changed. In other words, U is a perfect differential. If the state of a system is changed from an initial equilibrium state to a final equilibrium state through an infinitesimal reversible process, the change in internal energy is given as dU = TdS – PdV (from Eqn. (1))

(3)

In an adiabatic change of a gaseous system TdS = 0 and

dU = – PdV

Thus, in an adiabatic process, the work done by the gas is equal to the decrease in internal energy of the system (or the work done on the gas is equal to the increase in internal energy of the system).

5.23.2

Helmholtz Free Energy F

This is also called the thermodynamic potential at constant volume, and is defined by F = U – TS

(4)

The free energy F is also a unique function of the state of the system. It is a perfect differential. It is also an extensive property of the system. Consider an infinitesimal reversible isothermal change of a system from an initial equilibrium state 1 to the final equilibrium state 2, then F1 = U1 – TS1 F2 = U2 – TS2 (since T is constant) Then, increase in Helmholtz free energy F2 – F1 = (U2 – U1) – T (S2 – S1) or

(dF)T = (dU)T – TdS But from the second law of thermodynamics (dQ)R = TdS

\

(dF)T = (dU)T – (dQ)R But from first law, dQ = dU + dW

5.46 Heat and Thermodynamics \

(dF)T = – (dW)R

or

(dW)R = – (dF)T

(5)

i.e., decrease in Helmholtz free energy in a reversible isothermal process is equal to the work done by the system in that process. In other words, in a reversible isothermal process, the total external work is obtained from the Helmholtz free energy of the system. Hence, Helmholtz free energy of a system is the energy which is available for work in a reversible isothermal process. Thus, this energy is analogous to the potential energy of a system. From Eqn. (4), we have U = F + TS

(6)

Thus, the internal energy of a system has two parts: (a) Helmholtz free energy F (which is available for work in a reversible isothermal process) and (b) latent (or bound) energy TS which is unavailable for the useful work. As the entropy increases, the unavailable energy increases, hence the available energy for work decreases.

5.23.3

Enthalpy or Total Heat Function

It is defined by the equation H = U + PV

(7)

The enthalpy H is also a unique function of the state of the system. It is a perfect differential and also, an extensive property of the system. Consider an infinitesimal reversible isobaric change of a system from an initial equilibrium state to a final equilibrium state 2, then H1 = U1 + PV1 H2 = U2 + PV2

(at constant P)

Increase in enthalpy H2 – H1 = (U2 – U1) + P (V2 – V1) or

(dH)P = dU + PdV But

dQ = dU + PdV (from 1st law of thermodynamics)

Therefore, (dH)P = dQ

(8)

i.e., in a reversible isobaric process, the change in enthalpy of the system is equal to the heat extracted (or heat rejected) by the system. In an adiabatic throttling process, the total initial enthalpy of gas is equal to the total final enthalpy. In this process, the gas is passed through an insulated porous

The Second Law and Third Law of Thermodynamics 5.47

plug from a constant high pressure to a constant low pressure. Let Pi, Vi be the initial pressure and volume of the gas and Pf, Vf, the final pressure and volume. Net external work done by the gas will be (Pf Vf – Pi Vi) which is obtained from the internal energy of the gas (\ the gas is insulated from the surroundings). If Ui and Uf be the initial and final internal energies of the gas, then decrease in internal energy = Ui – Uf. But decrease in internal energy = net external work done by the gas or

Ui – Uf = Pf Vf – Pi Vi

or

Ui + Pi Vi = Uf + Pf Vf

or

Hf = Hi

(9)

i.e., enthalpy remains constant in an adiabatic throttling process. Enthalpy also remains constant in an isothermal change of an ideal gas, because dU = 0 i.e., U1 = U2 and from Boyle’s law, P1 V1 = P2 V2 \

U1 + P1 V1 = U2 + P2 V2

or

H1 = H2

in an isothermal change. The locus of points representing the equilibrium states of the same enthalpy of a thermodynamic system is called an iso-enthalpic curve.

5.23.4

Gibb’s Free Energy G

It is also called the thermodynamic potential at constant pressure. It is defined by the equation G = H – TS = U +PV – TS = F + PV

(10)

where the symbols have their usual meanings. G is also a unique function of the state of the system. It is a perfect differential and an extensive property of the system. Consider an infinitesimal reversible isothermal and isobaric change of a system from an initial equilibrium state 1 to a final equilibrium state 2. Then G1 = U1 + PV1 – TS1 G2 = U2 + PV2 – TS2 (at constant P and T).

5.48 Heat and Thermodynamics \ Increase in Gibb’s free energy, in this case is G2 – G1 = (U2 – U1) + P (V2 – V1) – T (S2 – S1) or

(dG)T, P = dU + PdV – TdS nd

But from 2 law (dQ)R = TdS st

and from 1 law \

(dQ)R = dU + (dW)R (dG)T, P = (dQ)R – (dW)R + PdV – TdS = – (dW)R + PdV

(11)

If in the above reversible process, the work is done by the system only in expansion at constant pressure then (dW)R = PdV and

(dG)T, P = 0

(12)

Thus, in a reversible change at a constant temperature and pressure, the Gibb’s free energy of a system remains constant if the work is done by the system only in expansion at constant pressure. But if, the process is natural (i.e., irreversible), the entropy increases and Gibb’s free energy of the system decreases (see Eqn. (10)).

5.24

MAXWELL’S RELATIONS

Consider a homogenous system whose volume V is the only external parameter of relevance. The starting point of our discussion is the fundamental thermodynamic relation for a quasi-static infinitesimal process dQ = TdS = dE + pdV

(1)

That this equation gives rise to a wealth of other relations will be seen subsequently.

1.

S and V as Independent Variables

Equation (1) can be written as dE = TdS – pdV

(2)

This shows how E depends on independent variations of the parameters S and V, i.e., E = E (S, V) fi

( )

∂E dE = ___ ∂S

( )

∂E dS + ___ ∂V V

dS S

(3)

The Second Law and Third Law of Thermodynamics 5.49

Since Eqns. (2) and (3) must be equal for all possible values of dS and dV, it follows that the corresponding coefficients of dS and dV must be the same. Hence,

( ) ( ) ∂E ___ ∂S

∂E ___ ∂V

=T

V

(4) =–p

S

The important content of Eqn. (2) is that the combination of parameters on the r.h.s. is always equal to the exact differential of a quantity which is energy E in this case. Hence, the parameters T, S, p and V which occur in the r.h.s. of Eqn. (2) cannot be varied completely arbitrarily; there must exist some combination between them to guarantee that their combination yields the differential dE. To obtain this connection, it is only necessary to note that the second derivatives of E must be independent of the order of differential i.e., ∂2E _____ ∂2E _____ = ∂V∂S ∂S∂V

( )( ) ( )( ) ∂ ___ ∂V

or

S

∂E ___ ∂S

∂ = ___ ∂S V

∂E ___ ∂V

V

S

Hence, using Eqn. (4), we get

( ) ( ) ∂T ___ ∂V

2.

∂p = – ___ ∂S S

(5) V

S and p as Independent Variables

We have pdV = d (pV) – Vdp We now return to Eqn. (2) and transform this into an expression involving dp rather than dV. Then dE = TdS – pdV = TdS – d (pV) + VdP or

d (E + pV) = TdS + Vdp We can write this as dH = TdS + Vdp

(6)

where we have used the definition H ∫ E + pV,

the enthalpy

(7)

5.50 Heat and Thermodynamics Considering S and p as independent variables H = H (S, p)

( )

∂H dH = ___ ∂S

\

p

( )

∂H dS + ___ ∂p

dp

(8)

S

Comparing Eqns. (6) and (8), we get

( ) ∂H ___ ∂S

p

( )

S

∂H ___ ∂p

=T (9) =V

Again, the important aspect of Eqn. (6) is the fact that the combination of parameters on the r.h.s. is equal to the exactly differential of a quantity, which has been designated as H. Then ∂2H ∂2H _____ _____ = ∂p∂S ∂S∂p fi

( ) ( ) ∂T ___ ∂p

∂V = ___ ∂S S

(10)

p

(using Eqn. (9).

3. T and V as Independent Variables We now transform Eqn. (2) into an expression involving dT rather than dS. We can write dE = TdS – pdV = d (TS) – SdT – pdV or

dF = – SdT – pdV

(11)

using F ∫ E – TS

(12)

The function F is called “Helmholtz free energy”. Considering T and V as independent variables F = F (T, V)

( )

∂F dF = ___ ∂T

( )

∂F dT + ___ ∂V V

T

dV

(13)

The Second Law and Third Law of Thermodynamics 5.51

Comparing Eqns. (11) and (13), we get

( ) ( ) ∂F ___ ∂T

∂F ___ ∂V

=–S

V

(14) =–p

T

The equality ∂2F _____ ∂ 2F _____ = ∂V∂T ∂T∂V then gives

( ) ( ) ∂S ___ ∂V

∂p = ___ ∂T T

(15) V

4. T and p Independent Variables We finally transform Eqn. (2) into an expression involving dT and dp (rather than dS and dV). Thus, we can write dE = TdS – pdV = d (TS) – SdT – d (pV) + Vdp or

dG = – SdT + Vdp

(16)

where we have introduced the definition G ∫ E – TS + pV

(17a)

called the Gibb’s free energy. or

G = H – TS

(17b) (using Eqn. (17))

or

G = F + pV

(17c)

(using Eqn. (12)) Considering T and p as independent variables. G = G (T, p) \

( )

∂G dG = ___ ∂T

p

( )

∂G dT + ___ ∂p

dp T

(18)

5.52 Heat and Thermodynamics Comparing Eqns. (16) and (18), we get

( ) ∂G ___ ∂T

( ) ∂G ___ ∂p

p

=–S (19) =V

T

The equality ∂2G _____ ∂2G _____ = ∂p∂T ∂T∂p then implies

( ) ( )

∂S – ___ ∂p

∂V = ___ ∂T T

(20)

p

Thus, we have derived the important relations in Eqns. (5), (10), (15) and (20) which are summarized as follows

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ∂T ___ ∂V

∂p = – ___ ∂S S

∂T ___ ∂p

∂V = ___ ∂S S

p

∂S ___ ∂V

∂p = ___ ∂T T

V

∂S – ___ ∂p

∂V = ___ ∂T T

p

(21a) V

(21b) (21c) (21d)

These are known as Maxwell’s relations.

5.25 1.

HOW TO REMEMBER THE MAXWELL’S RELATIONS ∂(p, V) ∂(T, S) ______ ______ = ∂(x, y) ∂(x, y)

2. We have Jacobian ∂(X, Y) ______ = ∂(x, y)

| | ( ) ( ) ∂X ___ ∂x ∂X ___ ∂y

y

∂Y ___ ∂x ∂Y ___ ∂y

y

( ) ( ) ( )( ) ( )( )

∂X = ___ ∂x

x

y

x

∂Y ∂X ___ – ___ ∂y x ∂y

x

∂Y ___ ∂x

y

The Second Law and Third Law of Thermodynamics 5.53

3. For x and y, we have four variables T, S, p, V. There are in all six possible conbinations (T, S), (T, p), (T, V) (S, p), (S, V) (p, V) 4. The pairs (p, V) and (T, S) for (x, y) lead to trivial relations, because from 1, ∂(p, V) ∂(T, S) ______ = 1 = ______ ∂(p, V) ∂(T, S) Therefore there remain four combinations (T, V), (T, p), (S, V) and (S, p) Thus we have four cases as follows. 5. Because from 1 and 2:

( )( ) ( )( ) ∂T ___ ∂x

y

∂S ∂T ___ – ___ ∂y x ∂y

∂S ___ ∂x

x

y

( )( ) ( )( )

∂p = ___ ∂x

∂p ∂V ___ – ___ ∂y x ∂y

y

x

∂V ___ ∂x

y

I. x = T, y = V This gives (using (5)).

( ) ( ) ∂S ___ ∂V

(

∂p = ___ ∂T T

(first relation) V

∂T ∂V because T and V are independent variables ___ = 0 = ___ ∂V ∂T

II. x = T, y = p This gives

( ) ( ) ∂S ___ ∂p

∂V = – ___ ∂T T

p

(second relation)

III. x = S, y = V

( ) ( )

∂T – ___ ∂V

or

∂p = ___ ∂S S

(using (5)) V

∂S ∂V ___ = 0 = ___ ∂V ∂S ∂p ∂T ___ = – ___ (third relation) ∂V S ∂S V

( ) ( )

)

5.54 Heat and Thermodynamics IV. x = S, y = p This gives (using (5))

( ) ( ) ∂T ___ ∂p

∂V = – ___ ∂S S

(fourth relation) p

T A(F) The four characteristic functions (U, H, F and V G) and associated Maxwell relations need to be remembered. A useful mnemonic device for this purpose (Fig. 5.29), is called as a VAT – VUS diagram because of the labels on the top and left U G side. A characteristic function is indicated at the midpoint of each side and its thermodynamic coordinates at the ends of the side. So for example, the Helmholtz function A (i.e., F) is a S H P function of thermodynamic coordinates V and T, and the internal energy function U is a function of Fig. 5.29 The VAT-VUS diagram thermodynamic coordinates V and S. Now, consider for example, the internal energy function U (V, S). The differential dU equals the sum of terms including dV and dS. The coefficient of dV is found by the arrow that connects V to P. Notice that the connection goes against the arrow, so the coefficient of dV is not P, but – P. Similarly, the coefficient of dS is found by going in the direction of the arrow that connects S to T (i.e., it is T).

\

dU = TdS – PdV The Maxwell relations are obtained by applying eqns. dZ = Mdx + Ndy

( ) ( ) ∂M ___ ∂y

∂N = ___ ∂x x

y

to each of the four thermodynamic potential functions. Consequently, we have 1.

dU = TdS – PdV; hence

2.

( ) ( ) ∂T ___ ∂V

∂p = – ___ ∂S S

V

dH = TdS + Vdp; hence

3.

( ) ( ) ∂T ___ ∂p

∂V = ___ ∂S S

p

dA = – SdT – pdV; hence

( ) ( ) ∂S ___ ∂V

∂p = ___ ∂T T

V

The Second Law and Third Law of Thermodynamics 5.55

4.

dG = – SdT + Vdp; hence

( ) ( ) ∂S ___ ∂p

∂V = – ___ ∂T T

p

(The four equations on the right are Maxwell’s relations).

5.26 THERMODYNAMIC VARIABLES IN TERMS OF THERMODYNAMIC POTENTIALS We know that dU = TdS – pdV

(1)

dF = – SdT – pdV

(2)

dH = TdS + Vdp

(3)

dG = – SdT + Vdp

(4)

From Eqn. (1)

( )

( )

∂U T = ___ ; ∂S V

∂U p = – ___ ∂V

( )

∂F S = – ___ ∂T

S

From Eqn. (2)

( )

∂F p = – ___ ; ∂V T

V

From Eqn. (3)

( )

( )

∂H V = ___ ∂p

( )

∂G S = – ___ ∂T

∂H T = ___ ; ∂S p

S

From Eqn. (4)

( )

∂G V = ___ ; ∂p T

p

Thus, the thermodynamic variables S, p, T, V can be obtained from differentiation of thermodynamic potentials (U, F, G, H) as

( ) ( ) ( ) ( ) ( ) ( )

∂F S = – ___ ∂T

∂G = – ___ ∂T V

∂U T = ___ ∂S

∂H = ___ ∂S

V

∂U p = – ___ ∂V

S

p

∂F = – ___ ∂V

( ) ( )

∂H V = ___ ∂p

∂G = ___ ∂p S

p

T

T

5.56 Heat and Thermodynamics 5.27

TdS EQUATIONS

We know that the thermodynamic coordinates S, p, T and V of a thermodynamic system are related to each other and any one of these coordinates can be expressed as a function of the other two. Thus, S = S (T, V) or

S = S (T, P) On this basis, two TdS equations can be obtained.

First TdS Equation

( )

∂P TdS = CV dT + T ___ ∂T

dV V

Second TdS Equation

( )

∂V TdS = CP dT – T ___ ∂T

dP

P

These can be derives as follows. Let entropy S of a system be a function of its temperature T and volume V: S = S (T, V)

( ) ( ) ( ) ( )

∂S dS = ___ ∂T

\

∂S dT + ___ ∂V V

∂S TdS = T ___ ∂T

or But and

∂S dT + ___ ∂V V

dV T

T∂S = ∂Q

( ) ∂Q ___ ∂T

V

= CV

( )

T

( )

V

∂S TdS = CV dT + T ___ ∂V

\

dV

T

dV

Now, from Maxwell’s relation

( ) ( ) ∂S ___ ∂V

T

∂P = ___ ∂T

V

Hence, ∂P TdS = CV dT + T ___ ∂T This is the first TdS equation.

dV

The Second Law and Third Law of Thermodynamics 5.57

Now, let the entropy S of a system be a function of its temperature T and pressure P.

5.28

EXPRESSIONS FOR C P C V

We know that if the molar specific heats at constant pressure and at constant volume are CP and CV respectively then, the heat required to raise the temperature of 1 mole of gas at constant pressure is dQ = CP dT and the heat required at constant volume is dQ = CP dT and the heat required at constant volume is dQ = CV dT then, by definition ∂Q ∂Q CP = ___ , CV = ___ ∂T P ∂T V ∂Q ∂Q and CP – CV = ___ – ___ (1) ∂T P ∂T V But, from second law of thermodynamics

( ) ( ) ( ) ( )

dQ = TdS Consequently, Eqn. (1) becomes

( ) ( )

∂S CP – CV = T ___ ∂T

∂S – T ___ ∂T P

(2) V

But entropy S = S (T, V), therefore

( ) ( ) ( ) ( ) ( )( ) ∂S dS = ___ ∂T

∂S ___ ∂T

or

∂S = ___ ∂T P

V

∂S dT + ___ ∂V

∂S + ___ ∂V V

T

dV

T

∂V ___ ∂T

P

Multiplying both sides by T, we get

( ) ( ) ( )( ) ( ) ( ) ( )( ) ∂S T ___ ∂T

or

∂S T ___ ∂T

P

∂S – T ___ ∂T

∂S = T ___ ∂T P

V

∂S = T ___ ∂V

∂S + T ___ ∂V V

T

∂V ___ ∂T

T

∂V ___ ∂T

P

(3)

P

From Eqns. (2) and (3)

( ) ( )

∂S CP – CV = T ___ ∂V

∂V – ___ ∂T T

(4) P

But from Maxwell’s relations

( ) ( ) ∂S ___ ∂V

T

∂P = ___ ∂T

(5) V

5.58 Heat and Thermodynamics

( )( )

∂P CP – CV = T ___ ∂T

\

V

∂V ___ ∂T

(6) P

The relations (5) and (6) are called the specific heat equations. Further, from first and second laws, TdS = dU + PdV

( ) ( )

∂S T ___ ∂V

\

T

∂U = ___ ∂V

+P T

Substituting this value in Eqn. (4), we get CP – C V =

(a)

[( ) ] ( ) ∂U ___ ∂V

+P

T

∂V ___ ∂T

(7) P

CP – CV For a Perfect Gas

For 1 mole of perfect gas, the equation of state is PV = RT

( ) ( )

\ and

∂P ___ ∂T

V

∂V ___ ∂T

R = __ P P

R = __ V

Hence, from Eqn. (6) R R CP – CV = T × __ × __ V P TR2 TR2 = ____ = ____ = R PV RT CP – CV = R

or for a perfect gas.

(b) CP – CV For a van der Waals Gas For 1 mole of a real (i.e., van der Waals) Gas,

( P + ___Va ) (V – b) = RT 2

where a and b are van der Waals constants. or

RT ( P + ___Va ) = _____ V–b 2

(8)

The Second Law and Third Law of Thermodynamics 5.59

Differentiating the above equation with respect to T at a constant volume

( ) ∂P ___ ∂T

R = _____ V–b V

(9)

Similarly, differentiating with respect to T at a constant pressure

( )

or

( )[ ∂V ___ ∂T

P

]

( )

or

( )

∂V 2a ∂V RT R 0 – ___3 ___ = – _______2 ___ + _____ V –b ∂T ∂T P P V (V – b) 2a RT R _______ – ___3 = _____ 2 V –b (V – b) V R ______ (V – b) ∂V ___ = _____________ 2a RT ∂T P _______ – ___3 2 (V – b) V

[

]

( ) ( )

∂P Substituting the values of ___ ∂T we get

V

∂V and ___ ∂T

(10)

from Eqns (9) and (10) in Eqn. (6), P

R ______ (V – b) R CP – CV = T ______ × _____________ (V – b) 2a RT _______ – ___3 2 (V – b) V R = ________________ 2 2a (V – b) 1 – ___3 × _______ RT V 2 Now, since b 0. But water contracts on melting i.e., DV < 0. The slope of the melting curve in this case is negative. The three curves OA, OB and OC are given by the Gibbs free energy equation per unit mass of each phase gliq – ggas = 0 (Vaporisation) gsolid – ggas = 0 (Sublimation) gsolid – gliq = 0 (Fusion or melting) These three equations must meet at the triple point. The triple point is independent of pressure. For water it is nearly 0°C. The dotted curve OA¢ is merely the continuation of OA. It represents the vapour pressure of super cooled liquid.

5.30

CLAUSIUS CLAPEYRON EQUATION

From Maxwell’s relations

or

( ) ( ) ( ) ( ) ∂S ___ ∂V

T

∂P = ___ ∂T

∂S T ___ ∂V

T

∂P = T ___ ∂T

(1)

V

(2) V

But from 2nd law of thermodynamics TdS = dQ

The Second Law and Third Law of Thermodynamics 5.61

therefore,

( ) ( ) ∂Q ___ ∂V

T

∂P = T ___ ∂T

(3) V

( )

∂Q Here ___ represents the quantity of heat absorbed or liberated per unit change ∂V T in volume at constant temperature. This means that at constant temperature, the heat absorbed or liberated at constant temperature must be the latent heat and change in volume must be due to a change of state. Considering a unit mass of the substance, let L be the latent heat when the substance changes in volume from V1 to V2 at constant temperature. Then dQ = L dV = V2 – V1

and

Substituting these in Eqn. (3), we get

( or or

L _______ V2 – V1

) ( ) ∂P = T ___ ∂T T

(4) V

dP L _______ = T ___ V2 – V1 dT dP _________ L ___ = dT T (V2 – V1)

(5)

This equation is called the Clausius-Clapeyron latent heat equation.

5.31 FIRST ORDER PHASE TRANSITION AND CLAUSIUS‐CLAPEYRON EQUATION We now consider the first order phase transition which are accompanied by the discontinuities in the first order derivatives of the Gibb’s function. The solidliquid and liquid-gas transitions are the first order phase transition. The thermodynamic state of a homogenous one component system is determined by two thermodynamic variables for example the pressure P and temperature T. Let us consider the two phases 1 and 2 in equilibrium, satisfying the condition G1 (P, T) = G2 (P, T) or

(1)

g (P, T) = G1 (P, T) – G2 (P, T) = 0

Equation (1) can be solved to give the phase-equilibrium curve of P as a function of T: (i) The melting curve in the case of solid-liquid transition or (ii) Pressure curve in the case of liquid-gas transition.

5.62 Heat and Thermodynamics It is also possible that the equilibrium of three phases is governed by the condition G1 (P, T) = G2 (P, T)

(2)

G2 (P, T) = G3 (P, T) These equations have solutions for specific pair of values (Ptr, Ttr) called ‘triple point’ where three phases are simultaneously present in equilibrium. This is depicted in Fig. 5.31 where regions 1, 2 and 3 are three phases of the same system. It is to be noted that the equilibrium of more than three phases of the same system is impossible. The molar Gibbs free energies of the two phases are plotted in Fig. 5.32 as function of temperature T at constant pressure. The curves intersect at the Fig. 5.31 Phase diagram of one-component temperature T0 at which the two phases system can exist in equilibrium. At the temperatures below T0, the 1st phase is stable and above T0, the 2nd phase is stable. This is due to the fact that the stable state is one where G is smaller. Since entropy is a positive quantity, G must be decreasing function of T. If the phase 1 is stable for T < T0, then at the point of interaction of the two phases.

( ) ( ) ∂G1 ____ ∂T

∂G2 > ____ ∂T P

(3) P

Let us change the temperature of each of the phases by dT and the pressure by dP in such a way that the two phases continue to be in equilibrium in new states. Then, in the new condition, the Gibbs free energies of the two phases will obviously be equal, i.e., g (P + dP, T + dT) = 0

(4)

g can be expanded in a Taylor series as

( )

∂g g (P + dP, T + dT) = g (P, T) + ___ ∂P

( )

∂g dP + ___ ∂T T

dT = 0 P

The function g (P, T) is always zero at equilibrium, and so

( ) ∂g ___ ∂P

( )

∂g dP + ___ ∂T T

dT = 0 P

(5)

The Second Law and Third Law of Thermodynamics 5.63

G

V

S

Phase 2

Phase 1

Phase 2

Phase 1

Phase 2

Phase 1

Cp

Phase 2

Phase 1 To

T

Fig. 5.32 The variation of G, V, S and Cp with T during the first order transition

Now, since

( ) ( )

∂g v = ___ ∂P

(6)

T

∂g s = – ___ (7) ∂T P where v and s are the molar volume and molar entropy respectively, Eqn. (5) gives the equation of the phase transition boundary and

(s2 – s1) dT = (v2 – v1) dP or

s2 – s1 Ds dP ______ ___ = v – v = ___ dT Dv 2 1

(8)

where s1, v1 and s2, v2 are the molar entropies and molar volumes of the two phases. Equation (8) is the Clausius-Clapeyron equation.

5.64 Heat and Thermodynamics The phase transition is accompanied by the evolution or absorption of a certain quantity of heat (called latent heat). We know that the pressure P and temperature T remain constant during the phase-transition, therefore, T (s2 – s1) ∫ (H2 – H1) ∫ L

(9)

where L = T (s2 – s1) is the latent heat per unit mass of the transition and the quantity (H2 – H1) is the difference between the enthalpies of the two coexisting phases. Substituting this in Eqn. (8), we get dP __ L 1 L ___ = × _______ = ____ dT T ( v2 – v1) TDv

(10)

Thus, the latent heat L and volume change Dv in a first order transition are isolated to the slope of the phase equilibrium curve by the Clausius Clapeyron equation. During the first order transition, the Gibbs function is continuous but the first order derivative of the Gibbs function changes discontinuously. These derivatives are ∂g (i) the volume v = ___ which changes from v1 to v2 and ∂P T ∂g (ii) the entropy s = – ___ which changes from s1 to s2. ∂T P These are demonstrated in Fig. 5.32. There is a jump in both specific volume v and entropies. We are also explain the behaviour of other thermodynamic properties e.g., specific heat Cp during the first order transition. We know that during the transition both P and T are constant. Thus, when P is constant, dT = 0 and the specific heat Cp is

( ) ( )

( )

∂s CP = T ___ ∂T

=• P

The variation of CP is also shown in the Fig. 5.32. It remains finite upto the transition temperature in the phase 1, becoming infinity at the transition temperature and then returns to finite value for phase 2. From Eqn. (3), it is clear that s2 > s1 and hence the latent heat is always positive. The Eqn. (10) may be used for melting, vaporization and sublimation.

5.32

SECOND ORDER PHASE TRANSITION

A phase change of first order is defined as one in which the first order derivatives of the Gibbs function change discontinuously at the transition. In first order transition.

The Second Law and Third Law of Thermodynamics 5.65

g2 – g1 = 0

( ) ( ) ( ) ( )

∂g2 – ___ ∂T

∂g1 + ___ ∂T P

L = s2 – s1 = __ T P

∂g2 ___ ∂P

∂g1 + ___ ∂P T

T

= v2 – v1

The investigations with liquid Helium by Ehrenfest (1933) have led to the findings that there are some phenomena where during the phase transition no latent heat is evolved and there is no change in volume. We then define the second order phase transition to be one that takes place at constant temperature and pressure with no change of entropy and volume. The second order phase change takes place in such a way that there is no discontinuity in the first order derivatives of the Gibbs function, but the second order derivatives of Gibbs function change discontinuously. Now, let us consider a phase change between two phases 1 and 2 in equilibrium at a pressure P and temperature T, and let g = G1 – G2 then

g (P, T) = 0

(1)

We now change the temperature of each phase by dT and the pressure by dP such that the two phases continue to be in equilibrium. Then g (P + dP, T + dT) = 0

(2) nd

Making a Taylor-series expansion upto the 2 order,

[

∂g ∂g g (P, T) + ___ dP + ___ dT ∂P ∂T

]

[

2 ∂2g ∂2g 1 ∂g + __ ____2 (dP)2 + 2 _____ dPdT + ___2 (dT)2 2 ∂P ∂P∂T ∂T

+…=0

] (3)

We know that the function g is always zero at equilibrium. If both ∂g/∂P and ∂g/∂T are finite but the sum of the first order terms is zero i.e.,

( ) ( )

∂g ∂g ___ dP + ___ dT = 0 ∂P ∂T

(4)

This leads to the Clausius Clapeyron equation and corresponds to the first ∂g ∂g order phase transition. However, when ___ and ___ are individually zero, we must ∂P ∂T

5.66 Heat and Thermodynamics ∂g ∂g consider the second order terms. When ___ and ___ are zero both at (P, T) and at ∂P ∂T (P + dP, T + dT), we have ∂2g ∂2g ∂g ____ ___ = 2 dP + _____ dT = 0 ∂P ∂P ∂T∂P

(5)

∂2g ∂2g ∂g _____ ___ = dP + ___2 dT = 0 ∂T ∂T∂P ∂T

(6)

Adding Eqns. (5) and (6), we find that the sum of the second order terms of Eqn. (3) is zero i.e., ∂2g ∂ 2g ∂2g 2 ____ _____ ___ (dP) + 2 (dT)2 = 0 (7) dPdT + 2 2 ∂T∂P ∂P ∂T This corresponds to a phase transition of the second order. If such an equality holds for an arbitrary dP and dT, following conditions must be satisfied: ∂2g ___ >0 ∂T2 (8) ∂2g ____ >0 ∂P2 and ∂2g ∂2g 2 ∂2g ____ ___ _____ – >0 (9) ∂T∂P ∂T2 ∂P2

( )

and Now, we know that

∂2G2 ∂2G1 _____ _____ π ∂T2 ∂T2

(10a)

∂2G1 _____ ∂2G2 _____ π ∂P2 ∂P2

(10b)

CP ∂s ___ = ___ ; T ∂T P

(( )

( )

∂G ___ ∂T

[ ( )]

∂G ∂ = ___ – ___ ∂T ∂T

P

∂2G = – ____2 ∂T

=–s

P

)

(11a)

isothermal compressibility

( ) ( ) ( )

∂v 1 ___ k = – __ v ∂P or

T

( ( ))

∂v kv = – ___ ; ∂P T

∂ ∂G = – ___ ___ ∂P ∂P

∂G v = ___ ∂P

T

2

T

∂G = – ____2 ∂P

(11b)

The Second Law and Third Law of Thermodynamics 5.67

Volume coefficient of expansion

( ) ( ) [( ) ]

∂V 1 ___ b = __ v ∂T or

∂V bv = ___ ∂T

P

∂ = ___ ∂T P

∂G ___ ∂P

T P

2

∂G = _____ ∂T∂P

(11c)

Further, Eqns. (5) and (6), give us two equations. for the phase transition boundary. – [ ∂(Dv)/∂T ]P dP ____________ ___ = (12a) dT [ ∂(Dv)/∂P ]T and

[ ∂(Ds)/∂T ]P dP __________ ___ = dT [ ∂(Dv)/∂T ]P

(12b)

where Ds = s2 – s1 and Dv = v2 – v1. With the help of Eqn. (11), Eqn. (12) can be expressed as v2 b2 – v1 b1 dP __________ ___ = dT v1 k1 – v2 k2 b2 – b1 = ______ k1 – k2

(13)

and CP – CP dP _____________ 2 1 ___ = dT T (v2 b2 – v1 b1) CP – CP 2 1 = __________ Tv (b2 – b1)

(14)

where v1 = v2 ∫ v for the change of the second order. Eqns. (13) and (14) are known as Ehrenfest equations. At a second order phase transition, not only the thermodynamic potentials are continuous but their first derivatives (∂G/∂P)T and (∂G/∂T)P are also continuous, however, their second derivatives (Eqns. 11) undergo a jump. These are shown graphically in Fig. 5.33. Thus, in the first order phase transition, the first derivatives of the Gibbs function change discontinuously while in the second order phase transition, the second derivatives of the Gibbs function change discontinuously.

5.68 Heat and Thermodynamics

Fig. 5.33 Variation of G, V, S and CP during the second order transitions

5.33

THE THIRD LAW OF THERMODYNAMICS

We already that an ordered state corresponds to a lower energy of the particles forming a body. If we cool a body to absolute zero, the maximum conceivable order would be established in the system and the minimum entropy would correspond to this state. On the basis of many experiments conducted at low temperatures, an important conclusion can be reached: “at absolute zero of temperature, any changes in state proceed without a change in entropy”. To remove the lack of uniqueness in the determination of entropy, Nernst, in 1905, formulated the following rule: “The entropy of a system at absolute zero is a universal constant which may be taken to be zero”. This rule is called the third law of thermodynamics. Thus, entropy of a body at a temperature T is T

dQ ST = Ú ___, 0 T

ST = 0 K = 0

Some interesting results follow from the third law.

(1)

The Second Law and Third Law of Thermodynamics 5.69

5.33.1

Heat Capacity Vanishes at Absolute Zero

This can be seen by uniting the heat capacity CR (T) associated with a reversible path R in the form

( ) (

∂S CR = T ___ ∂T

(R)

∂S = _____ ∂ ln T

)

(R)

(2)

As T Æ 0, ln T Æ – • and S Æ 0, so that the derivative or CR also tends to zero. Further, since T

CR S = Ú ___ dT 0 T

(3)

it follows that heat capacities must approach zero, at least, as rapidly as T for the integral to coverage and the entropy to become zero.

5.33.2 Absolute Zero is Unattainable by a Finite Change of Parameters Consider a system being cooled from Ti to Tf by varying a parameter x from xi to xf. From second law: Ti

ST (xi) – S0 (xi) = i

Cx = xi _____ dT T

(4)

Cx = xf _____ dT T

(5)

Ú 0 (initial state entropy)

and Tf

ST (xf) – S0 (xf) = f

Ú 0 (final state entropy)

Since heat capacities are always positive, we will achieve the greater cooling if we require the final entropy to be as small as possible. This means that the process should be adiabatic and reversible (e.g., isentropic demagnetization). Then, ST = ST i

f

and Tf Ti Cx = xf Cx = xi S0 (xi) – S0 (xf) = Ú _____ dT – Ú _____ dT T 0 0 T

(6)

By the third law S0 (xi) = S0 (xf) = 0 Therefore, Tf Cx f ___ dT Ú T dT = Ú ___ 0 0 T

Ti C xi

(7)

5.70 Heat and Thermodynamics If the process continues until Tf = 0, since each of the integrals converges, we have Ti Cxi dT = 0 (8) Ú ___ 0 T However, Cxi is greater than zero for Ti π 0, therefore, Eqn. (7) cannot be true i.e., there is no non-zero solution of the Eqn. (7) and so unattainability of absolute zero from Eqn. (7), consequently, the third law implies that “It is impossible to reduce the temperature of a system to absolute zero in any finite number of operations”.

5.34

APPARENT VIOLATIONS OF THE THIRD LAW

Some systems appear to violate the third law but on closer inspection, the violation is not real. Glass provides the classic example, Vitreous glass is basically SiO2. When liquid SiO2 is cooled, it undergoes a phase change at 290 K to become a crystalline solid. The entropy of this crystalline solid (quartz) decreases continuously to zero as the absolute zero is approached, in accordance with the third law. Molten SiO2 can also be made into glass (amorphous quartz) by rapid cooling. Figure 5.34 explains the difference between the amorphous and the crystalline forms of quartz; the former is in a metastable state while the latter is in the equilibrium state. The third law applies only to systems in equilibrium, and is, therefore, not applicable to glass. There is, therefore, no question of glass violating the third law.

Fig. 5.34 Schematic drawing to show that glass is in a metastable state as compared to a crystal which is in a state of stable equilibrium. Plotted here is the free energy of the system, as a function of what is called the configuration coordinate. The latter quantity is pretty complicated but we need not go into all that here. The figure is enough to illustrate that glass is not in an equilibrium state, and therefore, need not respect the third law. [After G. Venkataraman, “A Hot Story” Universities Press (India) Ltd. (1993)]

The Second Law and Third Law of Thermodynamics 5.71 Example 5.3 A reversible heat engine is working between temperatures 167°C and 57°C. if the work done in a cycle is 12 Joules, find how much heat is taken in at the higher temperature. Solution Let Q1 be the heat taken in at temperature T1 and Q2 be the heat rejected at temperature T2, where

T1 = 273 + 167 = 440°K T2 = 273 + 57 = 330°K The efficiency of the heat engine is T2 Q2 h = 1 – ___ = 1 – __ T1 Q1 330 1 = 1 – ____ = __ (i) 440 4 Now, the work done in a cycle is W = 12 Joules which is done by taking the amount of heat Q = Q1 – Q2. Thus, W = JQ W Q = Q1 – Q2 = __ J

or

12 Joules = _____________ = 2.87 Calories 4.18 Joules/Cal

(ii)

From (i) and (ii) Q1 – Q2 h = _______ Q1 or

Q1 – Q2 Q1 = _______ h = 2.87 × 4 = 11.48 Cal.

A Carnot engine working between 100°C and 300°C absorbs 750 J heat from the high temperature source. Calculate the work done by the engine, heat rejected and efficiency of the engine. Solution We have efficiency Example 5.4

W T2 – T1 h = ___ = ______ T2 Q2 573 – 373 200 = _________ = ____ = 0.349 573 573

5.72 Heat and Thermodynamics Work done by the engine W = 0.349 Q2 = 0.349 × 750 J = 262 J Let heat rejected by the engine be Q1. Then Q2 – Q1 T2 – T1 _______ = ______ T2 Q2 or

Q1 T1 ___ = __ Q2 T2

or

T1 Q2 Q1 = _____ T2 Thus, heat rejected by the engine 373 = ____ × 750 J 573 = 488 J

Example 5.5 Calculate the difference in temperature of water at the top and bottom of a rainfall of height 420 m (specific heat of water = 103 cal/kg°C) J = 4.2 J/Cal Solution Let m kg of water falls from a height h metres. The decrease in potential energy of water = increases in kinetic energy W = mgh joules. This, energy will convert into heat. If t°C is the rise in temperature of water,, heat developed Q = mct cal.

From W = JQ,

mgh = J × mcT gh 4.8 × 420 t = ___ = ________3 JC 4.2 × 10

\

= 0.98°C. A lead bullet with velocity 224 m/s strikes a target. If entire energy produced in impact remains with the bullet, calculate the rise in temperature, specific heat of lead = 0.03 cal/g°C, J = 4.2 Joule/cal. Solution The kinetic energy of bullet Example 5.6

= 1/2 mv2 joules = 1/2 mv2/J calories If it increases the temperature of the bullet by Dt°C, the heat produced Q = mCDt cal.

The Second Law and Third Law of Thermodynamics 5.73

From 1st law of thermodynamics, 2 1 mv __ ____ = mCDt 2 J 1 v2 Dt = __ _____ 2 J×C (224)2 1 = __ × _____________ = 199°C. 2 4.2 (0.03 × 103)

or

100 g steam at 100°C is converted into water at the same temperature. The latent heat of steam is 540 kcal/kg. Calculate the change in entropy.

Example 5.7 Solution

Given

m = 100 g = 0.1 kg T = 100°C = 373 K L = 540 kcal/kg

Decrease in entropy dQ Q mL DS = Ú ___ = __ = ___ T T T 0.1 × 540 = ________ 373 = 0.145 kcal/K. Two bodies of unit mass and specific heat C are at temperatures T1 and T2 respectively (T1 > T2). If they are used as a source of heat and sink for a reversible engine, show that the maximum work obtained from such arrangement is Example 5.8

_____

C [ T1 + T2 – 2 ÷T1 T2 ] Let T be the common temperature of two bodies such that T1 > T > T2. The change in entropy of the composite system is

Solution

T

T

mCdT mCdT DS = Ú ______ + Ú ______ T T T1 T2 = mC ln (T/T1) + mC ln (T/T2) = mC [ln (T/T1) + ln (T/T2)] where m is the mass of the body (here m = 1). Since the engine operates in a reversible cycle, the entropy change of the composite system is zero. Thus, C [ln (T/T1) + ln (T/T2)] = 0 or or

ln (T2/(T1 T2)) = 0 T2 e° = _____ T1 T2

5.74 Heat and Thermodynamics or

T2 _____ =1 T1 T2

_____

T = ÷T1 T2

or

Maximum work is obtained when engine operates as a reversible cycle. The work done is given by W = Heat received – Heat rejected = C (T1 – T) – C (T – T2) _____

= C [ T1 + T2 – 2 ÷T1 T2 ] = C [T1 + T2 – 2T] Example 5.9 Two bodies of equal mass m and specific heat C are at temperatures T1 and T2 (T1 > T2). They are brought in thermal contact such that they attain an equilibrium temperature. Show that the change in entropy is

[

(T1 + T2)/2 _____ 2mC ln _________ ÷T1 T2

]

Solution Let the equilibrium temperature be T such that T1 > T > T2. From the law of conservation of energy, we have

mC (T1 – T) = mC (T – T2) or

T 1 + T2 T = ______ 2 The total change in entropy in this case is T

(i)

T

mCdT mCdT DS = Ú ______ + Ú ______ T1 T T2 T = mC ln (T/T1) + mC ln (T/T2) = mC ln (T2/(T1 T2))

(ii)

Substituting (i) in (ii)

[ [

(T1 + T2)2 DS = mC ln ________ 4 T1 T2

]

(T1 + T2)/2 _____ = mC ln _________ ÷T1 T2

[

]

2

(T1 + T2)/2 _____ = 2mC ln _________ ÷T1 T2

]

The Second Law and Third Law of Thermodynamics 5.75 Example 5.10 A refrigerator operating on the reverse Carnot cycle removes 2000 J/min of heat from a reservoir at 35°C and rejects heat to a reservoir at 100°C. Calculate the heat rejected to the hot reservoir and the power required. Solution Efficiency of a Carnot engine

T2 Q2 h = 1 – ___ = 1 – __ T1 Q1 or Here

T1 Q1 = Q2 __ T2 T1 = 273 + 100 = 373 K T2 = 273 + 35 = 308 K Q2 = 2000 J/min

2000 100 = _____ J/s. = ____ J/s 3 60 So, the heat rejected to the hot reservoir is T1 Q1 = Q2 __ T2 100 373 = ____ × ____ Joules/s. 3 308 = 40.37 J/s Hence, the power required W = Q1 – Q2 = (40.37 – 33.33) J/s = 7.04 J/s = 7.04 Watt 7.04 = ____ HP 747

( 1 HP = 747 Watts)

= 0.009 HP A Carnot type engine is designed to operate between 480 K and 300 K. Assuming that the engine actually produces 1.2 kJ of mechanical energy per kilocalories of heat absorbed, compare the actual efficiency with a theoretical maximum efficiency. Solution The maximum efficiency Example 5.11

T h – Tc = ______ Th 480 – 300 = _________ = 37.5% 480

5.76 Heat and Thermodynamics Energy output Actual efficiency = ____________ Energy input 1.2 = ________ 1 × 4.184 = 28.68% A copper can of negligible heat capacity contains 1 kg of water just above the freezing point. A similar can contain 1 kg of water just below the boiling point. The two cans are brought into thermal contact. Find the change in entropy of the system. Solution For contents of either can Example 5.12

dQ = mCdT Specific heat of water C = 1 kcal/kg°K for water \

\ 1 kcal = 1 K × 4.184 J

1 cal = 4.184 J mC = 1 kg × 1 K × 4.184 J/kg°K = 4184 J/K

Since the heat capacities of the two masses are equal, the final temperature will be the average of initial temperatures T1i +T2i 273 K + 373 K T1f – T2f = _______ = _____________ = 323 K 2 2 The change in entropy is DS = DS1 + DS2 T1f

T

dT1 2f dT2 = Ú mC ___ + Ú mC ___ T T2 T1i T2i 1

( )

T1f T2i = mC ln ___ + mC ln ___ T1i T2f

( )

[ (

) (

323 K 323 K = (4184 J/K) ln ______ + ln ______ 273 K 373 K = 100 J/K Example 5.13

Prove that

( )

∂V dH = CP dT – T ___ ∂T

dP + V dP P

)]

The Second Law and Third Law of Thermodynamics 5.77 Solution

Differentiating S = f (P, T), we get

( )

T

( )

CP dT dP + _____ T T

∂S dS = ___ ∂P ∂S = ___ ∂P

( )

∂S dP + ___ ∂T

dT P

From Maxwell’s relations,

( ) ( ) ∂S ___ ∂P

∂V = – ___ ∂T T

P

Therefore, CP dT ∂V dS = _____ – ___ T ∂T

( )

dP P

Now, H = U + PV \

dH = dU + PdV + VdP = dQ + VdP = TdS + VdP

[

( ) ] ( )

CP dT ∂V = T _____ – ___ T ∂T ∂V = CP dT – T ___ ∂T

dP + VdP

P

dP + VdP

P

Deduce an expression for change in entropy of a van der Waals gas at constant volume and constant temperature. Solution For a van der Waals gas Example 5.14

( P + ___Va ) (V – b) = RT 2

Let 1 mole of van der Waals gas has pressure P, volume V at temperature T. If CV is the molar specific heat at constant volume and due to change in state, the temperature increases by dT and volume increases by dV, then increase in internal energy of the gas a dU = CV dT + ___2 dV V (since, internal energy changes in expansion of volume against the internal pressure a/V2)

5.78 Heat and Thermodynamics Work done by the gas dW = PdV \ Change in entropy dQ dU + dW dS = Ú ___ = Ú ________ T T a ___ CV dT + 2 dV + PdV V ____________________ =Ú T a ___ T2 V2 P + 2 dV V dT S2 – S1 = Ú CV ___ + Ú ___________ T V1 T T1

(

)

(

or

T2

)

V2

(

RdV dT = Ú CV ___ + Ú ______, T V1 (V – b) T1

a RT P + ___2 = _____ V –b V

)

( )

T2 V2 – b = CV loge __ + R loge ______ T1 V1 – b At a constant volume, T2 S2 – S1 = CV loge __ T1 At a constant temperature,

( )

V2 – b S2 – S1 = R loge ______ V1 – b

Example 5.15 Using the Maxwell’s thermodynamic relations prove that in reversible isothermal expansion of 1 mole of a van der Waals gas from volume V1

( )

V2 – b to V2, the amount of heat transferred is RT loge ______ . V1 – b Solution Let entropy S of the system is the function of temperature T and volume V, i.e.,

S = S (T, V)

( ) ( ) ( ) ( )

∂S dS = ___ ∂T or But

∂S dT + ___ ∂V V

dV

T

∂S ∂S TdS = T ___ dT + T ___ ∂T V ∂V ∂Q ∂S T ___ = ___ = CV ∂T V ∂T V

( ) ( )

T

dV

The Second Law and Third Law of Thermodynamics 5.79

and from Maxwell’s thermodynamic relations

( ) ( ) ∂S ___ ∂V

∂P = ___ ∂T T

V

( )

∂P TdS = CV dT + T ___ ∂T

\

dV

(i)

V

Now, for 1 mole of van der Waals gas

( P + ___Va ) (V – b) = RT 2

a RT P = _____ – ___2 V–b V ∂P _____ R ___ = V –b ∂T

or \

(ii)

From Eqns (i) and (ii) RT TdS = CV dT + _____ dV V–b But for isothermal expansion dT = 0 \

RT TdS = _____ dV V–b

\

RT dV Ú TdS = Ú ____ V1 V – b

V2

or

( )

V2 – b Q = RT loge ______ V1 – b

Example 5.16 0.1 kg of steam at 100°C is converted into ice at – 10°C. Calculate the change in entropy. Given, latent heat of ice = 80 cal/g, latent heat of steam = 540 cal/g, specific heat of ice = 0.5 cal/g°C, specific heat of water = 1 cal/g°C, log 2.63 = 0.4200, log 2.73 = 0.4363, log 3.73 = 0.5717. Solution This process is done in four steps: (i) steam at 100°C converts into water at 100°C (isothermal change) (ii) water at 100°C converts into water at 0°C (iii) water at 0°C converts into ice at 0°C (isothermal change) (iv) ice at 0°C converts into ice at – 10°C. Given mass of steam m = 0.1 kg = 100 g. First Step

Change in entropy dQ Q mL1 DS1 = Ú ___ = __ = ____ T T T1 100 × 540 = _________ = 145 cal/K (decrease) 373

5.80 Heat and Thermodynamics Second Step

Change in entropy Tf DS2 = mC1 loge __ Ti

( )

273 = 2.3026 × 100 × 1 × log10 ____ 373

= 2.3026 × 100 × 1 × [2.4363 – 2.5717] = – 31.18 cal/K or 31.18 cal/K (decrease) Change in entropy

Third Step

dQ Q mL2 DS3 = Ú ___ = __ = ____ T T2 T 100 × 80 = ________ = 29.30 cal/K (decrease) 273 Fourth Step

Change in entropy Tf¢ DS4 = mC2 loge ___ Ti¢

( )

263 = 2.3026 × 100 × 0.5 × log10 ____ 273 = 2.3026 × 100 × 0.5 × [2.4200 – 2.4363} = – 1.88 cal/K or 1.88 cal/K (decrease) \ Total decrease in entropy = (145 + 31.18 + 29.30 + 1.88) = 67.36 cal/K Example 5.17 1 kg water at 0°C is mixed with 1 kg water at 100°C. Calculate the change in entropy of the mixture. Given: specific heat of water = 1 kilo cal/kg°C. log 3.23 = 0.5092, log 2.73 = 0.4362, log 3.73 = 0.5717. Solution If t°C is the temperature of mixture, then heat given by water at 100°C = Heat taken by water at 0°C

or \

1 × 1 × (100 – t) = 1 × 1 × (t – 0) t = 50°C

Increase in entropy in heating 1 kg water at 0°C (= 273 K) to water at 50°C (= 323 K) Tf DS1 = mC loge __ Ti

()

The Second Law and Third Law of Thermodynamics 5.81

( )

323 = 1 × 1 × 2.3026 log10 ____ 273

= 2.3026 × (2.5092 – 2.4362) = 0.168 kilo.cal/K The decrease in entropy in cooling 1 kg water at 100°C (= 373 K) to 50°C (= 323 K) is Tf DS2 = mC loge __ Ti

()

( )

323 = 1 × 1 × 2.3026 log10 ____ 373

= 2.3026 × (2.5092 – 2.5717) = – 0.144 kilo-cal/K \ Net increase in entropy of the system DS = DS1 + DS2 = 0.168 – 0.144 = 0.024 kilo cal/K Example 5.18 Calculate the change in vapour pressure of water when its boiling point changes from 100°C to 110°C. Given: Volume of 1 g water vapour = 1640 cm3 and latent heat of vaporization = 540 cal g–1. Solution

dT = 10 K, T = 100°C = 373 K, V1 = volume of 1 g of water = 1 cm3 V2 = volume of 1 g water vapour = 1640 cm3

\

V2 – V1 = 1639 cm3 = 1639 × 10–6 m3. L = 540 cal/g = 540 × 4.2 = 2268 J/g

\

L dP = _________ dT T (V2 – V1) 2268 × 10 = _______________ 373 × 1639 × 10–6 = 0.371 × 105 N/m2 = 0.371 atmospheres

5.82 Heat and Thermodynamics

QUESTIONS 1. Explain the difference between reversible and irreversible processes. Explain with examples the physical conditions required for a process to be reversible. 2. What is a real engine? Define efficiency and deduce an expression for it. 3. Discuss the working of Carnot’s ideal engine on P-V diagram and deduce an expression for its efficiency. Why is it not possible in practice? 4. What is meant by a P-V diagram or indicator diagram? How is the work done on the system or the work done by the system (during a change of state) obtained from this graph? Show that the work done on the system or the work done by the system is the path function. 5. The efficiency of a Carnot engine can be increased in two ways (i) by decreasing the temperature of the sink T2, keeping the temperature of source T1 constant, or (ii) by increasing the temperature of source T1, keeping the temperature of sink T2 constant. Which of these methods is more effective and why? 6. What is a refrigerator? Show that if an ideal Carnot’s refrigerator working between the temperatures T1 and T2 (where T2 < T1) extracts heat Q2 at temperature T2, the work required to run the refrigerator is T1 – T2 W = Q2 ______ T1

(

7. 8. 9.

10. 11. 12. 13. 14. 15.

)

Obtain an expression for its coefficient of performance. Differentiate between the path function and point function. Define the term entropy. State its units and explain its physical significance. Show that the entropy is a point function of the state of the system i.e., the change in entropy of a substance in change of state does not depend on the reversible path chosen between the two states. Deduce an expression for the change in entropy of a perfect gas in terms of its pressure, volume and specific heats. State and prove Carnot’s theorem. Explain second law of thermodynamics. Explain the need of second law of thermodynamics. State its both statements and show their equivalence. Calculate the change in entropy of the universe in a reversible and irreversible processes. What do you mean by the temperature-entropy diagram? How are the isothermal and adiabatic curves on this diagram represented? Draw T-S

The Second Law and Third Law of Thermodynamics 5.83

16. 17. 18. 19. 20. 21.

diagram from the P-V diagram of a Carnot cycle and use it to calculate the efficiency of the cycle. What is thermodynamic scale of temperature? Explain absolute zero on this scale. Show that the area enclosed by the T-S diagram for a Carnot cycle represents the available energy. State the principle of increase in entropy and explain it with example. Describe briefly the thermodynamic potentials U, F, H and G. Deduce Maxwell’s thermodynamic relations from them. Establish thermodynamic variables S, T, P and V in terms of thermodynamic potentials U, F, H and G. Use Maxwell’s thermodynamic relations to derive the two TdS equations. Hence find the ratio and difference in two specific heats of a gas.

( ) ( )

∂P 22. Show that TdS = CV dT + T ___ ∂T

V

∂V 23. Show that TdS = CP dT – T ___ ∂T

P

dV. dP.

( ) ( )

∂S 24. Establish the thermodynamic relation ___ ∂V the Clausius-Clapeyron equation

T

∂P = ___ ∂T

and use it to deduce

V

dP _________ L ___ = dT T (v2 – v1) Use this expression to explain the effect of pressure on the melting point. 25. Establish Clausius-Clapeyron’s latent heat equation and explain the effect of increase in pressure on (i) freezing point and (ii) boiling point of water. 26. Explain the concept of negative temperatures. 27. What is the third law thermodynamics? Explain. 28. Discuss the fact that the Carnot cycle is the most efficient cycle operating between two fixed temperature reservoirs. 29. What is an internal combustion engine? Explain the working of an Otto engine with proper diagram and deduce the expression for its efficiency. What is the limit of its efficiency? 30. At atmospheric pressure (1.01 × 105 N/m2) and 100°C, 1 cm3 of water forms 1671 cm3 of steam. Latent heat of steam is 540 cal/gm. Calculate the external work needed to convert 1 g of water into steam at one atmospheric pressure. What is the increase in internal energy of water? Mechanical equivalent of heat = 4.2 J/cal. [Ans. 40.16 cal., 499.84 cal.]

5.84 Heat and Thermodynamics 31. Calculate the efficiency of Carnot engine working between 127°C and 27°C. [Ans. 25%] 32. What is the efficiency of a Carnot engine working between 300°C and 15°C? Calculate the work done by the engine if its absorbs 100 cal. if heat at the high temperature. [Ans. 0.497, 207.9 J] 33. An engine operating between 27°C and 227°C develops 74600 watt power. Calculate (i) the efficiency of the engine (ii) amount of heat absorbed (iii) amount of heat rejected [Ans. (1) 40%, (ii) 186500 J/s, (iii) 111900 J/s] 34. Assuming that a domestic refrigerator works as a reversible engine between the temperatures 0°C and 17°C, find the electrical energy required to freeze 1 kg of water at 0°C (latent heat of ice 80 cal/gm, 1 cal = 4.18 J) [Ans. 2.08 × 104 J] 35. 20 g water at 70°C is mixed with 10 g water at 40°C. Find the change in entropy of the system. (Given log 3.13 = 0.4786, log 3.33 = 0.5224, log 3.43 = 0.5353) [Ans. 0.4145 cal/K (increase)] 36. 1 kg of ice at 0°C is converted into water at the same temperature. What is the change in entropy? [Ans. 293 cal/K (increase)] 37. 10 g of steam at 100°C is converted into water at the same temperature. Find the change in entropy. [Ans. 14.48 cal/K (decrease)] 38. Calculate the change in entropy when 1 kg ice at – 10°C is changed into the steam at 100°C. Given specific heat of ice = 0.5 Kcal/kg°C, latent heat of ice = 3.4 × 105 J/kg, latent heat steam = 22.68 × 105 J/kg, J = 4.2 Joule/ cal) [Ans. 8.71 × 103 J/K (increase)] 39. 10 g water at 20°C is converted into ice at – 10°C. Calculate the change in entropy. (log 2.63 = 0.4200, log 2.73 = 0.43622, log 2.93 = 0.4669) [Ans. 3.824 cal/K (decrease)] 40. Give example(s) thermodynamics.

of

apparent

violation

of

the

OBJECTIVE TYPE QUESTIONS 1. The necessary condition for a reversible process is (a) the loss in energy in complete cycle must be zero. (b) the process should be done suddenly.

third

law

of

The Second Law and Third Law of Thermodynamics 5.85

2.

3.

4.

5.

6.

7.

8.

(c) the process should be done slowly so that the working substance remains in thermal and mechanical equilibrium with its surroundings in each intermediate state. (d) the loss in energy must be zero and the process must be quasistatic. The reversible process is (a) the flow of heat from one body to the other when the two bodies at different temperatures are kept in contact. (b) the slow expansion of a gas enclosed in an insulated cylinder kept in contact with a heat source. (c) the adiabatic compression of a gas. (d) the slow movement of frictionless piston out of the perfectly insulating cylinder in contact with a heat source. An ideal gas is compressed isothermally the process is (a) reversible (b) irreversible (c) either reversible or irreversible (d) nothing can be said till other restrictions ate known The area of P-V diagram of a Carnot cycle represents (a) rejected energy per cycle (b) absorbed energy per cycle (c) work done per cycle (d) change in volume per cycle If the temperature of source and sink and T1 and T2 respectively, the efficiency of Carnot engine is (a) 1 – T1/T2 (b) (T1 – T2)/T (d) 1 – T2/T1 (c) T1/T2 A Carnot engine works between ice point (0°C) and steam point (100°C) . Its efficiency is (a) 36.6% (b) 37.3% (c) 26.8% (d) 73.2% If the temperature of sink is 0°K, the efficiency of Carnot engine will be (a) 100% (b) 50% (c) 10% (d) 25% According to Clausius theorem dQ dQ (b) ___ < 0 (a) ___ > 0 T T dQ dQ (c) ___ = constant (d) ___ = 0 T T

5.86 Heat and Thermodynamics 9. The correct relation is dS (a) dQ = ___ T

(b) dQ = SdT

(c) dQ = dU + dW

dV (d) dQ = dU + ___ P

10. In an irreversible process (a) the internal energy of universe increases (b) the temperature of universe increases (c) the entropy of universe increases (d) the heat and internal energy of universe increases 11. The entropy change in a complete Carnot cycle is Q2 (a) zero (b) – ___ T2 Q1 (Q1 – Q2) (c) + ___ (d) ________ T1 T1 12. The entropy of a system in an reversible adiabatic process (a) increases (b) decreases (c) both (a) and (b) (d) none of these 13. The entropy of gas in irreversible adiabatic process (a) decreases (b) increases (c) remains unchanged (d) none of these 14. The entropy of a system in reversible cycle (a) remains unchanged (b) remains zero (c) increases (d) decreases 15. According to the principle of increase in entropy, in a natural process (a) DS = 0 (b) DS > 0 (c) DS < 0 (d) DS £ 0 16. Entropy remains constant in (a) adiabatic process (b) isothermal process (c) isobaric process (d) cyclic process 17. The second law of thermodynamics is dQ dQ (b) ___ > 0 (a) ___ ≥ 0 T T (c)

dQ ___ 0. Since no work is performed and no heat enters the system, dU(b) = 0. (c) Keeping the speck in the enclosure, compress adiabatically and reversibly back to Vi doing work W(c). For this process dS(c) = 0 and dU(c) = W(c). Let the internal energy and entropy now be Uf and Sf. Note that the pressure depends only on the total energy density Y. Therefore, pressure is the same function of volume during the compression as it was during the expansion, that is, W(a) = – W(c). Since no heat enters the system, Uf = Ui and Yf = Yi. The radiation is black both initially and finally. The energy density of black radiation depends only on the temperature (Stefan’s law). It follows that Ti = Tf. Thus, from the facts that in the initial and the final states (i) the volume is the same, (ii) the radiation is black, and (iii) the energy density is the same we can conclude that the initial and final states are identical. Then Sf = Si and dS(b) = 0. Therefore, there could have been no irreversible change in (b) and the radiation must have been black after the expansion. Hence, the black radiation after slow adiabatic expansion remains black, characteristic of the lower temperature.

6.22 Heat and Thermodynamics 6.17

WIEN’S LAW

The derivation of the law depends upon the existence of the Doppler effect which concerns the change in wavelength when radiation is reflected from a moving mirror or wall. Consider radiation moving in the direction MN and being reflected in the mirror at the position S1, which moves a distance z (position S2) with a velocity v perpendicular to its plane Fig. 6.22. We consider the reflection of the beginning A and the end B of the incident wave of wavelength l which is equal to the distance AB. For the incident wave, before reflection, the frequency is v = c/l, where c is the velocity of radiation (or light). S2

S1 M

B N Q

P

A q

q q

N¢

P¢ Q¢ Z

Fig. 6.22 Doppler effect for radiation

The time taken for A to reach P is NP/c, and for B to reach P¢ is

[

(

z z 2z sin2 q 1 1 __ __ _____ _____ ________ + + (l + NN¢ + N¢Q¢ + Q¢P¢) = l + NP – c c cos q cos q cos q

)]

1 = __ c (NP + l + 2z cos q) Therefore, the time interval between the arrival of successive wave beginnings at P and P¢ respectively is 1 __ c (l + 2z cos q) The corresponding frequency of the reflected radiation, v¢, is equal to the reciprocal of this time interval. Thus

Radiation 6.23

c v¢ = __________, l + 2z cos q c l¢ = __ = l + 2z cos q v¢ Equating the time taken for the mirror to move a distance z and for the radiation to move from B to N¢, we get l + (z/cos q) __z = ___________ v c Therefore, 2lv cos q l¢ = l + ___________ c – (v/cos q)

(1)

Since v > 1 for lT 1

8pkT 8pch ____ lkT _____ Y(Planck) = 4 Æ _____ 5 ◊ l ch l lT >> 1 l

6.21

DETERMINATION OF THE SOLAR CONSTANT

The fraction dI/I of the incident radiation I absorbed is experimentally found to be proportional to the thickness of the medium dz that is traversed. Thus – (dI/I) = m dz

(1)

where the constant m is called the absorption coefficient. Integration gives the Lambert law I = I0 exp (– mz)

b ec

H

Hs

where I0 is the value of I at z = 0. The solar radiation traverses the atmosphere before it reaches the earth’s surface. The absorption by the atmosphere must therefore be taken into account. When the zenith angle is b (angle made by the sun from the vertical), the length of the atmosphere traversed by the radiation before it reaches earth is H sec b, where H is the height of ‘equivalent’ constant density atmosphere, Fig. 6.29. The density of real atmosphere changes with height. Therefore, it is replaced by an equivalent atmosphere of height H and constant density. We thus have

(2)

90 – b

Horizon

Fig. 6.29 Equivalent constant density atmosphere and zenith angle

I = I0 exp (– m H sec b) = I0 [exp (– m H)]sec b = I0 asec b, a = exp (– m H)

(3)

where the constant a is called the transmission coefficient. In the absence of the atmosphere the mean value of solar constant would be S0 = 60I0. The corresponding quantity in the presence of atmosphere is given by S = S0 asec b log S = (log a) sec b + log S0

(4)

6.34 Heat and Thermodynamics The measurements with the help of a pyrheliometer of known exposure area yield values of S for different values of b. We can plot a straight line graph between log S and sec b, Fig. 6.30. The intercept of this line on the log S-axis gives the value log S0. Thus, S0 is determined. Its value is about 1.93 cal cm–2 minute–1 or about 1.34 × 106 ergs cm–2 sec–1.

6.22

log S

log So

TEMPERATURE OF THE SUN

sec b

O

Fig. 6.30

Log S vs. sec b graph

Knowledge of S0 enables us to estimate the temperature of the sun. The total energy radiated per second from the surface of the sun is S0 Q = 4pR02 ___ (1) 60 If r is the radius of the sun, then the total radiating surface area is 4pr2 and energy emitted per second per unit area is R02 S0 Q ____ ___ ◊ ___ = (2) 4pr2 r2 60 By Stefan’s law we must have Q ____ = s T4 (3) 4pr2 where T is the surface temperature of the sun. Thus R02 S0 1/4 (4) T = ____2 ◊ ___ s r 60 Here r/R0 is the mean angular radius of the sun at the earth (angle subtended by the solar radius at the earth). Using

(

)

S0 = 1.937 cal cm–2 minute–1 s = 5.735 × 10–5 erg sec–1 cm–2 °K–1 r/R0 = 959≤ = 4.65 × 10–3 radians we get 1.937 × 4.18 × 10–7 T = ___________________________ 60 × (4.65)2 × 10–6 × 5.735 × 10–5

(

)

1/4

= 5742°K where we have multiplied S0 by J = 4.18 × 107 ergs/cal to express it in ergs. This is the brightness temperature of the sun’s surface.

Radiation 6.35

The temperature of the sun can also be estimated by using Wien’s law lmT = constant = 0.293 Abbot’s measurements show that lm = 4753Å for the solar radiation. Hence 0.293 = 6060°K T = __________ 7553 × 10–8 This temperature, called the colour temperature, is nearly 300°K higher than the brightness temperature. This shows that the sun does not radiate like a blackbody. Example

The rate of cooling or warming is given by Newton’s law of cooling

as d __ (DT) = – KDT, dt where K is a constant. On what factors does K depend and what are its dimensions? If at some instant, the temperature difference is DT0, show that at a time t later DT = DT0 e–kT. Solution The constant K depends on the nature of the exposed surface and its area. The dimension of K is (Time)–1 as time enters the denominator of l.h.s. of the equation (and temperature is in the numerator of both sides). Now, d __ (DT) = – KDT dt d(DT) _____ = – Kdt DT

or Integrating

ln DT = – Kt + C where C is the constant of integration. Let DT = DT0 at t = 0. \

ln DT0 = C

and ln DT = – Kt + ln DT0 or

ln DT – ln DT0 = – Kt

or

DT ln ____ = – Kt DT0

or at a later time t.

( )

DT = DT0 e–Kt

6.36 Heat and Thermodynamics

QUESTIONS 1. What is meant by a blackbody? On what factors does the blackbody radiation depend and how? 2. Explain the meaning of emissive power and absorptive power. State the Kirchhoff’s law of blackbody radiation and explain its significance. 3. What is meant by the pressure of radiation? Show that the pressure of radiation of diffused radiations enclosed inside an enclosure is equal to 1/3rd of energy density. 4. Explain Stefan-Boltzmann’s law and prove it thermodynamically. Show that the Newton’s law of cooling is an approximation of it for low temperature difference. 5. State the Stefan’s law of radiation. Describe the experimental method of determination of Stefan’s constant. 6. What is solar constant? How can the surface temperature of sun be determined with its help? 7. State Wien’s distribution law and derive it. 8. For an adiabatic expansion of blackbody radiation in an enclosure, prove that TV1/3 = constant. 9. Calculate the number of modes of vibrations per unit volume of an enclosure in the frequency range v and v + dv. On this basis, derive Rayleigh-Jean’s law and discuss its drawbacks. 10. What is Planck’s quantum hypothesis of radiation? Obtain Planck’s formula for distribution of energy in the blackbody spectrum. 11. Show that the average energy of a Planck’s oscillator of frequency v in thermal equilibrium at an absolute temperature T is given as __

hv E = ________ hv/kT e –1 Hence show that as h tends to zero, the average energy of Planck’s oscillator becomes independent of frequency like the average energy of a classical oscillator. 12. Deduce (i) Wien’s law, (ii) Rayleigh-Jean’s law from the Planck’s distribution law. 13. Write short notes on (i) Kirchoff’s law of radiation (ii) Prevost’s theory of heat exchange (iii) Stefan’s law (iv) Wien’s displacement law

Radiation 6.37

(v) (vi) (vii) (viii)

Wien’s law Rayleigh-Jean’s law Energy of Planck’s oscillator Planck’s law of radiation

OBJECTIVE TYPE QUESTIONS 1. The transfer of heat energy by some sort of collision without gross movement of any part of the body due to temperature difference between the parts of the body is called (a) conduction (b) radiation (c) convection (d) none of these 2. Electromagnetic radiations are emitted from (a) only from the red hot body (b) the antenna of radio and TV (c) only the body at a high temperature present in the surrounding (d) all the above bodies 3. The speed of heat radiations is (b) 3 × 108 cm s–1 (a) 300 ms–1 (c) 3 × 108 ms–1 (d) 3 × 10–10 ms–1 4. The intensity of radiation measured by a thermopile placed at a distance d from the source is I. If the distance d of thermopile is doubled, the intensity of radiation will be I I (a) __ (b) __ 2 4 (c) 2I (d) 4I 5. A perfectly blackbody is (a) good absorber but bad reflector of heat radiation (b) good absorber and good reflector of heat radiation (c) only the good reflector of heat radiation (d) none of these 6. For a perfectly blackbody, the absorption power is (a) 1 (b) more than 1 (c) less than 1 (d) zero 7. Stefan Boltzmann’s law is (b) lmT = constant (a) E = sT 5 4 4 (d) E = sT 4 (c) E = s(T – T0 )

6.38 Heat and Thermodynamics 8. The value of solar constant is (a) 18.6 × 108 Jm–2 minute–2 (c) 8.4 × 104 Jm–2 minute–1 9. Wien’s displacement law is

(b) 2.04 × 104 Jm–2 minute–1 (d) none of these 1 (b) __ l/T = constant 3

(a) lT = constant

l lT (c) ___ = constant (d) ___ = constant TV V 10. In the spectrum of blackbody radiation, the energy distribution can be completely explained by (a) Wien’s law (b) Planck’s law (c) Stefan’s law (d) Rayleigh-Jeans law 11. The number of allowed vibration modes per unit volume in the wavelength range l and l + dl is 8pl2 dl 8p (b) _______ (a) ___4 dl l c3 4pl2 dl 4p (c) _______ (d) ___4 dl 3 c l 12. According to Planck’s hypothesis, the adsorption or emission of energy is (a) zero (b) only hv (c) continuous (d) discrete in integral multiple of hv 13. The average energy of Planck’s oscillator at an absolute temperature T and frequency v is (a) kT (b) hv hv (d) ________ hv/kT e –1

(c) nhv Answers 1. (a); 8. (c);

2. (d); 9. (a);

3. (c); 10. (b);

4. (b); 11. (a);

5. (a); 12. (d);

6. (a); 13. (d).

7. (c);

References

1. Mark W. Zemansky and Richard H. Dittman, Heat and Thermodynamics, McGraw Hill, Inc. (1981). 2. Francis W. Sears and Gerhard L. Salinger, Thermodynamics, Kinetic Theory and Statistical Thermodynamics, Narosa, New Delhi (1986). 3. Richard E. Sonntag, Claus Borgnakke and Gordon J. Van Wylen, Fundamentals of Thermodynamics, John Wiley & Sons, Inc. (2003). 4. G. Venkataraman, A Hot Story, Universities Press (India) Ltd. (1993). 5. B.K. Agarwal, Thermal Physics, Lokbharti Publications, Allahabad (India) (1988). 6. A.K. Saxena, An Introduction to Thermodynamics and Statistical Mechanics, Narosa, New Delhi (2010). 7. R.C. Srivastava, Subit K. Saha and Abhay K. Jain Thermodynamics, A Core Course, Prentice Hall of India, New Delhi (2007). 8. S.K. Roy, Thermal Physics and Statistical Mechanics, New Age International, New Delhi (2001).

Index

A Absolute 1.6 Absolute temperature scale 5.17 Absorption coefficient 6.6, 6.33 Adiabatic 3.43 Adiabatic demagnetization 3.30, 3.41, 3.43 Adiabatic expansion of gases 4.12 Adiabatic process 2.13 B Blackbody 6.5, 6.9 Bolometer 6.3 Boyle’s Law 1.12, 1.21 Boyle temperature 3.13 British thermal unit 1.28 Brownian motion 2.1 Btu 1.27 C Calorie 1.27 Carnot cycle 5.8 Carnot engine 5.7 Carnot steam engine 5.34 Cascade process 3.35 Celsius scale 1.2 Centigrade scale 1.3 Charle’s law 1.13, 2.12 Claude process 3.39 Clausius-clapeyron equation 5.63 Clausius-clapeyron latent heat equation 5.61

Clausius inequality 5.26 Clausius statement 5.5 Clausius theorem 5.24, 5.27 Closed system 1.18 Coefficient of linear expansion 1.30 Coefficient of performance 5.14 Coefficient of pressure 4.35 Coefficient of viscosity 2.41 Coefficient of volume expansion 2.37 Collision cross section 1.24 Compression (or expansion) ratio 5.39 Conduction 6.1 Constant volume gas thermometer 3.35 Continuity of the liquid and gaseous states 4.30 Convection 6.1 Covolume 5.58 CP – CV for a perfect gas 5.58 D Dalton’s Law of partial pressures 2.15 Degrees of freedom 2.17 Differential air thermometer 6.2 E Efficiency of carnot engine 5.11, 5.15 Efficiency of engine 5.7 Ehrenfest equations 5.67 Einstein’s theory of translational brownian motion 2.4 Enthalpy 2.20, 5.46 Entropy 5.27

I.2 Index Equivalence of the thermodynamic scale with the perfect gas scale 5.20 F Fahrenheit scale 1.2 First law of thermodynamics 4.5, 4.14 First TdS equation 5.56

Joule-Thomson coefficient 3.24 Joule-Thomson expansion 3.43 Joule-Thomson inversion temperature 3.23 Joule-Thomson’s effect 3.17

G Gases 3.10 Gas thermometer 3.44 Gibb’s free energy 5.47

K Kamerlingh-Onnes method 3.40 Kelvin scale 1.6 Kelvin statement 5.4 Kinetic theory of gases 2.8 Kirchhoff’s law 6.8

H Heat capacity 1.26, 4.15 Heat capacity at constant pressure 1.26 Heat capacity at constant volume 1.26 Heat engine 5.6 Heat waves 6.1 Helmholtz free energy 5.45, 5.46

L Langevin’s theory of Brownian Motion 2.2 Law of corresponding states 3.15 Linde process 3.39 Liquefaction 3.4 Liquid thermometer 3.44

I Indicator diagram 4.3 Infrared radiations 6.2 Intensive 4.6 Internal energy 4.7 Inversion curve 5.1 Irreversible process 1.20 Isobaric 1.18 Isobaric processes 1.18 Isochoric 4.25 Isochoric processes 1.20, 4.11 Isotherm 4.35 Isothermal 4.11

M Magnetic thermometer 3.45 Magnetocaloric effect 3.30 Maxwell Boltzmann law of distribution of velocities 2.36 Maxwell Boltzmann’s speed distribution function 1.27 Maxwell’s relations 5.52 Monochromatic emissive power 6.6

J Joule’s coefficient 3.17 Joule’s expansion 3.15 Joule’s expansion of an Ideal Gas 3.15 Joule’s expansion of van der Waal’s or Real Gas 3.16 Joule’s law 4.36

O Open system 1.18 Otto cycle 5.37

N Negative temperatures 5.43 Newton’s law of cooling 6.20

P Partial derivative 4.32 Perfect gas 1.12, 4.37

Index Perfect gas equation 2.14 Perfect (or ideal) gas 3.2 Perfect reflector 6.5 Perpetual-motion machine 5.4 Phase change of first order 5.27 Physical significance of entropy 3.18 Planck radiation law 6.32 Planck’s constant 6.30 Pressure law 5.41 Prevost’s theory of exchanges 6.6 Principle of degradation of energy 4.9 Q Quasi-static 4.2 Quasistatic irreversible process 5.2 Quasistatic process 1.20 R Radiation 6.1 Rankine cycle 5.35 Rayleigh-Jeans law 6.29 Real gas 3.1 Reciprocal theorem 4.34 Reciprocity theorem 3.43 Regenerative cooling 3.37 Regenerative cooling method 4.17 Relationship Between Cp and Cv 3.44 Resistance thermometer 2.28, 5.2 Ritchie’s experiment 6.5 S Second law of thermodynamics 5.4 Second order phase transition 5.67 Second TdS equation 5.56 Specific 4.15 Specific enthalpy 1.19 Specific heat capacity 1.3 Specific value 1.7 Standard atmosphere 4.34 Standard fixed point of thermometry 2.33

State function 1.20 State variables 1.17 Stefan-Boltzmann law 6.16 Stefan constant 6.16 Stern’s experiment 1.17 T TdS equations 5.56 Temperature 1.3 Temperature of inversion 3.19 Theorem of maximum work 5.43 Thermal conductivity coefficient 2.42 Thermocouple 1.11 Thermocouple thermometer 3.44 Thermodynamic coordinates 1.19 Thermodynamic potentials 5.44 Thermodynamic system 1.19 Thermopile 6.3 Third law of thermodynamics 5.68 Throttling process 3.17 Torr 1.7 Total differential 4.33 Total emissive power 6.6 Total emissivity 6.9 Total enthalpy 4.15 U Ultraviolet catastrophe 6.30 Universal gas constant 1.16 V van der Waal forces 3.15 van der Waals constants 3.10 van der Waals equation 3.8 van der Waals equation of state 1.24 van der Waals force 1.23 van der Waals interaction 1.22 Vapour 3.6, 3.10 Vapour pressure 3.6 Vapour pressure thermometer 3.44

I.3

I.4 Index Variation of specific heat of gases with temperature 4.22 VAT – VUS diagram 5.54 Virial coefficients 3.7 W Wien formula 6.26

Wien’s displacement law 6.25 Wien’s distribution law 6.25 Z Zartman and KO’s experiment 2.34 Zeroeth law 1.1 Zero 1.6