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IOP Concise Physics

Transformations of Materials

Dimitri D Vvedensky

Chapter 1

Overview of thermodynamics

When a solid is heated, the energy supplied by the heat enhances the vibrations ofthe constituent atoms, which eventually acquire enough energy to break away fromthe solid and form a liquid. When a liquid freezes, the reverse occurs. At sometemperature, the motion of the particles is slow enough for the forces of attraction tohold the particles in the positions of a solid. As the new bonds are formed, heatenergy is released. This illustrates a general principle for such changes of phase: thechange of the state (phase) of a substance requires the absorption or release of heat.During changes of state such as solid–liquid and liquid–vapor, work must beperformed against the cohesive forces of the molecules. The required energy entersthe system in the form of heat (heat of fusion and heat of vaporization). The sameamount of heat that must be provided to melt or evaporate a given quantity ofmaterial is released back in the reverse process of crystallization or condensation.

The transfer of heat and the performance of work on a system are key elements ofthermodynamics. Indeed, thermodynamics provides the driving force towardequilibrium in general, and phase transformations in particular. But the pathwaytoward equilibrium from a near-equilibrium or far-from-equilibrium initial state ismediated by kinetics, i.e. atomistic processes that determine the rate at whichequilibration occurs. For example, phase transformations in materials are oftenaccompanied by the diffusive transport of matter. Where a material has multipleatomic constituents, the solidification process may alter the equilibrium compositionof the solid if the diffusion of some species is pre-empted by too rapid a cooling rate.Regions of a material with different concentrations, and the interfaces between theseregions, are known collectively as the microstructure of the material. The micro-structure and composition of a material determine many of its physical, and all of itsmechanical, properties. Understanding the driving forces for transformations (e.g.thermal, chemical, elastic, interfacial) and the resulting microstructure is the maingoal of the following chapters.

doi:10.1088/2053-2571/ab191ech1 1-1 ª Morgan & Claypool Publishers 2019

https://doi.org/10.1088/2053-2571/ab191ech1

This chapter introduces basic thermodynamics with the goal of establishing amacroscopic theoretical framework for determining the conditions for equilibriumbetween phases. We then turn our attention to statistical mechanics, which providesa more microscopic basis for understanding thermodynamics. Both approaches willbe used in later chapters to study phase transformations and their kinetic pathways.

1.1 Basic concepts and terminology1.1.1 Systems and boundaries

System refers to a region of space that is enclosed by a surface called the boundary ofthe system. The region of space outside the boundary is called the surroundings orthe environment of the system. The system may be a solid, liquid, or gas, or a mixtureof these. The system in fact need not contain any matter at all. For example, anelectromagnetic field in what is otherwise a vacuum can be regarded as a system. Theboundary need not be a fixed surface either in shape or volume. A gas can expand(or contract) against a non-rigid boundary and, in doing so, the occupied volumeenclosed by the boundary increases (or decreases).

The boundary controls the interactions between the system and its surroundings.Apart from interactions that result from the deformation of the boundary, these takethe form of the exchange of matter and energy across the boundary. If there is noexchange of energy across the boundary, the system is said to be thermally isolatedand the boundary is said to be adiabatic. If such energy exchange is permitted, theboundary is said to be diathermal. If matter can be exchanged between the systemand its surroundings, the boundary is said to be permeable and the system is said tobe open. If no such interchange of matter is permitted, the boundary is said to beimpermeable and the system is said to be closed. Thus, there are four types ofboundaries defined in terms of the exchange of matter and energy: (i) adiabaticimpermeable, (ii) adiabatic permeable, (iii) diathermal impermeable and (iv) dia-thermal permeable.

1.1.2 Equilibrium and state variables

The fundamental physical observation that underlies thermodynamics is thetendency of systems to equilibrate, i.e. to come to rest to a ‘state’ defined by itsboundary. In this equilibrium state, the system is completely characterized bymacroscopic quantities known as state variables. These variables have threeimportant properties. Since an equilibrium system is defined to be macroscopicallyat rest, there can be no time-dependence revealed by any macroscopic measurementof that system. Hence, state variables are time-independent. Given this property, andthe fact that state variables are characteristic of the entire system, a system inequilibrium cannot exhibit macroscopic fluctuations in any of its state variables.Thus, state variables are spatially homogeneous. Finally, if the concept of a statevariable is to be practically useful, it must be independent of the manner in whichequilibrium was attained. For example, whether a system is shaken violently orgently perturbed, once equilibrium has been attained (with the same boundary andexternal conditions), all memory of the history of the system is lost and macroscopic


1-2

measurements on the system yield indistinguishable results. Therefore, state varia-bles depend only the equilibrium state of the system and not on the history ofthe system.

Examples of state variables are the pressure P, temperature T and volume V of anequilibrium system. An example of a quantity which is not a state variable is theenergy transferred into or out of a system as it changes from an initial equilibriumstate to an intermediate equilibrium state and then back to the original equilibriumstate. Even though the initial and final states are the same and the total energy(system plus surroundings) must be conserved, the energy exchange depends on theintermediate state and, hence, on the history of the system, which violates the thirdproperty of a state variable discussed above.

There are two types of state variables: extensive and intensive. Extensive variablesare proportional to the amount of material in a system. Intensive variables areindependent of the amount of material present and so have an essentially localcharacter. To determine whether a given state variable X is extensive or intensive,the system is notionally partitioned into a number of equal parts, within each ofwhich the value of the state variable is denoted as Xi. An extensive variable is onewhose value for the entire system is equal to the sum of its values within theindividual parts, i.e.

∑=X X . (1.1)i

i

Examples of extensive variables are volume and total mass. An intensive variable isone whose value within each part is the same as for the entire system, i.e.

=X X . (1.2)i

Examples of intensive variables are pressure and temperature.How many state variables does it take to uniquely define an equilibrium state?

We again appeal to experiment and observe that not all of the state variables areindependent, but that a subset of state variables determines the values of theremaining state variables. This implies that there exists a functional relation—calledan equation of state—among the state variables of the system.

1.1.3 Processes

One of the fundamental tenets of thermodynamics is that systems seek theequilibrium state determined by a particular set of external conditions (e.g. temper-ature and volume). If these conditions are changed, then the system responds byattaining the equilibrium state appropriate to these new conditions. The manner inwhich the system evolves between two equilibrium states is called a process.

A process that is carried out in such a way that the system departs onlyinfinitesimally from equilibrium between specified initial and final states is calledquasistatic. A quasistatic process is thus a succession of neighboring equilibriumstates. The departures from equilibrium are deemed to be small because thesuccession of equilibrium states are produced by small changes in the state variables.


1-3

If there are large departures from equilibrium, then a process is not quasistatic, but issaid to be nonstatic.

The most important processes in thermodynamics are reversible and irreversible.The direction of a reversible process can be ‘reversed’ by infinitesimal changes insome property of the surroundings, with no change in the total entropy (system plussurroundings). For an irreversible process, a system may still be capable of returningto its initial state, but the surroundings are not returned to the corresponding initialstate. An irreversible process increases the entropy of the Universe. All real processare irreversible, though they can sometimes be analyzed as though they arereversible. Processes that are irreversible include friction, free expansion, andchemical reactions.

1.2 The laws of thermodynamicsThermodynamics is concerned with how the properties of substances change withtemperature and how these properties affect the ability of substances to emit andabsorb heat and to do different types of work. The study of thermodynamics beganin earnest at the start of the industrial revolution in the early 19th century as the needarose to understand the conversion between heat and mechanical energy to improvethe efficiency of steam engines. As its name implies, thermodynamics was initiallydeveloped to study the relationship between thermal (i.e. heat) and dynamical(i.e. mechanical) behavior.

For most of the time that the laws of thermodynamics were being formulated, theexistence of atoms had yet to be conclusively demonstrated, so the properties ofmatter were expressed in terms of macroscopic quantities such as temperature,pressure and volume. No reference was made to the underlying atomic constituentsof matter. What emerged was a set of concise statements—the laws ofthermodynamics—from which the behavior of any substance, particularly itsresponse to changes of temperature, could be predicted once a relatively smallnumber of empirical relationships (which thermodynamics identifies) are establishedby experimental measurement. The zeroth, first, second, and third laws aresummarized in the following subsections.

1.2.1 The zeroth law of thermodynamics

Suppose two systems A and B are allowed to attain a state of mutual thermalequilibrium and that there is a third system C in thermal equilibrium with system A.Are B and C in thermal equilibrium as well? Experience says they are. Thisobservation is called the zeroth law of thermodynamics. As the numbering of thislaw implies, it was formulated after the first, second, and third laws, but its inclusionis necessary for logical consistency.

Zeroth law of thermodynamics. If two systems are each separately in thermalequilibrium with a third system, then they are also in equilibrium with one another.

This law enables us to introduce the concept of temperature as a way ofdetermining whether two systems are in thermal equilibrium with one anotherwithout the need to bring them into thermal contact. If we label the two systems in


1-4

question as A and B, then the thermometer (which we can take as a mercurythermometer) is labeled as system C. Thus, by bringing the thermometer intothermal contact with system A, the mercury is found to be at a certain level, whichwe mark. If systems A and B are in thermal equilibrium, then placing thethermometer into thermal contact with system B results in the same ‘reading’ ofthis thermometer.

1.2.2 The first law of thermodynamics

The first and second laws of thermodynamics describe systems during changes ofstate and are expressed as constraints on such changes that can occur in nature. Thefirst law is a conservation principle for different forms of energy:

First law of thermodynamics. In any process, the energy of a closed thermody-namic system can change only by an exchange of heat between the system and itssurroundings (heating or cooling) or by mechanical work being performed on or bythe system.

The energy U is a state function that is expressed mathematically as an exactdifferential, so its change during a process is determined only by the differencebetween the initial and end points of the process. The path between the initial andfinal states is immaterial. The changes in heat Q and work W done during a processare not state functions because their values depend on the details of the process.Accordingly, such variables are expressed as inexact differentials. This is expressedmathematically as

đ đ= −dU Q W , (1.3)

where ‘đ ’ denotes an inexact differential (i.e. a path-dependent quantity), andheat flow into the system and work done by the system are taken as positivequantities.

1.2.3 The second law of thermodynamics

The relationship between work and heat is part of everyday experience. Work can beconverted into heat (for example, through friction), but the conversion of heat intowork is more subtle, requiring heat engines. The first law of thermodynamics treatswork and heat as being completely interchangeable: one can be converted into theother as long as the total energy is conserved. Similarly, heat flow is unrestricted,whether it is from a hotter to a colder body or vice versa, again as long as the totalenergy is conserved. The physical basis of the second law is the observation thatthere are restrictions on the conversion between work and heat and on the flow ofheat. One form of this law states this explicitly:

Second law of thermodynamics (Clausius’ Statement). There exists no thermody-namic process whose sole effect is to extract a quantity of heat from the colder of tworeservoirs and deliver it to the hotter reservoir.

When formulated in terms of cycles (heat engines and heat pumps) theseobservations provide the basis for introducing a state variable associated


1-5

with reversible heat flow at a given temperature. This state variable is called theentropy1.

Second law of thermodynamics. For any process taking place in a thermallyisolated system (including any reservoirs), the entropy either increases (if the processis irreversible) or remains constant (if the process is reversible).

The fact that the entropy of a thermally isolated system can never decrease,

⩾dS 0, (1.4)

establishes an important distinction between processes that are consistent with thefirst law of thermodynamics (in that they conserve energy), but do not occur innature. Two bodies initially in equilibrium at different temperatures that are broughtinto thermal contact will evolve toward an equilibrium state in which the temper-ature is the same in both bodies. The reverse process, whereby two systems inthermal contact evolve into states with different temperatures is not observed, eventhough the first law of thermodynamics would not be violated.

The second law defines a ‘thermodynamic driving force’ toward an equilibriumdetermined by specified external constraints. Suppose we have a system initially in anon-equilibrium state. Then, for fixed external constraints (e.g. the volume andtemperature), the system evolves toward an equilibrium state determined by theseconstraints. During equilibration the entropy must increase because this process isirreversible. This version of the second law is called the entropy maximum principle.For these reasons the second law is sometimes said to provide an ‘arrow of time’ forthe Universe (regarded as a thermally isolated system).

1.2.4 The third law of thermodynamics

The Carnot cycle, introduced by Sadi Carnot as an ideal reversible heat engine,is used to (i) provide a theoretical upper bound for the efficiency of any heat engineoperating between reservoirs with fixed upper and lower temperatures, and(ii) establish the existence of an absolute temperature bounded from below byabsolute zero.An immediate question iswhether absolute zero is physically attainable.

Suppose that we take the lower temperature reservoir in a Carnot cycle to zero.The efficiency of the Carnot cycle then becomes unity, which implies the existence ofan engine that converts heat completely into work. This would enable us to toconstruct a perpetual motion machine, which would violate the second law ofthermodynamics. Thus, the validity of the second law implies the unattainability ofabsolute zero. This is formalized by one form of the third law of thermodynamics:

Third law of thermodynamics (first form). There is no process by which a systemcan attain absolute zero in a finite number of steps.

This is the ‘unattainability’ form of the third law. There are two other forms,whose equivalence with each other and with this form remains a matter of debate:

1 The name ‘entropy’, which was introduced by Rudolf Clausius, comes from the two Greek words, en,meaning ‘in’ and trope, meaning ‘turning’. Clausius meant to convey the idea of an engine converting heat intowork, i.e. entropy being a measure of the extent to which this is possible.


1-6

Third law of thermodynamics (second form). The entropy of any pure substanceapproaches zero at absolute zero.

Third law of thermodynamics (third form). As the temperature approachesabsolute zero, the entropy of a system approaches a constant.

The third form is clearly weaker than the second form, which regards entropy asan absolute quantity. For most applications, only the entropy differences during aprocess are of importance.

1.3 Fundamental equationsThe first and second laws of thermodynamics can be used to establish a relationshipbetween state variables:

= −dU TdS PdV . (1.5)

Like the first law of thermodynamics, this equation is valid for any process and anysystem. But only in the case of a reversible process can the term TdS be identifiedwith the heat absorbed by the system and the term PdV identified with themechanical work done by the system. Because of the central role this equationplays in the calculation of thermodynamic quantities, (1.5) is called the fundamentalequation of thermodynamics.

The fundamental equation is expressed with S and V as independent variables, sothe solution of the fundamental equation is an expression of the form U S V( , ).The coefficients of the differentials of S and V are then given in terms of derivativesof U by

⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛⎝

⎞⎠= ∂

∂= − ∂

∂T

US

PUV

, . (1.6)V S

These relations are termed equations of state.A word about thermodynamic notation. Partial derivatives of thermodynamic

functions are written with the variables that are being held constant as subscripts.The reason for this convention is that the same symbol is used for a thermodynamicfunction with different independent variables. For example, in the fundamentalequation, the internal energy =U U S V( , ), but U can also be expressed asU T V( , ).Although mathematically questionable, this convention avoids the proliferation offunctions that refer to the same physical quantity.

The entropy and volume are not the most convenient variables for consolidatingthe results of experiments. There is no practical instrument for measuring theentropy, but thermometers, thermocouples, etc for the measurement and control oftemperature are common laboratory equipment. Similarly vacuum and pressuregauges are used to measure pressure, and the volume of a system can be determineddirectly. Thus, instead of varying the entropy, one might regard the temperature asthe more ‘natural’ independent variable for performing particular measurements.The Gibbs free energy G T P( , ) is the thermodynamic potential that is minimizedwhen a system reaches equilibrium at constant pressure and temperature. Thisprovides a convenient criterion for the spontaneity of processes at constant pressure


1-7

and temperature. The fundamental equation can be reformulated in terms of otherpairs of independent variables, which leads to new thermodynamic functions: theHelmholtz free energy, F T V( , ) and the enthalpy, H S P( , ). The experimentalconditions associated with each of these functions is shown in figure 1.1. In the nexttwo sections, we will derive the Gibbs and Helmholtz free energies. The correspond-ing development for the enthalpy is taken up in exercise 6.

1.3.1 The Gibbs function

Consider a process for which the ‘natural’ variables are the temperature andpressure. We first transform the fundamental equation of thermodynamics into aform where the independent variables are the temperature T and pressure P. Thus,

= −= + − − − += − − +

dU TdS PdVTdS SdT SdT PdV VdP VdPd TS SdT d PV VdP( ) ( ) .

(1.7)

Bringing the total differentials to the left-hand side of this equation,

− + = − +d U TS PV SdT VdP( ) (1.8)

clearly shows that T and P are now the independent variables. The quantity whosedifferential is taken on the left-hand side of this equation is called the Gibbs functionor the Gibbs free energy, and is denoted by G:

= − +G U TS PV . (1.9)

The thermodynamic significance of the Gibbs function will be given below after weconsider some consequences of G being an exact differential.

Taking the differential of G T P( , ),

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟= ∂

∂+ ∂

∂dG

GT

dTGP

dP, (1.10)P T

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

Figure 1.1. Schematic illustration of how different thermodynamic variables are held constant in differentexperiments: (a) Constant volume and adiabatic boundaries. U S V( , ) is constant; S U V( , ) maximized inequilibrium. (b) Constant volume and diathermal boundaries. Helmholtz free energy F T V( , ) minimizedin equilibrium. (c) Constant pressure and diathermal boundaries. Gibbs free energy G T P( , ) minimized inequilibrium, and (d) constant pressure and adiabatic boundaries. Enthalpy H S P( , ) is constant (exercise 6).


1-8

and comparing with

= − +dG SdT VdP (1.11)

allows us to make the identifications

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

∂∂

= − ∂∂

=GT

SGP

V, . (1.12)P T

By observing that G is comprised of state variables and so G itself must be a statefunction, we conclude that the mixed second partial derivatives of G must be equal:

∂∂ ∂

= ∂∂ ∂

GT P

GP T

. (1.13)2 2

Taking the appropriate derivatives of the expressions in (1.12), then yields theMaxwell relation associated with the Gibbs function:

⎜ ⎟⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

∂∂

= − ∂∂

VT

SP

. (1.14)P T

The thermodynamic significance of the Gibbs function is demonstrated byconsidering a system at constant temperature and pressure. The differential changein G during such a process is given by

= − +dG dU TdS PdV . (1.15)

We now use the first law of thermodynamics (2.60) to substitute for dU in (1.15):

đ đ= − − −dG Q TdS W PdV( ) ( ). (1.16)

Since the temperature is held constant, we can relate the changes in the entropy and theheat during any isothermal process by using Clausius’ theorem: đ ⩽Q TdS . Thus, thequantity đ −Q TdS in (1.16) is less than or equal to zero, which implies that

đ⩽ − −dG W PdV( ), (1.17)

or,

đ− ⩾ −dG W PdV .

Thus, the decrease in the Gibbs function sets an upper limit to the ‘non-PdV ’ work(i.e. work not directly associated with the movement of the system boundary) in anyprocess between two equilibrium states at the same temperature and pressure whenthe system is in contact with a single heat reservoir and a single pressure reservoir.The term ‘free energy’ originates from the fact that the change in G sets a limit tohow much energy is available for conversion into ‘non-PdV ’ work, which includesother forms of work and irreversibility.

If, in (1.17), the volume is kept constant and there are no other forms of work, weobtain the inequality

⩽dG 0, (1.18)


1-9

i.e. the Gibbs free energy of a system in contact with a temperature and heatreservoir never increases. Since the inequality in (1.18) refers to irreversible processes,we see that irreversible processes at constant temperature and pressure decrease theGibbs free energy. Thus, in thermal equilibrium, the Gibbs free energy is a minimumwith respect to changes at constant T and P. This is analogous to the result that theentropy of a thermally isolated system never decreases and leads to the following:

Minimum principle for gibbs free energy. For a system at constant temperatureand pressure, the state of equilibrium is determined by the minimum of the Gibbsfree energy.

1.3.2 The Helmholtz function

We now consider the first law of thermodynamics for a general isothermal process.The first law is

đ đ= −dU Q W . (1.19)

By substituting the Clausius inequality, đ⩾dS Q T/ , into this equation, we obtain

đ⩽ −dU TdS W . (1.20)

Since the temperature is held constant, we can add with no effect the quantitySdT tothe left-hand side of this equation which, after a simple rearrangement, enables us towrite

đ− ⩽ −d U TS W( ) , (1.21)

or,

đ− − ⩾d U TS W( ) . (1.22)

This equation states the decrease in the quantity U − TS is an upper bound to thework done by a system during an isothermal process. To see the significance of theright-hand side of this equation, we take the differential and use the fundamentalequation:

− = − −= − − −= − −

d U TS dU TdS SdTTdS PdV TdS SdTSdT PdV

( )( )

.

(1.23)

This is an equation analogous to the fundamental equation, but written in terms of Tand V as the independent variables. The right-hand side of this equation is composedentirely of state variables and is, therefore, itself a state variable. This quantity,which is usually denoted by F (some books use A), is called theHelmholtz function orthe Helmholtz free energy2:

2 The mathematical operation we have carried out in changing the independent variable inU S V( , ) to obtainF T V( , ) is called a Legendre transformation. A very readable discussion of this transformation has been givenby Callen (1960).


1-10

= −F U TS. (1.24)

The thermodynamic content of the derivatives of F can be obtained by taking thedifferential of F T V( , ),

⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛⎝

⎞⎠= ∂

∂+ ∂

∂dF

FT

dTFV

dV , (1.25)V T

and comparing with (1.23), we deduce that the partial derivatives of F with respect toT and V are given by

⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛⎝

⎞⎠

∂∂

= − ∂∂

= −FT

SFV

P, . (1.26)V T

Then, since F is an exact differential (i.e. a function of state), we utilize therequirement that the mixed second partial derivatives of F must be equal to obtainthe Maxwell relation associated with this function:

⎜ ⎟⎛⎝⎜

⎞⎠⎟

⎛⎝

⎞⎠

∂∂

= ∂∂

SV

PT

. (1.27)T V

The thermodynamic interpretation of the Helmholtz free energy, as well as amotivation for the name ‘free energy’ is contained in equation (1.22): the decrease inthe Helmholtz free energy sets an upper limit to the work done by a system in anyprocess between two equilibrium states at the same temperature when the system isin contact with a single heat reservoir at this temperature. Since the Helmholtz freeenergy is a function of state, this conclusion holds for any process between specifiedinitial and final states as long as these states are at the same temperature. The term‘free energy’ is now seen to come from the fact that the change in F sets a limit onhow much energy is ‘freed’ to be available for conversion into work. Thus, thecapability of a system to do work under isothermal conditions during a particularprocess is determined by the Helmholtz free energy. Moreover, since the work doneunder these conditions is a measurable quantity, the change in the Helmholtzfunction is directly accessible in an experiment.

Another consequence of (1.22) is that if đ =W 0, then

⩽dF 0, (1.28)

i.e. the Helmholtz free energy never increases. This is analogous to the result that theentropy of a thermally isolated system never decreases:

Minimum principle for helmholtz free energy. For a mechanically isolated systemat constant temperature, the state of equilibrium is determined by the minimum ofthe Helmholtz free energy.

The four thermodynamic potentials, their fundamental equations, and theirequations of state are compiled in table 1.1.


1-11

1.4 Thermal, mechanical, and chemical equilibriaThe fundamental equation (1.5) is valid for systems with fixed particle number. Toallow for variable numbers of different types of particle, the fundamental equationbecomes

∑ μ= − +dU TdS PdV dN , (1.29)i

i i

where μi is the chemical potential for the ith species. The name ‘chemical potential’ isapposite, as μ measures of the potential of a substance for physical and chemicalchange in a system. This equation for dU S V N( , , ) states that there are three waysto increase the internal energy of a system: heat, mechanical work, and addingparticles. For future reference, we solve this equation for the differentialdS U V N( , , ):

μ= + −

∑dS

dUT

PdVT

dN

T, (1.30)i i i

Consider two systems with internal energies, volumes, and particle numbers U1,V ,1 N1 andU V N, ,2 2 2. Suppose that these two systems are isolated but separated by afixed diathermal impermeable boundary (section 1.1.1), so that only energy can beexchanged; the volumes and particle numbers are unchanged from their initialvalues. Using (1.30), the change in entropy during any exchange of energy (with

=dV 0 and = =dN dN 01 2 ) is

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟= ∂

∂+ ∂

∂dS

SU

dUSU

dU . (1.31)V N V N1 ,

12 ,

2

1 1 2 2

Since the total system is isolated, the total energy = +U U U1 2 remains constant, so= −dU dU2 1. Hence, using (1.30) to evaluate the partial derivatives in (1.31), the

differential entropy change is obtained as

Table 1.1. The fundamental equations for the internal energy U, Helmholtz F and Gibbs G free energies, andthe enthalpy H. Other thermodynamic variables have their usual meanings: the temperature T, entropy S,pressure P, and volume V.

Function Fundamental equation Equations of state

U = −dU TdS PdV ⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛⎝

⎞⎠= ∂

∂= − ∂

∂T

US

PUV

,V S

= −F U TS = − −dF SdT PdV ⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛⎝

⎞⎠= − ∂

∂= − ∂

∂S

FT

PFV

,V T

= − +G U TS PV = − +dG SdT VdP ⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛⎝

⎞⎠= − ∂

∂= ∂

∂S

GT

VGP

,P T

= +H U PV = +dH TdS VdP ⎜ ⎟ ⎜ ⎟⎛⎝

⎞⎠

⎛⎝

⎞⎠= ∂

∂= ∂

∂T

HS

VHP

,P S


1-12

⎛⎝⎜

⎞⎠⎟= − ⩾dS

T TdU

1 10, (1.32)

1 21

where we have invoked the inequality in (1.4). If, say, >T T2 1, then

− >T T1 1

0, (1.33)1 2

so >dU 01 to maintain the inequality. Hence, energy flows from the system with thehigher temperature to the system with the lower temperature. This energy transfercontinues until =T T2 1, at which point thermal equilibrium is reached.

Now suppose that the two systems are separated by a moveable diathermalimpermeable boundary. The two systems remain isolated. Again using (1.30) tocalculate the differential entropy change, yields

= + + +dSdUT

dUT

P dVT

P dVT

. (1.34)1

1

2

2

1 1

1

2 2

2

Since the system is isolated the total energy = +U U U1 2 and total volume= +V V V1 2 remain constant, so = −dU dU2 1 and = −dV dV2 1, whereupon we obtain

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟= − + −dS

T TdU

PT

PT

dV1 1

. (1.35)1 2

11

1

2

21

If the system is in thermal equilibrium, = =T T T1 2 , this reduces to

= − ⩾dSP P dV

T( )

0. (1.36)1 2 1

Thus, if >P P1 2, then >dV 01 to maintain the inequality. Thus, the volume increasesfor the system with the larger pressure. This process continues until =P P2 1, whichdefines mechanical equilibrium for this system.

Finally, suppose that the two systems, still isolated, are separated by a fixeddiathermal permeable boundary. The differentiable entropy change is, from (1.30),given by

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

μ μ= − + − +dS

T TdU

T TdN

1 1, (1.37)

1 21

1

1

2

21

where we have used the fact that = +U U U1 2 and = +N N N1 2 are constant for anisolated system, so = −dU dU2 1 and = −dN dN2 1. If the two systems are in thermalequilibrium, = =T T T1 2 , (1.37) reduces to

μ μ= − + ⩾dSdNT

( ) 0. (1.38)1 21

Hence, if μ μ>2 1, then >dN 01 , meaning that mass flows from the system with thelarger chemical potential to the system with the lower chemical potential. Thisprocess continues until μ μ=2 1, at which point chemical equilibrium is reached.


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A system in thermodynamic equilibrium is in thermal, mechanical, and chemicalequilibrium, which means that the temperature, pressure, and chemical potential areuniform throughout the system. With the internal energy given by (1.29), thecorresponding Gibbs free energy (1.11) is

∑ μ= − + +dG SdT VdP dN , (1.39)i

i i

which, for processes at constant temperature and pressure reduces to

∑ μ=dG dN . (1.40)i

i i

The chemical potential is an intensive variable, and the Gibbs free energy for asystem is obtained by summing the chemical potential of each atomic or molecularconstituent in proportion to the amount of that species in the system.

1.5 Phase equilibria1.5.1 The Clausius–Clapeyron equation

A phase is a region of a thermodynamic system within which all physical propertiesare uniform. Suppose we have two phases of a given substance. For specificity weconsider a liquid in equilibrium with its vapor. We consider this two-phase system tobe in equilibrium at atmospheric temperature and pressure, which are therebyregarded as fixed quantities. Neglecting any surface tension effects, i.e. neglectingthe thermodynamics of the liquid–vapor interface, we can write the total Gibbsfunction for the two-phase system as the sum of specific Gibbs functions for each inproportion to the number of particles present in each phase:

v v= +G N g N g .ℓ ℓ

Since T and P are held fixed, only the relative amounts of particles in the liquid andvapor phase can change, subject to the total number of particles being fixed:

v+ =N N N. (1.41)ℓ

Thus, the differential of G is given by

v v= +dG g dN g dN . (1.42)ℓ ℓ

But the differential of (1.41) requires that

v+ =dN dN 0.ℓ

Thus, (1.42) can be written as

v= −dG g g dN( ) .ℓ ℓ


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Since, in equilibrium, this differential must vanish, we find

v=g T P g T P( , ) ( , ), (1.43)ℓ

i.e. in equilibrium, the specific Gibbs functions of two phases must be the same.Equation (1.43) provides a relation between T and P. To determine this

relationship in the form of a differential equation, we take the differential of bothsides of (1.43) and use the fundamental equations for g T P( , )ℓ and vg T P( , ):

v vv v− + = − +s dT dP s dT dP,ℓ ℓ

where =s S N/i i i and v = V N/i i i, for v=i ℓ, . This equation can be rearranged as

v vv

v= −

−dPdT

s s(1.44)ℓ

ℓ

The difference in the specific entropies can be related to the latent heat of vapor-ization λ (which is defined as the amount of heat absorbed when a given amount ofthe liquid phase is transformed to the vapor phase) by

v λ− =T s s( ) ,ℓ

equation (1.44) becomes

v vv

λ=−

dPdT T ( )

. (1.45)ℓ

This is the Clausius–Clapeyron equation. Although we have derived this relation fora liquid–vapor system, it holds for any two phases in equilibrium. An immediateconsequence of (1.45) is that, since v vv > ℓ, the boiling temperature is raised byraising the pressure.

1.5.2 The Gibbs phase rule

There are many thermodynamic variables for a given system, such as pressure,temperature, density, internal energy, and so on. How many of these variables areneeded to specify the thermodynamic state of a system? Consider an ideal gas. Theequation of state for such a gas with N particles is =PV Nk TB , where kB isBoltzmann’s constant. Thus, we immediately see that any two of the variables P, V,and T uniquely specify the third because the equation of state provides a relationbetween these variables. But what about the more general case of a system withseveral components and phases?

Consider a system composed of c independent components and p coexistingphases. How many thermodynamic variables are there? We first observe that it issufficient to specify the amount, the temperature and pressure of each phase, i.e.

α… = …α α αα αN N N T P p, , , , , , 1, 2, , ,c1 2

and αNi is the number of particles (atoms or molecules) in the ith component of theαth phase. Instead of specifying the number of particles in each phase, we can specify


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the total number of particles and −c 1 of the αNi , or equivalently, the fraction αniof particles of each type in each phase and the total number αN of particles in theαth phase:

= = + + ⋯ +αα

αα α α αn

NN

N N N N, .ii

c1 2

To specify the state of this system, we do not need to specify the total amount ofmaterial nor the volume of the system. The equation of state of an ideal gas can bewritten as =P nk TB , where =n N V/ is the density of the system. This form of theequation of state is expressed entirely in terms of intensive variables: P, n, and T.Accordingly, with our multicomponent multiphase system, we need specify only

…α αα α α

−T P n n n, , , , , .c1 2 1

Thus, there are +c 1 variables for each phase, which means that there is a total of+p c( 1) variables, all of which are intensive.

Equilibrium provides constraints on these variables. Thermal equilibriumrequires that the temperature is the same in each phase:

= = ⋯ =T T T . (1.46)p1 2

Mechanical equilibrium requires that the pressure is the same in each phase:

= = ⋯ =P P P . (1.47)p1 2

Finally, chemical equilibrium requires that the chemical potential of each compo-nent in each phase is the same:

μ μ μ

μ μ μ

μ μ μ

= = ⋯ =

= = ⋯ =⋮ ⋮ ⋮

= = ⋯ =

,

,

.

(1.48)

p

p

c c cp

11

12

1

21

22

2

1 2

The number of constraints imposed by thermodynamic equilibrium is −p 1, from(1.46), plus −p 1, from (1.47), plus −c p( 1), from (1.48), for a total of

− +p c( 1)( 2). Hence, from +p c( 1) variables, the number of variables f that canbe independently specified is

= + − − + = − +f p c p c c p( 1) ( 1)( 2) 2. (1.49)

This relation, known as Gibbs’ phase rule, stipulates that the equilibrium state of asystem with c-components and p-phases is completely specified by f intensivevariables. For a single phase of a one-component system, such as an ideal gas,c = 1 and p = 1. Thus, the Gibbs phase rule (1.49) yields f = 2, which means that twointensive variables completely specify the thermodynamic state of a system. Thesecan be taken as the pressure and temperature.


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1.6 SummaryThis chapter has outlined some of the basic issues associated with phase trans-formations and has summarized the key elements of the main theoretical tools usedin their analysis. We began with the four laws of thermodynamics, which establishthe notions of temperature, energy, entropy, and the entropy at absolute zero. Thesolution of the fundamental equation, expressed in terms of the internal energy asthe dependent variable and entropy, volume, and particle numbers as the independentvariables, provides all of the thermodynamic information about a thermodynamicsystem. But other independent variables aremore closely related to quantities that canbemeasured and controlled, such as temperature, pressure, and particle numbers. Thefundamental equation can be formulated in terms of these quantities and the resultingthermodynamic function is theGibbs free energy. This leads naturally to the definitionof the chemical potential as a regulator of mass flow, just temperature regulates heatflow and pressure mechanical work.

Further readingThere are several excellent texts on thermodynamics that provide both theappropriate background to this course as well as coverage of some of themain topics of this course. Callen (1960) develops thermodynamics from a fewpostulates and concentrates on the foundations of the subject. Pippard (1957)provides a concise and very readable discourse on all of the main topics ofthermodynamics.

Further reading from recommended texts:Cottrell (1967): Chapters 5–7.Porter and Easterling (2004): Chapter 1.Balluffi, Allen, and Carter (2005): Chapters 1 and 2.Christian (2002): section 1, Chapter 4.

Exercises1. The British scientist and author C P Snow is often credited with the

following aphorism of the first, second, and third laws of thermodynamics:First law: You can’t win.Second law: You can’t break even.Third law: You can’t get out of the game.

Explain the relation between these statements and the conventional forms of thelaws of thermodynamics.

2. Consider two equal amount of substances A and B that are thermallyisolated and initially at temperatures TA and TB, respectively. The systemsare then brought into thermal contact and allowed to equilibrate, at whichpoint the temperature is Tf.

(a) Observing that no work is done during this process, show that theheat transfer between the two systems satisfies + =Q Q 0A B . Supposethe substances are gases that interact through a stationary diathermal


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wall, so the volume remains constant. By expressing the heat transferbetween the two systems in terms of the constant-volume specificheats CA and CB (which you can assume to be independent oftemperature), show that

= ++

TC T C T

C C.f

A A B B

A B

(b) By calculating the entropy change in each system show that the totalentropy change between the initial and final states is

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟Δ = +S C

T

TC

T

Tln ln .A

f

AB

f

B

Verify that Δ ⩾S 0.3. The transformation between the internal energy U S V( , ), the Helmholtz

free energy F T V( , ), the Gibbs free energy G T P( , ), and the enthalpyH S P( , ) is known as a Legendre transformation.

A function f x( ) is usually represented as a curve =y f x( ) in the x–yplane. An alternative representation is the envelope of a family of tangentlines ϕ m( ), where ϕ is the y-intercept of a line with slope m, whose equationis ϕ= +y mx m( ). In particular, the tangent line at any point x f x( , ( )) isgiven by ϕ = −m f x mx( ) ( ) , where

=mdfdx

.

The function ϕ m( ) is the Legendre transformation of f x( ). Use thisconstruction to represent the Legendre transformation of =f x2.

4. The internal energyU T V( , ) is a different function fromU S V( , ). In fact,U T V( , ) provides considerably less information than U S V( , ). Consider

these functions for a monatomic ideal gas, for which =U T V NkT( , ) 32

.Since = ∂ ∂T U S( / )V ,U T V( , ) can be interpreted as a differential equationforU S V( , ):

⎜ ⎟⎛⎝

⎞⎠= ∂

∂U S V

Nk US

( , )3

2.

V

Integrate this differential equation and show that, to within a constant ofintegration, this solution is the same as that obtained directly from thefundamental equation:

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥= −

−

U S V UVV Nk

S S( , ) exp2

3( ) .0

0

2/3

0


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5. The solution of the fundamental equation for a single-component systemhas the formU S V N( , , ).

(a) Determine the three equations of state associated with this equation.(b) Suppose that the system size is scaled by a positive factor λ. Show

how U S V N( , , ) transforms under the scaling of its independentvariables by λ.

(c) By taking the derivative with respect to λ of the equation obtained in(a), show that μ= − +U TS PV N .

(d) Explain the relationship between the equation derived in (b) and thefundamental equation in (1.5), and identify the independent varia-bles. Use the solution of the fundamental equation for an idealmonatomic gas

⎛⎝⎜

⎞⎠⎟=U S V N A

NV

SNk

( , , ) exp2

3,

B

5/3

2/3

where A is a constant, to illustrate your points.(e) Determine the behavior of G and its independent variables under the

scaling of the system size. Then, use the procedure in (b) to obtainμ=G N . Show that this is consistent with the fundamental equation

in (1.5).6. The procedure used in sections 1.3.1 and 1.3.2 to derive the Helmholtz

and Gibbs free energies from the fundamental equation of thermo-dynamics can be applied to obtain a state function H S P( , ), called theenthalpy.

(a) Determine the differential form dH of the enthalpy and express thisfunction in terms of state variables.

(b) Identify the equations of state associated with the enthalpy.(c) Show that the change of the enthalpy during a reversible isobaric

process is equal to the heat flow during that process.(d) Explain why the enthalpy is constant for the conditions in

figure 1.1(d).7. A section of the PT phase diagram of of water is shown below:

(a) Determine the number of degrees of freedom for the regions marked‘solid’, ‘liquid’, and ‘gas’.

(b) Determine the number of degrees of freedom along the coexistencelines, marked ‘A’, ‘B’, and ‘C’ in the phase diagram.

(c) Determine the number of degrees freedom at the triple point, marked‘D’ in the phase diagram.


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With increasing temperature and pressure, the liquid–gas coexistence curveends in a critical point ‘E’. At this point, the meniscus between the liquid andvapor disappears, and the densities of the liquid and vapor are equal. Beyondthis point the liquid and vapor form a single supercritical fluid phase.

(a) How many degrees of freedom are associated with the critical point?(b) Show that the Gibbs phase rule in (c) does not provide a consistent

description of the number of phases in equilibrium at the criticalpoint and the number of degrees of freedom needed to specify thatpoint.

(c) Show that a consistent picture can be obtained by including anadditional relation between intensive variables, namely, the equalityof the densities of two phases. This provides an additional constraintalong the liquid–gas coexistence curve.

8. The triple point of water, where the solid s( ), liquid ℓ( ), and vapor v( )coexist, occurs at temperature =T 273.16 K and pressure =P 611.73 Pa(0.006 04 atm), where =1 Pa 1 N m−2. The thermodynamic data for thetransformations between these phases of water are compiled in the tablebelow:

(a) Provide a qualitative microscopic explanation for the trends in λ andΔs for the three transformations.

Transformation λ (kJ mol−1) Δs (J mol−1 K−1) vΔ (m3 mol−1)

→s ℓ 5.98 22 −1.64 × 10−6

v→ℓ 44.9 165 0.022v→s 50.9 186 0.022


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(b) Use the Clausius–Clapeyron equation and the data in the table in(a) to estimate the slopes of the coexistence lines near the triple point.Sketch the phase diagram near the triple point, labeling the phasesseparated by the coexistence lines. You may find it convenient to use

Pln as a variable. The data are provided only as a guide for makingbroad estimates; detailed calculations are not required and the sketchneed only be schematic.

9. The Clausius–Clapeyron equation (section 1.5.1) for the transformation ofa material from the liquid to the gas phase is:

v vv

λ=−

dPdT T ( )

.ℓ

We expect that v vv ≫ ℓ, so that we can approximate v vv − ℓ by the specificvolume of the gas, vv. By assuming that the vapor behaves as monatomicideal gas, and that λ is a constant, obtain an expression for P T( ).

10. Consider the transformation between the graphite and diamond forms ofcarbon:

⟶C(graphite) C(diamond),

for which Δ =° −G 2.90 kJ mol 1 at room temperature (298 K) and pressure(1 bar).

(a) In which direction does this transformation occur spontaneously?Why does this ‘spontaneous’ reaction take geological times?

(b) Integrate the fundamental equation in (1.11) at constant (room)temperature T0 from room pressure P0 to P. Use this to obtain thechange ΔG T P( , )0 in the Gibbs free energy during the transformationfrom graphite to diamond:

∫Δ = Δ + Δ ′ ′°G T P G V T P dP( , ) ( , ) ,P

P

0 00

where ΔV T P( , )0 is the volume difference between diamond andgraphite.

(c) At room temperature and pressure,

Δ = − × − −V T P( , ) 1.9 10 m mol ,0 06 3 1

which may taken to be independent of pressure. The SI unit of pressureis the Pascal (Pa), defined as 1 Pa = 1 N m−2, with 1 atm ≈ 105 Pa.Hence, determine the pressure at which graphite and diamond coexist.

11. A relation analogous to Gibbs’ phase rule is Duhem’s theorem: The stateof a c-component p-phase closed system with given total amounts ofeach component is completely specified by two independent (intensive orextensive) variables. The numbers Ni of particles of the ith component

= …i c( 1, 2, , ) are given, but the numbers αNi of these particles in the αth


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phase α = … p( 1, 2, , ) are variables. The proof of this theorem is similar toour derivation of the phase rule.

(a) Given that the system is closed, so the volume is fixed, identifythe thermodynamic variables and show that there are +cp 2 suchvariables.

(b) Show that chemical equilibrium provides −c p( 1) constraints.(c) The total number of particles of each component is fixed, which

provides the remaining constraints. Determine the number of theseconstraints.

(d) Use the results of (a), (b), and (c) to show that the number ofvariables that can be independently specified is two.

ReferencesCallen H B 1960 Thermodynamics (New York: Wiley)Pippard A B 1957 The Elements of Classical Thermodynamics (Cambridge: Cambridge University

Press)


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