thermodynamic concepts and relationships...1.3.4 molar gibbs’ free energy of a pure metal 15 1.4...

1

Thermodynamic Conceptsand Relationships

1.1 Introduction 2

1.2 Thermodynamic Concepts and Relationships 2

1.2.1 First Law of Thermodynamics. Principle of Conservation of Energy 2

1.2.2 Enthalpy 2

1.2.3 Second Law of Thermodynamics. Entropy 3

1.2.4 Gibbs’ Free Energy 8

1.2.5 Intensive and Extensive Thermodynamic Quantities 9

1.3 Thermodynamics of Single-Component Systems 10

1.3.1 Clausius–Clapeyron’s Law 10

1.3.2 Equilibrium between Liquid and Solid Phases. Influence of Pressure

and Crystal Curvature on Melting Point

11

1.3.3 Equilibrium between Liquid and Gaseous Phases. Influence of Pressure on

Boiling Point. Bubble Formation in Melts

13

1.3.4 Molar Gibbs’ Free Energy of a Pure Metal 15

1.4 Thermodynamics of Multiple-Component Systems 16

1.4.1 Partial Molar Thermodynamic Quantities 16

1.4.2 Relative Thermodynamic Quantities and Reference States. Relative Partial

Molar Thermodynamic Quantities or Partial Molar Quantities of Mixing

18

1.4.3 Relative Integral Molar Thermodynamic Quantities or Integral Molar

Quantities of Mixing

19

1.4.4 Other Thermodynamic Functions and Relationships 21

1.5 Thermodynamics of Alloys 21

1.5.1 Heat of Mixing 22

1.5.2 Ideal and Non-Ideal Solutions 23

1.6 Thermodynamics of Ideal Binary Solutions 25

1.6.1 Molar Gibbs’ Free Energy of Ideal Binary Solutions 25

1.7 Thermodynamics of Non-Ideal Binary Solutions 26

1.7.1 Activities of Non-Ideal Solutions Raoult’s and Henry’s Laws 26

1.7.2 Excess Quantities of Non-Ideal Solutions 30

1.7.3 Molar Gibbs’ Free Energies of Non-Ideal Binary Solutions 32

1.8 Experimental Determination of Thermodynamic Quantities of Binary Alloys 34

1.8.1 Determination of Molar Heat of Mixing of Binary Alloys 34

1.8.2 Determination of PartialMolarGibbs’ Free Energy ofMixing ofBinaryAlloys 35

Summary 36

Further Reading 41

Solidification and Crystallization Processing in Metals and Alloys, First Edition. Hasse Fredriksson and Ulla A�kerlind.

� 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

COPYRIG

HTED M

ATERIAL

1.1 Introduction

Solidification or crystallization is a process where the atoms are transferred from the disordered

liquid state to the more ordered solid state. The rate of the crystallization process is described and

controlled by kinetic laws. These laws give information of the movements of the atoms during the

rearrangement. Inmost cases a driving force is involved that makes it possible to derive the rate of the

solidification process.

The aim of this book is to study the solidification processes in metals and alloys. The laws

of thermodynamics and other fundamental physical laws, which control the solidification, rule

these processes.

In this preparatory chapter the basic concepts and laws of thermodynamics are introduced.

They will be the tools in following chapters. In particular, this is true for the second chapter,

where the driving forces of solidification for pure metals and binary alloys are derived and

the relationship between energy curves of solid metals and metal melts and their phase diagrams

is emphasized.

1.2 Thermodynamic Concepts and Relationships

1.2.1 First Law of Thermodynamics. Principle of Conservation of Energy

One of the most fundamental laws of physics is the law of conservation of energy. So far, it is known

to be valid without any exceptions. When applied in thermodynamics it is called the first law of

thermodynamics and is written

Q ¼ UþW ð1:1Þ

where

Q ¼ heat energy added to a closed system

U ¼ the internal energy of the system or the sum of the kinetic and potential energies of

the atoms

W ¼ work done by the system.

Differentiating Equation (1.1) gives the relationship

dQ ¼ dUþ dW ¼ dUþ pdV ð1:2Þ

The added heat dQ is used for the increase dU of the internal energy of the system and for the

external work pdV against the surrounding pressure p when the volume of the system increases by

the amount dV.

1.2.2 Enthalpy

The enthalpy of a closed system is defined as

H ¼ Uþ pV ð1:3Þ

The enthalpy H can be described as the ‘heat content’ of the system. In the absence of phase

transformations we have

H ¼ðT

0

nCpdT ð1:4Þ

2 Solidification and Crystallization Processing in Metals and Alloys

where

H ¼ enthalpy of the system

n ¼ number of kmol

Cp ¼ molar heat capacity of the system (J/kmol).

When the heat content of a system is changed the enthalpy changes by the amount DH. When the

system absorbs heat from the surroundings the heat content of the system is increased and DHis positive.

When the system emits heat to the surroundings and reduces its heat content, DH is negative.

The amount of heat absorbed by the surroundings is (�DH), which is a positive quantity.

Differentiating Equation (1.3) gives

dH ¼ dUþ pdV þVdp ð1:5Þ

By use of Equation (1.2) we obtain

dH ¼ dQþVdp

At constant pressure, the heat absorbed by a system equals its enthalpy increase:

ðdHÞp ¼ ðdQÞp ð1:6Þ

1.2.3 Second Law of Thermodynamics. Entropy

Second Law of Thermodynamics

By experience it is known that heat always is transferred spontaneously from a warmer to a colder

body, never the contrary.

It is possible to transfer heat intomechanical work but never to 100%.Consider a closed system (no

energy exchange with the surroundings) which consists of an engine in contact with two heat

reservoirs. If an amountQ1 of heat is emitted from reservoir 1 at a temperature T1 to the engine and an

amount Q2 of heat is absorbed by reservoir 2 at temperature T2, the energy difference Q1�Q2 is

transferred into mechanical work in the ideal case.

The reversed process is shown in Figure 1.1. It can be described as follows.

. Heat can be transferred from a colder body to a warmer body only if work or energy is supplied.

This is one of many ways to express the second law of thermodynamics.

In practice, heat is transferred intomechanical work in heat engines. The process in such amachine

can theoretically be described by the Carnot cycle (Figure 1.2a). Fuel is burned and the combustion

gas runs through the following cycle in the engine

1. The gas absorbs the amount Q1 of heat and expands isothermally at temperature T1.

2. The gas expands adiabatically. The temperature decreases from T1 to T2.

3. The gas is compressed isothermally at temperature T2 and emits the amount Q2 of heat.

4. The gas is compressed adiabatically. The temperature changes from T2 back to T1.

Calculations of the expansion and compression works show that

Q1

Q2

¼ T1

T2ð1:7Þ

T2

T1

Engine W = Q1 – Q2

Q1

Q2

Figure 1.1 Engine in contactwith

two heat reservoirs. T1 > T2. The

energy law gives Q1¼Q2þW.

p

VT2

T1

Figure 1.2a The Carnot cycle.

The enclosed area represents the

work done by the heat engine dur-

ing a cycle.

Thermodynamic Concepts and Relationships 3

In the ideal case, the efficiency h of the Carnot cycle is

h ¼ Q1 �Q2

Q1

¼ T1 � T2

T1ð1:8Þ

Entropy

Avery useful and important quantity in thermodynamics is the entropy S. Entropy is defined by the

relationship

dS ¼ dQ

Tð1:9Þ

As an example we represent the Carnot cycle in a TS diagram (Figure 1.2b). Horizontal lines

represent the isothermal expansion and compression. The vertical lines illustrate the adiabatic steps

of the cycle. The area of the rectangle equals thework done by the gas. A reversible adiabatic process

is also called isentropic.

Entropy Change at an Isothermal Irreversible Expansion of a GasIn many solidification processes the entropy change is of great interest. As a first example we will

consider the isothermal, irreversible expansion of an ideal gas and calculate the entropy change.

Example 1.1

Calculate the change in entropy when n kmol of an ideal gas with the pressure p0 and the volume V1

expands irreversibly to the volume V1þV2 in the way shown in the figure.

Solution:

There are no forces between the molecules in the gas and therefore there is no change of internal

energy when the gas expands. No change of internal energy means no temperature change.

When the tap is opened, the gas expands isothermally from volume V1 to V1þV2. The first law of

thermodynamics and the definition of entropy give

DS ¼ðdq

T¼

ðV1 þV2

V1

pdV

Tð10Þ

Using the general gas law pV ¼ nRT to eliminate T we obtain

DS ¼ðV1 þV2

V1

nRdV

V¼ nR ln

V1 þV2

V1

ð20Þ

Answer: The entropy increases by the amount nR lnV1 þV2

V1

.

The final state in Example 1.1 is far more likely than the initial state. When the tap is opened the

molecules move into the empty container until the pressures in the two containers are equal rather

than that no change at all occurs. Hence, the system changes spontaneously from one state to another,

more likely state and the entropy increases.

S

T

T1T2

Figure 1.2b The Carnot cycle.

The enclosed area represents the

work done by the heat engine dur-

ing a cycle.

n p0 V1tap

0 0 V2

p V1tap

p V2


All experience shows that the result in Example 1.1 can be generally applied. The following

statements are generally valid:

. If a process is reversible the entropy change is zero.

DS ¼ 0:

. The entropy increases in all irreversible processes.

DS > 0:

Entropy of Mixtures

Entropy Change at Mixing Two GasesAs a second example of deriving entropy changes we will calculate the entropy change when two

gases mix.

Example 1.2

A short tube and a closed tap connect two gases A and B of equal pressures, each in a separate closed

container. When the tap is opened the two gases mix irreversibly. No changes in pressure and

temperature are observed.

Calculate the total change of entropy as a function of the initial pressure p and the final partial

pressures when the two gases mix. The data of the gases are given in the figure.

Solution:

When the tap is opened the two gasesmix by diffusion. The diffusion goes on until the composition of

the gas is homogeneous. It is far more likely that the gases mix by diffusion than that the gases remain

separate. Hence, the total entropy change is expected to be positive.

In a gas, the distances between the molecules are large and the interaction between them can be

neglected.Thediffusionofeachgasisindependentoftheother.Thetotalentropychangecanberegarded

as the sumof theentropychangeofeachgasafter its separatediffusionfromonecontainer into theother.

DSmix ¼ DSmixA þDSmix

B ð10Þ

nAkmolofgasAchange theirpressure fromp topAwherepA is itsfinal partial pressure. In the sameway

nB kmol of gas B change their pressure from p to pB.

The initial pressure and the final total pressure are equal as no pressure change is observed.

p ¼ pA þ pB ð20ÞUsing the result of Example 1.1 we obtain

DSmixA ¼ nAR ln

V1 þV2

V1

ð30Þ

and

DSmixB ¼ nBR ln

V1 þV2

V2

ð40Þ

The partial pressures can be calculated with the aid of Boyle’s law applied on each gas:

pV1 ¼ pAðV1 þV2Þ ð50Þ

and

pV2 ¼ pBðV1 þV2Þ ð60Þ

nA p V1tap

n = number of kmolp = pressureV = volumeThe temperature T is constant.

nB p V2

pA + pB pA + pBtap

After the diffusion the gases aremixed and the pressure is equal inthe two containers. Thetemperature T is constant.


The ratio of the volumes from Equations (50) and (60) are inserted into Equations (30) and (40) and weobtain

DSmixA ¼ nAR ln

p

pAð70Þ

and

DSmixB ¼ nBR ln

p

pBð80Þ

The total entropy change is

Smix ¼ DSmixA þDSmix

B ¼ nAR lnp

pAþ nBR ln

p

pBð90Þ

The ratio of the pressures is > 1 and the entropy change is therefore positive, as predicted

above.

Answer: The total entropy change when the gases mix is

nAR lnp

pAþ nBR ln

p

pB; where p ¼ pA þ pB:

Entropy Change at Mixing Two Liquids or SolidsDiffusion occurs not only in gases but also in liquids and solids. The entropy change DSmix or simply

Smix, owing to mixing of two compounds in a melt or a solid can be calculated if we make a minor

modification of Equation (90) in Example 1.2.

Instead of the partial pressures of the two gases we introduce the mole fractions xA and xB:

xA ¼ pA

pA þ pBand xB ¼ pA

pA þ pB

Equation (90) can then be written as

Smix ¼ � nAR lnxA � nBR lnxB ð1:10Þ

By use of the relationship n ¼ nA þ nB where n is the total number of kmole we obtain

Smix ¼ � nRðxAlnxA þ xBlnxBÞ ð1:11ÞThe molar entropy of mixing will then be

Smixm ¼ �RðxAlnxA þ xBlnxBÞ ð1:12Þ

Equations (1.11) and (1.12) are directly applicable onmixtures of gases but also on liquids and solids.

These applications will be discussed later.

Entropy and ProbabilityThe two examples given above indicate that entropy in some way is related to probability. The

probability function can be found by the following arguments.

Consider N molecules in a container with the volume V (Figure 1.3). The molecules do not

interact at all. Each molecule is free to move within the volume V. The probability of finding it

within a unit volume is the same everywhere. Hence the probability of finding a molecule within a

volume V1 is V1/V. The probability of finding two molecules within the same volume V1 equals the


product of their probabilities (V1/V)2. The probability of finding N molecules within a particular

volume V1 is (V1/V)N.

Equation (20) in Example 1.1 on page 4 and Equation (90) on page 6 give us the clue to relating

entropy and probability. We have seen above that the overall probability is the product of the

probabilities of independent events. We also know that the entropy of two systems is the sum of their

entropies. It is striking that the logarithmic function converts the multiplicative property of

probability to the additive property of entropy.

These arguments led the physicists Maxwell and Boltzmann in 1877 to suggest a relationship

between entropy and probability. Boltzmann interpreted entropy as a measure of the order, or rather

disorder, of a system. The more probable a state of a system is and the greater its disorder is, the

higher will be its entropy.

Boltzmann derived an alternative expression of entropy by using statistical thermodynamics.

S ¼ kBln P ð1:13Þ

where

kB ¼ Boltzmann’s constant

P ¼ probability of the state.

Equation (1.13) is the fundamental relationship between entropy and probability. P is called the

statistical weight of the state, i.e. its configuration.

Consider a binary system, a gas or a liquid, of twocomponentsBandC.NB atomsofBandNCatoms

of C are arranged at random among N¼NBþNC sites. This can be done in many different ways and

is equivalent tomixing the two components. The statistical weightP is the number of alternativeways

to arrange the B and C atoms among the N sites. Statistical considerations give the result

P ¼ N!

NB!NC!¼ N!

NB!ðN�NBÞ! ð1:14Þ

We use Stirling’s formula

lim N!!ffiffiffiffiffiffi2p

pNNþ 1=2e�N when n!1

for the very high numbers N, NB and NC¼N�NB and obtain

P ¼ NN þ 1=2

NNB þ 1=2B ðN�NBÞN �NB þ 1=2

e�N

e�NBe�ðN�NBÞ ð1:15Þ

The last factor in Equation (1.15) is equal to 1. The term 1/2 in the exponents can be neglected in

comparison with N and NB. Taking the logarithm of both sides of Equation (1.15) we obtain

ln P ¼ N lnN �NBlnNB �ðN �NBÞlnðN �NBÞ ð1:16ÞIf we introduce the mole fractions

xB ¼ NB

Nand xC ¼ N�NB

Nand the relationship xB þ xC ¼ 1

Equation (1.16) can, after some calculations, be transformed into

lnP ¼ Nð� xB lnxB � xC lnxCÞ ð1:17Þ

Instead of N we introduce n, the number of kmol with the aid of the relationship:

n ¼ N

NA

ð1:18Þ

V

V1

Figure 1.3 Volume element.


where NA is Avogadro’s number, i.e. the number of molecules of 1 kmol. Hence, we obtain

ln P ¼ nNAð� xBlnxB � xClnxCÞ ð1:19Þ

Instead of NA we introduce R/kB into Equation (1.18), where kB is Boltzmann’s constant.

kBlnP ¼ � nRðxBlnxB þ xClnxCÞ ð1:20Þ

According to Equation (1.13) kBlnP equals the entropy change Smix when the two components B and

C mix. Hence, we obtain

Smix ¼ � nRðxBlnxB þ xClnxCÞ ð1:21Þ

Equation (1.21) is valid for homogeneously mixed solids and liquids.

Equation (1.21) is identical with Equation (1.11) on page 6 derived in an entirely different way.

The conclusion is that the concept of entropy as a function of probability is in complete agreement

with the classical theory and does not lead to any contradictions.

Entropy Change during SolidificationFreezing and boiling causes changes in the molecular order and are therefore expected to lead to

changes in entropy.

At freezing and melting the system is in equilibrium and the temperature is constant. Energy is

transferred reversibly and isothermally in the shape of heat between the system and its surroundings.

At constant pressure the heat absorbed by a system equals its enthalpy increase:

ðDQtrÞp ¼ ðDHtrÞp ð1:22Þ

and the entropy change will be

DS ¼ DHtr

Ttrð1:23Þ

Subscript ‘tr’ stands for transformation.

At solidification the system emits heat and the phase transition is exothermic. In this case the

enthalpy change is negative and the entropy change is negative. The system changes from a

disordered state (liquid) to a more ordered state (solid). The disorder decreases.

In the case of melting the opposite is valid. The phase transition is endothermic and the entropy

change is positive. Heat has to be added to the system and its disorder increases.

1.2.4 Gibbs’ Free Energy

Consider a system in thermal equilibrium with its surroundings at the temperature T. A spontaneous

change of state of the system always leads to an increase of the entropy of the system. This can be

expressed mathematically by Clausius’ inequality

dS� dQreversible

T� 0 ð1:24Þ

If the process is reversible the sign of equality is valid. In the case of an irreversible process the upper

sign has to be used. Heat transfer at constant pressure is of special interest in chemistry

and metallurgy. If there is no other work than expansion work, dQrev¼ dH and Equation (1.24)

can be written as

TdS� dH � 0 ð1:25Þ


Equation (1.25) is valid if either of the following two conditions is fulfilled

1. The enthalpy of the system is constant.

dSH;p � 0 ð1:26Þ

2. The entropy of the system is constant.

dHS;p � 0 ð1:27Þ

The conditions 1 and 2 can be expressed in a simpler and understandable way by introduction of

another thermodynamic function, the Gibbs’ free energy or Gibbs’ function

G ¼ H� TS ð1:28Þ

When the state of the system changes at constant temperature the change of G equals

dG ¼ dH� TdS ð1:29Þ

Equation (1.25) can be expressed in terms of the Gibbs’ function as

dGT;p � 0 ð1:30Þ

. At constant temperature and pressure spontaneous processes always occur in the direction of

decreasing G.

The deviation of the thermodynamic function G of a system from its equilibrium value can be

regarded as the driving force in chemical and metallurgical reactions. At equilibrium

dGT;p ¼ 0 ð1:31Þ

and the driving force is zero. This condition will be applied later in this chapter.

Gibbs’ free energy is a most useful instrument for studying the driving forces of various processes,

for example chemical reactions, solution processes, melting/solidification and evaporation/

condensation processes.

1.2.5 Intensive and Extensive Thermodynamic Quantities

Consider a single-component system, for example a pure metal, which consists of a certain amount

of the pure substance. The system can be described by a number of quantities such as volume V,

pressure p, absolute temperature T, number of kmol n of the pure element, internal energy U and

entropy S.

The quantities of the system can be divided into two groups

. intensive quantities, which are independent of the amount of matter;

. extensive quantities, which are proportional to the amount of matter.

The pressure p, the temperature T are intensive quantities while volume, energy and entropy of the

system are examples of extensive quantities.

A single-component system may consist of several phases, e.g. solid and liquid phases. A single-

component system can be regarded as a special case of a multiple-component system. Phases in

single- and multiple-component systems will be discussed in Section 1.5.1.


Extensive quantities, which refer to 1 kmol, will normally be denoted by capital letters with no

index, otherwise with an index or by the product if the capital letter and the number of kmol.

In metallurgy an index ‘m’ for molar (1 kmol) is often used. We will also use this principle of

designation. As we strictly use the SI system the concept molar is used in the sense of 1 kmol.

Gibbs’ free energy G is an extensive thermodynamic quantity that will be frequently used in this

and the following chapters, especially in Chapter 2.

1.3 Thermodynamics of Single-Component Systems

A single-component system may consist of only one phase, for example a solid, a liquid or a gas

depending on the temperature. At certain temperatures discontinuous phase transformations occur

when the system changes from one phase to another.

Examples of such transformations are melting of solid metals and evaporation of liquids. In this

section wewill study this type of processes in single-component systems. The results are general but

will entirely be applied on the special single-component systems, which consist of pure metals.

At the transformation temperatures two phases are in equilibrium with each other. The conditions

for equilibriumwill be set up in terms of thermodynamics and the influence of someparameters on the

transformation temperatures will be studied.

1.3.1 Clausius–Clapeyron’s Law

The general expression for equilibrium between two arbitrary phases a and b in contact with each

other is

Ga ¼ Gb ð1:32Þ

The Gibbs’ free energies can be regarded as functions of the independent variables pressure p and

temperature T.

A change of one of the independent variables results in changes ofGa andGb. If the equilibrium is

to be maintained a simultaneous and corresponding change of the other independent variable must

occur. There must be a relationship between Dp and DT. In order to find it we use the condition for

maintaining the equilibrium of the system:

dGa ¼ dGb ð1:33Þ

Gibbs’ free energy is defined as

G ¼ H� TS ¼ Uþ pV � TS ð1:34Þ

The total differential of G is

dG ¼ dUþ pdV þVdp� TdS� SdT ð1:35Þ

By use of the first law of thermodynamics, dQ ¼ dUþ pdV and the definition of entropy,

dS ¼ dQ=T ¼ ðdUþ pdVÞ=T , Equation (1.35) can be reduced to

dG ¼ Vdp� SdT ð1:36Þ

Equation (1.36) is applied on the two phases a and b and the results are introduced into

Equation (1.33).

Vadp� SadT ¼ Vbdp� SbdT ð1:37Þ


or

dT

dp¼ Vb �Va

Sb � Sa¼ DVa! b

DSa! b Clausius--Clapeyron’s law ð1:38Þ

where

DVa! b ¼ difference between the volumes of the b and a phases, i.e. Vb�Va

DSa! b ¼ difference between the entropies of the b and a phases, i.e. Sb� Sa.

Equation (1.38) is the Clausius–Clapeyron’s law, which is the desired relationship between the

changes in pressure and temperature in a two-phase system at equilibrium. It will be applied on

the melting and boiling processes below.

1.3.2 Equilibrium between Liquid and Solid Phases. Influence of Pressureand Crystal Curvature on Melting Point

Influence of Pressure on Melting Point

Consider a solid metal at the melting point in equilibrium with the corresponding metal melt. If

the pressure p0 is increased by an amount dp to p¼ p0þDp, a new equilibrium is developed

and the temperature is changed by the amount dT.

The temperature change or the change of melting point as a function of the pressure change can be

found if we apply Clausius–Clapeyron’s law on the liquid–solid system. Solidification means that

a liquid phase a is transformed into a solid phase b at the melting-point temperature.

If we apply Equation (1.38) on 1 kmol of the liquid, which solidifies, we obtain

dT

dp¼ DVL! s

m

DSL! sm

ð1:39Þ

Subscript ‘m’ stands for molar (1 kmol).

If DSL! sm is substituted by DHL! s

m =TM (compare Equation (1.23) on page 8) in Equation (1.39)

we obtain

dT

dp¼ TMDVL! s

m

DHL! sm

¼ TM Vsm �VL

m

� ��Hfusion

m

ð1:40Þ

or

DTM ¼ TMðpÞ� TMðp0Þ ¼TM VL

m �Vsm

� �Hfusion

m

Dp ð1:41Þ

where

DTM ¼ change of melting point, i.e. TM(p)� TM(p0)

p0 ¼ normal pressure, usually 1 atm

p ¼ pressure p equal to p0þDpTM(p) ¼ melting-point temperature at pressure p

DVL! sm ¼ difference in molar solid and liquid volumes at temperature TM, i.e. V

s�V L

�DHL! sm ¼ molar heat of fusion H fusion

m ¼ �ðHs �HLÞ.

As the molar volume of a metal melt normally is larger than that of the solid metal and the heat of

fusion is positive, an increase of the pressure leads to an increase of the melting point.

The pressure dependence on the melting-point temperature in metal melts is very small. Moderate

changes of the pressure during the casting and solidification processes hardly influence the liquidus


temperature at all. The effect can normally be neglected except for pressures of the magnitude

�102 atm.

Influence of Crystal Curvature on Melting Point

It is a common practical situation that one phase has a much smaller volume than the other phase

while the phases still are at equilibrium with each other. An example of such a case is shown in

Figure 1.4a.

A small spherical a phase particle or a gas pore with radius r is floating freely in a melt of pressure

pL. The surface tension between particle and melt causes the pressure inside the particle to be higher

than outside in the melt. The pressure inside the melt is obtained by a mechanical equilibrium study,

which gives the result

pa ¼ pL þ 2s

rð1:42Þ

where s is the surface tension between the melt and the a phase.

Equation (1.42) is a special case of Laplace’s formula that states that there will be a pressure

difference, owing to surface tension at a point on a curved surface between two phases

pa ¼ pL þs1

rmax

þ 1

rmin

� �ð1:43Þ

where rmax and rmin are the maximum andminimum radii of curvature of the particle that is no longer

necessarily spherical. We introduce the concept curvature, which is denoted by K and defined by

the relationship

K ¼ 1

2

1

rmax

þ 1

rmin

� �ð1:44Þ

Equation (1.43) can then be written as

paK ¼ pL þ 2sK ð1:45Þ

Equation (1.45) is introduced into Figure 1.4b. At a planar surface K is zero and paK ¼ pL.

The pressure paK will therefore change when the solid surface is curved instead of planar. This

pressure change paK � pa0 is accompanied by a corresponding change of the melting point of the solid.

The difference in melting-point temperature can be calculated with the aid of Equation (1.41) while

the following conditions are assumed to be valid:

1. the a phase and the melt are in equilibrium with each other

2. the melt maintains a constant pressure pL independent of the particle curvature.

According to Equation (1.45) the pressure pa0 at a planar interface equals the constant liquid pressure

pL. It is also identical with the pressure p0 in Equation (1.41). Hence, we obtain

TMðpaKÞ� TMðpa0 Þ ¼TM Va

m �VLm

� �Hfusion

m

ðpL þ 2sKÞ� pL� �

or

TM paK� �� TM pa0

� � ¼ TM Vam �VL

m

� �Hfusion

m

2sK ð1:46Þ

Melt pL

α phase

rpLpα2σ

+=

Figure 1.4a Spherical particle in a

melt. Curvature K equals 1/r.

Melt pL

α phase

KpLp 2σK +=α

Figure 1.4b Spherical particle in a

melt. Curvature K equals 1/r.


where

TMðpaKÞ ¼ melting point at curvature K of the particle and constant liquid pressure pL

TMðpa0 Þ ¼ melting point at curvature 0 and constant liquid pressure pL

s ¼ surface tension

K ¼ average curvature of the particle

Vam ¼ molar volume of the a phase at the melting point

VLm ¼ molar volume of the melt at the melting point.

K is positivewhen the curvature is convex. This is the case for a growing dendrite tip in a metal melt.

As the volume change in Equation (1.46) is negative, there will be a decrease in melting-point

temperature in the neighbourhood of a growing dendrite tip. The more curved the dendrite tip is, the

lower will be the melting point.

1.3.3 Equilibrium between Liquid and Gaseous Phases.Influence of Pressure on Boiling Point. Bubble Formation in Melts

Influence of Pressure on Boiling Point

Clausius–Clapeyron’s law (Equation (1.38) on page 11) is also valid for evaporation and condensa-

tion. It can be used for calculation of the change in boiling-point temperature of a liquid and its vapour

in equilibrium with each other, caused by a change of pressure of the two phases. Boiling means that

a liquid phase L is transformed into a gaseous phase g at the boiling-point temperature.

If we apply Equation (1.38) on 1 kmol of the liquid, which evaporates, we obtain

dT

dp¼ DVL! g

m

DSL! gm

ð1:47Þ

If DSL! gm is substituted by DHL! g

m =TB in Equation (1.47) we obtain

dT

dp¼ TBDVL! g

m

DHL! gm

¼ TB Vgm �VL

m

� �H

evapm

ð1:48Þ

or

DTB ¼ TBðpÞ� TBðp0Þ ¼TB Vg

m �VLm

� �H

evapm

Dp ð1:49Þ

where

DTB ¼ change of boiling-point temperature

p0 ¼ normal pressure, usually 1 atm

p ¼ pressure equal to p0þDpDVL! g

m ¼ difference in molar gaseous and liquid volumes at boiling-point temperature TBDHL! g

m ¼ molar heat of evaporation Hevapm .

As the molar volume of the vapour is very much larger than that of the liquid, an increased pressure

corresponds to an increase of the boiling-point temperature and vice versa.

The boiling point change is of themagnitude 104 times larger than themelting point change for the

same change of pressure and can not be neglected.

Equation (1.49) can be simplified by neglecting the molar volume of the liquid in comparison with

that of the vapour. If we replace the quantities DT and Dp by the corresponding differentials and

assume, somewhat doubtfully, that the gas behaves like an ideal gas, we obtain for a temperature T

close to the boiling point

dT ¼ T

Hevapm

Vgmdp ¼ T

Hevapm

RT

pdp ð1:50Þ


or

dp

p� Hevap

m

R

dT

T2ð1:51Þ

It is reasonable to assume that the heat of evaporation is approximately independent of pressure and

temperature.

ðp

p0

dp

p� Hevap

m

R

ðT

TB

dT

T2ð1:52aÞ

or

lnp

p0¼ Hevap

m

R

1

TB� 1

T

� �ð1:52bÞ

Figure 1.5 shows ln p as a function of 1/T for some common elements. The straight lines in the figure

shows that the approximations we made above seem to be justified and acceptable.

Equation (1.52b) can alternatively be written as

DTB ¼ TBðpÞ� TBðp0Þ ¼ RTBðpÞTB p0ð ÞH

evapm

lnp

p0ð1:53Þ

which gives the change of boiling-point temperature as a function of pressure. The molar heat of

evaporation is positive. At p¼ p0 the boiling point is TB(p0). An increase of the pressure leads to a

positive value ofDT, i.e. the boiling point increases. A decrease of the pressure results in a decrease of

the boiling point.

Bubble Formation in Melts

The condition for bubble formation in a liquid is that the pressure inside the bubble� the pressure of

the surrounding liquid.

To treat bubble formationwe substitute thea phase in Figure 1.4a by a gas phase of pressure p ¼ pgK

(Figure 1.6a). The pressure inside a bubblewith the curvatureK (page 12) is derived in the sameway as

on page 12. In analogy with Equation (1.45) we obtain the total pressure inside the bubble equals

pgK ¼ pL þ 2sK ð1:54Þ

where pL ¼ p0 þ rgh.

Figure 1.5 Vapour pressure as a

function of 1/T for some common

elements.

The melting points of the metals

are marked by small black dots.

The rectangle represents four dif-

ferent axes. The functions and the

corresponding scales are found in

the corners. Reproduced from

Elsevier � 1983.

p0

Surface of the melt

Melt

Gas

pp LgK + 2σK= pL

Figure 1.6a Pressure inside a

spherical gas pore in a melt. Curva-

tureK equals 1/rwhere r is the radius

of the gas pore. p0 is the pressure

above the surface.


The smaller the bubble is when it is formed, the larger will be pgK, which facilitates bubble formation.

If we replace p by pgK in Equation (1.53) we obtain

DTB ¼ TBðpgKÞ � TBðp0Þ ¼ RTBðpgKÞTBðp0ÞH

evapm

lnpL þ 2sK

p0ð1:55Þ

The third term in Equation (1.55) is always positive. This leads to the obvious conclusion that a gas

bubble with a higher internal pressure than the surroundings can only be formed above the normal

boiling temperature TB (p0) of the melt. Energy for evaporation must be supplied, or else the boiling

will stop.

The equilibrium between phases will be discussed more generally in the following sections.

If the pressure p0 is kept low by pumping away the gas above the surface of the melt (Figure 1.6b),

the pressure pL, which equals p0þ rgh, will also go down and the boiling-point temperature becomes

strongly reduced. The consequence is that the bubble-formation condition is easier to fulfil and the

boiling starts at a much lower temperature than normally.

1.3.4 Molar Gibbs’ Free Energy of a Pure Metal

The Gibbs’ free energy of a single phase, for example a pure metal, is defined by Equation (1.28) on

page 9. If we apply this definition to the Gibbs’ free energy of 1 kmol of the pure metal and use an

index m, which stands for molar, we obtain

Gm ¼ Hm � TSm ð1:56Þ

Eachmetal has its own characteristic values of the thermodynamic quantities. Thevalues of themolar

enthalpy and the molar entropy of the solid metal are different from those of the liquid metal.

The values of Hm are fairly constant but Sm varies with the temperature for each phase.

Figures 1.7 and 1.8, which show the functions

Gsm ¼ Hs

m � TSsm ð1:57aÞ

GLm ¼ HL

m � TSLm ð1:57bÞ

confirm the statements above. We can conclude that the thermodynamic quantities vary with

temperature, as the curves are non-linear.

Often, a reference temperature Tref is used instead of zero and relative values DH and DS are

introduced:

DHim ¼

ðT

Tref

CpðTÞdT and DSim ¼ðT

Tref

CpðTÞT

dT ð1:58aþbÞ

To pump

↑

Surface of the melt p0

Melt

pL

Gas

pp LgK + 2σK=

Figure 1.6b Pressure inside a

spherical gas pore in amelt. Curvature

K equals 1/r where r is the radius of

the gas pore. h is the depth of the

bubble below the surface.

G

T

sm

Figure 1.8 The molar Gibbs’ free energy of a liquid

metal A as a function of temperature.

G

T

sm

Figure 1.7 The molar Gibbs’ free energy of a

solid metal A as a function of temperature.


Calculation of Molar Enthalpy and Entropy of Solids and Liquids

The curves in Figures 1.7 and 1.8 are calculated from Equations (1.57a) and (1.57b) in combination

with the relationships

Him ¼

ðT

0

CpðTÞdT and Sim ¼ðT

0

CpðTÞT

dT ð1:59aþbÞ

where i¼ s and L, respectively.

The entropy of the liquid phase is normally larger than that of the solid. For this reason the curve in

Figure 1.8 is steeper than that in Figure 1.7.

1.4 Thermodynamics of Multiple-Component Systems

Both single- and multiple-component systems may be homogeneous, i.e. appear as a single phase

with uniform composition and physical properties, or heterogeneous, i.e. consist of several phases

with different physical properties. At the phase boundaries these properties change abruptly. In

multiple-component systems the composition of each phase must be specified.

Below, we will introduce a number of new thermodynamic concepts and relationships between

them, which will be used on binary alloys in the remaining sections in this chapter and in Chapter 2.

They are generally valid for many quantities in multiple-component systems, for example volume V,

enthalpyH, entropy S and Gibbs’ free energyG and also for several other thermodynamic functions.

Tomake the description below as short and surveyable as possible and avoid boring repetitions we

use thevolumeV as a principle example throughout this section,when newconcepts and relationships

are introduced. However, we want to point out strongly that the concepts and relationships are also

valid for other thermodynamic quantities.

. In all definitions, concepts and relationships the enthalpyH, the entropy S or the Gibb’s free energy

G can replace the volume V.

Independent Thermodynamic Variables of Multiple-Component Systems

As a multiple-component system we will consider an alloy to make the description below realistic.

Alloys can be described by the following independent variables

. temperature;

. pressure;

. composition.

The thermodynamic quantities of a multiple-component system, for example an alloy, changes very

little when the pressure is changed and the influence of pressure can be neglected in most cases.

The influence of temperature on thermodynamic variables, for example the Gibbs’ free energy,

is often strong and must definitely be taken into consideration.

The composition of a system is expressed as concentration of its components. Usual measures are

number of kmol, mole fraction and atomic per cent (at%).

1.4.1 Partial Molar Thermodynamic Quantities

Consider a system, which consists of components A, B . . . c where c corresponds to the number of

components. Index i refers to component i of the system where i¼A, B, . . . c. The number of kmol

of component i in the alloy is ni.

Each system has its own characteristic thermodynamic quantities such as volume V and entropy S.

In addition, it is convenient and useful to define so-called partial molar thermodynamic quantities


or shortly molar quantities for multiple-component systems. A partial molar quantity is the cor-

responding molar quantity of each component instead the molar quantity of the total system. As an

example we use the volume V.

Partial molar quantity V i ¼ @V

@nið1:60Þ

The definition is general and can be applied on any thermodynamic quantity. It is applied in

Table 1.1 below on four of the most important and frequent basic thermodynamic quantities.

By use of basic mathematics we will derive some important and useful relationships. They are

derived below for the volume but are valid for all extensive thermodynamic quantities. Examples

are given in Table 1.1.

When the numbers of kmol of the components are changed by dnA, dnB . . . dnc, the volume dV of

the system changes by the amount

dV ¼ @V

@nAdnA þ @V

@nBdnB þ � � � þ @V

@ncdnc: ð1:61Þ

If we introduce the partial molar volume we obtain

dV ¼ VAdnA þVBdnB þ � � � þVcdnc � � � ð1:62Þ

Systems with Constant Composition

At constant pressure and temperature the intensive quantities V i remain constant. The total volume

of a system,which consists of nA, nB, . . . nc kmol of the components A, B . . . c, is found by integratingEquation (1.62).

V ¼ nAVA þ nBVB þ � � � þ ncVc ð1:63Þ

If Equation (1.63) is divided by the total number of kmol (nAþ nBþ � � � nc) the molar volume Vm of

the alloy (subscript m stands for molar) as a function of the mol fractions is obtained:

Vm ¼ xAVA þ xBVB þ � � � þ xcVc ð1:64Þ

Systems with Variable Composition

IfwedifferentiateEquation (1.63) for the case that thecompositionof the systemmaychangeweobtain

dV ¼ nAdVA þ nBdVB þ � � � þ ncdVc

� �þ VAdnA þVBdnB þ � � � þVcdnc� � ð1:65Þ

From a comparison between Equations (1.62) and (1.65) we can conclude that the first bracket in

Equation (1.65) must be zero. If we divide it with (nAþ nBþ � � � nc) we obtain

xAdVA þ xBdVB þ � � � þ xcdVc ¼ 0 ð1:66Þ

where dVA; dVB; . . . ; dVc are the changes in volume in the individual partial molar quantities when p

and T are constant.

Table 1.1 Some partial molar thermodynamic quantities (V, S, H and G)

Partial molar volume Partial molar entropy Partial molar enthalpy Partial molar Gibbs’ free energy

V i ¼ @V

@niSi ¼ @S

@niHi ¼ @H

@niGi ¼ @G

@ni


Graphical Construction of Partial Molar Quantities for Binary Systems

If we apply Equation (1.66) on a binary system, where xB is regarded as the independent composition

variable, we can derive some useful relationships. From Equation (1.66) it follows that

xA@VA

@xBþ xB

@VB

@xB¼ 0: ð1:67Þ

This relationship was first derived by Duhem. If we take the derivative of the binary version of

Equation (1.64) with respect to xB we obtain

@Vm

@xB¼ @xA

@xBVA þ xA

@VA

@xBþVB þ xB

@VB

@xB

Since xAþ xB¼ 1 and @xA/@xB¼�1 and Equation (1.67) is valid we obtain

@Vm

@xB¼ VB �VA ð1:68Þ

If we combine Equations (1.64) and (1.68) to an equation system we can solve the partial molar

volumes as functions ofxA,xB,Vm and its partial derivativeswith respect toxA andxB.The solution is

VA ¼ Vm � xB@Vm

@xB¼ Vm þð1� xAÞ @Vm

@xAð1:69Þ

VB ¼ Vm þð1� xBÞ @Vm

@xBð1:70Þ

where Vm ¼ xAVA þ xBVB according to Equation (1.64).

Equations (1.69) and (1.70) are called Gibbs–Duhem’s equations.

BothVm,VA andVB obviously varieswithxB.This is shown inFigure 1.9,which also shows a useful

graphical way to construct the partial molar volumes VA and VB when the xBVm curve is known.

The construction is based on Equations (1.69) and (1.70).

The intersections V0A, respectively V0

B, between the vertical axes and the Vm curve represent the

molar volumes of pure elements A and B, respectively, at a given temperature.

As we have emphasized before, the theory of partial molar quantities given above and completely

analogous formulas are valid for all thermodynamic functions.

1.4.2 Relative Thermodynamic Quantities and Reference States. RelativePartial Molar Thermodynamic Quantities or Partial Molar Quantitiesof Mixing

In the case of potential energy it is well known that a zero level has to be defined. This is also true for

thermodynamic quantities. An example is the entropy S. Its zero level appears as an integration

V B

V0B

V m

VA

V 0A x B

0 xB 1

A

V

B

Figure 1.9 Molar volume Vm of a

binary alloy as a function of the mole

fraction xB.

VA ¼ Vm � xB@Vm

@xB

VB ¼ Vm þð1� xBÞ @Vm

@xB


constant and is fixed by the third law of thermodynamics, which says that S¼ 0 for a pure crystalline

substance at T¼ 0K.

For the thermodynamic quantities specific reference states or standard states are defined.With the

aid of these reference states relative partial molar quantities can be defined. Normally, the state of

the pure substance is used as a reference state.

Lewis and Randall suggested that

A relative partial molar quantity¼ the difference between a partial molar quantity of a component in

a solution and the molar quantity of the pure substance.

The relative partial molar volume is also called the partial molar volume of mixing of substance i

and is designated by Vmixi .

Vmix

i ¼ V i �V0i ð1:71Þ

where

V i ¼ partial molar volume of component i in a solution, i.e. the volume change of an

infinitely large quantity of a solution of a given composition when 1 kmol of pure

substance i is added at specified pressure and temperature

V0i ¼ molar volume of the pure substance i in its standard state, i.e. at specified values

of p and T

Vmix

i ¼ the difference between the partial molar volume of component i in a solution and

the molar volume of pure substance i at fixed values of p and T.

The relative partial molar thermodynamic quantities are marked by a bar, a component subscript and

the designation ‘mix’.

Relationships between Partial Molar Quantities of Mixing in a Binary System

The relationships (1.61)–(1.70), which were derived for the partial molar volumeV, are also valid for

the relative molar thermodynamic quantities or simply for quantities with subscript ‘mix’.

In analogy with Equation (1.67) on page 18 there are partial differential equations that relate the

relative partial molar quantities of the components in a multiple component solution:

xA@V

mix

A

@xBþ xB

@Vmix

B

@xBþ � � � þ xc

@Vmix

c

@xc¼ 0 ð1:72Þ

If we applyEquation (1.72) on a binary system and use the relationshipxA¼ 1� xBwe obtainVmix

A as

a function of Vmix

B and xB:

Vmix

A ¼ �ðxB0

xB

1� xB

@Vmix

B

@xBdxB ð1:73Þ

Partial integration of Equation (1.73) gives

Vmix

A ¼ðxB0

Vmix

B

ð1� xBÞ2dxB � xBV

mix

B

1� xBð1:74Þ

1.4.3 Relative Integral Molar Thermodynamic Quantities orIntegral Molar Quantities of Mixing

In addition to the relative partialmolar quantitieswe have to define relative integralmolar quantities.

As an example we choose the relative integral molar volume Vmixm .


Vmixm ¼ the difference between the volume of 1 kmol of a solutionVm and the sum of the volumes of

the corresponding pure component substances

Vmixm ¼ Vm �ðxAV0

A þ xBV0B þ � � � þ xcV

0c Þ ð1:75Þ

With the aid of Equation (1.64) on page 18 and Equation (1.71) on page 19, Equation (1.75) can be

transformed into

Vmixm ¼ xAV

mix

A þ xBVmix

B þ � � � þ xcVmix

c ð1:76ÞThe relative integralmolar volume is thevolumeofmixing or thevolume of 1 kmol of the alloy formed

by xA kmol of pure substance Aþ xB kmol of pure substance Bþ . . .þ xC kmol of pure substance C.

For a binary system the relative integral molar quantities appear in a set of formulas, which are

analogous with Equations (1.69) and (1.70) on page 18. These equations give the partial molar

quantities of mixing as a function of the relative integral molar quantities.

The partial molar volumes of mixing for a binary system are:

Vmix

A ¼ Vmixm þð1� xAÞ @V

mixm

@xAGibbs--Duhem’s ð1:77Þ

Vmix

B ¼ Vmixm þð1� xBÞ @V

mixm

@xBequations ð1:78Þ

The integral molar volume of mixing of a binary alloy can be obtained as a function of the partial

molar free energy of mixing of component B by inserting the expression in Equation (1.74) for Vmix

A

into Equation (1.76). The result is

Vmixm ¼ ð1� xBÞ

ðxB0

Vmix

B

ð1� xBÞ2dxB ð1:79Þ

Graphical Construction of Partial and Integral Molar Quantities of Mixing for aBinary Alloy

Equations (1.77) and (1.78) are calledGibbs–Duhem’s equations. Vmix, Vmix

A and Vmix

B can be derived

graphicallyifthemolarvolumeVmofthebinaryalloyisknownasafunctionofcomposition.Theprinciple

is shown in Figure 1.10. The construction is based on Equations (1.71), (1.75), (1.77) and (1.78).

As an example of a thermodynamic quantity other than the volume, Figure 1.11 shows the molar

entropy of mixing of a binary system as a function of composition. Entropy can be regarded as a

measure of the disorder of a system. Mixing corresponds to disorder. Hence the entropy of mixing is

always positive, i.e. the curve is convex.

Molar enthalpy of mixing will be discussed in Section 1.5 in connection with ideal and non-ideal

solutions. The Gibbs’ free energy is discussed in Sections 1.6 and 1.7 and applied in Chapter 2.

BV0

BV

Vmmix

mV

AVmixAV

0AV xB

xB 1

A

0

B

V mixBV

Figure 1.10 The molar volume Vm of a system as a

function of the mole fraction xB.

The partial and integral molar volumes of mixing,

Vmix

A , VmixB and Vmix

m are marked in the figure.

Figure 1.11 Molar entropy of mixing of a binary

system as a function of composition.


1.4.4 Other Thermodynamic Functions and Relationships

The theory of thermodynamics of multiple-component systems is not complete without the theory of

ideal and non-ideal solutions.

These concepts, together with the concepts of activity and excess quantities, belong to the general

theory of thermodynamics but they will be introduced and penetrated in connection with the

thermodynamics of alloys in this book. The concepts are of utmost importance for the study of the

Gibbs’ free energy of alloys, which is the aim of Chapter 2.

1.5 Thermodynamics of Alloys

An alloy is a multiple-component metallic system. Depending on the pressure, temperature and

composition of the alloy it can either be homogeneous, i.e. consist of a single phase, or heteroge-

neous, i.e. consist of several phases.

When a secondmetallic component (solute) is added to a pure liquidmetal (solvent) a solution or a

single phase with a given composition is normally formed. When the solution solidifies a homoge-

neous solid solution is produced. If the metals can bemixed in any proportions, the miscibility is said

to be complete and either of the metals can be considered as solvent.

If the miscibility is not complete, the alloy becomes heterogeneous at certain temperatures and

compositions and a new phase is precipitated. The two immiscible phases have different composi-

tions, which depend on temperature. In a temperature–composition diagram there exists a regionwith

two phases.

The phase diagram of ametallic system describes the phases of the alloy and their compositions as

a function of temperature. These topics will be discussed in Chapter 2.

The so-called intermediate phases are a special type of phases. Their compositions correspond to

chemical compounds with integer proportions between the two kinds of metal atoms.

Classification of Phases

Phases of alloys may be classified according to their degree of atomic order.

. Liquid phase: A multiple-component phase with no long-range order.

. Disordered substitutional solid solution: Random distribution of the solute atoms on the lattice

sites of the solvent.. Interstitial solid solution: Random distribution of interstitial solute atoms in a solvent lattice.. Intermediate phase or chemical compound: Solvent and solute atoms, respectively, form two

independent but interpenetrating lattices.

In this chapter only binary alloys will be treated. In Chapter 2 a brief discussion of ternary alloys is

included in addition to the dominating binary alloys.

Binary Alloys

In the introduction to this chapter and this book the close connection between the driving force of

solidification and the solidification process inmetals and alloyswas pointed out. The driving forces of

solidification will be defined and calculated for pure metals and alloys in Chapter 2. They are closely

related to the Gibbs’ free energy of the solid and liquid phases.

It is fairly easy to find the driving force of solidification for puremetals. A binary system is farmore

complicated than a single-component system. It is necessary to start with a study of the thermody-

namics and stability of ideal and non-ideal solutions to find themolar Gibbs’ free energy as a function

of composition and temperature.When these functions are known, an expression of the driving force

of solidification of binary alloys can be derived.

In Section 1.4 the thermodynamics of multiple-component systems has been treated. By deriving

expressions of enthalpy and entropy of ideal and non-ideal solutions we obtain the contributions

required to find the molar Gibbs’ free energies for the liquid and solid phases of binary alloys as

functions of composition and temperature.


These functions are the basis of theGibbs’molar free-energy curves,which are drawn for the liquid

and solid phases in the next chapter.

1.5.1 Heat of Mixing

The enthalpy of mixing or, simpler, the heat of mixing is entirely a function of the forces between the

atoms in ametal or an alloy. These forces depend on the type of atoms involved and on the interatomic

distances. The temperature dependence of the heat ofmixing is thereforeweak. On the other hand the

heat of mixing of an alloy is strongly influenced by its composition.

Consider one kmol of a binary alloy, which contains NAB randomly distributed A–B atoms pairs

and an excess of either A–A or B–B atom pairs. The alloy has been formed by mixing 1/2 NAB

A–A atom pairs with 1/2 NAB B–B atom pairs. Figure 1.12 shows the energy changes of the system

when the alloy is formed, i.e. when the original A–A and B–B bonds are broken and pairs of

A–B bonds are formed instead. There is no energy change of the excess A–A or B–B atom pairs

before and after the mixing, as their bonds remain unchanged.

The molar heat of mixing is defined as

Hmixm ¼ the net energy, which has to be added to the system to mix NAB A atoms and NAB B atoms.

Figure 1.12a shows that the heat of mixing has the same magnitude as the enthalpy change of

mixing but opposite sign.

Hmixm ¼ �ðDHmÞmix ¼ NAB EAB � 1

2ðEAA þEBBÞ

ð1:80Þ

where

NAB ¼ number of atom pairs A–B per kmol of the alloy

�ðDHmÞmix ¼ molar enthalpy change at mixing

Eij ¼ binding energy between specified types of atoms.

When the forces between the A and B atoms are weaker than the forces between A–A atoms and

B–B atoms, the heat of mixing is positive. This is the situation in Figures 1.12a and b.

The solution process to split up strong A–A and B–B bonds and replace them with weaker

A–B bonds requires additional energy. Heat is absorbed by the system when the two components

mix. Primarily, the temperature of the alloy decreases and heat is successively transferred from the

surroundings.

Energy

Energy lost by thesystem to the sur-roundings when theA–B bonds are formed

Energy added to the system tobreak the A–A and B–B bonds

mixmH mix

initialm

finalm

mixm )(H H−=− ΔH

0

Initial state Final state:

½ NAB A–A atom pairsseparated from½ NAB B–B atom pairs

NAB A–B atom pairs

(NAB A atoms mixedwith NAB B atoms)

Figure 1.12a Enthalpy change and

heat of mixing when 1/2 NAB A–A

atom pairs mix with 1/2 NAB B–B

atom pairs.

Figure 1.12b. Molar heat of mix-

ing of a binary alloy as a function

of composition. Hmixm is positive if

the forces between unlike atoms are

weaker than the forces between like

atoms.


When the forces between the A and B atoms are stronger than the forces between A–A atoms and

B–B atoms, the heat of mixing is negative. This is the situation in Figures 1.12c and d.

The solution process to split up weak A–A and B–B bonds and replace them with stronger A–B

bonds releases energy. Heat is emitted by the system when the two components mix (Figure 1.12d).

Primarily, the temperature of the alloy increases and heat is gradually transferred to the surroundings.

The solution process is spontaneous and a stable alloy is formed. Apart from initial activation

energy the solution process does not require energy addition from outside.

Theoretical Model of Heat of Mixing

The numberNAB ofmixed atom pairs is proportional to the presence of the two types of atoms, i.e. the

concentrations of A and B. The heat of mixing Hmixm , based on the forces between neighbouring

atoms, can then be written

Hmixm ¼ �ðDHmÞmix ¼ LxAxB ð1:81Þ

where L is a constant, which can be derived by identification of Equations (1.80) and (1.81).

L includes both the proportionality constant and the bracket factor, as the binding energies of the three

molecule types are constant.

For ideal solutions L¼ 0 and the heat ofmixing is zero. The solubility of an ideal solution is total at

all proportions.

Depending on the interatomic forces and the sign of the bracket factor, the constant L and therefore

themolar heat ofmixingHmixm can be either positive or negative for non-ideal solutions (Figures 1.12).

Equation (1.81) will be used in next chapter in connection with a simple solution model.

1.5.2 Ideal and Non-Ideal Solutions

An ideal solution is defined as a solution in which

1. The homogeneous attractive forces (A–AandB–B) and the heterogeneous attractive forces (A–B)

are equal. From condition 1 it follows that

2. The heat of mixing is zero.

3. The solubility of the components is complete, i.e. the components dissolve at all proportions.

Solutions that do not fulfil the conditions 1–3, are non-ideal. The thermodynamics of non-ideal

solutions will be discussed in Section 1.7.

Energy

Energy lost by thesystem to the surroun-dings when the A–Bbonds are formed.

Energy added to the system tobreak the A–A and B–B bonds

0

Initial state mixmH mix

initialm

finalm

mixm )(H H−=− ΔH


Final state:


NAB A–B atom pairs(NAB A atoms mixedwith NAB B atoms)

Figure 1.12c Enthalpy change and

heat of mixing when 1/2 NAB A–A

atom pairs mix with 1/2 NAB B–B

atom pairs.

mixmH

0

xB

0 0.5 1

Figure 1.12d Molar heat of mixing

of a binary alloy as a function of

composition. Hmixm is negative if

the forces between unlike atoms are

stronger than the forces between like

atoms.


Survey of Concepts and Relationships of Gibbs’ Free Energy for Binary Solutions

Partial Molar Gibbs’ Free Energy

Gi ¼ @G

@nið10Þ

Gm ¼ xAGA þ xBGB ð20ÞSolutions with variable composition:

xAdGA þ xBdGB ¼ 0 ð30Þ

Gibbs–Duhem’s equations:

GA ¼ Gm þð1� xAÞ @Gm

@xAð40Þ

GB ¼ Gm þð1� xBÞ @Gm

@xBð50Þ

Relative Partial Molar Gibbs’ Free Energy or Partial Molar Gibbs’ Free Energy of Mixing

Gmix

i ¼ Gi �G0i ð60Þ

Gmix

i ¼ RT lnpi

p0ið70Þ

Duhem’s relationship:

xA@G

mix

A

@xBþ xB

@Gmix

B

@xB¼ 0 ð80Þ

Relationship between Gibbs’ free energy of mixing and the partial Gibbs’ free energy of mixing

Gmix

A ¼ �ðxB0

xB

1� xB

@Gmix

B

@xBdxB ð90Þ

and

Gmix

A ¼ðxB0

Gmix

B

ð1� xBÞ2dxB � xBG

mix

B

1� xBð100Þ

Relative Integral Molar Gibbs’ Free Energy or Integral Molar Gibbs’ Free Energy of Mixing

Gmixm ¼ Gm � xAV

0A þ xBV

0B

� � ð110Þ

Gmixm ¼ xAG

mix

A þ xBGmix

B ð120Þ

Gibbs–Duhem’s equations:

Gmix

A ¼ Gmixm þð1� xAÞ @G

mixm

@xAð130Þ

Gmix

B ¼ Gmixm þð1� xBÞ @G

mixm

@xBð140Þ

Relationship between the integral molar Gibbs’ free energy ofmixing and the partial molar Gibbs’

free energy of mixing:

Gmixm ¼ ð1� xBÞ

ðxB0

Gmix

B

ð1� xBÞ2dxB ð150Þ


1.6 Thermodynamics of Ideal Binary Solutions

Molar Enthalpy of Mixing

In an ideal solution of A and B atoms the A–B forces equals the A–A and B–B forces. From

Equation (1.80) on page 22we can conclude that themolar enthalpy ofmixingmust be zerowhen two

pure substances A and B mix. Hence, we have

Hmixm ¼ 0 ð1:82Þ

Molar Entropy of Mixing

In Section 1.2.3 we derived the desired expression. According to Equation (1.12) on page 6 themolar

entropy of mixing

Smixm ¼ �RðxAlnxA þ xBlnxBÞ ð1:12Þ

Molar Gibbs’ Free Energy of Mixing

Themolar Gibbs’ free energy ofmixing can easily be derivedwhenHmixm and Smix

m of an ideal solution

are given.

Gmixm ¼ Hmix

m � TSmixm ð1:83Þ

As Hmixm is zero we obtain

Gmixm ¼ RTðxAlnxA þ xBlnxBÞ ð1:84Þ

1.6.1 Molar Gibbs’ Free Energy of Ideal Binary Solutions

Mechanical Mixture of Two Components

In amechanical mixture of two pure elements there is no interaction between the two phases. In such

cases, overall thermodynamic quantities can be calculated as the sum of corresponding extensive

thermodynamic quantities of the individual components. The latter are often known and can be found

in tables.

Molar Gibbs’ Free Energy of an Ideal Binary Solution

If the two components form a solution instead of a mechanical mixture the interaction between the

atoms and the energy of mixing must be taken into consideration.

Consider a binary solution of components A and B. The molar Gibbs’ free energy Gm of the

solution is equal to the sumof themolar Gibbs’ free energies of the components plus themolar Gibbs’

free energy of mixing:

Gm ¼ xAG0A þ xBG

0B þGmix

m ð1:85Þ

where

Gm ¼ molar Gibbs’ free energy of the solution

G0i ¼ molar Gibbs’ free energy of the pure components (i¼A, B)

xi ¼ number of kmole of the components (i¼A, B)

Gmixm ¼ molar Gibbs’ free energy of mixing.


If the expression of Gmixm in Equation (1.84) is introduced in Equation (1.85) we obtain

Gm ¼ xAG0A þ xBG

0B þRTðxAlnxA þ xBlnxBÞ ð1:86Þ

As xA¼ 1� xB we obtain alternatively

Gm ¼ ð1� xBÞG0A þ xBG

0B þRT ð1� xBÞlnð1� xBÞþ xBlnxB½ � ð1:87Þ

Equation (1.86) is the desired expression for the Gibbs’ molar free energy Gm of an ideal binary

solution as a function of composition and temperature.

Equation (1.87) has been plotted in the Figures 1.13 for a given temperature. TheGm curvewill be

frequently used in Chapter 2.

The dotted line in Figure 1.13a represents the two first terms in Equation (1.87). The last term

is negative, as both mole fractions in Equation (1.87) are < 1 As the entropy of mixing Smixm is

always positive (Equation (1.12) on page 6 and Figure 1.11 on page 20), the term � TSmixm in

Equation (1.83) is always negative. Hence, theGm curve must be convex downwards and lie below

the dotted line.

1.7 Thermodynamics of Non-Ideal Binary Solutions

Before itmakes sense to discuss the corresponding quantities of non-ideal solutionswewill introduce

some new concepts, which are closely related to non-ideal solutions.

1.7.1 Activities of Non-Ideal Solutions Raoult’s and Henry’s Laws

Fugacity

Consider a pure liquid or solid substance in equilibrium with its vapour phase. The condition for

equilibrium is that the Gibbs’ free energies of the two phases are equal.

Giðl or cÞ ¼ GiðgÞ ð1:88Þ

We neglect the pressure dependence of Giðl or cÞ and assume that Giðl or cÞ � G0i ðl or cÞ where

G0i ðl or cÞ is the Gibbs’ free energy of the liquid or solid at standard pressure 1 atm and standard

temperature 273K. This is most reasonable at ordinary pressures.

Gm

0BG

0A GG mix

x B0

0 x B 1

A

G mixB

G mixA

G B

GA

B

Figure 1.13b The molar Gibbs’ free energy of ideal

binarysolutionsasafunctionofcomposition.Thepartial

and integral molar quantities are marked in the figure.

Gm0BG

0AG

xB

0

A B

1

Figure 1.13a The molar Gibbs’ free energy of ideal

binary solutions as a function of composition xB.


The Gibbs’ free energy of the gas Gi(g) depends on the pressure of the gas. We assume that the

vapour behaves like an ideal gas. On page 10 we derived the relationship

dG ¼ Vdp� SdT ð1:36Þ

For an ideal gas at constant temperatureV¼RT/p and dT¼ 0These expressions are introduced into

Equation (1.34) on page 10. After integration we obtain

ðGiðgÞ

G0iðgÞ

dG ¼ðpip0

RTdp

p

which gives

GiðgÞ ¼ G0i ðgÞþRT ln pi=p

0i ð1:89Þ

where

GiðgÞ ¼ Gibbs’ free energy of the ideal gas phase i

G0i ¼ standard Gibbs’ free energy, i.e. the Gibbs’ free energy at STP (273K, p0i ¼ 1 atm).

Nextwe consider a liquid or solid solution in equilibriumwith its gas phase. The pressure of the gas is

composed of the partial pressures of the component gases in equilibrium with the components of the

solution. The components of the liquid or solid solution are not assumed to be ideal. Equation (1.89) is

only valid for component i in a non-ideal liquid or solid solution if the pressure pi is replaced by a

corrected pressure or fugacity fi

Giðl or cÞ ¼ G0i ðgÞþRT ln fi= f

0i ð1:90Þ

where

Giðl or cÞ ¼ partial molar Gibbs’ free energy of component i in the solid or liquid solution

G0i ðgÞ ¼ standard Gibbs’ free energy of pure component i in the gas phase (273K, 1 atm).

fi ¼ fugacity or corrected partial pressure of component i in the gas phase over the

solution

f 0i ¼ fugacity of component i over the pure condensed phase.

If the solution is replaced by a pure condensed phase of component i the equilibrium condition

will be

Giðl or cÞ � G0i ðl or cÞ ¼ GiðgÞ ¼ G0

i ðgÞþRT ln f 0i ð1:91Þ

where the pressure of the gas has been replaced by the fugacity f 0i over the pure condensed phase.

If we subtract Equation (1.91) from Equation (1.90) we obtain by definition the partial molar

Gibbs’ free energy of mixing of component i in the solution

Giðl or cÞ ¼ G0i ðl or cÞþRT ln fi= f

0i

or

Gmix

i ðl or cÞ ¼ RT ln fi= f0i ð1:92Þ

Analogously, Equation (1.89) can be written

Gmix

i ðgÞ ¼ RT lnpi=p0i ð1:93Þ


As the gas and the solution are in equilibrium with each other, we can conclude that Gmix

i ðl or cÞ ¼G

mix

i ðgÞ and therefore

fi

f 0i¼ pi

p0ið1:94Þ

Definitions of Activity and Activity Coefficient

ActivityThe activity of component i in a solution is defined as the ratio of the fugacity fi of component i over

the solution to the fugacity of the standard state equal to the pure stable condensed phase at the

pressure 1 atm and temperature 273K.

ai ¼ fi= f0i ð1:95Þ

Giðl or cÞ ¼ G0i ðl or cÞþRT ln ai ð1:96Þ

The activity always has the value 1 at the standard state.

Activity CoefficientThe activity coefficient gi or fi is defined by the relationship

ai ¼ gixi or ai ¼ fixi ð1:97Þ

where xi is the mole fraction of component i in the solution. At low pressures the fugacities can be

replaced by the pressures

ai gixi � pi=p0i or ai fixi � pi=p

0i ð1:98Þ

Raoult’s Law

Raoult’s law is the statement that the activity and the mole fraction are equal, e.g. the activity

coefficient equals 1.

If we designate the activity coefficient fi, Raoult’s law is a special case of Equation (1.97) a¼ fixiand can be written

ai ¼ xi or fi ¼ 1 ð1:99Þ

It has been observed that Raoult’s law is obeyed in the limiting case when xi approaches unity. This

is equivalent to the statement that the solvent in a dilute solution obeys Raoult’s law when the

concentration of the solute approaches zero.

limxi ! 1

asolventi ¼ xi or limxi ! 1

fi ¼ 1 ð1:100Þ

which gives

Gi ¼ G0i þRT lnxi ð1:101Þ

The deviation fromRaoult’s law for real solutions (I and III) (Figure 1.14) can be either positive or

negative. Case II corresponds to an ideal solution. Equation (1.101) is only valid for the solvent in

dilute solutions. For a binary solution index i equals A.

solventia

1.0

Raoult’s law a i = x i

I II

III

x i

0 1

Figure 1.14 Raoult’s law. Activity

of solute component as a function of

mole fraction.


Henry’s Law

The activity asolutei ¼ gixi of the solute is proportional, but not equal, to the mole fraction when xiapproaches zero, i.e. in dilute solutions. The designation gi is used here for the activity of the solute tomake it possible to distinguish between the activities of the solvent and the solute. In a binary solution

index i¼B

limxi ! 0

asolutei ¼ limxi ! 0

gixi ¼ g0i xi or limxi ! 0

gi ¼ g0i ð1:102Þ

where g0i is a constant.

Hence, at low pressures

asolutei fi

f 0i� g0i xi and as

fi

f 0i� pi

p0i

we obtain

pi ¼ p0i g0i xi ¼ const xi ð1:103Þ

. The partial pressure p over a dilute solution of component i is proportional to itsmole fraction in the

solution.

This is Henry’s law, which has been known formore than 200 years. As Figure 1.15 showsHenry’s

law is not valid for largevalues ofxi. The activity gi of the solute is generally described by the constantg0i and a relative activity coefficient, defined by the relationship

greli ¼ gig0i

ð1:104Þ

WhenHenry’s law is valid the value of greli is 1.When Henry’s law is not valid for solute component i

its activity can generally be written

asolutei ¼ gixi or asolutei ¼ greli g0i xi ð1:105Þ

By use of Equations (1.96) and (1.105) the Gibbs’ free energy of component i in the solution can be

written as

Gi ¼ G0i þRT ln g0i þRT ln greli xi ð1:106Þ

It can be seen from Figure 1.16 that a positive deviation from Raoult’s law corresponds to a negative

deviation from Henry’s law and vice versa.

It should be noticed that the activities and activity coefficients are not the same in Raoult’s and

Henry’s laws. They differ by a factor g0i . This is shown in Table 1.2.

solutei

a

0i

Henry’s law

i0i

solutei

xa =

Non-idealsolution

0

x i

1

γ

γ

Figure 1.15 Henry’s law. For a

binary solution i¼B.

The deviation from Henry’s law

for real solutions can be either posi-

tive or negative.

Activity

Henry’slawa solute

a solvent

Raoult´s law

Non-ideal solution

xi

0 1

Figure 1.16 Activity as a function

of component i in a solution.

Table 1.2 Comparison between Raoult’s and Henry’s laws.

Law Activity Gibb’s free energy

General law ai¼ fixi Gi ¼ G0i þRT lnai

Raoult’s law, valid for solvents

in dilute solutions

ai¼ xi and fi¼ 1

valid when xi ! 1

Gi ¼ G0i þRT lnxi

General laws ai ¼ gixi ¼ greli g0i xi Gi ¼ G0i þRT lng0i þRT lngreli xi

Henry’s law, valid for solutes

in dilute solutions

ai ¼ g0i xi and greli ¼ 1

valid when xi ! 0

Gi ¼ G0i þRT lng0i þRT lnxi


Calculation of Activity Coefficients in Binary Solutions

With the aid of Equation (1.98) on page 28 Equation (1.93) on page 27 can be written as

Gmix

i ¼ RT lnpi

p0i¼ RT ln gixi ð1:107Þ

These expressions are introduced into Equation (80) on page 24.

After division with RT we obtain:

xA@ln gAxA

@xBþ xB

@ln gBxB@xB

¼ 0 ð1:108Þor

xA@ðln gA þ lnxAÞ

@xBþ xB

@ðln gB þ lnxBÞ@xB

¼ 0

Partial derivation in combination with the conditions xA¼ 1� xB and @xA/@xB¼�1 gives

xA@ln gA@xB

� 1þ xB@ln gB@xB

þ 1 ¼ 0

or

xA@ln gA@xB

þ xB@ln gB@xB

¼ 0 ð1:109Þ

If we replace xA by (1� xB) and integrate Equation (1.109) we obtain

ln gA ¼ �ðxB0

xB

1� xB

@ln gB@xB

@xB ð1:110Þ

By partial integration we obtain

ln gA ¼ � xB

ð1� xBÞ2þ

ðxB0

ln gBð1� xBÞ2

@xB ð1:111Þ

If gB is known as a function of xB it is possible to calculate gA. The accuracymay be poor if xB is close

to unity. In this case both ln gB and (1� xB) approach zero.

1.7.2 Excess Quantities of Non-Ideal Solutions

The concept excess quantity of a non-ideal solution is closely related to activity and activity

coefficient. The definition of an excess quantity XEx is

ðXÞnon-ideal solution ¼ ðXÞideal solution þXEx ð1:112Þ

. Excess quantity¼ a quantity of a non-ideal solution minus the corresponding quantity of an

ideal solution.

The quantityX can for example beH, SorG or anyother thermodynamic functions of these quantities.

By definition all excess quantities are zero for ideal solutions. TheGibbs–Duhem’s equations and the

other relationships given in Section 1.4 are also valid for excess quantities.

As examples of excess quantities of non-ideal solutions we choose the excess partial molar

quantities and the excess integral molar quantities of enthalpy, entropy and Gibbs’ free energy to

illustrate the definition.


Excess Enthalpy

AsHmixi ¼ 0 for an ideal solution, the partial molar enthalpy of mixingH

mix

i of component i of a non-

ideal solution is simply equal to the excess enthalpy

As ðHmix

i Þnon ideal ¼ 0þHEx

i we obtain

HEx

i ¼ Hmix

i ð1:113ÞThe Equation (1.112) on page 30 is valid both for partial and integralmolar quantities. Hencewe have

for a non-ideal solution

HExm ¼ Hmix

m

� �non-ideal

ð1:114Þ

Excess Entropy

According to Equation (1.12) on page 6 the integral molar entropy of mixing or simply the molar

entropy of mixing can be written as

ðSmixm Þideal ¼ �RðxAln xA þ xBlnxB þ � � �Þ ð1:12Þ

According to Equation (1.112) on page 30 the molar entropy of mixing of a non-ideal solution can

therefore be written as

Smixm

� �non ideal

¼ �RðxAln xA þ xBlnxB þ � � �Þþ SExm ð1:115Þ

or

SExm ¼ Smixm

� �non ideal

þRðxAln xA þ xBlnxB þ � � �Þ ð1:116Þ

Analogously to Equation (1.64) on page 17 we obtain

SExm ¼ xASEx

A þ xBSEx

B þ � � � ð1:117Þ

Smixm ¼ xAS

mix

A þ xBSmix

B þ � � � ð1:118Þ

The excess partial molar entropy of component i of a non-ideal solution will be

SEx

i ¼ Smix

i

� �non-ideal

þR ln xi ð1:119Þ

Excess Gibbs’ Free Energy

The excess partial molar Gibbs’ free energy of component i is found with the aid of the basic

relationship G¼H� TS (Equation (1.28) on page 9):

GEx

i ¼ HEx

i � TSEx

i ð1:120ÞWith the aid of Equations (1.113) and (1.120) we obtain

GEx

i ¼ Hmix

i � TðSmix

i þR ln xiÞ

or

GEx

i ¼ Gmix

i �RT lnxi ð1:121ÞBy use of the relationship

GExm ¼ xAG

Ex

A þ xBGEx

B þ � � � þ xcGEx

c ð1:122Þ


and Equation (1.121) we obtain the excess molar Gibbs’ free energy:

GExm ¼ Gmix

m

� �non ideal

�RTðxAlnxA þ xBlnxB þ � � �Þ ð1:123Þ

Gibbs–Duhem’s Equations for a Binary Non-Ideal SolutionGibbs–Duhem’s equations, analogous to Equations (130) and (140) on page 24 are also valid for theexcess partial molar Gibbs’ free energies and the excess integral molar Gibbs’ free energy:

GEx

A ¼ GExm þð1� xAÞ @G

Exm

@xAð1:124Þ

GEx

B ¼ GExm þð1� xBÞ @G

Exm

@xBð1:125Þ

The excess molar Gibbs’ free energy of a binary solution as a function of the excess partial molar

Gibbs’ free energy of component B is, in analogy with Equation (150) on page 25,

GExm ¼ ð1� xBÞ

ðxB0

GEx

B


The excess Gibbs’ free energy

GEx

i ¼ HEx

i � TSEx

i ð1:120Þ

is a very important quantity. It is used for accurate calculations of the molar Gibbs’ free energies of

non-ideal liquid and solid binary alloys as functions of temperature.

Relationship between the Excess Molar Gibbs’ Free Energy and ActivityWith the aid of Equation (1.121) and Equation (1.107) on page 31 we obtain

GEx

i ¼ RT ln gixi �RT ln xi ¼ RT ln gi ð1:127Þ

With the aid of Equations (1.122) and (1.127) we obtain

GExm ¼ RTðxAln gA þ xB ln gB þ � � �Þ ð1:128Þ

1.7.3 Molar Gibbs’ Free Energies of Non-Ideal Binary Solutions

Calculation of accurate functions, which describe the molar Gibbs’ free energy of non-ideal liquid

and solid metals, is very important. These functions are the basis of the phase diagrams of alloy

systems.

The Gibbs’ molar free energy of a non-ideal solution is the sum of that of an ideal solution plus the

excess Gibbs’ molar free energy.With the aid of Equation (1.86) on page 27 and Equation (1.112) on

page 31 we obtain

ðGmÞnon-ideal ¼ xAG0A þ xBG

0B þRTðxAlnxA þ xBlnxBÞþGEx

m ð1:129Þ

or


0B þRTðxAlnxA þ xBlnxBÞþHEx

m � TSExm ð1:130Þ


According to Equation (1.114) on page 31 we obtain


0B þRTðxAln xA þ xBln xBÞþ Hmix

m

� �non-ideal

� TSExm ð1:131Þ

Equation (1.131) is the desired expression for Gibbs’ molar free energy of a non-ideal solution as a

function of composition and temperature.

Equation (1.131) is not as easy to calculate and show graphically as Equation (1.86) on page 20

because of the ðHmixm Þnon-ideal and SExm terms.

On page 24 we discussed a theoretical model for the molar heat of mixing. It has been

checked experimentally because the ðHmixm Þnon-ideal term can be determined experimentally, as

we will see in next section. The molar heat of mixing is therefore comparatively easy to

determine.

The difficulties to make accurate calculations of (Gm)non-ideal for non-ideal metal solutions

emanate from the SExm term in Equation (1.131). There exists no satisfactory theoretical model of

the excess molar entropy and it cannot be measured directly by experiments.

The entropy of an ideal solid solution consists of two terms, one due to lattice vibrations and one

originating from the binding energies of the component atoms.

The entropy of a pure metal is essentially due to lattice vibrations. The vibrational entropy of an

ideal solution of metals is approximately the same as that of pure metals and does not contribute to

the excess entropy.

No satisfactory model for the exchange energy and therefore also for the excess entropy has

been found so far. However, there is very convincing experimental evidence, described in

Figure 1.17, that the molar heat of mixing and the excess entropy of non-ideal solutions are

proportional. This seems reasonable as both ðHmixm Þnon-ideal and SExm depend on the binding energy of

the component atoms.

SExm ¼ A Hmixm

� �non-ideal

ð1:132Þ

The proportionality constant A is obtained as the slope of the line in Figure 1.17. Then SExm can be

calculated if ðHmixm Þnon-ideal is known.

kJ)( non–ideal

mixmH

−60 −40 −20 0 20

ExmS (J/K)

0

T = 3000 K

50

−50

−100

Figure 1.17 Molar heat of mixing

as a function of excess molar entropy

of liquid and solid alloys. Repro-

duced from Elsevier � 1983.


1.8 Experimental Determination of Thermodynamic Quantitiesof Binary Alloys

In Section 1.4.2 theoretical formulas of the relative thermodynamic quantities S, H andG have been

derived for alloys and graphical methods to construct partial molar and relative partial molar

quantities have been indicated (pages 19 and 20).

Only two of the relative thermodynamic quantities can be measured experimentally

. molar heat of mixing Hmixm

. partial molar Gibbs’ free energy of mixing Gmix

i .

From experimental values of these quantities it is possible to calculate other thermodynamic

quantities with the aid of for example Equation (1.132) on page 33 and equations and relationships

given in Section 1.4 and other parts of the text. An example is given on page 35.

1.8.1 Determination of Molar Heat of Mixing of Binary Alloys

The best way to measure enthalpy quantities is direct measurement of heat, i.e. calorimetric

measurements. As most metals have high melting points, measurements of heat when two metal

melts are mixed, require high-temperature calorimetry. Such measurements are hard to perform and

require accurate knowledge of heat capacities of the metals, good insulation arrangements and

vessels with low heat capacities.

Determination of the Molar Heat of Mixing

Method 1

A crucible with a given amount of solute metal A is placed in a calorimeter at room temperature.

A known quantity of molten solvent metal B is poured into the crucible. The alloy is formed and

allowed to cool to room temperature and the total heat is measured by the calorimeter. A blind

experiment is then performed. The same experiment is carried out without metal A. From the

difference in total heat and the known weights of the metal samples Hmixm of the alloy can be

calculated.

Themethod is not very accurate, owing to variation of the heat losses between the two experiments

and the risk of contamination, but it is fairly rapid. TheHmixm values ofmany liquid iron–silicon alloys

have been determined by this method.

Method 2

A flask, which is surrounded by a salt bath, contains a given amount of metal A, which is molten by

electrical heating until a steady temperature is achieved (Figure 1.18). A sample of the solute metal B

is placed in the sample tube and cooled by a water-ice bath to 0 �C. The sample tube is rotated into

a vertical position and the sample is dropped into the molten metal A. The alloy is formed and

heated until the molten alloy reaches the initial steady temperature. The heat of mixing can be

calculated when the masses of tin and sample metal, temperature, supplied amount of heat and

material constants are known.

The accuracy of the method has been successively improved and is of the magnitude 5%. In recent

years differential and much more accurate thermal methods have been developed. They will be

discussed in Chapter 7 (thermal analysis).

Derivation of the Partial Molar Heat of Mixing

Hmixm is determined as a function of composition by experiments on a series of alloys with different xB

values. An xBHmixm curve is drawn and the partial molar heat of mixing for each component can be

calculated fromEquations (1.86) and (1.87) on page 27 or derived graphically from the diagram in the

way shown in Figure 1.9 on page 19.

Ice-water bath

Gas control Sampletube

To thermo regulator

Main heaters

Heat regulator

Salt bath

Dewar vesselLiquid metal

Figure 1.18 Liquid-metal calorim-

eter. Reproduced from Elsevier �1979.


1.8.2 Determination of Partial Molar Gibbs’ Free Energy of Mixingof Binary Alloys

Method 1

The partial molar Gibbs’ free energies Gmix

i can be determined experimentally by measuring the

partial vapour pressure of component i in equilibrium with the solution and the vapour pressure in

equilibrium with the corresponding pure element under equivalent conditions.

If the pressure is so low that the ideal gas laws are valid and if there is no change in the chemical

composition of the vapour, Gmix

i can be calculated from the ratio of the measured pressures with the

aid of Equation (1.93) on page 28

Gmix

i ¼ RT lnpi

p0ið1:93Þ

This is the case for most metallic elements, for example Al, In, Pb, Cr, Mn, Ni, Cu, Ag, Au and Zn,

which havemonoatomic vapours. This is not the case for semiconductor elements, covalent elements

andmost metals in group III andVin the periodic table, for example C, Si, Ge, Sn, Sb, Bi, P, As and S.

In these cases polyatomic molecules are formed in the vapour.

Method 2

AlternativelyGmix

i can be determined by use of a galvanic cell. The positive pole of the cell consists

of an electrode made of the alloy with component i. The negative pole consists of pure element i.

The electrolyte is an ionic conductor with ions of metal i. It is illustrated by the scheme

�� pure metal i

ionic conductor

with ions of metal i

þalloy with component i

The work that is required to transport 1 kmol of ions reversibly from one electrode to the other when

the cell is closed by an outer conductor equals

Gmix

i ¼ Fzie ð1:133Þ

where

F ¼ Faraday’s constant, 96 500 103 Coulomb/kmol

zi ¼ valence of the charge carrier in the electrolyte

e ¼ reversible emf of the cell.

Calculation of the Molar Gibbs’ Free Energy of Mixing

In the case of a binary solution the partial molar Gibbs’ free energy ofmixingGmix

B is determined for a

series of alloys of various compositions bymethod 1 or 2 above. For each composition xB the value of

Gmixm can be calculated with the aid of Equation (15’) on page 24.

Gmixm ¼ ð1� xBÞ

ðxB0

Gmix

B


Determination of Entropy Quantities for Binary Alloys

Entropy cannot be measured directly and has to be calculated indirectly from other measurable

quantities.

If Hmixm and Gmix

m are known the entropy of mixing Smixm can be calculated from the relationship

Gmixm ¼ Hmix

m � TSmixm ð1:135Þ


If theGmix

i values aremeasured at various temperatures it is possible to calculateGmixm as a function of

temperature. Then Smixm can be calculated from the relationship

Smixm ¼ � @Gmix

m

@Tð1:136Þ

Today, it is also possible to determine Gmixm from measurements of the compositions of different

phases in equilibriumwith each other. The equilibrium is achieved by heat treatment of the specimen.

A method to determine the excess molar Gibbs’ free energy GExm is described on page 36.

Summary

& Thermodynamic Concepts and Relationships

Q ¼ UþW U ¼ Internal energy

H ¼ Uþ pV ðdHÞp ¼ ðdQÞp Enthalpy

dS ¼ dQ

TS ¼ kBln P Entropy

Molar entropy of mixing (solids, liquids and gases):

Smixm ¼ �RðxAlnxA þ xBlnxBÞ

G ¼ H� TS Gibbs’ free energy

At constant temperature and pressure spontaneous processes always occur in the direction of

decreasing G.

The deviation of the thermodynamic function G of a system from its equilibrium value can be

regarded as the driving force in chemical and metallurgical reactions.

& Thermodynamics of Single-Component Systems

Equilibrium Condition between two Phases a and b

Ga ¼ Gb

Change of Transformation Temperature as a Function of Change of Pressure

dT

dp¼ Vb �Va

Sb � Sa¼ DVa!b

DSa!bClausius--Clapeyron’s law

This law can be applied on melting and vaporization processes.

Change of Melting Point owing to Pressure

DTM ¼ TMðpÞ� TMðp0Þ ¼TM VL

m �V sm

� �Hfusion

m

Dp

The melting points of metals increase with increasing pressure.

Change of Melting Point owing to Crystal Curvature

DTM ¼ TM paK� �� TM pa0

� � ¼ TM Vam �VL

m

� �Hfusion

m

2sK

K ¼ 1

2

1

rmax

þ 1

rmin

� �Dp ¼ 2sK


An application is the decrease in temperature, which occurs in the neighbourhood of a growing

dendrite tip.

Change of Boiling Point owing to Pressure

DTB ¼ TBðpÞ� TBðp0Þ ¼TB Vg

m �VLm

� �H

evapm

Dp

or

DTB ¼ TBðpÞ� TBðp0Þ ¼ RTBðpÞTBðp0ÞH

evapm

lnp

p0

Bubble Formation in Melts

DTB ¼ TBðpgKÞ� TBðp0Þ ¼ RTBðpgKÞTBðp0ÞH

evapm

lnpL þ 2sK

p0

� �

& Gibbs’ Free Energy of a Pure Metal

Gsm ¼ Hs

m � TSsmGL

m ¼ HLm � TSLm

Consider Figures 1.7 and 1.8 on page 15.

The entropy of the liquid phase is normally larger than that of the solid. For this reason the liquid curve

is steeper than the solid curve.

& Thermodynamics of Multiple-Component Systems

Designations of Components:

Multiple-Component Systems: Binary Systems:

i¼A, B, . . . i¼A, B

Concepts and RelationshipsThe concepts and relationships below are valid for many thermodynamic quantities, among them

Volume V

Enthalpy H Common designation:

Entropy S

Gibbs’ free energy G X¼V, H, S, G

Partial Molar Quantities (X ¼ V, H, S and G)

X ¼ nAXA þ nBXB þ � � �

dX ¼ XAdnA þXBdnB þ � � �

Definition: Xi ¼ @X

@ni

Systems with Constant Composition:

Xm ¼ xAXA þ xBXB þ � � � ð1 kmolÞ

Systems with Variable Composition:

xAdXA þ xBdXB þ � � � ¼ 0


Graphical Construction of the Partial Molar Quantities of a Binary SystemConsider Figure 1.9 on page 19.

The construction is based on the relationships

xA þ xB ¼ 1

Xm ¼ xAXA þ xBXB

XA ¼ Xm þð1� xAÞ @Xm

@xAGibbs--Duhem’s

XB ¼ Xm þð1� xBÞ @Xm

@xBequations

X is plotted as a function of xB. At point (xB, X) a tangent to the curve is drawn. The intersections

between the tangent and the lines xB¼ 0 and xB¼ 1 give XA and XB, respectively.

Relative Partial Molar Quantities (X ¼ V, H, S and G)Partial Molar Quantities of Mixing

Definition:

A relative partial molar quantity¼ the difference between a partial molar quantity of a component in

a solution and the molar quantity of the pure substance.

Relative partial molar quantities are also called partial molar quantities of mixing of substance i and

are designated by Xmixi .

Xmix

i ¼ Xi �X0i

Relationships between Partial Molar Quantities of Mixing in a Binary System

Xmix

A ¼ �ðxB0

xB

1� xB

@Xmix

B

@xBdxB or X

mix

A ¼ðxB0

Xmix

B

ð1� xBÞ2dxB � xBX

mix

B

1� xB

Relative Integral Molar Quantities (X ¼ V, H, S and G)

Integral Molar Quantities of Mixing

Definition:

Xmixm ¼ difference between a molar quantity Xm of a solution and the sum of the quantities of the

corresponding pure component substances.

Xmixm ¼ Xm �ðxAX0

A þ xBX0B þ � � � þ xcX

0c Þ

Relationships between Partial and Integral Molar Quantities of Mixing

Xmixm ¼ xAX

mix

A þ xBXmix

B þ � � � þ xcXmix

c

Xmix

A ¼ Xmixm þð1� xAÞ @X

mixm

@xAGibbs--Duhem’s

Xmix

B ¼ Xmixm þð1� xBÞ @X

mixm

@xBequations

Xmixm ¼ ð1� xBÞ

ðxB0

Xmix

B

ð1� xBÞ2dxB


Graphical Construction of Partial and Integral Molar Quantities of Mixing

X is plotted as a function of xB.

Graphical construction of the above quantities is described in the text.

Study Figure 1.10 on page 21 carefully.

& Thermodynamics of Alloys

An alloy is a multiple-component metallic system. An alloy of a given pressure, temperature and

composition can either be homogeneous, e.g. consist of a single phase, or heterogeneous, e.g. consist

of several phases.

Heat of Mixing

Hmixm ¼ �ðDHmÞmix ¼ NAB WAB � 1

2ðWAA þWBBÞ

Theoretical Model of Heat of Mixing

Hmixm ¼ �ðDHmÞmix ¼ LxAxB

If L > 0 Hmixm ¼ �DHmix

m is positive

If L < 0 Hmixm ¼ �DHmix

m is negative:

Ideal and Non-Ideal Solutions of Two Components

– A–A and B–B and A–B forces are equal.

– The heat of mixing is zero.

– The solubility is complete.

All other solutions are non-ideal.

& Thermodynamics of Binary Ideal Solutions

Enthalpy of Mixing: Hmixm ¼ 0

Entropy of Mixing: Smixm ¼ �RðxAlnxA þ xBlnxBÞ

Gibbs’ Free Energy of Mixing:

Gmixm ¼ Hmix

m � TSmixm ¼ RTðxAlnxA þ xBlnxBÞ

Gibbs’ Free Energy of a Binary Ideal Solution

Gm ¼ xAG0A þ xBG

0B þGmix

m

or

Gm ¼ xAG0A þ xBG

0B þRTðxAlnxA þ xBlnxBÞ

Gm is plotted as a function of xB.

Graphical construction of the quantities in the equations above is given in the text. See Figure 1.13b

on page 27.


& Thermodynamics of Non-Ideal Solutions

Activities and Activity CoefficientsThe activity of component i in a solution is defined as the ratio of the fugacity of component i over

the solution to the fugacity of the standard state equal to the pure stable condensed phase at the

pressure 1 atm and temperature 273K.

ai ¼ fi= f0i

Giðl or cÞ ¼ G0i ðl or cÞþRT lnai

Raoult’s law: ai ¼ xi or fi¼ 1

Henry’s law: limxi ! 0

asolutei ¼ limxi ! 0

gixi ¼ g0i xi or limxi ! 0

gi ¼ g0i

See Figure 1.14 on page 29 and Figures 1.15 and 1.16 on page 29.

Law Activity Gibbs’ free energy

General law ai¼ fixi Gi ¼ G0i þRT lnai

Raoult’s law, valid for solvents

in dilute solutions

ai¼ xi and fi¼ 1

valid when xi ! 1

Gi ¼ G0i þRT lnxi

General law ai ¼ gixi ¼ greli g0i xi Gi ¼ G0i þRT lng0i þRT lngreli xi

Henry’s law, valid for solutes

in dilute solutions

ai ¼ g0i xi and greli ¼ 1

valid when xi ! 0

Gi ¼ G0i þRT lng0i þRT lnxi

& Excess Quantities of Non-Ideal Solutions

Definition:

Excess quantity¼ relative partial molar quantity of a non-ideal solution

minus that of an ideal solution.

ðXÞnon-ideal solution ¼ ðXÞideal solution þXEx

Relationships between Partial and Molar Excess Quantities andThermodynamic Quantities of Mixing

HEx

i ¼ Hmix

i HExm ¼ Hmix

m

� �non-ideal

SEx

i ¼ Smix

i �R lnxi SExm ¼ Smixm

� �non-ideal

þRðxAlnxA þ xBlnxB þ � � �Þ

GEx

i ¼ Gmix

i �RT lnxi GExm ¼ Gmix

m

� �non-ideal

�RTðxAlnxA þ xBlnxB þ � � �

For ideal solutions the excess quantities are zero.

The excess Gibbs’ free energy

GEx

i ¼ HEx

i � TSEx

i

is a very important quantity. It is used for accurate calculations of the molar Gibbs’ free energies of

non-ideal liquid and solid binary alloys as functions of temperature.


& Gibbs’ Free Energy of a Non-Ideal Solution

Gibbs–Duhem’s equations, analogous to Equations (130) and (140) on page 25 are also valid for theexcess partial molar Gibbs’ free energies and the excess integral molar Gibbs’ free energy:

GEx

A ¼ GExm þð1� xAÞ @G

Exm

@xA

GEx

B ¼ GExm þð1� xBÞ @G

Exm

@xB

GExm ¼ ð1� xBÞ

ðxB0

GEx

B

ð1� xBÞ2dxB

Relationships between Partial Excess Gibbs’ Free Energy and Activity Coefficients:

GEx

i ¼ RT lngixi �RT lnxi ¼ RT lngi

GExm ¼ RTðxAlngA þ xBlngB þ � � � þ xclngcÞ

Molar Gibbs’ Free Energies of Non-Ideal Binary Solutions

Calculation of accurate functions, which describe themolarGibbs’ free energy of non-ideal liquid and

solid metals, is very important. These functions are the basis of the phase diagrams of alloy systems.


0B þRTðxAlnxA þ xBlnxBÞþGEx

m


0B þRTðxAlnxA þ xBlnxBÞþHEx

m � TSExm


0B þRTðxAlnxA þ xBlnxBÞþ Hmix

m

� �non-ideal � TSExm

The only measurable thermodynamic quantities are

– molar heat of mixing Hmixm

– partial molar Gibbs’ free energy of mixing Gmix

i

However, as both Hmixm

� �non-ideal

and SExm depend on the binding energy of the component atoms, it is

reasonable to assume that

SExm ¼ A Hmixm

� �non-ideal

Calculation of other thermodynamic quantities are described briefly in the text.

Further Reading

A.H. Carter, Classical and Statistical Thermodynamics, Upper Saddle River, New Jersey, US, Prentice Hall,

2001.

L. Darken, R.W. Gurry, Physical Chemistry of Metals, New York, McGraw-Hill Book Company, 1953.

D. R. Gaskell, Introduction to Metallurgical Thermodynamics, Bristol, US, Taylor & Francis, 1981.

G. Grimvall, Thermophysical Properties of Materials, Amsterdam, Elsevier Science B.V. 1999.

M. Kaufman, Principles of Thermodynamics, New York, Marcel Dekker Inc, 2002.


thermodynamic concepts and relationships...1.3.4 molar gibbs’ free energy of a pure metal 15 1.4...

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