Hillert M. Phase Equilibria, Phase Diagrams and Phase Transformations: Their Thermodynamic Basis

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1.11 Characteristic state functions 25

(U − TS) has been given its own name and symbol, Helmholtz energy, F,

F = U − TS. (1.61)

Under experimental conditions of constant T, V and N we obtain

dF =−Ddξ. (1.62)

The equilibrium condition can then be written as

D =−(∂ F/∂ξ)

T,V,N

= 0. (1.63)

In an experiment under constant T, V and N there may be spontaneous changes until F

has approached a minimum.

In the same way we may introduce a state variable H = U + PV obtaining

dH = d(U + PV) = T dS + V dP + G

dN − Ddξ. (1.64)

This may be regarded as the combined law for the variables S, P and N. The new variable H

is regarded as the characteristic state function for these variables and it is called enthalpy.

In fact, it has already been introduced in connection to the ﬁrst law in Section 1.3.

The equilibrium condition under constant S, P and N is

D =−(∂ H/∂ξ )

S,P,N

= 0. (1.65)

In an experiment under constant S and P there may be spontaneous internal changes until

H has approached a minimum.

By applying both modiﬁcations we can deﬁne U − TS+ PV as a new state variable,

G, obtaining

dG = d(U − TS+ PV) =−SdT + V dP + G

dN − Ddξ. (1.66)

This may be regarded as the combined law for the variables T, P and N and the character-

istic state function for these variables is called Gibbs energy, G. This characteristic state

function is of particular interest because T and P are the variables, which are most easily

controlled experimentally and they are both potentials. G may decrease spontaneously

to a minimum under constant T, P and N and the equilibrium condition is

D =−(∂ G/∂ξ )

P,T ,N

= 0. (1.67)

It may be mentioned that the mathematical operation, we have used in order to introduce

a potential instead of an extensive variable, is called Legendre transformation.An

important aspect is that no information is lost during such a transformation, as will be

discussed in Section 2.1.

Exercise 1.13

Suppose that it would be practically possible to keep H, P and N constant during an

internal reaction in a system. What state function should then be used in order to predict

the state of equilibrium?

26 Basic concepts of thermodynamics

Hint

Find a form of the combined law which has dH,dP and dN on the right-hand side.

Solution

dH = T dS + V dP + G

dN − Ddξ ; T dS = dH − V dP −G

dN + Ddξ .Wethus

obtain D = T (∂ S/∂ξ)

H,P,N

> 0 for spontaneous reactions. Equilibrium is where S(ξ)

has its maximum.

Exercise 1.14

In Exercise 1.6 we considered an internal reaction in a closed system giving the fol-

lowing internal production of entropy under isothermal conditions. 

S =−ξ K /T −

R[ξ ln ξ − (1 + ξ ) ln(1 +ξ )]. Now, suppose the heat of reaction under constant T, V and

N is Q = ξ K . Derive an expression for Helmholtz energy F and use it to calculate the

equilibrium value of ξ . Compare with the previous result obtained by maximizing the

internal production of entropy.

Hint

Use F = U − (TS), where U is obtained from the ﬁrst law and S from the

second law.

Solution

Under constant T, V and N, the heat of reaction must be compensated by heat ﬂow from

the surroundings, Q = ξ K . Since the volume is constant U = Q = ξ K .

The total increase of entropy is S = Q/T + 

S = ξ K / T + 

S =−R[ξ ln ξ −

(1 + ξ ) ln(1 +ξ )] and thus F = U − (TS) = ξ K + RT[ξ ln ξ − (1 + ξ ) ln(1 +

ξ)].

This is identical to −T 

S.For this particular system, we thus get the same result if

we minimize F or maximize 

S under constant T, V and N.

1.12 Entropy

Before ﬁnishing the present discussion of basic concepts of thermodynamics, a few

words regarding entropy should be added. No attempt will here be made to explain the

nature of entropy. However, it is important to realize that there is a fundamental difference

between entropy and volume in spite of the fact that these two extensive state variables

appear in equivalent places in many thermodynamic equations, for instance in the forms

of the combined law deﬁning dU or dG.For volume there is a natural zero point and

1.12 Entropy 27

one can give absolute values of V.Asaconsequence, the change of G due to a variation

of P, (∂G/∂ P)

T,N

,isawell deﬁned quantity because it is equal to V. One may thus

compare the values of G of two systems at different pressures.

For internal energy or enthalpy there is no natural zero point but in practical appli-

cations it may be convenient to choose a point of reference for numerical values. The

same is true for entropy although it is quite common to put S = 0 for a well-crystallized

substance at absolute zero. That is only a convention and it does not alter the fact that

the change of the Gibbs energy G due to a variation of T, (∂G/∂T )

P,N

, cannot be given

an absolute value because it is equal to −S.Asaconsequence, it makes no sense to

compare the values of G of two systems at the same pressure but different temperatures.

The interaction between such systems must be based upon kinetic considerations, not

upon the difference in G values. The same is true for the Helmholtz energy F because

(∂ F/∂T )

V,N

is also equal to −S.

The convention to put S = 0atabsolute zero is useful because the entropy difference

between two crystalline states of a system of ﬁxed composition goes to zero there accord-

ing to Nernst’s heat theorem, sometimes called the third law.Itshould be emphasized

that the third law deﬁned in this way only applies to states, which are not frozen in a

disordered arrangement.

Statistical thermodynamics can provide answers to some questions, which are beyond

classical thermodynamics. It is based upon the Boltzmann relation

S = klnW, (1.68)

where k is the Boltzmann constant (= R/N

, where N

is Avogadro’s number) and W

is the number of different ways in which one can arrange a state of given energy. 1/W is

thus a measure of the probability that a system in this state will actually be arranged in a

particular way. Boltzmann’s relation is a very useful tool in developing thermodynamic

models for various types of phases and it will be used extensively in Chapters 19–22.

It will there be applied to one physical phenomenon at a time. The contribution to the

entropy from such a phenomenon will be denoted by S or more speciﬁcally by S

and we can write Boltzmann’s relation as

S

= k ln W

, (1.69)

where W

and S

are evaluated for this phenomenon alone. Such a separation of the

effects of various phenomena is permitted because W = W

· W

·...

S = k ln W = k ln(W

· W

· ...) = k(ln W

+ ln W

+···)

= S

+ S

+ ··· (1.70)

Finally, we should mention here the possibility of writing the combined law in a form

which treats entropy as a characteristic state function, although this will be discussed in

much more detail in Chapters 3 and 6.From Eq. (1.63)weget

− dS =−(1/T )dU − (P/T )dV + (G

/T )dN − (D/T )dξ. (1.71)

28 Basic concepts of thermodynamics

This formalism is sometimes called the entropy scheme and the formalism based

upon dU is then called the energy scheme. See Section 3.5 for further discussion.

With the entropy scheme we have here introduced new pairs of conjugate variables,

(−1/T, U ), (−P/T , V ) and (G

/T, N ). We may also introduce H into this formalism

by the use of dU = dH − V dP,

dS =−(1/T )dH + (V /T )dP + (G

/T )dN − (D/T )dξ. (1.72)

Two more pairs of conjugate variables appear here, (−1/ T, H) and (−P, V /T ). It is

evident that S is the characteristic state function for H, P and N as well as for U, V and

N.Bysubtracting dS from d(H/T )wefurther obtain

d(H/T − S) = H d(1/T ) + (V /T )dP + (G

/T )dN − (D/T )dξ. (1.73)

This is equal to d(G/T ) because G = U − TS+ PV = H − TS. This form of the

combined law has the interesting property that it yields directly the following useful

expression for the enthalpy,

H =



∂(G/T )

∂(1/T )



P,N

. (1.74)

Exercise 1.15

Equation (1.66) shows that S can be calculated from G as −(∂ G/∂ T )

P,N

.Itmaybe

tempting to try to derive this relation as follows: G = H − TS; ∂G/∂T =−S.How-

ever, that derivation is not very satisfactory. Show that a correction should be added and

then prove that the correction is zero under some conditions.

Hint

Remember that H and S may depend upon T.

Solution

G = H − TS gives strictly (∂G/∂T )

P,N

= (∂ H/∂T )

P,N

− T (∂ S/∂ T )

P,N

− S.How-

ever, the sum of the ﬁrst two terms (the contributions from the T-dependence of

H and S)iszero under reversible conditions and constant P and N because then

dH = T dS + V dP + G

dN − Ddξ = T dS according to Eq. (1.64).

Exercise 1.16

We have discussed the consequences for G and F of the fact that S has no natural zero

point. Actually, nor does U.Find a quantity for which this fact has a similar consequence.

1.12 Entropy 29

Hint

Look for a form of the combined law which has U as a coefﬁcient just as S is a coefﬁcient

in dG =−SdT + V dP + G

dN − Ddξ .

Solution

Equation (1.71)gives dS = (1/T )dU + (P/T )dV − (G

/T )dN + (D/ T )dξ ;

d(−F/ T ) = d(S − U/T ) = dS − (1/ T )dU − U d(1/T ) =−U d(1/T ) +(P/T )dV −

/T )dN + (D/T )dξ ;(∂[F/ T ]/∂[1/ T ])

V,N,ξ

= U .

Thus, one cannot compare the values of F/T at different temperatures.

Manipulation of thermodynamic

quantities

2.1 Evaluation of one characteristic state function from another

For the sake of simplicity we shall only consider closed systems in this chapter and thus

omit terms in dN.Inthe ﬁrst chapter we have deﬁned some characteristic state functions,

U, F, H and G in addition to S. Each one was introduced through a particular form of the

combined law. The independent variables in each form may be regarded as the natural

variables for the corresponding characteristic state function. In integrated form these

functions can thus be written as

U = U (S, V ,ξ) (2.1)

H = H(S, P,ξ) (2.2)

F = F(T, V,ξ) (2.3)

G = G(T, P,ξ). (2.4)

All these expressions are regarded as fundamental equations because all thermody-

namic properties of a substance can be evaluated from any one of them. This is because

the combined law in its various forms shows how the values of all the dependent variables

can be calculated for any given set of values of the independent variables. As an example,

from the combined law for the variables S, V and ξ , Eq. (1.49), we get

T = (∂U/∂ S)

V,ξ

(2.5)

−P = (∂U/∂ V )

S,ξ

(2.6)

−D = (∂U/∂ξ )

S,V

. (2.7)

As a consequence, we can now calculate the value for any other of the characteristic state

functions at a given set of values of S, V and ξ, for instance

G = U − TS+ PV = U − S(∂U/∂ S)

V,ξ

− V (∂U/∂V )

S,ξ

. (2.8)

It should be noted that the calculation of the value of G from U is only possible because

one knows an expression for U as a function of its natural variables. As a consequence of

the same principle, even if G can thus be obtained as an analytical expression from U(S,

V, ξ)bythe use of the above relation, the result is not a fundamental equation because

G(S, V, ξ ) does not allow the dependent variables to be calculated. They can only be

calculated from G as G(T, P, ξ) through S =−(∂ G/∂ T )

P,ξ

and V = (∂ G/∂ P)

T,ξ

.Itis

2.2 Internal variables at equilibrium 31

thus necessary ﬁrst to replace S and V by T and −P as a new set of independent variables,

which is seldom possible to do analytically. If that replacement is not done, then some

information has been lost in the calculation of G from U.

It should again be emphasized that an expression for a characteristic state function

will be a fundamental equation only if expressed as a function of its natural variables.

Exercise 2.1

Show that G = (∂[H/S]/∂[1/S])

Hint

Evidently, no internal reaction is considered. The natural variables of H are S and P. It

should thus be possible to express G in terms of H and its derivatives.

Solution

Without any internal reaction we have for a closed system from Eq. (1.64): dH = T dS +

V dP; T = (∂ H/∂ S)

We get (∂[H/S]/∂[1/S])

= H + (1/S)(∂ H/∂[1/S])

= H − S(∂ H/∂ S)

= U +

PV − ST = G.

2.2 Internal variables at equilibrium

We have already emphasized that ξ is a dependent variable if the system is to remain

in internal equilibrium. Since D = 0insuch a state, the equilibrium value of ξ can be

evaluated from a fundamental equation for any one of the characteristic state functions,

for instance U(S, V, ξ) from Eq. (1.49),

− D =

(

∂U/∂ξ

)

S,V

= 0. (2.9)

By applying this to the equation for U one obtains a relation for the equilibrium value

of ξ for the prescribed values of S and V,

ξ = ξ

(

S, V

)

, (2.10)

where ξ has thus become a dependent variable since we now consider states of equi-

librium. Equation (2.10) can be used to eliminate ξ from the equation for U(S, V, ξ )to

yield an equation for states of internal equilibrium. This may be inserted in U(S, V, ξ)

in order to yield

U = U (S, V ). (2.11)

It may be of practical importance to calculate the values of various state variables at

equilibrium. That would be straightforward if a fundamental equation under equilibrium

conditions is available. If only a fundamental equation containing ξ is available, then one

32 Manipulation of thermodynamic quantities

should ﬁrst apply the equilibrium condition to ﬁnd how the equilibrium value of ξ varies

with the other independent variables. We have just seen how one, in principle, can use that

information to eliminate ξ and obtain the wanted fundamental equation. In practice that

may be difﬁcult or even impossible to do analytically. In that case one can evaluate the

corresponding derivative as a function of ξ from the fundamental equation available and

insert the equilibrium value of ξ.Weshall now see that this method will give the correct

value.

The proof is based on the equilibrium condition for the relevant characteristic state

function (v



, v



,...,ξ), where v



, v



etc. are the natural variables for  and



∂

∂ξ



= 0. (2.12)

The subscript v

indicates that all the natural variables v



, v



etc. are kept constant.

Any derivative of the function for equilibrium states with respect to one of the natural

variables, v

, can be expressed through derivatives of the initial function containing ξ as

an independent variable,



∂

∂v





∂

∂v



,ξ



∂

∂ξ





∂ξ

∂v





∂

∂v



,ξ

(2.13)

when ξ has its equilibrium value. The subscript v

indicates that all natural variables

except v

are kept constant. This proof depends on the use of Eq. (2.12)which is valid

only when the variables v

that are kept constant are the natural variables of .Asan

example



∂U

∂ξ



T,V

= 0, (2.14)

because the natural variables of U are S and V. That is why C

, the heat capacity at

constant volume, is different when ξ is frozen-in and when ξ has time to adjust to

internal equilibrium (see Eq. (1.21)).

Furthermore, it should be emphasized that this method of calculation can only be

applied to the ﬁrst derivatives of the characteristic state function, and not to higher-order

derivatives, since, in general

∂

∂ X

∂ξ

= 0. (2.15)

Exercise 2.2

Avery simple model for the magnetic disordering of a ferromagnetic element gives

the following expression at a constant pressure, G = ξ (1 −ξ )K + RT[ξ ln ξ +

(1 − ξ ) ln(1−ξ )] where K is a constant and ξ is the fraction of spins being disordered.

The degree of magnetic disorder varies with T,ξ= ξ (T ). Derive an expression for

the contribution to the enthalpy due to magnetic disordering, H . Then calculate, at

the temperature where ξ = 1/4, the corresponding contribution to the heat capacity at

constant P which is deﬁned as C

(

∂H/∂T

)

2.3 Equations of state 33

Hint

A magnetic state cannot be frozen-in. ξ will always have its equilibrium value. It can

be found from (∂G/∂ξ)

T,P

= 0, which yields (1 − 2ξ )K + RT ln[ξ/(1 − ξ )] = 0.

Unfortunately, this does not give ξ as an analytical function of T and we cannot replace ξ

by T in G. Thus, it is convenient to make use of the fact that

(

∂G/∂ξ

)

T,P

= 0atequi-

librium, which gives −S = (∂G/∂T )

= (∂G/∂T )

P,ξ

according to Eq. (2.13).

We can then evaluate H from G + T S.

Solution

S =−(∂G/∂T )

P,ξ

=−R[ξ ln ξ + (1 − ξ) ln(1 − ξ)]; H = G + T S =

ξ(1 − ξ )K + RT[ξ ln ξ + (1 − ξ ) ln(1 − ξ )] − RT[ξ ln ξ + (1 − ξ ) ln(1 − ξ )] =

ξ(1 − ξ )K ;(∂H/∂T )

P,ξ

= 0. It is evident that C

cannot be evaluated in this

way. The reason is that

(

∂H/∂ξ

)

T,P

= 0. The natural variables for H are not T and P

but S and P. Thus, we must use the basic form of Eq. (2.13), C

≡ (∂H/∂T )

(∂H/∂T )

P,ξ

+ (∂H/∂ξ )

T,P

(∂ξ/∂T )

From the relation between ξ and T at equilibrium, given in the hint, we get:

(∂T /∂ξ )

=−(K/R){−2/ ln[ξ/(1 − ξ )] − (1 − 2ξ )[1/ξ + 1/(1 − ξ)]/(ln[ξ/(1 −

ξ)])

}; C

= 0 + (1 − 2ξ )KR/K

{

2/ ln[ξ/(1 − ξ)] + (1 − 2ξ)/ξ (1 − ξ )(ln[ξ/

(1 − ξ )])



= 1.29R for ξ = 1/4.

2.3 Equations of state

If a characteristic state function for a particular substance is given with a different set of

variables than the natural one, then it describes some of the properties but not all of them.

Such an equation is often regarded as a state equation and not a fundamental equation.

As an example,

U = U (T, P)

is often called the caloric equation of state. Some of the quantities which are usually

regarded as variables may also be represented with an equation between other variables,

for instance.

V = V (T, P).

This is sometimes called the thermal equation of state. The practical importance of

some state equations stems from the fact that they can be evaluated fairly directly

from measurable quantities and can thus be used to rationalize the results of mea-

surements on a particular substance. As an example, the derivatives of V with respect

to T and P can be obtained by measuring the thermal expansivity and the isothermal

compressibility, respectively. It is much more laborious to evaluate the fundamental

equation for a substance.

34 Manipulation of thermodynamic quantities

A major problem in the evaluation of the fundamental equation or an equation of state

is the choice of mathematical form. The form is not speciﬁed by thermodynamics but

must be chosen from a knowledge of the physical character of the particular substance

under consideration. A considerable part of the present text will be concerned with the

modelling of the fundamental equation for various types of substances.

Exercise 2.3

An ideal classical gas is deﬁned by two equations of state. For one mole, they are PV =

RT and U = A + BT where A and B are two constants. Try to derive a fundamental

equation.

Hint

Try to ﬁnd F(T, V). Use U = F + TS = F − T (∂ F/∂T )

which yields (∂U/∂T )

−T (∂

F/∂T

)

. Also use P =−(∂ F/∂V )

Solution

(∂

F/∂T

)

=−(∂U/∂T )

/T = B/T .

Integration yields (∂ F/∂T )

=−B ln T + K

; F =−BT ln T + K

T + K

where K

and K

are independent of T but may depend upon V. That dependency is

obtained from RT/V = P =−(∂ F/∂V )

=−T (∂ K

/∂V )

− (∂ K

/∂V )

Thus, (∂ K

/∂V )

= 0 and (∂ K

/∂V )

=−R/ V . K

is independent of V and K

−R ln V + K

We get F =−BT ln T − RT ln V + K

T + K

To determine K

: A + BT = U = F + TS = F − T

(

∂ F/∂T

)

= BT + K

.We

thus ﬁnd K

= A.Wecannot determine K

from the information given. Our result

is F = A + K

T − BT ln T − RT ln V which is a fundamental equation, F(T, V ). We

can also derive the Gibbs energy G = F + PV = F − V

(

∂ F/∂V

)

= F + RT but

in order to have a fundamental equation in G we must obtain G(T, P)byreplacing

V with P and T which is possible in the present case where V = RT/P.Wethus get

G = A + (K

+ R − R ln R)T − (B + R)T ln T + RT ln P.

2.4 Experimental conditions

By experimental conditions we here mean the way an experiment is controlled from the

outside. It primarily concerns variables which we may regard as external. Let us ﬁrst

consider the pair of conjugate variables −P and V. Either one of them can be controlled

from the outside without any knowledge of the properties of the system. In the pair

of conjugate variables T and S, one can control the value of T from the outside but the

control of S requires knowledgeof the properties of the system or extremely slow changes.

In practice, it may even be difﬁcult to keep S constant when another variable is changed.