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

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19.4 Heat capacity due to thermal vibrations 425

liquid

solid

glass

m.p.

Figure 19.1 Schematic diagram comparing the heat capacities of a crystallized metal and its

amorphous state.

Solution

C

= a + bT where a + bT

m.p.

= 0. A second relation between a and b is obtained

by integrating for the entropy between T

m.p.

/3 and T

m.p.

: R =



(C

/T )dT =

a ln 3 + (2/3)bT

m.p.

; a = R/(ln 3 − 2/3) = 2.3R; b =−2.3R/T

m.p.

.H

changes

between T

m.p.

/3 and T

m.p.



C

dT = a(T

m.p.

− T

m.p.

/3) + 0.5b(T

m.p.

− T

m.p.

/9)

= 2.3RT

m.p.

(2/3 − 4/9) = 0.51RT

m.p.

Thus we get, at T

m.p.

/3aswell as at 0 K,H

= RT

m.p.

− 0.51RT

m.p.

= 0.49RT

m.p.

19.4 Heat capacity due to thermal vibrations

According to quantum mechanics, the energy of heating can be added only as quanta.

Foranharmonic oscillator of frequency v, the magnitude of the quanta is hν where h

is Planck’s constant. Einstein constructed a simple model of a crystal by assuming that

each atom vibrates independent of all the others and has three directions of movement.

Acrystal with N atoms could thus be regarded as consisting of 3N linear oscillators,

all of them with a frequency ν. The question, how many quanta such a crystal should

have at equilibrium, is closely related to our previous question how many vacancies a

crystal should have at equilibrium. If there are n quanta, the energy increase is U = nhν

compared to the conditions at absolute zero. In order to evaluate the entropy contribution

we must ﬁnd in how many ways n quanta can distribute themselves on 3N oscillators. By

numbering the oscillators a

... a

and the quanta k

... k

we can describe

a particular distribution by ﬁrst giving the number of a certain oscillator and then the

quanta which are placed there, e.g. a

....Ifall the permutations

of these elements represent possible distributions there should be (3N + n)! different

distributions. However, we must start with an oscillator and the ﬁrst factor should thus

be 3N instead of 3N + n. Furthermore, it would be impossible to distinguish between

many of the distributions since all the oscillators are identical and so are all the quanta.

426 Modelling of disorder

The number of distinguishable distributions is thus

W =

3N + n

(3N + n)!

(3N)!n!

. (19.14)

This is almost identical to the result for thermal vacancies except for the ﬁrst factor,

which is of no importance for large N and n when we take the logarithm, and except

for the fact that N has been replaced by 3N .Wecan thus use the ﬁnal result from the

previous derivation but this time we shall apply it to the Helmholtz energy because the

frequency, and thus the energy hν, will vary with the distance between the atoms, i.e.

the volume. Thus we should like to keep the volume constant. We obtain

F = nhν + 3kNT



3N + n



. (19.15)

The equilibrium number of quanta under constant T and V is obtained from ∂F/∂n = 0,

3N + n

= exp(−hν/kT). (19.16)

Again it will be possible to eliminate the internal variable at equilibrium and we ﬁnd

F = 3RT ln[1 − exp(−hν/kT)]. (19.17)

By standard methods we can calculate the heat capacity due to thermal vibrations,

=−T (∂

F/∂ T

)

= 3R



hν



exp(hν/kT)

[exp(hν/kT) − 1]

. (19.18)

The only parameter characteristic of the particular material under consideration is the

frequency, v.Italways appears in the dimensionless combination hν/kT. The combina-

tion of constants hν/k is of the dimension temperature and we can thus introduce a new

material constant instead of the frequency, the Einstein temperature  = hν/k.

= 3R(/T )

. exp(/T )[exp(/T ) − 1]

. (19.19)

There have been many attempts to improve Einstein’s model by removing the assumption

that the atoms vibrate independent of each other. An elegant method was proposed by

Debye. If the atoms cooperate when they vibrate, it means that the vibrating units are

larger and the frequency should be lower. For each possible frequency, one should be

able to apply Einstein’s model and by a summation over all the frequencies one may

obtain the proper result. Debye considered a whole spectrum of oscillators down to the

mechanical vibrations of the crystal. His spectrum thus extends all the way from the high

frequency of individual atoms and down to the acoustic range. He further assumed that

the number of oscillators is still 3N and that they distribute themselves over the range of

frequencies proportional to ν

.From Debye’s model one does not get C

as an analytical

expression of T but C

is now available in tables using the parameter x = /T where

 is equal to hν

max

/k and ν

max

is the maximum frequency, i.e. the vibrational frequency

of an atom. One often tabulates not only C

but also S and U. Debye’s model agrees

fairly well with measurements for many materials.  is evaluated as a material constant

to give the best agreement between model and experiment.

19.4 Heat capacity due to thermal vibrations 427

Einstein

Debye

/3R

T/Θ

or T/Θ

or 0.77T/Θ

Figure 19.2 Comparison between the Einstein and Debye theories of C

made at temperatures

given relative to the characteristic temperature ,asdeﬁned in each theory. In addition, the thin

line is for Einstein’s theory plotted versus 0.77T /.

We can calculate C

from C

by means of the Gr¨uneisen constant. With the approx-

imate value γ = 2wewould get from Eq. (2.38)

= C

(1 + 2αT ). (19.20)

S as a function of T at constant volume is obtained by integrating C

/T .Onthe other

hand, if one should like to have S as a function of T at constant pressure it could be

obtained by integrating C

/T which yields

S = S(x) + 2αU (x), (19.21)

if α is a constant. Furthermore, after some manipulations one obtains approximately

H = U (x) + (5/2)RαT

+ (9/8)R + H

. (19.22)

Notice that H

is obtained as an unknown constant of integration, and it represents the

cohesive energy. The term (9/8)R comes from quantum mechanics which says that a

linear oscillator of frequency v has a zero point energy of hν/2.

A comparison between Einstein’s and Debye’s models is given in Fig. 19.2. Both

models predict a C

value of 3R at very high temperatures but the diagram seems

to indicate that the theories give quite different results up to fairly high temperatures.

However, if  is regarded as an empirical constant, to be evaluated from experimental

information, then the two models would work with different  values. These empirical

constants are called Einstein’s and Debye’s temperatures. Reasonably good agreement can

thus be obtained from high down to fairly low temperatures and for many applications it is

reasonable to use the equations derived by Einstein. The value of the Einstein temperature

can be estimated from tabulated values of the Debye temperature by multiplication with a

factor. The factor is slightly different depending on which quantity one is most interested

428 Modelling of disorder

in. For heat capacity, internal energy, entropy and Helmholtz energy one should use the

values 0.77, 0.73, 0.71 and 0.72, respectively. The thin line in Fig. 19.2 shows how well

Einstein’s theory agrees with Debye’s if the same temperature scale is used but different

 values. The difference is negligible above 0.3

The fact that C

approaches the same value of 3R at high temperatures, independent

of the material constant , does not mean that all crystalline substances would have the

same stability at high temperatures. For a given element with two possible structures,

the structure with the highest S value will increase its relative stability towards high

temperatures more than the other one. The S value is the result of integration over C

from absolute zero. S will thus be larger, the faster C

approaches the asymptotic value

of 3R, i.e. the lower  is. A difference in C

between two structures at low temperatures

will be very important at high temperatures because T appears in the denominator of the

integrand. A structure with a low  will thus have high stability at high temperature.

The fact that the S term dominates over the U term in F at high temperatures can be

demonstrated by a calculation for some temperature T

F = U − T

S = U



CdT − T



(C/T )dT

= U

−



[C(T

− T )/T ]dT . (19.23)

The integrand is always positive since C and T

− T are both positive. We can make a

similar calculation for G and obtain a similar result.

At low temperatures where T   one would obtain, according to Einstein

= 3R







· exp



−





. (19.24)

This would imply that C

decreases rapidly as one approaches the absolute zero. Exper-

iments showed a slower decrease and that was the reason why one wanted to improve

Einstein’s theory. Debye’s theory instead yields the following expression at low tempera-

ture, and it agrees fairly well with experiments if the electronic contribution is subtracted

for metallic conductors.

= 234R(T /)

. (19.25)

Exercise 19.3

From Debye’s theory, derive the term in H originating from the zero-point energy.

Hint

Firstneglectthe difference betweenU and H. According to Einstein, the zero-point energy

for all oscillators is the same, yielding U = 3N(hν/2) = 3Nk · /2 = (3/2)R.

However, according to Debye there is a distribution of frequencies z(v) = K ν

and

3N =

∫

z(ν)dν =

∫

K ν

dν = K ν

max

/3 and thus K = 9N/ν

max

19.5 Magnetic contribution to thermodynamic properties 429

saturation

magnetization

Figure 19.3 Schematic diagram showing the variation of the saturation magnetization as a

function of temperature.

Solution

H

∼

U =

∫

(hν/2) · z(ν)dν =

∫

(hν/2) ·(9N/ν

max

) · ν

dν = (9N/ν

max

) ·

(hν

max

/8) = (9/8) · Nk(hν

max

/k) = (9/8) · R

Debye

19.5 Magnetic contribution to thermodynamic properties

The atoms in a ferromagnetic substance contain unpaired electron spins which give each

atom a certain magnetic moment. It can have different directions and will thus give rise

to magnetic disorder. According to the localized spin model of magnetism, a disordered

arrangement gives a contribution to the entropy which can be evaluated from Boltzmann’s

relation. Due to its simplicity we shall only use this model. W is the number of different

arrangements for the whole system. Let w be the number of possible directions for each

atom. If the direction of each atom in the disordered state is independent of all the others,

we can write W = w

for one mole of the substance. Thus

S = k ln W = k ln w

= kN ln w = R ln w . (19.26)

According to quantum mechanics,

w = β + 1 = 2s + 1, (19.27)

where β is the number of unpaired electron spins and s is the resulting spin. For a free

electron s = 1/2.

There is a critical temperature, the Curie temperature T

, below which the spins will

position themselves parallel to each other in a ferromagnetic substance. For a perfectly

ordered state, the above contribution to the entropy should disappear completely but, in

practice, there will be some disorder left. However, it will disappear as one approaches

absolute zero and the saturation magnetization will thus increase (see Fig. 19.3). The

degree of magnetic order can be measured by measuring the saturation magnetization.

Note that an externally applied magnetic ﬁeld usually does not change the ordering

appreciably. It simply makes the magnetization of the various magnetic domains align

along one direction. From the saturation magnetization at low temperatures, one can

get direct information on the magnetic moment of the atoms, i.e. the values of s and β.

430 Modelling of disorder

5000 1000 1500

real bcc

non-magnetic

bcc

J/mol

Figure 19.4 Heat capacity (C

)ofbcc-Fe. Notice that a considerable part of the magnetic effect

occurs above T

The magnetic moment per atom is usually given in units of Bohr magnetons which is

identical to 2s, i.e. to β according to our simple model. Such measurements indicate that

β is seldom an integer. This casts some doubt on the localized spin model and one may

wonder if the entropy of magnetic disorder can be evaluated as shown above. Despite

this, the method is often used and may be regarded as a convenient approximation.

The disordered state above the Curie temperature is called paramagnetic. It is evident

that it has a high entropy and should thus grow even more stable at higher temperatures.

The reason why the ferromagnetic state becomes stable below a critical temperature must

be a lower energy for that state. When the disorder increases as the temperature is raised,

energy must thus be added and the magnetic transformation is revealed in the curve for

the heat capacity (see Fig. 19.4).

The curve shows that the transformation occurs gradually up to T

but a small part

of the order still exists at T

.Itdoes not disappear until well above T

and it thus yields

a contribution to the heat capacity above T

.Wecan evaluate the magnetic enthalpy

by integrating the abnormal contribution to the curve. As always, the transformation

temperature is determined by the balance between enthalpy and entropy. This is best

demonstrated by approximating the magnetic transformation (which is actually a second-

order transition (see Section 15.1)) with a sharp transformation. For such a transformation

we have

G = H − T S. (19.28)

With G = 0weﬁnd a transformation temperature T

= H/S. The larger the

energy gain is at the magnetic ordering, the higher the transformation temperature will

be.

Figure 19.5 illustrates the difference in the Gibbs energy between the most stable

state at each temperature, G

, and the completely disordered state. The slope of the

curve at low temperature gives the entropy decrease due to complete order of the spins.

19.6 A simple physical model for the magnetic contribution 431

∆G

ordering

= G

− G

dis

slope =−∆S

ordering

−5

−10

2000

Curie

T (K)

Figure 19.5 The effect of magnetic ordering on the Gibbs energy of bcc-Fe. The reference state is

the completely disordered state at each temperature.

The thin line gives the difference between the completely ordered and the completely

disordered states at each temperature. The intersection with the abscissa gives the tem-

perature, T

, for the hypothetical, sharp transformation. In reality, the ferromagnetic state

is stable to a higher temperature, T

, because it lowers its Gibbs energy by disordering

gradually.

Exercise 19.4

Evaluate β for bcc–Fe from an estimate of the magnetic entropy of bcc–Fe using

Fig. 18.2.

Hint

The sharp bend in the curve for

bcc

−

fcc

is due to the magnetic entropy of bcc–Fe

and is not inﬂuenced by the magnetic entropy of fcc–Fe which changes at much lower

temperatures.

Solution

The slopes for the paramagnetic and ferromagnetic states can be estimated where the

curve is reasonably ﬂat. We obtain S =−d(

param.bcc

−

fcc

)/dT +d(

ferrom.bcc

−

fcc

)/dT = 2.4 +9.1 = 11.5 = 1.4R = R ln(β + 1); β

∼

19.6 A simple physical model for the magnetic contribution

The magnetic disorder remaining below the Curie temperature can be treated in about

the same way as we treated vacancies. We have already discussed the contribution to the

entropy. In order to develop a complete model for the magnetic transformation we must

432 Modelling of disorder

also formulate the contribution to the enthalpy. The calculations will be especially simple

if the magnetic moment of the atoms is 1/2 since this value givestwo possible orientations

with a spin difference of 1, according to the requirements of quantum mechanics. Let us

suppose that n atoms in a system of totally N atoms have their spins directed opposite

to the majority which consists of N − n atoms. We may guess that the extra energy

connected with the misorientation of the n atoms can be represented by the following

simple expression,

H = K

n(N − n) = K

x(1 − x)N

, (19.29)

where x represents the fraction of atoms with the wrong orientation, x = n/N. This

expression gives H a maximum at x = 1/2, i.e. for the disordered, demagnetized state.

Since x = 0 represents the fully magnetized state, we can write

H = x(1 − x) · 4H

dis

, (19.30)

where H

dis

is the enthalpy increase on complete disordering.

Let us now turn to the conﬁgurational entropy. The two types of atoms can mix with

each other in a number of ways,

W =

N !

n!(N − n)!

. (19.31)

Boltzmann’s relation will thus yield

S/k = ln N ! − ln n! − ln(N − n)!

∼

N ln N − n ln n − (N − n) ln(N − n)

=−n ln

− (N − n)ln

N − n

(19.32)

S/R =−

−

N − n

=−x ln x − (1 − x) ln(1 − x). (19.33)

By combination of H and S we obtain for the Gibbs energy of disordering

G = x(1 − x) · 4H

dis

+ RT[x ln x + (1 − x) ln(1 − x)]. (19.34)

We want to determine the equilibrium value of x under constant T and P and shall thus

consider a process by which x is increased. By identifying x with an internal variable ξ

we obtain, at equilibrium,

− D =



∂G

∂ξ



T,P

dG

= (1 − 2x) · 4H

dis

+RT



ln x +

− ln(1 − x) −

1 − x



= 0 (19.35)

1 − x

=−(1 − 2x) · 4H

dis

/RT. (19.36)

The solutions of Eq. (19.36) are plotted against temperature in Fig. 19.6. Below a

critical temperature there are two identical states, x

= 1 − x

. They represent ordered

states but the degree of order varies from perfect at absolute zero (x = 0or1)and

decreases gradually as the temperature is raised. Finally it disappears completely at

19.6 A simple physical model for the magnetic contribution 433

T/T

Figure 19.6 Magnetic ordering as a function of temperature according to a simple model. x = 0.5

represents the disordered state. The dashed line represents unstable equilibria.

the critical temperature where x = 0.5. The states above the critical temperature are

disordered because x = 0.5. The value of the critical temperature is obtained by inserting

the value of x = 0.5, which primarily yields 0/0. The limiting value is obtained as

4H

dis

2x − 1

ln[x/(1 − x)]

4H

dis

1/x + 1/(1 − x)

4H

dis

· 2x(1 − x) = 2H

dis

/R. (19.37)

Above the critical temperature there is only one solution showing that the disordered

state (x = 0.5) is the stable state. The equation also has the solution x = 0.5 below the

critical temperature but there it represents an unstable state. That is why it was there

drawn with a dashed line. The fact that the disordered state is unstable there, can be

tested with a stability condition from Eq. (6.54). We obtain negative values for x = 0.5

and T < T

= 2H

dis

B =



∂

∂x



T,P

=−2 · 4H

dis

+ RT[1/x + 1/(1 − x)]

=−8H

dis

+ 4RT < 0. (19.38)

Let us now assume incorrectly that the magnetic transformation is a sharp one and that

a completely disordered state becomes completely ordered at T

. The entropy of the

completely disordered state is S

dis

= R ln 2 and for a sharp transformation we should

thus have expected the following transition temperature,

= H

dis

/S

dis

= H

dis

/Rln2= 1.44H

dis

/R = 0.72T

. (19.39)

The fact that T

is higher than T

was illustrated in Fig. 19.5. That result demonstrates

that the possibility for some of the atoms to disorder within the ordered state increases

the stability of the ordered state and raises the transformation temperature from the value

predicted for a sharp transformation. On the other hand, it should also be pointed out

434 Modelling of disorder

that the stability of the disordered state is increased by the existence of some short-range

order which has not been considered in our model. This is the factor that produces the tail

above the C

maximum which was shown in Fig. 19.4.Itmakes the difference between

and T

smaller than indicated by the above model.

The model we have used here for the magnetic order–disorder transformation is based

upon the assumption that each individual atom has its own magnetic moment. This is

called the localized spin model and is regarded as a rather crude approximation. Mag-

netism can be treated in a more satisfactory way by the electron band theory. However,

it is much more complicated and thus difﬁcult to apply, in particular at high temperature

and in alloys.

Exercise 19.5

Use the C

curve in Fig. 19.4 to evaluate the enthalpy of magnetic disordering. Then

estimate the number of unpaired electrons in bcc-Fe.

Hint

The number of unpaired electrons could be estimated from the magnetic entropy which,in

turn, could be evaluated from the enthalpy and the estimated transformation temperature

for a sharp transformation.

Solution

A rough graphical integration yields H

dis

= 7000 J/mol. We know T

= 1043 K and

accepting the result of the simple model we can estimate T

= 0.72T

, and thus we get

S

dis

= H

dis

= 7000/(0.72 · 1043) = 9.3; ln(β +1) = S/R = 1.12;

β = 2.1.

19.7 Random mixture of atoms

Before leaving the discussion of models of disordering phenomena we should mention

disorder in solution phases. Actually, most substances can have a variable composition

and we shall call such substances solution phases. On the atomic scale they consist of a

mixture of different species, in the simplest case atoms. In a crystal the atomic sites are

arranged in a regular pattern, a lattice, but the distribution of different kinds of atoms

on the sites is generally determined by chance to some extent. This situation is often

described as chemical disorder or conﬁgurational disorder.For a complete description

of such a case one needs a model with an internal variable representing the degree of

order. Such a model will be further discussed in the Chapters 21 and 22.Inorder to

prepare for that discussion it is convenient now to discuss the entropy of solution phases

with the maximum chemical disorder, so called random mixtures. The degree of order

can be used as an internal variable but will not be introduced until later.