CHAPTER 4: PHASE TRANSITIONS 4.1 INTRODUCTION A region within which all properties are uniform consists of a distinct phase. For example, solid ice, liquid water, and gaseous water vapor are separate phases of the same chemical species ( )2H O . Each phase can be distinguished with the density ρ of the constituent. For example, a portion of the Arctic Ocean in vicinity of the North Pole is frozen and consists of ice in a top layer and liquid water beneath it. The atmosphere above the ice contains some water vapor. The density in these three layers are different, since water exists in these layers separately in some combination of three (solid, liquid and gaseous) phases. Similarly. A vessel containing immiscible oil and water contains only liquid. We have two distinct phases in this case as oil waterρ ρ≠ . Similarly, in metallurgical applications, various phases may exist within the solid state, since the density may differ over a solid region that is at a uniform temperature and pressure. In liquid mixtures that are miscible at a molecular level (such as those of alcohol and water for which molecule of one species are uniformly intermixed with those of the other), even though the mixture may contain several chemical components, a single phase exists, since the system properties are macroscopically uniform throughout a given volume. Air, for instance, consists of two major components (molecules of oxygen and nitrogen) that are chemically distinct, but consists of a single phase, since they are well mixed. The distinguishing characteristic of a phase transition is an abrupt sudden change in one or more physical properties, in particular the heat capacity, with a small change in a thermodynamic variable such as the temperature. Examples of phase transitions are:

• The transitions between the solid, liquid, and gaseous phases (boiling, melting, sublimation, etc.)

• The transition between the ferromagnetic and paramagnetic phases of magnetic materials at the Curie point.

• The emergence of superconductivity in certain metals when cooled below a critical temperature.

• Quantum condensation of bosonic fluids, such as Bose-Einstein condensation and the superfluid transition in liquid helium.

• The breaking of symmetries in the laws of physics during the early history of the universe as its temperature cooled.

60

The first-order phase transitions are those that involve a latent heat. During such a transition, a system either absorbs or releases a fixed (and typically large) amount of energy. Because energy cannot be instantaneously transferred between the system and its environment, first-order transitions are associated with "mixed-phase regimes" in which some parts of the system have completed the transition and others have not. This phenomenon is familiar to anyone who has boiled a pot of water: the water does not instantly turn into gas, but forms a turbulent mixture of water and water vapor bubbles ( Fig. 4.1). Mixed-phase systems are difficult to study, because their dynamics are violent and hard to control. However, many important phase transitions fall in this category, including the solid/liquid/gas transitions.

Fig. 4.1 Phase change of water

The second class of phase transitions are the continuous phase transitions, also called second-order phase transitions. These have no associated latent heat. Examples of second-order phase transitions are the ferromagnetic transition, the superfluid transition, and Bose-Einstein condensation.

Several transitions are known as the infinite-order phase transitions. They are continuous but break no symmetries. The most famous example is the Berezinsky-Kosterlitz-Thouless transition in the two-dimensional XY model. Many quantum phase transitions in two-dimensional electron gases belong to this class. Apartfrom this, there is another phase transition known as Lambda phase transition. 4.2 FIRST ORDER PHASE TRANSITION The first order phase transition occurs between the triple point and critical point (excluding the critical point). A first order phase transformation should satisfy the following two requirements:

61

1. There are changes in entropy and volume 2. The first order derivatives of Gibbs function change discontinuously.

Consider a system consisting of two phases of a pure substance at equilibrium. Under these conditions, the two phases are at the same pressure and temperature. Consider the change of state associated with a transfer of moles from phase 1 to phase 2 (

dn,p T remaining constant). That is,

It is the phase transition between the triple point and the critical point in a T s diagram ( Fig. 4.2).

−

Critical point

p

p+dp

p

S

T

T.P

X (p, T)

y (p+dp ,T+dp)

Fig. 4.2 First order phase transition

dg sdT vdp= − + (4.1) For a reversible isothermal isobaric phase transition

0d g = (4.2)

g =constant=molar Gibbs function ( kJ/kgmol) Hence at state ( ),X p T in Fig. 4.2,

( ) ( )i g fg= (4.3) At state , ( ),Y p dp T dT+ +

( ) ( ) ( ) ( ) i i fg dg g dg+ = + f (4.4)

62

Subtracting Eq. (4.3) from Eq. (4.4) we get,

( ) ( )i fdg dg= (4.5)( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( )

( ) ( )

( ) ( )

i i f f

i f i f

f i

f i

s v dp s v dp

v v dp s dT s dT

dp s sdT v v

⇒ − + = − +

⇒ − = −

−⇒ =

−

(4.6)

From the relation

Tds dq= (4.7) We get

( ) ( )f i

dp ldT T v v

=⎡ ⎤−⎣ ⎦

(4.8)

Eq. (4.8) is known as Clausius- Clapeyron’s equation and can be used to estimate the latent heat if ,p v and T are known. The vapor pressure curve can be represented by the following correlation,

2

ln ln

1

Bp A C T DT

dp B C Dp dT T T

= + + +

⇒ = − + +

T

(4.9)

The three different phase transition processes for water are discussed in the following subsections (Fig. 4.3).

63

Vapour

dp

dT

Triple point

Sublimationcurve

Vaporization curve

Critical point

LiquidSolid

(Negativeslop)

Forwater

Fusion curve

For any other substance(positive slope)

p1

T1T

p

Fig. 4.3 Phase diagram for a pure substance on p T− coordinates

4.2.1 FUSION It is the solid to liquid phase transition.

[ ]" 'fuldp

dT T v v=

−

(4.10)

where ful is the latent heat of fusion

' (Prime) indicates the saturated solid state " ( Double prime) indicates the saturated liquid state

For water, (Indicating contraction) ⇒ "v v< ' dp vedT

→ −

For most other substances ( expansion) "v v> ' dp vedT

⇒ → +

4.2.2 VAPORIZATION It is the liquid to vapor phase transformation.

[ ]"' "vapldp

dT T v v=

−

(4.11)

64

Usually, "' "v v>Assuming vapor to behalf as an ideal gas

_

"' RTvp

≅ (4.12)

_ _ _2 2

vap vapvap

l p ldp p ldT RT RT RTT

p

∴ = = = (4.13)

Clausius- Clapeyron’s equation can also be used to estimate approximately the vapor pressure of a liquid at any arbitrary temperature in conjunction with a relation for latent heat of a substance, known as Trouton’s rule. 4.2.2.1 TROUTON’S RULE

_

88 /fg

B

hkJ kgmol K

T≅ −

(4.14)

where latent heat of vaporization, kJ/kgmol _

fgh →and boiling point at 1.013 bar (N.B.P.) BT → From Eq. (4.10) and (4.14),

_2

88 Bdp p TdT RT

= (4.15)

_ 2101.325

88

B

p TB

T

Tdp dTp TR

⇒ =∫ ∫

_88 1 1 ln

101.325B

B

TpT TR

⎛ ⎞⇒ = − ⎜ ⎟

⎝ ⎠−

_88 101.325 exp 1 BTp

TR

⎡ ⎤⎛ ⎞⇒ = −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

(4.16)

where p is the vapor pressure in kPa for any temperature T .

65

4.2.3 SUBLIMATION It is the solid to vapor phase transformation.

'"'subldp

dT T v v=

⎡ ⎤−⎣ ⎦ (4.17)

where latent heat of sublimation. subl →

v "' 'v >>∵ and vapor pressure is low, _

"' RTvp

= (4.18)

_2

subl pdpdT T R

∴ = (4.19)

( )_ log 2.303

1sub

d pl R

dT

⇒ = −⎛ ⎞⎜ ⎟⎝ ⎠

(4.20)

At the triple point:

sub vap fusl l l= + (4.21) Now,

_2

tp vap

vaptp

p ldpdT RT

= (4.22)

and

_2

tp sub

subtp

p ldpdT RT

= (4.23)

Since, sub vapl l> at the triple point

sub vap

dp dpdT dT

> (4.24)

66

4.3 SECOND ORDER PHASE TRANSITION The second order phase transition occurs at the critical point where saturated liquid flashes into vapor. There is no distinction of phases at triple point. However, there is distinct variation of the value of specific heat . Following are the assumptions for second order phase transition

pC

1. There are no changes of entropy and specific volume 2. Molar Gibbs function is continuous g3. First order derivative of is continuous g

;

;

; ;

f ff f

g gg g

f g f g f g

g gs v

T pg g

s vT pg g g g

g gT T p p

∂ ∂= − =

∂ ∂∂ ∂

= − =∂ ∂∂ ∂ ∂ ∂

= =∂ ∂ ∂ ∂

=

(4.25)

;

;

; ;

f ff f

g gg g

f g f g f g

g gs v

T pg g

s vT pg g g g

g gT T p p

∂ ∂= − =

∂ ∂∂ ∂

= − =∂ ∂∂ ∂ ∂ ∂

= =∂ ∂ ∂ ∂

=

(4.26)

4. Second order derivatives of Gibbs function changes discontinuously. g

2

2 0 0 at critical pointT

g v pp p v

⎛ ⎞∂ ∂ ∂⎡ ⎤= >> =⎜ ⎟ ⎢ ⎥∂ ∂ ∂⎣ ⎦⎝ ⎠∵

(4.27)

At critical point the latent heat of vaporization, l is zero.

(Remember: Fluid containing one component exhibits a single critical point.)

67

4.3.1 ANALYSIS OF SECOND ORDER PHASE TRANSITION

pp

sC TT

∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠

(4.28)

or,

2

2p

p pp

C g gT T T T

⎡ ⎤ ⎛ ⎞∂ ∂ ∂⎛ ⎞= − = −⎢ ⎥ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠⎢ ⎥ ⎝ ⎠⎣ ⎦

(4.29)

From the definition of isothermal compressibility,

2

2 TT TT

g pvp p p

κ⎡ ⎤ ⎛ ⎞⎛ ⎞∂ ∂ ∂

= − −⎢ ⎥ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

(4.30)

Again, following the definition of volume expansivity,

2

T p

g gvT p T

β⎡ ⎤⎛ ⎞∂ ∂ ∂

= =⎢ ⎥⎜ ⎟∂ ∂ ∂ ∂⎢ ⎥⎝ ⎠⎣ ⎦ p

(4.31)

For reversible isothermal isobaric process

( ) ( ) [ ] at ,i fs s p T= (4.32)( ) ( ) ( ) ( ) ( ) i i f fs ds s ds at p dp T dT,⎡ ⎤⇒ + = + + +⎣ ⎦ (4.33)

( ) ( ) i fTds Tds⇒ = (4.34)

p pp

vTds C dT T dp C dT T vdpT

β∂⎛ ⎞= − = −⎜ ⎟∂⎝ ⎠∵

(4.35)

[ 2nd T-dS relation]

( ) ( ) ( ) ( ) i i f fp pC dT T vdp C dT T vdpβ β∴ − = − (4.36)

or, ( ) ( )

( ) ( )( )f i

p pf i

C CdpdT Tv β β

−=

−

(4.37)

As there is no change in volume

( ) ( ) ( ) at ,i fv v p T⎡ ⎤= ⎣ ⎦ (4.38)

Further, at point [ ],p dp T dT+ + ,

68

( ) ( ) ( ) ( )i i fv dv v dv+ = + f (4.39)( ) ( ) i fdv dv⇒ =

( ) ( ) ( ) ( ) i i f fT TvdT vdp vdT vdpβ κ β κ⇒ − = −

( ) ( ) ( ) ( ) f i f iT Tvdp vdp vdT vdTκ κ β β⇒ − = −

( ) ( )

( ) ( ) f i

f iT T

dpdT

β βκ κ

−⇒ =

−

(4.40)

4.3.2 APPLICATION OF SECOND ORDER PHASE TRANSITION

1. Transition of a super conductor from super conducting to the normal state in zero magnetic field (It is a true second order transition).

2. Ferromagnetic to paramagnetic transition in a simple model. 3. Order-disorder transformation.

4.4 LAMBDA TRANSITION It is the third type of phase transition between the two liquid phases of , ordinary liquid helium l, and super fluid helium ll***. This transition can occur at any point along the line separating these two liquid phases in Fig.4.4 .

4HeHe

Fig. 4.4 Phase diagram of helium

69

From this figure, it is evident that l and ll can coexist in equilibrium over a range of temperature and pressures, and l can be converted to ll either by lowering the temperature, provided pressure is not too great, or by reducing the pressure, provided that the temperature is below 2.18 K. A graph of versus T for the two phases has the general shape shown in Fig.4.5, and the transition takes its name from the resemblance of the curve to the shape of the Greek letter

He HeHe He

pC

λ . The value of does not change discontinuously, but its variation with temperature is different in the two phases.

pC

1.5 2.0 2.5 3.0

10

20

30

40

50 x 103

0T (K)

c (J

kilo

mol

e k

)

v-1

-1

Fig. 4.5 Lambda transition

4.5 PROPERTIES OF SUPERFLUID Superfluids have the unique quality that their atoms are in the same quantum state. This means they all have the same momentum, and if one moves, they all move. This allows superfluids to move without friction through the tiniest of cracks, and superfluid helium will even flow up the sides of a jar and over the top. This apparant defiance of gravity comes from a special type of surface wave present in superfluid helium, which in effect pushes this extremely thin film up the sides of the container. It was discovered in 1962 by Tisza, who named the phenomenon third sound. Another unusual result of third sound is the fountain effect, where superfluid excited by photons will form a fountain vertically upward off of its surface. Superfluids also have an amazingly high thermal conductivity. When heat is introduced to a normal system, it diffuses through the system slowly. In a superfluid, heat is transmitted so fast that thermal waves become possible. This

70

fourth kind of wave found in superfluids is called second sound, quite improperly becuase they involve no pressure variations. Following are some of the remarkable properties of superfluid He II: 1. Ability to flow through microscopic passages with no apparent friction 2. The quantization of vortices 3. The ability to support four wave modes: (a) Sound which is analogous to sound in ordinary gases and liquids, (b) Sound which carries temperature and entropy perturbations with virtually no pressure variations (c) Sound which are waves on thin films (d) Sound which are acoustic-like waves in the "superfluid" component of He II. As in the case of both classical and high temperature superconductivity, superfluidity is a manifestation of quantum mechanical effects at the macroscopic level. *** Helium vapor compressed isothermally between 5.25 K to 2.18 K, condenses to liquid helium l. When vapor is compressed below 2.18 K, a liquid called super fluid helium ll results.

4HeHe

He 4.6 MIXTURE OF VARIABLE COMPOSITION Let us consider a system containing a mixture of substances 1,2, 3,…….k. If some quantities of a substance are added to the system, the energy of the system will increase. Thus for a system with variable composition, the internal energy depends not only on and V , but also on the number of moles (or mass) of various constituents of the system.

S

( )1 2, , , ,.........., kU U S V n n n= (4.41)

where, are the number of moles of substances 1,2,3,……k. 1 2 3, , ,........, kn n n n The composition may change not only due to addition or subtraction, bur also due to chemical reaction and inter phase mass transfer. For a small change in U , assuming the function to be continuous,

1 2 1 2 2

1 1 1

1, , ,....., , , ,....., 1 , , ,.....,

22 , , ,....., , , ,.....,

............

k k k

k k

V n n n S n n n V S n n

kkV S n n V S n n

U U UdU dS dV dnS V n

U Udn dnn n

−

⎛ ⎞∂ ∂ ∂⎛ ⎞ ⎛ ⎞= + + ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞∂ ∂+ + + ⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(4.42)

71

1, , , ,i i j

k

ii iV n S n S V n

x x UdU dS dV dny y n=

⎛ ⎞⎛ ⎞ ⎛ ⎞∂ ∂ ∂= + + ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑ (4.43)

where the subscript i any substance =

other substance except the one whose number j = of moles is changing.

If composition does not change,

dU TdS pdV= − (4.44)

,

iV n

U TS

∂⎛ ⎞∴ =⎜ ⎟∂⎝ ⎠

(4.45)

and

, iS n

U pV

∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠

(4.46)

1 , ,

j

k

ii i S V n

UdU TdS pdV dnn=

⎛ ⎞∂∴ = − + ⎜ ⎟∂⎝ ⎠

∑ (4.47)

Eq. (4.47) can be written as

1

k

i ii

dU TdS pdV dnµ=

= − + ∑ (4.48)

where, , , j

ii S V n

Un

µ⎛ ⎞∂

= ⎜ ⎟∂⎝ ⎠ is the molar chemical potential. It signifies the change in

internal energy per unit mole of component i when and number of moles of all other components are constant. The chemical potential drives mass (or species) similar to the thermal potential that drives heat transfer from higher to lower temperature.

,S V

Ref. to Eq. (4.41), we can write in a similar manner,

( )1 2, , , ,........, kG G p T n n n= (4.49)

or,

72

1,, , ,ii j

k

iip n iT n T p n

G G GdG dp dT dnp T n=

⎛ ⎞⎛ ⎞∂ ∂ ∂⎛ ⎞= + + ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠∑

(4.50)

1 , , j

k

Ii i T p n

GVdp SdT dnn=

⎛ ⎞∂= − + ⎜ ⎟∂⎝ ⎠

∑ (4.51)

Since,

G U pV TS= + − (4.52)

( )1 , , j

k

ii T p n

xd U pV TS Vdp SdT dny=

⎛ ⎞∂+ − = − + ⎜ ⎟∂⎝ ⎠

∑ (4.53a)

or,

1 , , j

k

ii i T p n

GdU pdV Vdp TdS SdT Vdp SdT dnn=

⎛ ⎞∂+ + − − = − + ⎜ ⎟∂⎝ ⎠

∑

1 , , j

k

ii i T p n

GdU TdS pdV dnn=

⎛ ⎞∂= − + ⎜ ⎟∂⎝ ⎠

∑ (4.53b)

Comparing Eq. (4.47) and Eq.(4.51), we get,

1 1, , , ,j j

k k

i ii ii iS V n S V n

U Gdn dnn n= =

⎛ ⎞ ⎛ ⎞∂ ∂=⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

∑ ∑ (4.54)

, , , ,

j j

ii iS V n S V n

U Gn n

µ⎛ ⎞ ⎛ ⎞∂ ∂

∴ = =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

(4.55)

Eq. (4.51) can be written as

1

k

i ii

dG Vdp SdT dnµ=

= − + ∑ (4.56)

Hence,

1

k

i ii

dU TdS pdV dnµ=

= − + ∑ (4.57)

1

k

i ii

dG Vdp SdT dnµ=

= − + ∑ (4.58)

1

k

i ii

dH TdS Vdp dnµ=

= + + ∑ (4.59)

73

1

k

i ii

dF SdT pdV dnµ=

= − − + ∑(4.60)

Now,

, , , , , , , ,j j ji i i iS V n S V n S V n S V n

U G H Fn n n n

µ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂

= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠j

(4.61)

4.6.1 GIBBS-DUHEM EQUATION Let us consider a homogeneous phase of a multi-component system for which

1

k

i ii

dU TdS pdV nµ=

= − + ∆∑ (4.62)

If the phase is enlarged in size, and V will increase in size, while and ,U S ,T pµ will remain the same. Thus,

1

k

i ii

U T S p V nµ=

∆ = ∆ − ∆ + ∆∑ (4.63)

Let the system be enlarged to times the original size. k

( ) 1U kU U k U∴ ∆ = − = − (4.64a)

( )1S kS S k S∆ = − = − (4.64b)

( )1V kV V k V∆ = − = − (4.64c)

( )1i i in kn n k n∆ = = = − i

n

(4.64d)

Substituting Eqs. (4.64a-d) in Eq. (4.63), we get

( ) ( ) ( )1

1 1 1k

i ii

k U T k S p k V µ=

− = − − − + ∑ (4.65)

1

k

i ii

U TS pV nµ=

∴ = − + ∑ (4.66)

1

k

i ii

U pV TS nµ=

⇒ + − = ∑ (4.67)

,1

k

T p i ii

G nµ=

= ∑ (4.68)

Differentiating Eq. (4.68), we get

74

,1 1

k k

T p i i i ii i

dG n d dnµ µ= =

= +∑ ∑ (4.69)

(at constant temperature and pressure) When temperature and pressure changes,

1

k

i ii

dG SdT Vdp dnµ=

= − + + ∑ (4.70)

Comparing Eq. (4.69) and Eq. (4.70), we get,

1 1 1

k k k

i i i i i ii i i

n d dn SdT Vdp dnµ µ µ= = =

+ = − + +∑ ∑ ∑ (4.71)

or,

10

k

i ii

SdT Vdp n dµ=

− + − =∑ (4.72)

Eq. (4.72) is known as Gibbs-Duhem Equation. It represents simultaneous changes of and ,T p µ . Now,

, 1 1 2 21

........k

T p i i k ki

G n n n nµ µ µ µ=

= = + + +∑ (4.73)

For a phase consisting of single constituent,

G nµ= G = gn

µ∴ = (4.74)

Hence, chemical potential is the molar Gibbs function and is a function of T and p . For a single phase, iµ is a function of and mole fraction ,T p ix . 4.7 CONDITIONS OF EQUILIBRIUM OF A HETEROGENEOUS SYSTEM Let us consider a heterogeneous system of volume V with several homogeneous phases ( existing in equilibrium. Let us further

suppose that each phase consists of ), , ,......a b c rφ =

( )1, 2,3,.......,i i c= constituents. Further, number of constituents in any phase is different from others.

75

With each phase, a change in internal energy is accompanied by a change in entropy , volume V and composition. S

( )1

c

i ii

dU T dS p dV dnφ φ φ φ φ φµ

=

= − + ∑ (4.75)

A change in the internal energy of the entire system can therefore be expressed as,

( )1

r r r r c

i ia a a a idU T dS p dV dnφ φ φ φ φ φ

φ φ φ φ

µ= = = = =

= − +∑ ∑ ∑ ∑∑ (4.76)

Now, change in internal energy of the entire system involves changes in the internal energy of the constituent phases,

..........r

a b ra

dU dU dU dU dUφφ =

= + + + = ∑ (4.77)

Similarly, changes in volume, entropy or chemical composition can be written as

........r

a b ra

dV dV dV dV dVφφ =

= + + + = ∑ (4.78)

.......r

a b ra

dS dS dS dS dSφφ =

= + + + = ∑ (4.79)

........r

a b ra

dn dn dn dn dnφφ =

= + + + = ∑ (4.80)

For a close system in equilibrium,

0dU dV dS dn= = = = (4.81)

( ).(4.77) .......r

a b c rj b

Eq dU dU dU dU dU=

⇒ = − + + + = −∑ j

j

(4.82)

( ).(4.78) ........r

a b c rj b

Eq dV dV dV dV dV=

⇒ = − + + + = −∑ (4.83)

( ).(4.79) ......r

a b c rj b

Eq dS dS dS dS dS=

⇒ = − + + + = − j∑ (4.84)

76

( ).(4.80) .......r

a b c rj b

Eq dn dn dn dn dn=

⇒ = − + + + = − j∑ (4.85)

Energy Equation (Eq. (4.76)) for the heterogeneous system is,

( )1

0r r r r c

i ia a a a idU T dS p dV dnφ φ φ φ φ φ

φ φ φ φ

µ= = = = =

= − +∑ ∑ ∑ ∑∑ =

=

(4.86)

( ) ( )1 1

0r r r c c

a a j j a a j j i i i ij aj b j b j b i i

T dS T dS p dV p dV dn dnµ µ= = = = =

+ − − + +∑ ∑ ∑∑ ∑ (4.87)

( ) ( ) ( ) ( ) 0r r r r r r r r

a j j j a j j j j j j i jj j jj b j b j b j b j b j b j b j b

T dS T dS p dV p dV dn dnµ µ= = = = = = = =

⇒ − + + − − + =∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑(4.88)

( ) ( ) ( )( ) 0j a j j a j ij ia i j

j j j iT T dS p p dV dnµ µ⇒ − − − + − =∑ ∑ ∑∑ (4.89)

Since, and ( are independent and the expression is zero (Eq.

(4.89)), their coefficients must be zero individually.

,jdS dVj )i jdn

j aT T∴ = (4.90) Thermal Equilibrium⇒

j ap p= (4.91) Mechanical Equilibrium⇒

ij iaµ µ= (4.92) Chemical Equilibrium⇒ 4.8 GIBBS PHASE RULE Let us consider a heterogeneous system in which constituents ( ) are present in all the phases. Initially we consider no chemical reaction to avoid the complexity. Let us assume that there are

c 1,2,3,.....,i c=

φ phases. In each phase, all the constituents are present, i.e.,

1 21, constituents are , ,....., cn n nφ =

1 22, constituents are , ,....., cn n nφ =

........................................... 1 25, constituents are , ,......, cn n nφ =

(4.93)

77

For ease of description we will identify the constituents by subscript and phase by superscript. The Gibbs function of the whole heterogeneous system is

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1 1 2 2 3 3,

1 1 1 1.......

c c c c

T p i i i i i i i ii i i i

G n n n n φ φµ µ µ= = = =

= + + + +∑ ∑ ∑ ∑ µ (4.94)

( ), , iG G T p n= (4.95)

Since there is no chemical reaction, only way in which s may change is by the transport of the constituents from one phase to another. In this case, the total number of moles of each constituent will remain constant.

'n

( ) ( ) ( ) ( )1 2 31 1 1 1............ constantn n n n φ+ + + + = ( ) ( ) ( ) ( )1 2 32 2 2 2............ constantn n n n φ+ + + + =

...........................................

( ) ( ) ( ) ( )1 2 3 ............ constantc c c cn n n n φ+ + + + =

(4.96)

At chemical equilibrium, G will be rendered a minimum at constant subject to the aforementioned equations of constraints.

& T p

, 0T pdG∴ = (4.97)

Therefore, from the condition of equilibrium of a heterogeneous system (chemical equilibrium),

ij iaµ µ= (4.98)( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 2 31 1 1 1

1 2 32 2 2 2

1 2 33 3 3 3

1 2 3

.............

.............

...............................................................

.............c c c c

φ

φ

φ

φ

µ µ µ µ

µ µ µ µ

µ µ µ µ

µ µ µ µ

= = = =

= = = =

= = = =

= = = =

(4.99)

Eqs.(4.99) , which specify the condition of phase equilibrium are called the equations of phase equilibrium. There are ( )1k φ − numbers of such equations. Now the composition of each phase containing constituents is fixed if k 1k − constituents are known, since sum of the mole fractions of each constituent in the phase must equal unity. Therefore, for φ phases, there are a total of

78

( 1kφ − ) variables, in addition to temperature and pressure, which must be

specified. There are, then, ( )1 2kφ − + variables altogether. If the number of variables is equal to the number of equations, then whether or not we can actually solve the equations, the temperature, pressure, and composition of each phase is determined. The system is then called nonovariant and said to have zero variance. If the number of variables is one greater than the number of equations, an arbitrary value can be assigned to one of the variables and the remainder are completely determined. The system is then called monovariant and said to have a variance of 1. In general, the variance f is defined as the excess of the number of variables over the numbers of equations.

( ) ( )1 2 1 2f k k kφ φ⎡ ⎤ ⎡ ⎤= − + − − = − +⎣ ⎦ ⎣ ⎦ φ (4.100)

This equation is called the Gibbs phase rule. For chemical reactions, Gibbs phase rule is modified as

( ) 2f c r φ= − − + (4.100)

where, number of independent reversible chemical reaction r =

φ = number of phases constituent k =

f = variance or degree of freedom Consider the diagram for water. At any point on the saturated vapor line, T s−

1, 1c φ= = . Hence, . This indicates that 2 (two) independent thermodynamic properties are needed to fix up the state of the system at equilibrium. Similarly, at a location within the saturated liquid and vapor line,

2f =

1, 2 1c fφ= = ⇒ = . Hence, only one thermodynamic property is sufficient to fix up the state. At the triple point, 1, =3 =0c fφ= ⇒ . It is a unique state where all the three phases exist in equilibrium. Exercise 4.1: To make baking soda ( , a concentrated aqueous solution of is saturated with . The reaction is given as

) 3

3

3NaHCO 2Na CO

2CO

3 2 22 2Na CO H O CO NaHCO+ −+ + + ⇔

79

Thus ions and ions, and are present in arbitrary amounts, except that all the

Na+3CO−

2 ,H O CO2 2NaHCONa+ and 3CO− are from . Find the number of

degree of freedom of this system. 2Na CO3

Exercise 4.2:

Determine the number of degree of freedom for the system at each lettered point and state the variables for a cadmium-bismuth system (Fig. 4.6)

400

300

200

100

Liquid solution Cd+Bi

A

271 C

Liquidsolution andsolid Bi C

B

D

Solid Cd + solid BiE

1 44 C

321 C

Liquid solution andsolid Cd

20 40 60 80 100Weight % Cd

Fig. 4.6 Phase diagram of cadmium-bismuth system

Note: We can have several separate solid phases in a system, and the same is true for (immiscible) liquids. On the other hand there can be no more than one gas phase, because all gases freely intermingle. 4.9 THIRD LAW OF THERMODYNAMICS Marching towards absolute zero temperature. 4.9.1 MOTIVATION

Application of first law and second law of thermodynamics to reactive systems become difficult due to the non availability of a standard reference entropy value of various substances. There is a need to have a reference entropy value for all substances for evaluating the efficiency of a reactive system. Third law of thermodynamics provides a base value for the entropy.

80

The third law of thermodynamics was formulated during the early part of twentieth century. The initial work was done primarily by W. H. Nernst [1864-1941] and Max Planck [1858-1947]. 4.9.2 ATTAINING LOW TEMPERATURE

1. Below 5 K is possible by Joule Kelvin expansion, by producing liquid helium.

2. Still lower temperature can be attained by adiabatic demagnetization of a paramagnetic salt.

3. Temperature as low as 0.001 K has been achieved by magnetic cooling. 4.9.3 MAGNETIC PROPERTY

1. Diamagnetic: substance is repelled by magnet 2. Paramagnetic: substance attracted by magnet, such as Iron, Gadolinium sulphate

Adiabatic Demagnetization of a paramagnetic salt:

Gadolinium sulphate is used here. Salt is hung by a fine nylon thread inside the salt tube such that it does not

touch the sides.

Magnet

Liquid helium

Liquid nitrogen

To pump

Radiation shield

Salt

Dewar flask

Measuring coilsSupporting threads

Exhaust gas

Fig: 4.7 Adiabatic demagnetization of a paramagnetic salt

81

The salt is first cooled slightly below 1 K by surrounding it with liquid helium boiling under reduced pressure. Next, the salt is exposed to a strong magnetic field of about 25000

Gauss. Heat produced due to magnetization of the salt is transferred to the liquid helium without causing an increase in salt temperature. With the magnetic field still present, the inner chamber containing the

salt is evacuated of gaseous helium. Finally, the magnetic field is removed. The molecules disalign

themselves, which require energy. This energy is obtained by the salt getting cooler in the process.

Repetition of the process lowers down the temperature of the salt.

Temperature as low as 0.001 K have been achieved so far. 4.9.4 MEASUREMENT OF TEMPERATURE In the neighborhood of absolute zero, all ordinary methods of temperature measurement fail. Curie’s law gives the most convenient method for measurement of at low temperature (approximately).

cT

χ = (4.101)

where χ is the magnetic susceptibility of the salt

T is the absolute temperature c is the Curie’s constant.

The fundamental features of all cooling process are that the lower the temperature achieved, the harder it is to go still lower.

Interpretations of adiabatic demagnetization:

Final temperature f iT T∞ (initial temperature)

First demagnetization produces a temperature half that at start 12f iT T⎛ ⎞=⎜ ⎟

⎝ ⎠

Second demagnetization produces a temperature 1 1 12 2 4f i iT T⎛ ⎞= =⎜ ⎟

⎝ ⎠T

Third demagnetization produces a temperature 1 1 12 4 8f i iT T⎛ ⎞= =⎜ ⎟

⎝ ⎠T

……………………………………………………………………………………

demagnetization produces a temperature rth 12f irT T=

82

Infinite number of adiabatic demagnetizations will be required to attain absolute zero temperature.

FOWLER-GUGGENHEIM STATEMENT OF THIRD LAW: It is impossible by any procedure, no matter how idealized, to reduce any condensed system to the absolute zero of temperature in a finite number of operations.

Any isothermal magnetization from to (magnetic intensity) such as 0 H 1k i− ,

1 2f i− etc. is associated with the release of heat, i.e., a decrease in entropy (Fig. 4.8).

k

f1

f2

f3 H

T

i

i

i

(0,0) (0, H )i

1

2

3

T

SS (0,0) S (0 ,H )i

l

l

l k

f

f

f

1

2

32

3

H= 0 H =Hi

1

H =H H = 0i

kl

lf

f

1

2

1

2

T = 0

S ( 0,H )S (0, 0)

i

i

S (0, H ) - S (0,0) = 0i

{s

T

Fig. 4.8 T-H and T-S diagrams of a paramagnetic substance to show the

equivalence of three statements of the third law The processes etc. represent reversible adiabatic demagnetizations with temperatures getting lower and lower. Repeated cycles of isothermal magnetization and adiabatic demagnetization would bring about a very low temperature. It is seen that

1 1 2 2 3, , i f i f i f− − − 3

( ) ( ),S T H S T− ,0⎡ ⎤⎣ ⎦ decreases as the temperature decreases, i.e., III II IS S S∆ < ∆ < ∆ . Thus entropy change associated

83

with an isothermal reversible process of a condensed system approaches zero. It is called Nernst-Simon statement of third law of thermodynamics. NERNST-SIMON STATEMENT:

The entropy change associated with any isothermal reversible process of a condensed system approaches zero.

0lim 0TT

S→

∆ =

Another statement of third law can be given like this: It is impossible by any procedure, no matter how idealized, to reduce the entropy of a system to zero point value in a finite number of operations.

Physical and chemical facts which substantiate the third law: 1. For any phase change that takes place at low temperature, Clausius-Clayperon

equation hold good, f i

f i

s sdpdT v v

−=

−

(4.102)

From third law of thermodynamics, ( )

0lim 0f iT

s s→

− = , since f iv v− is not zero. It

shows that

0lim 0x

dpdTδ →

= (4.103)

This is substantiated by all known sublimation curves.

H

G

H

HHG

T

Fig.4.9 The temperature dependence of the change in the Gibbs function and in

the enthalpy for an isobaric process

84

2.

G H T∆ = ∆ − ∆s (4.104)At low temperature, T is very small. s∆

G H∴ ∆ = ∆ (4.105)Which confirms that .

0lim 0T

s→

∆ =

3. From the Gibbs-Helmholtz equation

p

GG H TT

∂∆⎡ ⎤∆ = ∆ + ⎢ ⎥∂⎣ ⎦

(4.106)

( )

0 lim 0

TG H

→∆ − ∆ →∵ (Fig. 4.9)

0

0 0

0 0

0

lim 0

lim lim

lim lim lim

lim 0

T

T T

pT T

pT

GTG H G

T TG H G CT T T

C

→

→ →

→ →

→

∂∆⎡ ⎤ =⎢ ⎥∂⎣ ⎦∆ − ∆ ∂∆

⇒ =∂

∂ ∂ ∂⎡ ⎤⇒ − = −⎢ ⎥∂ ∂ ∂⎣ ⎦ 0T →

∴ =

(4.107)

Similarly, using Gibbs Helmholtz equation containing U and F

0lim 0 as 0vT

C T→

= → (4.108)

0 0p vC C as T∴ = = → (4.109)

85