MLZ 325
SOLUTION
THERMODYNAMICS
PHASE EQUILIBRIUM IN ONECOMPONENT SYSTEMS
P
1
2
3
1
H2O
Slush (ice/water)
(state of matter is different)
…
2
Ca-ZrO2
(cubic zirconia)
CCl4-H2O
(nonpolar;
do not dissolve in each other)
3
White Gold
(Au-Ag-Ni)
SiO2-MgO-Fe3O4
(obsidian-crystalline)
C

Unary: pure H2O(l), white gold, air, cubic zirconia

Phase diagrams: stability maps



…
…
Phase: region of uniform chemical composition; physically
distinct (bounded); mechanically separable (multiphase systems)
Eqm: No net reaction; lowest energy state; properties invariant
over time; attainable by multiple paths.
Component: Measurement of chemical complexity; number of
chemically distinguishable constituents
PHASE EQUILIBRIUM IN ONECOMPONENT SYSTEMS


In a closed system of fixed composition, e.g.,
a one-component system, equilibrium, at the
temperature, T, and the pressure, P, occurs
when the system exists in that state which has
the minimum value of G.
The equilibrium state can thus be determined
by means of examination of dependence of G
on P and T.
THE VARIATION OF G WITH T AT
CONSTATNT P

Case 1:
At a total pressure of 1 atm, ice and water are
in equilibrium with one another at
0 C, and hence, for these values of T and P, G’
of the system has its minimum value.
If,
by
the
addition
of
heat,
1 mole of ice is melted, then for the change of
state
T  273 K
H 2O s  
P 1atm
H 2Ol 
G s l   GH 2O l   GH 2O  s   0
THE VARIATION OF G WITH T AT
CONSTATNT P

Thus, at the state of equilibrium between ice
and water,
GH 2O  l   GH 2O  s 

The Gibbs free energy of the system, G’, is
G   n H 2O  s   GH 2O  s   n H 2O  l   GH 2O  l 

Thus, at 0 C and 1 atm, the value of G’ is
independent of the proportions of the ice
phase and the water phase present.
THE VARIATION OF G WITH T AT
CONSTATNT P

Case 2:
If the ice + water system is at 1 atm and some
T > 0 C, then the system is not stable and the
ice spontaneously melts.
This process decreases G of the system, and
equilibrium is attained when all of the ice has
melted:
T  273 K
H 2O s  
P 1atm
H 2Ol 
G m  G H 2 Ol   G H 2 Os   0
GH 2O  l   GH 2O  s 
THE VARIATION OF G WITH T AT
CONSTATNT P

Case 3:
If the ice + water system is at 1 atm and some
T < 0 C, then the system is not stable and the
water spontaneously solidifies.
This process decreases G of the system, and
equilibrium is attained when all of the water
has solidified:
T  273 K
H 2O s  
P 1atm
H 2Ol  Gm  G H Ol   G H Os   0
2
2
GH 2O  l   GH 2O  s 
THE VARIATION OF G WITH T AT
CONSTATNT P
Fig. 7.1
P = 1 atm
L
G
Fig. 7.2
P = 1 atm
GSL 0
S
Tm
-Sm
Tm
T
 G 
slope  
  S  0
 T  P
  2G 
   S    c P  0
curvature  
 T 2 
T
 T  P

P
Tm
T
 Gm 
slope  
  S m  0

T

P
 S H 2Ol   S H 2Os 
Negative slope & negative curvatureconcave down & slope increases with increasing T
THE VARIATION OF G WITH T AT
CONSTATNT P
THE VARIATION OF H WITH T AT
CONSTANT P
G  H T S
(for convenience)
 Gm  H m  T  S m
eqmT  Tm   Gm  0
P = 1 atm


 H m T   H m Tn.m. p  273 
H m T 
0
H (J)
 H m  Tm  S m
H=H(T)P

H m Tn.m. p
273

298
Fig. 7.3
T
 cP
Tn.m. p  273
 H m T   6008  37.44  T  273
T
 dT
THE VARIATION OF S WITH T AT
CONSTATNT P
S=S(T)P
T·S (J/K)
P = 1 atm
S m T 

S m Tn.m. p
273
 S m T  

298
Fig. 7.4
H m 273 T cP
 S m T  
 
 dT
273
T
273
T
6008
 T 
 37.44  ln

273
 273 
THE VARIATION OF G WITH T AT
CONSTATNT P
G  H T S
Hm, T· Sm, Gm
(J/K)
P =1 atm
T·Sm
Hm
-T·Sm
0
Gm
273
Fig. 7.5
 Eqm
(G)
between
2
phases is a compromise
betwee enthalpy (H) and
entropy (S).
 G is minimum at fixed
values of T & P.
 Minimization
of
G
requires that H be small
and S be large.
 In the absence of S,
H(l) > H(s) solid always
stable w.r.t. liquid.
 In the absence of H,
S(l) > S(s)  liquid
always stable w.r.t. solid.
THE VARIATION OF G WITH T AT
CONSTATNT P
However, contribution of S to G is dependent
on T
 A unique temperature Tm exists;
 T > Tm
 S-contribution to G outweighs Hcontribution to G. G(liquid) < G (solid).
 T = Tm
 S-contribution to G exactly balances Hcontribution to G; G (solid) = G(liquid).
 T < Tm
 H-contribution to G outweighs Scontribution to G. G (solid) < G(liquid).
THE VARIATION OF G WITH P AT
CONSTATNT T
dG  V  dP  S  dT
 G l 

 V l 


P

T
 G s 
 P   V s 
T
 Gm 

  Vm

P

T
For H2O,
Vm (273 K, 1 atm) < 0
 slope < 0
 Ice melts when P > 1 atm
THE VARIATION OF G WITH P AT
CONSTATNT T
G=G(P)T
Gm=Gm (P)T
T =273 K
solid
T = 273 K
liquid
Gm
G V(l)
V(s)
1 atm
0
Vm
1 atm
P
P
Fig. 7.6
THE VARIATION OF G WITH T & P
Figs. 7.1 & 7.6
 It is possible to maintain eqm between the solid
and liquid phase by simultaneously varying T and P
in such a manner that Gm remains zero.
P = 1 atm
L
G
-S(s)
S
T =273 K
solid
G V(l)
V(s)
-S(l)
T
P
Fig. 7.1
Fig. 7.6
liquid
THE VARIATION OF G WITH T & P
For eqm to be maintained:
Gm  0
 Gl  Gs
 dGl  dGs
dGl   Sl  dT  Vl  dP
dGs   S s  dT  Vs  dP
  Sl  dT  Vl  dP   S s  dT  Vs  dP
  Sl  dT  S s  dT  Vs  dP  Vl  dP
 S s  Sl   dT  Vs  Vl   dP
 Sl  S s   dT  Vl  Vs   dP
S  S s 
 dP 

  l
 dT  eqm Vl  Vs 
S
 dP 

  m
 dT  eqm Vm
H m
 dP 



Eqn (7.5) Clapeyron Eqn
 dT  eqm Tm  Vm
THE CLAPEYRON EQUATION
H m
 dP 
 

 dT eqm Tm  Vm
Vm  0 for H 2O
H m  0 for all materials
T 0
H m

 0 for H 2O
Tm  Vm
 dP 

 0 for H 2O

 dT eqm
 slope  0 for H 2O
}
P2
 dP 


dT

eqm
P
P1
An increase in P
decreases the eqm Tm.
 Ice-scating is possible.
Tm(T2)
Tm(T1)
T
THE CLAUSIUS-CLAPEYRON EQUATION
(7.5 Eqm between the vapour phase and a condensed phase)
V  Vevaporation ; H  H evaporation
1 mole Fe  VFe  s   7.1 cm 3
or V  Vsub lim ation ; H  H sub lim ation
1 mole gas STP   22.414 lt  22,414 cm 3
 V  Vvapour  Vcondensed
 V  Vg  VFe  s   22,407 cm 3
Vvapour  Vcondensed
 V  Vvapour
 V  Vg error  0.03% 
H
 dP 

 
 dT eqm T  V
H
 dP 
 

 dT eqm T  Vg
Assuming ideal gas behaviour (& n =1 mole):
P V  n  R  T
H
P  H
 dP 
 


 dT eqm T  R  T R  T 2
P

dP H

 dT
P R T 2
d ln P 
Eqn (7.6)
n
H dT Clausius-Clapeyron Eq

R T2
THE CLAUSIUS-CLAPEYRON EQUATION
Case-1:
cP  a  b  T  c  T 2 
d

T
cP  0
 cP g   cP condensed phase 
cP(g)
H
T
H T
dT
  d ln P 
 
R T known  T 2
P 1 atm
P
2
cP(s)
1
H T
 ln P  
 const.
R T
Eqn
(7.7)
As eqm is maintained between
the vapour phase and the
condensed phase, the value of
P at any T in Eqn. (7.7) is the
saturated vapour pressure
exerted by the condensed
H
phase at T.
Eqn (7.7) indicates that the
saturated vapour pressure
exerted by a condensed phase
increases exponentially with
increasing T.
T1
T2
T
H (T1)
H (T2)
cP=0
T1
T
T2
THE CLAUSIUS-CLAPEYRON EQUATION
Case-2:
cP  a  b  T  c  T 2 
cP  a  0
 cP  const.
d

T
(independent of T)
 H T  H 298 
H (T2)
cP(g)
H
T
 cP  const.  dT
cP(s)
298 K
 H T  H 298  cP 
T
 dT
T
298 K
 H T  H 298  cP  T  298
 H T  H 298  298  cP   cP  T
H 298  298  cP   cP  T   dT
H T

dT

H
R T 2
R T 2
P
T
T

H 298  298  cP 
cP  T
  d ln P  
 dT  
 dT
2
2
R

T
R

T
1 atm
298 K
298 K
 d ln P 
cP = const ≠ 0
298
T
THE CLAUSIUS-CLAPEYRON EQUATION
H 298  298  cP  
 ln P 
R
dT cP
dT

 
2

T
R 298 K T
298 K
T
T
 ln P 
H 298  298  cP    1   
 ln P 
298  cP  H 298   cP  ln T   H 298  298  cP   cP  ln298
1  cP  T 
 ln

 

R
 298 
 298 
 T

H 298  298  cP    1   H 298  298  cP    1   cP  ln T  cP  ln298
 ln P  
 


R
R
R
T 
 298  R
R

R T
R
R  298

298  cP  H 298   cP  ln T  const.
 ln P 
R T
R
A
 ln P   B  ln T  C
T
R


THE CLAUSIUS-CLAPEYRON EQUATION
cP(g)
H
Case-3:
cP  const;
cP(s)
it is dependent on T
cP  a  b  T  c  T 2 
 H T  H 298 
d

T
T
T
 cP  dT
298 K
 H T  H 298 
d


2

a


b

T


c

T




  dT
 
T

298 K
T
H
cP=f (T)
298
T
7.6 Graphical Representation of Phase Equilibria
in a One-Component System
Example (Case-3):
In an eqm between a liquid phase and a vapour, the Tn.b.p. of the
liquid is defined as that T at which the saturated vapour
pressure exerted by the liquid is 1 atm.
Knowledge of c H O g  , c H O l  , and the molar heat of evaporation
at any one T, and Tn.b.p., allows the saturated vapour pressure –
T relationship to be determined for any material.
For example, for H2O:
P
2
P
2
cP H 2O g   30  10.7 10 3  T  0.33 105  T 2
cP H 2Ol   75.44
J
K
298  2500 K
J
K
 cP  45.44  10.7 10 3  T  0.33 105  T 2
273  373 K
J
K
273  373 K
H evap., T  41,090 J at Tn.b. p  373 K
T
 H evap., T  H evap., 373   cP l  g   dT
373 K
T


 H evap., T  41,090    45.44  10.7 10 3  T  0.33 105  T 2  dT
373 K
7.6 Graphical Representation of Phase Equilibria
in a One-Component System
  45.44  10.7 10
T
H evap., T  41,090 
3

 T  0.33 105  T  2  dT
373 K
 H evap., T  57,383  45.44  T  5.35 10 3  T 2 
d ln P 
H evap, T
2
0.33 105
T
 dT
R T
5

 3 2 0.33 10 
57,383  45.44  T  5.35 10  T 

T


 d ln P  
 dT
2
R T
 d ln P 
57,383
R T 2
 dT 
45.44
5.35 10 3
0.33 105
 dT 
 dT 
 dT
3
R T
R
R T
57,383 45.44
5.35 10 3
0.33 105
ln P  

 ln T 
T 
 const.
R T
R
R
2  R T 2
7.6 Graphical Representation of Phase Equilibria
in a One-Component System
At Tn.b. p  373 K ; P  1 atm
57,383 45.44
5.35  10 3
0.33  105
 ln 1  

 ln 373 
 373 
 const.
2
R  373
R
R
2  R  373

 const.  50.62
Variation of the saturated
57,383 45.44
5.35  10
0.33  10
vapour pressure of H2O
 ln P  

 ln T 
T 

50
.
62
with T in the range of 273R T
R
R
2  R T 2
373 K.
3
5
Curve-fitting of experimentally measured vapour pressure of liquid water leads to an
expression of the form
ln P 
A
 B  ln T  C
T
Eqn (7.10)
The Structure of Unary Phase Diagrams
1. The simplest representation of regions of stability of the phases in a
unary system is obtained when the phase diagram is plotted in (P, T)
coordinates.
The lines on these phase diagrams are called phase boundaries and
represent
the
limits of stability of each of the phase forms that
the system exhibits.
2. The domain of stability of each single phase is represented by an area.
3. The domain of stability for two phases coexisting in eqm is a line.
4. The domain of stability for three phases existing simultaneously in eqm
is a triple point, at which 3 single-phase areas and 3 two-phase lines
meet.
5. There are no regions where more than three phases may coexist at eqm.
Schematic Representation of the Phase Diagram for H2O
1. Within the single-phase areas, the P exerted on the phase and the T of
the phase can be independently varied without upsetting the one-phase
eqm.
F=C–P+2
 2 degrees of freedom (T and P)
F=1–1+2=2
2. OA, OB, OC:
Along which two phases coexist in eqm, and for any of these equilibria
only one variable can be independently varied.
 1 degree of freedom (P or T)
F=1–2+2=1
2a. Line AOA’:
Variation of P with T required for phase eqm beween solid and liquid
phases.
 Clapeyron Eqn [Eqn (7.5)]
2b. Line BOB’:
Variation with T, of saturated vapour pressure of liquid.
 phase eqm between liquid and gas.
 Clausius-Clapeyron Eqn [Eqn (7.6)]
2c. Line COC’:
Variation with T, of saturated vapour pressure of solid.
 phase eqm between solid and gas.
 Clausius-Clapeyron Eqn [Eqn (7.6)]
Schematic Representation of the Phase Diagram for H2O
3. Triple Point:
State represented by invariant values of P and T at which solid, liquid,
and gas are in eqm with each other.
 0 degrees of freedom.
A
m
1
P (atm)
F=1–3+2=0
solid
.006
B’
b
B
liquid
C’
O
gas
A’
C
0 .0075
T (C)
100