PHASE EQUILIBRIA 611

(D) In this case, C:2, P: I so that F=3

The choice of variables is T, P and concentration of the solution.
(c) In this case, C=2. Since P= l, hence F:3. The choice of variables is I, P and concentration of the solution.
(d) The numhr of components will depend upon the method used to prepare the system and the tinal status of the systenr.
Thus, if we are dealing with a mixture of three gases, C=3. Since P= 1, therelbre, F=C-P+2=3-I +2=4. This means that 7",
P and the concentrations of two of tlre ttrree components must be specified.

If we .are dealing with a mixture of gases produced by the decomposition of water vapour
lHzO(d Hz(S) + Ql2)Oz@|, then we need to specify only one component because the information regarding the
--
other two can be obtained from the equilibrium constant expression
Pu2o
^ -;'z"aoJ'|T
andtheknownstoichiometry,i.e., Pa2=2por.Therefore,C=l.Since,P=l,hence F:C-P+2:1-1+2=2.
This implies that only T and P need to be specified.
(e/ The number of components will depend upon the method used to prepare the system and the final status of the
system.Thus,ifwearegivenamixtureofthreespecies,C=3.SinceP:3,henceF=C-P+2:3-3+2=2.
Thus, we need to specify two variables such as land P.
If we are dealing with a mixture produced by the decomposition of CaCO3, only two components need to be
specified since the third can be calculated from the stoichiometry of the reaction. Thus, C :2 and since P : 3,
therefore, F : C - P + 2 : 2 - 3 + 2 : I so that only lor Pneed to be specified.
(/) This systan is analogous to the system considered in (d) above. The answer is C=1, 2 or 3 so drat F : 2,3 or 4.
(g) Here C = 3. Since there is an equilibrium, only one gaseous component must be specified along with the two solid
phases since the second gaseous component can be calculated from the equilibrium constant expression

K = pu2lpu2o
Here P = 3. Hence, F = C - P + 2 :J,- 3 + 2 = 2. This means that I and P define the system. Other
possibility is that ? and the concentration of one of the gases define the system.
(ft) This system is analogous to the system considered in (e) above. This answer is C : 2 or 3 so that F : I or 2.

ONE-COMPONENT SYSTEMS
Since the minimum number of phases in any system is l, it is evident that
F:C-P+2:1-l*2:2
The one-component system is, thus, bivariant. It can be completely defined by specifying two
variables. The two variables are temperature and pressure. The composition or concentration remains
invariably 100 per cent because there is only one component.
If a one-component system has two phases in contact with each other,
F=C-P+2:l-2*2:l
The system is then said to be monovariant. It c_?n be completely defined by specifying only one
variable, vtz., either temperature or pressure. If terhperature is fixed, pressure of the system is fixed
automatically and vice-versa. Thus, the system consisting of a liquid in contact with its vapour, e.8.,
HzO(r) szo(g) #
is monovariant. If
temperature is fixed, the pressure of the vapour is fixed automatically. This is
because at each temperature, there can be one and only one vapour pressure.

THE WATER SYSTEM

Water can exist in three phases, namely, solid, liquid and vapour. Hence, there can be three fonns of
equilibria, vrz.,
1. Liquid $ yxpsul 2. Solid i Vapour 3. Solid fr Liquid. Each equilibrium involves two
612 PHASE EQUILIBRIA

phases. The phase diagram for the water system is shown in Fig . 2.

218 atm - Critical pressure - - aA

C.)
c:I
b
L)
(t)
o 2 Liquid
water
o
Solid \9o
c c--- {- -j
z ice
2
;

r! \z
, ttrl

&
>)
a
(t) 1 atm
r\
;----'\----:%;---2,
x g2 -q$
:
trl s
& I I

O.
!\{
lo\./ I
l1

4'58 mm -)V rripre i water

I

A'- ' .X t potnt I r"pour

I

Sublimation i
CUfVe lll ./ I I I
a,/arrr I

L,- i-/^;: rCrit I

critical
-K-t
n-t(-t :', : : :
B--I\',' ii i 'em'erature
0 0.0075 100 374
TEMPERATURE, 'C (Not to scale)

Fig. 2. The phase diagram for the watersystem.

The curve OA represents the equilibrium between liquid water and vapour at different temperatures.
Itis called the vapour pressure curve of water as it gives the vapour pressure of water at different
temperatures. It can be seen that, for any given temperature, there exists one and only one vapour
pressure. Similarly, for each vapour pressure, only one temperature can be maintained. Thus, the
degree of freedom of the system is one, i.e., the system is univariant :
F:C-P+2:I-242=l
At 100'C, the vapour pressure of water equals the pressure of the atmosphere (760 mm Hg). This is.
therefore, the boiling point of water. The carve OA extends as far as the critical temperature of water
(374"C) since above this temperature liquid water cannot exist.
The variation of vapour pressure with temperafure is quantitatively given by Clapeyron-Clausius
equation, vrz., .

dP
dT

Exampte 5. At 100'C, the specitic volurnes of water aird steam are, respectively, I c.c. and 1673 c.c. Calculatr
the change in vapour pressure of the system by l'C change in temperature. The molar heat of vaporisation trf
water in this range may be taken as 9.70 kcal.
Solution :
Motar volume of liquid water, V1 = 18 cm3 mol-l = l8x 10-6 m3 mol-l
Molar volume of steam, Ve = 18 x 1673 cm3 mol-l = 30114 x 19-6 rn3 to1-l
Heat of vaporisation, AIfu :9700 cat mol-l x 4'184 J cal-l : 40584'8 J mol-l

dP = M'*
dr T(vs _U)
Ihthis case, dT = lKand I = 273 + 100 = 373 K
 PHASE EQUILIBRIA 613

dP= 40584.8Jmol-r xlK

373K(301 14 - l8) x 10-6t3 *tF
= 0.@361 x 106 N m-2 (J = N m)
: 0.03561 atm = 27.08 mm of Hg (l atm : 101.325 N m-?)
Thus, the vapour pressure of water increases by about 27 mm of Hg by l'C rise in temperature, at 100"C.

Example 6. The vapour pressure of water at 95oC is found to be 634 mm. What would be the vapour pressure
rt a tempe;ature of 1006C ? 'ihe heat of vapourisation in this range of temperature may be taken as 40593 J mol-l.
Solution : The integrated form of Clapeyron-Clausius eguation,for Liquid 3 Vapour equilibrium is

tnPz =M"*.{e=l}
R lT[z)
Pt
Inthiscase, Tr=273 + 95 = 368K, Pr = 634mm:Tz=273 + 100 = 373K, P2=?
1n--!:- 4o593Jryol-t -
| t*
rnus. 634mm- 8.314JK-t mol-l LSOSx, fztf lJ
Pz = 759'8 mm
Thus, the vapour pressure of water increases from 634 mm to about 760 mm of Hg when the temperature changes
o
:iom 95 to 100'c.
The eurve OB represents the equilibrium between ice and vapour. It is called the vapour pressure
:urve of ice or sublimation curve of ice. Its lower end I extends to absolute zero. Again, as can be
seen, for each temperature there can be one and only one pressure and similarly for each pressure, one
and only one temperature can be maintained. In other words, the degree of freedom is 1.
The variation of vapour pressure of ice.rvith temperature, is quantitatively given by Clapeyron-
Ciausius equation, viz.,

dP
dr ==,f'*--
r(vs_%) ornL
A-
pt-Msub lrr-!\
I:IEI
Since, along OA, the two phases are liquid and vapour and along OB, the two phases are solid and
vapour, at the point O, where the two curves meet, three phases, namely, solid, liquid and vapour, will
ctxxist. Such a point is known as the triple point. The temperature and pressure at the triple point of
\\arer are 0.0075"C and 4'58 mm of Hg, respectively. According to the phase rule, O is an invariant point:

F:C-P+2:l-3+2=O
The curve OC represents the equilibrium between ice and water. It is called the fusion curve of ice
:-s ir indicates the temperatures and pressures at which the solid (ice) and liquid (water) can coexist in
equitibriurn. In other words, this curve shows the effect.of pressure on the melting point of ice. As can
'oe seen, the line OC is inclined towards the pressure a*is which indicates that the melting point of ice is
io*'ered by increase of pressure. This follows from Clapeyron-Clausius equation, viz.,
dP
or
dr _ rtur(y, -%)
dT 7n" (V -Vt) dP AI/tr,
Since density of ice is less than that of water, V, is greater than yl. In other words, the expression
--: the right hand side of the above equation is negative. Hence, dTldP should also have a negative sign.
This means that the increase of pressure must lower and decrease of pressure must raise the freezing
r.rnt of t'arer. It can be easily shown with the help of the above equation that freezing point of water is
.clu€rrd bv 0.0075"C by I atm increase in pressure. Thus, while the tieezing point of water at a
rr:ssure of -1.58 mm is *0.0075oC, at a higher pressure of 760 mm (r'.e., I atm) it is reduced to OoC.
614 PHASE EQUILIBRIA

Example 7. The specific volumes of ice and water at OoC are 1.(D07 cm3 and 1.0fi)l cm3, respectiveiy. Wbar
would be tbe change in melting point of ice per atm increase of pressure ? Heat of fusion of ice = 79.8 cal g-1.
Solution:
Molar volume of ice, Vs : 18x 1'0907 x 10-6 m3
Molar volume of water, V1 : 18x 1'0001 x 10-6 m3
Temperature, Tfu. : 273 K
Molarheatof fusionof ice,Afil6r,= lSgmol-l x 79'8calg-l x 4'184Jcal-l = 6fi)9'9 Imol-l
Increase ofpressure, dP = | atm : 101325 11 m-2; aF: ?
dT 273Kx ( - 0.0906) x 18 x l0{m3
dP 60D:9 J mol-l

273K x 18 x 10-6(- 0.0906)13 x t0t325 N m-2 : - 0.0075 K

dT= (J:N m)
6009 .9 J mol-
The negative sign indicates that the melting point of ice (or freezing point of water) decreases by increase {
pressure. Alternatively, the melting point of ice increases by decrease ofpressure.
Along the curve OC, there are two phases, namely, ice and water. Therefore, according to tbe
Phase rule, the system should be univariant, i.e., the degree of freedom should be 1. This is seen to be
actually the case as, for any given pressure, melting point will have one definite value (d.e., solid ad
liquid phases will coexist in equilibrium at one particular temperature only). The curve OC must meet
the other two curves at the triple point O.
Significance of Areas or Regions Between the Lines. Suppose, the state of equilibrium is
represented by a point .r on the line OA. The two phases in contact are liquid and vapour. If tbc
temperature is kept constant and pressure increased, the v3llour will be compressed wholly into tb
liquid phase. This change is represented by the dotted line .rD in Fig. 2. Similarly, if pressure is keg
constant and temperature is decreased, the vapour will change into liquid again. This change k
represented by the dotted line.rc. Thus, the area above the curve Oz{ represents exclusively the liquid
phase, as shown.
Coming back to the point r, if the pressure is maintained constant and the temperature is increased
the liquid will change completely into vapour. The change is represented by the dotted line 11
Similarly, if the temperature is kept constant and the pressure is diminished, liquid will again bt
converted completely into the vapour phase as represented by the doned line xz. Thus, the ?r€d ur(
region below the curve OA represents the vapour phase only.
Similarly, if a system represented by a point s on the solid-vapour equilibrium curve (OB) u
subjected to increase of temperature at a constant pressure (along sf) or decrease of pressure a r
constant temperature (along sa), it will change completely into vapour. Thus, any point lying belou tu
curve OB also represents the vapour phase only.
Lastly, if the system at s is subjected to increase of pressure at constant temperature (along Jrrr r
decrease of temperature at constant pressure (along sn), the vapour will condense completely into scbs
phase. Thus, the area lying above the curve OB will represent the solid phase exclusively.
It may be noted that if a point representing the state of equilibrium of the system lies wifrir r
particular region or area, the number of phases is I ; if on a line, the number of phases is 2 and if rT. r
point where the lines meet, the number of phases is 3. Accordingly, the system will be biyarirc.
univariant and non-variant, respectively.
Metastable Equilibrium. Sometimes it is possible with due care to cool water (or, as a rnas lr
fact, any liquid) below its freezing temperature without the separation of ice. The water is then sed u
be supercooled and can be kept as such almost indefinitely if the presence of ice or any ofur riir.
phase is carefully avoided. The vapour pressure curve of liquid water AO can, therefore, ctlr
Lelow the point O, as shown by the dotted cuwe OA'. The liquid ir vapour system along rb cr.c
 PHASE EQUTLTBRTA 615

O.4' is said to be in metastable equilibrium because as soon as a small particle of ice is brought in
contact u'ith the supercooled liquid, the entire liquid solidifies. It will be seen from the phase diagram
rhet the curve OA' lies above the curve OB. Thus, the metastable system has a higher vapour pressure
than the stable one at the same temperature.

Effect of Change of Temperature and Pressure. The significance of the equilibrium diagram can
be further understood by following the changes that occur on altering the temperature or pressure of the
s!'stem. Suppose, it is desired to know the effect of heating ice when it is under a pressure of 1 atm and
at a certain temperature represented by the point X in the figure. As the system is bivariant, the
temperature can have any value at the same pressufe. Therefore, heating the ice slowly, at constant
pressure, will shift the system along the line XIZ At lfusion takes place. The liquid phase also appears.
The system has now two phases and, therefore, one degree of freedom. Now, the temperature cannot
alter without the alteration of pressure. The effect of continued heating will be simply to calse fusion of
ice at constant temperature. When fusion is complete, i.e., when ice has changed completely into water,
the system again has only one phase and becomes bivariant. The temperature begins to rise along YZ.
At Z, vaporisation begins, i.e., the vapour phase also makes its appearance. The number of phases rises
to 2 and the system becomes monovariant. The continued heating will not alter the temperature as the
pressure has been kept constant. The only effect will be to convert more and more of the liquid into
vapour. When the liquid phase disappears completely, the system will have only one phase (vapour) and
u'ill become again bivariant. The temperature of the vapour will rise along the line Z' .
The Triple Point. It will be interesting to note the effect of heating and applying pressure on the
system when all the three phases coexist as at the triple point O. The effect of heat will be simply to
cause more and more of the ice to melt but there will be no rise in temperature or pressure until the
whole of the ice has completely changed into liquid. When this happens, the system has only two phases
(liquid and vapour). The system changes from non-variant to univariant. Therefore, further addition of
heat will cause a rise of temperature. The equilibrium will shift along the curve Oz{.
Now, suppose, pressure is applied tdihe system at the triple point. There can be no change in
pressure or temperature as long as all the three phases are present. The only effect of applying pressure
will be to cause condensation of vapour to liquid or solid phase. Ultimately, the vapour phase will
disappear and only two phases, solid and liquid, will stay and further application of pressure
will cause
increase of pressure with change of temperature along the curve OC.

THE CARBON DIOXIDE SYSTEM

The phase diagram for carbon dioxide system is shown in Fig. 3. It has three distinct areas in
which carbon dioxide can exist either as solid, liquid or gas.
z4B is the sublimation curve along which solid carbon dioxide is in equilibrium with the gas, BD is
the vaporisation curve along which liquid carbon dioxide is in equilibrium with the gas while BC is the
fusion curve along which solid and liquid carbon dioxide are in equilibrium with each other.
B is the triple point at which all the three phases of carbon dioxide coexist in equilibrium with one
another. The temperature of the system at this point is-57"C while the pressure is 5'2 atm. A slight
variation in temperature or pressure at this point may result in the disappearance of one of the two
phases. For example, a slight increase in temperature will result in the disappearance of the solid phase
and the equilibrium will shift along the curve BD while a slight decrease in temperature will result in
rhe disappearance of the liquid phase and the equilibrium will shift along the curve 8,4. Keeping the
temperature constant, if the pressure is increased, the gaseous phase will disappear and the equilibrium
will shift along the curve BC.
The phase diagram of carbon dioxide resembles that of water (Fig. 2) in showing three distinct
areas for solid, liquid and gaseous phases. But it differs essentially from the latter in several respects.
In the first place, the fusion curve slopes away from the pressure axis. Ihis indicates that increase of
pressure raises the melting point of solid carbon dioxide, i.e., the factor dTldP of the Clapyron-Clausius