Sy second term, phase rule

Phase Equilibria
by
Dr. B. R. THORAT
DEPT OF CHEMISTRY,
I. Y. COLLEGE OF ARTS, SCI. AND COMM.

Terms used to explain phase ruleTerms used to explain phase rule
System
A system is part of physical
universe isolated from all
others for the specific study
Homogeneous
system:- System with
only one phase
Heterogeneous
system:- System with
more than one phases
Phase
A phase (Greek word –
appearance) is a homogeneous,
physically distinct and
mechanically separable portion or
part of system
Homogeneous means system has
identical composition and physical
properties throughout the whole system
Physically distinct means a phase should have definite boundary surface (two immiscible
liquids, two different crystals of same or different solid materials, precipitate formed in
solution, etc)
Mechanically separable means the phases can be separated by simple mechanical operation
such as filtration, decantation, hand picking, separation by using separating funnel, etc
Number of Phases (P)
It is number of physically distinct states such as solids, liquids and
gaseous present in the system at equilibrium.

Terms used to explain phase ruleTerms used to explain phase rule
Components:- Chemically independent constituents of a system
Constituents of system:- A chemical species (ions or molecules) which are present in system
Number of Components (C)
The term component does not mean constituents of the system.
It is used to explain chemical composition of each phase of the system. “It is the smallest
number of independent variable, chemical constituents by means of which the
compositions of each phase can be explained either directly or by means of chemical
equation.”
1. One component system (C = 1):-
Solid water (ice) ↔ Liquid water (P = 2)
Solid water (ice) ↔ Water vapour (P = 2)
Liquid water ↔ Water vapour (P = 2)
Solid water (ice) ↔ Liquid water ↔ Water vapour. (P = 3)
2. Two component System (C = 2):-
CaCO3 (s) ↔ CaO (s) + CO2 (g).
During decomposition three phases are co-exist along with three chemical constituents.
But whole system can be expressed completely by two chemical constituents in the form
of equation of each phase hence it is example of two components system.

There are three ways in which two chemical constituents can be selected to express every
phase
i. Chemical constituents: CaO and CO2 are choosen.
Phase Composition
x CaCO3 x CaO + x CO2
y CaO y CaO + 0 CO2
z CO2 0 CaO + z CO2
ii. Chemical constituents: CaCO3 and CaO are choosen.
Phase Composition
x CaCO3 x CaCO3 + 0 CaO
y CaO 0 CaCO3 + y CaO
z CO2 z CaCO3 - z CaO
ii. Chemical constituents: CaCO3 and CO2 are choosen.
Phase Composition
x CaCO3 x CaCO3 + 0 CO2
y CaO y CaCO3 - y CO2
z CO2 0 CaCO3 + z CO2
Any phase of the system can be completely expressed by two chemical constituents, hence
it is example of two component system

A system under study gets influenced by the change in pressure, temperature and
composition of the system. All these variables are used to determined equilibrium
conditions.
Number of degree of freedom or variance (F):
The number of degree of freedom or variance of a system is the minimum number of
independent variables such as temperature, pressure and concentration (conc. of each
constituent) that must be specific in order to define the system completely at
equilibrium.
The degree of freedom is also called as number of degree of variance. A system having
one degree of freedom is called univariant or monovariant system
A system having two degree of freedom is called bivariant system and so on. The system
having zero degree of freedom is called as non-variant or invariant system

This can be explaining with help of following examples.
i. If only one phase is present (solid, liquid or gas) then we have to describe values of at
least two parameters namely pressure and temperature to determine the equilibrium
condition of the system. e.g. H2O system, C = 1, P = 1 therefore F = C – P + 2 = 2 i.e. two
parameters under different conditions are determined to define equilibrium. It is example
bivariant system.
ii. In case of two phases which are in equilibrium to each others, the only one variant is
needed to explain equilibrium while other is fixed automatically.
Liquid ↔ Vapour
(P = 2, C = 1). F = C – P + 2 = 1
For example boiling point of water is 373 K, pressure is automatically fixed 76 cm of
Hg.
iii. In case of triple point (all three phases solid, liquid and gas) of water, where all the
three phases are in equilibrium with each others.
Solid water (ice) ↔ Liquid water ↔ Water vapour.
(P = 3, C = 1). F = C – P + 2 = 0.
This equilibrium will be arises at 273.0098 K and pressure at 4.58 mm of Hg i.e. at fixed
temperature and pressure.

PHASE RULEPHASE RULE
In 1876, an American physicist J. Willard Gibbs explains the concept phase rule to study the
effect of pressure, temperature and composition of heterogeneous system
StatementStatement
The phase rule gives relationship between the degree of freedom (F), the number of
components (C), and number of phases (P) to explain phase equilibrium in heterogeneous
system
F = C – P + 2
The phase rule is used for the study of physical equilibrium between different or same states
of matter which are made up of one or more chemical constitutes.
e.g. vaporization, fusion, sublimation, allotropes transition, solubility of solids, liquids and
gases with each others, vapour pressure of solution, chemical transitions, distribution of
solute in two immiscible liquids, etc.
These equilibrium are studied by using various empirical laws, Raoult,s law, Henry’s law,
distribution law, etc.
It is does not directly used for the quantitative study of each phase.

For any one component in two phases, it is possible to write one equation for variables. For three
phases, two equations are required to explain. In general for P phases, (P – 1) equations are
required for one component. But P phases of the system containing C components, therefore
there are total C (P – 1) equations.
The number of equations are equal to number of variables, therefore degree of freedom (F) will
be
= [P (C – 1) + 2] - [C (P – 1)]
= PC – P + 2 - PC + C
F = C – P + 2.
F(variables) =
Total number of variables
that need to be specified
-
Total number of equations
that are available
Mathematical derivation of Gibbs phase rule
Consider non-reactive heterogeneous system having C components distributed in P phases.
In each phase, C components having C different concentrations.
In closed system, if we arbitrarily chosen concentrations of (C – 1) components then
concentration of last is automatically fixed i.e. there are (C – 1) composition variables in each
phase. For P phases, total numbers of composition variables are P (C – 1) along with two
important variables as temperature and pressure.
Total no. of variables = no of composition variables + temp. + pressure.
= P (C – 1) + 1 + 1.
= P (C – 1) + 2.
At equilibrium, the partial molal free energy (μ) of each component of any phase of system is
equal to the partial molal free energy of the same component in any other phase. It is the
function of temperature, pressure and (C – 1) concentration variables.

Consider a system containing minimum two independent
components. The number of phases exist in equilibrium are depends
on the conditions such as temperature and pressure.
“These conditions are determined experimentally and plot a diagram
is called as phase diagram. The phase diagram is plot of pressure
against temperature or composition.”
Liquid – Liquid mixture
When two liquids (which are not reacting chemically) are brought in
contact with each others resulting three types of binary liquid
systems.
i. Both liquids are completely miscible with each other in all
proportions. E.g. Water + Ethyl alcohol, Benzene + Toluene, etc.
ii. The liquid pairs are partially miscible with each others.
e.g. Water + Phenol.
iii. Liquid pairs are practically immiscible with each other such as
Water + Benzene, Carbon disulphide + Water.

a. Completely miscible liquid pairsa. Completely miscible liquid pairs
The solution of two components which are completely miscible withThe solution of two components which are completely miscible with
each other iseach other is ideal or non-idealideal or non-ideal..
Solutions which fill full the following conditions is said to be ideal.Solutions which fill full the following conditions is said to be ideal.
i. It shouldi. It should obey the Raoult’s law of vapour pressure at allobey the Raoult’s law of vapour pressure at all
compositions.compositions.
ii.ii. The chemical and physical nature of the components should not beThe chemical and physical nature of the components should not be
change and the activity of each component should be equal to its molechange and the activity of each component should be equal to its mole
fraction under all conditions of temperature, pressure andfraction under all conditions of temperature, pressure and
concentration.concentration.
iii.iii. There should be no change in volume (∆Vmixing = 0) on mixing.There should be no change in volume (∆Vmixing = 0) on mixing.
iv.iv. There should be no change in enthalpy (∆Hmixing = 0) on mixing.There should be no change in enthalpy (∆Hmixing = 0) on mixing.
The solution which does not obey these conditions is called as non-The solution which does not obey these conditions is called as non-
ideal solutions.ideal solutions. The phase diagram for liquid-liquid mixtures is theThe phase diagram for liquid-liquid mixtures is the
plot of vapour pressure against composition or temperature againstplot of vapour pressure against composition or temperature against
composition.composition.

Raoult’s lawRaoult’s law
Consider binary solution containingConsider binary solution containing two components A and Btwo components A and B. Both the components are. Both the components are
volatilevolatile formingforming ideal solutionideal solution..
SupposeSuppose pA and pB are the partial vapour pressurespA and pB are the partial vapour pressures of A and B at equilibrium andof A and B at equilibrium and p is thep is the
total vapour of the solutiontotal vapour of the solution. According to the Dalton’s law, the total vapour pressure of the. According to the Dalton’s law, the total vapour pressure of the
solution is-solution is-
p = pA + pB.p = pA + pB.
The partial vapour pressure of any component at equilibrium is depends on total vapourThe partial vapour pressure of any component at equilibrium is depends on total vapour
pressure of solution and composition of vapour.pressure of solution and composition of vapour. IfIf x’A and x’B are the mole fractionx’A and x’B are the mole fraction ofof
component A and B in the vapour mixture.component A and B in the vapour mixture.
pA = x’A x p and pB = x’B x p.pA = x’A x p and pB = x’B x p.
The partial pressure can be varying with the nature of the components, their compositions inThe partial pressure can be varying with the nature of the components, their compositions in
the mixture and temperature.the mixture and temperature.
Statement-Statement-
Partial vapour pressure of any component of the mixture is equal to the multiplication of
mole fraction (xA) of it in solution and vapour pressure (pA0) of it in pure state.
pA = xA x pA0 and pB = xB x pB0.pA = xA x pA0 and pB = xB x pB0.
Where, xA and xB are the mole fractions of A and B in liquid mixtures respectively and pA0Where, xA and xB are the mole fractions of A and B in liquid mixtures respectively and pA0
and pB0 be the vapour pressure of the pure component at same temperature. p is theand pB0 be the vapour pressure of the pure component at same temperature. p is the totaltotal
vapour pressure is sum of partial vapour pressures of components of mixture.vapour pressure is sum of partial vapour pressures of components of mixture.
The variation of partial and total vapour pressure with composition of liquid mixture atThe variation of partial and total vapour pressure with composition of liquid mixture at
constant temperature is shown below-constant temperature is shown below-

Vapour
pressure
A B
PA
0
P
0
B
mole fraction
xA
Bx
= 1
= 0
xA
Bx = 1
= 0
constant temperature
total pressure
partial vapour pressure
of liquid B
of liquid A

Raoult’s lawRaoult’s law
Deviations from the Raoult’s law (non-ideal solutions):-Deviations from the Raoult’s law (non-ideal solutions):-
Some mixtures (solutions) of two volatile liquid components A and BSome mixtures (solutions) of two volatile liquid components A and B
havinghaving different physical and chemical propertiesdifferent physical and chemical properties than the pure liquids.than the pure liquids.
Such solutions are called asSuch solutions are called as non-ideal solutionsnon-ideal solutions. These solutions do. These solutions do notnot
obey Raoult’s law.obey Raoult’s law.
They show two types of deviations as-They show two types of deviations as-
i.i.Positive deviations and,Positive deviations and,
ii.ii.Negative deviation from Raoult’s law.Negative deviation from Raoult’s law.

Solution (Raoult’s law) showsSolution (Raoult’s law) shows Positive deviationsPositive deviations
If the forces of attraction between unlike molecules i.e. between A and B are weaker than
the forces of attraction between like molecules i.e. between A & A and B & B. Such
solutions show positive deviations from Raoult’s law. In such cases, the vapour pressure
curves of the constituents and the mixture lie above those of ideal dotted lines.
The positive deviation indicates that the tendency of each molecule to escape from the
solution to vapour is greater in non-ideal solutions than that in ideal solution.
The extend of positive deviation is depends on many factors, some of them are listed as-
i. Difference in polarity of the molecules.
ii. Difference in length of hydrocarbon chain or groups in the molecules.
iii. Difference in intermolecular forces of attraction.
iv. Association of the constituents in the liquid state.
e.g. Carbon tetrachloride + Heptane, Diethy ether + Acetone, Heptane + Ethyl alcohol.
Vapour
pressure
A B
PA
0
P
0
B
mole fraction
xA
Bx
= 1
= 0
xA
Bx = 1
= 0
total pressure
of liquid B
of liquid A

Solution (Raoult’s law) showsSolution (Raoult’s law) shows Negative deviationsNegative deviations
If the forces of attraction between unlike molecules i.e. between A and B are stronger
than the forces of attraction between like molecules i.e. between A & A and B & B. Such
solutions show negative deviations from Raoult’s law. In such cases, the vapour pressure
curves of the constituents and the mixture lie below those of ideal dotted lines.
The negative deviation indicates that the tendency of each molecule to escape from the
solution to vapour is relatively lower in non-ideal solutions than that in ideal solution.
e.g. Pyridine + Formic acid, Chloroform + Acetone (due to partial association between
the molecules through hydrogen bonding), Haloacid + Nitric acid.
Vapour
pressure
A B
PA
0
P
0
B
mole fraction
xA
Bx
= 1
= 0
xA
Bx = 1
= 0
total pressure
of liquid B
of liquid A

Vapour pressure of LiquidVapour pressure of Liquid
Consider a liquid taking in a vessel with some free space over the liquid. The particlesConsider a liquid taking in a vessel with some free space over the liquid. The particles
of every state are inof every state are in continuous motioncontinuous motion i.e. molecules in liquid state are moving withi.e. molecules in liquid state are moving with
different kinetic energies. Thedifferent kinetic energies. The molecules that possess energy above the averagemolecules that possess energy above the average
kinetic energy can overcome the intermolecular forces that hold them in liquid state.kinetic energy can overcome the intermolecular forces that hold them in liquid state.
These energetic molecules are escape from the liquid surface as vapour.These energetic molecules are escape from the liquid surface as vapour. The processThe process
in which molecules of a liquid go into vapour state (or gaseous state) is called asin which molecules of a liquid go into vapour state (or gaseous state) is called as
vaporization or evaporisation.vaporization or evaporisation. The reverse process where by vapour molecules areThe reverse process where by vapour molecules are
converted into liquid state is called condensation.converted into liquid state is called condensation.
liquid surface
Balanced intermolecular
forces at the interier part
of liquid
unbalaced intermolecular forces at the
surface of liquid

A stage come when the number of molecules escaping from the liquid is equal to the number ofA stage come when the number of molecules escaping from the liquid is equal to the number of
molecules returning to the liquid i.e. rate of condensation and rate of evaporation are equal.molecules returning to the liquid i.e. rate of condensation and rate of evaporation are equal.
HenceHence dynamic equilibriumdynamic equilibrium is established between the liquid and vapour at givenis established between the liquid and vapour at given
temperature.temperature.
““The pressure exerted by the vapour in equilibrium with the liquid at a fixed temperature is
called as vapour pressure.”
TheThe vapour pressurevapour pressure of liquid is depends on the nature of liquid particularlyof liquid is depends on the nature of liquid particularly strength ofstrength of
intermolecular forcesintermolecular forces.. If intermolecular forces increase, vapour pressure of the liquidIf intermolecular forces increase, vapour pressure of the liquid
decreases.decreases. e.g. Ethanol having weaker intermolecular forces (hydrogen bonding) than watere.g. Ethanol having weaker intermolecular forces (hydrogen bonding) than water
at any given temperature therefore vapour pressure of ethanol is higher than water.at any given temperature therefore vapour pressure of ethanol is higher than water.
If the temperature of liquid is increased, average kinetic energy of liquid molecules increasesIf the temperature of liquid is increased, average kinetic energy of liquid molecules increases
therefore rate of evaporation increased and increased the vapour pressure because boththerefore rate of evaporation increased and increased the vapour pressure because both rate ofrate of
evaporation and the kinetic energy are proportional to temperature.evaporation and the kinetic energy are proportional to temperature.

London forces
Dipole-dipole forces

Vapour pressure of Liquid -Vapour pressure of Liquid - Clapeyron equationClapeyron equation
For any pure substance in a single phase such as liquid or gaseous, are in contact with
each other then any variation in free energy is given by the equation.
dG = VdP – SdT (1)
At equilibrium of two phases - dG = 0 at constant pressure and temperature
“A useful thermodynamic equation is applicable to a system consisting of two phases of
same substance in equilibrium is called as Clapeyron equation.”
Consider the transition of a pure substance from one phase into another phase e.g.-
1. Solid convert to liquid- at melting point of solid- melting or fusion.
2. Liquid convert to vapour- at boiling point of liquid- boiling or vaporization.
3. Solid convert to vapour- at sublimation point of solid- sublimation.
4. Allotropic transitions at the transition temperature of the two allotropic forms.
e.g. S(rhombic) is in equilibrium with S(monoclinic).

Clapeyron equationClapeyron equation
Consider equilibrium of compound A as -
For which ΔG is given as- ΔG = G2 –G1 (2)
Where, G2 and G1 are the molar free energies of the substances in final and initial states respectively.
At equilibrium state ΔG = 0, at constant P and T
G1 = G2.
All such transformations will be in equilibrium at constant pressure and at constant temperature, when
molar free energies of the substances are identical in both phases.
A1
A2
ΔG = G2 –G1 = 0.
G1 = G2 Or dG1 = dG2 (3)
The variation of free energies for pure substances with respect to P and T in single phase is
given by the equation-
dG1 = V1dP – S1dT (4)
dG2 = V2dP – S2dT (5)
From equation (3), (4) and (5)-
V2dP – S2dT = V1dP – S1dT
(V2 – V1) dP = (S2 – S1) dT
dP/dT = (S2 – S1)/(V2 – V1) = ΔS/ΔV (6)
But- ΔG = ΔH – TΔS and at equilibrium, ΔG = 0
ΔH = TΔS or ΔS = ΔH/T (7)
From equation (6) and (7)-
dP/dT = ΔH/T.ΔV (8)
Where, ΔH is the change in enthalpy for reversible transformation occurring at temperature T. this
equation is called as Clapeyron equation.

Clapeyron EquationClapeyron Equation
The Clapeyron equation gives relation between the temperature and pressure of the system
containing two phases of a pure substance which are in equilibrium with each others.
This equation also shows that dP/dT is directly proportional to the enthalpy of the transition and
inversely proportional to temperature and the volume change accompanying during the transition
i.e. ΔH and ΔV are the functions of temperature and or pressure.
The equation (8) can be modified as-
(P2 – P1)/(T2 – T1) = ΔH/T.ΔV (9)(9)
Where, T is taken as average of T2 and T1 or equation (8) can be integrated with the proper limits
of T and P by assuming that ΔH and ΔV are constant.
P2 T2
∫dP = ΔH/ΔV ∫ dT/T
P1 T1
(P2 – P1) = (ΔH/ΔV) ln T2/T1.
(P2 – P1) = (2.303ΔH/ΔV) ln T2/T1. (10)(10)

the Clausius-Clapeyron equation

Vapour pressure of Liquid - the Clausius-Clapeyron equation
The variation of vapour pressure with temperature can be expressed thermodynamically by
means of Clausius-Clapeyron equation.
Let us consider a system- Liquid vapour
For the transition of liquid to vapour, P is the vapour
pressure at given temperature T;
ΔHv is the heat of vaporization of liquid.
The volume of given liquid is Vl and that of vapour is Vg.
Therefore, the Clepeyron equation can be written as- dP/dT = [ΔHv /T(Vg – Vl)] (1)
But molar volume of the substance in vapour phase is very large as compared to volume of same
liquid in liquid state. (Vg – Vl) ≈ Vg. (18 ml water liquid = 22400 ml water vapour)
Therefore, the Clepeyron equation becomes- dP/dT = ΔHv/TVg. (2)
If we assume that the vapour behaves ideally then Vg per mole of substance can be calculated as-
Vg = RT/P (3)
Substituting (3) in equation (2) as-
dP/P = (ΔHv/R). dT/T2
(5)
d(lnP) = (ΔHv/R). dT/T2
(6)
The equation (6) is known as the Clausius-Clapeyron equation. The heat of vaporization is a function
of temperature. Assume that ΔHv will be independent on the temperature (constant over the range of
temperature and pressure).
dP/dT = PΔHv/RT2
(4)

(b) Integration with limits:-
Integrate equation (6) with the limits of pressure P1 and P2 corresponding to the temperature T1 and T2, then-
P2 T2
∫d(lnP) = (ΔHv/R) ∫dT/T2
P1 T1
P2 T2
[lnP] = (ΔHv/R) [-1/T]
P1 T1
ln(P2/P1) = (ΔHv/R) [(T2 – T1)/T1T2] (9)
log(P2/P1) = (ΔHv/2.303R) [(T2 – T1)/T1T2] (10)
The equation (10) can be used to calculated ΔHv from the values of the vapour pressure at any two temperatures or
when ΔHv is known, P at some desired temperature can be calculated from a single available vapour pressure at a
given temperature.
Then equation (6) can be integrated-
(a) Integrated without limits of temperature and pressure:-
∫d(lnP) = (ΔHv/R) ∫dT/T2
+ C (C is constant of integration)
(lnP) = (ΔHv/R) . (-1/T) + C
(logP) = (-ΔHv/2.303RT) + C’ (7)
The equation (7) - shows that logarithm of the vapour pressure should be a function of the reciprocal of the
absolute temperature T. It is also a equation of straight line of the type, y = -mx+c; slope = m = (-ΔHv/2.303R).
From the slope of the graph, heat of vaporization of the various liquids can be calculated as-
m = (-ΔHv/2.303R) (8)
ΔHv = (-2.303R . m) = -4.576 . m cal/mol.

Example 01: The vapor pressure of 1-propanol is 10.0 torr at 14.70
C. Calculate the vapor pressure at 52.80
C. Given:
Heat of vaporization of 1-propanol = 47.2 kJ/mol.
Example 02: The vapor pressure of water is 1.0 atm at 373 K, and the enthalpy of vaporization is 40.7 kJ mol-1
. Estimate
the vapor pressure at temperature 363 and 383 K respectively.

Phase Diagram of systemPhase Diagram of system
The phase diagram of a substance is a map showing the conditions of temperature and
pressure at which its various phases are thermodynamically most stable and in equilibrium.
At point A - the vapour phase of the substance is
thermodynamically the most stable.
At C the liquid phase is the most stable.
The boundaries between regions in a phase diagram at B and
D, which are called phase boundaries, show the values of p and
T at which the two neighboring phases are in equilibrium.
Phase Diagram
For one component system
(water & sulfur system)
For two component system
(Pb-Ag, Zn-Mg, Na-K system)
Component which reacts to
form compounds
Components which are
completely miscible but do
not react (Eutectic mixture)
Pb-Ag system
System having
congruent melting point
Zn-Mg system
System having
Incongruent melting point
Na-K system
Depending upon the component present

Phase diagram of one component systemPhase diagram of one component system
Phase diagram of a pure substance shows the dependence of equilibrium pressure on equilibrium
temperature at - boiling, melting, sublimation and crystalline transformation points.
The temperature at which the surface disappears is the critical temperature, Tc. The vapour
pressure at the critical temperature is called the critical pressure, Pc, and the critical temperature
and critical pressure together identify the critical point of the substance.
boiling temperature
melting temperature
freezing temperature
There is a set of conditions under which three different phases (typically solid, liquid, and
vapour) all simultaneously coexist in equilibrium. It is represented by the triple point, where the
three phase boundaries meet.
triple point
from the phase rule: F = C – P + 2.
= 1 – 1 + 2 = 2.
The system is thus bivariant

Phase diagrams of pure water systemsPhase diagrams of pure water systems
Water exists in three forms – solid (ice), liquid (water) and gas (vapour).
The diagram consist of three curses OA, OB and OC.
These curves divide the entire region into three regions or areas AOC, AOB and BOC.

Curve OA: vapour
pressure curve or
vaporization curves of
liquid water
As temperature increases, vapour pressure also increase and at 1000
C and 760 mm
pressure, liquid water boils. The vapor and liquid phases are in equilibrium along the curve
OA. F = C – P + 2.
= 1 – 2 + 2 = 1
one value of temperature has only one value of pressure.
As temperature increases, the curve OA extend to the point A, which is critical point of
water, beyond which only one phase i.e. vapour will exist.

Curve OB : the vapour
pressure curve or
sublimation curve of ice
Two phases – solid ice and vapour are in equilibrium along OB curve which is univariant
(F = C-P+2 = 1-2+2 = 1).

curve OC : freezing point
curve of water or fusion
curve of ice.
Two phases – ice and water are in equilibrium with others at various pressures. This is
univariant curve (F = C-P+2 = 1-2+2 = 1).
As pressure decreases, the melting point of water increases and reaches to triple point. The
effect of pressure on melting or freezing is very small.

Two curves OA and OB meet at the point
O. Since along OA, two phases present are
water and vapour and along OB the two
phases presents are ice and vapour,
therefore at O the three phases, ice, water
and vapour will co-exist in equilibrium.
This point is known as triple point.
At this point, solid, liquid and vapour are in equilibrium under pressure equal to their vapour
pressure. The point is non-variant (F = C-P+2 = 1-3+2 = 0) i.e. in order to define the system
at O, we have not to mention any of the variable factors i.e. system is self-defined.

Under certain conditions, water can be
cooled to -90
C without the separation of ice
at 00
C i.e. below its freezing point is known
as supercooled water. But as soon as the
equilibrium is disturbed either by stirring or
by adding a small piece of ice, supercooled
water immediately changes to ice. Such an
equilibrium is known as metastable
equilibrium.
An equilibrium which in itself is stable but become unstable on being disturbed by stirring
or adding a piece of the solid phase.

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