Clapeyron discovered an important fundamental equation which helps in finding the extensive application in one component, two phase system and by Clausius from second law of thermodynamics. Hence this fundamental equation is known as Clapeyron-Clausius Equation. One of the uses of this equation is to determine if a phase transition will occur in a given situation

The equilibrium can in any of the following two phases:

(i) Solid and Liquid, S = L, at the melting point of the solid.

(ii) Liquid and Vapor, L = V, at the boiling point of the solid.

(iii) Solid and Vapor, S = V, at the sublimation temperature of the solid.

(iv)One crystalline form and another crystalline form as, for example, rhombic and monoclinic sulphur, S_{R} = S_{M}.

Let us consider any two phases (say, solid and liquid) which belongs to same substance and are in equilibrium with each other at a given temperature and pressure. We know that in a reversible reaction, the reaction can be change from one phase to another by proceeding infinitesimally slow. For example, by providing heat infinitesimally slowly to a system containing any substance like solid can be changed into any form like liquid at the same temperature and pressure. Similarly, by withdrawing heat infinitesimally slow form system we can get back the solid from the liquid without change in temperature and pressure.

As the system remains in a state of equilibrium, the free energy change of either process will be zero.

Hence we conclude that in the two phases at equilibrium with each other, equal amount of a given substance must have exactly the same free energy.

Let us suppose a pure substance which is initially in phase A and then changes to phase B. Both the phases are in equilibrium with each other at given temperature and pressure.

Let G_{A} be the free energy per mole of the substance in the initial phase A and G_{B} be the free energy per mole of the substance in the final phase B and we know that the two phases at equilibrium with each other having equal amount of a given substance have exactly the same free energy.

G_{A} = G_{B}

Hence, there will be no free energy change, i.e.

∆G = G_{B} – G_{A} = 0

If the temperature of the above system is raised from T to T + dT as well as the pressure from P to P + dP so that equilibrium will maintained. The relationship between dT and dP can be obtained from thermodynamics.

At the new temperature and pressure let the free energy per mole of the substance in phase A be G_{A} + dG_{A} and that in phase B be G_{B} + dG_{B}. As the two phases are still in equilibrium, hence their free energy will be equal to each other i.e.

G_{A} + dG_{A} = G_{B} + dG_{B} ………………….. (1)

According to thermodynamics,

dG = VdP – SdT ……………….. (2)

The above equation gives the change of free energy when a system undergoes a change of temperature dT and a change of pressure dP.

For phase A the equation (2) can be written as:

dG _{A} = V_{A} dP – S_{A} dT

And

For phase B can be written as:

dG _{B} = V_{B} dP – S_{B} dT

as G_{A} = G_{B}, hence, from equation (1),

dG_{A} = dG_{B}

therefore, V G_{A} dP – S_{A} dT = V_{B} dP – S_{B} dT

dP / dT = (S_{B} – S_{A}) / (V_{B} – V_{A})

as V_{A} and V_{B} are molar volumes of the pure substance in the two phases A and B respectively. Hence difference between them i.e. (V_{B} – V_{A}) represents the change in volume when one mole of the substance passes from initial phase A to final phase B.

It can be represented as ∆V = V_{B} – V_{A}. Similarly entropy change would be given as ∆S = S_{B} – S_{A}. Hence,

dP / dT = ∆S / ∆V

if the heat exchanged reversibly per mole of the substance during phase transformation be q at temperature T , then the change in entropy i.e. ∆S will be given as:

∆S = q / T

Hence, dP / dT = q / T ∆V

**Thus, dP / dT = q / T (V _{B} – V_{A})**

This equation is known as Clapeyron-Clausius Equation.

Needs more explanation on application of clausius-clpeyron equation.