# Clausius-Clapeyron Equation

## Clausius-Clapeyron Equation

The Clausius-Clapeyron equation was initially proposed by a German physics Rudolf Clausius in 1834 and further developed by French physicist Benoît Clapeyron in 1850. This equation is extremely useful in characterizing a discontinuous phase transition between two phases of a single constituent.We know that-

dG = VdP − SdT -----(equation-1)

Let us consider a single-constituent equilibria-

**𝑃ℎ𝑎𝑠𝑒-1 ⇌ 𝑃ℎ𝑎𝑠𝑒-2**

Where phase-1 may be solid, liquid, or gas; whereas phase-2 may be liquid or vapor depending upon the nature of the transition whether it is melting, vaporization or sublimation, respectively.

For phase-1, change in free energy is-

dG

_{1}= V

_{1}dP − S

_{1}dT -----(equation-2)

and for Phase-2, change in free energy is-

dG

_{2}= V

_{2}dP − S

_{2}dT -----(equation-3)

At equilibrium- dG

_{1}= dG

_{2}(i.e. ΔG = 0

so, V

_{2}dP − S

_{2}dT = 𝑉

_{1}dP − S

_{1}dT

V

_{2}dP − V

_{1}dP = S

_{2}dT − S

_{1}dT

(V

_{2}− V

_{1})dP = (S

_{2}− S

_{1})dT

ΔV. ΔP = ΔS. dT

dP/dT = ΔS/ΔV -----(equation-4)

Now, if the ΔH is the latent heat of phase transformation takes place at temperature (𝑇), then the entropy change is-

ΔS = ΔH/T -----(equation-5)

Now, putting the value of 𝛥𝑆 from (equation-5) into (equation-4), we get-

dP/dT = ΔH/T.ΔV -----(equation-6)

The (equation-6) is known as Calpeyron equation.

Now if phase-1 is solid while phase-2 is vapor (i.e. solid ⇌ melt), then the equation-6 becomes-

dP/dT = Δ

_{fus}H/T

_{f}.ΔV -----(equation-7)

where Δ

_{fus}is latent heat of fusion and T

_{f}is melting point.

For vaporisation, equilibrium (i.e. liquid ⇌ vapour),

dP/dT = Δ

_{vap}H/T.V

_{v}-----(equation-8)

If the vapor act as an ideal gas-

then,V = RT/P

so, the above equation becomes-

dP/dT = Δ

_{vap}H.P/RT

^{2}-----(equation-9)

or, 1/P(dP/dT) = Δ

_{vap}H/RT

^{2}-----(equation-10)

or, dlnP/dT = Δ

_{vap}H/RT

^{2}-----(equation-11)

The equation-11 is known as the Clausius-Clapeyron equation.

Another form of

**Clausius-Clapeyron equation**-

from equation-11

dlnP = (Δ

_{vap}H/RT

^{2}).dT

If the temperature changes from T

_{1}to T

_{2}and pressure is varied from P

_{1}to P

_{2}, then -

∫dlnP = ∫(Δ

_{vap}H/RT

^{2}).dT

lnP

_{2}/P

_{1}= Δ

_{vap}H/R ∫dT/T

^{2}

lnP

_{2}/P

_{1}= (Δ

_{vap}H/R) [1/T

_{1}− 1/T

_{2}] -----(equation-12)

The equation-12 is another form of Clausius-Clapeyron equation.

converting ln into log-

2.303 logP

_{2}/P

_{1}= (Δ

_{vap}H/R) [1/T

_{1}− 1/T

_{2}] -----(equation-13)

or, logP

_{2}/P

_{1}= (Δ

_{vap}H/ 2.303 R) [1/T

_{1}− 1/T

_{2}] -----(equation-14)

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