# Clausius-Clapeyron Equation

The Clausius-Clapeyron equation was proposed by a German physics Rudolf Clausius in 1834 and later on developed by French physicist Benoรฎt Clapeyron in 1850. The Clausius-Clapeyron Equation describes the relationship between the vapor pressure and temperature of a pure substance. 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-

๐โ๐๐ ๐-1 ⇌ ๐โ๐๐ ๐-2

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. ฮG = 0

so, V_{2}dP − S_{2}dT = ๐_{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

ฮV. ฮP = ฮS. dT

dP/dT = ฮS/ฮV -----(equation-4)

Now, if the ฮH is the latent heat of phase transformation takes place at temperature (๐), then the entropy change is-

ฮS = ฮH/T -----(equation-5)

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

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 = ฮ_{fus}H/T_{f}.ฮV -----(equation-7)

where ฮ_{fus} is latent heat of fusion and T_{f} is melting point.

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

dP/dT = ฮ_{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 = ฮ_{vap}H.P/RT^{2} -----(equation-9)

or, 1/P(dP/dT) = ฮ_{vap}H/RT^{2} -----(equation-10)

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

Another form of Clausius-Clapeyron equation-

from equation-11

dlnP = (ฮ_{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 = ∫(ฮ_{vap}H/RT^{2}).dT

lnP_{2}/P_{1} = ฮ_{vap}H/R ∫dT/T^{2}

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

Converting ln into log-

2.303 logP_{2}/P_{1} = (ฮ_{vap}H/R) [1/T_{1} − 1/T_{2}] -----(equation-13)

or, logP_{2}/P_{1} = (ฮ_{vap}H/ 2.303 R) [1/T_{1} − 1/T_{2}] -----(equation-14)

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