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**Clausirs Clapeyron Equation:** The two phases in equilibrium may be any of the following types:

(i) Solid and Liquid, S L, at the melt ing point of the solid.

(ii) Liquid and vapour, L V, at the boiling point of the liquid.

(iii) Solid and vapour, S V; at the sublimat ion temperature of the solid. “The equal amounts of a given substance must have exactly the same free energy in the two phases at equilibrium with each other.”

Consider t ype equation A B at the given temperature and pressure. If G_{A }& G_{B} be the free energy per mole of the substance A & B in initial.

Then, G_{A }= G_{B}

Hence ΔG = GB - G_{A} = 0 ΔG = G_{B} - G_{A} = 0

If temperature of the system raise from T to T + dT then pressure raise from P to P + dt, in order to maintain the equilibrium. The free energy of substance raise from G_{A} to G_{A} + dG_{A }and G_{B} + dG_{B}. Since the two phase are still in equilibrium hence,

C_{A} = dG_{A} = G_{B} + dG_{B } [∵ G_{A} = G_{B}]

then dG_{A} = dG_{B} …(a)

According to thermodynamics, dG = VdP – SdT

then dGA = VAdP – SAdT …(b)

dG_{B} = V_{B}dP – S_{B}dT

Using equation (a) and (b)

we get

dG_{A} = dG_{B}

V_{A}dP – S_{A}dT = V_{B}dP – S_{B}dT

V_{A }& V_{B} be the volume of A and B.

If q heat exchanged reversibly per mo le of substance during the phase transformat ion at temperature t, then change in entropy. Δ

Hence,

Thus,

This is Clapeyron equation.**Claypeyron equation for liquid vapour equilibrium **

where

ΔH_{V }= heat of vapourisation

V_{g} = volume of gas

V_{l} = volume of liquid

⇒ [∵ Vg >> Vl]

& PV = RT (per mole)

⇒

⇒

⇒

⇒

⇒

This is Clasusius-Clapeyron equation.**Claypeyron equation for solid vapour equilibria Solid Vapour **

ΔH

V_{g} >> V_{S }&

Similarly the above equation, we get

⇒

**Claypeyron equation for solid liquid equilibria**

**Solid Liquid **

ΔH_{f }= heat of fusion

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