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 GA & GB be the free energy per mole of the substance A & B in initial.
Then, GA = GB
Hence ΔG = GB - GA = 0 ΔG = GB - GA = 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 GA to GA + dGA and GB + dGB. Since the two phase are still in equilibrium hence,
CA = dGA = GB + dGB [∵ GA = GB]
then dGA = dGB …(a)
According to thermodynamics, dG = VdP – SdT
then dGA = VAdP – SAdT …(b)
dGB = VBdP – SBdT
Using equation (a) and (b)
dGA = dGB
VAdP – SAdT = VBdP – SBdT
VA & VB 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
This is Clapeyron equation.
Claypeyron equation for liquid vapour equilibrium
ΔHV = heat of vapourisation
Vg = volume of gas
Vl = volume of liquid
⇒ [∵ Vg >> Vl]
& PV = RT (per mole)
This is Clasusius-Clapeyron equation.
Claypeyron equation for solid vapour equilibria
ΔHS = heat of sublimation
Vg >> VS &
Similarly the above equation, we get
Claypeyron equation for solid liquid equilibria
ΔHf = heat of fusion