Question

Derive Clapeyron equation and Clausius modification for liquid-vapor equilibrium conditions.

Answer 1

In equilibrium:

dGa = Va*dp -Sa*dT

dGb = Vb*dp -Sb*dT

so

dGa =dGb (in phase equilibriua)

then

Va*dp -Sa*dT = Vb*dp -Sb*dT

solve

(Va-Vb)*dP = (Sa-Sb)*dT

which is

dV*dP = dS*dT

so

dP/dT = dS/dV

since

dG = dH - T*dS = 0

then

dG = Ga- Gb = 0

so

dS = dH/T

then

dP/dT = dS/dV

turns to

dP/dT = dS/dV = dH/(T*dV)

dP/dT = dH/(T*dV)

we know that

dP= dH/(T*dV)dT

can be integrated if we assume dH is constnat i.e. vapor enthalpy and dV is change of vapor

since vapor volume >> liqudi volume, assume dV = Vgas

dP= Henthalpy/(T*Vgas)dT

(P2-P1) = Henthalpy/(Vgas)( lnT1 - lnT2)

so

Vg = RT/p (for ideal gas)

substitute

dP / DT = P*dH/(R*T^2)

then

d(lnp) = dH/R*dT/(T^2)

finally

ln(p2) - ln(p1) = dH/R*(1/T1- 1/T2)

which is clasisus clapeyron equation