Question

Consider the equilibrium process of liquid helium vaporizing to a gas. Show that the ratio of the latent heat of vaporization to the entropy of vaporization (∆H/∆S) is approximately equal to the boiling temperature. Comment on the closeness of the approximation for helium.

Answer 1

Using one of the relation between thermodynamics properties H = U+PV

nH= nU+ pnV   (for ‘n’ moles)

differentiation gives

d(nH)=d(nU)+ pd(nV) + (nV)dP

since d(nU)=Td(nS) –pd(nV) --------------------- (1)

Equation (1) can be derived from d(nU)= dQ – dW , 1st law of thermodynamics for closed system for ‘n’ moles. Using dQ/T = d(nS) and dW= Pd(nV), (for reversible process.) in above equation of 1st law. Hence, d(nU)=Td(nS) –Pd(nV) But above equation is valid for irreversible also as ‘U’, ‘V’ and ‘S’ depend upon the state alone, not on the type of process.

hence, d(nH)= Td(nS) + (nV)dP ---------------------(2)

for vaporisation betwwen two phase at constant T and P

d(nH) = Td(nS)

integration of above equation from phase change L to V

ΔHLV = T ΔSLV

ΔHLV / ΔSLV = T ---------------------- (A)