Question

 The enthalpy change for vaporization of liquid ‘A’ at 200 K and 1 atm is 22kJ 1mol. Find out ∆S_vaporization for liquid ‘A’ at 200 K? The normal boiling point of liquid ‘A’ is 300 K.

A.(l) [200 K 1 atm] → A (g) [200 k 1 atm]

Given: C_p.m(A, g) = 30 J 1 mol-K, C_{p,, m}(A, l) = 40 J1mol – K; Use: In (312) = 0.405 

Answer 1

= 22/3 + 10 × 0.405 = 74.05 J/k

V = constant