If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and 100°C is 41kJ mol–1. Calculate the internal energy change, when ?

Hello Student
(i) The change : H2O (l) → H2O(g)
ΔH = ΔU + ΔngRT
or ΔU = ΔH - ΔngRT

Substituting the values, we get ΔU = 41.00 kJ/mol - 8.3 J/mol/K X 373 K/1000
= 41.00 kJ mol−1 − 3.096 kJ mol−1
= 37.904 kJ/mol

(ii) The change : H2O (l) → H2O(s)
There is negligible change in volume, so, we can put pΔV = ΔngRT ≈ 0.
So, ΔH ≅ ΔU
or ΔU = 41.00kJ mol−1