Standard enthalpy of vapourisation D _vapH ^Q for water at 100° C is 40.66 kJmol^–1. The internal energy of vapourisation of water at 100°C (in kJmol^–1) is ?

Dear student!
We know that, for gaseous reactant and products, we have a relation between standard enthalpy of vaporization( Δ_vapH⁰)  and standard internal energy( ΔU⁰) as :

 Δ_vapH⁰ =  ΔU⁰ +  Δn_g RT 

where,Δn_g =  n₂- n₁ i.e. difference between no. of moles of reactant and product.

Here, for vaporization of water ,

H₂O (l) -> H₂O (g)

So, change in no. of moles ,Δn_g = 1-0 =1

Thus,  Δ_vapH⁰ =  ΔU⁰ + RT 

 ΔU⁰ = Δ_vapH⁰ - RT  = 44.66 - ( 8.314 x 10^-3 KJ/Kmol  x 373K) = 40.66 - 3.10 KJ/mol = 37.55 KJ/mol