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 ?
We know that, for gaseous reactant and products, we have a relation between standard enthalpy of vaporization( ΔvapH0) and standard internal energy( ΔU0) as :
ΔvapH0 = ΔU0 + Δng RT
where,Δng = 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 ,Δng = 1-0 =1
Thus, ΔvapH0 = ΔU0 + RT
ΔU0 = ΔvapH0 - RT = 44.66 - ( 8.314 x 10-3 KJ/Kmol x 373K) = 40.66 - 3.10 KJ/mol = 37.55 KJ/mol