∆Hvap=∆U+∆ngRT where, ∆Hvap = enthalpy change of vaporisation ∆U= internal energy ∆ng=n2−n1= difference between number of moles of reactant and product. H2O(l)⟶H2O(g) H2O(l)⟶H2O(g) ∴∆ng=1−0=1 T=127∘C=127+273=400K R=8.3JK−1mol−1 ∴∆Hvap =∆U+∆ngRT ⇒∆U=∆Hvap−∆ngRT =40×103−1(8.3)(400)=36680Jmol−1 =36.68kJmol−1 ≈37kJmol−1.