Standard enthalpy of vaporisation o vap H for water at 100°C is 40.66 kJ mol–1. The internal energy of vaporisation of water at 100°C (in kJ mol–1) is (2012)

o vap H = 40.66 kJ mol–1 T = 100 + 273 = 373 K, ∆U = ? ∆H = ∆ U + ∆ngRT ⇒ ∆ U = ∆H – ∆ngRT ∆ng = number of gaseous moles of products – number of gaseous moles of reactants H₂O(l) ⇌ H₂O(g) ∆ng = 1 – 0 = 1 ∆ U = ∆H – RT ∆ U = (40.66 × 103) – (8.314 × 373) = 37559 J/mol or 37.56 kJ/mol