H2O(l)→H2O(g) ∵ΔH=ΔU+Δng‌RT ∆H = enthalpy of vaporisation ∆U = change in internal energy Δng= number of moles of water vapour R=8.3Jmol−1K−1 T = 373 K ∴41‌kJ=ΔU+RT ΔU=41−8.3×373×10−3=41−3.095 =37.90‌kJ‌mol−1=38‌kJ‌mol−1