The enthalpy change for vaporization of 1 mol of water at 373 K is 40.79 kJ/mol. What is Δ U if water vapor behaves as a
For H₂O(l) → H₂O(g) , Δ ng = 1 . Using Δ H = Δ U + Δ ng RT , where RT = 8.314 × 373 × 10⁻³ = 3.1012 kJ , Δ U = 40.79 - 3.1012 = 37.69 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Enthalpy Changes - Reaction Formation Combustion and Hess's Law