Standard enthalpy of vaporization _ vap H^0 for water at 100^ C is 40.66 kJ mol ⁻¹ . What is the internal energy of vaporization of water at 100^ C (in kJ mol ⁻¹ )?

Step-by-step solution

The vaporization of water is represented by the equation: H ₂ O (l) H ₂ O (g) The change in the number of moles of gas is n_g = 1 - 0 = 1 . Using the relation between enthalpy change and internal energy change: _ vap H^ = _ vap U^ + n_g RT Rearranging for _ vap U^ : _ vap U^ = _ vap H^ - n_g RT Substituting the given values ( T = 100^ C = 373 K , R = 8.314 J K ⁻¹ mol ⁻¹ ): _ vap U^ = 40.66 - (1 8.314 10⁻³ 373) _ vap U^ = 40.66 - 3.101 _ vap U^ = +37.559 kJ mol ⁻¹ Rounding to two decimal places, we get +37.56 kJ mol