A certain liquid boils at a temperature of 300 K under a constant pressure of 1 bar . The standard molar entropy of vaporization for the liquid is 125 J K ⁻¹ mol ⁻¹ . Assuming the vapor behaves as an ideal gas, the standard molar internal energy of vaporization of the liquid at 300 K is ________ kJ mol ⁻¹ . [Given: R = 25 3 J K ⁻¹ mol ⁻¹ ]

Step-by-step solution

At the boiling point, the liquid and its vapor are in equilibrium, so the change in Gibbs free energy, G = 0 . Using the relation G = H_ vap - T_b S_ vap : H_ vap = T_b S_ vap H_ vap = 300 125 = 37500 J mol ⁻¹ = 37.5 kJ mol ⁻¹ For the vaporization process, Liquid Gas , the change in gaseous moles per mole of liquid is n_g = 1 . Using the first law relation: H_ vap = U_ vap + n_g RT U_ vap = H_ vap - RT Substituting the values: U_ vap = 37.5 - ( 25 3 300 10⁻³ ) U_ vap = 37.5 - 2.5 = 35 kJ mol ⁻¹ Answer: 35