A 240 g sample of a volatile liquid (molar mass = 60 g mol ⁻¹ ) is completely vaporized at a constant temperature of 300 K . The total change in internal energy for this process is found to be 130 kJ . Assuming the vapor behaves as an ideal gas, the standard molar enthalpy 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

Number of moles of the liquid vaporized: n = 240 60 = 4 mol The total internal energy change is given as U = 130 kJ . Molar internal energy of vaporization: U_ molar = 130 4 = 32.5 kJ mol ⁻¹ For the vaporization process, Liquid Gas , the change in gaseous moles per mole of liquid is n_g = 1 . Using the relation between enthalpy and internal energy: H_ molar = U_ molar + n_g RT Substituting the values: H_ molar = 32.5 + 1 ( 25 3 300 10⁻³ ) H_ molar = 32.5 + 2.5 = 35 kJ mol ⁻¹ Answer: 35