JEE MainChemistryThermodynamics (C)
A volatile liquid has a molar mass of 60 g mol ⁻¹ and boils at 350 K . If its standard entropy of vaporization is 90 J K ⁻¹ mol ⁻¹ , the total heat required to vaporize 120 g of this liquid at its boiling point is _______ kJ . (Nearest Integer)
Correct answer
63
Step-by-step solution
At the boiling point, the liquid and vapor phases are in equilibrium, meaning G = 0 . The molar enthalpy of vaporization is calculated as: H_ vap = T_ b S_ vap H_ vap = 350 90 = 31500 J mol ⁻¹ = 31.5 kJ mol ⁻¹ Next, calculate the number of moles ( n ) of the liquid to be vaporized: n = Mass Molar mass = 120 60 = 2 mol The total heat required is the enthalpy change for 2 moles: Total heat = n H_ vap = 2 31.5 = 63 kJ Answer: 63