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JEE MainChemistryThermodynamics (C)

A certain liquid boils at 350 K under standard atmospheric pressure. If its molar enthalpy of vaporisation is 28 kJ mol ⁻¹ , the molar entropy of vaporisation of the liquid is ______ J K ⁻¹ mol ⁻¹ .

Correct answer

80

Step-by-step solution

At the normal boiling point, the liquid and its vapour are in equilibrium, which means the change in Gibbs free energy is zero ( G = 0 ). From the Gibbs free energy equation: G = H_ vap - T_ b S_ vap = 0 Rearranging for the entropy of vaporisation: S_ vap = H_ vap T_ b Substitute the given values, ensuring units are consistent by converting kJ to J: S_ vap = 28 1000 J mol ⁻¹ 350 K S_ vap = 80 J K ⁻¹ mol ⁻¹ Answer: 80

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