Enthalpy of vapourization of liquid at 127° c is 50kJ/mole. Change in Internal energy during this process would be

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Detailed Solution

Let the liquid be ‘X’Xl → XgΔn = 1-0 = 1ΔH = ΔE +ΔnRT50 = ΔE +18.3×10-3kJ K-1mole-140050 = ΔE +3.32ΔE = 46.68 kJ

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