Latent heat of vapourisation of a liquid at 500 K and 1atm pressure is 10 K.Cal/mole. What is the change in internal energy when 3 mole liquid vapourised at the same temperature will be

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Detailed Solution

Xl→Xg,ΔHV = 10K cal/mole;3Xl→3Xg, ∆H=30 kcals ; ∆n=3, ∆H=∆E+∆nRT ; R=2×10-3 kcal mole-1K-1;30 = ΔE + 3 x 2 x 10-3 x 500ΔE = 30 - 3 = 27 Kcal

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