The latent heat of vaporization of a liquid at 500 K and 1 atm pressure is 10 kcal/mol. What will be the change in internal energy if 3 moles of the liquid changes to vapour state at the same temperat

Question

The latent heat of vaporization of a liquid at 500 K and 1 atm pressure is 10 kcal/mol. What will be the change in internal energy if 3 moles of the liquid changes to vapour state at the same temperature and pressure?

Prashant Kotian · Accepted Answer

Correct answer: A. For 3 moles, Δ H = 3 * 10 = 30 kcal. Δ H = Δ U + Δ n_g * R * T => 30 = Δ U + 3 * (2 * 10^-3) * 500 => 30 = Δ U + 3 => Δ U = 27 kcal.

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