The latent heat of vaporization of a liquid at 500 K and 1 atm pressure is 10 kcal/mol. What will be — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

The total enthalpy of vaporization $\Delta H = 3 \text{ moles} \times 10 \text{ kcal/mol} = 30 \text{ kcal}$. For vaporization $A(l) \to A(g)$, the change in gaseous moles $\Delta n_g = 3$. Utilizing $\Delta H = \Delta U + \Delta n_g RT$, we find $\Delta U = 30 \text{ kcal} - 3 \times (2 \times 10^{-3} \text{ kcal/K}) \times 500 \text{ K} = 30 - 3 = 27 \text{ kcal}$.