ThermodynamicsHard

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?

Options

A.27 kcal
B.13 kcal
C.−27 kcal
D.−13 kcal

Solution

$q = n.L_{m} = 3 \times 10 = 30\text{ kcal}$

$w = - P\left( V_{\text{vap}} - V_{\text{liq}} \right) \approx - P.V_{\text{vap}} = - nRT $$${= - 3 \times \frac{2}{1000} \times 500 = - 3\text{ k cal} }{\therefore\Delta U = q + w = 27\text{ kcal}}$$

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