For water
Question:

For water $\Delta_{\text {vap }} \mathrm{H}=41 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $373 \mathrm{~K}$ and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is $\mathrm{kJ} \mathrm{mol}^{-1}$

$\left[\right.$ Use $\left.: \mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right]$

Solution:

Given equation is

$\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}): \Delta \mathrm{H}=41 \frac{\mathrm{kJ}}{\mathrm{mol}}$

$\Rightarrow$ From the relation $: \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$

$\Rightarrow 41 \frac{\mathrm{kJ}}{\mathrm{mol}}=\Delta \mathrm{U}+(1) \times \frac{8.3}{1000} \times 373$

$\Rightarrow \mathrm{DU}=41-3.0959$

$=38 \mathrm{~kJ} / \mathrm{mol}$