Question:
The internal energy change (in J) when $90 \mathrm{~g}$ of water undergoes complete evaporation at $100^{\circ} \mathrm{C}$ is_____________
(Given: $\Delta \mathrm{H}_{\text {vap }}$ for water at $373 \mathrm{~K}=41 \mathrm{~kJ} / \mathrm{mol}, \mathrm{R}=8.314 \mathrm{JK}^{-}$ ${ }^{1} \mathrm{~mol}^{-1}$ )
Solution:
$(189494)$
$\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta n_{\mathrm{g}} \mathrm{RT}$
$n=\frac{90}{18}=5 \mathrm{~mol}$
$\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad \Delta n=1$
$41000=\Delta \mathrm{U}+1 \times 8.314 \times 373$
$\Rightarrow \Delta \mathrm{U}=37898.875 \mathrm{~J}$
For 5 moles, $\Delta \mathrm{U}=37898.87 \times 5=189494 \mathrm{~J}$