Question:
The internal energy change (in $\mathrm{J}$ ) 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:
$\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad 90 \mathrm{gm}$ of $\mathrm{H}_{2} \mathrm{O}$
$\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \quad \Rightarrow 5$ moles of $\mathrm{H}_{2} \mathrm{O}$
$5 \times 41000 \mathrm{~J}=\Delta \mathrm{U}+1 \times 8.314 \times 373 \times 5$
$\Delta U=189494.39$ Joule