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For water at 100°C and 1 bar,

Question:

For water at $100^{\circ} \mathrm{C}$ and $1 \mathrm{bar}$,

$\Delta_{\text {vap }} \mathrm{H}-\Delta_{\text {vap }} \mathrm{U}=\ldots 10^{2} \mathrm{~J} \mathrm{~mol}^{-1}$

(Round off to the Nearest Integer)

[Use : $\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ]

[Assume volume of $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ is much smaller than volume of $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$. Assume $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ treated as an ideal gas]

Solution:

$\mathrm{H}_{2} \mathrm{O}_{(\ell)} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{(\mathrm{V})}$

$\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$

for 1 mole waters;

$\Delta \mathrm{n}_{\mathrm{g}}=1$

$\therefore \Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}=1 \mathrm{~mol} \times 8.31 \mathrm{~J} / \mathrm{mol}-\mathrm{k} \times 373 \mathrm{~K}$

$=3099.63 \mathrm{~J} \cong 31 \times 10^{2} \mathrm{~J}$

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