The internal energy change


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 $373 \mathrm{~K}=41 \mathrm{~kJ} / \mathrm{mol}, \mathrm{R}=8.314 \mathrm{JK}^{-}$ ${ }^{1} \mathrm{~mol}^{-1}$ )



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

$n=\frac{90}{18}=5 \mathrm{~mol}$

$\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad \Delta n=1$

$41000=\Delta \mathrm{U}+1 \times 8.314 \times 373$

$\Rightarrow \Delta \mathrm{U}=37898.875 \mathrm{~J}$

For 5 moles, $\Delta \mathrm{U}=37898.87 \times 5=189494 \mathrm{~J}$

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