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The $\Delta \mathrm{H}$ for vaporization of a liquid is $20 \mathrm{~kJ} / \mathrm{mol}$. Assuming ideal behaviour, the change in internal energy for the vaporization of $1 \mathrm{~mol}$ of the liquid at $60^{\circ} \mathrm{C}$ and 1 bar is close to -
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$17.2 \mathrm{~kJ} / \mathrm{mol}$
$\begin{array}{l}
\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\
20=\Delta \mathrm{E}+8.314 \times 10^{-3} \times 333 \\
\Delta \mathrm{E}=17.2 \mathrm{~kJ} / \mathrm{mol}
\end{array}$
\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\
20=\Delta \mathrm{E}+8.314 \times 10^{-3} \times 333 \\
\Delta \mathrm{E}=17.2 \mathrm{~kJ} / \mathrm{mol}
\end{array}$
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