Standard enthalpy of vaporisation $\Delta_{\mathrm{vap}} H^{\mathrm{s}}$ for water at $100^{\circ}…

Standard enthalpy of vaporisation $\Delta_{\mathrm{vap}} H^{\mathrm{s}}$ for water at $100^{\circ} \mathrm{C}$ is $40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The internal energy of vaporisation of water at $100^{\circ} \mathrm{C}$ (in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ is (Assume water vapour to behave like an ideal gas).
  1. +37.56
  2. -43.76
  3. +43.76
  4. +40.66

Solution

$\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \stackrel{100^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{H}_2 \mathrm{O}(g)$
$\Delta_{\text {vap }} H^{\mathrm{s}}=\Delta_{\text {vap }} E^{\mathrm{s}}+\Delta n_g R T$
For the above reaction,
$\begin{gathered}
\Delta n_g=n_p-n_r=1-0=1 \\
\therefore 40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}=\Delta_{\mathrm{vap}} E^{\mathrm{s}}+1 \times 8.314 \\
\quad \times 10^{-3} \times 373 \\
\Delta_{\mathrm{vap}} E^{\mathrm{s}}=40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}-3.1 \mathrm{~kJ} \mathrm{~mol}{ }^{-1} \\
=+37.56 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{gathered}$

Asked in: NEET 2012 (Screening)

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