Standard enthalpy of vaporisation Δ_vap H^s for water at 100^° C is 40.66 ~kJ ~mol^-1 . The internal energy of vaporisation of water at 100^° C (in .kJ mol^-1 ) is (Assume water vapour to behave like an ideal gas).

Correct answer: (A) +37.56. H_2 O(l) stackrel100^° Clongrightarrow H_2 O(g) Δ_vap H^s=Δ_vap E^s+Δ n_g R T For the above reaction, Δ n_g=n_p-n_r=1-0=1 ∴ 40.66 ~kJ ~mol^-1=Δ_vap E^s+1 × 8.314 × 10^-3 × 373 Δ_vap E^s=40.66 ~kJ ~mol^-1-3.1 ~kJ ~mol^-1 =+37.56 ~kJ ~mol^-1

Q62 / 200 · NEET 2012 (Screening) · Chemistry

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).

Correct answer
A
+37.56
B
-43.76
C
+43.76
D
+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, $\begin{gathered} \Delta n_g=n_p-n_r=1-0=1 \\ \therefore 40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}=\Delta_{\mathrm{vap}} E^{\mathrm{s}}+1 \times 8.314 \\ \quad \times 10^{-3} \times 373 \\ \Delta_{\mathrm{vap}} E^{\mathrm{s}}=40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}-3.1 \mathrm{~kJ} \mathrm{~mol}{ }^{-1} \\ =+37.56 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{gathered}$

Hence, the correct answer is option: 'A'.

Solution

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