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) 100^ C H_2 O(g) Δ_ vap H^ s =Δ_ vap E^ s +Δ n_g R T For the above reaction, gathered 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 gathered

NEET 2012

Exam: NEET
Year: 2012
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2012 · Chemistry · Thermodynamics (C)

Standard enthalpy of vaporisation Δ_ vap H^ s for water at 100^ C...

Standard enthalpy of vaporisation

\Delta_{\mathrm{vap}} H^{\mathrm{s}}

for water at

100^{\circ} \mathrm{C}

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).

Options

A+37.56
B-43.76
C+43.76
D+40.66

Correct answer

A. +37.56

Step-by-step 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}

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