Chemistry · Thermodynamics & Thermochemistry

JEE Main 2024 — 8 April, Shift 2 — Question 78

$\Delta_{\text {vap }} \mathrm{H}^{\ominus}$ for water is $+40.49 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at 1 bar and $100^{\circ} \mathrm{C}$ . Change in internal energy for this vapourisation under same condition is $\qquad$ kJ $\operatorname{mol}^{-1}$ . (Integer answer) (Given $\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

Answer: 38

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad \Delta \mathrm{H}_{\text {vap }}^{0}=40.79 \mathrm{~kJ} /$ mole $\Delta \mathrm{H}_{\text {vap }}^{0}=\Delta \mathrm{U}_{\text {vap }}^{0}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$

$40.79=\Delta \mathrm{U}_{\text {vap }}^{0}+\frac{1 \times 8.3 \times 373.15}{1000}$

$\Delta \mathrm{U}_{\text {vap }}^{0}=40.79-3.0971$ $=37.6929$

$\Delta \mathrm{U}_{\text {vap }}^{0} \simeq 38$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Internal Energy and the First Law