JEE Main 20248 AprilEvening Shift
- Exam
- JEE Main
- Year
- 2024
- Date
- 8 April
- Shift
- Evening Shift
- Subject
- Chemistry
- Chapter
- Thermodynamics (C)
JEE Main 2024 · 8 April · Evening Shift · Chemistry · Thermodynamics (C)
(Δ_ vap H^ ) for water is (+40.79 kJ mol^ -1 ) at 1 bar and (100 ^...
(\Delta_{\text{vap }} \mathrm{H}^{\ominus}) for water is (+40.79 \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) (\mathrm{kJ} \mathrm{mol}^{-1}). (Integer answer) (Given (\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{ mol}^{-1}))
Correct answer
38
Step-by-step solution
\begin{aligned} & \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} / \mathrm{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\end{aligned}