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Thermodynamics question

2021 · 26 Aug · Shift 2 · Q15
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Thermodynamics question

2021 · 26 Aug · Shift 2 · Q15

JEE MainChemistryThermodynamicsNumerical+4 / −1
For water Δ\DeltaΔ vap H = 41 kJ mol −-− 1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ‾\underline{\hspace{2cm}}​ kJ mol −-− 1 [Use : R = 8.3 J mol −-− 1 K −-− 1]
Numerical answer
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Correct answer: 38

  1. For vaporization, ΔH=ΔU+Δ(PV)\Delta H = \Delta U + \Delta(PV)ΔH=ΔU+Δ(PV)

  2. Given that water vapour behaves as an ideal gas and vapour volume is much larger than liquid volume, Δ(PV)≈PVgas=nRT\Delta(PV) \approx P V_{\text{gas}} = nRTΔ(PV)≈PVgas​=nRT for 111 mol.

  3. Hence, ΔU=ΔH−RT\Delta U = \Delta H - RTΔU=ΔH−RT

  4. Substitute the values: ΔH=41 kJ mol−1\Delta H = 41\,\text{kJ mol}^{-1}ΔH=41kJ mol−1 R=8.3 J mol−1K−1R = 8.3\,\text{J mol}^{-1}\text{K}^{-1}R=8.3J mol−1K−1 T=373 KT = 373\,\text{K}T=373K

  5. Compute RTRTRT: RT=8.3×373=3095.9 J mol−1=3.096 kJ mol−1RT = 8.3 \times 373 = 3095.9\,\text{J mol}^{-1} = 3.096\,\text{kJ mol}^{-1}RT=8.3×373=3095.9J mol−1=3.096kJ mol−1

  6. Therefore, ΔU=41−3.096=37.904 kJ mol−1\Delta U = 41 - 3.096 = 37.904\,\text{kJ mol}^{-1}ΔU=41−3.096=37.904kJ mol−1

  7. As an integer, ΔU≈38 kJ mol−1\Delta U \approx 38\,\text{kJ mol}^{-1}ΔU≈38kJ mol−1

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