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Thermodynamics question

2007 · Shift 0 · Q14
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Thermodynamics question

2007 · Shift 0 · Q14

JEE MainChemistryThermodynamicsMCQ+4 / −1
Assuming that water vapour is an ideal gas, the internal energy change (ΔU)\left( {\Delta U} \right)(ΔU) when 111 mol of water is vapourised at 111 bar pressure and 100∘C{100^ \circ }C100∘C(Given : molar enthalpy of vapourisation of water at 111 bar and 373K=41 kJ mol−1 373K = 41\,kJ\,mo{l^{ - 1}}\,373K=41kJmol−1 and R=8.3 J mol−1 K−1R = 8.3\,J\,mo{l^{ - 1}}\,{K^{ - 1}}R=8.3Jmol−1K−1 )
  1. A
    41.00 kJ mol−141.00\,kJ\,mo{l^{ - 1}}41.00kJmol−1
  2. B
    4.100 kJ mol−14.100\,kJ\,mo{l^{ - 1}}4.100kJmol−1
  3. C
    3.7904 kJ mol−13.7904\,kJ\,mo{l^{ - 1}}3.7904kJmol−1
  4. D
    37.904 kJ mol−137.904\,kJ\,mo{l^{ - 1}}37.904kJmol−1
View written solutionFree

Correct answer: D

  1. Use the relation between enthalpy and internal energy

For any process,

ΔH=ΔU+Δ(PV)\Delta H = \Delta U + \Delta(PV)ΔH=ΔU+Δ(PV)

So,

ΔU=ΔH−Δ(PV)\Delta U = \Delta H - \Delta(PV)ΔU=ΔH−Δ(PV)
  1. For vapourisation of water

Given:

  • ΔHvap=41 kJ mol−1\Delta H_{\text{vap}} = 41\,\text{kJ mol}^{-1}ΔHvap​=41kJ mol−1
  • T=373 KT = 373\,\text{K}T=373K
  • Water vapour behaves as an ideal gas

During vapourisation, the volume of liquid water is negligible compared to the volume of vapour. Hence,

Δ(PV)≈(PV)vapour−(PV)liquid≈nRT\Delta(PV) \approx (PV)_{\text{vapour}} - (PV)_{\text{liquid}} \approx nRTΔ(PV)≈(PV)vapour​−(PV)liquid​≈nRT

For 111 mol,

Δ(PV)=RT=8.3×373 J mol−1\Delta(PV) = RT = 8.3 \times 373\,\text{J mol}^{-1}Δ(PV)=RT=8.3×373J mol−1
  1. Calculate RTRTRT
RT=8.3×373=3095.9 J mol−1RT = 8.3 \times 373 = 3095.9\,\text{J mol}^{-1}RT=8.3×373=3095.9J mol−1 RT=3.0959 kJ mol−1RT = 3.0959\,\text{kJ mol}^{-1}RT=3.0959kJ mol−1
  1. Now calculate ΔU\Delta UΔU
ΔU=ΔH−RT\Delta U = \Delta H - RTΔU=ΔH−RT ΔU=41.0−3.0959=37.9041 kJ mol−1\Delta U = 41.0 - 3.0959 = 37.9041\,\text{kJ mol}^{-1}ΔU=41.0−3.0959=37.9041kJ mol−1

Thus,

ΔU≈37.904 kJ mol−1\boxed{\Delta U \approx 37.904\,\text{kJ mol}^{-1}}ΔU≈37.904kJ mol−1​
  1. Check options
  • A: 41.00 kJ mol−141.00\,\text{kJ mol}^{-1}41.00kJ mol−1 ❌
  • B: 4.100 kJ mol−14.100\,\text{kJ mol}^{-1}4.100kJ mol−1 ❌
  • C: 3.7904 kJ mol−13.7904\,\text{kJ mol}^{-1}3.7904kJ mol−1 ❌
  • D: 37.904 kJ mol−137.904\,\text{kJ mol}^{-1}37.904kJ mol−1 ✅

Therefore, the correct option is D.

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