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Thermodynamics question

2024 · 9 Apr · Shift 2 · Q27
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Thermodynamics question

2024 · 9 Apr · Shift 2 · Q27

JEE MainChemistryThermodynamicsNumerical+4 / −1
When ΔHvap=30 kJ/mol\Delta \mathrm{H}_{\mathrm{vap}}=30 \mathrm{~kJ} / \mathrm{mol}ΔHvap​=30 kJ/mol and ΔSvap=75 J mol−1 K−1\Delta \mathrm{S}_{\mathrm{vap}}=75 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}ΔSvap​=75 J mol−1 K−1, then the temperature of vapour, at one atmosphere is ‾\underline{\hspace{2cm}}​ K.
Numerical answer
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Correct answer: 400

  1. At the boiling point under 111 atm, liquid and vapour are in equilibrium, so
ΔGvap=0\Delta G_{\text{vap}} = 0ΔGvap​=0
  1. Using the relation
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

we get at equilibrium:

0=ΔHvap−TΔSvap0 = \Delta H_{\text{vap}} - T\Delta S_{\text{vap}}0=ΔHvap​−TΔSvap​

So,

T=ΔHvapΔSvapT = \frac{\Delta H_{\text{vap}}}{\Delta S_{\text{vap}}}T=ΔSvap​ΔHvap​​
  1. Substitute the given values:
ΔHvap=30 kJ mol−1=30000 J mol−1\Delta H_{\text{vap}} = 30\,\text{kJ mol}^{-1} = 30000\,\text{J mol}^{-1}ΔHvap​=30kJ mol−1=30000J mol−1 ΔSvap=75 J mol−1K−1\Delta S_{\text{vap}} = 75\,\text{J mol}^{-1}\text{K}^{-1}ΔSvap​=75J mol−1K−1

Therefore,

T=3000075=400 KT = \frac{30000}{75} = 400\,\text{K}T=7530000​=400K
  1. Hence, the temperature of vapour at one atmosphere is
400 K\boxed{400\,\text{K}}400K​
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