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Thermodynamics question

2022 · 28 Jul · Shift 1 · Q2
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  5. /2022 · 28 Jul · Shift 1 · Q2

Thermodynamics question

2022 · 28 Jul · Shift 1 · Q2

JEE MainChemistryThermodynamicsMCQ+4 / −1
Which of the following relation is not correct?
  1. A
    ΔH=ΔU−PΔV\Delta \mathrm{H}=\Delta \mathrm{U}-\mathrm{P} \Delta \mathrm{V}ΔH=ΔU−PΔV
  2. B
    ΔU=q+W\Delta \mathrm{U}=\mathrm{q}+\mathrm{W}ΔU=q+W
  3. C
    ΔSsys +ΔSsurr ⩾0\Delta \mathrm{S}_{\text {sys }}+\Delta \mathrm{S}_{\text {surr }} \geqslant 0ΔSsys ​+ΔSsurr ​⩾0
  4. D
    ΔG=ΔH−TΔS\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}ΔG=ΔH−TΔS
View written solutionFree

Correct answer: A

  1. We check each thermodynamic relation.

  2. Option A: Enthalpy is defined as H=U+PVH = U + PVH=U+PV Hence, ΔH=ΔU+Δ(PV)\Delta H = \Delta U + \Delta(PV)ΔH=ΔU+Δ(PV) At constant pressure, ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta VΔH=ΔU+PΔV So the given relation ΔH=ΔU−PΔV\Delta H = \Delta U - P\Delta VΔH=ΔU−PΔV has the wrong sign.

    Therefore, Option A is incorrect.

  3. Option B: By the first law of thermodynamics, ΔU=q+W\Delta U = q + WΔU=q+W when WWW is taken as work done on the system. So this relation is correct.

  4. Option C: For any spontaneous process, ΔSuniv=ΔSsys+ΔSsurr≥0\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \ge 0ΔSuniv​=ΔSsys​+ΔSsurr​≥0 Equality holds for a reversible process. So this is correct.

  5. Option D: Gibbs free energy is defined as G=H−TSG = H - TSG=H−TS Therefore, for changes at constant temperature, ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS So this is correct in the usual thermodynamic context.

  6. Hence, the relation which is not correct is A.

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