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Thermodynamics question

2012 · Shift 1 · Q17
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Thermodynamics question

2012 · Shift 1 · Q17

JEE AdvancedChemistryThermodynamicsMultiple correct+4 / −2
For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is(are) correct? (Take Δ\DeltaΔ S as change in entropy and WWW as work done) IIT-JEE 2012 Paper 1 Offline Chemistry - Thermodynamics Question 15 English
  1. A
    ΔSX→Z=ΔSX→Y+ΔSY→Z\Delta {S_{X \to Z}} = \Delta {S_{X \to Y}} + \Delta {S_{Y \to Z}}ΔSX→Z​=ΔSX→Y​+ΔSY→Z​
  2. B
    ΔWX→Z=ΔWX→Y+ΔWY→Z\Delta {W_{X \to Z}} = \Delta {W_{X \to Y}} + \Delta {W_{Y \to Z}}ΔWX→Z​=ΔWX→Y​+ΔWY→Z​
  3. C
    WX→Y→Z=WX→Y{W_{X \to Y \to Z}} = {W_{X \to Y}}WX→Y→Z​=WX→Y​
  4. D
    ΔSX→Y→Z=ΔSX→Y\Delta {S_{X \to Y \to Z}} = \Delta {S_{X \to Y}}ΔSX→Y→Z​=ΔSX→Y​
View written solutionFree

Correct answer: A, C

  1. Key ideas

For an ideal gas:

  • Entropy change ΔS\Delta SΔS is a state function, so it depends only on initial and final states.
  • Work WWW is a path function, so it depends on the process/path taken.
  • For only PPP-VVV work, W=∫P dVW = \int P\,dVW=∫PdV so if a step occurs at constant volume, then dV=0dV=0dV=0 and hence W=0W=0W=0 for that step.

From the figure (two-step path via YYY), the segment Y→ZY \to ZY→Z is a vertical line on the PPP-VVV diagram, hence it is isochoric (constant volume). Therefore, WY→Z=0W_{Y\to Z}=0WY→Z​=0

  1. Check option A

ΔSX→Z=ΔSX→Y+ΔSY→Z\Delta S_{X\to Z} = \Delta S_{X\to Y} + \Delta S_{Y\to Z}ΔSX→Z​=ΔSX→Y​+ΔSY→Z​

Since entropy is a state function, the entropy change over consecutive steps is additive: ΔSX→Z=SZ−SX\Delta S_{X\to Z} = S_Z - S_XΔSX→Z​=SZ​−SX​ ΔSX→Y=SY−SX,ΔSY→Z=SZ−SY\Delta S_{X\to Y} = S_Y - S_X, \qquad \Delta S_{Y\to Z} = S_Z - S_YΔSX→Y​=SY​−SX​,ΔSY→Z​=SZ​−SY​ Adding, (SY−SX)+(SZ−SY)=SZ−SX=ΔSX→Z(S_Y-S_X)+(S_Z-S_Y)=S_Z-S_X=\Delta S_{X\to Z}(SY​−SX​)+(SZ​−SY​)=SZ​−SX​=ΔSX→Z​

So A is correct.

  1. Check option B

ΔWX→Z=ΔWX→Y+ΔWY→Z\Delta W_{X\to Z} = \Delta W_{X\to Y} + \Delta W_{Y\to Z}ΔWX→Z​=ΔWX→Y​+ΔWY→Z​

This statement is not correct as written for thermodynamics. Work is not a state function, so one cannot speak of "ΔW\Delta WΔW" between two states in the same way as entropy.

Even if interpreted loosely as work along a path, work depends on path, so work from XXX to ZZZ along one path need not equal the sum associated with some other interpretation. Thus this option is not accepted.

So B is incorrect.

  1. Check option C

WX→Y→Z=WX→YW_{X\to Y\to Z} = W_{X\to Y}WX→Y→Z​=WX→Y​

Work along the composite path is WX→Y→Z=WX→Y+WY→ZW_{X\to Y\to Z}=W_{X\to Y}+W_{Y\to Z}WX→Y→Z​=WX→Y​+WY→Z​ But Y→ZY\to ZY→Z is at constant volume, so WY→Z=∫P dV=0W_{Y\to Z}=\int P\,dV=0WY→Z​=∫PdV=0 Hence, WX→Y→Z=WX→Y+0=WX→YW_{X\to Y\to Z}=W_{X\to Y}+0=W_{X\to Y}WX→Y→Z​=WX→Y​+0=WX→Y​

So C is correct.

  1. Check option D

ΔSX→Y→Z=ΔSX→Y\Delta S_{X\to Y\to Z} = \Delta S_{X\to Y}ΔSX→Y→Z​=ΔSX→Y​

But entropy change over the full path from XXX to ZZZ is ΔSX→Y→Z=SZ−SX\Delta S_{X\to Y\to Z}=S_Z-S_XΔSX→Y→Z​=SZ​−SX​ whereas ΔSX→Y=SY−SX\Delta S_{X\to Y}=S_Y-S_XΔSX→Y​=SY​−SX​ These are equal only if SZ=SYS_Z=S_YSZ​=SY​, which is not generally true. In fact, ΔSX→Y→Z=ΔSX→Y+ΔSY→Z\Delta S_{X\to Y\to Z} = \Delta S_{X\to Y} + \Delta S_{Y\to Z}ΔSX→Y→Z​=ΔSX→Y​+ΔSY→Z​ So D is incorrect.

  1. Final answer

The correct options are: A, C\boxed{A,\ C}A, C​

  1. Comparison with stored answer

Stored correct answer: A,CA, CA,C

Our derived answer matches the stored answer exactly.

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