JEE Challenger
More from Thermodynamics
ChemMedAdv 2012 • Paper 1

Thermodynamic State Functions and Work Done Along Different Paths

For an ideal gas, consider only P-VP\text{-}V work during a transition from an initial state XX to a final state ZZ. State ZZ can be reached via either of the two paths depicted in the figure. Which of the following choice(s) is (are) correct? [Let ΔS\Delta S denote entropy change and ww denote work done]

Question Diagram 1

Options

A

ΔSx→z=ΔSx→y+ΔSy→z\Delta S_{x \to z} = \Delta S_{x \to y} + \Delta S_{y \to z}

Correct
B

wx→z=wx→y+wy→zw_{x \to z} = w_{x \to y} + w_{y \to z}

C

wx→y→z=wx→yw_{x \to y \to z} = w_{x \to y}

Correct
D

ΔSx→y→z=ΔSx→y\Delta S_{x \to y \to z} = \Delta S_{x \to y}

Topics & Concepts

ThermodynamicsEntropy and Second LawFirst Law of Thermodynamics

Step-by-Step Solution

To determine the correct choice(s), we analyze the thermodynamic properties (state functions and path functions) for the given process on the P-VP\text{-}V diagram.

1. Entropy Change (ΔS\Delta S)

Entropy (SS) is a state function, which means that the change in entropy between any two state points depends only on the initial and final states, independent of the path taken.

  • For the process occurring along the path X→Y→ZX \to Y \to Z, the total entropy change is the sum of entropy changes in each step: ΔSX→Y→Z=ΔSX→Y+ΔSY→Z\Delta S_{X \to Y \to Z} = \Delta S_{X \to Y} + \Delta S_{Y \to Z}

  • Since state ZZ is reached in both paths (X→ZX \to Z and X→Y→ZX \to Y \to Z) starting from the same initial state XX: ΔSX→Z=ΔSX→Y→Z\Delta S_{X \to Z} = \Delta S_{X \to Y \to Z}

  • Equating the two expressions gives: ΔSX→Z=ΔSX→Y+ΔSY→Z\Delta S_{X \to Z} = \Delta S_{X \to Y} + \Delta S_{Y \to Z}

Thus, Option (A) is correct.

2. Work Done (ww)

Work done (ww) is a path function, given by w=−∫P dVw = -\int P \, dV (using IUPAC convention). The magnitude of work corresponds to the area under the curve on a P-VP\text{-}V diagram.

  • For the two-step path X→Y→ZX \to Y \to Z: wX→Y→Z=wX→Y+wY→Zw_{X \to Y \to Z} = w_{X \to Y} + w_{Y \to Z}

  • Step Y→ZY \to Z is an isochoric process (constant volume, dV=0dV = 0), so no work is done along this step: wY→Z=0w_{Y \to Z} = 0

  • Substituting wY→Z=0w_{Y \to Z} = 0: wX→Y→Z=wX→Y+0=wX→Yw_{X \to Y \to Z} = w_{X \to Y} + 0 = w_{X \to Y}

Thus, Option (C) is correct.

3. Evaluation of Options (B) and (D)

  • Option (B): wX→Zw_{X \to Z} is the work done along the direct straight-line path X→ZX \to Z. The area under line XZXZ is less than the area under line XYXY (which represents wX→Yw_{X \to Y}). Therefore: wX→Z≠wX→Y+wY→Zw_{X \to Z} \neq w_{X \to Y} + w_{Y \to Z} Hence, Option (B) is incorrect.

  • Option (D): Since YY and ZZ are at different thermodynamic states (different pressures at the same volume), ΔSY→Z≠0\Delta S_{Y \to Z} \neq 0. Therefore: ΔSX→Y→Z=ΔSX→Y+ΔSY→Z≠ΔSX→Y\Delta S_{X \to Y \to Z} = \Delta S_{X \to Y} + \Delta S_{Y \to Z} \neq \Delta S_{X \to Y} Hence, Option (D) is incorrect.

Conclusion

The correct choices are A and C.

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