JEE Advanced · Chemistry · 6. Thermodynamics (C)
For an ideal gas, consider only P-V work in going from an initial state \(\mathrm{X}\) to the final state \(\mathrm{Z}\). The final state \(\mathrm{Z}\) can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct?
[Take \(\Delta \mathrm{S}\) as change in entropy and w as work done].
- A \(\Delta \mathrm{S}_{x \rightarrow z}=\Delta \mathrm{S}_{x \rightarrow y}+\Delta \mathrm{S}_{y \rightarrow z}\)
- B \(\mathrm{w}_{x \rightarrow z}=\mathrm{w}_{x \rightarrow y}+\mathrm{w}_{y \rightarrow z}\)
- C \(\mathrm{w}_{x \rightarrow y \rightarrow z}=\mathrm{w}_{x \rightarrow y}\)
- D \(\Delta \mathrm{S}_{x \rightarrow y \rightarrow z}=\Delta \mathrm{S}_{x \rightarrow y}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{w}_{x \rightarrow y \rightarrow z}=\mathrm{w}_{x \rightarrow y}\)
Step-by-step Solution
Detailed explanation
\(\Delta S_{X \rightarrow Z}=\Delta S_{X \rightarrow Y}+\Delta S_{Y \rightarrow Z}\) [Entropy is a state function, hence additive]
\(w_{X \rightarrow Y \rightarrow Z}=w_{X \rightarrow Y}\) [Work done in \(Y \rightarrow Z\) is zero because it is an isochoric process].
\(w_{X \rightarrow Y \rightarrow Z}=w_{X \rightarrow Y}\) [Work done in \(Y \rightarrow Z\) is zero because it is an isochoric process].
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