Q11 / 60 · JEE Advanced 2012 (Paper 1) · Chemistry

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]. Figure

  • Correct answer
    A
    $\Delta \mathrm{S}_{x \rightarrow z}=\Delta \mathrm{S}_{x \rightarrow y}+\Delta \mathrm{S}_{y \rightarrow z}$
  • B
    $\mathrm{w}_{x \rightarrow z}=\mathrm{w}_{x \rightarrow y}+\mathrm{w}_{y \rightarrow z}$
  • Correct answer
    C
    $\mathrm{w}_{x \rightarrow y \rightarrow z}=\mathrm{w}_{x \rightarrow y}$
  • D
    $\Delta \mathrm{S}_{x \rightarrow y \rightarrow z}=\Delta \mathrm{S}_{x \rightarrow y}$

Solution

$\Delta S_{X \rightarrow Z}=\Delta S_{X \rightarrow Y}+\Delta S_{Y \rightarrow Z}$ [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].

Hence, the correct answer is options: 'A', 'C'.

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