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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Advanced 2012 · Paper 1 · Chemistry · Thermodynamics (C)

For an ideal gas, consider only P-V work in going from an initial...

For an ideal gas, consider only P-V work in going from an initial state X\mathrm{X} to the final state Z\mathrm{Z}. The final state Z\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 ΔS\Delta \mathrm{S} as change in entropy and w as work done]. 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

Options

  1. AΔSx→z=ΔSx→y+ΔSy→z\Delta \mathrm{S}_{x \rightarrow z}=\Delta \mathrm{S}_{x \rightarrow y}+\Delta \mathrm{S}_{y \rightarrow z}
  2. Bwx→z=wx→y+wy→z\mathrm{w}_{x \rightarrow z}=\mathrm{w}_{x \rightarrow y}+\mathrm{w}_{y \rightarrow z}
  3. Cwx→y→z=wx→y\mathrm{w}_{x \rightarrow y \rightarrow z}=\mathrm{w}_{x \rightarrow y}
  4. DΔSx→y→z=ΔSx→y\Delta \mathrm{S}_{x \rightarrow y \rightarrow z}=\Delta \mathrm{S}_{x \rightarrow y}

Correct answer

A. ΔSx→z=ΔSx→y+ΔSy→z\Delta \mathrm{S}_{x \rightarrow z}=\Delta \mathrm{S}_{x \rightarrow y}+\Delta \mathrm{S}_{y \rightarrow z}

Step-by-step solution

ΔSX→Z=ΔSX→Y+ΔSY→Z\Delta S_{X \rightarrow Z}=\Delta S_{X \rightarrow Y}+\Delta S_{Y \rightarrow Z} [Entropy is a state function, hence additive] wX→Y→Z=wX→Yw_{X \rightarrow Y \rightarrow Z}=w_{X \rightarrow Y} [Work done in Y→ZY \rightarrow Z is zero because it is an isochoric process].
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