IIT-JEE 2012 · Paper 1 · Chemistry Q35

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ChemistryMultiple correct+4Question 35 of 60

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 $\Delta S$ as change in entropy and $w$ as work done]

P (atmosphere) against V (liter): path X → Y is horizontal (isobaric expansion), Y → Z is vertical (isochoric pressure drop); a dashed direct path goes from X to Z.

\Delta S_{x\to z}=\Delta S_{x\to y}+\Delta S_{y\to z}

w_{x\to z}=w_{x\to y}+w_{y\to z}

w_{x\to y\to z}=w_{x\to y}

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

Answer and worked solution

Official answer: A, C

(A) Entropy is a state function, so $\Delta S$ from X to Z is the same by any path: $\Delta S_{x\to z}=\Delta S_{x\to y}+\Delta S_{y\to z}$ . True.
(B) Work depends on the path. The direct path X → Z (the dashed line) encloses a different area under the curve from X → Y → Z. False in general.
(C) Y → Z is at constant volume, so $w_{y\to z}=0$ and $w_{x\to y\to z}=w_{x\to y}$ . True.
(D) $\Delta S_{y\to z}\ne0$ (the temperature changes on the isochoric step), so $\Delta S_{x\to y\to z}\ne\Delta S_{x\to y}$ . False.

(A), (C)

Topics:Entropy First law and enthalpy

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IIT-JEE 2012 · Paper 1, Chemistry question 35