Thermodynamics

JEE Main level, single correct, medium

64 of 190

Question 64 of 190

In the following p−Vp-V diagram the equation of state along the curved path is given by (V−2)2=4ap(V-2)^{2}=4 a p where aa is a constant. The total work done in the closed path is

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Answer: (B) −13a-\frac{1}{3 a}

From the equation of state (V−2)2=4ap(V-2)^2 = 4ap, we get p=(V−2)24ap = \frac{(V-2)^2}{4a}.
At V = 1: p=14ap = \frac{1}{4a} and at V = 3: p=14ap = \frac{1}{4a}.
The curved path A→B→C follows the given equation, while C→A is a straight line at constant pressure.
Work along the curved path:
WABC=∫13(V−2)24adV=14a[(V−2)33]13=14a⋅23=16aW_{ABC} = \int_1^3 \frac{(V-2)^2}{4a} dV = \frac{1}{4a}\left[\frac{(V-2)^3}{3}\right]_1^3 = \frac{1}{4a} \cdot \frac{2}{3} = \frac{1}{6a}.
Work along C→A at constant pressure:
WCA=14a(1−3)=−12aW_{CA} = \frac{1}{4a}(1-3) = -\frac{1}{2a}.
Total work: Wtotal=16a−12a=−13aW_{total} = \frac{1}{6a} - \frac{1}{2a} = -\frac{1}{3a}.

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