JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2026
- Medium
- MCQ
In the following \(p-V\) diagram the equation of state along the curved path is given by \((V-2)^2=4\) ap where \(a\) is a constant. The total work done in the closed path is

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Answer & Solution
Correct Answer
(D) \(-\frac{1}{3a}\)
Step-by-step Solution
Detailed explanation
\(w =\) Area of parabola
\(=\frac{2}{3}(\) Area of rectangle AC 31 A\()\)
\(=\frac{2}{3} P _0(3-1)=\frac{4 P _0}{3}\)
When \(V =1\)
\((1-2)^2=4 aP _0\)
\(P _0=\frac{1}{4 a }\)
\(w =\frac{4}{3} P _0=\frac{4}{3} \frac{1}{4 a }=\frac{1}{3 a }\)
\(w _{ gas }=\frac{-1}{3 a }\)
\(=\frac{2}{3}(\) Area of rectangle AC 31 A\()\)
\(=\frac{2}{3} P _0(3-1)=\frac{4 P _0}{3}\)
When \(V =1\)
\((1-2)^2=4 aP _0\)
\(P _0=\frac{1}{4 a }\)
\(w =\frac{4}{3} P _0=\frac{4}{3} \frac{1}{4 a }=\frac{1}{3 a }\)
\(w _{ gas }=\frac{-1}{3 a }\)
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