JEE Mains · Physics · STD 11 - 11. thermodynamics
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
- A \(-\frac{1}{a}\)
- B \(+\frac{1}{3a}\)
- C \(\frac{1}{2a}\)
- D \(-\frac{1}{3a}\)
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 \(V =1\) \((1-2)^2=4 aP _0\) \(P _0=\frac{1}{4 a }\) \(w =\frac{4}{3} P _0=\frac{4}{3} \frac{1}{4 a }=\frac{1}{3 a }\) \(w _{ gas }=\frac{-1}{3 a }\)
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