JEE Advanced · Physics · 13. Thermodynamics
The figure shows the \(p-V\) plot of an ideal gas taken through a cycle \(A B C D A\). The part \(A B C\) is a semi-circle and \(C D A\) is half of an ellipse. Then,

- A the process during the path \(A \rightarrow B\) is isothermal
- B heat flows out of the gas during the path \(B \rightarrow C \rightarrow D\)
- C work done during the path \(A \rightarrow B \rightarrow C\) is zero
- D positive work is done by the gas in the cycle \(A B C D A\).
Answer & Solution
Correct Answer
(D) positive work is done by the gas in the cycle \(A B C D A\).
Step-by-step Solution
Detailed explanation
(A) \(p-V\) graph is not rectangular hyperbola. Therefore, process \(A-B\) is not isothermal.
(B) In process \(B C D\), product of \(p V\) (therefore temperature and intemal energy) is decreasing. Further, volume is decreasing. Hence, work done is also negative. Hence, \(Q\) will be negative or heat will flow out of the gas.
(C) \(W_{A B C}=\) positive
(D) For clockwise cycle on \(p-V\) diagram with \(P\) on \(y\)-axis, net work done is positive.
(B) In process \(B C D\), product of \(p V\) (therefore temperature and intemal energy) is decreasing. Further, volume is decreasing. Hence, work done is also negative. Hence, \(Q\) will be negative or heat will flow out of the gas.
(C) \(W_{A B C}=\) positive
(D) For clockwise cycle on \(p-V\) diagram with \(P\) on \(y\)-axis, net work done is positive.
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