ExamBro
ExamBro
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,

  1. A the process during the path \(A \rightarrow B\) is isothermal
  2. B heat flows out of the gas during the path \(B \rightarrow C \rightarrow D\)
  3. C work done during the path \(A \rightarrow B \rightarrow C\) is zero
  4. D positive work is done by the gas in the cycle \(A B C D A\).
Verified Solution

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.
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app