Q44 / 57 · JEE Advanced 2009 (Paper 2) · Physics

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, Figure

  • A
    the process during the path $A \rightarrow B$ is isothermal
  • Correct answer
    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
  • Correct answer
    D
    positive work is done by the gas in the cycle $A B C D A$.

Solution

(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.

Hence, the correct answer is options: 'B', 'D'.

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