Heat and Thermodynamics 5 Question 37

38. The figure shows the $p-V$ plot 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,

(2009)

(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 $A B C D A$

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Answer:

Correct Answer: 38. $(b, d)$

Solution:

  1. (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 internal 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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