Heat And Thermodynamics Ques 157
- 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 $A B C D A$
Show Answer
Answer:
Correct Answer: 157.$(b, d)$
Solution:
Formula:
- (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.
BETA
