Q. $P$-$V$ diagram of an ideal gas is as shown in figure. Work done by the gas in process $ABCD$ is
A) $4P_{0}V_{0}$
B) $2P_{0}V_{0}$
C) $3P_{0}V_{0}$
D) $P_{0}V_{0}$
Explanation
(c) $W_{AB} = - P_{0}V_{0},\ W_{BC} = 0$ and $W_{CD} = 4P_{0}V_{0}$
$\Rightarrow W_{ABCD} = - P_{0}V_{0} + 0 + 4P_{0}V_{0} = 3P_{0}V_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.