Q.
An ideal gas is taken via path $ABCA$ as shown in figure. The net work done in the whole cycle is
A) $6P_{1}V_{1}$
B) Zero
C) $3P_{1}V_{1}$
D) $- 3P_{1}V_{1}$
Explanation
(d) The work done in a $PV$ diagram is the area enclosed.
The work done $= \frac{1}{2}\left( 3V_{1} - V_{1} \right).(4P_{1} - P_{1})$
$\Rightarrow W = - 3P_{1}V_{1}$. If the direction of change is clockwise it is positive. Since here it is anticlockwise, work done is negative
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.