Question

A thermodynamic system undergoes cyclic process $ABCDA$ as shown in figure. The work done by the system during one complete cycle is

• A. $P_0{V}_0$
• B. $2P_0{V}_0$
• C. $\frac{P_0{V}_0}{2}$
• D. Zero

Answer 1

The correct option is D Zero

In $ABCDA$ process, $BCO$ is a clockwise cycle and $AOD$ is an anticlockwise cycle.
So, we can write that
$W_{BCOB}=-\text{Area of triangle BCO}=-\frac{1}{2}(3P_0-2P_0)(2V_0-V_0)=-\frac{P_0{V}_0}{2}$
$W_{AODA}=\text{Area of triangle AOD}=\frac{1}{2}(2P_0-P_0)(2V_0-V_0)=+\frac{P_0{V}_0}{2}$
Net work done in cycle $ABCDA$ is given by
$W=W_{BCOB}+W_{AODA}$
$=\frac{P_0{V}_0}{2}-\frac{P_0{V}_0}{2}=0$
Hence, option (d) is the correct answer.