A thermodynamical system undergoes cyclic process $\mathrm{ABCDA}$ as shown in the figure. Work done by the system is

    

• Correct answer
  A

  Zero
• B

  $2P_0{V}_0$
• C

  $P_0{V}_0$

   
• D

  $\frac{3}{2}{P}_0{V}_0$

Solution

As can be seen from figure,

   

${\text{W}}_{\text{BCOB}}=-\text{area of}\Delta \text{BCO}=-\frac{{\text{P}}_0{\text{V}}_0}{2}$

${\text{W}}_{\text{AODA}}=+\text{area of}\Delta \text{AOD}=+\frac{{\text{P}}_0{\text{V}}_0}{2}$

$\therefore {\text{W}}_{\text{net}}={\text{W}}_{\text{BCOB}}+{\text{W}}_{\text{AODA}}=-\frac{{\text{P}}_0{\text{V}}_0}{2}+\frac{{\text{P}}_0{\text{V}}_0}{2}=\text{zero}$

Hence, the correct answer is option: 'A'.