Question

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

• A. Zero
• B. $2P_0{V}_0$
• C. $P_0{V}_0$
• D. $\frac{3}{2}{P}_0{V}_0$

Answer 1

Answer: (A)

As can be seen from figure,

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