A thermodynamic system is taken through the cycle $\mathrm{ABCD}$ as shown in the figure. Heat rejected by the gas during the cycle is

    

In a cyclic process $\Delta \text{U}=0$. In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.

               


$$\begin{gathered} \therefore \Delta \text{W}=-\text{Area of rectangle ABCD} \\ =-\text{P}\left(2\text{V}\right) \\ =-2\text{PV} \end{gathered}$$

According to the first law of thermodynamics

$\Delta \text{Q}=\Delta \text{U}+\Delta \text{W}$     or   $\Delta \text{Q}=\Delta \text{W }\left(\text{As }\Delta \text{U}=0\right)$

i.e., heat supplied to the system is equal to the work done

So heat absorbed, $\Delta \text{Q}=\Delta \text{W}=-2\text{PV}$

$\therefore$Heat rejected by the gas = $2PV$