An ideal gas is taken through the cycle (A arrow) (B arrow C arrow...

An ideal gas is taken through the cycle (A \rightarrow B \rightarrow C \rightarrow A) as shown in the figure,
From the graph, $$\begin{aligned} W_{A B} & =P \times d V \\ & =10 \times(2-1)=10 \mathrm{~J} \end{aligned}$$) Similarly, (W_{B C}=P \times(0)=0 \mathrm{~J}) According$to$the first law of thermodynamics (\Delta Q=\Delta U+W) Here, (\Delta U = 0)$$\begin{aligned} & \therefore \quad Q=W \\ & 5=W_{A B}+W_{B C}+W_{C A} \\ & W_{C A}=5-W_{A B}-W_{B C}=5-10-0 \\ & W_{C A}=-5 \mathrm{~J} \\ & \therefore \quad\left|W_{C A}\right|=5 \mathrm{~J} \\ \end{aligned}$$) Hence, the magnitude of work done during the process (C \rightarrow A) is (5 \mathrm{~J}).