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 \(5 \mathrm{~J}\).