An ideal gas is taken through A \(\rightarrow\) B \(\rightarrow\) C \(\rightarrow\) A, as shown in the given figure. If the et heat supplied to the gas in the cycle is 15 J, the work done by the gas in the process C \(\rightarrow\) A is : 

-5 J

dU = 0

therefore, by First Law of Thermodynamics : dQ_cyclic = dW_cyclic

Since B \(\rightarrow\) C is an isochoric process 

\(\Rightarrow dW_{B \rightarrow C} = 0\)

\(\Rightarrow 15 = dW_{A \rightarrow B} + dW_{B \rightarrow C} + dW_{C \rightarrow A} = 0\)

\(\Rightarrow 15 = 10(2-1) + 0 + dW_{C \rightarrow A}\)

\(dW_{C \rightarrow A} = -5 \)J