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 :
Answer & explanation
Correct answer: option 1
dU = 0
therefore, by First Law of Thermodynamics : dQcyclic = dWcyclic
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