An ideal gas is taken through the cycle \( A\rightarrow B\rightarrow C\rightarrow A \) as shown in figure. If the net heat supplied to the gas in the cycle is 10 J, the work done in the process \( C\rightarrow A \) is:
Show Hint
The net work done during a cyclic process on a P-V diagram is equal to the area enclosed by the loop. Since the cycle goes counter-clockwise here, the net work must be negative, which serves as a quick sanity check for your signs.
Concept:
According to the First Law of Thermodynamics for a complete operating cycle, the net change in internal energy over a closed loop is zero \( (\Delta U_{\text{net}} = 0) \), meaning the net heat supplied equals the net work done:
\[
Q_{\text{net}} = W_{\text{net}} = W_{AB} + W_{BC} + W_{CA}
\]
Step 1: Analyzing the individual process paths from the graph.
Looking at the V-P coordinates in the diagram:
• Process \( A \rightarrow B \): The pressure stays constant at \( P = 20 \, \text{Nm}^{-2} \) while the volume increases from 1 to 2. This is an isobaric process:
\[ W_{AB} = P \cdot \Delta V = 20 \times (2 - 1) = 20 \, \text{J} \]
• Process \( B \rightarrow C \): The volume stays constant at \( V = 2 \) while the pressure drops. This is an isochoric process, which does zero work:
\[ W_{BC} = 0 \]
Step 2: Applying the cyclic work identity to find \( W_{CA} \).
We are given that the total net heat supplied is \( Q_{\text{net}} = 10 \, \text{J} \):
\[
10 = W_{AB} + W_{BC} + W_{CA}
\]
Substitute our derived values:
\[
10 = 20 + 0 + W_{CA} \implies W_{CA} = 10 - 20 = -10 \, \text{J}
\]
Thus, the work done during process \( C \rightarrow A \) is \( -10 \, \text{J} \).
