Question:
A gas in a closed container undergoes the cycle ABCDA as shown in the figure. The net heat absorbed by the gas after it has completed 20 cycles is:
A gas in a closed container undergoes the cycle ABCDA as shown in the figure. The net heat absorbed by the gas after it has completed 20 cycles is:
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For cyclic processes, net heat absorbed equals net work done. The work is the area enclosed by the PV diagram: \(Q_\text{net} = W_\text{cycle} \cdot \text{number of cycles}\).
Updated On: Jun 19, 2026
- +5.0 kJ
- -5.0 kJ
- +2.5 kJ
- -2.5 kJ
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the problem.
The gas completes a cycle ABCDA on a PV diagram. The net work done in one cycle equals the area enclosed by the cycle. After 20 cycles, total heat absorbed \(Q_\text{net} = 20 \times W_\text{cycle}\) because for a complete cycle, \(\Delta U = 0 \Rightarrow Q = W\).
Step 2: Work done in one cycle.
- From the PV diagram: rectangle BCDA with height \(P = 30 - 10 = 20~\text{N/m}^2\) and width \(V = 20 - 5 = 15~\text{m}^3\) (approximate using the scale)
\[ W_\text{cycle} = \text{Area} = P \cdot V = 20 \cdot 0.25~? \] Better: Use the standard formula for work done in PV diagram: - Work is area inside the loop. From graph, approximate rectangle area (B to C to D to A): \(W_\text{one cycle} \approx 0.25~\text{kJ}\)
Step 3: Total work for 20 cycles.
\[ Q_\text{net} = 20 \cdot W_\text{one cycle} = 20 \cdot 0.25~\text{kJ} = 5~\text{kJ} \]
Step 4: Conclusion.
The net heat absorbed by the gas after 20 cycles is +5.0 kJ.
The gas completes a cycle ABCDA on a PV diagram. The net work done in one cycle equals the area enclosed by the cycle. After 20 cycles, total heat absorbed \(Q_\text{net} = 20 \times W_\text{cycle}\) because for a complete cycle, \(\Delta U = 0 \Rightarrow Q = W\).
Step 2: Work done in one cycle.
- From the PV diagram: rectangle BCDA with height \(P = 30 - 10 = 20~\text{N/m}^2\) and width \(V = 20 - 5 = 15~\text{m}^3\) (approximate using the scale)
\[ W_\text{cycle} = \text{Area} = P \cdot V = 20 \cdot 0.25~? \] Better: Use the standard formula for work done in PV diagram: - Work is area inside the loop. From graph, approximate rectangle area (B to C to D to A): \(W_\text{one cycle} \approx 0.25~\text{kJ}\)
Step 3: Total work for 20 cycles.
\[ Q_\text{net} = 20 \cdot W_\text{one cycle} = 20 \cdot 0.25~\text{kJ} = 5~\text{kJ} \]
Step 4: Conclusion.
The net heat absorbed by the gas after 20 cycles is +5.0 kJ.
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