Question:
A gas in a closed container undergoes the cycle ABCA as shown in the figure. Find the net heat absorbed by the gas after it has completed 10 cycles.
A gas in a closed container undergoes the cycle ABCA as shown in the figure. Find the net heat absorbed by the gas after it has completed 10 cycles.
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For cyclic PV processes, net work done is area enclosed in the PV diagram; heat absorbed over cycle equals work done: \(Q = W_{\text{cycle}}\). Sign depends on cycle direction.
Updated On: Jul 18, 2026
- -1.5 kJ
- +1.5 kJ
- +2.25 kJ
- -2.25 kJ
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The Correct Option is A
Solution and Explanation
Step 1: Recall net work done in a PV cycle.
For a gas undergoing a cyclic process, the net work done is the area enclosed by the cycle in the PV diagram. The heat absorbed is related to work done by the first law: \(\Delta U = Q - W\). Over a complete cycle, \(\Delta U = 0 \implies Q = W\).
Step 2: Determine work done per cycle.
From the PV diagram, the area of the cycle ABCA is negative (clockwise cycle), indicating net work done by the gas is negative: \(W_{\text{cycle}} = -0.15 \, \text{kJ}\) per cycle.
Step 3: Consider number of cycles.
Net heat absorbed after 10 cycles:
\[ Q_{\text{total}} = 10 \cdot W_{\text{cycle}} = 10 \cdot (-0.15) = -1.5 \, \text{kJ} \]
Step 4: Verify sign convention.
Negative sign indicates the gas releases heat overall; consistent with clockwise PV cycle.
Step 5: Dimensional consistency.
Work/heat in kJ, number of cycles dimensionless; calculation consistent.
Step 6: Final conclusion.
Hence, the net heat absorbed by the gas after 10 cycles is:
\[ \boxed{-1.5 \, \text{kJ}} \]
For a gas undergoing a cyclic process, the net work done is the area enclosed by the cycle in the PV diagram. The heat absorbed is related to work done by the first law: \(\Delta U = Q - W\). Over a complete cycle, \(\Delta U = 0 \implies Q = W\).
Step 2: Determine work done per cycle.
From the PV diagram, the area of the cycle ABCA is negative (clockwise cycle), indicating net work done by the gas is negative: \(W_{\text{cycle}} = -0.15 \, \text{kJ}\) per cycle.
Step 3: Consider number of cycles.
Net heat absorbed after 10 cycles:
\[ Q_{\text{total}} = 10 \cdot W_{\text{cycle}} = 10 \cdot (-0.15) = -1.5 \, \text{kJ} \]
Step 4: Verify sign convention.
Negative sign indicates the gas releases heat overall; consistent with clockwise PV cycle.
Step 5: Dimensional consistency.
Work/heat in kJ, number of cycles dimensionless; calculation consistent.
Step 6: Final conclusion.
Hence, the net heat absorbed by the gas after 10 cycles is:
\[ \boxed{-1.5 \, \text{kJ}} \]
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