Question:
A gas in a closed container undergoes the cycle \(ABCA\) as shown in the figure. The net heat released by the gas after it has undergone \(20\) cycles is
A gas in a closed container undergoes the cycle \(ABCA\) as shown in the figure. The net heat released by the gas after it has undergone \(20\) cycles is
Show Hint
In a cyclic process,
\[
\Delta U=0
\]
so the net heat equals the net work. On a \(P-V\) graph, work done equals the area enclosed by the cycle.
Updated On: Jun 25, 2026
- \(3\text{ kJ}\)
- \(2\text{ kJ}\)
- \(1.5\text{ kJ}\)
- \(4.5\text{ kJ}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Identify the coordinates from the \(P-V\) graph.
From the graph: \[ A=(5,10),\quad B=(20,30),\quad C=(5,30) \]
Step 2: Find the area enclosed by the cycle.
The enclosed region is a right triangle.
Base along volume axis: \[ 20-5=15\text{ m}^3 \] Height along pressure axis: \[ 30-10=20\text{ N m}^{-2} \] Area: \[ \frac12\times 15\times 20=150\text{ J} \]
Step 3: Relate area with heat released.
For one complete cycle, \[ \Delta U=0 \] So, \[ Q=W \] The cycle \(ABCA\) is anticlockwise, so work done by the gas is negative. Hence, heat is released by the gas.
Heat released in one cycle: \[ 150\text{ J} \]
Step 4: Find heat released in \(20\) cycles.
\[ Q=20\times 150 \] \[ Q=3000\text{ J} \] \[ Q=3\text{ kJ} \]
Step 5: Final conclusion.
Therefore, the net heat released by the gas is \[ \boxed{3\text{ kJ}} \]
From the graph: \[ A=(5,10),\quad B=(20,30),\quad C=(5,30) \]
Step 2: Find the area enclosed by the cycle.
The enclosed region is a right triangle.
Base along volume axis: \[ 20-5=15\text{ m}^3 \] Height along pressure axis: \[ 30-10=20\text{ N m}^{-2} \] Area: \[ \frac12\times 15\times 20=150\text{ J} \]
Step 3: Relate area with heat released.
For one complete cycle, \[ \Delta U=0 \] So, \[ Q=W \] The cycle \(ABCA\) is anticlockwise, so work done by the gas is negative. Hence, heat is released by the gas.
Heat released in one cycle: \[ 150\text{ J} \]
Step 4: Find heat released in \(20\) cycles.
\[ Q=20\times 150 \] \[ Q=3000\text{ J} \] \[ Q=3\text{ kJ} \]
Step 5: Final conclusion.
Therefore, the net heat released by the gas is \[ \boxed{3\text{ kJ}} \]
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