A thermodynamic system is taken though the cycle $A B C D$ as shown in figure. Heat rejected by the gas during the cycle is

2 P V

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\frac{1}{2} P V

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P V

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4 P V

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Solution

The correct option is A $2 P V$
Given: Cyclic process $A B C D A$

We know that,
In cyclic process
$Δ U = 0$
So, from the first law of the thermodynamics,
$Δ Q = Δ W =$ area under the $P V$ curve.
$Δ W = (2 P - P) (3 V - V) = 2 P V$
$∴ Heat rejected = 2 P V$
Hence, option $(C)$ is correct.

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