A thermodynamic system is taken though the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is
A
2PV
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B
12PV
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C
PV
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D
4PV
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Solution
The correct option is A2PV Given: Cyclic process ABCDA
We know that,
In cyclic process ΔU=0
So, from the first law of the thermodynamics, ΔQ=ΔW= area under the PV curve. ΔW=(2P−P)(3V−V)=2PV ∴Heat rejected=2PV
Hence, option (C) is correct.