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Question

One mole of a monoatomic ideal gas is taken along ABCA as shown in the diagram. The net heat absorbed by the gas in the given cycle is
217389_4fc396bbcf5c4ec28939d1482bfd5a22.png

A
PV
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B
PV2
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C
2PV
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D
4PV
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Solution

The correct option is B PV2
Work done by gas = Area under PV graph for the process.
w=PdV
For the given graph
w=12(2PP)(2VV)=12PV
The first law of thermodynamics gives us the relation:
q=ΔU+w
For a cyclic process, ΔU=0
Hence q = w = 12PV
Hence option B is correct.

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