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Question

Two moles of an ideal mono-atomic gas is taken through a cyclic process as shown in the PT diagram. In the process BC,PT2 = constant. Then the ratio of heat absorbed and heat released by the gas during the process AB and process BC respectively is:-
1130325_85e989ed769e442ca28a0376880dbd5c.png

A
2
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B
3
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C
5:6
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D
6
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Solution

The correct option is C 5:6
taken=2f=3AB:w=2R(To)=2RT0Δu=f2nRΔT=32×2×rTo=3RT0BC:PT2=constantp.(pvπR)2=constantp3v2=constantw=nR(T0250)123=3nRT0Δu=3nRT0ΔQ=6nRT0Hence,Ratio:56Hence,theoptionCiscorrectanswer.

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