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Question

One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in the figure. Calculate the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB.
1011110_5c9ebd8b55e64bbaab5d46db4991ffde.png

A
72P0V0.
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B
82P0V0.
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C
32P0V0.
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D
52P0V0.
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Solution

The correct option is D 52P0V0.
Given :
solution :
QAB=nCVΔT

n(32R)(TBTA)

32(pBVBpAVA)

3p0V0

QCA=nCpδT

n(52R)(TATC)

52(pAVApCVC)

52(p0V02p0V0)=52p0V0


hence for the b part question answer is 52p0V0
hence the correct opt : D

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