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Question

In the process shown in the figure for an ideal diatomic gas, the value of q and ΔH respectively are:

A
79.5P1V1 and 94.5P1V1
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B
54.5P1V1 and 94.5P1V1
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C
9P1V1 and 0
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D
79.5P1V1 and defined (P varies)
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Solution

The correct option is A 79.5P1V1 and 94.5P1V1
ω = work under PV curve.
E=nCvT=PfVfPiViγ1
q=Eω
H=γ×E=γγ1(PfVfPiVi)
γ for diatomic =1+25=75
Here, |ω|= area of trapezium
=12×3V1(P1+7P1)=12P1V1
Expansion Hence, ω=12P1V1
E=28P1V1P1V1751=52×27P1V1
=67.5P1V1
q=67.5P1V(12P1V1)
=79.5 P1V1
H=γE=75×52×27 P1V1=94.5 P1V1
[A]

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