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Question

One mole of a monoatomic ideal gas undergoes the process AB in the given PV diagram. The specific heat for this process is


A
3R/2
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B
13R/6
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C
5R/2
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D
2R
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Solution

The correct option is B 13R/6
W=vfviPdV
Work done is area under PV graph, ie area of trapezium,hence
W=12(6P0+3P0)4V0
W=18P0V0
ΔU=PfVfPiVi5/31=30P0V03P0V02/3
=81P0V02
ΔQ=W+ΔU=1172P0V0
nC(TfTi)=1172P0V0
C(6P0×5V0R3P0V0R)=1172P0V0
C=1172P0V027P0V0R=13R6

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