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Question

The PV diagram of 2 gm of helium gas for a certain process is shown in the figure. Find the amount of heat given to the gas during the process ?


A
4P0V0
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B
6P0V0
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C
4.5P0V0
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D
2P0V0
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Solution

The correct option is B 6P0V0
Since, Helium is a monoatomic gas,
Number of degrees of freedom (f)=3
To find the change in internal energy from A B, we use
Δ U=f2μRΔ T
where μ is the number of moles of helium gas.
Given, mass of helium (m)=2 gm
Thus, the number of moles of helium gas
μ= MassMolecular weight=22=1
Using ideal gas equation, we can write Δ(PV)=μRΔT
i.e Δ U=f2(PfVfPiVi)
Substituting the data given in the diagram,
ΔU=32(2P0×2V0P0×V0)=92P0V0
Work done during process A B is equal to the area enclosed by the graph above the volume axis i.e.
WAB=12(P0+2P0)×(2V0V0)=32P0V0
Hence, using first law of thermodynamics,
ΔQ=ΔU+ΔW=92P0V0+32P0V0=6P0V0
Hence, option (b) is the correct answer.

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