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Question

At 27 C two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V.
Given (12)23=0.63.

A
The final temperatre of gass is 500 K.
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B
Change in internal energy will be 2767 J.
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C
The work done by the gas during process 2767 J.
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D
The work done by the gas during process 1383.5 J.
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Solution

The correct options are
B Change in internal energy will be 2767 J.
C The work done by the gas during process 2767 J.
Given, T1=27C=300K , n=2 moles, γ=5/3
V1=V,V2=2V,CV=3R2 for monatomic gas

i) To find final temperature: T2
In adiabatic process, PVγ = constant,
TVγ1 = constant,
P1γTγ=constant.
T1Vγ11=T2Vγ12T2=T1(V1V2)γ1
or T2=300(V2V)531=300(12)2/3=300×0.63
or T2=189 K

ii) To find change in internal energy =ΔU.
ΔU=nCVΔT
or ΔU=3×8.31×(189300)
or ΔU=2767J

iii) Work done
The process is adiabatic, so ΔQ=0
According to first law of thermodynamics
ΔQ=ΔW+ΔU0=ΔW+(2767)
or ΔW=2767J.

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