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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=27∘C=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, ∴TVγ−1 = constant, P1−γTγ=constant. ∴T1Vγ−11=T2Vγ−12⇒T2=T1(V1V2)γ−1 or T2=300(V2V)53−1=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×(189−300) or ΔU=−2767J iii) Work done The process is adiabatic, so ΔQ=0 According to first law of thermodynamics ΔQ=ΔW+ΔU∴0=ΔW+(−2767) or ΔW=2767J.

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