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Question

# At 27∘C, three moles of an ideal diatomic gas occupy a volume V. If the gas expands adiabatically to a volume 3V, calculate the work done by the gas during the process. (R=8.31 J/mol-K,30.4=1.55)

A
2767 J
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B
6669 J
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C
6000 J
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D
6583 J
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Solution

## The correct option is B 6669 JGiven that, Initial temperature, T1=27∘C=27+273=300 K Initial volume, V1=V Final volume, V2=3V γ=1.4 (for diatomic gas) For adiabatic process, TVγ−1=constant T1V1γ−1=T2V2γ−1 T2=T1(V1V2)γ−1 ∴T2=300(V3V)1.4−1=193 K As we know that, change in internal energy ΔU=nCvΔT =3×(52R)ΔT =3×52×8.31×(193−300) =−6669 J Negative sign means internal energy will decrease. From first law of thermodynamics, ΔQ=ΔU+ΔW ⇒ΔW=−ΔU=−(−6669)=6669 J(ΔQ=0 for adiabatic process) Alternative: Work done for an adiabatic process, W=nR(T1−T2)γ−1 W=3×8.31(300−193)1.4−1≈6669 J Tips: In a closed adiabatic process, work done is equal to the change in internal energy of the system.

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