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Question

# Two moles of an ideal monoatomic gas occupies a volume V at 27∘C. The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

A
189 K, 2.7 kJ
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B
195 K, 2.7 kJ
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C
189 K, 2.7 kJ
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D
195 K, 2.7 kJ
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Solution

## The correct option is C 189 K, −2.7 kJFor adiabatic process relation of temperature and volume is, T2Vγ−12=T1Vγ−11⇒T2(2V)2/3=300(V)2/3[γ=53 for monoatomic gases ]⇒T2=30022/3≈189K Also, in adiabatic process, ΔQ=0,ΔU=−ΔW or ΔU=−nR(ΔT)γ−1=−2×32×253(300−189) ≈−2.7kJT2≈189K,ΔU≈−2.7kJ

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