3TV5/3-1 = T2 ((2)V)5/3-1
T = T2(22/3)
300/(22/3) = T2
2R/(2/3) (T2-T)
At 27 degree C two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to volume 2V
calculate the final temp, change in internal energy and work done.
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10 Answers
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21Point!
its not given that Constant Ext. Pressure or nethin lyk dat,
Then v can take REVERSIBLE, jus lyk IITIM... took
1hmm.......
T1=300K. V1=V. V2=2V. T2=?
mnoatomic => Cv=3/2, Cp=5/2 , γ=5/3
TVγ-1 = constant.
300.V2/3 = T2. (2V)2/3
=> [300/T2] = 3√4
=> T2 = 300/ 3√4 K.
1ΔU= nCvΔT. = 2X3/2X (300 - 300/3√4)
= 3.(300 - 300/3√4)
work done = nR(T1-T2)/(γ-1) all the terms we know already.
11FOR TEMPERATURE USE
T1VIγ-1=T2V2γ-1
WE GET P1=23/5P2
AND 22/5T1=T2
T1=300K
SO
300 X 22/5=T2
ΔE =nCvdt
=3/2 x 2 x R x (300 X 22/5-300)
as it is adiabatic so ΔQ=0
by applying ΔE=ΔQ-ΔW
we can find the work done