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Question

When 1 mole of a mono atomic ideal gas a T K undergoes an adiabatic change under a constant external pressure of 1 atm, its volume changes from 1L to 2L. The final temperature in Kelvin is:

A
T22/3
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B
T+23×0.0821
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C
T
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D
T23×0.0821
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Solution

The correct option is D T23×0.0821
In an adiabatic process, Q = 0
From first law of thermodynamics,
ΔU=Q+W
ΔU=W...1
Now W = P(V2V1) (given Pext=1atm
= -1(2-1) atm L = -1 atm L
For calculating ΔU we need to calculate Cv
Cv = Rγ1
for monoatomic gas γ=53
Hence Cv=3R2
Now using eq. 1
ΔU=nCvdT
Now using eq. 1
1×3R2(TfTi)=1atmL
Assuming final temperature as Tf and solving for it we get
Tf=Ti23R

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