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Question

When one mole of monoatomic ideal gas at temperature T undergoes adiabatic change reversibly, change in volume is from 1 L to 2L, the final temperature in Kelvin would be

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 A T22/3
We know the relation for an adiabatic reversible process :
T1T2=(V2V1)γ1
where, T1= initial temperature
T2= final temperature
V1= initial volume
V2= final volume
γ=CpCv=53 (for monoatomic gas)
so,
T1T2=(21)(5/3)1=2(2/3)
T2=T12(2/3)

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