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Question

When one mole of monoatomic ideal gas at T K undergoes adiabatic change under a constant external pressure of 1 atm changes volume from 1 litre to 2 litres. 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.

For an adiabatic process,

TVγ1=constant
Therefore,
T1V1γ1=T2V2γ1

γ=53(the gas is monoatomic)

T1V1531=T2V2531

T1V123=T2V223

Given that:-
V1=1L
V2=2L
T1=TK

T(1)23=T2(2)23

T2=T223

Hence the final temperature will be T22/3.

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