One mole of an ideal monoatomic gas at temperature T and volume 3 L expands to 6 L against a constant external pressure of 2 atm under adiabatic conditions. Then the final temperature of the gas will be:
A
T+302R
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B
T−123R
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C
T+203R
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D
T253+1
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Solution
The correct option is BT−123R Let the final temperature be T2.
For an adiabatic process, q = 0.
From the first law of thermodynamics, △U=q+w △U=w △U = nCvdT and w=−Pext(V2−V1) for an adiabatic process.
For an ideal monoatomic gas Cv=3R2 nCv(T2−T)=−P×(V2−V1) 32×R×(T2−T)=−2×(6−3) =32×R×(T2−T)=−6