One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be
(1) (T + 2.4) K
(2) (T – 2.4) K
(3) (T + 4) K
(4) (T – 4) K


Dear Student,

workdone=nRTi-Tfγ-1Ti=T K  we have to find the final temp.n=1,γ=5/36R=nRTi-Tfγ-16=T-Tf53-1T-Tf=10-6=4Tf=T-4  final tempRegards

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