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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

$workdone=\frac{nR\left({T}_{i}-{T}_{f}\right)}{\gamma -1}\phantom{\rule{0ex}{0ex}}{T}_{i}=TKwehavetofindthefinaltemp.n=1,\gamma =5/3\phantom{\rule{0ex}{0ex}}6R=\frac{nR\left({T}_{i}-{T}_{f}\right)}{\gamma -1}\phantom{\rule{0ex}{0ex}}6=\frac{\left(T-{T}_{f}\right)}{{\displaystyle \frac{5}{3}}-1}\phantom{\rule{0ex}{0ex}}T-{T}_{f}=10-6=4\phantom{\rule{0ex}{0ex}}{T}_{f}=T-4finaltemp\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Regards$

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