One mole of an ideal gas at an initial temperature of $T\, K$ does $6\, R\, joules$ of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $\frac{5}{3}$ , the final temperature of gas will be
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$\Delta \mathrm{U}=\mu \mathrm{C}_{\mathrm{V}} \Delta \mathrm{T} $ and $ 0=\mathrm{W}+\Delta \mathrm{U}$
$\Rightarrow \Delta \mathrm{U}=-6 \mathrm{R} \quad(\because \mathrm{W}=6 \mathrm{R})$
Therefore $-6 R=1\left(\frac{R}{\gamma-1}\right) \Delta T=\frac{3}{2} R \Delta T$
$\Rightarrow \Delta \mathrm{T}=-4 \Rightarrow \mathrm{T}_{\mathrm{final}}=(\mathrm{T}-4) \mathrm{K}$
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