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1 mole of an ideal gas at an initial temperature of $T \mathrm{~K}$ does $6 R$ joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$, then final temperature of the gas will be
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The correct answer is:
$(T-4) \mathrm{K}$
Work done by gas in adiabatic process,
$\begin{array}{rlrl}
& & W & =\frac{n R\left(T_i-T_f\right)}{\gamma-1} \\
\therefore & 6 R & =\frac{1 \times R\left(T-T_2\right)}{\left(\frac{5}{3}-1\right)} \\
& \text { or } & 6 & =\frac{T-T_2}{2 / 3} \\
& \text { or } & T & -T_2=4 \text { or } T_2=(T-4) \mathrm{K}
\end{array}$
$\begin{array}{rlrl}
& & W & =\frac{n R\left(T_i-T_f\right)}{\gamma-1} \\
\therefore & 6 R & =\frac{1 \times R\left(T-T_2\right)}{\left(\frac{5}{3}-1\right)} \\
& \text { or } & 6 & =\frac{T-T_2}{2 / 3} \\
& \text { or } & T & -T_2=4 \text { or } T_2=(T-4) \mathrm{K}
\end{array}$
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