Solution:
In an adiabatic process, $\Delta U = -W = -6R$. Also, $\Delta U = n C_v \Delta T = \frac{R}{\gamma - 1}(T_2 - T)$. With $\gamma = 5/3$, we get $\frac{3}{2}(T_2 - T) = -6$, leading to $T_2 = T - 4\text{ K}$.
One mole of an ideal gas at an initial temperatures of $T\text{ K}$ does $6\text{ R}$ 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 temperatures of gas will be: (2004)
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