Two moles of an ideal monoatomic gas occupies a volume $$V$$ at $${27^ \circ }C.$$  The gas expands adiabatically to a volume $$2V.$$ Calculate $$(a)$$ the final temperature of the gas and $$(b)$$ change in its internal energy.

Solution :
In an adiabatic process
$$\eqalign{ & T{V^{\gamma - 1}} = {\text{Constant}} \cr & {\text{or, }}{T_1}{V_1}^{\gamma - 1} = {T_2}{V_2}^{\gamma - 1} \cr} $$
For monoatomic gas $$\gamma = \frac{5}{3}$$
$$\eqalign{ & \left( {300} \right){V^{\frac{2}{3}}} = {T_2}{\left( {2\,V} \right)^{\frac{2}{3}}} \cr & \Rightarrow \,\,{T_2} = \frac{{300}}{{{{\left( 2 \right)}^{\frac{2}{3}}}}} \cr} $$
$${T_2} = 189K$$   (final temperature)
Change in internal energy $$\Delta U = n\frac{f}{2}R\,\Delta T$$
$$\eqalign{ & = 2\left( {\frac{3}{2}} \right)\left( {\frac{{25}}{3}} \right)\left( { - 111} \right) \cr & = - 2.7\,kJ \cr} $$