Two moles of an ideal monoatomic gas occupy a volume V at 27^o C. The gas expands adiabatically to a volume 2V. Calculate (i) the final temperature of the gas and (ii) change in its internal energy. (given $2^{2/3}=1.59$)

For adiabatic process relation of temperature and volume is 

$\begin{array}{l}{\mathrm{T}}_2{\mathrm{V}}_2^{\mathrm{\gamma }-1}={\mathrm{T}}_1{\mathrm{V}}_1^{\mathrm{\gamma }-1}\\ \Rightarrow {\mathrm{T}}_2(2\mathrm{V}{)}^{2/3}=300(\mathrm{V}{)}^{2/3} \mathrm{For} \mathrm{monoatomic} \mathrm{gasses},\\ \left[\mathrm{\gamma }=\frac{5}{3}\right]\\ \Rightarrow {\mathrm{T}}_2=\frac{300}{2^{2/3}}=189\mathrm{K} \mathrm{Also}, \mathrm{in} \mathrm{adiabatic} \mathrm{process},\\ \mathrm{\Delta Q}=0,\mathrm{\Delta U}=-\mathrm{\Delta W}\\ \text{ or }\mathrm{\Delta U}=\frac{-\mathrm{nR}\left(\mathrm{\Delta T}\right)}{\mathrm{\gamma }-1}\\ =-2\times \frac{3}{2}\times \frac{25}{3}(300-189)\\ =-2.7\mathrm{KJ}\end{array}$