A monoatomic gas $\left(C_V=\frac{3}{2}R\right)$ is allowed to expand 2 adiabatically and reversibly from initial volume of 8 L at 300 K to a volume ${\mathrm{V}}_2$ at 260 K ${\mathrm{V}}_2$ is :

It is known that,

$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1};$ 

Given,

$\gamma =\frac{5}{3}$;

$T_1=300 \mathrm{K}$,

$V_1=8\mathrm{L}$,

$T_2=250 \mathrm{K}$

Put these values in the above equation we get the value of final volume as 10.5 L.