Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. 
If V₂ = 2V₁, then the ratio of temperature $\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}$ is

Here  ${\mathrm{P}}_1\text{ and }{\mathrm{P}}_2,{\mathrm{T}}_1\text{ and }{\mathrm{T}}_2,{\mathrm{V}}_1\text{ and }{\mathrm{V}}_2$ are initial and final pressures temperatures and volumes, respectively
Given ${\mathrm{V}}_2=2{\mathrm{V}}_1 ; P{V}^{1/2}=$ constant  From graph, $\gamma =$adiabatic constant = 1/2
$\Rightarrow {\mathrm{P}}_1{\mathrm{V}}_1^{1/2}=P_2{\mathrm{V}}_2^{1/2}\Rightarrow \frac{{\mathrm{P}}_1}{P_2}={\left(\frac{2{\mathrm{V}}_1}{{\mathrm{V}}_1}\right)}^{1/2}=2^{1/2}$  {For adiabatic process}

Also, ${\mathrm{P}}^{1-\mathrm{\gamma }}{T}^{\gamma }=\mathrm{constant}$

$\begin{array}{l}\Rightarrow {\mathrm{P}}_1^{1-\mathrm{\gamma }}{\mathrm{T}}_1^{\mathrm{\gamma }}={\mathrm{P}}_2^{1-\mathrm{\gamma }}{\mathrm{T}}_2^{\mathrm{\gamma }}\Rightarrow {\left(\frac{P_1}{P_2}\right)}^{1-\mathrm{\gamma }}={\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^{\mathrm{\gamma }}\Rightarrow {\left(\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right)}^{\frac{1-\mathrm{\gamma }}{\mathrm{\gamma }}}=\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\\ \therefore {\mathrm{T}}_2/{\mathrm{T}}_1={\left(2^{1/2}\right)}^{\frac{1-0.5}{0.5}}=\sqrt{2}\end{array}$