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 T₂/T₁ is

Given ${\mathrm{V}}_2=2{ \mathrm{V}}_1; P{V}^{1/2}=\text{ constant }$

From graph, $\gamma =\text{ adiabatic constant }=1/2$

$\Rightarrow {\mathrm{P}}_1{ \mathrm{V}}_1^{1/2}=P_2{ \mathrm{V}}_2^{1/2}\Rightarrow \frac{{\mathrm{P}}_1}{P_2}={\frac{2{ \mathrm{V}}_1}{{ \mathrm{V}}_1}}^{1/2}=\sqrt{2}\text{ {For adiabatic process} }$

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

$\Rightarrow {\mathrm{P}}_1^{1-\gamma }{\mathrm{T}}_1^{\gamma }={\mathrm{P}}_2^{1-\gamma }{\mathrm{T}}_2^{\gamma }\Rightarrow {\frac{P_1}{P_2}}^{1-\gamma }={\frac{{\mathrm{T}}_2}{{ \mathrm{T}}_1}}^{\gamma }\Rightarrow {\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}}^{\frac{1-\gamma }{\gamma }}=\frac{{\mathrm{T}}_2}{{ \mathrm{T}}_1}$

$\therefore {\mathrm{T}}_2/{\mathrm{T}}_1={\left(\sqrt{2}\right)}^{{ }^{\frac{1-0.5}{0.5}}}=\sqrt{2}$