For one-mole ideal gas, a process follows the relation $\mathrm{P}=\left(\frac{{\mathrm{V}}^2}{{\mathrm{V}}^2+{\mathrm{V}}_0^2}\right)\left({\mathrm{P}}_0\right),$ where P₀ and V₀ are constants. If the volume the gas changes from V₀ to 2V₀. What is the change in temperature?

PV = nRT

where n = number of moles and R = gas constant
By putting the value of volume V = V₀ in the equation given in the problem, for ideal gas we get,

$\mathrm{P}=\frac{{\mathrm{P}}_0}{2};$

${\mathrm{RT}}_1=\left(\frac{{\mathrm{P}}_0}{2}\right)\left({\mathrm{V}}_0\right)\dots \left(1\right)$

and with volume V = 2V₀ the given equation provides the pressure value,

$\mathrm{P}=\frac{4{\mathrm{P}}_0}{5}$

${\mathrm{RT}}_2=\left(\frac{4{\mathrm{P}}_0}{5}\right)\left(2{\mathrm{V}}_0\right)\dots \left(2\right)$

Subtracting equation (1) from (2)

$\therefore {\mathrm{T}}_2-{\mathrm{T}}_1=\frac{11{\mathrm{P}}_0{\mathrm{V}}_0}{10\mathrm{R}}$