Since the P-V graph of the process is a straight line and two points (V₀, 2P₀) and (2V₀, P₀) are known, its equation will be

$\left(\mathit{\text{P}}-{\text{P}}_0\right)=\frac{\left(2{\text{P}}_0-{\text{P}}_0\right)}{\left({\text{V}}_0-2{\text{V}}_0\right)}\left(\mathit{\text{V}}-2{\text{V}}_0\right)=\frac{{\text{P}}_0}{{\text{V}}_0}\left(2{\text{V}}_0-\mathit{\text{V}}\right)$

$\therefore \mathit{\text{P}}=3{\text{P}}_0-\frac{{\text{P}}_0\mathit{\text{V}}}{{\text{V}}_0}$

According to the equation for an ideal gas,

$\mathit{\text{T}}\mathit{=}\frac{\mathit{\text{pV}}}{\mathit{\text{nR}}}$

  $=\left(3{\text{P}}_0-\frac{{\text{P}}_0\mathit{\text{V}}}{{\text{V}}_0}\right)\frac{\mathit{\text{V}}}{\mathit{\text{nR}}}$

  $=\frac{3{\text{P}}_0{\text{V}}_0\mathit{\text{V}}-{\text{P}}_0{\mathit{\text{V}}}^2}{\mathit{\text{nR}}{\text{V}}_0}$                ...(i)

For $T$ to maximum, $\frac{\mathit{\text{dT}}}{\mathit{\text{dV}}}=0$

            $3{\text{P}}_0{\text{V}}_0-2{\text{P}}_0\mathit{\text{V}}=0$

or   $\mathit{\text{V}}=\frac{3{\text{V}}_0}{2}$                ...(ii)

Putting this value in Eq. (i), we get

${\mathit{\text{T}}}_{\text{max}}=\frac{3{\text{P}}_0{\text{V}}_0\left(\frac{3{\text{V}}_0}{2}\right)-{\text{P}}_0\left(\frac{9}{4}{\text{V}}_0^2\right)}{\mathit{\text{nR}}{\text{V}}_0}=\frac{9{\text{P}}_0{\text{V}}_0}{4\mathit{\text{nR}}}$