One mole of a gas expands obeying the relation as shown in the P-V diagram. The maximum temperature in this process is equal to

The slope of the graph is, $m=\frac{P_0}{2V_0}$
The equation of this graph is given by, 
$\begin{array}{l}P=-\frac{P_{0}V}{2V_0}+\frac{3P_0}{2}\\ =\frac{nRT}{V}=-\frac{P_0}{2V_0}+\frac{3P_0}{2}\end{array}$

$\begin{array}{l}\Rightarrow T-\frac{P_0}{2nR{V}_0}{V}^2+\frac{3P_0}{2nR}V\\ =\frac{dT}{dV}=-\frac{2P_{0}V}{2nR{V}_0}+\frac{3P_0}{2nR}=0\end{array}$

$\Rightarrow V=\frac{3}{2}{V}_0$

$ATV=\frac{3V_0}{2},$ the temperature will be maximum.
$[n=1]$
$T_{\max }=-\frac{P_0}{2R{V}_0}\times \frac{9V_0^2}{4}+\frac{3P_0}{2R}\times \frac{3V_0}{2}$

$\begin{array}{l}\Rightarrow -\frac{9}{8}\frac{P_0{V}_0}{2R}+\frac{9}{4}\frac{P_0{V}_0}{R}\\ \Rightarrow \frac{9}{8}\frac{P_0{V}_0}{R}\end{array}$