and minima.

since the graph shown here is a straight line, we can write it's equation in the form of $:$

$y=m x+c$

here $y=P, x=V, m=\frac{P_{2}-P_{1}}{V_{2}-V_{1}}=\frac{\frac{P_{0}}{2}-P_{0}}{2 V_{0}-V_{0}}$

$\Rightarrow m=\frac{-P_{0}}{2 V_{0}}$

$c$ is the intercept and $c=\frac{3 P_{0}}{2}$

We can find $c$ by extra polating the graph Now

$P=\frac{-P_{0}}{2 V_{0}} V+\frac{3 P_{0}}{2}$        $...(1)$

Also, from equation of state, we know the $P V=n R T$

for $1$ mole $P V=R T$

or $P=\frac{R T}{V}$

substituting this in equation $( 1)$

$\frac{R T}{V}=\frac{-P_{0}}{2 V_{0}} V+\frac{3 P_{0}}{2}$

$T=\frac{-\bar{P}_{0}}{2 R V_{0}} \cdot V^{2}+\frac{3 P_{0}}{2 R} V$

To find the minimum temperature, we will differentiate the above equation

i.e., $\frac{d T}{d V}=-\frac{2 P_{0}}{2 R V_{0}} \cdot V+\frac{3 P_{0}}{2 R}$

for $\frac{d T}{d V}=0 \Rightarrow \frac{P_{0}}{R V_{0}} \cdot V=\frac{3 P_{0}}{2 R}$

$\Rightarrow V=\frac{3 V_{0}}{2}$

This is the critical point.

$\frac{d^{2} T}{d V^{2}}=-\frac{P_{0}}{R V_{0}}$

which is negative since pressure and volume can't be negative.

$\Rightarrow$ pt of maximum

Now at $V=\frac{3 V_{0}}{2}$

maximum temperature $=T_{\max }=\frac{-P_{0}}{2 R V_{0}} \frac{9 V_{0}^{2}}{4}+\frac{3 P_{0}}{2 R} \cdot \frac{3 V_{0}}{2}$

$\Rightarrow T_{\max }=\frac{9 P_{0} V_{0}}{8 R}$