n moles of an ideal gas undergo a thermodynamic process in which pressure of gas varies linearly with the volume of gas as shown in figure. If $\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{\mathrm{nR}}=200\mathrm{K}$, find the maximum temperature of the gas during the process (in K).

Gas pressure is given as

$\begin{array}{l} Slope=\frac{2{\mathrm{P}}_0-{\mathrm{P}}_0}{{\mathrm{V}}_0-2{\mathrm{V}}_0}=-\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0} \\ \left(\mathrm{P}-2{\mathrm{P}}_0\right)=-\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0}\left(\mathrm{V}-{\mathrm{V}}_0\right)\\ \Rightarrow \mathrm{P}=3{\mathrm{P}}_0-\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0}\mathrm{V}\\ \mathrm{T}=\frac{\mathrm{PV}}{\mathrm{nR}}=\frac{3{\mathrm{P}}_0}{\mathrm{nR}}\mathrm{V}-\frac{{\mathrm{P}}_0}{{\mathrm{nRV}}_0}{\mathrm{V}}^2\end{array}$

T is maximum when $\frac{\mathrm{dT}}{\mathrm{dV}}=0$

$\begin{array}{l} \frac{\mathrm{dT}}{\mathrm{dV}}=\frac{3{\mathrm{P}}_0}{\mathrm{nR}}-\frac{2{\mathrm{P}}_0}{{\mathrm{nRV}}_0}\mathrm{V}=0\text{ or, }\mathrm{V}=\frac{3{\mathrm{V}}_0}{2}\\ \Rightarrow {\mathrm{T}}_{max}=\frac{3{\mathrm{P}}_0}{\mathrm{nR}}\left(\frac{3{\mathrm{V}}_0}{2}\right)-\frac{{\mathrm{P}}_0}{{\mathrm{nRV}}_0}{\left(\frac{3}{2}{\mathrm{V}}_0\right)}^2\\ =\frac{9}{2}\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{\mathrm{nR}}-\frac{9}{4}\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{\mathrm{nR}}=\frac{9{\mathrm{P}}_0{\mathrm{V}}_0}{4\mathrm{nR}}\\ \Rightarrow {\mathrm{T}}_{max}=\frac{9}{4}\times 200=450\mathrm{K}\end{array}$