‘n’ moles of an ideal gas undergoes a process A → B as shown in the figure. The maximum temperature of the gas during the process will be

Equation of line AB is given by $\mathrm{y}-{\mathrm{y}}_1=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}\left(\mathrm{x}-{\mathrm{x}}_1\right);\mathrm{P}-{\mathrm{P}}_0=\frac{2{\mathrm{P}}_0-{\mathrm{P}}_0}{{\mathrm{V}}_0-2{\mathrm{V}}_0}\left(\mathrm{V}-2{\mathrm{V}}_0\right) \mathrm{or},$
$\mathrm{P}=-\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0}\mathrm{V}+3{\mathrm{P}}_0 \mathrm{or}, \mathrm{PV}=-\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0}{\mathrm{V}}^2+3{\mathrm{P}}_0\mathrm{V} \mathrm{or}, \mathrm{nRT}=-\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0}{\mathrm{V}}^2+3{\mathrm{P}}_0\mathrm{V} \mathrm{or},$
$\mathrm{T}=\frac{1}{\mathrm{nR}}\left(-\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0}{\mathrm{V}}^2+3{\mathrm{P}}_0\mathrm{V}\right)$........(i)

For maximum value of T,$\frac{{\displaystyle \mathrm{dT}}}{{\displaystyle \mathrm{dV}}}=0\text{ or, }-\frac{{\displaystyle {\mathrm{P}}_0}}{{\displaystyle {\mathrm{V}}_0}}\left(2\mathrm{V}\right)+3{\mathrm{P}}_0=0$
$\therefore \mathrm{V}=\frac{3}{2}{\mathrm{V}}_0$

$\mathrm{So},\mathrm{from} \mathrm{equation} \left(\mathrm{i}\right), {\mathrm{T}}_{max}=\frac{1}{\mathrm{nR}}\left[-\frac{{\mathrm{P}}_0}{{\mathrm{V}}_0}\times \frac{9}{4}{\mathrm{V}}_0^2+\frac{9}{2}{\mathrm{P}}_0{\mathrm{V}}_0\right]=\frac{9}{4}\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{\mathrm{nR}}$