n moles of an ideal gas undergo a process A and B as shown in the figure. The maximum temperature of the gas during the process will be

As, T will be maximum temperature where product of pV is maximum 

Equation of line AB, we have

$\begin{array}{l}\mathrm{Y}-{\mathrm{Y}}_1=\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}\left(\mathrm{X}-{\mathrm{X}}_1\right)\\ \mathrm{p}-{\mathrm{p}}_0=\frac{2{\mathrm{p}}_0-{\mathrm{p}}_0}{{\mathrm{v}}_0-2{\mathrm{v}}_0}\left(\mathrm{V}-2{\mathrm{V}}_0\right)\\ \Rightarrow \mathrm{p}-{\mathrm{p}}_0=\frac{-{\mathrm{p}}_0}{{\mathrm{V}}_0}\left(\mathrm{V}-2{\mathrm{V}}_0\right)\\ \Rightarrow \mathrm{p}=\frac{-{\mathrm{p}}_0}{{\mathrm{V}}_0}\mathrm{V}+3{\mathrm{p}}_0\\ \mathrm{pV}=\frac{-{\mathrm{p}}_0}{{\mathrm{V}}_0}{\mathrm{V}}^2+3{\mathrm{p}}_0\mathrm{V}\\ \mathrm{nRT}=\frac{-{\mathrm{p}}_0}{{\mathrm{V}}_0}{\mathrm{V}}_2+3{\mathrm{p}}_0\mathrm{V}\\ \mathrm{T}=\frac{1}{\mathrm{nR}}\left(\frac{-{\mathrm{p}}_0}{{\mathrm{V}}_0}{\mathrm{V}}^2+3{\mathrm{p}}_0\mathrm{V}\right)\end{array}$

For maximum temperature,

$\begin{array}{l}\frac{\mathrm{\partial }\mathrm{T}}{\mathrm{\partial }\mathrm{V}}=0\\ \frac{-{\mathrm{p}}_0}{{\mathrm{V}}_0}\left(2\mathrm{V}\right)+3{\mathrm{p}}_0=0\\ \frac{-{\mathrm{p}}_0}{{\mathrm{V}}_0}\left(2\mathrm{V}\right)=-3{\mathrm{p}}_0\Rightarrow \mathrm{V}=\frac{3}{2}{\mathrm{V}}_0\end{array}$

(condition for maximum temperature) Thus, the maximum temperature of the gas during the process will be

$\begin{array}{l}{\mathrm{T}}_{max}=\frac{1}{\mathrm{nR}}\left(\frac{-{\mathrm{p}}_0}{{\mathrm{V}}_0}\times \frac{9}{4}{\mathrm{V}}_0^2+3{\mathrm{p}}_0\times \frac{3}{2}{\mathrm{V}}_0\right)\\ =\frac{1}{\mathrm{nR}}\left(-\frac{9}{4}{\mathrm{p}}_0{\mathrm{V}}_0+\frac{9}{2}{\mathrm{p}}_0{\mathrm{V}}_0\right)\\ =\frac{9}{4}\frac{{\mathrm{p}}_0{\mathrm{V}}_0}{\mathrm{nR}}\end{array}$

Alternate solution

Since, initial and final temperature are equal, hence maximum temperature is at the middle of line.

i.e. $\mathrm{pV}=\mathrm{nRT}=\frac{\left(\frac{3}{2}{\mathrm{p}}_0\right)\left(\frac{3{\mathrm{V}}_0}{2}\right)}{\mathrm{nR}}$

${\mathrm{T}}_{max}\Rightarrow \frac{9{\mathrm{p}}_{\mathrm{o}}{\mathrm{V}}_0}{4\mathrm{nR}}$