'\(n\)' moles of an ideal gas undergoes a process \(A \rightarrow B\) as shown in the figure. The maximum temperature of the gas during the  process will be

\(P = \frac{{ - {P_0}}}{{{V_0}}}V + 3P\)

\([slope = \frac{{ - {P_0}}}{{{V_0}}},c = 3{P_0}]\)

              \(P{V_0} + {P_0}V = 3{P_0}{V_0}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\)

\(But\,\,\,\,\,\,\,PV = nRT\,\)

\(\therefore P = \frac{{nRT}}{V}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\)

\(From\left( i \right)\& \left( {ii} \right)\frac{{nRT}}{V}{V_0} + {P_0}V = 3{P_0}{V_0}\)

\(\therefore nRT{V_0} + {P_0}{V^2} = 3{P_0}{V_0}\)

\(...\left( {iii} \right)\)

For temperature to be maximum \(\frac{{dT}}{{dV}} = 0\)

Differentiating \(e.q.(iii)\,by\,'v'\,we\,get\)

\(nR{V_0}\frac{{dT}}{{dV}} + {P_0}\left( {2v} \right) = 3{P_0}{V_0}\)

\(\therefore nR{V_0}\frac{{dT}}{{dV}} = 3{P_0}{V_0} - 2{P_0}V\)

\(\frac{{dT}}{{dV}} = \frac{{3{P_0}{V_0} - 2{P_0}V}}{{nR{V_0}}} = 0\)

\(V = \frac{{3{V_0}}}{2}\,\,\,\,\,\,\,\,\,\therefore P = \frac{{3{P_0}}}{2}\)          \([From (i)]\)

\(\therefore \,{T_{\max }} = \frac{{9{P_0}{V_0}}}{{4nR}}\,\,\left[ {From\,\left( {iii} \right)} \right]\)