'$n$' 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
JEE MAIN 2016, Diffcult
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The euatione for the line is
$P = \frac{{ - {P_0}}}{{{V_0}}}V + 3P$
$[slope = \frac{{ - {P_0}}}{{{V_0}}},c = 3{P_0}]$
$P{V_0} + {P_0}V = 3{P_0}{V_0}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$
$But\,\,\,\,\,\,\,PV = nRT\,$
$\therefore P = \frac{{nRT}}{V}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$
$From\left( i \right)\& \left( {ii} \right)\frac{{nRT}}{V}{V_0} + {P_0}V = 3{P_0}{V_0}$
$\therefore nRT{V_0} + {P_0}{V^2} = 3{P_0}{V_0}$
$...\left( {iii} \right)$
For temperature to be maximum $\frac{{dT}}{{dV}} = 0$
Differentiating $e.q.(iii)\,by\,'v'\,we\,get$
$nR{V_0}\frac{{dT}}{{dV}} + {P_0}\left( {2v} \right) = 3{P_0}{V_0}$
$\therefore nR{V_0}\frac{{dT}}{{dV}} = 3{P_0}{V_0} - 2{P_0}V$
$\frac{{dT}}{{dV}} = \frac{{3{P_0}{V_0} - 2{P_0}V}}{{nR{V_0}}} = 0$
$V = \frac{{3{V_0}}}{2}\,\,\,\,\,\,\,\,\,\therefore P = \frac{{3{P_0}}}{2}$ $[From (i)]$
$\therefore \,{T_{\max }} = \frac{{9{P_0}{V_0}}}{{4nR}}\,\,\left[ {From\,\left( {iii} \right)} \right]$
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