One mole of an ideal monoatomic gas is taken along the path $ABCA$ as shown in the $PV$ diagram. The maximum temperature attained by the gas along the path $BC$ is given by
JEE MAIN 2018, Diffcult
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Equation of the $BC$
$p = {p_0} - \frac{{2{P_0}}}{{{V_0}}}\left( {V - 2{V_0}} \right)$
using=$PV=nRT$
Temperature,$T = \frac{{{P_0}V - \frac{{2{P_0}{V^2}}}{{{V_0}}} + 4{P_0}V}}{{1 \times R}}$
$\left( {n = 1\,mole\,given} \right)$
$T = \frac{{{P_0}}}{R}\left[ {5V - \frac{{2{V^2}}}{{{V_0}}}} \right]$
$\frac{{dT}}{{dV}} = 0 \Rightarrow 5 - \frac{{4V}}{{{V_0}}} = 0 \Rightarrow V = \frac{5}{4}{V_0}$
$T = \frac{{{P_0}}}{R}\left[ {5 \times \frac{{5{V_0}}}{4} - \frac{2}{{{V_0}}} \times \frac{{25}}{{16}}V_0^2} \right] = \frac{{25}}{8}\frac{{{P_0}{V_0}.}}{R}$
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