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Q.
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 :
We know that $P V=R T$. Therefore,
At Point $B: 3 P_{0} V_{0}=R T_{ B }$
$\Rightarrow T_{ B }=\frac{3 P_{0} V_{0}}{R}$
point $C: P_{0} 2 V_{0}=R T_{ C }$
$\Rightarrow T_{ C }=\frac{3 P_{0} V_{0}}{R}$
For process $BC,\, PV$ equation is given by
Slope of $BC,\, m=\frac{3 P_{0}-P_{0}}{2 V_{0}-V_{0}}=-2 \frac{P_{0}}{V_{0}}$
Using $y=mx +c$, we get
$P=\frac{-2 P_{0}}{V_{0}}\left(V-2 V_{0}\right)+P_{0}$
Multiply both sides by $V$, we get
$P V=-\frac{2 P_{0} V}{V_{0}}\left(V-2 V_{0}\right)+P_{0} V$
$\Rightarrow R T=P_{0} V-\frac{2 P_{0} V^{2}}{V_{0}}+4 P_{0} V$
$\Rightarrow T=\frac{P_{0} V}{R}-\frac{2 P_{0} V^{2}}{R V_{0}}+\frac{4 P_{0} V}{R}$
$=\frac{5 P_{0} V}{R}-\frac{2 P_{0} V^{2}}{R V_{0}}$
$\Rightarrow T=\frac{P_{0}}{R}\left[5 V-\frac{2 V^{2}}{V_{0}}\right]$
For temperature to be maximum: $\frac{d T}{d V}=0 .$ Therefore,
$\frac{d T}{d V} =\frac{P_{0}}{R}\left[5-\frac{4 V}{V_{0}}\right]$
$\Rightarrow 5-\frac{4 V}{V_{0}}=0 ; V=\frac{5}{4} V_{0}$
$T =\frac{P_{0}}{R}\left[\frac{5 \times 5 V_{0}}{4}-\frac{2}{V_{0}} \times \frac{25}{16} V_{0}^{2}\right]$ $\Rightarrow \frac{P_{0}}{R}\left[\frac{25 V_{0}}{4}-\frac{25 V_{0}}{8}\right]$
$T =\frac{25}{8} \frac{P_{0} V_{0}}{R}$
