Question

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 :

• A. $\frac{25}{16}\frac{P_o{V}_o}{R}$
• B. $\frac{25}{8}\frac{P_o{V}_o}{R}$
• C. $\frac{25}{4}\frac{P_o{V}_o}{R}$
• D. $\frac{5}{8}\frac{P_o{V}_o}{R}$

Answer 1

Answer: (B)

We know that $PV=RT$. Therefore,
At Point $B:3P_0{V}_0=R{T}_B$
$\Rightarrow T_B=\frac{3P_0{V}_0}{R}$
point $C:P_{0}2V_0=R{T}_C$
$\Rightarrow T_C=\frac{3P_0{V}_0}{R}$
For process $BC,PV$ equation is given by
Slope of $BC,m=\frac{3P_0-P_0}{2V_0-V_0}=-2\frac{P_0}{V_0}$
Using $y=mx+c$, we get
$P=\frac{-2P_0}{V_0}\left(V-2V_0\right)+P_0$
Multiply both sides by $V$, we get
$PV=-\frac{2P_{0}V}{V_0}\left(V-2V_0\right)+P_{0}V$
$\Rightarrow RT=P_{0}V-\frac{2P_0{V}^2}{V_0}+4P_{0}V$
$\Rightarrow T=\frac{P_{0}V}{R}-\frac{2P_0{V}^2}{R{V}_0}+\frac{4P_{0}V}{R}$
$=\frac{5P_{0}V}{R}-\frac{2P_0{V}^2}{R{V}_0}$
$\Rightarrow T=\frac{P_0}{R}\left[5V-\frac{2V^2}{V_0}\right]$
For temperature to be maximum: $\frac{dT}{dV}=0.$ Therefore,
$\frac{dT}{dV}=\frac{P_0}{R}\left[5-\frac{4V}{V_0}\right]$
$\Rightarrow 5-\frac{4V}{V_0}=0;V=\frac{5}{4}{V}_0$
$T=\frac{P_0}{R}\left[\frac{5\times 5V_0}{4}-\frac{2}{V_0}\times \frac{25}{16}{V}_0^2\right]$ $\Rightarrow \frac{P_0}{R}\left[\frac{25V_0}{4}-\frac{25V_0}{8}\right]$
$T=\frac{25}{8}\frac{P_0{V}_0}{R}$