Question

$n$ moles of an ideal gas undergoes a process $A\to B$ as shown in the figure. Maximum temperature of the gas during the process is:

• A. $\frac{3P_o{V}_o}{2nR}$
• B. $\frac{9P_o{V}_o}{4nR}$
• C. $\frac{9P_o{V}_o}{2nR}$
• D. $\frac{9P_o{V}_o}{nR}$

Answer 1

Answer: (B)

$\frac{9P_o{V}_o}{4nR}$

Equation of line AB ($2$-point form)
$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$
$P-P_0=\frac{2P_0-P_0}{V_0-2V_0}(V-2V_0)$
$P-P_0=\frac{-P_0}{V_0}(V-2V_0)$
$P-P_0=\frac{-P_{0}V}{V_0}+2P_0$
$P=P_0+\frac{-P_{0}V}{V_0}+2P_0$
$P=3P_0-\frac{P_{0}V}{V_0}$
Multiply with V
$PV=3P_{0}V-\frac{P_0{V}^2}{V_0}$
Since, $PV=nRT$
$nRT=3P_{0}V-\frac{P_0{V}^2}{V_0}$
$T=\frac{1}{nR}(3P_{0}V-\frac{P_0{V}^2}{V_0})⟶1$
For maximum temperature,
$\frac{dT}{dV}=0$
Differentiating the above equation $1$ w.r.t., V
$\frac{dT}{dV}=\frac{{-P}_0}{V_0}\times 2V+3P_0$
$\therefore 3P_0-2\frac{P_0}{V_0}V=0$
$3P_0=2\frac{P_0}{V_0}V$
We get, $V=\frac{3}{2}{V}_0$
For max. temperature, substitute it in $1$,
$T_{max}=\frac{1}{nR}(\frac{-P_0}{V_0}\times \frac{9}{4}{V_0}^2+3P_0\times \frac{3}{2}{V}_0)$
$T_{max}=\frac{1}{nR}(-\frac{9}{4}{P}_0{V}_0+\frac{9}{2}{P}_0{V}_0)$
$T_{max}=\frac{P_0}{nR}{V}_0(\frac{9}{2}-\frac{9}{4})$
$T_{max}=\frac{9}{4}\frac{P_0{V}_0}{nR}$