Question

${}^{\prime }{n}^{\prime }$ moles of an ideal gas undergoes a process $A\to B$ as shown in the figure. The maximum temperature of the gas during the process will be :

• A. $\frac{9P_0{V}_0}{4nR}$
• B. $\frac{3P_0{V}_0}{2nR}$
• C. $\frac{9P_0{V}_0}{2nR}$
• D. $\frac{9P_0{V}_0}{nR}$

Answer 1

Answer: (A)

Equation of line $AB$
$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$
$P-P_0=\frac{2P_0-P_0}{V_0-2V_0}$
$\left(V-2V_0\right)=-\frac{P_0}{V_0}\left(V-2V_0\right)$
$P=-\frac{P_0}{V_0}V+3P_0$
$PV=-\frac{P_0}{V_0}{V}^2+3P_{0}V$
$nRT=-\frac{P_0}{V_0}{V}^2+3P_{0}V$
$T=\frac{1}{nR}\left(\frac{P_0}{V_0}{V}^2+3P_{0}V\right)$
$\frac{dT}{dV}=0$ (For maximum temperature)
$\frac{P_0}{V_0}2V+3P_0=0$
$\frac{P_0}{V_0}2V=-3P_0$
$V=\frac{3}{2}{V}_0$ (Condition for maximum temperature)
$T_{max}=\frac{1}{nR}\left(-\frac{P_0}{V_0}\times \frac{9}{4}{V}_0^2+3P_0\times \frac{3}{2}{V}_0\right)$
$=\frac{1}{nR}\left(-\frac{9}{4}{P}_0{V}_0+\frac{9}{2}{P}_0{V}_{0}0\right)=\frac{9}{4}\frac{P_0{V}_0}{nR}$0