Question

An ideal gas is taken along the process $AB$ as shown in the $P-V$ diagram. Find the volume of the gas where the temperature becomes maximum.

• A. $\left(\frac{3V_0}{2}\right)$
• B. $\frac{4}{5}{V}_0$
• C. $\frac{5}{3}{V}_0$
• D. $\frac{3}{5}{V}_0$

Answer 1

The correct option is A $\left(\frac{3V_0}{2}\right)$

From the curve, we can deduce that, the graph is a straight line having a negative slope.
Using the equation of straight line
$y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)x-x_1$
we can write that,
$P-2P_0=\left(\frac{2P_0-P_0}{V_0-2V_0}\right)(V-V_0)$
$P-2P_0=\frac{-P_0}{V_0}(V-V_0)$
$P=3P_0-\frac{P_{0}V}{V_0}$
$\Rightarrow PV=3P_{0}V-\frac{P_0{V}^2}{V_0}..........\left(1\right)$
From the ideal gas equation we know that, $PV=nRT.........\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ we get,
$nRT=3P_{0}V-\frac{P_0{V}^2}{V_0}.............\left(3\right)$
so for $T_{max},\frac{dT}{dV}=0$ and $\frac{d^{2}T}{d{V}^2}<0$
Differentiating $\left(3\right)$ with respect to $V$ we get,
$nR\frac{dT}{dV}=3P_0-\frac{2P_{0}V}{V_0}.........\left(4\right)$
Using, $\frac{dT}{dV}=0$ we get,
$\Rightarrow 3P_0-\frac{2P_{0}V}{V_0}=0$
$\Rightarrow V=\frac{3V_0}{2}$
Differentiating $\left(4\right)$ again with respect to $V$ we get,
$nR\frac{d^{2}T}{d{V}^2}=-\frac{2P_0}{V_0}$
$\Rightarrow \frac{d^{2}T}{d{V}^2}=-\frac{2P_0}{nR{V}_0}<0$
When the volume of the gas is $V=\frac{3V_0}{2}$ the temperature of the ideal gas becomes maximum.
Thus, option (a) is the correct answer.