Question

$P-V$ diagram of $n$ moles of an ideal gas is as shown in the figure. Find the maximum temperature attained between A and B.

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

Answer 1

The correct option is B $\frac{9P_0{V}_0}{4nR}$

For given number of moles of a gas, $T\propto PV$
Although $(PV{)}_A=(PV{)}_B$ or $T_A=T_B$, the above process is not isothermal. Because in isothermal process, $P-V$ graph is a rectangular hyperbola, while here it is a straight line.
From the graph, the $P-V$ equation can be written as,
$P=-\frac{P_0}{V_0}V+3P_0$
Multiplying by $V$ on both sides,
$PV=-\left(\frac{P_0}{V_0}\right)V^2+3P_{0}V$
$\Rightarrow nRT=3P_{0}V-\left(\frac{P_0}{V_0}\right)V^2$
[as $PV=nRT$]
$\Rightarrow T=\frac{1}{nR}[3P_{0}V-\left(\frac{P_0}{V_0}\right)V^2]$
This is the required $T-V$ equation. This is quadratic in $V$. Hence, $T-V$ graph is a parabola.
Now, to find maximum or minimum value of $T$ we should set $\frac{dT}{dV}=0$
$\Rightarrow 3P_0-\left(\frac{2P_0}{V_0}\right)V=0$
$\Rightarrow V=\frac{3}{2}{V}_0$
Further $\frac{d^{2}T}{d{V}^2}=-\frac{2P_0}{V_0}$ which is negative at $V=\frac{3}{2}{V}_0$
Hence, $T$ is maximum at $V=\frac{3}{2}{V}_0$ and this maximum value is,
$T_{max}=\frac{1}{nR}\left[\right(3P_0)\left(\frac{3V_0}{2}\right)-\left(\frac{P_0}{V_0}\right){\left(\frac{3V_0}{2}\right)}^2]$
i.e $T_{max}=\frac{9P_0{V}_0}{4nR}$