Question

One mole of an ideal gas is expanded from the state $A(P_0,V_0)$ to final state $B$ having volume $V$. The process follows a path represented by a straight line on the P-V diagram (see figure).
If final temperature at $B$ is half the maximum temperature during the process, find volume $\left(V\right)$ ?

• A. $\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)V_0$
• B. $\frac{V_0}{\sqrt{2}}$
• C. $\sqrt{2}{V}_0$
• D. $(\sqrt{2}-1)V_0$

Answer 1

The correct option is A $\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)V_0$

Line AB is seen to be passing from $(2P_0,0)\&(P_0,V_0)$
so, equation of $AB$ can be written as,
$P-P_0=\frac{2P_0-P_0}{-V_0}(V-V_0)$
$P-P_0=\frac{-P_{0}V}{V_0}+P_0$
$P=\frac{-P_{0}V}{V_0}+2P_0$
So, equation of the given straight line is the process is
$P=-\left(\frac{P_0}{V_0}\right)V+2P_0$
For an ideal gas $PV=nRT$
$\therefore [-\frac{P_0}{V_0}V+2P_0]V=nRT$
$\Rightarrow T=-\frac{P_0}{V_{0}R}{V}^2+\frac{2P_0}{R}V[\because n=1]...\left(ii\right)$
This is equation of a parabola.
The vertex of parabola corresponds to $\frac{dT}{dV}=0$
$-\frac{2P_0}{V_{0}R}V+\frac{2P_0}{R}=0\Rightarrow V=V_0$
Temperature in this state is
$T_0=\frac{P_0{V}_0}{nR}$
$\because n=1\therefore T_0=\frac{P_0{V}_0}{R}$
This is the maximum temperature during the process.

If final temperature at B is $\frac{T_0}{2}=\frac{P_0{V}_0}{2R}$ then volume can be obtained using (ii)
$\frac{P_0{V}_0}{2R}=-\frac{P_0}{R{V}_0}{V}^2+\frac{2P_0}{R}V$
$\Rightarrow \frac{V_0}{2}=-\frac{V^2}{V_0}+2V$
$\Rightarrow 2V^2-4V_{0}V+V_0^2=0$
$V=\frac{4V_0\pm \sqrt{16V_0^2-8V_0^2}}{4}=\frac{4V_0\pm \sqrt{8}{V}_0}{4}=V_0\pm \frac{V_0}{\sqrt{2}}=\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)V_0$
$\because V>V_0$