Question

The given graph shows a hypothetical speed distribution for a sample of $N$ gas particles [Given that, for $V>V_0;\frac{dN}{dV}=0),\frac{dN}{dV}$ is change in number of particles with change in velocity]

Select the correct choice (s):

• A. The value of $a{V}_0$ is $2N$
• B. The ratio $\frac{V_{avg}}{V_0}$ is equal to $\frac{2}{3}$
• C. The ratio $\frac{V_{rms}}{V_0}$ is equal to $\frac{1}{\sqrt{2}}$
• D. Three fourth of the total particles has a speed between $0.5V_0$ and $V_0$

Answer 1

Answer: (A), (B), (C), (D)

Three fourth of the total particles has a speed between $0.5V_0$ and $V_0$

From the given graph we can see that slope of curve $\left(\frac{dN}{dV}\right)$ vs V is $\frac{a}{V_0}$.

The graph is a straight line passing through origin, thus of format $y=mx$

$\Rightarrow \frac{dN}{dV}=\frac{a}{V_0}\times V$
or $dN=\frac{a}{V_0}VdV$

Integrating with limits $N=0$ to $N=N$ and $V=0$ to $V=V_0$
${\int }_0^{N}dN=\frac{a}{V_0}{\int }_0^{V_0}VdV$
or, $N=\frac{a}{V_0}\left(\frac{V_0^2}{2}\right)$
or, $N=\frac{a{V}_0}{2}.....\left(i\right)$
Thus $a{V}_0=2N$
$\therefore$ option (a) is correct.

Thus rms value of $V$ can be written in form of integration as,
$V_{rms}^2=\frac{1}{N}{\int }_0^{V_0}{V}^{2}dN$
$\Rightarrow V_{rms}^2=\frac{1}{N}{\int }_0^{V_0}{V}^2\left(\frac{a}{V_0}\right)VdV$
or, $V_{rms}^2=\frac{a}{N{V}_0}{\int }_0^{V_0}{V}^{3}dV$
or, $V_{rms}^2=\frac{a}{N{V}_0}\left[\frac{V_0^4}{4}\right]=\frac{a{V}_0^3}{4N}$

Substituting $a{V}_0=2N$
$V_{rms}^2=\frac{\left(2N\right)V_0^2}{4N}=\frac{V_0^2}{2}$

$\Rightarrow {\left(\frac{V_{rms}}{V_0}\right)}^2=\frac{1}{2}$
$\therefore \frac{V_{rms}}{V_0}=\frac{1}{\sqrt{2}}$

Now the average velocity can be given as,
$V_{avg}=\frac{1}{N}{\int }_0^{V_0}VdN$
$\Rightarrow V_{avg}=\frac{1}{N}{\int }_0^{V_0}V\left(\frac{a}{V_0}\right)VdV$
or, $V_{avg}=\frac{a}{N{V}_0}\left[\frac{V_0^3}{3}\right]=\frac{a{V}_0^2}{3N}$

$\because a{v}_0=2N$
$\Rightarrow V_{avg}=\frac{\left(2N\right)V_0}{3N}=\frac{2V_0}{3}$
$\therefore \frac{V_{avg}}{V_0}=\frac{2}{3}$
Thus options (b) & (c) are correct.

For the number of particle having speed from $0.5V_0$ to $V_0;$
${\int }_0^{N^{\prime }}dN={\int }_{0.5V_0}^{V_0}\left(\frac{a}{V_0}\right)VdV$
or, $N^{\prime }=\frac{a}{V_0}{\left[\frac{V^2}{2}\right]}_{0.5V_0}^{V_0}$
$\Rightarrow N^{\prime }=\frac{a}{V_0}\times \frac{1}{2}[V_0^2-(0.5V_0{)}^2]$
$\Rightarrow N^{\prime }=\frac{a{V}_0^2}{2V_0}[1-\frac{1}{4}]$
or, $N^{\prime }=\frac{3}{4}\times \left(\frac{a{V}_0^2}{2V_0}\right)=\frac{3}{4}\left[\frac{a{V}_0}{2}\right]$
Substituting $\frac{a{V}_0}{2}=N$
$\therefore N^{\prime }=\frac{3}{4}N$
Therefore three fourth of total number of particles have speed between $0.5V_0$ and $V_0$.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: Area under the curve of}\frac{dN}{dV}\\ \text{vs}V\text{gives the total number of gas particles.}\\ \text{In order to solve this problem we need to find}\\ \text{the value of rate of charge of number of particles}\\ \text{w.r.t change in velocity, then expressing in terms}\\ \text{of the relation between the}{V}_0\text{and number of}\\ \text{particles}N\text{.}\end{array}$