Figure shows a hypothetical speed distribution for a sample of N gas particles (note that $P\left(v\right)=0$for speed $v>2v_0$, ).

a) What is$a{v}_0$?

b) What is$v_{avg}/v_0$?

c) What is$v_{rms}/v_0$?

d) What fraction of the particles has a speed between$1.5v_0$and$2.0v_0$?

In the given problem graph of$P\left(v\right)$ Vs$v$ is shown.

At, $v=v_0$, $P\left(v\right)=a$

$v=2v_0$,$P\left(v\right)=a$

We use the concept rms speed, average speed, and speed distribution function to calculate the different quantities.

Formulae:

$\int P\left(v\right)dv=1$

$v_{avg}=\int vP\left(v\right)dv$

${\text{v}}_{rms}^2=\int {\text{v}}^2\text{P}\left(\text{v}\right)\text{dv}$

$\text{No.ofparticles}=N\underset{v_1}{\overset{v_2}{}}P\left(v\right)dv$

$\text{Frac.ofparticles}=\underset{v_1}{\overset{v_2}{}}P\left(v\right)dv$

The distribution function gives the fraction of particles with speeds between $\text{v}$and$\text{v}+\text{dv}$, so its integral over all speeds is unity.

$\int \text{P}\left(\text{v}\right)\text{dv}=1$

Evaluate the integral by calculatingthearea under the curve. The area of the triangular part is half the product of base and height, i.e.,$a{v}_0$.

So,

$\begin{array}{c}\int P\left(v\right)dv=\frac{1}{2}a{v}_0+a{v}_0\\ \int P\left(v\right)dv=\frac{3}{2}a{v}_0\\ 1=\frac{3}{2}a{v}_0\end{array}$

$\Rightarrow a{v}_0=\frac{2}{3}$……. (i)

$a{v}_0=0.67$

Therefore the value of$a{v}_0$ is$0.67$ .

The average speed is given by

$v_{avg}=\int vP\left(v\right)dv$

For the triangular part of the distribution,

$P\left(v\right)=av/v_0$

And the involvement of this portion is

$\begin{array}{c}{v}_{avg}={\frac{a}{\text{v}}}_0\underset{0}{\overset{v_0}{}}{v}^{2}dv\\ ={\frac{a}{3v}}_0{v}_0^3\end{array}$

Substituting the value of$a{v}_0$from equation (i), we get

$v_{avg}=\frac{2}{9}{v}_0$

For the rectangular part,$P\left(v\right)=a$so the contribution of this part is

$\begin{array}{c}{v}_{avg}=\int vP\left(v\right)dv\\ v_{avg}=a\underset{v_0}{\overset{2v_0}{}}vdv\end{array}$

$\begin{array}{c}{v}_{avg}=\frac{a}{2}(4v_0^2-v_0^2)\\ v_{avg}=\frac{3a{v}_0^2}{2}\end{array}$

Substituting the value of $a=\frac{2}{3v_0}$from equation (i) in the above equation, we get

$v_{avg}=v_0$

Therefore,

$\begin{array}{l}{v}_{avg}=\frac{2}{9}{v}_0+v_0\\ v_{avg}=\frac{11}{9}{v}_0\end{array}$

$\begin{array}{c}{v}_{avg}=1.22v_0\\ \frac{v_{avg}}{v_0}=1.22\end{array}$

Therefore the ratio,$\frac{v_{avg}}{v_0}=1.22$ .

The mean square speed is given by

$v_{rms}^2=\int v^{2}P\left(v\right)dv$

The contribution of triangular section is

$\begin{array}{l}P\left(v\right)=\frac{av}{v_{\text{0}}}\\ v_{rms}^2=\frac{a}{v_{\text{0}}}\underset{0}{\overset{v_0}{}}{v}^{3}dv\end{array}$

$v_{rms}^2=\frac{\frac{a}{v_0}{v}_0^4}{4}$

Substituting the value of$\text{a}=2/3{\text{v}}_0$from (i), we get

$v_{rms}^2=\frac{1}{6}{v}_0^2$

The contribution of the rectangular portion is

$\text{P}\left(\text{v}\right)\text{dv}=\text{a}$

$\begin{array}{c}{v}_{rms}^2=\text{a}\underset{v_0}{\overset{2v_0}{}}{v}^{2}dv\\ =\frac{\text{a}}{3}(8v_0^3-v_0^3)\\ =\frac{a}{{\text{v}}_0}\underset{0}{\overset{v_0}{}}{v}^{3}dv\\ =\frac{14}{9}{v}_0^2\end{array}$

Thus,

$\begin{array}{c}{v}_{rms}=\sqrt{\frac{1}{6}{v}_0^2+\frac{14}{9}{v}_0^2}\\ v_{rms}=1.31v_0\\ \frac{v_{rms}}{v_0}=1.31\end{array}$

Therefore the ratio, $\frac{v_{rms}}{v_0}=1.31$

The number of particles with speed between $1.5v_0$to $2v_0$is given by

$\text{No.ofparticles}=N\underset{1.5v_0}{\overset{2V_0}{}}P\left(v\right)dv$

Since $P\left(v\right)=a$throughout the range of integration, the number of particles with the speeds in the given range is

$\text{No.ofparticles}=Na(2v_0-1.5v_0)$

$\text{No.ofparticles}=\frac{N}{3}$

Where, $a=\frac{2}{3v_0}$is substituted in the above.

The fraction of particles with speeds between$1.5v_0$ and$2v_0$ is given by

$\text{frac}=\underset{1.5v_0}{\overset{2v_0}{}}P\left(v\right)dv$

$\begin{array}{l}\text{frac}=\frac{1}{3}\\ \text{frac}=0.33\end{array}$

Therefore the required fraction is$0.33$ .