91Ó°ÊÓ

Maxwell speed distribution:

$\frac{dP}{dv}=\sqrt{\frac{2}{\mathrm{Ï€}}}{\left(\frac{m}{k_{B}T}\right)}^{3/2}{v}^2{e}^{-\frac{1}{2}m{v}^2/k_{B}T}$

The differential probability per unit speed is

$\frac{dP}{dv}=\sqrt{\frac{2}{\mathrm{π}}}{\left(\frac{m}{k_{B}T}\right)}^{3/2}{v}^2{e}^{-\frac{1}{2}m{v}^2/k_{B}T}$ ……. (1)

To find the probability of finding a particle with speed greater than , we integrate the above expression yielding:

$P\left(v\right)={∫}_v^{∞}\sqrt{\frac{2}{\mathrm{π}}}{\left(\frac{m}{k_{B}T}\right)}^{3/2}{v}^2{e}^{-\frac{1}{2}m{v}^2/k_{B}T}dv$ ……. (2)

As we can see, there is no exact solution to the above integral over an arbitrary range. Thus, our strategy is to approximate P by using a numerical software. First however, let us perform a change of variables to simplify eq. (2) by letting:

$v=a{v}_{\text{rms}}$

$dv=v_{\text{rms}}da$

Here, a is an integer which will later correspond to the coefficients of the rms speed, namely .$a=2,6,10$

Recall that the rms speed of a molecule of an ideal gas is:

$v_{\text{rms}}=\sqrt{\frac{3k_{B}T}{m}}.$ ……. (3)

Put equation (3) into (2) and performing a change of variables from v to weget:

$\begin{array}{rcl}P\left(a\right)& =& {∫}_v^{∞}\sqrt{\frac{2}{\mathrm{π}}}{\left(\frac{m}{k_{B}T}\right)}^{3/2}{v}^2{e}^{-\frac{1}{2}m{v}^2/k_{B}T}dv\\ & =& \sqrt{\frac{2}{\mathrm{π}}}{\left(\frac{m}{k_{B}T}\right)}^{3/2}{∫}_v^{∞}{v}^2{e}^{-\frac{1}{2}m{v}^2/k_{B}T}dv\\ & =& \sqrt{\frac{2}{\mathrm{π}}}{\left(\frac{m}{k_{B}T}\right)}^{3/2}{∫}_a^{∞}{\left(a{v}_{\text{rms}}\right)}^2{e}^{-\frac{1}{2}m{\left(a{v}_{\text{mma}}\right)}^2/k_{B}T}{v}_{\text{rms}}da\\ & & \end{array}$

Further solving the above,

$\begin{array}{rcl}P\left(a\right)∂& =& \sqrt{\frac{2}{\mathrm{π}}}{\left(\frac{m}{k_{B}T}\right)}^{3/2}{v}_{\text{rms}}^2{∫}_a^{∞}{a}^2{e}^{-a^2\frac{m}{2k_{B}T}{v}_{{\max }^2}}da\\ & =& \sqrt{\frac{2}{\mathrm{π}}}{\left(\frac{m}{k_{B}T}\right)}^{3/2}{\left(\frac{3k_{B}T}{m}\right)}^{3/2}{∫}_a^{∞}{a}^2{e}^{-a^2\frac{m}{k_{B}T}\frac{3s_{B}T}{m}}da\\ & =& \sqrt{\frac{54}{\mathrm{π}}}{∫}_a^{∞}{a}^2{e}^{-\frac{3}{2}{a}^2}da\\ & & \end{array}$

The speed greater than $2v_{\text{rms}}$

$\begin{array}{rcl}{P}_{a=2}& =& \sqrt{\frac{54}{\mathrm{π}}}{∫}_a^{∞}{a}^2{e}^{-\frac{3}{2}{a}^2}da\\ & ≈& 0.0074\\ & & \end{array}$

The speed greater than$6v_{\text{rms}}$

$\begin{array}{rcl}{P}_{(a=6})& =& \sqrt{\frac{54}{\mathrm{π}}}{∫}_6^{∞}{6}^2{e}^{-\frac{3}{2}{\left(6\right)}^2}da\\ & ≈& 2.956×{10}^{-23}.\\ & & \end{array}$

The speed greater than$10v_{\text{rms}}$

$\begin{array}{rcl}{P}_{(a=10)}& =& \sqrt{\frac{54}{\mathrm{π}}}{∫}_{10}^{∞}{10}^2{e}^{-\frac{3}{2}{\left(10\right)}^2}da\\ & ≈& 9.949×{10}^{-65}.\\ & & \end{array}$

We have therefore shown that by numerically computing the probability obtained from the Maxwell speed distribution, the fraction of molecules faster than $2v_{\text{rms}}$is$0.0074~0.01$ , faster than $6v_{\text{rms}}$is$2.956×$${10}^{-23}~{10}^{-23}$ , and faster than$10v_{\text{rms}}$ is $9.949×{10}^{-65}~{10}^{-64}$.