Figure 9.8 cannot do justice to values at the very highspeed end of the plot. This exercise investigates how small it really gets. However, although integrating the Maxwell speed distribution over the full range of speeds from 0 to infinity can be carried out (the so-called Gaussian integrals of Appendix K), over any restricted range, it is one of those integrals that. unfortunately. cannot be done in closed form. Using a computational aid of your choice. show that the fraction of molecules moving faster thanis; faster than$6v_{\text{rms}}$is$-{10}^{-23}$; and faster than$10{\text{v}}_{\text{ms}}$is$~{10}^{-64}$. where$v_{rms}$" from Exercise 41, is$\sqrt{3k_{\text{B}}T/m}$. (Exercise 48 uses these values in an interesting application.)

Maxwell speed distribution:

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

The differential probability per unit speed is

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

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

$P\left(v\right)={\int }_v^{\infty }\sqrt{\frac{2}{\pi }}{\left(\frac{m}{k_{B}T}\right)}^{3/2}{v}^2{e}^{-\frac{1}{2}m{v}^2/k_{B}T}dv$ ……. (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:

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

Further solving the above,

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

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

$\begin{array}{rcl}{P}_{a=2}& =& \sqrt{\frac{54}{\pi }}{\int }_a^{\infty }{a}^2{e}^{-\frac{3}{2}{a}^2}da\\ & \approx & 0.0074\\ & & \end{array}$

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

$\begin{array}{rcl}{P}_{(a=6})& =& \sqrt{\frac{54}{\pi }}{\int }_6^{\infty }{6}^2{e}^{-\frac{3}{2}{\left(6\right)}^2}da\\ & \approx & 2.956\times {10}^{-23}.\\ & & \end{array}$

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

$\begin{array}{rcl}{P}_{(a=10)}& =& \sqrt{\frac{54}{\pi }}{\int }_{10}^{\infty }{10}^2{e}^{-\frac{3}{2}{\left(10\right)}^2}da\\ & \approx & 9.949\times {10}^{-65}.\\ & & \end{array}$

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\times$${10}^{-23}~{10}^{-23}$ , and faster than$10v_{\text{rms}}$ is $9.949\times {10}^{-65}~{10}^{-64}$.