Physics Friday 68

Maxwell Speed Distribution

Previously, we described the Boltzmann factor. Multiplying a normalization constant to give an actual probability, we obtain the Boltzmann distribution:
$f(E)=Ae^{-\frac{E}{kT}}$ .
Now, suppose that we have a system of tiny non-interacting particles (an ideal gas), with the energy being purely kinetic. Thus, the probability that a particle of mass m has a speed v is proportional to $e^{-\frac{mv^2}{2kT}}$ .
However, in 3-dimensional velocity space, the velocity vectors that give the same speed form a sphere, with radius v; the higher the speed v, the larger the number of possible velocity vectors there are. Thus, the distribution of speeds is proportional to $4\pi{v^2}$ , the surface area of the sphere in velocity space. Combining, we have
$f(v)=A\cdot4\pi{v^2}e^{-\frac{mv^2}{2kT}}$ . Normalizing, we have
$\begin{eqnarray}1&=&\int_0^{\infty}f(v)\,dv\\&=&4\pi{A}\int_0^{\infty}v^2e^{-\frac{mv^2}{2kT}}\,dv\\&=&4\pi{A}\frac{\operatorname{\Gamma}\left(\frac{3}{2}\right)}{2\left(\frac{m}{2kT}\right)^{3/2}}\\1&=&A\left(\frac{2\pi{k}T}{m}\right)^{3/2}\\A&=&\left(\frac{m}{2\pi{k}T}\right)^{3/2}\end{eqnarray}$ ,
(where we’ve used the fact that $\int_0^{\infty}x^ne^{-ax^2}\,dv=\frac{\operatorname{\Gamma}\left(\frac{n+1}{2}\right)}{2a^{(n+1)/2}}$ )
so
$f(v)=\left(\frac{m}{2\pi{k}T}\right)^{3/2}4\pi{v^2}e^{-\frac{mv^2}{2kT}}$ .
This is the Maxwell speed distribution. Now, we can find three important speeds with this:
I. Most probable speed:
This is v_p>0 where $\frac{df}{dv}(v_p)=0$ . The derivative is
$\frac{df}{dv}=\left(\frac{m}{2\pi{k}T}\right)^{3/2}4\pi\left[2v-\frac{mv^3}{kT}\right]e^{-\frac{mv^2}{2kT}}$ ,
and this is zero for v_p>0 when
$\begin{eqnarray}2-\frac{mv_p^2}{kT}&=&{0}\\\frac{mv_p^2}{kT}&=&2\\v_p&=&\sqrt{\frac{2kT}{m}}\end{eqnarray}$ .

II. Mean speed
$\begin{eqnarray}\bar{v}&=&\int_{0}^{\infty}vf(v)\,dv\\&=&\left(\frac{m}{2\pi{k}T}\right)^{3/2}4\pi\int_{0}^{\infty}v^3e^{-\frac{mv^2}{2kT}}\,dv\\&=&\left(\frac{m}{2\pi{k}T}\right)^{3/2}4\pi\frac{1}{2\left(\frac{m}{2kT}\right)^2}\\\bar{v}&=&\sqrt{\frac{8kT}{\pi{m}}}\end{eqnarray}$ .

III. Root-mean-square speed:
$\begin{eqnarray}v_{rms}&=&\sqrt{\int_{0}^{\infty}v^2f(v)\,dv}\\&=&\sqrt{\left(\frac{m}{2\pi{k}T}\right)^{3/2}4\pi\int_{0}^{\infty}v^4e^{-\frac{mv^2}{2kT}}\,dv}\\&=&\sqrt{\left(\frac{m}{2\pi{k}T}\right)^{3/2}4\pi\frac{\operatorname{\Gamma}\left(\frac{5}{2}\right)}{2\left(\frac{m}{2kT}\right)^{5/2}}}\\v_{rms}&=&\sqrt{\frac{3kT}{m}}\end{eqnarray}$ .

The last of these gives us the average kinetic energy of ideal gas molecules:
$\bar{K.E.}=\frac{1}{2}mv_{rms}^2=\frac{3}{2}kT$

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Tags: Boltzmann Factor, Friday Physics, Maxwell Speed Distribution, physics, Statistical Mechanics, Thermodynamics

This entry was posted on April 17, 2009 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “Physics Friday 68”

Physics Friday 69 « Twisted One 151’s Weblog Says:
April 24, 2009 at 12:08 am | Reply
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Physics Friday 70 « Twisted One 151’s Weblog Says:
May 1, 2009 at 1:00 pm | Reply
[...] Friday 70 By twistedone151 We noted previously that for an ideal gas, the molecule speeds should follow the Maxwell speed distribution; from this, [...]
Physics Friday 72 « Twisted One 151’s Weblog Says:
May 15, 2009 at 12:06 am | Reply
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