The Kinetic Theory of Gases represents the first statistical theory in physics, and is thus extremely significant. It also does a pretty good job of describing ideal gases!

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We can work out the pressure exerted by an ideal gas on it's container by starting from the change in momentum of a molecule when it strikes the container wall

The average force due to one molecule is then

$F=\frac{\Delta(mv)}{\Delta t}=\frac{2mv_{x}}{2l/v_{x}}=\frac{mv_{x}^{2}}{l}$

The net force on the wall will be the sum of the forces from all $N$ molecules

$F=\frac{m}{l}\Sigma_{i=1..N} v_{xi}^{2}$

$\frac{\Sigma_{i=1..N} v_{xi}^{2}}{N}=\bar{v_{x}^{2}}$ → $F=\frac{m}{l}N\bar{v_{x}^{2}}$

$v^{2}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}$ → $\bar{v^{2}}=\bar{v_{x}^{2}}+\bar{v_{y}^{2}}+\bar{v_{z}^{2}}=\bar{3v_{x}^{2}}$

$F=\frac{m}{l}N\frac{\bar{v^{2}}}{3}$

$P=\frac{F}{A}=\frac{1}{3}\frac{Nm\bar{v^{2}}}{Al}=\frac{1}{3}\frac{Nm\bar{v^{2}}}{V}$

$PV=\frac{2}{3}N(\frac{1}{2}m\bar{v^{2}})=NkT$

$\bar{K}=\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$

If we look at a simulation of particles moving according to the kinetic theory we can get some idea of the distribution of the speeds of the particles.

The Maxwell-Boltzmann Distribution gives the probability $f(v)$ that a particular particle in an ideal gas has a given speed $v$. It was originally derived by James Clerk Maxwell based on symmetry arguments, later Ludwig Boltzmann derived it on a more general basis. We will follow Maxwell's reasoning.

Suppose we have a gas of $N$ particles. The number of particles which have a velocity in the x direction between $v_{x}$ and $v_{x}+dv_{x}$ will be $Nf(v_{x})dv_{x}$

$\int_{-\infty}^{\infty}f(v_{x})dv_{x}=1$

The same function $f$ should describe the $y$ and $z$ direction as well, so

$Nf(v_{x})f(v_{y})f(v_{z})dv_{x}dv_{y}dv_{z}$

gives the number of particles with velocity between $v_{x}$ and $v_{x}+dv_{x}$,$v_{y}$ and $v_{y}+dv_{y}$, and, $v_{z}$ and $v_{z}+dv_{z}$

Maxwell realized that all directions are equivalent the distribution function should depend only on the total speed of the particle

$f(v_{x})f(v_{y})f(v_{z})=F(v_{x}^2+v_{y}^2+v_{z}^2)$

This implies that the form of $f$ must be

$f(v_{x})=Ae^{-Bv_{x}^2}$

Which is a Gaussian Function.

The fact that the velocity in a given direction is given by the function

$f(v_{x})=Ae^{-Bv_{x}^2}$

implies that the mean velocity in any given direction is zero. Does this also mean the mean particle speed is zero?

We now know that

$f(v_{x})=Ae^{-Bv_{x}^2}$

But we need to find out what A and B are, and we also would like to express the function in terms of the speed of the molecules rather than the velocity in a given direction.

Starting from the number of particles with velocity between $v_{x}$ and $v_{x}+dv_{x}$,$v_{y}$ and $v_{y}+dv_{y}$, and, $v_{z}$ and $v_{z}+dv_{z}$

$Nf(v_{x})f(v_{y})f(v_{z})dv_{x}dv_{y}dv_{z}=NA^{3}e^{-B(v_{x}^2+v_{y}^2+v_{z}^2)}dv_{x}dv_{y}dv_{z}$ $=NA^{3}e^{-Bv^2}dv_{x}dv_{y}dv_{z}$

To transform $dv_{x}dv_{y}dv_{z}$ to an integral in the speed we consider a sphere in velocity space and can reason that the equivalent volume to $dv_{x}dv_{y}dv_{z}$ is $4\pi v^{2}dv$

So the probability distribution as a function of speed is

$f(v)dv=4\pi v^2A^{3}e^{-Bv^{2}}dv$

We now have

$f(v)dv=4\pi v^2A^{3}e^{-Bv^{2}}dv$

We can use some of the information about the system to find out what $A$ and $B$ have to be. All particles have to have a velocity between 0 and $\infty$, so

$\int_{0}^{\infty}f(v)dv=1$

We saw before that

$\bar{K}=\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$

so

$\int_{0}^{\infty}\frac{1}{2}mv^{2}f(v)dv=\frac{3}{2}kT$

We will need to use the standard integral of a gaussian function.

$\int_{-\infty}^{\infty}e^{-ax^{2}}=\sqrt{\frac{\pi}{a}}$

This is explained quite nicely on this page

The integrals we need to do

$\int_{0}^{\infty}f(v)dv$

and

$\int_{0}^{\infty}\frac{1}{2}mv^{2}f(v)dv$

can be found using integration by parts.

$\int u\,dv = uv - \int v\,du$

$\int_{0}^{\infty}f(v)dv=1$

$4\pi A^{3}\int_{0}^{\infty}v^{2}e^{-Bv^{2}}dv=1$

$\int u\,dv = uv - \int v\,du$

$u=v$ $dv=ve^{-Bv^{2}}dv$

$du=dv$ $v=-\frac{1}{2B}e^{-Bv^{2}}$

$4\pi A^{3}\int_{0}^{\infty}v^{2}e^{-Bv^{2}}dv=4\pi A^{3}([(-\frac{v}{2B}e^{-Bv^{2}})]_{0}^{\infty}+\int_{0}^{\infty}\frac{1}{2B}e^{-Bv^{2}}dv)$

$=\frac{4\pi A^{3}}{2B}\int_{0}^{\infty}e^{-Bv^{2}}dv=\frac{4\pi A^{3}}{2B}\frac{1}{2}\sqrt{\frac{\pi}{B}}=1$

$4\pi A^{3}=\frac{4B^{3/2}}{\sqrt{\pi}}$

$\bar{K}=\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$

$\int_{0}^{\infty}\frac{1}{2}mv^{2}f(v)dv=\frac{3}{2}kT$

$2m\pi A^{3}\int_{0}^{\infty}v^{4}e^{-Bv^{2}}dv=\frac{3}{2}kT$

$u=v^{3}$ $dv=ve^{-Bv^{2}}dv$

$du=3v^{2}dv$ $v=-\frac{1}{2B}e^{-Bv^{2}}$

$2m\pi A^{3}\int_{0}^{\infty}v^{4}e^{-Bv^{2}}dv=2m\pi A^{3}([(-\frac{v^3}{2B}e^{-Bv^{2}})])_{0}^{\infty}+\int_{0}^{\infty}\frac{3v^2}{2B}e^{-Bv^{2}}dv)$

$=\frac{3m}{4B}(4\pi A^3)\int_{0}^{\infty}v^{2} e^{-Bv^{2}}dv=\frac{3m}{4B}$

$\frac{3m}{4B}=\frac{3}{2}kT$

$B=\frac{m}{2kT}$

$4\pi A^{3}=\frac{4B^{3/2}}{\sqrt{\pi}}$

$B=\frac{m}{2kT}$

$f(v)dv=4\pi v^2A^{3}e^{-Bv^{2}}dv$

$f(v)=4\pi (\frac{m}{2\pi k T})^{\frac{3}{2}}v^{2}e^{-\frac{1}{2}\frac{mv^{2}}{kT}}$

Now that we have this we can find various useful quantities, such as the average velocity

$\bar{v}=\int_{0}^{\infty}vf(v)dv=2\sqrt{\frac{2}{\pi}}\sqrt{\frac{kT}{m}}\approx 1.6 \sqrt{\frac{kT}{m}}$

or the most probable speed of a particle, from the condition $\frac{df}{dv}=0$ which gives

$v=\sqrt{\frac{2kT}{m}}\approx 1.41 \sqrt{\frac{kT}{m}}$

A third measure of velocity that we could already obtain before deriving the Maxwell Boltzmann distribution, is the root-mean-square, or rms velocity. Here we take the square root of the average of the squared velocity

$\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$ → $\sqrt{\bar{v^{2}}}=\sqrt{\frac{3kT}{m}}\approx 1.73 \sqrt{\frac{kT}{m}}$

The distribution of speeds is highly temperature dependent, as can be seen in simulations and is predicted by the Maxwell distribution.

The plot below is for He atoms at various temperatures.