At constant temperature, it is found that the product of the pressure and volume of an ideal gas are constant

$PV=\mathrm{constant}$

This is named Boyle's Law, after Robert Boyle who formulated it in 1662.

Joesph Louis Gay-Lussac published Charles' Law in 1802, attributing it to unpublished work of Jacques Charles in the 1780's (Gay-Lussac has his own law..though it's not clear he should!).

Charles' Law states that at constant pressure the volume of a gas is proportional to the temperature.

$V\propto T$

Gay Lussac's Law states that for a fixed volume the pressure is proportional to the temperature

$P\propto T$

The combination of the previous 3 laws implies that

$PV\propto T$

Our previous laws were for systems of constant mass, but we can see that the amount of mass should effect the volume (at a given pressure) or the pressure (at a given volume).

$PV\propto mT$

Measuring the amount of mass in moles will allow us to write the ideal gas law in terms of a universal constant. A mole of gas is a given number of molecules, Avagadro's number, $N_{A}=6.02\times 10^{23}$. If we have a certain mass $m$ of a gas which has a certain molecular mass (measured in atomic mass units, $\mathrm{u}$, which are also the number of grams per mole.), the the number of moles $n$ is given by

$n=\frac{m[\mathrm{g}]}{\textrm{molecular mass}[\mathrm{g/mol}]}$

and

$PV=nRT$ where $R=8.314\mathrm{J/(mol.K)}$

This equation is the ideal gas law

The ideal gas law can also be written in terms of the number of molecules $N$

$PV=nRT=\frac{N}{N_{A}}RT=NkT$

where $k=\frac{R}{N_{A}}=\frac{8.314\mathrm{J/(mol.K)}}{6.02\times 10^{23}}=1.38\times 10^{-23}\mathrm{J/K}$ is the Boltzmann Constant.

If $PV=nRT$ the absolute zero temperature occurs when $P=0$. In practice most gases will liquefy before this point, but we can measure the pressure of fixed volume of gas at a couple of reference points and extrapolate down to zero pressure to get an estimate for absolute zero.

There are a number of conditions which must be satisfied for a gas to be considered ideal

- There must be a large number of molecules and they should move in random directions with a range of different speeds.
- The spacing between molecules should be much greater than the size of the molecules.
- Molecules are assumed to interact only through collisions.
- The collisions are assumed to be elastic.

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

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

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

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

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

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

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

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

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

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