# Kinetic theory of gases

Using only a few simple assumptions:

• A gas consists of "billiard-ball"-like particles,
• There is a function $f(v)$ which is *not* a function of time, such that at equilibrium, the number of particles with speeds between $v$ and $v+dv$ is given by $$f(v)\,dv.$$ [We don't actually need to know *what* the function $f$ is. Just that it's time independent.]
• Pressure is the force on container walls due to the particles bouncing off the walls--due to their change in momentum.

We will find..

1. Pressure is proportional to the average kinetic energy of the particles.
2. A consequence of these assumptions is the ideal gas law $$PV=nRT$$ if we identify the temperature with the speeds of the particles according to: $$\frac 32 k_BT=\frac 12m\overline{v^2},$$ where $k_B$ is a new constant of nature.

### What is pressure?

Isaac Newton's guess was that a static repulsion between pieces of matter was responsible for this force (per unit area). [Jello as 'springs'.] No motion of "atoms" is needed.

A number of scientists in the 19th century developed a kinetic theory which posits that pressure is due to collisions of constantly moving atoms.

Though we can't see atoms directly, Brownian motion in which pollen or other small particles seem to jiggle around would seem to be consistent with the idea of constantly moving atoms.

Can we account for pressure, temperature, and even the ideal gas law with a few simple assumptions?

### Assumptions of kinetic theory - an ideal gas

1. A. Greg claims these circles and relative spacings are to scale for He atoms compressed to 1950 atmospheres.
Molecules are far apart from each other. That is, typical distances are large compared to molecular sizes, and to equilibrium separations in solids.

One implication of this is that disturbances in the medium propagate with the typical speed of the molecules. So, the speed of sound ~ a typical speed molecular speed. In air at STP, the speed of sound ~ 700 mph.

2. The only intramolecular forces are associated with collisions.

3. Collisions are elastic.

4. Molecules are evenly spread in space throughout a container.

5. Molecular velocities are evenly spread in direction.

6. We have a large number of molecules. We can use statistics without worrying too much about inhomogeneities.

### Statistics

At equilibrium, we shall assume that there is a time-independent function, $f(v)$, such that $f(v)\,dv$ is the fraction of all the molecules with speeds between $v$ and $v+dv$.

The fraction of the molecules with speeds inclusively between 0 and $\infty$ is 100% = 1.0, that is: $$\int_0^{\infty} f(v)\,dv=1.$$

Since $f(v)\,dv$ is a fraction (dimensionless), $f(v)$ must have units of $v^{-1}$. You might think of it as "probability per speed interval".

The mean speed is $$\overline{v} \equiv \int_0^{\infty} vf(v)\,dv.$$

The mean square speed is $$\overline{v^2} \equiv \int_0^{\infty} v^2f(v)\,dv.$$

This is usually something quite different from $\bar{v}^2$.

The $n$th moment of the probability distribution $f(v)$ is: $$\overline{v^n} \equiv \int_0^{\infty} v^nf(v)\,dv.$$

The root mean square speed or rms speed is: $$v_{rms}=\sqrt{\,\overline{v^2}}.$$

### Molecular flux

How many molecules hit a unit area--a portion of the wall of a container of gas--in a unit time? This is the molecular flux, $\Phi$, which has units of [molecules]$/(m^2\cdot s)$.

${\frak n} = N / V\$
is the density of molecules (number per unit volume), written as ${\frak n}$ to distinguish it from $n$ (kilomoles).

$(dA\,\cos\theta)(vdt)$
is the volume of the slanted cylinder shown.

$f(v)dv$
is the fraction of all molecules with speeds in the range $v$ to $v+dv$.

If *all* of the molecules in the speed range $v$ to $v+dv$ in the cylinder shown were moving in the direction $-\theta, -\phi$ towards $dA$, they would all hit $dA$. This number of molecules is: $${\frak n}f(v)dv\,dA\,\cos\theta\,vdt={\frak n}v f(v) \cos\theta dv\,dA\, dt.$$

But only a fraction of these atoms are actually headed in the right direction. Now, we calculate that fraction (from symmetry) by asking what the chances are that a molecule is moving in a direction with spherical coordinates between $\theta$ and $\theta+d\theta$ and between $\phi$ and $\phi +d\phi$. This probability depends on $\theta$ and $\phi$...

Since the directions of the velocity vectors are evenly spread over all directions, the fraction of atoms with directions between $\theta$ and $\theta+d\theta$ and $\phi$ to $\phi+d\phi$ is the same as the ratio of the area swept out in the figure to the right $d\sigma=(v\cos\theta\,d\phi)(v\,d\theta)$ divided by the total surface area of the sphere of radius $v$ which is $\pi v^2$.

That ratio is: $$\frac{(vd\theta)(v\sin\theta\,d\phi)}{4\pi v^2} = \frac{\sin\theta\,d\theta\, d\phi}{4\pi}.$$

So, the total number of particles in the cylinder oriented in direction $\theta, \ \phi$ with the right speed $v$ to cross $dA$ in time interval $dt$ is $dN(v,\theta,\phi)$. $$dN = {\frak n}v f(v) \cos\theta dv\,dA\, dt \frac{\sin\theta\,d\theta\, d\phi}{4\pi}.$$

The flux is the number crossing $dA$, per unit area, per unit time. $$d\Phi (v,\theta,\phi) = \frac{dN}{dA\,dt}.$$

To find the total flux, we need to integrate over all the cylinder directions $\phi$: $0\to 2\pi$ and $\theta$: $0 \to \pi/2$, and over all the speeds $v$: $0\to \infty$: $$\Phi = \frac{{\frak n}}{4\pi}\int_0^{\infty}dv\, v f(v) \int_0^{\pi/2}d\theta\, \cos\theta \, \sin\theta \int_0^{2\pi} d\phi = \frac{1}{4}{\frak n}\overline{v}.$$

[Where we've used $d(\sin \theta) = \cos\theta\,d\theta$ to write... $$\begineq \int_0^{\pi/2}\sin\theta \cos\theta \,d\theta &=& \int_0^{\pi/2}\sin\theta\,d(\sin\theta)\\ &=&\int F\,dF=F^2/2\\ &=&\left.\sin^2\theta/2\right|_0^{\pi/2} = (1-0)/2=1/2.\endeq$$ ]

### Pressure in the kinetic theory

From Newton's law: $$F=ma = m\frac{d}{dt}v=\frac{d}{dt}(mv) =\frac{d}{dt}p.$$

If many molecules are bombarding a small area $dA$, the pressure is the total change of momentum per unit area: $$P = \frac{dp}{dA\,dt}.$$

Consider a molecule bouncing elastically against a wall: $|\myv{v}| =|\myv{v'}|$ and $\theta=\theta'$. The change of momentum of the molecule is perpendicular to the wall, of magnitude: $$mv\,cos\theta -(-mv\, cos\theta)=2mv\,cos\theta.$$

To get the total change in momentum for the area $dA$ in the time interval $dt$, we should integrate this momentum change, a function of $v$, over all velocities and directions... \begin{align}dp & =\int 2mv\,\cos\theta\,dN\\ & =...\\ & = \frac{1}{3}{\frak n}m\overline{v^2} dA\,dt.\end{align}

Using the expression above for $P=\frac{dp}{dA\,dt}$: $$P=\frac{1}{3} {\frak n} m \overline{v^2} =\frac{1}{3} \frac{N}{V} m \overline{v^2}.$$

Multiply by $V$ to get something suggestive of the ideal gas law: $$PV = \frac{1}{3} Nm \overline{v^2}=\frac{2}{3}N$\frac{1}{2}m\overline{v^2}$.$$

We can write this even more suggestively as... $$PV = \frac{N}{N_A}R$\frac{2}{3}\frac{N_A}{R} \frac{1}{2}m\overline{v^2}$=nR$\frac{2}{3}\frac{N_A}{R}\frac{1}{2}m\overline{v^2}$.$$

This *is* the ideal gas law, as long as the quantity in the brackets is the temperature, confirming our frequent assertion that temperature is proportional to average kinetic energy.

The ratio $$\frac{R}{N_A}=\frac{8.31\times10^3 {\rm J/kilomole\cdot K}}{6.02\times 10^{26} {\rm molecules/kilomole}}\equiv k_B {\rm [J/K]},$$

is the Boltzmann constant. $$\frac{3}{2}k_BT = \frac{1}{2}m\overline{v^2}.$$

#### image credits

Kevin Dooley, Andrea Pacheco