Kinetic Theory of Gases

In the previous lesson you tracked heat and work going in and out of a gas without ever asking what the gas was made of. A cylinder of air was a black box with a pressure, a volume and a temperature painted on the side. The first law told you how those three bookkeeping numbers had to balance, and that was enough to solve a great many problems.

Now open the box. Inside are something like $10^{25}$ molecules per cubic metre, each one flying in a straight line until it hits something, each one obeying nothing more exotic than $F = ma$ and conservation of momentum. Nothing in there knows what "pressure" or "temperature" means. So where do those quantities come from? If you can answer that (if you can get from bouncing billiard balls to $PV = Nk_BT$ using only mechanics you already have) then temperature stops being a reading on a thermometer and becomes something you can picture.

Pressure is a rain of impacts

Hold your palm out. Air pushes on it with about $10^5$ newtons per square metre, roughly the weight of a small car spread over the palm, and you feel nothing because the same push acts on the other side. That push is not a fluid leaning on you. It is roughly $10^{27}$ separate collisions per second, each one far too small and far too brief to notice individually. Averaged over any time you can perceive, the hail of impacts is perfectly steady, and you call the average force per unit area pressure.

The word average is doing the real work there. Individual molecules arrive at all speeds and all angles. What you want is the mean rate at which they deliver momentum to a wall, and momentum delivery per unit time is force: that is Newton's second law written as $F = \Delta p / \Delta t$, and it is the single tool this whole derivation runs on.

Counting the hits

Take a cube of side $L$, volume $V = L^3$, holding $N$ identical molecules each of mass $m$. Pick the wall on the right, the one perpendicular to the $x$ axis, with area $A = L^2$. Assume the molecules bounce elastically off the walls and, for now, that they do not hit each other. (They do, constantly, but a collision between two molecules just swaps their momenta around inside the gas and does not change the total delivered to the wall.)

Follow one molecule with velocity components $v_x$, $v_y$, $v_z$. When it strikes the right wall it reverses $v_x$ and leaves $v_y$ and $v_z$ alone. Its momentum in the $x$ direction goes from $+mv_x$ to $-mv_x$, a change of $-2mv_x$. By Newton's third law the wall receives $+2mv_x$.

How often does that happen? After bouncing off the right wall the molecule crosses to the left wall and comes back, a round trip of distance $2L$ at constant $x$ speed $v_x$. So the time between successive hits on the right wall is $\Delta t = 2L / v_x$, and the average force this one molecule exerts is

Add up all $N$ molecules. They have different $v_x$, so what appears is the sum of $v_x^2$ over the whole gas, which is $N$ times the average value $\overline{v_x^2}$:

Now use symmetry. There is no reason a gas in a box should prefer one direction, so $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}$. Since the speed satisfies $v^2 = v_x^2 + v_y^2 + v_z^2$, averaging gives $\overline{v^2} = 3\overline{v_x^2}$, so $\overline{v_x^2} = \tfrac13 \overline{v^2}$. Divide the force by the area $A = L^2$ and note $L^2 \cdot L = V$:

The combination $Nm/V$ is just the mass density $\rho$. So

That is a real result and it is worth pausing on. Pressure rises if you pack in more mass per unit volume, and rises with the square of the molecular speed: one factor because each impact is harder, one because impacts come more often.

What temperature actually measures

Rewrite the pressure result by multiplying and dividing by 2:

where $\bar K = \tfrac12 m \overline{v^2}$ is the average translational kinetic energy of one molecule. So $PV$ is, up to a factor of $\tfrac23$, the total translational kinetic energy of the gas. Nothing experimental has entered yet; this is pure mechanics.

Now bring in the one measured fact: gases obey $PV = Nk_BT$, with $k_B = 1.38 \times 10^{-23}$ J/K. Set the two expressions for $PV$ side by side and the $N$ cancels:

Temperature is average translational kinetic energy per molecule, in different units. That is the whole content of the word. A thermometer is an energy meter with an awkward scale.

Solve for the speed. The square root of $\overline{v^2}$ is called the root-mean-square speed, written $v_{\text{rms}}$: square the speeds, take the mean, take the root, in that order:

The second form uses per-mole quantities: $M$ is the molar mass in kg/mol and $R = 8.31$ J/mol·K. Both forms give the same number, because $R = N_A k_B$ and $M = N_A m$. Use whichever matches the data you were given, and never mix them: putting a molar mass into the $k_B$ formula is the most common way to be off by $10^{23}$.

Example 1: how fast is the air in this room

Nitrogen, $M = 0.0280$ kg/mol, at $T = 300$ K.

The numerator is $3 \times 8.31 \times 300 = 7479$. Divide by 0.0280 to get $2.671 \times 10^5$. The square root is 517, so $v_{\text{rms}} = 517$ m/s.

Carry that number. Air molecules move at roughly 500 m/s, a little faster than the 343 m/s speed of sound, which is exactly as it should be, since sound is carried by those molecules and cannot outrun them.

Example 2: a molecular beam that sticks

A beam of molecules, each of mass $3.00 \times 10^{-26}$ kg, hits a surface head-on at 600 m/s, at a rate of $5.00 \times 10^{20}$ per second, spread over $2.00 \times 10^{-4}$ m². The molecules stick instead of bouncing. Find the pressure.

Start where the derivation started: momentum per impact. A sticking molecule goes from $mv$ to zero, so it delivers $mv$, not $2mv$.

Force is momentum per second, so multiply by the arrival rate:

Divide by the area:

Note what did not appear: no temperature, no $k_B$, no $\tfrac13$. The one third came from averaging over random directions, and a beam is not random: it all goes one way. When a problem gives you a rate of arrival, go back to $F = \Delta p/\Delta t$ and build the answer from scratch.

Three different average speeds

Molecules do not all move at $v_{\text{rms}}$. Their speeds are spread out in the Maxwell distribution, and three different averages of that spread get used, all within a few percent of each other:

• the most probable speed, $\sqrt{2RT/M}$ (the peak of the curve,
• the mean speed $\bar v = \sqrt{8RT/(\pi M)}$) used for effusion and mean free path,
• the rms speed $\sqrt{3RT/M}$: the one tied to energy, since energy needs $\overline{v^2}$.

They are in the ratio $\sqrt{2} : \sqrt{8/\pi} : \sqrt{3}$, that is $1.41 : 1.60 : 1.73$. The rms speed is always the largest, because squaring weights the fast molecules more heavily before you average.

How far a molecule gets: mean free path

A nitrogen molecule at 500 m/s does not cross the room in a hundredth of a second. It crosses about 60 nanometres and hits something. Picture a molecule of diameter $d$ sweeping out a tube: it collides with any molecule whose centre lies within $d$ of its own centre line, so the tube has cross-section $\pi d^2$. Travelling a length $\lambda$ sweeps a volume $\pi d^2 \lambda$, and that volume contains one other molecule when $\pi d^2 \lambda n = 1$, where $n = N/V$ is the number density. Because the targets are moving too, a careful treatment adds a factor of $\sqrt{2}$:

Since $P = nk_BT$, you can swap $n$ for $P/k_BT$ and get $\lambda = k_BT/(\sqrt{2}\pi d^2 P)$: the same statement, useful when a problem hands you a pressure instead of a density. The collision rate follows immediately: distance per second divided by distance per collision,

Example 3: a molecular diameter from a measurement

A gas with $n = 2.70 \times 10^{25}$ m⁻³ has a measured mean free path of $9.00 \times 10^{-8}$ m. What is the molecular diameter?

Rearrange for $d^2$ first, then take the root at the very end:

The denominator: $\sqrt{2}\,\pi = 4.443$; times $9.00 \times 10^{-8}$ gives $3.999 \times 10^{-7}$; times $2.70 \times 10^{25}$ gives $1.080 \times 10^{19}$. So $d^2 = 9.26 \times 10^{-20}$ m², and

Three ångströms, which is the right size for a small molecule. Whenever you compute an atomic length, check it against that scale: if you get $10^{-5}$ m or $10^{-15}$ m you have dropped a factor somewhere.

Effusion: leaking through a pinhole

Punch a hole so small that its width is much less than $\lambda$. Molecules do not flow through as a gust; they wander through one at a time, whenever a molecule happens to be heading that way. Counting how many cross a plane per unit area per unit time gives the effusion flux

The $n\bar v$ is a rate per area from the same "how many are close enough and moving the right way" argument as before; the quarter comes from averaging over all the directions that count as towards the hole. Multiply by the hole area $A$ to get molecules per second: $\dot N = \tfrac14 n\bar v A$.

The consequence worth remembering is that $\bar v \propto 1/\sqrt{M}$, so the lighter gas leaks faster:

and since a fixed amount of gas takes a time inversely proportional to the rate, $t \propto \sqrt{M}$. That single square root is how uranium was enriched: $^{235}$UF₆ and $^{238}$UF₆ differ by less than one percent in mass, so one stage separates them by a factor of only $\sqrt{0.352/0.349} = 1.0043$, and you must cascade hundreds of stages, each multiplying the enrichment.

Recognising when this is the technique

Reach for kinetic theory when a problem mentions anything on the molecular scale: a molecule's mass, a number density in per cubic metre, a molecular diameter, a count of impacts per second, a hole gas leaks through. Thermodynamics alone can never produce those; only a molecular picture can.

Three more specific signals:

If the problem gives a temperature and asks for a speed (or the reverse), you want $v_{\text{rms}} = \sqrt{3RT/M}$. Check first whether you have molar mass with $R$, or molecular mass with $k_B$.

If it gives a rate of molecules arriving and asks for force or pressure, do not use any formula from this lesson. Go back to $F = \Delta p/\Delta t$ and decide whether each molecule delivers $mv$ (sticks) or $2mv$ (bounces).

If it compares two gases at the same temperature, the answer is almost always a square root of a mass ratio: for speeds, for effusion rates, for leak times. Equal kinetic energies, unequal speeds. And if it asks about kinetic energy rather than speed, the ratio is 1.

The one trap that catches good students is dimensional carelessness: $M$ in kg/mol, not g/mol, and $T$ in kelvin, always. Nitrogen is 0.0280 kg/mol, not 28.

You now have the microscopic machinery. Next you will see what happens when molecules can store energy in ways other than flying in a straight line (rotating, vibrating) and how the equipartition theorem turns that count of degrees of freedom into heat capacities you can measure. That is Ideal Gas Law and Equipartition.