Kinetic Theory of Gases and Gas Pressure

So far the molecular picture has been used qualitatively: internal energy is molecular kinetic plus potential energy, temperature measures mean molecular kinetic energy. This chapter makes the picture quantitative for the simplest possible case — a gas dilute enough that its molecules barely interact except in brief collisions. From a short list of mechanical assumptions and nothing more than Newton's laws, we will derive the pressure a gas exerts on its container, obtaining

and then, by comparing it with the experimentally established gas law, pin down exactly what temperature is at the molecular level: $\bar{E}_k = \tfrac{3}{2}k_BT$, the result quoted without proof in Chapter 1.

This is one of the landmark derivations of classical physics — the first time anyone showed that a bulk, measurable property (pressure) could be computed from the mechanics of invisible particles. Follow it with pen and paper; the algebra is not hard, but each step encodes a physical idea worth understanding.

Reading note. The derivation in Section 2 is examinable in full. Learn the sequence of ideas — one molecule's momentum change, its collision rate, the sum over all molecules, the switch from one direction to three — rather than memorising the final line.

1. The Assumptions of the Kinetic Model

An ideal gas is a gas that obeys the following model exactly. Real gases obey it closely when they are dilute — low pressure, and a temperature well above the point at which they would liquefy.

1. A gas consists of a very large number $N$ of identical molecules in continual random motion, with a wide distribution of speeds and directions.
2. The molecules are point-like: their total volume is negligible compared with the volume of the container.
3. There are no intermolecular forces except during collisions. Between collisions each molecule moves in a straight line at constant velocity, and the intermolecular potential energy is negligible — so the internal energy of an ideal gas is purely kinetic.
4. All collisions — molecule with wall, molecule with molecule — are perfectly elastic: kinetic energy is conserved. The duration of a collision is negligible compared with the time between collisions.
5. The molecules obey Newtonian mechanics.

2. Deriving the Pressure of an Ideal Gas

Consider $N$ molecules, each of mass $m$, in a cubical box of side $L$ (volume $V = L^{3}$). Take one wall perpendicular to the $x$-axis and follow a single molecule with $x$-component of velocity $v_x$.

Step 1 — momentum change in one collision. The molecule strikes the wall, and because the collision is elastic its $x$-velocity reverses from $+v_x$ to $-v_x$ while $v_y$ and $v_z$ are unchanged. The change in the molecule's momentum is

so, by Newton's third law, the molecule delivers momentum $+2mv_x$ to the wall.

Step 2 — how often it hits. After bouncing off this wall the molecule crosses to the opposite wall, a distance $L$ away, bounces back, and returns — a round trip of $2L$ in the $x$-direction, taking

between successive collisions with the same wall.

Step 3 — force from one molecule. The average force this molecule exerts on the wall is the momentum it delivers per unit time:

Step 4 — sum over all molecules. Adding the contributions of all $N$ molecules, with their different $x$-velocities $v_{x,1}, v_{x,2}, \dots$,

where $\overline{v_x^{2}} = \dfrac{1}{N}\sum v_{x,i}^{2}$ is the mean square $x$-velocity.

Step 5 — three directions instead of one. The motion is random, so no direction is special:

The speed $c$ of a molecule satisfies $c^{2} = v_x^{2} + v_y^{2} + v_z^{2}$, so averaging gives $\overline{c^{2}} = 3\,\overline{v_x^{2}}$, i.e.

Step 6 — pressure. Pressure is force per unit area, and the wall has area $L^{2}$:

Rearranging, and introducing the gas density $\rho = \dfrac{Nm}{V}$:

That factor of $\tfrac{1}{3}$ carries a physical meaning: it is the share of the molecular motion directed at any one pair of walls.

3. The Root-Mean-Square Speed

Because the molecules move in every direction, the mean velocity of the gas is zero — it is going nowhere as a whole. The average that matters is the mean square speed $\overline{c^{2}}$, and its square root is the root-mean-square (rms) speed:

The rms speed is a particular kind of average, slightly larger than the plain mean speed and than the most probable speed, and it is the one that appears naturally in the pressure result:

4. Connecting to Temperature and Internal Energy

Rewrite the pressure result by inserting a factor of $2$ top and bottom, so that the mean molecular kinetic energy $\bar{E}_k = \tfrac{1}{2}m\overline{c^{2}}$ appears:

Experiment — the gas laws of the next chapter — establishes independently that $pV = Nk_BT$ for an ideal gas, where $k_B = 1.38\times 10^{-23}\ \dfrac{\text{J}}{\text{K}}$. Setting the two expressions for $pV$ equal:

This is the microscopic meaning of temperature, now derived rather than asserted. Two immediate consequences:

(using $M = N_A m$ for the molar mass and $R = N_A k_B$), and the internal energy of an ideal gas:

exactly as quoted in Chapter 1.

5. Worked Examples

Example 1 — rms speed of nitrogen. Find the rms speed of nitrogen molecules ($M = 28.0\times 10^{-3}\ \dfrac{\text{kg}}{\text{mol}}$) at $300\ \text{K}$.

Hydrogen molecules ($M = 2.0\times 10^{-3}$) at the same temperature move $\sqrt{14}\approx 3.7$ times faster, near $1900\ \dfrac{\text{m}}{\text{s}}$ — fast enough that a small fraction always exceeds Earth's escape velocity, which is why the atmosphere retains nitrogen but has lost almost all its hydrogen.

Example 2 — pressure from density and speed. Air near the ground has density $\rho = 1.2\ \dfrac{\text{kg}}{\text{m}^{3}}$ and an rms molecular speed of about $500\ \dfrac{\text{m}}{\text{s}}$. Estimate the pressure.

which recovers atmospheric pressure — a strong consistency check on the whole model.

6. What the Model Leaves Out

The kinetic model is an idealisation, and it is worth knowing where it breaks:

• All molecules are not identical in speed. They follow the Maxwell–Boltzmann distribution; $c_{\text{rms}}$ is one summary of a broad spread.
• Real molecules have finite size, so at high density the "available" volume is less than $V$, raising the pressure above the ideal prediction.
• Real molecules attract one another, so at low temperature the attractions pull molecules back slightly as they approach a wall, lowering the pressure below the ideal prediction — and eventually causing the gas to condense.
• Diatomic and polyatomic molecules store energy in rotation and vibration as well as translation, so their internal energy is more than $\tfrac{3}{2}nRT$ and $\bar{E}_k = \tfrac{3}{2}k_BT$ refers only to the translational part.

Common Exam Pitfalls

• Using the mean velocity instead of $\overline{c^{2}}$. The mean velocity of a gas at rest is zero; pressure depends on the mean square speed.
• Losing or misplacing the factor $\tfrac{1}{3}$. It comes from resolving the random motion equally among three perpendicular directions.
• Confusing $c_{\text{rms}}$ with the mean speed or most probable speed. They are close but not equal; only $c_{\text{rms}}$ appears in $p = \tfrac{1}{3}\rho\,c_{\text{rms}}^{2}$.
• Mixing $\sqrt{3k_BT/m}$ with $\sqrt{3RT/M}$. Use molecular mass $m$ with $k_B$, or molar mass $M$ with $R$ — never one from each pair.
• Thinking pressure comes from molecules colliding with each other. In this model pressure is entirely due to collisions with the walls; intermolecular collisions only redistribute speeds.
• Substituting a Celsius temperature into $\bar{E}_k = \tfrac{3}{2}k_BT$ or $c_{\text{rms}} = \sqrt{3k_BT/m}$. Absolute temperature only.
• Applying $\bar{E}_k = \tfrac{3}{2}k_BT$ to the total energy of a diatomic gas. It gives the translational part only.

Summary

• The kinetic model treats a gas as many identical point molecules in random motion, with no intermolecular forces except in brief elastic collisions, obeying Newtonian mechanics.
• Following one molecule's momentum change ($2mv_x$ per bounce), its collision rate ($v_x/2L$), summing over all molecules, and resolving into three equal directions gives $$pV = \tfrac{1}{3}Nm\,\overline{c^{2}} \qquad p = \tfrac{1}{3}\rho\,\overline{c^{2}}$$
• The rms speed $c_{\text{rms}} = \sqrt{\overline{c^{2}}}$ is the average that appears in the pressure result; the mean velocity of a static gas is zero.
• Writing $pV = \tfrac{2}{3}N\bar{E}_k$ and comparing with the experimental $pV = Nk_BT$ gives $$\bar{E}_k = \tfrac{3}{2}k_BT, \qquad c_{\text{rms}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}}, \qquad U = \tfrac{3}{2}nRT$$
• At a given temperature, lighter molecules have higher rms speeds ($c_{\text{rms}} \propto 1/\sqrt{m}$), which is why planets retain heavy gases more easily than light ones.
• The model fails at high density (finite molecular size raises $p$) and low temperature (intermolecular attraction lowers $p$, then condenses the gas); $\bar{E}_k = \tfrac{3}{2}k_BT$ counts translational energy only.