Pressure of an Ideal Gas

The pressure of a gas on its container walls is the macroscopic manifestation of countless tiny molecular impacts. Using only Newton's laws and elementary statistics, we can derive a beautiful formula relating pressure to mean-square molecular speed.

Concept

Consider $N$ molecules of mass $m$ in a cubic container of side $L$ (volume $V = L^3$). Let one molecule have velocity components $(v_x, v_y, v_z)$.

The pressure is defined as force per unit area on the walls:

$P = \frac{F}{A}$

We compute the time-averaged force exerted on a wall and divide by its area.

Derivation

Step 1 — Momentum change per collision. A molecule moving with $x$-velocity $v_x$ strikes the right wall (perpendicular to the $x$-axis), bounces back elastically: its $x$-velocity reverses, $v_x \to -v_x$.

Momentum change of the molecule: $\Delta p = -m v_x - (m v_x) = -2 m v_x$.

By Newton's third law, the wall receives momentum $+2 m v_x$ per impact.

Step 2 — Frequency of collisions. Between successive collisions of this molecule with the right wall, it travels a distance $2L$ (across to the left wall and back). Time between collisions:

$\Delta t = \frac{2L}{v_x}$

Step 3 — Average force from one molecule. Average force on the wall by this single molecule:

$F_1 = \frac{\text{momentum delivered}}{\text{time}} = \frac{2 m v_x}{2L/v_x} = \frac{m v_x^2}{L}$

Step 4 — Sum over all molecules. Total force on the right wall:

$F = \sum_{i=1}^{N} \frac{m v_{xi}^2}{L} = \frac{m}{L}\sum_{i=1}^{N} v_{xi}^2 = \frac{m N}{L} \langle v_x^2\rangle$

where $\langle v_x^2\rangle = \tfrac{1}{N}\sum v_{xi}^2$ is the mean of $v_x^2$.

Step 5 — Isotropy. Because molecular motion is random and isotropic,

$\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle = \tfrac{1}{3}\langle v^2\rangle$

where $\langle v^2\rangle = \langle v_x^2 + v_y^2 + v_z^2\rangle$ is the mean-square speed.

Step 6 — Pressure. Wall area is $A = L^2$, so

$P = \frac{F}{A} = \frac{m N \langle v_x^2\rangle}{L \cdot L^2} = \frac{m N \langle v^2\rangle}{3 V}$

Introducing the mass density $\rho = m N / V$:

$\boxed{P = \frac{1}{3}\rho\, \langle v^2\rangle}$

Equivalently, $PV = \tfrac{1}{3} N m \langle v^2\rangle$.

Worked Example

Q: Oxygen ($M = 32$ g/mol) is at STP ($P = 1.013 \times 10^5$ Pa, $T = 273$ K). Density of O$_2$ at STP is $\rho \approx 1.43$ kg/m³. Estimate the RMS speed.

A: From $P = \tfrac{1}{3}\rho v_{\text{rms}}^2$:

$v_{\text{rms}} = \sqrt{\frac{3 P}{\rho}} = \sqrt{\frac{3 \times 1.013\times 10^5}{1.43}} \approx \sqrt{2.125 \times 10^5} \approx 461 \text{ m/s}$

This is consistent with the value calculated from $v_{\text{rms}} = \sqrt{3 k_B T / m}$.

Common Confusions

• Average speed is not RMS speed. The derivation gives $\langle v^2\rangle$, not $\langle v\rangle^2$. In general $\langle v^2\rangle > \langle v\rangle^2$.
• The cube was used for convenience — the result holds for any container shape since pressure is an intensive property.
• The wall doesn't need to be rigid in the limit, only that collisions be elastic on average.

Key Takeaways

• Pressure arises from molecular momentum transfer to walls.
• The central result is $P = \tfrac{1}{3}\rho \langle v^2\rangle$, equivalently $PV = \tfrac{1}{3} N m \langle v^2\rangle$.
• Isotropy of motion is the crucial assumption that connects $\langle v_x^2\rangle$ to $\tfrac{1}{3}\langle v^2\rangle$.