understand that the root-mean-square speed cr.m.s. is given by c<>
Kinetic Theory of Gases
⚛️ What’s a gas doing at the microscopic level?
Imagine a crowded playground where every child (molecule) runs around, bumping into each other and the walls. The motion of these “kids” determines the pressure, temperature and volume of the gas.
Root‑Mean‑Square Speed ($c_{\text{rms}}$)
🚀 Key Formula
The root‑mean‑square speed is the square root of the average of the squares of all molecular speeds:
$$c_{\text{rms}}=\sqrt{\frac{3kT}{m}}$$
Where $k$ = Boltzmann constant, $T$ = absolute temperature (K), and $m$ = mass of one molecule (kg).
Why “root‑mean‑square”?
📐 Analogy
Think of a group of runners who finish a race at different speeds. If you square each speed, average those squares, and then take the square root, you get a number that’s always larger than the simple average speed. That’s exactly what $c_{\text{rms}}$ does for gas molecules.
Because faster molecules contribute more to the pressure, $c_{\text{rms}}$ is a better indicator of the “effective” speed that matters in gas behaviour.
Quick Derivation (Sketch)
- Start with the kinetic energy of one molecule: $$E_k=\frac{1}{2}m\langle c^2\rangle$$
- For an ideal gas, the equipartition theorem gives: $$E_k=\frac{3}{2}kT$$
- Set the two expressions equal: $$\frac{1}{2}m\langle c^2\rangle=\frac{3}{2}kT$$
- Solve for the mean square speed: $$\langle c^2\rangle=\frac{3kT}{m}$$
- Take the square root to obtain $c_{\text{rms}}$.
Practical Examples
| Gas | Molar Mass (g mol⁻¹) | Mass per Molecule (kg) | $c_{\text{rms}}$ at 300 K (m s⁻¹) |
|---|---|---|---|
| Nitrogen (N₂) | 28 | 4.65 × 10⁻²⁶ | ≈ 520 |
| Helium (He) | 4 | 6.63 × 10⁻²⁷ | ≈ 1 200 |
| Oxygen (O₂) | 32 | 5.31 × 10⁻²⁶ | ≈ 480 |
Exam Tips & Quick Checks
?? Remember the conversion
When you see a molar mass in g mol⁻¹, first convert it to kg per molecule:
$$m = \frac{M}{N_A}$$
Where $M$ is molar mass and $N_A$ = Avogadro’s number.
⚠️ Don’t forget to keep the temperature in Kelvin!
🧠 Quick mental check: For a lighter gas (smaller $m$), $c_{\text{rms}}$ will be higher.
📌 If the question asks for the average speed, use:
$$c_{\text{avg}}=\sqrt{\frac{8kT}{\pi m}}$$
and remember that $c_{\text{rms}} > c_{\text{avg}}$.
Revision
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