The Kinetic Molecular Theory pictures a gas as molecules in constant motion. How fast do they actually go? Faster than a passenger jet. And lighter molecules go faster still, which you can use, with Square Roots and Perfect Squares, to tell gases apart.
Key idea
At a given temperature every gas has the same average kinetic energy per mole, $\tfrac{3}{2}RT$. Because $KE = \tfrac{1}{2}mv^2$, lighter molecules must move faster: $$u_\text{rms} = \sqrt{\frac{3RT}{\mathcal{M}}}$$ ($R = 8.314\ \text{J/(mol·K)}$, $\mathcal{M}$ in kg/mol, $u$ in m/s). Graham's law for effusion through a tiny hole: $$\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{\mathcal{M}_2}{\mathcal{M}_1}}.$$
Why it works
Temperature is a measure of the average kinetic energy of the molecules, so two gases at the same temperature share the same average $\tfrac{1}{2}mv^2$. If one molecule is 16 times heavier, its $v^2$ must be 16 times smaller, so its speed is 4 times smaller: speed goes as $1/\sqrt{\mathcal{M}}$. Effusion is molecules escaping through a pinhole one by one, so the escape rate is proportional to how fast they arrive at the hole, which is proportional to their speed. That is Graham's law.
Try it in Gas in a Box: open the pinhole and count which molecules escape first.
Worked example: How fast is nitrogen?
Find the root-mean-square speed of $\mathrm{N_2}$ molecules ($28.02\ \text{g/mol}$) at $25^\circ\text{C}$.
- $\mathcal{M} = 0.02802\ \text{kg/mol}$; $T = 298.15\ \text{K}$.
- $$u_\text{rms} = \sqrt{\dfrac{3(8.314)(298.15)}{0.02802}} = \sqrt{2.654 \times 10^{5}} = 515\ \text{m/s}$$
- Check the units: J/kg $=$ m²/s², whose square root is m/s.
Answer: $515\ \text{m/s}$, about 1150 miles per hour.
Worked example: Comparing effusion rates
How many times faster does $\mathrm{H_2}$ ($2.02\ \text{g/mol}$) effuse than $\mathrm{O_2}$ ($32.00\ \text{g/mol}$)?
- Graham's law with the lighter gas on top: $\dfrac{\text{rate}_{\mathrm{H_2}}}{\text{rate}_{\mathrm{O_2}}} = \sqrt{\dfrac{\mathcal{M}_{\mathrm{O_2}}}{\mathcal{M}_{\mathrm{H_2}}}}$.
- $\sqrt{\dfrac{32.00}{2.02}} = \sqrt{15.84} = 3.98$: about 4, because oxygen molecules are about 16 times heavier.
Answer: About 3.98 times faster.
Worked example: Finding a molar mass
An unknown gas effuses $0.707$ times as fast as $\mathrm{O_2}$ under the same conditions. Find its molar mass.
- $\dfrac{\text{rate}_x}{\text{rate}_{\mathrm{O_2}}} = \sqrt{\dfrac{32.00}{\mathcal{M}_x}} = 0.707$.
- Square both sides: $\dfrac{32.00}{\mathcal{M}_x} = 0.500$, so $\mathcal{M}_x = 64.0\ \text{g/mol}$ (possibly $\mathrm{SO_2}$).
Answer: $64.0\ \text{g/mol}$
Common mistake
Thinking heavier molecules have more kinetic energy at the same temperature, or leaving out the square root in Graham's law. All gases at one temperature share the same average kinetic energy; heavy ones simply move slower. A gas 16 times heavier effuses 4 times slower, not 16. Check: the lighter gas always effuses faster, by the square root of the mass ratio.