From $\frac{1}{2}$m\langle \(c^{2}\) \rangle = $\frac{3}{2}$kT, rearrange to get \langle
Thermal Physics - OCR A-Level Physics
- From $\frac{1}{2}m\langle c^{2} \rangle = \frac{3}{2}kT$, rearrange to get $\langle c^{2} \rangle = \frac{3kT}{m}$.
- So $c_{rms} = \sqrt{\frac{3kT}{m}}$ for one molecule, or $c_{rms} = \sqrt{\frac{3RT}{M}}$ for molar quantities.
- $m$ is the mass of ONE molecule in kg. $M$ is the molar massThe mass of one mole of a substance, in kg mol⁻¹ (or g mol⁻¹). E.g. nitrogen N₂ = 0.028 kg mol⁻¹. in $\text{kg mol}^{-1}$ (not $\text{g mol}^{-1}$).
- For nitrogen ($N_2$) at room temperature (293 K): $M = 0.028 \text{ kg mol}^{-1}$, $c_{rms} \approx 510 \text{ m s}^{-1}$.
- For hydrogen ($H_2$) at room temperature: $M = 0.002 \text{ kg mol}^{-1}$, $c_{rms} \approx 1900 \text{ m s}^{-1}$.
Worked Example
Calculate the rms speed of helium atoms ($M = 4.0 \times 10^{-3} \text{ kg mol}^{-1}$) at 300 K.
Show Solution
1
Use $c_{rms} = \sqrt{\frac{3RT}{M}}$.
2
$c_{rms} = \sqrt{\frac{3 \times 8.31 \times 300}{4.0 \times 10^{-3}}}$.
3
$c_{rms} = \sqrt{\frac{7479}{4.0 \times 10^{-3}}}$.
4
$c_{rms} = \sqrt{1.87 \times 10^6}$.
5
$c_{rms} = 1370 \text{ m s}^{-1} \approx 1400 \text{ m s}^{-1}$.
Answer
$c_{rms} \approx 1400 \text{ m s}^{-1}$
Common Mistake
MEDIUM
Wrong: Using molar mass in $\text{g mol}^{-1}$ instead of $\text{kg mol}^{-1}$ in $c_{rms} = \sqrt{3RT/M}$.
Right: Always convert molar mass to $\text{kg mol}^{-1}$. For helium: $M = 4.0 \text{ g mol}^{-1} = 4.0 \times 10^{-3} \text{ kg mol}^{-1}$.
Right: Always convert molar mass to $\text{kg mol}^{-1}$. For helium: $M = 4.0 \text{ g mol}^{-1} = 4.0 \times 10^{-3} \text{ kg mol}^{-1}$.