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 $kg mol^{-1}$ (not g $mol^{-1}$).
- For nitrogen (N_2) at room temperature (293 K): M = 0.028 $kg mol^{-1}$, $c_{rms} \approx$ 510 $m s^{-1}$.
- For hydrogen (H_2) at room temperature: M = 0.002 $kg mol^{-1}$, $c_{rms} \approx$ 1900 $m s^{-1}$.
Worked Example
Calculate the rms speed of helium atoms (M = $4.0 \times 10^{-3}$ $kg mol^{-1}$) 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
$c_{rms} = 1370 m s^{-1} \approx 1400 m s^{-1}.$
Answer
c_{rms} ≈$ 1400 $m \(s^{-1}\)
Common Mistake
MEDIUM
Wrong: Using molar mass in g mol^{-1} instead of kg mol^{-1} in c_{rms} = \sqrt{3RT/M}.
Right: Always convert molar mass to kg mol^{-1}. For helium: M = 4.0 g mol^{-1} = 4.0 \times \(10^{-3}\) kg mol^{-1}.
Right: Always convert molar mass to kg mol^{-1}. For helium: M = 4.0 g mol^{-1} = 4.0 \times \(10^{-3}\) kg mol^{-1}.