Molecular Speeds

In an ideal gas the molecules move with a range of speeds described by the Maxwell–Boltzmann distribution. Three particular speeds are commonly used to characterise this distribution.

Concept

The Maxwell–Boltzmann speed distribution gives the fraction of molecules with speed between $v$ and $v + dv$:

$f(v)\, dv = 4\pi N \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 e^{-mv^2/(2 k_B T)}\, dv$

The distribution is asymmetric — a long tail toward high speeds — so its peak, mean, and root-mean-square values are all different.

Three characteristic speeds

1. Most probable speed $v_p$ — the speed at which $f(v)$ is maximum. Setting $df/dv = 0$:

$v_p = \sqrt{\frac{2 k_B T}{m}} = \sqrt{\frac{2 R T}{M}}$

2. Mean (average) speed $\langle v\rangle$:

$\langle v\rangle = \int_0^\infty v f(v)\,dv = \sqrt{\frac{8 k_B T}{\pi m}} = \sqrt{\frac{8 R T}{\pi M}}$

3. Root-mean-square speed $v_{\text{rms}}$:

$v_{\text{rms}} = \sqrt{\langle v^2\rangle} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}}$

Numerical comparison

$v_p : \langle v\rangle : v_{\text{rms}} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3} \approx 1.414 : 1.596 : 1.732$

So $v_p < \langle v\rangle < v_{\text{rms}}$ always. The differences are small (order 20%) but conceptually important.

Which one to use?

• For pressure and translational KE: use $v_{\text{rms}}$ because pressure depends on $\langle v^2\rangle$.
• For collision frequency / effusion rates: use $\langle v\rangle$.
• For peak of distribution / threshold reactions: use $v_p$.

Worked Example

Q: Calculate the three characteristic speeds for nitrogen ($M = 28$ g/mol) at $T = 300$ K.

A: $RT/M = (8.314)(300)/(0.028) = 8.908 \times 10^4$ J/kg.

• $v_p = \sqrt{2 \times 8.908 \times 10^4} \approx 422$ m/s
• $\langle v\rangle = \sqrt{8 \times 8.908 \times 10^4 / \pi} \approx 476$ m/s
• $v_{\text{rms}} = \sqrt{3 \times 8.908 \times 10^4} \approx 517$ m/s

Ordering checks: $422 < 476 < 517$ as expected.

Q2: By what factor does $v_{\text{rms}}$ change when $T$ doubles?

A: $v_{\text{rms}} \propto \sqrt{T}$, so it increases by $\sqrt{2} \approx 1.414$.

Common Confusions

• $\langle v^2\rangle \ne \langle v\rangle^2$. Always $\langle v^2\rangle \ge \langle v\rangle^2$.
• The "most probable" speed is the peak of the speed distribution $f(v)$, not of the velocity distribution (which peaks at zero).
• All three speeds scale as $\sqrt{T}$ and $1/\sqrt{m}$ — only the numerical prefactor differs.

Key Takeaways

• $v_p : \langle v\rangle : v_{\text{rms}} \approx 1 : 1.13 : 1.22$.
• All three increase as $\sqrt{T}$ and decrease as $1/\sqrt{m}$.
• Pressure and translational KE involve $v_{\text{rms}}$; mean free path involves $\langle v\rangle$.