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Maxwell-Boltzmann Distribution of Molecular Speeds

Physical Chemistry #8

§1 Interactive Simulation

Speed Distribution
Particle Simulation
Energy Distribution
Compare Gases
3D Velocity Space
f(v) Speed PDF
Cumulative CDF
f(E) Energy PDF
v_rms vs T
Escape Fraction
v_mp (m/s)
⟨v⟩ (m/s)
v_rms (m/s)
T (K)
⟨KE⟩ (kJ/mol)
M (g/mol)



Show v_mp, ⟨v⟩, v_rms
Show comparison T₂
Show escape fraction
Color by speed

§2 The Idea, Step by Step

Start simple: a crowd of molecules, all at different speeds

Picture the air in this room. Every molecule is zipping around, bumping into its neighbours billions of times each second. They do not all move at the same speed — a few crawl, most jog along in the middle, and a small number sprint. If you tagged every molecule, measured its speed, and made a bar chart of "how many molecules move this fast," you would get a lopsided hump: few slow ones, a big pile near the middle, and a long thin tail of very fast ones. That hump is the Maxwell-Boltzmann distribution. Heating the gas does not make every molecule faster — it shifts and spreads the whole hump toward higher speeds.

Put numbers on it

Two things set the shape: the temperature $T$ (how much energy the crowd shares) and the molar mass $M$ (how heavy each molecule is). Hotter means faster; heavier means slower. The handiest single speed is the most-probable speed $v_{mp}$ — the location of the peak:

$$v_{mp} = \sqrt{\frac{2RT}{M}}$$

For nitrogen ($M = 0.028$ kg/mol) at room temperature ($T = 300$ K): $v_{mp} = \sqrt{2(8.314)(300)/0.028} \approx 420$ m/s — faster than a cruising jet. Lighter hydrogen at the same temperature peaks near $1580$ m/s. That mass dependence is exactly why the lightest gases leak out of a planet's atmosphere first.

The precise picture

The full distribution gives the fraction of molecules whose speed lies between $v$ and $v+dv$:

$$f(v) = 4\pi\left(\frac{M}{2\pi RT}\right)^{3/2} v^2\, e^{-Mv^2/2RT}$$

The $v^2$ factor comes from counting directions in three-dimensional velocity space — a thin spherical shell of radius $v$ has area $\propto v^2$ — while $e^{-Mv^2/2RT}$ is the Boltzmann penalty against high energy. Because the curve is skewed, its three "averages" differ, $v_{mp} < \langle v\rangle < v_{rms}$, yet all scale the same way, as $\sqrt{T/M}$. The Temperature and Molar Mass sliders move $T$ and $M$ directly, so you can watch the peak and the width respond in real time.

Try this in the sim above
1. Slide Temperature from 300 K to 1200 K: the peak slides right and flattens, but quadrupling $T$ only doubles the speed, because $v \propto \sqrt{T}$.
2. Switch the gas from H₂ to UF₆ (or push Molar Mass up): the heavy gas bunches into a tall, narrow spike at low speed while the light gas spreads into a broad, fast curve.
3. Turn on the escape-fraction shading and lower the Escape Velocity slider — notice that only the thin high-speed tail ever exceeds it, which is how gases slowly leak away from planets.

§3 Equation Derivation

Maxwell-Boltzmann Speed Distribution

$$f(v) = 4\pi\left(\frac{M}{2\pi RT}\right)^{3/2} v^2\, e^{-Mv^2/2RT}$$ $$v_{mp} = \sqrt{\frac{2RT}{M}}, \quad \langle v\rangle = \sqrt{\frac{8RT}{\pi M}}, \quad v_{rms} = \sqrt{\frac{3RT}{M}}$$

Symbol Definitions

SymbolMeaningUnit
$f(v)$Speed distribution function (probability density)s/m (so that $\int_0^\infty f(v)\,dv = 1$)
$M$Molar masskg/mol
$R$Gas constant8.314 J·mol⁻¹·K⁻¹
$T$TemperatureK
$v_{mp}$Most probable speed (peak of f(v))m/s
$\langle v \rangle$Mean speedm/s
$v_{rms}$Root-mean-square speedm/s

Derivation from Boltzmann Statistics

Step 1 — 1D Maxwell distribution
From Boltzmann statistics, the probability that a molecule has x-velocity between $v_x$ and $v_x+dv_x$ is: $$g(v_x)\,dv_x = \left(\frac{m}{2\pi k_BT}\right)^{1/2} e^{-mv_x^2/2k_BT}\,dv_x$$ This is a Gaussian (normal) distribution centered at zero.
Step 2 — 3D joint distribution
Since $v_x$, $v_y$, $v_z$ are independent, the joint distribution is: $$g(v_x)g(v_y)g(v_z) = \left(\frac{m}{2\pi k_BT}\right)^{3/2} e^{-m(v_x^2+v_y^2+v_z^2)/2k_BT}$$
Step 3 — Convert to speed (shell volume)
Speed $v = \sqrt{v_x^2+v_y^2+v_z^2}$. In spherical velocity space, the volume shell $dv_x\,dv_y\,dv_z = 4\pi v^2\,dv$. Therefore: $$f(v)\,dv = 4\pi v^2 \left(\frac{m}{2\pi k_BT}\right)^{3/2} e^{-mv^2/2k_BT}\,dv$$
Step 4 — Replace m and k_B with M and R
Since $m = M/N_A$ and $k_B = R/N_A$: $m/k_B = M/R$. Substituting: $$\boxed{f(v) = 4\pi\left(\frac{M}{2\pi RT}\right)^{3/2} v^2\, e^{-Mv^2/2RT}}$$
Step 5 — Three characteristic speeds
Setting $df/dv = 0$ for $v_{mp}$; computing $\langle v\rangle = \int_0^\infty v\,f(v)\,dv$ (Gamma function); computing $v_{rms} = \sqrt{\langle v^2\rangle}$: $$v_{mp} = \sqrt{\frac{2RT}{M}} < \langle v\rangle = \sqrt{\frac{8RT}{\pi M}} < v_{rms} = \sqrt{\frac{3RT}{M}}$$ Numerically at 300 K for N₂ (M=0.028): $v_{mp}$=422 m/s, $\langle v\rangle$=475 m/s, $v_{rms}$=517 m/s.

Simulation Variable Mapping

SliderSymbolEffect on Distribution
Temperature T$T$Higher T → flatter, broader curve shifted to higher v
Molar Mass M$M$Higher M → sharper, narrower curve at lower v
Escape Velocity$v_{esc}$Fraction of molecules faster than $v_{esc}$: planetary gas retention

Worked Example

Problem: Calculate $v_{mp}$, $\langle v\rangle$, and $v_{rms}$ for O₂ (M = 32 g/mol) at 25°C (298 K).

$v_{mp} = \sqrt{2 \times 8.314 \times 298 / 0.032} = \sqrt{154,862} = \mathbf{394}$ m/s
$\langle v\rangle = \sqrt{8 \times 8.314 \times 298 / (\pi \times 0.032)} = \sqrt{196,856} = \mathbf{444}$ m/s
$v_{rms} = \sqrt{3 \times 8.314 \times 298 / 0.032} = \sqrt{232,293} = \mathbf{482}$ m/s

Ratio check: $v_{mp} : \langle v\rangle : v_{rms} = 1 : 1.128 : 1.225$ — consistent with $\sqrt{2}:\sqrt{8/\pi}:\sqrt{3}$.
📚 Reference: Atkins & de Paula — Physical Chemistry, 11th Ed., Chapter 1B: "The Kinetic Theory". McQuarrie & Simon — Physical Chemistry: A Molecular Approach, Chapter 17. Levine — Physical Chemistry, 6th Ed., §15.2.

§4 Frequently Asked Questions

📚 FAQ Reference: LibreTexts Chemistry — "Maxwell-Boltzmann Distributions" (chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry/Supplemental_Modules/Kinetics/Maxwell_Boltzmann_Distributions); Khan Academy — "Maxwell-Boltzmann Distribution"

§5 Common Misconceptions

📚 Misconceptions Reference: Taber — Chemical Misconceptions, RSC, 2002; Nakhleh — J. Chem. Educ. 69, 191 (1992); Loverude, Kautz & Heron — Am. J. Phys. 70, 137 (2002) "Student understanding of the first law of thermodynamics"