Maxwell-Boltzmann Distribution

The distribution in symbols

For an ideal gas of identical molecules of mass m at thermal equilibrium at temperature T, the probability density that a randomly chosen molecule has speed v (the magnitude of its velocity, ignoring direction) is:

f(v) = 4π × (m / 2πkT)^(3/2) × v² × exp(−mv²/2kT)

where k is Boltzmann's constant 1.381 × 10⁻²³ J/K. The function f(v) has units of 1/(m/s); integrating it from v₁ to v₂ gives the fraction of molecules with speeds in that range. The full integral from 0 to ∞ is 1.

The shape is set by a competition. The factor v² is geometric — it counts the surface area of a spherical shell in velocity space, so faster speed means more directions in which a velocity vector can point. The factor exp(−mv²/2kT) is the Boltzmann weight, the relative probability of a microstate with kinetic energy ½mv² at temperature T. At low v the v² factor wins and f rises; at high v the exponential wins and f plummets. Where they balance is the peak.

Three characteristic speeds

From f(v) we can compute three different "typical speeds", each useful in different contexts:

Most-probable speed   v_p   = √(2kT/m)            (location of f(v) peak)
Mean speed            ⟨v⟩   = √(8kT/πm)           (linear average)
Root-mean-square      v_rms = √(3kT/m)            (energy-weighted average)

For any gas at any temperature, the ratios are fixed:

v_p : ⟨v⟩ : v_rms = √2 : √(8/π) : √3 ≈ 1.000 : 1.128 : 1.225

The mean is 12.8% larger than the most-probable, and the RMS is 22.5% larger. The asymmetric high-speed tail explains why: a few molecules with very high v contribute disproportionately to ⟨v⟩, and even more disproportionately to ⟨v²⟩. The relation ½m⟨v²⟩ = (3/2)kT is just equipartition — three translational degrees of freedom, each carrying ½kT of energy.

Worked example: oxygen at 300 K

Take molecular oxygen O₂ at room temperature. Its molar mass is 32 g/mol = 0.032 kg/mol, so the per-molecule mass m = 0.032 / 6.022 × 10²³ = 5.31 × 10⁻²⁶ kg. With T = 300 K and k = 1.381 × 10⁻²³ J/K:

kT/m = (1.381e-23 × 300) / 5.31e-26 = 7.80e4 m²/s²

v_p   = √(2 × 7.80e4)        = √1.56e5  = 395 m/s
⟨v⟩   = √(8/π × 7.80e4)      = √1.99e5  = 446 m/s
v_rms = √(3 × 7.80e4)        = √2.34e5  = 484 m/s

For nitrogen N₂ (m = 4.65 × 10⁻²⁶ kg) at the same temperature:

v_p   = 422 m/s
⟨v⟩   = 476 m/s
v_rms = 517 m/s

For helium He (m = 6.65 × 10⁻²⁷ kg) at the same temperature:

v_p   = 1116 m/s
⟨v⟩   = 1259 m/s
v_rms = 1366 m/s

Helium is 8 times less massive than oxygen, so its speeds are √8 = 2.83 times larger — exactly the factor m^(−1/2) the formulas predict. This is why helium leaks out of latex balloons, why Saturn's moon Titan keeps its nitrogen atmosphere but Earth doesn't keep helium, and why mass spectrometers separate isotopes by accelerating them through the same potential and timing them.

How fast is fast? Sound travels at about 343 m/s through 300 K air, which is roughly the most-probable molecular speed. This is not a coincidence: pressure waves propagate as collective rearrangements of the same molecular motions, and the speed of sound c_s = √(γkT/m) differs from v_rms only by a factor of √(γ/3) where γ is the heat-capacity ratio.

Speeds across gases and temperatures

Gas	m (kg)	T (K)	v_p (m/s)	v_rms (m/s)
Hydrogen H₂	3.35 × 10⁻²⁷	300	1572	1925
Helium He	6.65 × 10⁻²⁷	300	1116	1366
Nitrogen N₂	4.65 × 10⁻²⁶	300	422	517
Oxygen O₂	5.31 × 10⁻²⁶	300	395	484
Argon Ar	6.63 × 10⁻²⁶	300	352	431
Carbon dioxide CO₂	7.31 × 10⁻²⁶	300	337	412
Xenon Xe	2.18 × 10⁻²⁵	300	195	239
Nitrogen N₂	4.65 × 10⁻²⁶	1000	770	944
Hydrogen H₂	3.35 × 10⁻²⁷	1000	2870	3514
Hydrogen H₂ (exobase)	3.35 × 10⁻²⁷	1500	3514	4302

Two patterns: doubling T multiplies all three speeds by √2 ≈ 1.414, and quadrupling the molecular mass halves the speeds. Both follow directly from the m and T appearing only as the ratio T/m inside square roots.

Why this exact form: a sketch of the derivation

Maxwell's original 1860 argument used spatial isotropy: the velocity distribution must factor into independent functions of vₓ, vy and vz, and must depend only on speed |v|. The only function with both properties is a Gaussian, e^(−αv²). Boltzmann (1871) gave a more general statistical-mechanics derivation. He showed that maximising the entropy of a gas at fixed total energy and particle number, under the constraint of energy conservation, produces the exponential form exp(−E/kT) for the probability of a state of energy E. For a free particle E = ½mv², so:

f(vₓ, vy, vz) ∝ exp(−m(vₓ² + vy² + vz²) / 2kT)

This 3D Gaussian over the velocity vector is the velocity distribution. To get the distribution of speeds (regardless of direction), integrate over the spherical shell of radius v: the volume element 4πv² dv supplies the v² factor that turns a 3D Gaussian into the familiar Maxwell-Boltzmann speed distribution.

The constants out front are fixed by normalization: ∫f(v)dv = 1. Carrying out the Gaussian integral with the standard substitutions reproduces the prefactor 4π(m/2πkT)^(3/2).

Where the distribution shows up

• Atmospheric escape. Earth's exobase (~500 km altitude, T ≈ 1000–1500 K) is the surface from which molecules above escape velocity (10800 m/s) sail away. Hydrogen has a fat enough tail above this threshold to lose ~10⁸ atoms/cm²/s; oxygen's tail is negligible. Mars, with weaker gravity (escape ~5000 m/s), loses oxygen too — the dry surface is partly a Maxwell-Boltzmann consequence.
• Reaction rates and Arrhenius. The fraction of molecules with kinetic energy above the activation barrier E_a is roughly exp(−E_a/kT). Because E_a is typically 50–200 kJ/mol while kT at 300 K is only 2.5 kJ/mol, doubling T from 300 to 600 K can speed reactions up by factors of 10⁵ or more.
• Evaporation and the cooling of sweat. Liquid water doesn't have a single boiling temperature — molecules in the high-speed tail of its Maxwell-Boltzmann-like distribution always have enough energy to escape, even at 25 °C. Removing those fast molecules cools what remains, which is why sweat evaporation dumps about 580 cal of latent heat per gram.
• Thermal velocity in plasmas. In a deuterium-tritium fusion plasma at 150 million K (13 keV), v_rms for deuterium is 1.3 × 10⁶ m/s. The high-speed tail is what overcomes Coulomb repulsion and tunnels through the barrier — fusion rates are dominated by molecules many sigma above ⟨v⟩.
• Doppler broadening of spectral lines. Atoms moving toward or away from a spectrometer shift their emitted wavelengths by ±v/c. The line shape is the convolution of the natural linewidth with the Maxwell-Boltzmann velocity distribution along the line of sight, giving a Gaussian Doppler width Δλ/λ = √(2kT/mc²). Stellar temperatures are routinely measured this way.

Variants and extensions

• Bose-Einstein distribution. Replaces the Maxwell-Boltzmann factor for indistinguishable bosons with f = 1/(exp((E−μ)/kT) − 1). At low temperatures, large fractions of bosons collapse into the ground state — the Bose-Einstein condensate. Reduces to Maxwell-Boltzmann in the dilute, classical limit.
• Fermi-Dirac distribution. For indistinguishable fermions: f = 1/(exp((E−μ)/kT) + 1). Pauli exclusion caps each state at one particle. Conduction electrons in metals follow this distribution — at room temperature kT ≈ 25 meV is much smaller than the Fermi energy ~5 eV, so most electrons are locked in deep states with no Maxwell-Boltzmann freedom.
• Maxwell-Jüttner distribution. Relativistic generalization for hot plasmas where v approaches c. The kinetic energy ½mv² is replaced by (γ−1)mc² with γ the Lorentz factor; the distribution rolls off more slowly than the classical Gaussian at very high speeds.
• Boltzmann velocity distribution in 1D and 2D. The same exp(−mv²/2kT) factor without the v² spherical-shell weight. f(vₓ) is a Gaussian centred at zero; f(vₓ, vy) has a 2πv shell factor. Useful when only one or two components matter, like flow through a slit.
• Effusion and Graham's law. The flux of molecules through a small hole is proportional to ⟨v⟩ ∝ 1/√m. Two gases mixing through the same hole therefore segregate according to √(m₂/m₁) — Graham's law of effusion, an immediate consequence of Maxwell-Boltzmann.

Common pitfalls

• Confusing speed and velocity distributions. The 3D velocity distribution has no v² factor; the speed distribution does. Plotting the wrong one gives the wrong peak position and the wrong total area.
• Using molar mass M instead of per-particle m. The formulas as written use the mass of a single molecule. To use molar mass, replace k with R = kN_A and remember M is in kg/mol, not g/mol. v_rms = √(3RT/M) with M in kg/mol is the molar version.
• Forgetting that T must be absolute. Use kelvins. The 300 K used in worked examples is room temperature; using 27 °C as 27 in the formula understates speeds by a factor of √(300/27) ≈ 3.3.
• Treating the distribution as static. Maxwell-Boltzmann is the equilibrium distribution. Out of equilibrium — strong shocks, lasers depositing energy faster than collisions thermalize it, plasmas with separate electron and ion temperatures — the actual velocity distribution can be far from Maxwell-Boltzmann. Watch for assumptions of "thermal equilibrium" before quoting the formulas.
• Underestimating the tail's importance. The high-speed tail looks negligible on a linear plot but contains the molecules that drive reaction rates, atmospheric escape, and tunneling. Always plot in log scale or compute fractional integrals when the question is "how many molecules above some threshold?"