Maxwell–Boltzmann Speed Distribution

Introduction

Gas molecules at one temperature do not all move at one speed. Some are slow, many lie near a characteristic range and a smaller fraction move much faster. The Maxwell–Boltzmann distribution predicts this spread for a classical ideal gas. Its shape comes from Boltzmann weighting of kinetic energy and from the number of velocity directions available at a given speed.

Core explanation

For a molecule of mass m at temperature T, velocity components v x, v y and v z are independently Gaussian in the classical ideal-gas model, each with zero mean and variance kT/m. The vector velocity distribution depends on exp[−m(v x²+v y²+v z²)/(2kT)]. Speed v = √(v x²+v y²+v z²) is nonnegative. The number of velocity vectors with magnitude between v and v+dv is proportional to the spherical-shell factor 4πv²dv. Multiplying by the Gaussian factor and normalising gives f(v) = 4π[m/(2πkT)]³ᐟ² v² exp[−mv²/(2kT)] for v ≥ 0, with probability f(v)dv in a small interval.

Three characteristic speeds differ. Maximising f(v) gives most probable speed v mp = √(2kT/m). Integrating v f(v) gives mean speed ⟨v⟩ = √(8kT/(πm)). Integrating v²f(v) and taking the square root gives v rms = √(3kT/m). Their order is v mp < ⟨v⟩ < v rms. The rms speed is linked to mean translational kinetic energy: ⟨(1/2)mv²⟩ = (1/2)m v rms² = 3kT/2.

At the same T, heavier molecules have lower characteristic speeds proportional to m⁻¹ᐟ². Raising T scales them as T¹ᐟ² and broadens the speed distribution. The curve starts at zero at v = 0 because the shell of velocity vectors at exactly zero speed has vanishing volume; this does not mean an individual molecule cannot momentarily have near-zero speed. A speed-density value is not itself a probability until multiplied by a speed interval or integrated over a range.

The law assumes classical thermal equilibrium and dilute conditions where the ideal-gas velocity statistics are appropriate. It can describe many ordinary gases well even when the molecules have internal rotational or vibrational energy; those internal modes do not replace the translational distribution. In a driven beam, plasma or highly degenerate quantum gas, the equilibrium classical form may not apply.

Step-by-step reasoning

Identify whether the question asks for a velocity component or nonnegative speed. For speed, include the v² shell factor and use the full f(v). Choose v mp, mean or rms according to the wording, and use mass per molecule with k or molar mass with R consistently. Check ordering and scaling before trusting the number.

Visual explanation

Plot f(v) against v as a skewed curve beginning at zero, rising to a peak and trailing to the right. Mark v mp at the peak, mean speed to its right and v rms farther right. Draw a second broader curve for higher T and a narrower lower-speed curve for a heavier gas at the same T.

Real-world analogy

In a crowd, many people may walk at moderate speeds while a few hurry much faster, producing an asymmetric speed histogram. The mean is pulled toward the fast tail and can lie above the most common speed. Molecular speeds follow a precise equilibrium law, unlike human choices, but the histogram intuition helps distinguish peak, mean and rms.

Real-world example

Helium and xenon in the same room have the same temperature but very different molecular masses. Helium's characteristic speeds are higher because each particle needs a larger speed to carry the same average translational kinetic energy 3kT/2. This mass dependence matters for effusion and gas-transport calculations.

Why?

Boltzmann weighting penalises high kinetic energy as exp(−mv²/2kT), while the number of directions associated with speed v grows as v². Their competition creates a peak away from zero and a decaying high-speed tail. The three characteristic speeds weight this tail differently, producing their distinct values.

Common misconception

Temperature does not set one universal molecular speed. At fixed T, different masses have different speed distributions even though mean translational kinetic energy per particle is the same. Another error is to use f(v) as a probability for an exact speed; continuous-variable probabilities require an interval and integration.

Worked example

At one T, compare particles of masses m and 4m. Each characteristic speed has the form a√(kT/m) with the same numerical factor a for a chosen definition. Replacing m by 4m halves v mp, ⟨v⟩ and v rms. Mean kinetic energy remains 3kT/2 for both, because the heavier mass is offset by the smaller mean squared speed.

Quick check

1. Which is larger for one classical ideal gas at a given temperature, mean speed or rms speed? Answer: v rms is larger: √(3kT/m) exceeds √(8kT/(πm)). Squaring places extra weight on the high-speed tail.

Exam focus

Distinguish mass per molecule m from molar mass M: k/m and R/M give equivalent speed formulas only with matching units. Memorise or derive the order v mp < ⟨v⟩ < v rms. Include the v² shell factor when writing the speed density and state the equilibrium classical assumption.

Advanced insight

The velocity-component distribution is symmetric about zero, but the speed distribution is asymmetric because speed cannot be negative and three-dimensional shell volume grows with v². Transforming to translational kinetic energy yields another distribution with its own density-of-states factor; the peak energy is not generally 3kT/2, which is the mean.

Summary

The Maxwell–Boltzmann speed density combines a v² state-counting factor with exp(−mv²/2kT). Its most probable, mean and rms speeds are √(2kT/m), √(8kT/πm) and √(3kT/m). Heavier particles are slower at the same T, while mean translational kinetic energy remains 3kT/2 per particle.

Practice questions

1. A gas temperature is quadrupled at fixed molecular mass. How do all three characteristic speeds change? Answer: Each is proportional to √T, so each doubles. The entire distribution shifts toward higher speeds and broadens. 2. Why does f(0) = 0 not imply that near-stationary molecules are forbidden? Answer: f is a density for intervals of speed. The three-dimensional velocity shell at exactly zero has vanishing volume, but a finite near-zero interval can contain molecules with nonzero probability. 3. Show that rms speed gives the expected mean translational kinetic energy. Answer: v rms² = 3kT/m, so (1/2)m⟨v²⟩ = (1/2)m v rms² = 3kT/2.