The Maxwell–Boltzmann Distribution
~9 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the kinetic theory of gases, specifically the derivation of pressure from molecular collisions, and you must understand the concept of the root-mean-square speed as a statistical average.
Why this matters
The Maxwell–Boltzmann distribution resolves the fundamental gap between the macroscopic pressure of a gas and the microscopic motion of its individual molecules.
In the previous lecture, we established that the root-mean-square speed, \(v_{\text{rms}}\), is the speed that correctly accounts for the kinetic energy and pressure of a gas. However, this single number is a statistical abstraction; it does not tell us how many molecules are moving slowly, how many are moving at that average speed, and how many are moving extremely fast. In a real gas at any given temperature, molecules possess a wide range of speeds due to constant collisions. The Maxwell–Boltzmann distribution provides the precise mathematical function that describes this spread. It is the bridge between the discrete, chaotic world of individual molecular collisions and the smooth, predictable laws of thermodynamics. Without this distribution, we could not explain why a small fraction of molecules in a cold gas can still have enough energy to overcome activation barriers in chemical reactions, or why evaporation occurs even below the boiling point.
The Spread of Speeds in Plain Language
Imagine a crowded ballroom where thousands of people are dancing. If you stopped the music and asked everyone to state their speed, you would not get a single number. Most people would be moving at a moderate pace, a few would be standing still or moving very slowly, and a very small number would be sprinting around the room. The Maxwell–Boltzmann distribution is the mathematical description of this crowd. It tells us that in a gas at a fixed temperature, the speeds of molecules are not uniform; they are distributed according to a specific curve. This curve starts at zero, because the chance of a speed very close to zero is vanishingly small, rises sharply to a peak, and then falls off gradually towards higher speeds. The peak of this curve represents the most probable speed, \(v_p\), which is the speed shared by the largest number of molecules. However, because the curve is skewed to the right (it has a long tail extending to high speeds), the average speed, \(\langle v \rangle\), is slightly higher than the most probable speed. Furthermore, the root-mean-square speed, \(v_{\text{rms}}\), which we derived from pressure, is even higher than the mean speed. This is because the squaring operation in the RMS calculation gives disproportionate weight to the fast-moving molecules in the tail of the distribution. Therefore, for any gas at a given temperature, the order of these speeds is always \(v_p < \langle v \rangle < v_{\text{rms}}\). This distinction is crucial: the most probable speed is where the curve peaks, the mean speed is the arithmetic average, and the RMS speed is the energy-weighted average. Understanding this hierarchy allows us to interpret experimental data correctly, as different physical phenomena depend on different aspects of this distribution.
Deriving the Distribution Function
The derivation of the Maxwell–Boltzmann distribution relies on the assumption that the velocity components \(v_x\), \(v_y\), and \(v_z\) are independent and follow a Gaussian distribution. The probability of a molecule having a specific velocity vector is proportional to the Boltzmann factor, \(e^{-E/kT}\), where \(E\) is the kinetic energy. Since the kinetic energy is \(\frac{1}{2}m(v_x^2 + v_y^2 + v_z^2)\), the probability density in velocity space is a three-dimensional Gaussian. To find the distribution of speeds (the magnitude of velocity, \(v\)), we must integrate over all directions in velocity space. This involves converting from Cartesian coordinates to spherical coordinates, where the volume element is \(4\pi v^2 \, dv\). The factor \(v^2\) arises from the geometry of the spherical shell in velocity space; as speed increases, the number of possible velocity vectors with that magnitude increases proportionally to the surface area of the sphere. Combining the Gaussian decay \(e^{-mv^2/2kT}\) with the geometric factor \(v^2\) yields the characteristic shape of the distribution. The normalization constant is determined by ensuring the total probability integrates to one. The resulting function, \(f(v)\), represents the fraction of molecules with speeds between \(v\) and \(v+dv\). It is important to note that \(f(v)\) is not a probability in the strict sense for a single molecule, but a probability density. The actual probability is found by integrating \(f(v)\) over a range of speeds. This function is universal for all ideal gases, depending only on the mass \(m\) and temperature \(T\), not on the specific chemical identity of the molecule.
Characteristic Speeds and Their Ratios
From the distribution function \(f(v)\), we can extract three distinct characteristic speeds, each with a specific physical meaning. The most probable speed, \(v_p\), is found by setting the derivative \(\frac{df(v)}{dv}\) to zero. This identifies the peak of the curve, where the number of molecules is greatest. Solving this condition yields \(v_p = \sqrt{\frac{2kT}{m}}\). The mean speed, \(\langle v \rangle\), is the first moment of the distribution, calculated by integrating \(v f(v)\) over all speeds. This gives \(\langle v \rangle = \sqrt{\frac{8kT}{\pi m}}\). The root-mean-square speed, \(v_{\text{rms}}\), is the square root of the second moment, \(\sqrt{\langle v^2 \rangle}\), which we already know from kinetic theory to be \(\sqrt{\frac{3kT}{m}}\). Comparing these three expressions reveals a fixed ratio that is independent of temperature and mass. Dividing each speed by \(v_p\), we find the ratio \(v_p : \langle v \rangle : v_{\text{rms}} = 1 : \sqrt{\frac{4}{\pi}} : \sqrt{\frac{3}{2}}\). Numerically, this is approximately \(1 : 1.128 : 1.225\). This means that the mean speed is about 12.8% higher than the most probable speed, and the RMS speed is about 22.5% higher. This hierarchy is a direct consequence of the asymmetry of the Maxwell–Boltzmann curve. The long tail of high-speed molecules pulls the mean and RMS values above the peak. In experimental contexts, if you measure the average speed of molecules passing through a hole (effusion), you are measuring a value related to the mean speed, whereas if you measure the pressure, you are effectively measuring the RMS speed. Confusing these values can lead to significant errors in calculating reaction rates or diffusion coefficients.
Physical Interpretation and Applications
The Maxwell–Boltzmann distribution is not just a mathematical curiosity; it has profound physical implications. One key application is in chemical kinetics. For a reaction to occur, molecules must collide with energy greater than the activation energy, \(E_a\). The fraction of molecules with speeds corresponding to energies above \(E_a\) is given by the area under the tail of the distribution curve beyond the threshold speed. Because the distribution is exponential, even a small increase in temperature significantly increases the area of this tail, leading to an exponential increase in reaction rate. This explains the Arrhenius equation. Another application is in the study of effusion and diffusion. The rate at which gas molecules escape through a small hole is proportional to the mean speed, not the RMS speed. This is because the flux of particles through a surface depends on the average velocity component perpendicular to the surface, which is related to the mean speed. In contrast, the pressure exerted by the gas depends on the momentum transfer, which is related to the square of the speed, hence the RMS speed. Understanding which characteristic speed applies to which phenomenon is essential for accurate modelling.
The result
The distribution of molecular speeds
kinetic-theory-pressure found the mean-square molecular speed but not how speeds are actually distributed among individual molecules. Treating each velocity component as an independent, Boltzmann-weighted Gaussian and converting to spherical (speed) coordinates gives \(f(v)=4\pi n\left(\tfrac{m}{2\pi kT}\right)^{3/2}v^2e^{-mv^2/2kT}\), from which three distinct characteristic speeds emerge in the fixed ratio \(v_p:\langle v\rangle:v_{\text{rms}}=\sqrt2:\sqrt{8/\pi}:\sqrt3\approx1:1.128:1.225\) — the most probable speed \(v_p\), the mean speed \(\langle v\rangle\), and the root-mean-square speed \(v_{\text{rms}}\) (matching kinetic-theory-pressure's result exactly).
A common misconception is to assume that the most probable speed \(v_p\) is the "average" speed of the gas. In fact, \(v_p\) is the speed at which the probability density is highest, but because the distribution is skewed, the arithmetic mean \(\langle v \rangle\) is higher. If you calculate the average speed by simply taking the peak of the curve, you will underestimate the true mean speed by about 12%. This error can propagate into calculations of mean free path and collision frequency, leading to incorrect predictions of transport properties. Always distinguish between the mode (most probable), the mean (arithmetic average), and the RMS (energy-weighted average) when interpreting gas kinetic data. C
Worked example
Problem. For oxygen (\(M = 32.00\ \mathrm{g\,mol^{-1}}\)) at 300 K, find the most probable speed \(v_p\). How high is the distribution at \(2v_p\) compared with its peak, and what fraction of molecules move faster than \(2v_p\)? For speeds in units of \(v_p\), \(x = v/v_p\), the fraction faster than \(x\) is \(\operatorname{erfc}(x) + \tfrac{2x}{\sqrt{\pi}} e^{-x^2}\).
- \(v_p = \sqrt{2RT/M} = \sqrt{2 \times 8.314 \times 300 / 0.03200} = 394.8\ \mathrm{m\,s^{-1}}\), so \(2v_p = 790\ \mathrm{m\,s^{-1}}\).
- In terms of \(x\), \(f \propto x^2 e^{-x^2}\). At the peak \(x = 1\); at \(x = 2\): \(\dfrac{f(2v_p)}{f(v_p)} = \dfrac{4e^{-4}}{e^{-1}} = 4e^{-3} = 0.199\).
- Fraction faster than \(2v_p\): \(\operatorname{erfc}(2) + \tfrac{4}{\sqrt{\pi}}e^{-4} = 0.00468 + 2.257 \times 0.01832 = 0.00468 + 0.04133\).
- So the fraction is 0.0460, about 4.6% of the molecules.
- Interpret: the curve at \(2v_p\) is still a fifth of its peak height, but because it then falls off as \(e^{-x^2}\), less than 5% of molecules lie beyond it. At \(3v_p\) the fraction is already below 0.1%.
Units check. \(\sqrt{\mathrm{J\,mol^{-1}}/\mathrm{kg\,mol^{-1}}} = \mathrm{m\,s^{-1}}\); the height ratio and the fraction are pure numbers.
Three quick tests
- Why is the mean speed greater than the most probable speed in the Maxwell–Boltzmann distribution?
Answer
The distribution is asymmetric with a long tail extending to higher speeds. This tail pulls the arithmetic mean above the peak (mode) of the distribution. - If the temperature of a gas is doubled, by what factor does the root-mean-square speed increase?
Answer
The RMS speed is proportional to the square root of the temperature. Therefore, doubling the temperature increases the speed by a factor of \(\sqrt{2} \approx 1.414\). - Which characteristic speed is directly related to the pressure exerted by an ideal gas?
Answer
The root-mean-square speed (\(v_{\text{rms}}\)) is directly related to the pressure, as pressure depends on the average kinetic energy, which is proportional to \(v^2\).
Back to CU-104 →
Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.