Physics · Kinetic theory of gases

Maxwell–Boltzmann distribution of molecular speeds

Q: How can I check a calculation involving Maxwell–Boltzmann Molecular Speed Distribution?

For $M=0.044\,\mathrm{kg\,mol^{-1}}$ and $T=330\,\mathrm K$, $v_{\mathrm{mp}}=a\simeq353.1528554\,\mathrm{m\,s^{-1}}$. The class 0.8a to 1.2a spans about 282.52–423.78 m/s. Using $F(x)=\operatorname{erf}(x)-\frac{2x}{\sqrt\pi}e^{-x^2}$ gives fraction $F(1.2)-F(0.8)\simeq0.3233884261$. This is 32.3388 %, not dimensional density. The midpoint approximation $g(1)\cdot0.4$ gives 33.2086 %, exceeding it by 0.869757 percentage points. Integration is therefore appropriate for this broad class. Independently check that total area over all nonnegative speeds is 1 and the fraction lies within 0–1. Enter M in kg/mol; using kB would instead require particle mass m. To calculate mean energy, also use the second speed moment rather than squaring the ordinary average of magnitudes.

Q: How do I check units and scale for Maxwell–Boltzmann Molecular Speed Distribution?

The prefactor combined with $v^2$ produces density in s/m, making $f(v)\,dv$ dimensionless. In the molar speed scale $a=\sqrt{\frac{2RT}{M}}$, M must be in kg/mol; 32 g/mol becomes 0.032 kg/mol. Temperature is in kelvin. Quadrupling T for the same species quadruples squared characteristic speeds and doubles their magnitudes. The density then becomes twice as wide horizontally and half as high vertically; total area stays 1. Density values alone are not percentages. At one exact speed continuous probability is zero. A finite measurement class needs its width or both integration limits, together with compatible units.

At thermal equilibrium, the Maxwell–Boltzmann density describes how molecular speed magnitudes are distributed in a classical ideal gas; it is a density over v≥0.

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Formula

Calculate

f (v) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v^{2} exp (- \frac{m v ^{2}}{2 k _{B} T})

v in m/s; m is mass per particle in kg; T in K; kB in J/K; f(v) in s/m so f(v)dv is dimensionless.

Variables & units

Variables & units
Symbol	Meaning	Unit
$f(v)$	Probability density of thermal speed magnitude	s/m
$v$	Thermal speed relative to local mean gas flow	m/s
$m$	Mass of one molecule	kg
$T$	Absolute equilibrium temperature	K
$k_{\mathrm{B}}$	Boltzmann constant	J/K

LaTeX

f(v)=4\pi\left(\frac{m}{2\pi k_{\mathrm B}T}\right)^{3/2}v^2\exp\!\left(-\frac{mv^2}{2k_{\mathrm B}T}\right)

QUANTA STUDYquanta-study.de

Maxwell density 4x²exp(−x²)/√π from x = 0 to 3.5, peak at 1 with the right tail retained in the displayed range. Area 0.8 to 1.2 is shaded. — Correctly normalized dimensionless speed density with x=v/vmp. Area from 0.8 to 1.2 is 32.3388 % of particles; density height is not a probability at one exact speed.

Maxwell-Boltzmann distribution calculator

Find the most probable, mean and root mean square molecular speeds, and the height of the distribution curve, for an ideal gas in equilibrium.

Calculation model

Find

Temperature T

Absolute temperature of the gas, e.g. 300 K, about room temperature.

Molar mass of one species M

Molar mass of the gas, e.g. 28 g/mol for nitrogen N₂ or 32 g/mol for oxygen O₂.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for vₘₚ, substitute, evaluate

1. Rearrange for v_mp
$v_{\mathrm{mp}}=\sqrt{\frac{2RT}M}$
2. Substitute values (in SI units)
- M = 28 g/mol = 0.028 kg/mol
$\sqrt{\frac{2\cdot 8.31446\cdot 300}{0.028}}$
3. Result
v_mp = 422.1 m/s

The inverse calculation explicitly uses the most probable speed, not the mean or RMS.

Result: Most probable speed v_mp

v_mp = 422.1 m/s

With T = 300 K and M = 28 g/mol, v_mp = 422.1 m/s.

Quick tips

Remember the order v_mp < ⟨v⟩ < vᵣ_ms, always in the ratio √2 : √(8/π) : √3, about 1 : 1.13 : 1.22.
Speeds grow with √T and fall with √M. Four times the temperature in kelvin doubles the speed; a gas four times heavier is half as fast.

Assumptions and limits

One classical nonrelativistic ideal species in thermal equilibrium without relevant quantum degeneracy. Speeds relative to local mean gas flow; positive temperature and molar mass. Not a laboratory beam or mixture represented by a mean substitute mass. R=kB·NA from exact defining SI constants; finite arithmetic precision.

Derivation & background

$f(v)$ is normalised thermal speed density relative to local gas motion, in s/m. A fraction comes from $\int f(v)\,dv$ , not a density value alone. Three isotropic classical Gaussian components and shell factor $4\pi v^2$ produce speed density. With $a=\sqrt{\frac{2k_{\mathrm B}T}{m}}$ , the characteristics are $v_{\mathrm{mp}}=a$ , $\langle v\rangle=\frac{2a}{\sqrt\pi}$ and $v_{\mathrm{rms}}=\sqrt{\frac32}a$ . Classical nonrelativistic equilibrium is required; mixtures and directed fluxes need separate weighting.

Exam blueprint

Validity range

Classical nonrelativistic thermal equilibrium without relevant quantum degeneracy. Speed relative to local mean flow; m particle mass or M in kg/mol. f(v) is density, not point probability. Mixtures and particle fluxes cannot be equated unchecked with one equilibrium speed density.

Derivation steps

The density f(v) tells what fraction of a classical gas at rest lies in a speed interval. It describes magnitudes, not directions. Its characteristic speeds and areas can be calculated from one normalised distribution.

1Consider classical equilibrium motion relative to local mean gas motion.
2Each component has a Gaussian density with factor $e^{-\frac{mv_i^2}{2k_{\mathrm B}T}}$ .
3The product of the three independent densities gives isotropic vector density $P(\mathbf v)$ .
4The speed class from $v$ to $v+dv$ contains a velocity-space shell of measure $4\pi v^2\,dv$ .
5This shell factor gives normalised speed density $f(v)$ , with units s/m.
6Calculate fractions using $\int f(v)\,dv$ . With $a=\sqrt{\frac{2k_{\mathrm B}T}{m}}$ and $x=\frac va$ , $F(x)=\operatorname{erf}(x)-\frac{2x}{\sqrt\pi}e^{-x^2}$ .

Rearrangements

Most probable speed

v_{\mathrm{mp}}=\sqrt{\frac{2k_{\mathrm B}T}{m}}

Maximum of density f(v), not the mean or root-mean-square speed.

Task variant

Rounded nitrogen model: particle mass 4.65·10⁻²⁶ kg at 300 K. Find most probable speed and explain how a 10 m/s bin near the peak gives a particle fraction.

The rounded nitrogen model at 300 K has $m\simeq4.65\cdot10^{-26}\,\mathrm{kg}$ , corresponding to about 28.0 u.
Its peak is at $v_{\mathrm{mp}}=\sqrt{\frac{2k_{\mathrm B}T}{m}}\simeq422\,\mathrm{m\,s^{-1}}$ .
Normalised total area is one.
For a narrow 10 m/s bin near the peak, $f(v_{\mathrm{mp}})\Delta v$ approximates probability, not the density value alone.

Common mistakes

Reading a density directly as percentage probability at an exact speed.

Probability is area $\int_{v_1}^{v_2}f(v)\,dv$ ; density has inverse-speed units. An exact single point has zero probability.

Exam context

Gas speed distributions, kinetic theory, diffusion, effusion and temperature dependence of molecular processes.

Formula cluster

Kinetic theory of gases

Interval areas give particle fractions; the first two moments connect the distribution to mean and RMS speed and energy.

vrms = √(3kBT/m)

Root-mean-square molecular speed

⟨Ekin⟩ = 3/2 kBT

Mean translational kinetic energy per gas molecule

pV = 1/3 Nm⟨v²⟩

Pressure from kinetic theory

Worked example

The rounded nitrogen model at 300 K has $m\simeq4.65\cdot10^{-26}\,\mathrm{kg}$ , corresponding to about 28.0 u. Its peak is at $v_{\mathrm{mp}}=\sqrt{\frac{2k_{\mathrm B}T}{m}}\simeq422\,\mathrm{m\,s^{-1}}$ . Normalised total area is one. For a narrow 10 m/s bin near the peak, $f(v_{\mathrm{mp}})\Delta v$ approximates probability, not the density value alone.

Applications

Gas speed distributions, kinetic theory, diffusion, effusion and temperature dependence of molecular processes.

How do you calculate with Maxwell–Boltzmann Molecular Speed Distribution?

Here is how to work through a typical Maxwell–Boltzmann Molecular Speed Distribution (f(v) = 4π(m/2πkBT)³ᐟ² v² e^(−mv²/2kBT)) task step by step:

1
Task
Rounded nitrogen model: particle mass 4.65·10⁻²⁶ kg at 300 K. Find most probable speed and explain how a 10 m/s bin near the peak gives a particle fraction.
Solution path
1. The rounded nitrogen model at 300 K has $m\simeq4.65\cdot10^{-26}\,\mathrm{kg}$ , corresponding to about 28.0 u.
2. Its peak is at $v_{\mathrm{mp}}=\sqrt{\frac{2k_{\mathrm B}T}{m}}\simeq422\,\mathrm{m\,s^{-1}}$ .
3. Normalised total area is one.
4. For a narrow 10 m/s bin near the peak, $f(v_{\mathrm{mp}})\Delta v$ approximates probability, not the density value alone.

Understand and apply the equation

Molecules in a gas do not all have the same speed. The Maxwell–Boltzmann distribution describes the fraction in a speed range. Curve height is a density; only area over an interval is probability. Starting from that picture, learn the formula, its speed scale and the distinction between most probable speed, mean speed and RMS. Examples cover interval fractions, fast particles and a mixture.

A density is not yet a probability

f(v)=4\pi\left(\frac{m}{2\pi k_{\mathrm B}T}\right)^{\frac{3}{2}}v^2e^{-\frac{mv^2}{2k_{\mathrm B}T}},\quad v\ge0

v is thermal speed relative to local mean gas motion; m is mass per particle.[1]

Function $f(v)$ has units s/m, or inverse speed. For a small interval, $f(v)\,dv$ gives a particle fraction; integrate for a finite range. An exact single value has zero probability in the continuous distribution even where density is positive. The peak marks the densest narrow speed class, not every particle’s speed. Normalisation across all nonnegative speeds gives total probability one.[1]

How to read the area

A hotter gas assigns more weight to higher speeds. Its density curve can become lower while still accounting for all particles. Width and height change together; total area stays one. For a specific range from $v_1$ to $v_2$ , the question is therefore not “How high is the curve?” but “What area lies between these boundaries?”[1]

P(v_1\le v\le v_2)=\int_{v_1}^{v_2}f(v)\,dv

A finite speed class requires two boundaries.

A lower curve can have the same total area

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Original model curves for 0.0280 kg/mol. Density is plotted in thousandths s/m, so displayed area must be multiplied by 10⁻³ to obtain probability. Normalisation extends to infinity, not merely across the visible plot.[1]

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Three components produce a spherical shell

\begin{aligned}P(\mathbf v)&=\left(\frac{m}{2\pi k_{\mathrm B}T}\right)^{\frac{3}{2}}e^{-\frac{m(v_x^2+v_y^2+v_z^2)}{2k_{\mathrm B}T}}\\{} f(v)dv&=P(\mathbf v)4\pi v^2dv\end{aligned}

The isotropic density is constant on a velocity sphere; shell area supplies v².[1][2]

In the local rest frame each velocity component has a zero-mean Gaussian distribution. The product of the three component densities depends only on $v_x^2+v_y^2+v_z^2$ . However, the requested quantity is speed, independent of direction. Combine all vectors with nearly the same magnitude in a thin shell in velocity space. Its volume factor $4\pi v^2\,dv$ produces the additional quadratic factor. This is a mathematical space of possible velocities, not a physical sphere in the gas sample.[1]

Explore further: scaled distribution and cumulative function

\begin{aligned}a&=\sqrt{\frac{2k_{\mathrm B}T}{m}}=\sqrt{\frac{2RT}{M}}\\x&=\frac va\\g(x)&=a f(ax)=\frac4{\sqrt\pi}x^2e^{-x^2}\end{aligned}

a is most probable speed; M is in kg/mol.[1]

\begin{aligned}\int_0^\infty g(x)\,dx&=1\\F(x)&=\operatorname{erf}(x)-\frac{2x}{\sqrt\pi}e^{-x^2}\end{aligned}

F is cumulative probability in scaled speed; $F(0)=0$ and $F(\infty)=1$ .[1]

Speed scale $a$ contains substance and temperature. Dividing all speeds by $a$ gives the same dimensionless shape for every classical equilibrium gas. Cumulative distribution $F(x)$ follows by integration by parts, with $\operatorname{erf}(x)=\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,dt$ . An interval fraction is $F(\frac{v_2}{a})-F(\frac{v_1}{a})$ . On ordinary speed axes, heating broadens the curve and lowers its height without changing normalisation.[1]

Keep mode, mean and RMS separate

\begin{aligned}v_{\mathrm{mp}}&=a\\\langle v\rangle&=\frac{2a}{\sqrt\pi}\\v_{\mathrm{rms}}&=\sqrt{\frac32}\,a\end{aligned}

The mode follows from $\frac{d}{dx}(x^2e^{-x^2})=0$ ; mean and RMS from first and second moments.[1]

Most probable speed is at the density peak. Ordinary mean speed is slightly higher; RMS from the second moment is higher still. None is mean velocity vector, which vanishes in the rest frame because directions cancel. Energy uses $\frac12m\langle v^2\rangle=\frac32k_{\mathrm B}T$ , not the square of ordinary mean speed. A heavier species at the same temperature has lower typical speeds but the same mean translational energy.[1][3]

Root-mean-square molecular speed connects the distribution with its second moment, $v_{\mathrm{rms}}^2=\langle v^2\rangle$ . Use it for mean translational energy; the most probable speed at the curve maximum is not a substitute.[1][3]

Actual intervals rather than single density values

Use the cumulative distribution for intervals and tails. In the third example the same calculation is carried out separately for each species before weighting by particle fractions.[1]

A speed class in model gas

Given: $M=0.044\,\mathrm{kg\,mol^{-1}}$ , $T=330\,\mathrm K$ ; classical ideal equilibrium. Interval $0.8v_{\mathrm{mp}}$ to $1.2v_{\mathrm{mp}}$ .
Find: Calculate the class bounds and particle fraction.

Approach: A speed interval uses the area under the distribution. $a=\sqrt{\frac{2RT}{M}},\quad P(v_1<v<v_2)=F(\frac{v_2}{a})-F(\frac{v_1}{a})$ . Find speed scale.
$a=\sqrt{\frac{2RT}{M}}=353.1528554\,\mathrm{m/s}$
Evaluate CDF at both bounds.
$P=F(1.2)-F(0.8)=0.3233884261$
Compare with the midpoint approximation.
$g(1)(1.2-0.8)=0.3320859979$

Result: The class about 282.52–423.78 m/s contains 32.3388 % of particles. The broad-bin midpoint approximation gives 33.2086 %.

Check: The approximation exceeds the result by 0.869757 percentage points; density alone would have units. Integration of f(v) with dv in m/s independently checks the exact class.

A fast distribution tail

Given: Same model gas; seek $v>2v_{\mathrm{mp}}$ . A sample contains $N=2.00\cdot10^{21}$ particles.
Find: Find fraction and expected count, then compare doubled speed scale.

Approach: For the fast tail use the complement of the cumulative distribution. $P(v>v_*)=1-F(\frac{v_*}{a})$ . Use the CDF complement.
$P(v>2a)=1-F(2)=0.04601170569$
Find expected count from fraction.
$\langle N_{tail}\rangle=NP=9.202341138\times10^{19}$
Quadrupling temperature doubles a.
$a(4T)=2a(T)$

Result: About 4.60117 % exceed 706.31 m/s. At fourfold T, the same absolute threshold has scaled value $x=1$ rather than 2.

Check: The fraction above the original threshold then becomes $1-F(1)=0.5724067045$ . Above the respective new 2a threshold, however, the same 4.60117 % fraction remains.

Fast particles in a mixture

Given: Ideal classical equilibrium mixture at 300 K in the gas rest frame. Equal particle counts of two model species with molar masses 0.00400 and 0.0400 kg/mol. Both are nonrelativistic and nondegenerate. Find the fraction with speed above 700 m/s.
Find: Find both species fractions and the fraction of the whole mixture.

Approach: each species has its own scale. Calculate their tail probabilities first, then weight by particle fractions.
$\begin{aligned}a_j&=\sqrt{\frac{2RT}{M_j}}\\P_j&=1-F\!\left(\frac{700\,\mathrm{m\,s^{-1}}}{a_j}\right)\\P_{\mathrm{mix}}&=\frac12P_1+\frac12P_2\end{aligned}$
Insert both scales and dimensionless thresholds.
$\begin{aligned}a_1&\simeq1116.767\,\mathrm{m\,s^{-1}},&x_1&\simeq0.626809\\a_2&\simeq353.153\,\mathrm{m\,s^{-1}},&x_2&\simeq1.982145\end{aligned}$
Evaluate the fractions using the cumulative function, then mix them.
$\begin{aligned}P_1&\simeq0.852865\\P_2&\simeq0.0490439\\P_{\mathrm{mix}}&\simeq0.450955\end{aligned}$

Result: About 85.29% of the light species and 4.904% of the heavy species exceed 700 m/s. Together this is about 45.10% of all particles.

Check: The mixture fraction lies between the species fractions and is their arithmetic mean for equal particle counts. Mean molar mass would be 0.0220 kg/mol, but its single Maxwell curve would not be the mixture curve. Equal mean translational energy at equal temperature does not imply equal probability above a speed threshold.

Distinguish local temperature and a laboratory beam

The formula uses velocities relative to local mean gas motion $\mathbf u$ . In the laboratory the vector distribution is shifted by that motion, so the laboratory speed distribution is not automatically the same Maxwell curve. A small opening selects particles differently too: an effusing flux is weighted by normal speed and passage rate. A beam measurement therefore cannot automatically be interpreted as the equilibrium distribution of the entire gas sample.[2][1][5]

In an ideal mixture, species share equilibrium temperature but have different masses. First calculate each species’ density or requested fraction. Weight them by particle fractions to obtain the total distribution. One mean molar mass generally does not reproduce this mixture. Equal mass fractions would also give different weights from equal particle fractions.[1]

Five exercises with solutions

Calculate · Foundation

A model gas has $M=0.020\,\mathrm{kg\,mol^{-1}}$ and $T=600\,\mathrm K$ . Find vmp.

Show hint

Use the molar scale.

Learning goal: Find most probable speed using molar mass in SI.

Show solution and calculation

Approach: $v_{\mathrm{mp}}=\sqrt{\frac{2RT}{M}}$ .
The supplied model values give about 706.306 m/s.

Check: compared with the first example at 0.044 kg/mol and 330 K, $\frac{T}{M}$ is four times as large, so most probable speed doubles.

v_{\mathrm{mp}}=\sqrt{\frac{2\cdot8.31446261815324\cdot600}{0.020}}\,\mathrm{m\,s^{-1}}\simeq706.305711\,\mathrm{m\,s^{-1}}

Interpret · Foundation

What probability corresponds to exactly $v=500\,\mathrm{m\,s^{-1}}$ in a continuous Maxwell density?

Show hint

Zero interval width.

Learning goal: Distinguish positive density from exact single-value probability.

Show solution and calculation

Approach: probability is the integral over a measurement interval; a single point has zero interval width.
Zero. f(500) can be positive, but one point has no area.
A finite measurement bin needs its bounds and an integral.

P(v=500\,\mathrm{m/s})=0

Calculate · Challenge

Find the scaled-model fraction below vmp.

Show hint

Use F(1), not a rectangle area.

Learning goal: Find cumulative speed probability up to the density mode.

Show solution and calculation

Approach: at most probable speed the scaled threshold is $x=1$ .
Insert it in the cumulative function.
The area calculated below is about 42.7593%.

Check: it lies between zero and one. The density peak is not the median; it does not split probability into equal halves.

\begin{aligned}P(v<a)&=F(1)\\&=\operatorname{erf}(1)-\frac2{\sqrt\pi}e^{-1}\\&\simeq0.4275932955\end{aligned}

Interpret · Application

For the same classical equilibrium gas, temperature increases from T to 4T. Two normalized Maxwell–Boltzmann curves then show a broader and lower distribution. Compare most probable speed and peak density height. Does the lower curve contain less total probability?

Show hint

The thermal speed scale grows with square root of temperature. Area under a normalized density must remain one.

Learning goal: Understand temperature scaling of a normalized speed density.

Show solution and calculation

Approach: the new speed scale is twice as large.
Stretching the horizontal axis requires halving density height to preserve area.
Most probable speed doubles and peak density halves.

Check: substitution $v=2u$ makes the new distribution’s integral equal one again. Density height is not the probability of an exact single speed.

\begin{aligned}v_{\mathrm{mp}}(4T)&=2v_{\mathrm{mp}}(T)\\f_{4T}(v)&=\frac12f_T\!\left(\frac v2\right)\\f_{4T,\max}&=\frac12f_{T,\max}\\\int_0^\infty f_{4T}(v)\,\mathrm dv&=\int_0^\infty f_T(u)\,\mathrm du=1\end{aligned}

[1][2]

Error analysis · Application

In a classical equilibrium gas, the normalized Maxwell speed density over the narrow interval from 490 to 510 m/s is approximately constant; take the rounded possible Maxwell-model value $f=0.0016\,\mathrm{s\,m^{-1}}$ . A solution reads this as “0.16% probability at exactly 500 m/s”. Calculate approximate interval probability and identify the probability of that exact single value.

Show hint

Probability is area, here approximated by density times interval width. A single point in a continuous distribution has zero width.

Learning goal: Translate probability density into interval probability with correct units.

Show solution and calculation

Approach: integrate the approximately constant density over the 20 m/s interval.
This gives probability 0.032 or 3.2%.
Probability of exactly 500 m/s is zero even though density there is positive.

Check: inverse-speed units of density cancel the speed units of interval width. The number 0.0016 alone is not a dimensionless probability; a more accurate interval probability would require integrating the actual Maxwell curve.

\begin{aligned}P(490\,\mathrm{m\,s^{-1}}\le v\le510\,\mathrm{m\,s^{-1}})&\approx f\Delta v\\{}&=(0.0016\,\mathrm{s\,m^{-1}})(20\,\mathrm{m\,s^{-1}})=0.032\\{}P(v=500\,\mathrm{m\,s^{-1}})&=0\end{aligned}

[1]

When the distribution must be replaced

The assumptions are classical nonrelativistic translation, thermal equilibrium and no relevant quantum degeneracy. Anisotropy, nonequilibrium or an accelerated beam require another description. Relevant relativistic effects already invalidate quadratic kinetic energy. Likewise, the classical curve is not an exact continuation of real gases down to $T=0$ .[2]

The existing rounded nitrogen model with $m=4.65\cdot10^{-26}\,\mathrm{kg}$ at 300 K still peaks at about 422.08 m/s. For a bin only 10 m/s wide near this maximum, $f(v_{\mathrm{mp}})\Delta v$ can be a useful approximation. Wider bins and ranges with strongly varying density are better integrated. A visible curve extending to three times most probable speed does not include the entire area normalised to infinity.[1]

Equipartition of energy assigns the same mean contribution to each suitable quadratic degree of freedom at classical equilibrium. The three translational components give $\langle E_{\mathrm{kin}}\rangle=3k_{\mathrm B}T/2$ . Which internal degrees of freedom actually contribute is a separate modelling question; a Maxwell speed curve does not decide it alone.[3][2]

In brief

Read a Maxwell curve as a density: probability belongs to area over an interval. Mass and temperature determine speed scale; different moments give different characteristics. The formula applies in the local gas rest frame. Beam measurements and mixtures require their own weighting. A higher density peak does not mean greater total probability.

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Quanta · Updated October 2, 2026

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October 2, 2026Added relevant scientific connections; preserved existing field and energy balances, derivations, examples, exercises, figures and model limits.
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October 1, 2026Reworked the introduction and explanations; added a third complete example, an original quantitative diagram and a short summary. Retained existing exercises, models and sources; clarified mathematical typography and model conditions.

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Frequently asked questions about Maxwell–Boltzmann Molecular Speed Distribution

What does Maxwell–Boltzmann Molecular Speed Distribution describe physically?+

The Maxwell–Boltzmann density describes thermal speed magnitudes of a classical ideal gas at equilibrium, relative to local mean flow. It is not a directional distribution or probability at a single point. f(v) has units s/m; only $\int f(v)\,dv$ over an interval gives a dimensionless particle fraction. Its maximum identifies the most probable speed of a narrow class. Mean and RMS occupy different positions. Mean velocity vector in the rest frame nevertheless vanishes because opposite directions cancel. All positive characteristic magnitudes are compatible with that. A flowing laboratory beam does not automatically have the same speed density; separate its mean motion before thermal analysis.

How can I check a calculation involving Maxwell–Boltzmann Molecular Speed Distribution?+

For $M=0.044\,\mathrm{kg\,mol^{-1}}$ and $T=330\,\mathrm K$ , $v_{\mathrm{mp}}=a\simeq353.1528554\,\mathrm{m\,s^{-1}}$ . The class 0.8a to 1.2a spans about 282.52–423.78 m/s. Using $F(x)=\operatorname{erf}(x)-\frac{2x}{\sqrt\pi}e^{-x^2}$ gives fraction $F(1.2)-F(0.8)\simeq0.3233884261$ . This is 32.3388 %, not dimensional density. The midpoint approximation $g(1)\cdot0.4$ gives 33.2086 %, exceeding it by 0.869757 percentage points. Integration is therefore appropriate for this broad class. Independently check that total area over all nonnegative speeds is 1 and the fraction lies within 0–1. Enter M in kg/mol; using kB would instead require particle mass m. To calculate mean energy, also use the second speed moment rather than squaring the ordinary average of magnitudes.

What assumptions and limits apply to Maxwell–Boltzmann Molecular Speed Distribution?+

The model assumes classical nonrelativistic translation and thermal equilibrium. Speeds are described relative to local mean gas flow. Quantum degeneracy, directional anisotropy or a nonequilibrium beam require other distributions. In mixtures each species has its own mass and, at shared T, its own speed scale. One mean molar mass generally does not reproduce the complete mixture density. Moreover an escaping particle flux is weighted by passage rate and normal velocity; it is not automatically the same equilibrium volume distribution. A visible curve stopping at $3v_{\mathrm{mp}}$ is only an excerpt, although the mathematical density is normalised over the full range to infinity.

How do I check units and scale for Maxwell–Boltzmann Molecular Speed Distribution?+

The prefactor combined with $v^2$ produces density in s/m, making $f(v)\,dv$ dimensionless. In the molar speed scale $a=\sqrt{\frac{2RT}{M}}$ , M must be in kg/mol; 32 g/mol becomes 0.032 kg/mol. Temperature is in kelvin. Quadrupling T for the same species quadruples squared characteristic speeds and doubles their magnitudes. The density then becomes twice as wide horizontally and half as high vertically; total area stays 1. Density values alone are not percentages. At one exact speed continuous probability is zero. A finite measurement class needs its width or both integration limits, together with compatible units.

When should I combine Maxwell–Boltzmann Molecular Speed Distribution with another relation?+

The normalised density supplies moments and interval fractions. Its maximum gives $v_{\mathrm{mp}}=a$ ; ordinary mean is $\frac{2a}{\sqrt\pi}$ and $v_{\mathrm{rms}}=\sqrt{\frac32}a$ . Using $\frac12m\langle v^2\rangle$ yields mean translational energy $\frac32k_{\mathrm B}T$ . For isotropic ideal pressure, additionally combine the second moment with $p=\frac13\rho\langle v^2\rangle$ . A fixed threshold $v>2a$ has 4.60117 % in the fast tail. Quadrupling temperature turns that same absolute threshold into $x=1$ ; 57.2407 % then exceed the old threshold. A 2a threshold moving with the new speed scale instead keeps the original fraction. That distinction matters more for temperature comparisons than simply locating the peak.

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Which formula describes Maxwell–Boltzmann Molecular Speed Distribution?

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How do you rearrange Maxwell–Boltzmann Molecular Speed Distribution for Most probable speed?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol.2, §2.4: Distribution of Molecular Speeds.

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P = \sigma \cdot A \cdot T^4

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Create a curated FSRS exam set for f(v) = 4π(m/2πkBT)³ᐟ² v² e^(−mv²/2kBT): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Maxwell–Boltzmann distribution of molecular speeds

At thermal equilibrium, the Maxwell–Boltzmann density describes how molecular speed magnitudes are distributed in a classical ideal gas; it is a density over v≥0.

AdvancedUniversity/collegeBachelor foundations

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Symbol

Meaning

Unit

f(v)

Probability density of thermal speed magnitude

s/m

v

Thermal speed relative to local mean gas flow

m/s

m

Mass of one molecule

T

Absolute equilibrium temperature

k_{\mathrm{B}}

Boltzmann constant

J/K

Derivation & background

Exam blueprint

Validity range

Derivation steps

1Consider classical equilibrium motion relative to local mean gas motion.
2Each component has a Gaussian density with factor $e^{-\frac{mv_i^2}{2k_{\mathrm B}T}}$ .
3The product of the three independent densities gives isotropic vector density $P(\mathbf v)$ .
4The speed class from $v$ to $v+dv$ contains a velocity-space shell of measure $4\pi v^2\,dv$ .
5This shell factor gives normalised speed density $f(v)$ , with units s/m.
6Calculate fractions using $\int f(v)\,dv$ . With $a=\sqrt{\frac{2k_{\mathrm B}T}{m}}$ and $x=\frac va$ , $F(x)=\operatorname{erf}(x)-\frac{2x}{\sqrt\pi}e^{-x^2}$ .

How do you calculate with Maxwell–Boltzmann Molecular Speed Distribution?

Here is how to work through a typical Maxwell–Boltzmann Molecular Speed Distribution (f(v) = 4π(m/2πkBT)³ᐟ² v² e^(−mv²/2kBT)) task step by step:

Task

Rounded nitrogen model: particle mass 4.65·10⁻²⁶ kg at 300 K. Find most probable speed and explain how a 10 m/s bin near the peak gives a particle fraction.

Solution path

The rounded nitrogen model at 300 K has $m\simeq4.65\cdot10^{-26}\,\mathrm{kg}$ , corresponding to about 28.0 u.
Its peak is at $v_{\mathrm{mp}}=\sqrt{\frac{2k_{\mathrm B}T}{m}}\simeq422\,\mathrm{m\,s^{-1}}$ .
Normalised total area is one.
For a narrow 10 m/s bin near the peak, $f(v_{\mathrm{mp}})\Delta v$ approximates probability, not the density value alone.

Understand and apply the equation

A density is not yet a probability

f(v)=4\pi\left(\frac{m}{2\pi k_{\mathrm B}T}\right)^{\frac{3}{2}}v^2e^{-\frac{mv^2}{2k_{\mathrm B}T}},\quad v\ge0

v is thermal speed relative to local mean gas motion; m is mass per particle.[1]

How to read the area

P(v_1\le v\le v_2)=\int_{v_1}^{v_2}f(v)\,dv

A finite speed class requires two boundaries.

A lower curve can have the same total area

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Three components produce a spherical shell

\begin{aligned}P(\mathbf v)&=\left(\frac{m}{2\pi k_{\mathrm B}T}\right)^{\frac{3}{2}}e^{-\frac{m(v_x^2+v_y^2+v_z^2)}{2k_{\mathrm B}T}}\\{} f(v)dv&=P(\mathbf v)4\pi v^2dv\end{aligned}

The isotropic density is constant on a velocity sphere; shell area supplies v².[1][2]

Explore further: scaled distribution and cumulative function

\begin{aligned}a&=\sqrt{\frac{2k_{\mathrm B}T}{m}}=\sqrt{\frac{2RT}{M}}\\x&=\frac va\\g(x)&=a f(ax)=\frac4{\sqrt\pi}x^2e^{-x^2}\end{aligned}

a is most probable speed; M is in kg/mol.[1]

\begin{aligned}\int_0^\infty g(x)\,dx&=1\\F(x)&=\operatorname{erf}(x)-\frac{2x}{\sqrt\pi}e^{-x^2}\end{aligned}

F is cumulative probability in scaled speed; $F(0)=0$ and $F(\infty)=1$ .[1]

Keep mode, mean and RMS separate

\begin{aligned}v_{\mathrm{mp}}&=a\\\langle v\rangle&=\frac{2a}{\sqrt\pi}\\v_{\mathrm{rms}}&=\sqrt{\frac32}\,a\end{aligned}

The mode follows from $\frac{d}{dx}(x^2e^{-x^2})=0$ ; mean and RMS from first and second moments.[1]

Actual intervals rather than single density values

Use the cumulative distribution for intervals and tails. In the third example the same calculation is carried out separately for each species before weighting by particle fractions.[1]

A speed class in model gas

Given: $M=0.044\,\mathrm{kg\,mol^{-1}}$ , $T=330\,\mathrm K$ ; classical ideal equilibrium. Interval $0.8v_{\mathrm{mp}}$ to $1.2v_{\mathrm{mp}}$ .
Find: Calculate the class bounds and particle fraction.

Approach: A speed interval uses the area under the distribution. $a=\sqrt{\frac{2RT}{M}},\quad P(v_1<v<v_2)=F(\frac{v_2}{a})-F(\frac{v_1}{a})$ . Find speed scale.
$a=\sqrt{\frac{2RT}{M}}=353.1528554\,\mathrm{m/s}$
Evaluate CDF at both bounds.
$P=F(1.2)-F(0.8)=0.3233884261$
Compare with the midpoint approximation.
$g(1)(1.2-0.8)=0.3320859979$

Result: The class about 282.52–423.78 m/s contains 32.3388 % of particles. The broad-bin midpoint approximation gives 33.2086 %.

Check: The approximation exceeds the result by 0.869757 percentage points; density alone would have units. Integration of f(v) with dv in m/s independently checks the exact class.

A fast distribution tail

Given: Same model gas; seek $v>2v_{\mathrm{mp}}$ . A sample contains $N=2.00\cdot10^{21}$ particles.
Find: Find fraction and expected count, then compare doubled speed scale.

Approach: For the fast tail use the complement of the cumulative distribution. $P(v>v_*)=1-F(\frac{v_*}{a})$ . Use the CDF complement.
$P(v>2a)=1-F(2)=0.04601170569$
Find expected count from fraction.
$\langle N_{tail}\rangle=NP=9.202341138\times10^{19}$
Quadrupling temperature doubles a.
$a(4T)=2a(T)$

Result: About 4.60117 % exceed 706.31 m/s. At fourfold T, the same absolute threshold has scaled value $x=1$ rather than 2.

Check: The fraction above the original threshold then becomes $1-F(1)=0.5724067045$ . Above the respective new 2a threshold, however, the same 4.60117 % fraction remains.

Fast particles in a mixture

Given: Ideal classical equilibrium mixture at 300 K in the gas rest frame. Equal particle counts of two model species with molar masses 0.00400 and 0.0400 kg/mol. Both are nonrelativistic and nondegenerate. Find the fraction with speed above 700 m/s.
Find: Find both species fractions and the fraction of the whole mixture.

Approach: each species has its own scale. Calculate their tail probabilities first, then weight by particle fractions.
$\begin{aligned}a_j&=\sqrt{\frac{2RT}{M_j}}\\P_j&=1-F\!\left(\frac{700\,\mathrm{m\,s^{-1}}}{a_j}\right)\\P_{\mathrm{mix}}&=\frac12P_1+\frac12P_2\end{aligned}$
Insert both scales and dimensionless thresholds.
$\begin{aligned}a_1&\simeq1116.767\,\mathrm{m\,s^{-1}},&x_1&\simeq0.626809\\a_2&\simeq353.153\,\mathrm{m\,s^{-1}},&x_2&\simeq1.982145\end{aligned}$
Evaluate the fractions using the cumulative function, then mix them.
$\begin{aligned}P_1&\simeq0.852865\\P_2&\simeq0.0490439\\P_{\mathrm{mix}}&\simeq0.450955\end{aligned}$

Result: About 85.29% of the light species and 4.904% of the heavy species exceed 700 m/s. Together this is about 45.10% of all particles.

Distinguish local temperature and a laboratory beam

Five exercises with solutions

Calculate · Foundation

A model gas has $M=0.020\,\mathrm{kg\,mol^{-1}}$ and $T=600\,\mathrm K$ . Find vmp.

Show hint

Use the molar scale.

Learning goal: Find most probable speed using molar mass in SI.

Show solution and calculation

Approach: $v_{\mathrm{mp}}=\sqrt{\frac{2RT}{M}}$ .
The supplied model values give about 706.306 m/s.

Check: compared with the first example at 0.044 kg/mol and 330 K, $\frac{T}{M}$ is four times as large, so most probable speed doubles.

v_{\mathrm{mp}}=\sqrt{\frac{2\cdot8.31446261815324\cdot600}{0.020}}\,\mathrm{m\,s^{-1}}\simeq706.305711\,\mathrm{m\,s^{-1}}

Interpret · Foundation

What probability corresponds to exactly $v=500\,\mathrm{m\,s^{-1}}$ in a continuous Maxwell density?

Show hint

Zero interval width.

Learning goal: Distinguish positive density from exact single-value probability.

Show solution and calculation

Approach: probability is the integral over a measurement interval; a single point has zero interval width.
Zero. f(500) can be positive, but one point has no area.
A finite measurement bin needs its bounds and an integral.

P(v=500\,\mathrm{m/s})=0

Calculate · Challenge

Find the scaled-model fraction below vmp.

Show hint

Use F(1), not a rectangle area.

Learning goal: Find cumulative speed probability up to the density mode.

Show solution and calculation

Approach: at most probable speed the scaled threshold is $x=1$ .
Insert it in the cumulative function.
The area calculated below is about 42.7593%.

Check: it lies between zero and one. The density peak is not the median; it does not split probability into equal halves.

\begin{aligned}P(v<a)&=F(1)\\&=\operatorname{erf}(1)-\frac2{\sqrt\pi}e^{-1}\\&\simeq0.4275932955\end{aligned}

Interpret · Application

For the same classical equilibrium gas, temperature increases from T to 4T. Two normalized Maxwell–Boltzmann curves then show a broader and lower distribution. Compare most probable speed and peak density height. Does the lower curve contain less total probability?

Show hint

The thermal speed scale grows with square root of temperature. Area under a normalized density must remain one.

Learning goal: Understand temperature scaling of a normalized speed density.

Show solution and calculation

Approach: the new speed scale is twice as large.
Stretching the horizontal axis requires halving density height to preserve area.
Most probable speed doubles and peak density halves.

Check: substitution $v=2u$ makes the new distribution’s integral equal one again. Density height is not the probability of an exact single speed.

\begin{aligned}v_{\mathrm{mp}}(4T)&=2v_{\mathrm{mp}}(T)\\f_{4T}(v)&=\frac12f_T\!\left(\frac v2\right)\\f_{4T,\max}&=\frac12f_{T,\max}\\\int_0^\infty f_{4T}(v)\,\mathrm dv&=\int_0^\infty f_T(u)\,\mathrm du=1\end{aligned}

[1][2]

Error analysis · Application

In a classical equilibrium gas, the normalized Maxwell speed density over the narrow interval from 490 to 510 m/s is approximately constant; take the rounded possible Maxwell-model value $f=0.0016\,\mathrm{s\,m^{-1}}$ . A solution reads this as “0.16% probability at exactly 500 m/s”. Calculate approximate interval probability and identify the probability of that exact single value.

Show hint

Probability is area, here approximated by density times interval width. A single point in a continuous distribution has zero width.

Learning goal: Translate probability density into interval probability with correct units.

Show solution and calculation

Approach: integrate the approximately constant density over the 20 m/s interval.
This gives probability 0.032 or 3.2%.
Probability of exactly 500 m/s is zero even though density there is positive.

\begin{aligned}P(490\,\mathrm{m\,s^{-1}}\le v\le510\,\mathrm{m\,s^{-1}})&\approx f\Delta v\\{}&=(0.0016\,\mathrm{s\,m^{-1}})(20\,\mathrm{m\,s^{-1}})=0.032\\{}P(v=500\,\mathrm{m\,s^{-1}})&=0\end{aligned}

[1]

When the distribution must be replaced

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Added relevant scientific connections; preserved existing field and energy balances, derivations, examples, exercises, figures and model limits.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Reworked the introduction and explanations; added a third complete example, an original quantitative diagram and a short summary. Retained existing exercises, models and sources; clarified mathematical typography and model conditions.

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Quanta. Maxwell–Boltzmann Molecular Speed Distribution. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/maxwell-boltzmann-geschwindigkeitsverteilung

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Retain Maxwell–Boltzmann Molecular Speed Distribution for exams

Create a curated FSRS exam set for f(v) = 4π(m/2πkBT)³ᐟ² v² e^(−mv²/2kBT): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Maxwell–Boltzmann distribution of molecular speeds

Variables & units

Maxwell-Boltzmann distribution calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Kinetic theory of gases

Worked example

Applications

How do you calculate with Maxwell–Boltzmann Molecular Speed Distribution?

A density is not yet a probability

How to read the area

A lower curve can have the same total area

Three components produce a spherical shell

Explore further: scaled distribution and cumulative function

Keep mode, mean and RMS separate

Actual intervals rather than single density values

A speed class in model gas

A fast distribution tail

Fast particles in a mixture

Distinguish local temperature and a laboratory beam

Five exercises with solutions

A model gas has M=0.020 kg mol−1M=0.020\,\mathrm{kg\,mol^{-1}}M=0.020kgmol−1 and T=600 KT=600\,\mathrm KT=600K. Find vmp.

What probability corresponds to exactly v=500 m s−1v=500\,\mathrm{m\,s^{-1}}v=500ms−1 in a continuous Maxwell density?

Find the scaled-model fraction below vmp.

For the same classical equilibrium gas, temperature increases from T to 4T. Two normalized Maxwell–Boltzmann curves then show a broader and lower distribution. Compare most probable speed and peak density height. Does the lower curve contain less total probability?

When the distribution must be replaced

In brief

Sources for this explanation

Frequently asked questions about Maxwell–Boltzmann Molecular Speed Distribution

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Maxwell–Boltzmann Molecular Speed Distribution for exams

Maxwell–Boltzmann distribution of molecular speeds

Variables & units

Maxwell-Boltzmann distribution calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Kinetic theory of gases

Worked example

Applications

How do you calculate with Maxwell–Boltzmann Molecular Speed Distribution?

A density is not yet a probability

How to read the area

A lower curve can have the same total area

Three components produce a spherical shell

Explore further: scaled distribution and cumulative function

Keep mode, mean and RMS separate

Actual intervals rather than single density values

A speed class in model gas

A fast distribution tail

Fast particles in a mixture

Distinguish local temperature and a laboratory beam

Five exercises with solutions

A model gas has M=0.020 kg mol−1M=0.020\,\mathrm{kg\,mol^{-1}}M=0.020kgmol−1 and T=600 KT=600\,\mathrm KT=600K. Find vmp.

What probability corresponds to exactly v=500 m s−1v=500\,\mathrm{m\,s^{-1}}v=500ms−1 in a continuous Maxwell density?

Find the scaled-model fraction below vmp.

For the same classical equilibrium gas, temperature increases from T to 4T. Two normalized Maxwell–Boltzmann curves then show a broader and lower distribution. Compare most probable speed and peak density height. Does the lower curve contain less total probability?

When the distribution must be replaced

In brief

Sources for this explanation

Frequently asked questions about Maxwell–Boltzmann Molecular Speed Distribution

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Maxwell–Boltzmann Molecular Speed Distribution for exams

A model gas has $M=0.020\,\mathrm{kg\,mol^{-1}}$ and $T=600\,\mathrm K$ . Find vmp.

What probability corresponds to exactly $v=500\,\mathrm{m\,s^{-1}}$ in a continuous Maxwell density?

A model gas has $M=0.020\,\mathrm{kg\,mol^{-1}}$ and $T=600\,\mathrm K$ . Find vmp.

What probability corresponds to exactly $v=500\,\mathrm{m\,s^{-1}}$ in a continuous Maxwell density?