Physics · Kinetic theory of gases

Thermal RMS speed of gas molecules

Q: What does Root-mean-square molecular speed describe physically?

RMS speed is the square root of mean squared thermal speed: $c_{\mathrm{rms}}=\sqrt{\langle c^2\rangle}$. It connects the distribution with mean translational energy because $\langle E_{\mathrm{tr}}\rangle=\frac12mc_{\mathrm{rms}}^2$. Classical equilibrium gives $c_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm B}T}{m}}=\sqrt{\frac{3RT}{M}}$. This is neither ordinary average ⟨c⟩ nor mean velocity vector, which vanishes in gas at rest. In the Maxwell distribution, crms exceeds both mean and most probable speed. Speeds are measured relative to the local gas rest frame. Shared flow can increase laboratory RMS speed without representing the same increase in thermal temperature. The sample size cancels from this characteristic speed relation.

Q: How can I check a calculation involving Root-mean-square molecular speed?

For $M=0.032\,\mathrm{kg\,mol^{-1}}$ and $T=400\,\mathrm K$, $\sqrt{\frac{3RT}{M}}$ gives about 558.384 m/s. Independently check with $m=\frac M{N_{\mathrm A}}$: $\frac12mc_{\mathrm{rms}}^2$ must equal $\frac32k_{\mathrm B}T=8.283894\cdot10^{-21}\,\mathrm J$. In the Maxwell model, $c_{\mathrm{mp}}\simeq455.918\,\mathrm{m\,s^{-1}}$ and $\langle c\rangle\simeq514.449\,\mathrm{m\,s^{-1}}$; ordering $c_{\mathrm{mp}}<\langle c\rangle<c_{\mathrm{rms}}$ must hold. Also check $\frac{c_{\mathrm{rms}}}{c_{\mathrm{mp}}}=\sqrt{\frac32}$. The result does not depend on whether the sample contains one mole or another amount. Before inferring temperature from measured speeds, check whether mean flow was subtracted. Uncorrected laboratory RMS speed would otherwise give an excessive thermal value. At equal energy, a heavier particle must have a smaller squared speed.

Q: What assumptions and limits apply to Root-mean-square molecular speed?

The temperature relation assumes classical nonrelativistic translation in thermal equilibrium. Speed is evaluated relative to mean gas motion. Quantum degeneracy or strongly interacting dense systems may require a different distribution. Away from equilibrium, the three component moments need not be equal. In a mixture, use the appropriate particle or molar mass separately for each species; an average mass does not automatically give their respective RMS values. Formulas for most probable and arithmetic mean speed additionally require a Maxwell–Boltzmann distribution. crms remains a statistical quantity and specifies neither every molecule’s speed nor sound-wave propagation speed.

Q: How do I check units and scale for Root-mean-square molecular speed?

The particle form needs m in kg per molecule; the molar form needs M in kg/mol. Thus 32 g/mol becomes 0.032 kg/mol. $\frac{RT}{M}$ has units $\mathrm{J\,kg^{-1}}=\mathrm{m^2\,s^{-2}}$, and only the square root gives m/s. Inserting unconverted 32 instead of 0.032 makes speed too small by $\sqrt{1000}$. Temperature is in kelvin. At unchanged species, crms scales with $\sqrt T$; at equal T it scales with $\frac1{\sqrt M}$. A ten-times-heavier species therefore has only $\frac1{\sqrt{10}}$ of the RMS speed but the same mean translational energy per particle.

Q: When should I combine Root-mean-square molecular speed with another relation?

For temperature inference, rearrange $T=\frac{Mc_{\mathrm{rms}}^2}{3R}$. At $M=0.040\,\mathrm{kg\,mol^{-1}}$ and thermal $c_{\mathrm{rms}}=600\,\mathrm{m\,s^{-1}}$, $T\simeq577.307\,\mathrm K$. For pressure, combine crms with $p=\frac13\rho c_{\mathrm{rms}}^2$ in an isotropic ideal gas. For flowing samples, use $\mathbf v=\mathbf u+\mathbf c$ and $\langle v^2\rangle=u^2+\langle c^2\rangle$ to subtract shared motion. For example, $u=100\,\mathrm{m\,s^{-1}}$ and thermal $c_{\mathrm{rms}}=300\,\mathrm{m\,s^{-1}}$ give laboratory RMS speed 316.228 m/s, not 400 m/s. The energy equation still uses thermal speed 300 m/s. If arithmetic mean speed is required, use its own Maxwell moment instead of merely renaming crms. In mixtures, evaluate each species using its own particle mass.

The RMS speed is the square root of the mean squared speed of an ideal-gas molecule.

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Formula

Calculate

v_{rms} = \frac{3 k _{B} T}{m}

v in m/s; kB in J/K; T in K; m in kg per molecule.

Variables & units

Variables & units
Symbol	Meaning	Unit
vrms	Square root of mean squared thermal speed relative to local mean gas flow	m/s
$kB$	Boltzmann constant	J/K
$T$	Absolute temperature	K
$m$	Mass of one molecule	kg

LaTeX

v_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm B}T}{m}}

QUANTA STUDYquanta-study.de

Normalised Maxwell density. Maximum at x equal to 1, mean speed at 1.1284 and RMS at 1.2247. Separate label leaders identify their closely spaced actual positions. — Three distinct characteristics of one Maxwell distribution at equilibrium at rest: cmp is most probable speed, ⟨c⟩ arithmetic mean and crms the square root of its second moment. Normalised density g(x) = 4x²exp(−x²)/√π with x = c/cmp.

Gas molecule RMS speed calculator

Find the root mean square speed of a gas from temperature and mass, plus the most probable and the mean speed from the Maxwell distribution.

Calculation model

Find

Molar mass M

Molar mass of the gas from the periodic table, e.g. 28 g/mol for nitrogen N₂.

Absolute temperature T

Temperature of the gas, e.g. 300 K; °C is converted to kelvin for you.

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

Working · rearrange for crms, substitute, evaluate

1. Rearrange for crms
$c_{\rm rms}=\sqrt{\frac{3RT}{M}}$
2. Substitute values (in SI units)
- M = 28 g/mol = 0.028 kg/mol
$\sqrt{\frac{3\cdot 8.31446\cdot 300}{0.028}}=516.963$
3. Result
crms = 517 m/s

Result: Thermal RMS speed crms

crms = 517 m/s

With M = 28 g/mol and T = 300 K, crms = 517 m/s.

Quick tips

Four times the temperature, twice the speed: c is proportional to √T. At the same temperature a molecule 16 times heavier is only a quarter as fast.
Remember the order: most probable < mean < RMS speed, in the ratio √2 : √(8/π) : √3. For nitrogen at 300 K about 422, 476 and 517 m/s.

Assumptions and limits

Dilute classical nonrelativistic ideal gas in thermal equilibrium at positive absolute temperature, with three translational degrees of freedom and no relevant quantum degeneracy or interaction energy. Thermal motion is measured relative to local mean gas motion. This is neither full internal energy of polyatomic molecules nor a single particle’s temperature. The specified mass belongs to the same individual species, not an averaged mixture mass. Speed is thermal with mean flow already removed.

Derivation & background

The temperature formula applies to classical equilibrium motion relative to the local gas rest frame. With $m=\frac M{N_{\mathrm A}}$ and $R=N_{\mathrm A}k_{\mathrm B}$ , $c_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$ for any sample size. Use $M$ in kg/mol. RMS is the square root of mean squared speed, not arithmetic mean or most probable speed. Remove common flow before inferring temperature.

Exam blueprint

Validity range

Classical nonrelativistic ideal gas in thermal equilibrium. RMS concerns thermal speeds relative to local mean flow. m is particle mass; M in kg/mol applies to any sample size. Other Maxwell characteristic speeds are distinct quantities.

Derivation steps

Find the second speed moment from translational energy, then take its square root.

1Evaluate thermal speeds relative to mean gas motion and define $c_{\mathrm{rms}}=\sqrt{\langle c^2\rangle}$ .
2Three classical translational contributions give $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T$ .
3The same energy average is $\frac12m\langle c^2\rangle$ .
4Equating gives $\langle c^2\rangle=\frac{3k_{\mathrm B}T}{m}$ .
5The positive square root gives $c_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm B}T}{m}}$ , a magnitude.
6Using $m=\frac M{N_{\mathrm A}}$ and $R=N_{\mathrm A}k_{\mathrm B}$ gives $c_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$ . Use kg/mol; sample size is arbitrary.
7Arithmetic mean and density peak are distinct characteristics requiring their own distribution moments.

Rearrangements

Absolute temperature

T=\frac{m v_{\mathrm{rms}}^2}{3k_{\mathrm B}}

m is single-molecule mass; v_rms here denotes the thermal value relative to local mean gas motion, not uncorrected laboratory RMS speed.

Task variant

Nitrogen model in the gas rest frame: molar mass 0.0280 kg/mol, temperature 300 K. Calculate thermal RMS speed and state its meaning.

For the nitrogen model with $M=0.0280\,\mathrm{kg\,mol^{-1}}$ at 300 K, $c_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}\simeq517\,\mathrm{m\,s^{-1}}$ .
This is thermal speed in the gas rest frame and is independent of sample size.

Common mistakes

Combining molar mass with Boltzmann’s constant or interpreting laboratory bulk flow as thermal motion.

Particle mass belongs with $k_{\mathrm B}$ , molar mass with $R$ . Temperature requires thermal RMS speed relative to local mean flow.

Exam context

Estimate molecular speeds from temperature and molar mass.

Formula cluster

Kinetic theory of gases

The second speed moment connects temperature, translational energy and gas pressure; the mean and density mode are different characteristics.

⟨Ekin⟩ = 3/2 kBT

Mean translational kinetic energy per gas molecule

pV = 1/3 Nm⟨v²⟩

Pressure from kinetic theory

U = 3/2 nRT

Internal energy of a monatomic ideal gas

Worked example

For the nitrogen model with $M=0.0280\,\mathrm{kg\,mol^{-1}}$ at 300 K, $c_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}\simeq517\,\mathrm{m\,s^{-1}}$ . This is thermal speed in the gas rest frame and is independent of sample size.

Applications

Estimate molecular speeds from temperature and molar mass.

How do you calculate with Root-mean-square molecular speed?

Here is how to work through a typical Root-mean-square molecular speed (vrms = √(3kBT/m)) task step by step:

1
Task
Nitrogen model in the gas rest frame: molar mass 0.0280 kg/mol, temperature 300 K. Calculate thermal RMS speed and state its meaning.
Solution path
1. For the nitrogen model with $M=0.0280\,\mathrm{kg\,mol^{-1}}$ at 300 K, $c_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}\simeq517\,\mathrm{m\,s^{-1}}$ .
2. This is thermal speed in the gas rest frame and is independent of sample size.

Understand and apply the equation

How can many different molecular speeds be represented by one number? Mean kinetic energy requires mean squared speed. RMS means squaring, averaging and then taking the square root. In classical equilibrium, temperature gives $c_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm B}T}{m}}$ . Distinguish this energy-related measure from ordinary mean speed, the most probable speed and common flow.

Square, average, then take the square root

c_{\mathrm{rms}}=\sqrt{\langle c^2\rangle}=\sqrt{\frac{3k_{\mathrm B}T}{m}}

c is speed relative to the local mean gas motion. The upper notation vrms denotes this thermal magnitude.[1]

The order is crucial. For the three speeds 100, 200 and 300 m/s, ordinary averaging gives 200 m/s. The RMS procedure averages their squares first; taking the square root then gives about 216.025 m/s. Faster particles influence the energy average more strongly. These three values are an illustrative sample, not a Maxwell distribution or proof of an equilibrium temperature.

\begin{aligned}\overline c&=\frac{100+200+300}{3}\,\mathrm{m\,s^{-1}}=200\,\mathrm{m\,s^{-1}}\\c_{\mathrm{rms}}&=\sqrt{\frac{100^2+200^2+300^2}{3}}\,\mathrm{m\,s^{-1}}\\&\simeq216.025\,\mathrm{m\,s^{-1}}\end{aligned}

Mean speed differs from the square root of mean squared speed.

Energy supplies the second moment

\begin{aligned}\frac12m\langle c^2\rangle&=\frac32k_{\mathrm B}T\\\langle c^2\rangle&=\frac{3k_{\mathrm B}T}{m}\\c_{\mathrm{rms}}&=\sqrt{\frac{3k_{\mathrm B}T}{m}}\end{aligned}

Three translational degrees of freedom in classical equilibrium supply the factor 3.[2]

Start with mean translational energy in classical equilibrium. Divide by $\frac m2$ to obtain mean squared speed. Only taking the square root gives a speed again. In isotropic equilibrium the three components contribute equally to the mean square, even though mean velocity vector is zero. The formula claims neither identical instantaneous particle speeds nor a direction for a particular molecule.[1]

\begin{aligned}\langle c_x^2\rangle=\langle c_y^2\rangle=\langle c_z^2\rangle&=\frac{k_{\mathrm B}T}{m}\\\langle\mathbf c\rangle&=0\end{aligned}

In isotropic equilibrium at rest, component means vanish but their second moments are positive.[5]

How do you use particle or molar mass and solve for temperature?

\begin{aligned}m&=\frac M{N_{\mathrm A}}\\R&=N_{\mathrm A}k_{\mathrm B}\\c_{\mathrm{rms}}&=\sqrt{\frac{3RT}{M}}\end{aligned}

The molar form applies to any sample size; it does not require exactly one mole. Insert M in kg/mol.[6]

Which mass is specified? Use individual particle mass $m$ with $k_{\mathrm B}$ , and molar mass $M$ with $R$ . A model value of 32 g/mol becomes 0.032 kg/mol. The ratio $\frac{RT}{M}$ has units J/kg, or square metres per square second. Inserting 32 instead of 0.032 loses a factor of $\sqrt{1000}$ in speed. The entire sample’s mass does not belong in the single-particle formula.

\begin{aligned}T&=\frac{Mc_{\mathrm{rms}}^2}{3R}\\M&=\frac{3RT}{c_{\mathrm{rms}}^2}\end{aligned}

These rearrangements concern classical equilibrium and thermal RMS speed of the same species.[1]

Distinguish three characteristic magnitudes

\begin{aligned}c_{\mathrm{mp}}&=\sqrt{\frac{2RT}{M}}\\\langle c\rangle&=\sqrt{\frac{8RT}{\pi M}}\\c_{\mathrm{rms}}&=\sqrt{\frac{3RT}{M}}\end{aligned}

These three values belong to the three-dimensional Maxwell–Boltzmann distribution.[3]

Three distinct characteristics are useful for a Maxwell distribution. $c_{\mathrm{mp}}$ is at the peak of the speed density. $\langle c\rangle$ is the ordinary mean of speeds. $c_{\mathrm{rms}}$ comes from the second moment and is largest: $c_{\mathrm{mp}}<\langle c\rangle<c_{\mathrm{rms}}$ . All are positive even when the gas as a whole is at rest. Their temperature formulas assume a Maxwell equilibrium distribution; the RMS definition itself also applies to other samples.[3]

View diagram: three characteristic speeds

Three characteristics of the same distribution

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Original Maxwell model curves for 0.032 kg/mol. At 400 K the values are about 455.92, 514.45 and 558.38 m/s. These curves show characteristics, not individual molecular trajectories.[3]

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Three complete calculations of thermal and laboratory RMS speed

All three calculations assume classical equilibrium motion. The third explicitly includes mean flow, so its contribution is subtracted from the laboratory mean square before inferring temperature.[1][7]

Three characteristic speeds of a model gas

Given: $M=0.032\,\mathrm{kg\,mol^{-1}}$ , $T=400\,\mathrm K$ ; classical ideal gas at equilibrium at rest.
Find: Find cmp, ⟨c⟩ and crms.

Approach: For a classical ideal gas at rest in equilibrium, three distinct characteristic values describe the same speed distribution. Use the same temperature and molar mass, keeping their different factors separate.
$\begin{aligned}c_{\mathrm{mp}}&=\sqrt{\frac{2RT}{M}}\\{}\langle c\rangle&=\sqrt{\frac{8RT}{\pi M}}\\{}c_{\mathrm{rms}}&=\sqrt{\frac{3RT}{M}}\end{aligned}$
Use the molar RMS form.
$c_{\mathrm{rms}}=\sqrt{\frac{3(8.31446261815324)(400)}{0.032}}\approx558.384\,\mathrm{m/s}$
Determine the other characteristic values from their respective moments.
$\begin{aligned}c_{\mathrm{mp}}&\approx455.918\,\mathrm{m/s}\\{} \langle c\rangle&\approx514.449\,\mathrm{m/s}\end{aligned}$

Result: The most probable speed is about 455.92 m/s, ordinary mean 514.45 m/s and RMS speed 558.38 m/s.

Check: With $m=\frac M{N_A}$ , $\frac12m c_{\mathrm{rms}}^2$ equals $\frac32k_BT=8.283894\cdot10^{-21}\,\mathrm J$ . $\frac{c_{\mathrm{rms}}}{c_{\mathrm{mp}}}=\sqrt{\frac32}\approx1.224745$ independently checks both speeds.

Temperature from a thermal RMS measurement

Given: $M=0.040\,\mathrm{kg\,mol^{-1}}$ ; thermal $c_{\mathrm{rms}}=600\,\mathrm{m\,s^{-1}}$ in the local rest frame, classical equilibrium distribution.
Find: Find T and check using particle energy.

Approach: The measured rms speed is thermal in the local rest frame. Solve the equilibrium relation for kelvin temperature, then check particle energy with mass derived from molar mass.
$\begin{aligned}T&=\frac{Mc_{\mathrm{rms}}^2}{3R}\\{}\langle E_{\mathrm{tr}}\rangle&=\frac12\frac M{N_A}c_{\mathrm{rms}}^2\end{aligned}$
Solve for T.
$T=\frac{(0.040)(600)^2}{3(8.31446261815324)}\approx577.3073\,\mathrm K$
Insert particle mass and calculate energy.
$\langle E_{\mathrm{tr}}\rangle=\frac12\frac{0.040}{N_{\mathrm A}}(600)^2=\frac{7200}{N_{\mathrm A}}\,\mathrm J$

Result: The model gives T≈577.31 K, about 304.16 °C. The 600 m/s must not be uncorrected laboratory RMS speed of a flowing gas.

Check: Multiplying particle energy by $N_A$ gives 7200 joules per mole. $\frac32RT$ gives the same value. Reinserting temperature gives 600 metres per second.

Removing flow from a laboratory measurement

Given: Classical ideal model gas of molar mass 0.0280 kg/mol. Measured laboratory RMS speed 650 m/s; local common gas speed 250 m/s. Thermal distribution is isotropic relative to gas motion.
Find: Find thermal RMS speed and equilibrium temperature.

Approach: the squares add. Subtract the bulk contribution from the laboratory square, then calculate temperature using the thermal part.
$\begin{aligned}c_{\mathrm{rms}}&=\sqrt{v_{\mathrm{rms,lab}}^2-u^2}\\T&=\frac{Mc_{\mathrm{rms}}^2}{3R}\end{aligned}$
First insert the speed values.
$c_{\mathrm{rms}}=\sqrt{650^2-250^2}\,\mathrm{m\,s^{-1}}=600\,\mathrm{m\,s^{-1}}$
Use molar mass with the gas constant.
$T=\frac{0.0280\cdot600^2}{3\cdot8.31446261815324}\,\mathrm K\simeq404.115\,\mathrm K$

Result: Thermal RMS speed is 600 m/s; temperature is about 404.12 K or 130.97 °C.

Check: $\sqrt{250^2+600^2}=650$ reconstructs the laboratory value. Inserting 650 m/s directly would interpret bulk kinetic energy as thermal motion. A laboratory value below 250 m/s would already contradict the stated assumptions.

Separate laboratory motion from the thermal part

\begin{aligned}\mathbf v&=\mathbf u+\mathbf c\\\langle\mathbf c\rangle&=0\\v_{\mathrm{rms,lab}}^2&=u^2+c_{\mathrm{rms}}^2\end{aligned}

u is local mean flow. The cross term vanishes because $\langle\mathbf c\rangle=0$ .[4][7]

Do not insert a measured laboratory RMS value into the temperature formula without checking. Common speed $u$ adds to its square on top of thermal motion. Subtract that contribution first. Sound speed is also a different quantity: it describes propagation of small pressure disturbances and requires a thermodynamic process assumption. In a mixture, each species at shared temperature has its own mass-dependent distribution. Mean translational energy per molecule still uses only the thermal part relative to local gas flow.[4][3][7]

Five exercises with solutions

Calculate · Foundation

A model gas has $M=0.020\,\mathrm{kg\,mol^{-1}}$ and $T=300\,\mathrm K$ . Find crms.

Show hint

Use the molar form with kg/mol.

Learning goal: Calculate thermal rms speed.

Show solution and calculation

Approach: in the local gas rest frame, $c_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$ , with $M$ in kg/mol.
$c_{\mathrm{rms}}=\sqrt{\frac{3R\cdot300\,\mathrm K}{0.020\,\mathrm{kg\,mol^{-1}}}}\approx611.678688\,\frac{\mathrm m}{\mathrm s}$ .
Molar translational energy is $\frac12M c_{\mathrm{rms}}^2\approx3741.50818\,\frac{\mathrm J}{\mathrm{mol}}$ .

c_{\mathrm{rms}}\approx611.679\,\mathrm{m/s}

Transfer · Application

At unchanged species, absolute temperature is quadrupled. How do all three Maxwell characteristic speeds change?

Show hint

Every characteristic speed scales with √T.

Learning goal: Scale Maxwell speeds with temperature.

Show solution and calculation

Approach: all three Maxwell characteristics contain factor $\sqrt T$ at fixed mass.
With $T_2=4T_1$ , that factor doubles.
Most probable speed, mean speed and RMS therefore all double.
Mean translational energy is proportional to $T$ and quadruples.

Check: ratios between the three characteristics remain constant.

\frac{c_2}{c_1}=2

Calculate · Application

A uniform flow has $u=100\,\mathrm{m\,s^{-1}}$ and thermal $c_{\mathrm{rms}}=300\,\mathrm{m\,s^{-1}}$ . Find laboratory RMS speed.

Show hint

Add squares, not magnitudes.

Learning goal: Separate bulk motion from thermal motion.

Show solution and calculation

Approach: $v_{\mathrm{rms,lab}}=\sqrt{u^2+c_{\mathrm{rms}}^2}$ .
Substituting the squares below gives about 316.228 m/s.
The temperature formula still uses thermal speed 300 m/s.
Adding magnitudes to obtain 400 m/s is wrong because the ensemble cross term vanishes.

Check: squaring the result recovers the sum of the two input squares.

v_{\mathrm{rms,lab}}=\sqrt{100^2+300^2}\,\mathrm{m\,s^{-1}}\simeq316.227766\,\mathrm{m\,s^{-1}}

Interpret · Application

Four equal-mass particles move in one dimension at −3, −1, +1 and +3 m/s. This is an illustrative sample rather than a full Maxwell gas. Find mean velocity, mean speed and rms; which determines mean kinetic energy?

Show hint

For rms square first, average, then take the square root.

Learning goal: Interpret rms from a discrete sample.

Show solution and calculation

Approach: Signs cancel in mean velocity.
Mean speed is 2 m/s; mean squared speed is 5 square metres per square second and rms is about 2.23607 m/s.
Mean energy uses that mean square, not the square of an ordinary mean.
Zero mean velocity therefore does not imply zero particle energy.

\begin{aligned}\overline v&=0,&\overline{|v|}&=2\,\mathrm{m\,s^{-1}}\\{}v_{\mathrm{rms}}&=\sqrt{\frac{9+1+1+9}{4}}=\sqrt5\,\mathrm{m\,s^{-1}}\\{}\langle E\rangle&=\frac12m\langle v^2\rangle\end{aligned}

[1]

Error analysis · Application

For an ideal gas with $M=0.028\,\mathrm{kg\,mol^{-1}}$ and $T=300\,\mathrm K$ someone inserts M into the particle form with kB. Which mass belongs with kB and which with R? Give the correct speed using $R=8.314462618\,\mathrm{J\,mol^{-1}\,K^{-1}}$ .

Show hint

The two forms are related by Avogadro’s constant.

Learning goal: Avoid mixing molar and particle mass.

Show solution and calculation

Approach: Molar mass belongs with the molar gas constant.
Boltzmann’s constant requires single-particle mass, obtained by dividing molar mass by Avogadro’s constant.
The correct speed is about 516.963 m/s.
The incorrect combination leaves an extra mole dimension under the square root and is not dimensionally a speed.

\begin{aligned}m&=\frac M{N_A},&R&=N_Ak_B\\{}c_{\mathrm{rms}}&=\sqrt{\frac{3RT}{M}}\approx516.963\,\mathrm{m\,s^{-1}}\end{aligned}

[1]

Not a universal speed formula

The temperature formula describes classical nonrelativistic translation in thermal equilibrium. Relevant quantum degeneracy, strongly interacting dense systems or directional anisotropy require a suitable alternative description. A mixture’s average molar mass is not automatically the right parameter: each species has its own RMS speed at the shared temperature.[5]

RMS characterises an ensemble. It does not claim every molecule has exactly that speed or strikes a wall at it. Wall impacts are weighted by the normal component of motion and particle flux. A pressure derivation must account for this selection rather than replace the gas with identical arrows of a single speed.[1]

In brief

RMS means square first, average second, take the square root last. In classical gas this measure connects mean translational energy to temperature. Particle mass belongs with Boltzmann’s constant; molar mass belongs with the gas constant. Separate thermal motion from common flow, and distinguish RMS from ordinary mean speed and sound speed.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Made particle/molar mass selection and temperature inversion easier to find with a question heading; aligned the example heading with the three existing calculations and added distribution/particle-energy links.
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. Root-mean-square molecular speed. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/rms-geschwindigkeit-gasmolekuel

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Frequently asked questions about Root-mean-square molecular speed

What does Root-mean-square molecular speed describe physically?+

RMS speed is the square root of mean squared thermal speed: $c_{\mathrm{rms}}=\sqrt{\langle c^2\rangle}$ . It connects the distribution with mean translational energy because $\langle E_{\mathrm{tr}}\rangle=\frac12mc_{\mathrm{rms}}^2$ . Classical equilibrium gives $c_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm B}T}{m}}=\sqrt{\frac{3RT}{M}}$ . This is neither ordinary average ⟨c⟩ nor mean velocity vector, which vanishes in gas at rest. In the Maxwell distribution, crms exceeds both mean and most probable speed. Speeds are measured relative to the local gas rest frame. Shared flow can increase laboratory RMS speed without representing the same increase in thermal temperature. The sample size cancels from this characteristic speed relation.

How can I check a calculation involving Root-mean-square molecular speed?+

For $M=0.032\,\mathrm{kg\,mol^{-1}}$ and $T=400\,\mathrm K$ , $\sqrt{\frac{3RT}{M}}$ gives about 558.384 m/s. Independently check with $m=\frac M{N_{\mathrm A}}$ : $\frac12mc_{\mathrm{rms}}^2$ must equal $\frac32k_{\mathrm B}T=8.283894\cdot10^{-21}\,\mathrm J$ . In the Maxwell model, $c_{\mathrm{mp}}\simeq455.918\,\mathrm{m\,s^{-1}}$ and $\langle c\rangle\simeq514.449\,\mathrm{m\,s^{-1}}$ ; ordering $c_{\mathrm{mp}}<\langle c\rangle<c_{\mathrm{rms}}$ must hold. Also check $\frac{c_{\mathrm{rms}}}{c_{\mathrm{mp}}}=\sqrt{\frac32}$ . The result does not depend on whether the sample contains one mole or another amount. Before inferring temperature from measured speeds, check whether mean flow was subtracted. Uncorrected laboratory RMS speed would otherwise give an excessive thermal value. At equal energy, a heavier particle must have a smaller squared speed.

What assumptions and limits apply to Root-mean-square molecular speed?+

The temperature relation assumes classical nonrelativistic translation in thermal equilibrium. Speed is evaluated relative to mean gas motion. Quantum degeneracy or strongly interacting dense systems may require a different distribution. Away from equilibrium, the three component moments need not be equal. In a mixture, use the appropriate particle or molar mass separately for each species; an average mass does not automatically give their respective RMS values. Formulas for most probable and arithmetic mean speed additionally require a Maxwell–Boltzmann distribution. crms remains a statistical quantity and specifies neither every molecule’s speed nor sound-wave propagation speed.

How do I check units and scale for Root-mean-square molecular speed?+

The particle form needs m in kg per molecule; the molar form needs M in kg/mol. Thus 32 g/mol becomes 0.032 kg/mol. $\frac{RT}{M}$ has units $\mathrm{J\,kg^{-1}}=\mathrm{m^2\,s^{-2}}$ , and only the square root gives m/s. Inserting unconverted 32 instead of 0.032 makes speed too small by $\sqrt{1000}$ . Temperature is in kelvin. At unchanged species, crms scales with $\sqrt T$ ; at equal T it scales with $\frac1{\sqrt M}$ . A ten-times-heavier species therefore has only $\frac1{\sqrt{10}}$ of the RMS speed but the same mean translational energy per particle.

When should I combine Root-mean-square molecular speed with another relation?+

For temperature inference, rearrange $T=\frac{Mc_{\mathrm{rms}}^2}{3R}$ . At $M=0.040\,\mathrm{kg\,mol^{-1}}$ and thermal $c_{\mathrm{rms}}=600\,\mathrm{m\,s^{-1}}$ , $T\simeq577.307\,\mathrm K$ . For pressure, combine crms with $p=\frac13\rho c_{\mathrm{rms}}^2$ in an isotropic ideal gas. For flowing samples, use $\mathbf v=\mathbf u+\mathbf c$ and $\langle v^2\rangle=u^2+\langle c^2\rangle$ to subtract shared motion. For example, $u=100\,\mathrm{m\,s^{-1}}$ and thermal $c_{\mathrm{rms}}=300\,\mathrm{m\,s^{-1}}$ give laboratory RMS speed 316.228 m/s, not 400 m/s. The energy equation still uses thermal speed 300 m/s. If arithmetic mean speed is required, use its own Maxwell moment instead of merely renaming crms. In mixtures, evaluate each species using its own particle mass.

Quanta exam set

Curated exam set for "Root-mean-square molecular speed":

Question (front)

Which formula describes Root-mean-square molecular speed?

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How do you rearrange Root-mean-square molecular speed for Absolute temperature?

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Question (front)

Which common mistake happens with Root-mean-square molecular speed?

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[1]OpenStax, University Physics Vol. 2, §2.2: Pressure, Temperature, and RMS Speed

Related formulas

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)

Maxwell–Boltzmann Molecular Speed Distribution

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\eta_{Carnot} = 1 - \frac{T_{kalt}}{T_{warm}}

Carnot Efficiency

Q = m \cdot c \cdot \Delta T

Heat Capacity: Q = m·c·ΔT

\Delta U = Q - W

First Law of Thermodynamics

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

Stefan-Boltzmann Law

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Retain Root-mean-square molecular speed for exams

Create a curated FSRS exam set for vrms = √(3kBT/m): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Thermal RMS speed of gas molecules

Variables & units

Gas molecule RMS speed 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 Root-mean-square molecular speed?

Square, average, then take the square root

Energy supplies the second moment

How do you use particle or molar mass and solve for temperature?

Distinguish three characteristic magnitudes

Three characteristics of the same distribution

Three complete calculations of thermal and laboratory RMS speed

Three characteristic speeds of a model gas

Temperature from a thermal RMS measurement

Removing flow from a laboratory measurement

Separate laboratory motion from the thermal part

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=300 KT=300\,\mathrm KT=300K. Find crms.

At unchanged species, absolute temperature is quadrupled. How do all three Maxwell characteristic speeds change?

A uniform flow has u=100 m s−1u=100\,\mathrm{m\,s^{-1}}u=100ms−1 and thermal crms=300 m s−1c_{\mathrm{rms}}=300\,\mathrm{m\,s^{-1}}crms​=300ms−1. Find laboratory RMS speed.

Four equal-mass particles move in one dimension at −3, −1, +1 and +3 m/s. This is an illustrative sample rather than a full Maxwell gas. Find mean velocity, mean speed and rms; which determines mean kinetic energy?

Not a universal speed formula

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More formulas: Heat and thermodynamics

Retain Root-mean-square molecular speed for exams

A model gas has $M=0.020\,\mathrm{kg\,mol^{-1}}$ and $T=300\,\mathrm K$ . Find crms.

A uniform flow has $u=100\,\mathrm{m\,s^{-1}}$ and thermal $c_{\mathrm{rms}}=300\,\mathrm{m\,s^{-1}}$ . Find laboratory RMS speed.