Physics · Kinetic theory of gases

Average translational kinetic energy of a gas molecule

Q: How do I check units and scale for Mean translational kinetic energy per gas molecule?

$k_{\mathrm B}=1.380649\cdot10^{-23}\,\mathrm{J\,K^{-1}}$ is an exact SI constant. Multiplication by T in kelvin gives joules per particle. The small numerical value is not unusually tiny energy for a whole sample: at 300 K, 10²³ particles give about 621.292 J total translational energy. Per mole, $\frac32RT\simeq3741.508\,\mathrm{J\,mol^{-1}}$. Thus use kB for individual particles and R for molar quantities. Do not insert Celsius temperature directly. Also check scaling: doubling absolute temperature doubles mean energy but increases thermal RMS speed of the same species only by $\sqrt2$.

Q: When should I combine Mean translational kinetic energy per gas molecule with another relation?

To find speed, combine $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T$ with $\langle E_{\mathrm{tr}}\rangle=\frac12m\langle c^2\rangle$. This gives $c_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm B}T}{m}}$, not arithmetic mean speed. For a gas sample, multiply by N or use $N=nN_{\mathrm A}$. A monatomic ideal gas without other active internal excitations then has $U=\frac32nRT$. Pressure additionally follows from $pV=Nk_{\mathrm B}T$. In a polyatomic gas, however, $\frac32nRT$ describes only the translational contribution. In flowing gas, $\mathbf c=\mathbf v-\mathbf u$; shared motion is accounted for separately and cannot be used without correction to determine thermal temperature. At equal T, different particle masses change only their corresponding mean squared speeds.

The mean translational kinetic energy of an ideal-gas molecule depends only on absolute temperature.

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Formula

Calculate

⟨ E_{kin} ⟩ = \frac{3}{2} k_{B} T

E in J; kB in J/K; T in K.

Variables & units

Variables & units
Symbol	Meaning	Unit
⟨Ekin⟩	Mean translational kinetic energy per molecule	J
$kB$	Boltzmann constant	J/K
$T$	Absolute temperature	K

LaTeX

\langle E_{\mathrm{kin}}\rangle=\frac{3}{2}k_{\mathrm B}T

QUANTA STUDYquanta-study.de

Normalised mean translational energy versus normalised temperature. Values 0.5, 1, 1.5 and 2 lie on the same line; T₀ is 300 kelvin. No state at zero kelvin is plotted. — At equal temperature, mean translational energy in classical equilibrium is the same for light and heavy particles. Normalisation: T₀ = 300 K and E₀ = 3kBT₀/2; the plotted line covers 150–600 K.

Mean kinetic energy of a gas molecule calculator

Find the mean translational kinetic energy of a gas particle from the temperature, or the other way round, for an ideal gas in equilibrium.

Calculation model

Find

Absolute temperature T

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

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

Working · rearrange for ⟨Etr⟩, substitute, evaluate

1. Rearrange for ⟨Etr⟩
$\langle E_{\rm tr}\rangle=\frac32k_BT$
2. Substitute values (in SI units)
$\frac{3\cdot 1.38065\times10^{-23}\cdot 300}2=6.21292\times10^{-21}$
3. Result
⟨Etr⟩ = 0.03878 eV

Result: Mean energy per particle ⟨Etr⟩

⟨Etr⟩ = 0.03878 eV

With T = 300 K, ⟨Etr⟩ = 0.03878 eV.

Quick tips

The mean translational energy depends only on T, not on the mass: helium and nitrogen at the same temperature have the same value; heavier particles simply move more slowly.
Always use kelvin: going from 20 °C to 40 °C does not double the energy, it rises by only about 7 %, from 293 K to 313 K.

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.

Derivation & background

In classical ideal equilibrium each of three translational directions contributes $\frac12k_{\mathrm B}T$ . Average thermal velocities relative to local gas motion. Bulk centre-of-mass motion is a separate contribution. At equal temperature, mean translational energy is mass-independent. Polyatomic rotation and vibration provide additional contributions; relevant quantum degeneracy is excluded.

Exam blueprint

Validity range

Classical nonrelativistic ideal gas in thermal equilibrium; three translational degrees of freedom, negligible interaction and no relevant quantum degeneracy. Energy concerns random motion relative to local mean flow, not complete energy of polyatomic molecules.

Derivation steps

Three classical quadratic translational terms each supply equilibrium mean $\frac12k_{\mathrm B}T$ . Average in the local gas rest frame.

1Define thermal velocity relative to local gas motion: $\mathbf c=\mathbf v-\mathbf u$ .
2Translational energy has three quadratic terms: $E_{\mathrm{tr}}=\frac m2(c_x^2+c_y^2+c_z^2)$ .
3Equipartition gives $\frac m2\langle c_x^2\rangle=\frac12k_{\mathrm B}T$ , likewise for the other two components.
4Add the averages: $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T$ .
5Known mean energy gives $T=\frac{2\langle E_{\mathrm{tr}}\rangle}{3k_{\mathrm B}}$ , in kelvin.
6For the sample add $N$ contributions. With $N=nN_{\mathrm A}$ and $R=N_{\mathrm A}k_{\mathrm B}$ , the sum is $\frac32nRT$ .

Rearrangements

Absolute temperature

T=\frac{2\langle E_{\mathrm{kin,tr}}\rangle}{3k_{\mathrm B}}

Use mean thermal translational energy per particle in joules; exclude common gas motion and internal rotational or vibrational contributions.

Task variant

A classical ideal gas at equilibrium at 300 K. Calculate mean thermal translational energy per molecule; common flow has been subtracted.

At $T=300\,\mathrm K$ , $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T=6.2129205\cdot10^{-21}\,\mathrm J$ per particle in classical equilibrium.
Rotation and vibration are not included in this translational contribution.

Common mistakes

Using the square of mean velocity to find mean kinetic energy.

Square each speed first, then average: $\langle E_{\mathrm{tr}}\rangle=\frac12m\langle c^2\rangle$ . Subtract common flow before thermal analysis.

Exam context

Connect microscopic particle motion with temperature and heat.

Formula cluster

Kinetic theory of gases

Mean translational energy gives thermal RMS speed and ideal pressure; total internal energy can include further contributions.

U = 3/2 nRT

Internal energy of a monatomic ideal gas

vrms = √(3kBT/m)

Root-mean-square molecular speed

pV = 1/3 Nm⟨v²⟩

Pressure from kinetic theory

⟨E⟩=f kBT/2; U=f nRT/2

Equipartition of Energy

Worked example

At $T=300\,\mathrm K$ , $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T=6.2129205\cdot10^{-21}\,\mathrm J$ per particle in classical equilibrium. Rotation and vibration are not included in this translational contribution.

Applications

Connect microscopic particle motion with temperature and heat.

How do you calculate with Mean translational kinetic energy per gas molecule?

Here is how to work through a typical Mean translational kinetic energy per gas molecule (⟨Ekin⟩ = 3/2 kBT) task step by step:

1
Task
A classical ideal gas at equilibrium at 300 K. Calculate mean thermal translational energy per molecule; common flow has been subtracted.
Solution path
1. At $T=300\,\mathrm K$ , $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T=6.2129205\cdot10^{-21}\,\mathrm J$ per particle in classical equilibrium.
2. Rotation and vibration are not included in this translational contribution.

Understand and apply the equation

Molecules move continuously even in gas at rest. Their directions can cancel on average; their kinetic energies cannot. In classical equilibrium, temperature determines the mean energy of this random translation: $\langle E_{\mathrm{kin,tr}}\rangle=\frac32k_{\mathrm B}T$ . This is an average over the gas sample, not a fixed energy of every molecule. That distinction explains why heavier molecules can move more slowly at the same temperature.

What does mean translational energy per molecule mean?

\langle E_{\mathrm{kin,tr}}\rangle=\frac12m\langle c^2\rangle=\frac32 k_{\mathrm B}T

Here c is speed relative to the gas at rest or its local mean flow. The v in the short form uses this thermal reference.[1]

Calculate each particle’s energy first, then average. Brackets denote this ensemble or suitable equilibrium average. Different particles have different instantaneous speeds. Thus $\frac12m\langle c^2\rangle$ is not generally equal to $\frac12m\langle c\rangle^2$ . Even two equal-mass particles with equal and opposite velocities have zero mean velocity vector but positive mean kinetic energy.[3]

Three quadratic contributions

The factor three comes from the three spatial directions. Each contains a quadratic energy term, such as $\frac12mc_x^2$ . In classical thermal equilibrium, each independent term supplies mean $\frac12k_{\mathrm B}T$ . Add the three directional contributions. This statistical equilibrium assumption determines the average; the familiar kinetic energy of one particle alone does not establish a temperature relation.[2][5]

\begin{aligned}\langle E_x\rangle=\langle E_y\rangle=\langle E_z\rangle&=\frac12k_{\mathrm B}T\\\langle E_{\mathrm{kin,tr}}\rangle&=\frac32k_{\mathrm B}T\end{aligned}

One temperature sets all three averages, rather than three independent temperatures.[2]

Explore further: one contribution from the Gaussian distribution

To examine the half contribution more closely, use the Gaussian density of one component. Set $a=\frac{m}{2k_{\mathrm B}T}$ , so density is proportional to $e^{-ac_x^2}$ . Integration by parts relates its second moment to the normalisation integral and gives $\langle c_x^2\rangle=\frac1{2a}$ . The boundary term vanishes at infinity because of the exponential factor. Multiplying by $\frac m2$ again gives the half energy contribution.[5]

Explore further: one contribution from the Gaussian distribution

\frac{\int_{-\infty}^{\infty}c_x^2e^{-ac_x^2}\,dc_x}{\int_{-\infty}^{\infty}e^{-ac_x^2}\,dc_x}=\frac1{2a}=\frac{k_{\mathrm B}T}{m}

Multiplication by $\frac m2$ gives mean energy contribution $\frac12k_{\mathrm B}T$ . The Gaussian form assumes equilibrium; it is not a model for every nonequilibrium distribution.[5]

How do you use kelvin and distinguish particle energy from sample energy?

T=\frac{2\langle E_{\mathrm{kin,tr}}\rangle}{3k_{\mathrm B}},\qquad k_{\mathrm B}=1.380649\times10^{-23}\,\mathrm{J/K}

kB is exact in SI. Precision of measured energy or temperature nevertheless depends on input data.[7]

The formula needs absolute temperature: $T[\mathrm K]=\vartheta[{}^\circ\mathrm C]+273.15$ . At 20 °C, use 293.15 K. Warming to 40 °C therefore does not double energy. Only temperature differences have the same numerical value in kelvin and Celsius. Product $k_{\mathrm B}T$ gives joules per particle. For the whole sample, additionally multiply by particle count; alternatively use amount of substance and the gas constant.[4]

\begin{aligned}N&=nN_{\mathrm A}\\R&=N_{\mathrm A}k_{\mathrm B}=8.31446261815324\,\mathrm{J\,mol^{-1}\,K^{-1}}\end{aligned}

NA=6.02214076·10²³ mol⁻¹ is exact, so R is exact as well. We use rounded R≈8.31446 where appropriate.[7]

Convert temperature and compare gases

Evaluate the energy relation per particle first, then for the whole sample. The two species illustrate mass independence of mean energy. The additional flow example separates laboratory-frame energy. Thermal RMS speed accordingly uses the mean squared motion relative to gas flow.[1][8]

Thermal translation at 20 °C

Given: Classical dilute gas in equilibrium at 20.0 °C; $N=3.00\cdot10^{22}$ particles and no shared flow.
Find: Find mean translational energy and total translational energy.

Approach: For three classical translational degrees of freedom without bulk flow, mean particle energy depends only on absolute temperature. Particle count then gives total translational energy.
$\begin{aligned}\langle E_{\mathrm{tr}}\rangle&=\frac32k_BT\\{}E_{\mathrm{tr,total}}&=N\langle E_{\mathrm{tr}}\rangle\end{aligned}$
First use absolute temperature.
$T=20.0+273.15=293.15\,\mathrm K$
Evaluate energy per particle.
$\langle E_{\mathrm{kin,tr}}\rangle=\frac32(1.380649\times10^{-23})(293.15)=6.07105881525\times10^{-21}\,\mathrm J$
Multiply by particle count.
$E_{\mathrm{tr,total}}=N\langle E_{\mathrm{kin,tr}}\rangle\approx182.132\,\mathrm J$

Result: About 6.07·10⁻²¹ J per particle and 182 J of sample translation. For a polyatomic gas this is not automatically its entire internal energy.

Check: $n=\frac N{N_A}\approx0.0498162\,\mathrm{mol}$ ; $\frac32nRT$ gives the same 182.132 J. Joules per kelvin times kelvin give joules per particle, not joules per mole.

Equal energies, different speeds

Given: Two classical ideal gas species at $T=600\,\mathrm K$ with model molar masses $M_1=0.00400\,\mathrm{kg\,mol^{-1}}$ and $M_2=0.0400\,\mathrm{kg\,mol^{-1}}$ . Both are described relative to mean flow.
Find: Compare mean translational energy and find corresponding RMS speeds.

Approach: Equal equilibrium temperature gives equal mean thermal translational energy per particle. Different molar masses nevertheless give different rms speeds in the local rest frame.
$\begin{aligned}\langle E_1\rangle&=\langle E_2\rangle=\frac32k_BT\\{}c_{\mathrm{rms},i}&=\sqrt{\frac{3RT}{M_i}}\end{aligned}$
Energy does not require species mass.
$\langle E_1\rangle=\langle E_2\rangle=\frac32k_{\mathrm B}(600)=1.2425841\times10^{-20}\,\mathrm J$
Mass-dependent speed follows from $\frac12m\langle c^2\rangle$ .
$\begin{aligned}c_{\mathrm{rms},1}&=\sqrt{\frac{3RT}{M_1}}\approx1934.30\,\mathrm{m/s}\\{} c_{\mathrm{rms},2}\approx611.679\,\mathrm{m/s}\end{aligned}$

Result: Both species have the same mean translational energy. The ten-times-heavier species has only $\frac1{\sqrt{10}}$ of the RMS speed.

Check: Molar energy $\frac12M c_{\mathrm{rms}}^2$ is $\frac32RT\approx7483.01636\,\frac{\mathrm J}{\mathrm{mol}}$ for each species. Tenfold mass is compensated by tenfold smaller squared speed.

Moving sample: separate temperature from laboratory energy

Given: 0.300 mol classical ideal model gas with molar mass 0.0280 kg/mol at 300 K. The sample moves uniformly as a whole at 200 m/s relative to the laboratory. No rotation of the sample; only translational energy is requested.
Find: Find thermal, bulk and total translational laboratory energies.

Approach: first find energy relative to mean gas motion, then add common centre-of-mass energy.
$\begin{aligned}E_{\mathrm{thermal}}&=\frac32nRT\\m_{\mathrm{sample}}&=nM\\E_{\mathrm{bulk}}&=\frac12nMu^2\\E_{\mathrm{lab}}&=E_{\mathrm{thermal}}+E_{\mathrm{bulk}}\end{aligned}$
Insert thermal energy and sample mass.
$\begin{aligned}E_{\mathrm{thermal}}&=\frac32\cdot0.300\cdot8.31446261815324\cdot300\,\mathrm J\\&\simeq1122.45245\,\mathrm J\\m_{\mathrm{sample}}&=0.300\cdot0.0280\,\mathrm{kg}=0.00840\,\mathrm{kg}\end{aligned}$
Account separately for common motion.
$\begin{aligned}E_{\mathrm{bulk}}&=\frac12\cdot0.00840\cdot200^2\,\mathrm J=168.0\,\mathrm J\\E_{\mathrm{lab}}&\simeq1290.45245\,\mathrm J\end{aligned}$

Result: Translational laboratory energy is about 1.290 kJ: 1.122 kJ is thermal and 168 J is common motion. Temperature remains 300 K.

Check: In the comoving frame $u=0$ , leaving only thermal energy. Treating all laboratory energy as thermal would incorrectly give about 344.90 K. This frame change is not heating and does not dissipate the 168 J.

A flow does not automatically make gas hotter

\begin{aligned}\mathbf c_i&=\mathbf v_i-\mathbf u\\\langle\mathbf c\rangle&=0\\\frac12m\langle v^2\rangle&=\frac12mu^2+\frac12m\langle c^2\rangle\end{aligned}

u is shared mean velocity in the considered uniform gas volume. Thermal energy uses relative motion c.[4][8]

Here $\mathbf u$ denotes common mean gas motion and $\mathbf c$ random motion relative to it. Squaring initially produces a cross term, but its average vanishes because $\langle\mathbf c\rangle=0$ . Bulk kinetic energy and thermal translational energy therefore add. Merely changing the observer’s frame does not increase thermodynamic temperature. Actual friction, compression or shocks can redistribute energy, however, requiring a physical energy balance.[4][6][8]

View diagram: thermal and bulk energy

Common motion offsets laboratory energy

QUANTA STUDYquanta-study.de

Original model curves for molar mass 0.0280 kg/mol and fixed bulk speed 200 m/s. Vertical separation is bulk kinetic energy, not additional temperature.[4][8]

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Separate translation from internal molecular motion

The three translational contributions also apply to a polyatomic ideal gas in classical equilibrium. Molecular rotation and vibration can carry additional internal energy; they do not replace translation’s $\frac32$ factor. For total internal energy, add those further active contributions. The monatomic model without electronic excitation instead contains only translation. The internal energy of a monatomic ideal gas sums the individual translational contributions of the entire sample under these conditions.[2]

An appropriate mean energy can determine an equilibrium temperature. A single particle does not possess thermodynamic temperature in the same sense. A mean alone also does not decide whether a chemical reaction threshold is crossed: the distribution contains particles above and below its mean. Such questions additionally need the distribution and a reaction model.[3]

Five exercises with solutions

Calculate · Foundation

At $T=200\,\mathrm K$ , find mean translational energy of a classical ideal particle.

Show hint

Three degrees of freedom, energy per particle.

Learning goal: Calculate thermal translational energy per particle using kB.

Show solution and calculation

Approach: $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T$ .
Substitution below gives about $4.142\cdot10^{-21}\,\mathrm J$ per particle.

Check: J/K multiplied by K gives J. Another species at the same temperature does not change this translational average in the specified classical model.

\langle E_{\mathrm{tr}}\rangle=\frac32\cdot1.380649\cdot10^{-23}\cdot200\,\mathrm J=4.141947\cdot10^{-21}\,\mathrm J

Rearrange · Application

In the appropriate classical equilibrium model, $\langle E_{\mathrm{tr}}\rangle=8.283894\cdot10^{-21}\,\mathrm J$ . Find temperature.

Show hint

Solve for T; do not divide by R.

Learning goal: Find absolute temperature from mean particle energy.

Show solution and calculation

Approach: rearrange the thermal translational relation for absolute temperature.
$T=\frac{2\langle E_{\mathrm{tr}}\rangle}{3k_B}=400\,\mathrm K$ .
Substitution gives the same energy.
R belongs to energy per mole, not this per-particle energy.

T=400\,\mathrm K

Interpret · Application

The same species warms from 240 to 360 K. How do ⟨Etr⟩ and crms change?

Show hint

Energy scales with T; RMS speed with √T.

Learning goal: Separate linear energy scaling from square-root thermal-speed scaling.

Show solution and calculation

Approach: energy is proportional to $T$ , while RMS speed at fixed mass is proportional to $\sqrt T$ .
Temperature ratio is $\frac{360}{240}=1.50$ .
Mean energy therefore increases by 50%, but RMS speed only by about 22.5%.

Check: squaring the speed factor recovers the energy factor.

\begin{aligned}\frac{\langle E_2\rangle}{\langle E_1\rangle}&=\frac{360}{240}=1.50\\\frac{c_{\mathrm{rms},2}}{c_{\mathrm{rms},1}}&=\sqrt{1.50}\simeq1.224745\end{aligned}

Interpret · Application

Two classical ideal gas species are in thermal equilibrium at the same temperature. Molecular mass of B is nine times that of A. Compare mean translational kinetic energy per molecule and RMS speed. A sketch shows shorter velocity arrows for B; does this mean less mean translational energy?

Show hint

Mean thermal energy depends only on T and translational degrees of freedom. Speed additionally depends on molecular mass.

Learning goal: Explain equal translational energy for different molecular masses.

Show solution and calculation

Approach: use the same mean translational energy for both gases and derive the mean-square speed ratio.
Energies agree; B has only one third of the RMS speed.

Check: nine times the mass multiplied by one ninth of the mean-square speed gives the same energy. Shorter velocity arrows therefore do not automatically imply less kinetic energy when mass differs.

\begin{aligned}\langle E_A\rangle=\langle E_B\rangle&=\frac32k_BT\\\frac{\langle c_B^2\rangle}{\langle c_A^2\rangle}&=\frac{m_A}{m_B}=\frac19\\\frac{c_{\mathrm{rms},B}}{c_{\mathrm{rms},A}}&=\frac13\end{aligned}

[2][1]

Error analysis · Foundation

In the rest frame of an isotropic equilibrium gas at 300 K, mean molecular velocity as a vector is zero. Someone inserts this zero vector into $E=\frac12m|\langle\vec c\rangle|^2$ and calls mean translational energy zero. Correct the averaging procedure and calculate energy per molecule using $k_{\mathrm B}=1.380649\cdot10^{-23}\,\mathrm{J\,K^{-1}}$ .

Show hint

Opposite velocity vectors can cancel; their squares are both positive. Square first, then average.

Learning goal: Distinguish the square of a mean from the mean of a square.

Show solution and calculation

Approach: form each particle’s kinetic energy and then average.
This requires mean squared speed, not the square of mean velocity.
At 300 K the result is about 6.213·10⁻²¹ J per molecule.

Check: energy is positive although the gas bulk rests. Even two particles with equal opposite velocities demonstrate the difference between these averaging operations.

\begin{aligned}\langle\vec c\rangle&=0\\\langle E\rangle&=\frac12m\langle|\vec c|^2\rangle=\frac32k_BT\\&=\frac32(1.380649\cdot10^{-23})(300)\,\mathrm J\\&=6.2129205\cdot10^{-21}\,\mathrm J\end{aligned}

[1][2][7]

A classical equilibrium value with clear limits

The calculation assumes a sufficiently dilute, nondegenerate gas in thermal equilibrium. Relevant interaction energy and relativistic particle motion are neglected. Very low temperature or high phase-space density may require quantum statistics; an idealised noninteracting gas is not automatically classical.[5]

Away from equilibrium, directions or species may have different distributions. One temperature then need not describe every second velocity moment. Formal continuation to $T=0$ is therefore not a complete account of a real ground state. Read the linear plot as a physical model only within the stated classical regime.[5]

In brief

Mean energy is the average of individual energies. The three classical translational contributions give the same energy average at equal temperature, regardless of molecular mass. Greater mass therefore means lower typical speed. Use kelvin and distinguish energy per particle, per mole and for the whole sample. Common gas motion is accounted for separately from thermal motion.

Sources for this explanation

Quanta · Updated October 2, 2026

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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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Change history

October 2, 2026Highlighted energy per particle, kelvin and the whole sample in question headings; added paragraph connections to monatomic internal energy and thermal RMS speed.
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. Mean translational kinetic energy per gas molecule. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/mittlere-kinetische-energie-molekuel

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Frequently asked questions about Mean translational kinetic energy per gas molecule

What does Mean translational kinetic energy per gas molecule describe physically?+

The relation describes mean translational energy per particle in a classical ideal gas at thermal equilibrium: $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T$ . Each of three spatial directions contributes $\frac12k_{\mathrm B}T$ . It concerns random motion relative to gas at rest or its local mean flow. An individual particle need not possess that energy. At equal temperature, light and heavy species share the same translational mean but have different speeds. At 300 K it is 6.2129205·10⁻²¹ J per particle. Molecular rotation, vibration and shared centre-of-mass motion are excluded from this value; those require separate energy contributions.

How can I check a calculation involving Mean translational kinetic energy per gas molecule?+

First check absolute temperature and the energy reference. At 20 °C, $T=293.15\,\mathrm K$ , not 20 K. Then $\frac32k_{\mathrm B}T$ gives 6.07105881525·10⁻²¹ J per particle. For $N=3.00\cdot10^{22}$ particles, total translational energy is 182.131764 J. Independently check using $n=\frac N{N_{\mathrm A}}$ and $\frac32nRT$ ; both results must agree. For a polyatomic gas, this sum is not necessarily its entire internal energy. Subtract shared flow before evaluating thermal motion, rather than counting laboratory bulk kinetic energy as a temperature increase. Finally, distinguish this energy average from an energy assigned to every individual particle.

What assumptions and limits apply to Mean translational kinetic energy per gas molecule?+

Assume classical nonrelativistic particles in thermal equilibrium with negligible interaction contributions. Statistical equipartition supplies the factor 3/2 for three translational degrees of freedom. An ideal gas is not automatically classical: low temperature or high phase-space density can make quantum effects important. Away from equilibrium, directions can have different second velocity moments; one temperature need not describe the entire distribution. Molecular rotation and vibration remain additional internal contributions. The translational equation therefore is not a universal law for total energy in arbitrary materials, nor a complete account of their actual ground states.

How do I check units and scale for Mean translational kinetic energy per gas molecule?+

$k_{\mathrm B}=1.380649\cdot10^{-23}\,\mathrm{J\,K^{-1}}$ is an exact SI constant. Multiplication by T in kelvin gives joules per particle. The small numerical value is not unusually tiny energy for a whole sample: at 300 K, 10²³ particles give about 621.292 J total translational energy. Per mole, $\frac32RT\simeq3741.508\,\mathrm{J\,mol^{-1}}$ . Thus use kB for individual particles and R for molar quantities. Do not insert Celsius temperature directly. Also check scaling: doubling absolute temperature doubles mean energy but increases thermal RMS speed of the same species only by $\sqrt2$ .

When should I combine Mean translational kinetic energy per gas molecule with another relation?+

To find speed, combine $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T$ with $\langle E_{\mathrm{tr}}\rangle=\frac12m\langle c^2\rangle$ . This gives $c_{\mathrm{rms}}=\sqrt{\frac{3k_{\mathrm B}T}{m}}$ , not arithmetic mean speed. For a gas sample, multiply by N or use $N=nN_{\mathrm A}$ . A monatomic ideal gas without other active internal excitations then has $U=\frac32nRT$ . Pressure additionally follows from $pV=Nk_{\mathrm B}T$ . In a polyatomic gas, however, $\frac32nRT$ describes only the translational contribution. In flowing gas, $\mathbf c=\mathbf v-\mathbf u$ ; shared motion is accounted for separately and cannot be used without correction to determine thermal temperature. At equal T, different particle masses change only their corresponding mean squared speeds.

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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, §2.2: Pressure, Temperature, and RMS Speed

More formulas: Heat and thermodynamics

\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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Symbol

Meaning

Unit

⟨Ekin⟩

Mean translational kinetic energy per molecule

kB

Boltzmann constant

J/K

T

Absolute temperature

Derivation & background

Exam blueprint

Validity range

Derivation steps

Three classical quadratic translational terms each supply equilibrium mean $\frac12k_{\mathrm B}T$ . Average in the local gas rest frame.

1Define thermal velocity relative to local gas motion: $\mathbf c=\mathbf v-\mathbf u$ .
2Translational energy has three quadratic terms: $E_{\mathrm{tr}}=\frac m2(c_x^2+c_y^2+c_z^2)$ .
3Equipartition gives $\frac m2\langle c_x^2\rangle=\frac12k_{\mathrm B}T$ , likewise for the other two components.
4Add the averages: $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T$ .
5Known mean energy gives $T=\frac{2\langle E_{\mathrm{tr}}\rangle}{3k_{\mathrm B}}$ , in kelvin.
6For the sample add $N$ contributions. With $N=nN_{\mathrm A}$ and $R=N_{\mathrm A}k_{\mathrm B}$ , the sum is $\frac32nRT$ .

How do you calculate with Mean translational kinetic energy per gas molecule?

Here is how to work through a typical Mean translational kinetic energy per gas molecule (⟨Ekin⟩ = 3/2 kBT) task step by step:

Task

A classical ideal gas at equilibrium at 300 K. Calculate mean thermal translational energy per molecule; common flow has been subtracted.

Solution path

At $T=300\,\mathrm K$ , $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T=6.2129205\cdot10^{-21}\,\mathrm J$ per particle in classical equilibrium.
Rotation and vibration are not included in this translational contribution.

Understand and apply the equation

\langle E_{\mathrm{kin,tr}}\rangle=\frac32k_{\mathrm B}T

. This is an average over the gas sample, not a fixed energy of every molecule. That distinction explains why heavier molecules can move more slowly at the same temperature.

What does mean translational energy per molecule mean?

\langle E_{\mathrm{kin,tr}}\rangle=\frac12m\langle c^2\rangle=\frac32 k_{\mathrm B}T

Here c is speed relative to the gas at rest or its local mean flow. The v in the short form uses this thermal reference.[1]

Three quadratic contributions

\begin{aligned}\langle E_x\rangle=\langle E_y\rangle=\langle E_z\rangle&=\frac12k_{\mathrm B}T\\\langle E_{\mathrm{kin,tr}}\rangle&=\frac32k_{\mathrm B}T\end{aligned}

One temperature sets all three averages, rather than three independent temperatures.[2]

Explore further: one contribution from the Gaussian distribution

\frac{\int_{-\infty}^{\infty}c_x^2e^{-ac_x^2}\,dc_x}{\int_{-\infty}^{\infty}e^{-ac_x^2}\,dc_x}=\frac1{2a}=\frac{k_{\mathrm B}T}{m}

Multiplication by $\frac m2$ gives mean energy contribution $\frac12k_{\mathrm B}T$ . The Gaussian form assumes equilibrium; it is not a model for every nonequilibrium distribution.[5]

How do you use kelvin and distinguish particle energy from sample energy?

T=\frac{2\langle E_{\mathrm{kin,tr}}\rangle}{3k_{\mathrm B}},\qquad k_{\mathrm B}=1.380649\times10^{-23}\,\mathrm{J/K}

kB is exact in SI. Precision of measured energy or temperature nevertheless depends on input data.[7]

\begin{aligned}N&=nN_{\mathrm A}\\R&=N_{\mathrm A}k_{\mathrm B}=8.31446261815324\,\mathrm{J\,mol^{-1}\,K^{-1}}\end{aligned}

NA=6.02214076·10²³ mol⁻¹ is exact, so R is exact as well. We use rounded R≈8.31446 where appropriate.[7]

Convert temperature and compare gases

Thermal translation at 20 °C

Given: Classical dilute gas in equilibrium at 20.0 °C; $N=3.00\cdot10^{22}$ particles and no shared flow.
Find: Find mean translational energy and total translational energy.

Approach: For three classical translational degrees of freedom without bulk flow, mean particle energy depends only on absolute temperature. Particle count then gives total translational energy.
$\begin{aligned}\langle E_{\mathrm{tr}}\rangle&=\frac32k_BT\\{}E_{\mathrm{tr,total}}&=N\langle E_{\mathrm{tr}}\rangle\end{aligned}$
First use absolute temperature.
$T=20.0+273.15=293.15\,\mathrm K$
Evaluate energy per particle.
$\langle E_{\mathrm{kin,tr}}\rangle=\frac32(1.380649\times10^{-23})(293.15)=6.07105881525\times10^{-21}\,\mathrm J$
Multiply by particle count.
$E_{\mathrm{tr,total}}=N\langle E_{\mathrm{kin,tr}}\rangle\approx182.132\,\mathrm J$

Result: About 6.07·10⁻²¹ J per particle and 182 J of sample translation. For a polyatomic gas this is not automatically its entire internal energy.

Check: $n=\frac N{N_A}\approx0.0498162\,\mathrm{mol}$ ; $\frac32nRT$ gives the same 182.132 J. Joules per kelvin times kelvin give joules per particle, not joules per mole.

Equal energies, different speeds

Given: Two classical ideal gas species at $T=600\,\mathrm K$ with model molar masses $M_1=0.00400\,\mathrm{kg\,mol^{-1}}$ and $M_2=0.0400\,\mathrm{kg\,mol^{-1}}$ . Both are described relative to mean flow.
Find: Compare mean translational energy and find corresponding RMS speeds.

Approach: Equal equilibrium temperature gives equal mean thermal translational energy per particle. Different molar masses nevertheless give different rms speeds in the local rest frame.
$\begin{aligned}\langle E_1\rangle&=\langle E_2\rangle=\frac32k_BT\\{}c_{\mathrm{rms},i}&=\sqrt{\frac{3RT}{M_i}}\end{aligned}$
Energy does not require species mass.
$\langle E_1\rangle=\langle E_2\rangle=\frac32k_{\mathrm B}(600)=1.2425841\times10^{-20}\,\mathrm J$
Mass-dependent speed follows from $\frac12m\langle c^2\rangle$ .
$\begin{aligned}c_{\mathrm{rms},1}&=\sqrt{\frac{3RT}{M_1}}\approx1934.30\,\mathrm{m/s}\\{} c_{\mathrm{rms},2}\approx611.679\,\mathrm{m/s}\end{aligned}$

Result: Both species have the same mean translational energy. The ten-times-heavier species has only $\frac1{\sqrt{10}}$ of the RMS speed.

Check: Molar energy $\frac12M c_{\mathrm{rms}}^2$ is $\frac32RT\approx7483.01636\,\frac{\mathrm J}{\mathrm{mol}}$ for each species. Tenfold mass is compensated by tenfold smaller squared speed.

Moving sample: separate temperature from laboratory energy

Given: 0.300 mol classical ideal model gas with molar mass 0.0280 kg/mol at 300 K. The sample moves uniformly as a whole at 200 m/s relative to the laboratory. No rotation of the sample; only translational energy is requested.
Find: Find thermal, bulk and total translational laboratory energies.

Approach: first find energy relative to mean gas motion, then add common centre-of-mass energy.
$\begin{aligned}E_{\mathrm{thermal}}&=\frac32nRT\\m_{\mathrm{sample}}&=nM\\E_{\mathrm{bulk}}&=\frac12nMu^2\\E_{\mathrm{lab}}&=E_{\mathrm{thermal}}+E_{\mathrm{bulk}}\end{aligned}$
Insert thermal energy and sample mass.
$\begin{aligned}E_{\mathrm{thermal}}&=\frac32\cdot0.300\cdot8.31446261815324\cdot300\,\mathrm J\\&\simeq1122.45245\,\mathrm J\\m_{\mathrm{sample}}&=0.300\cdot0.0280\,\mathrm{kg}=0.00840\,\mathrm{kg}\end{aligned}$
Account separately for common motion.
$\begin{aligned}E_{\mathrm{bulk}}&=\frac12\cdot0.00840\cdot200^2\,\mathrm J=168.0\,\mathrm J\\E_{\mathrm{lab}}&\simeq1290.45245\,\mathrm J\end{aligned}$

Result: Translational laboratory energy is about 1.290 kJ: 1.122 kJ is thermal and 168 J is common motion. Temperature remains 300 K.

A flow does not automatically make gas hotter

\begin{aligned}\mathbf c_i&=\mathbf v_i-\mathbf u\\\langle\mathbf c\rangle&=0\\\frac12m\langle v^2\rangle&=\frac12mu^2+\frac12m\langle c^2\rangle\end{aligned}

u is shared mean velocity in the considered uniform gas volume. Thermal energy uses relative motion c.[4][8]

View diagram: thermal and bulk energy

Common motion offsets laboratory energy

QUANTA STUDYquanta-study.de

Original model curves for molar mass 0.0280 kg/mol and fixed bulk speed 200 m/s. Vertical separation is bulk kinetic energy, not additional temperature.[4][8]

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Separate translation from internal molecular motion

Five exercises with solutions

Calculate · Foundation

At $T=200\,\mathrm K$ , find mean translational energy of a classical ideal particle.

Show hint

Three degrees of freedom, energy per particle.

Learning goal: Calculate thermal translational energy per particle using kB.

Show solution and calculation

Approach: $\langle E_{\mathrm{tr}}\rangle=\frac32k_{\mathrm B}T$ .
Substitution below gives about $4.142\cdot10^{-21}\,\mathrm J$ per particle.

Check: J/K multiplied by K gives J. Another species at the same temperature does not change this translational average in the specified classical model.

\langle E_{\mathrm{tr}}\rangle=\frac32\cdot1.380649\cdot10^{-23}\cdot200\,\mathrm J=4.141947\cdot10^{-21}\,\mathrm J

Rearrange · Application

In the appropriate classical equilibrium model, $\langle E_{\mathrm{tr}}\rangle=8.283894\cdot10^{-21}\,\mathrm J$ . Find temperature.

Show hint

Solve for T; do not divide by R.

Learning goal: Find absolute temperature from mean particle energy.

Show solution and calculation

Approach: rearrange the thermal translational relation for absolute temperature.
$T=\frac{2\langle E_{\mathrm{tr}}\rangle}{3k_B}=400\,\mathrm K$ .
Substitution gives the same energy.
R belongs to energy per mole, not this per-particle energy.

T=400\,\mathrm K

Interpret · Application

The same species warms from 240 to 360 K. How do ⟨Etr⟩ and crms change?

Show hint

Energy scales with T; RMS speed with √T.

Learning goal: Separate linear energy scaling from square-root thermal-speed scaling.

Show solution and calculation

Approach: energy is proportional to $T$ , while RMS speed at fixed mass is proportional to $\sqrt T$ .
Temperature ratio is $\frac{360}{240}=1.50$ .
Mean energy therefore increases by 50%, but RMS speed only by about 22.5%.

Check: squaring the speed factor recovers the energy factor.

\begin{aligned}\frac{\langle E_2\rangle}{\langle E_1\rangle}&=\frac{360}{240}=1.50\\\frac{c_{\mathrm{rms},2}}{c_{\mathrm{rms},1}}&=\sqrt{1.50}\simeq1.224745\end{aligned}

Interpret · Application

Two classical ideal gas species are in thermal equilibrium at the same temperature. Molecular mass of B is nine times that of A. Compare mean translational kinetic energy per molecule and RMS speed. A sketch shows shorter velocity arrows for B; does this mean less mean translational energy?

Show hint

Mean thermal energy depends only on T and translational degrees of freedom. Speed additionally depends on molecular mass.

Learning goal: Explain equal translational energy for different molecular masses.

Show solution and calculation

Approach: use the same mean translational energy for both gases and derive the mean-square speed ratio.
Energies agree; B has only one third of the RMS speed.

\begin{aligned}\langle E_A\rangle=\langle E_B\rangle&=\frac32k_BT\\\frac{\langle c_B^2\rangle}{\langle c_A^2\rangle}&=\frac{m_A}{m_B}=\frac19\\\frac{c_{\mathrm{rms},B}}{c_{\mathrm{rms},A}}&=\frac13\end{aligned}

[2][1]

Error analysis · Foundation

In the rest frame of an isotropic equilibrium gas at 300 K, mean molecular velocity as a vector is zero. Someone inserts this zero vector into $E=\frac12m|\langle\vec c\rangle|^2$ and calls mean translational energy zero. Correct the averaging procedure and calculate energy per molecule using $k_{\mathrm B}=1.380649\cdot10^{-23}\,\mathrm{J\,K^{-1}}$ .

Show hint

Opposite velocity vectors can cancel; their squares are both positive. Square first, then average.

Learning goal: Distinguish the square of a mean from the mean of a square.

Show solution and calculation

Approach: form each particle’s kinetic energy and then average.
This requires mean squared speed, not the square of mean velocity.
At 300 K the result is about 6.213·10⁻²¹ J per molecule.

Check: energy is positive although the gas bulk rests. Even two particles with equal opposite velocities demonstrate the difference between these averaging operations.

\begin{aligned}\langle\vec c\rangle&=0\\\langle E\rangle&=\frac12m\langle|\vec c|^2\rangle=\frac32k_BT\\&=\frac32(1.380649\cdot10^{-23})(300)\,\mathrm J\\&=6.2129205\cdot10^{-21}\,\mathrm J\end{aligned}

[1][2][7]

A classical equilibrium value with clear limits

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

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, 2026Highlighted energy per particle, kelvin and the whole sample in question headings; added paragraph connections to monatomic internal energy and thermal RMS speed.
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. Mean translational kinetic energy per gas molecule. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/mittlere-kinetische-energie-molekuel

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Create a curated FSRS exam set for ⟨Ekin⟩ = 3/2 kBT: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Average translational kinetic energy of a gas molecule

Variables & units

Mean kinetic energy of a gas molecule 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 Mean translational kinetic energy per gas molecule?

What does mean translational energy per molecule mean?

Three quadratic contributions

How do you use kelvin and distinguish particle energy from sample energy?

Convert temperature and compare gases

Thermal translation at 20 °C

Equal energies, different speeds

Moving sample: separate temperature from laboratory energy

A flow does not automatically make gas hotter

Common motion offsets laboratory energy

Separate translation from internal molecular motion

Five exercises with solutions

At T=200 KT=200\,\mathrm KT=200K, find mean translational energy of a classical ideal particle.

In the appropriate classical equilibrium model, ⟨Etr⟩=8.283894⋅10−21 J\langle E_{\mathrm{tr}}\rangle=8.283894\cdot10^{-21}\,\mathrm J⟨Etr​⟩=8.283894⋅10−21J. Find temperature.

The same species warms from 240 to 360 K. How do ⟨Etr⟩ and crms change?

Two classical ideal gas species are in thermal equilibrium at the same temperature. Molecular mass of B is nine times that of A. Compare mean translational kinetic energy per molecule and RMS speed. A sketch shows shorter velocity arrows for B; does this mean less mean translational energy?

A classical equilibrium value with clear limits

In brief

Sources for this explanation

Frequently asked questions about Mean translational kinetic energy per gas molecule

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Mean translational kinetic energy per gas molecule for exams

Average translational kinetic energy of a gas molecule

Variables & units

Mean kinetic energy of a gas molecule 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 Mean translational kinetic energy per gas molecule?

What does mean translational energy per molecule mean?

Three quadratic contributions

How do you use kelvin and distinguish particle energy from sample energy?

Convert temperature and compare gases

Thermal translation at 20 °C

Equal energies, different speeds

Moving sample: separate temperature from laboratory energy

A flow does not automatically make gas hotter

Common motion offsets laboratory energy

Separate translation from internal molecular motion

Five exercises with solutions

At T=200 KT=200\,\mathrm KT=200K, find mean translational energy of a classical ideal particle.

In the appropriate classical equilibrium model, ⟨Etr⟩=8.283894⋅10−21 J\langle E_{\mathrm{tr}}\rangle=8.283894\cdot10^{-21}\,\mathrm J⟨Etr​⟩=8.283894⋅10−21J. Find temperature.

The same species warms from 240 to 360 K. How do ⟨Etr⟩ and crms change?

Two classical ideal gas species are in thermal equilibrium at the same temperature. Molecular mass of B is nine times that of A. Compare mean translational kinetic energy per molecule and RMS speed. A sketch shows shorter velocity arrows for B; does this mean less mean translational energy?

A classical equilibrium value with clear limits

In brief

Sources for this explanation

Frequently asked questions about Mean translational kinetic energy per gas molecule

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Mean translational kinetic energy per gas molecule for exams

At $T=200\,\mathrm K$ , find mean translational energy of a classical ideal particle.

In the appropriate classical equilibrium model, $\langle E_{\mathrm{tr}}\rangle=8.283894\cdot10^{-21}\,\mathrm J$ . Find temperature.

At $T=200\,\mathrm K$ , find mean translational energy of a classical ideal particle.

In the appropriate classical equilibrium model, $\langle E_{\mathrm{tr}}\rangle=8.283894\cdot10^{-21}\,\mathrm J$ . Find temperature.