Physics · Kinetic theory

Equipartition of energy

Q: What does this relationship describe?

Equipartition assigns mean $\frac12k_{\mathrm B}T$ to each independent quadratic energy contribution in classical thermal equilibrium. f counts such terms, rather than merely names of motions. Translation has three momentum terms. One harmonic vibrational mode instead has kinetic and potential terms, together kBT. Summation for a fixed ideal sample gives $U=\frac f2nRT$ relative to the chosen energy zero. Bulk sample motion is excluded. A linear molecule with two active rotations and frozen vibration has $f=5$, for example. This describes a model regime, not full activation of every possible motion at every temperature. Electronic excitation, dissociation and interaction energy require additional contributions.

Q: What does a checked numerical calculation look like?

For $n=0.600\,\mathrm{mol}$, $T=360\,\mathrm K$ and constant $f=5$, $U=\frac52nRT\simeq4489.809814\,\mathrm J$. Appropriate molar heat capacity is $C_{V,m}=\frac52R\simeq20.78615655\,\mathrm{J\,mol^{-1}\,K^{-1}}$. Heating by 80 K within the same model regime gives $\Delta U=nC_{V,m}\Delta T\simeq997.7355142\,\mathrm J$. In a rigid closed vessel without other work or external-energy change this is also heat input. Independently, $\frac U{nT}$ returns the same Cv,m; moreover $\frac{\Delta U}{U}=\frac{80}{360}$. If thermal activation changes during heating, integrate an appropriate temperature-dependent heat capacity. Correct arithmetic alone does not establish that the substance stays within the assumed $f=5$ regime throughout the process.

Q: Which common mistake changes the result?

A common counting error assigns only $\frac12k_{\mathrm B}T$ to one harmonic vibrational mode. It has two quadratic energy terms and supplies kBT in the fully classical limit. Also distinguish f from speed density f(v), and energy per particle from energy per mole. The former uses kB, the latter R. T must be absolute in kelvin; only temperature differences may use Celsius differences. Equipartition alone does not imply $Q=\Delta U$: that requires a suitable closed-system process balance without net work or external-energy change. In a transition regime, temperature-dependent f cannot be treated as constant when differentiating U.

In classical thermal equilibrium, each quadratic degree of freedom contributes an average $\frac12k_{\mathrm B}T$ per particle.

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Formula

Calculate

⟨ E ⟩ = \frac{f}{2} k_{B} T, U = \frac{f}{2} n RT

E and U in J; kB in J/K; T in K; n in mol; R in J/(mol·K); f is dimensionless.

Variables & units

Variables & units
Symbol	Meaning	Unit
⟨E⟩	mean energy per particle from quadratic degrees of freedom	J
$f$	number of active quadratic degrees of freedom	dimensionless
$kB$	Boltzmann constant	J/K
$T$	absolute temperature	K
$U$	internal energy of the ideal particle system	J
$n$	amount of substance	mol
$R$	molar gas constant	J/(mol·K)

LaTeX

\langle E\rangle=\frac{f}{2}k_BT,\quad U=\frac{f}{2}nRT

QUANTA STUDYquanta-study.de

Three stacked bars total 1.5, 2.5 and 3.5 RT on one scale. The final mode has separate kinetic and potential half-contributions. — Classical active model limits per mole: three translations give 1.5 RT; two rotations add RT; one harmonic mode adds two half RT terms. No universal temperature threshold.

Equipartition energy calculator

Find the mean energy per particle, the thermal energy of a gas sample or the molar heat capacity from the number of degrees of freedom, treated classically.

Calculation model

Find

Degrees of freedom f (quadratic terms)

Number of degrees of freedom, e.g. 3 for a monatomic gas such as helium; a vibration counts twice.

Absolute equilibrium temperature T

Absolute temperature, e.g. 300 K at room temperature; °C is converted to kelvin.

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

Working · rearrange for ⟨E⟩, substitute, evaluate

1. Rearrange for ⟨E⟩
$\langle E\rangle=\frac f2k_{\rm B}T$
2. Substitute values (in SI units)
$\frac{3\cdot 1.38065\times10^{-23}\cdot 300}2=6.21292\times10^{-21}$
3. Result
⟨E⟩ = 6.213 · 10⁻²¹ J

Result: Mean energy per particle ⟨E⟩

⟨E⟩ = 6.213 · 10⁻²¹ J

With f = 3 and T = 300 K, ⟨E⟩ = 6.213 · 10⁻²¹ J.

Quick tips

Each degree of freedom adds ½ kT per particle and ½ RT per mole: monatomic gases have f = 3, diatomic ones without vibration f = 5.
Unit check: kT is an energy per particle, RT an energy per mole. Mixing up k and R puts you off by a factor of 6·10²³.

Assumptions and limits

Classical thermal equilibrium with an explicitly specified constant count of independent active quadratic energy terms. f counts terms rather than names of motions: one harmonic mode has two. T>0 absolute; quantum freeze-out, excitation, dissociation and changing composition excluded. No automatic material/temperature assignment; three translation terms remain independent of further internal terms. Energy is relative to model zero, with bulk motion separate.

Derivation & background

Each independent accessible classical quadratic energy term has equilibrium mean $\frac12k_{\mathrm B}T$ . Translation counts three momentum terms; a fully classical harmonic mode counts kinetic and potential terms. A fixed ideal sample has $U=\frac f2nRT$ relative to its chosen zero. Bulk motion and other energy contributions are separate. Only constant $f$ gives $C_{V,m}=\frac f2R$ . Quantum activation depends on substance and temperature and is not a universal sharp staircase.

Exam blueprint

Validity range

Classical thermal equilibrium with accessible independent quadratic energy terms. $U=\frac f2nRT$ uses an ideal fixed particle model, fixed zero and no other relevant contributions. $C_{V,m}=\frac f2R$ requires constant f; quantum activation and changing composition need extensions.

Derivation steps

Equipartition counts quadratic energy terms: three for translation and two per harmonic vibration. Each classically active term supplies $\frac12k_{\mathrm B}T$ , explaining heat-capacity limits without assuming all possible motions are active at every temperature.

1Specify classical equilibrium probabilities and independent accessible quadratic energy terms.
2For energy $aq^2$ with $a>0$ , density is proportional to $e^{-\frac{aq^2}{k_{\mathrm B}T}}$ .
3The normalised second Gaussian moment gives $\langle q^2\rangle=\frac{k_{\mathrm B}T}{2a}$ .
4Multiplying by $a$ gives $\langle aq^2\rangle=\frac12k_{\mathrm B}T$ . Add the $f$ active contributions.
5Using $N=nN_{\mathrm A}$ and $R=N_{\mathrm A}k_{\mathrm B}$ gives $U=\frac f2nRT$ relative to the fixed zero.
6Only constant $f$ gives $C_{V,m}=\frac f2R$ . One harmonic mode counts two terms; partial quantum activation needs a different model.

Rearrangements

Energy per particle

\langle E\rangle=\frac f2k_{\mathrm B}T

f counts active quadratic degrees of freedom.

Task variant

One mole of monatomic classical ideal gas at 300 K. Find sample thermal energy and check against energy per particle; electronic excitations are negligible.

One mole of monatomic classical ideal gas at 300 K: approach $U=\frac32nRT$ .
Substitution gives about 3.74 kJ.
Per particle, $\langle E\rangle=\frac32k_{\mathrm B}T\simeq6.21\cdot10^{-21}\,\mathrm J$ .
Multiplication by Avogadro’s constant reconstructs molar energy.

Common mistakes

Counting a harmonic vibrational mode as only one half-energy contribution.

A classical harmonic mode contains one kinetic and one potential quadratic term; together they give mean $k_{\mathrm B}T$ .

Exam context

Problems in Kinetic theory assess formula choice, units, model assumptions and a brief result check.

Formula cluster

Kinetic theory

Connects counted active energy terms with internal energy and heat capacity; translation remains a distinct contribution.

⟨Ekin⟩ = 3/2 kBT

Mean translational kinetic energy per gas molecule

U = 3/2 nRT

Internal energy of a monatomic ideal gas

Q = m·c·ΔT

Heat Capacity: Q = m·c·ΔT

Worked example

One mole of monatomic classical ideal gas at 300 K: approach $U=\frac32nRT$ . Substitution gives about 3.74 kJ. Per particle, $\langle E\rangle=\frac32k_{\mathrm B}T\simeq6.21\cdot10^{-21}\,\mathrm J$ . Multiplication by Avogadro’s constant reconstructs molar energy.

Applications

Explain heat capacities and the temperature dependence of molecular degrees of freedom.

How do you calculate with Equipartition of Energy?

Here is how to work through a typical Equipartition of Energy (⟨E⟩=f kBT/2; U=f nRT/2) task step by step:

1
Task
One mole of monatomic classical ideal gas at 300 K. Find sample thermal energy and check against energy per particle; electronic excitations are negligible.
Solution path
1. One mole of monatomic classical ideal gas at 300 K: approach $U=\frac32nRT$ .
2. Substitution gives about 3.74 kJ.
3. Per particle, $\langle E\rangle=\frac32k_{\mathrm B}T\simeq6.21\cdot10^{-21}\,\mathrm J$ .
4. Multiplication by Avogadro’s constant reconstructs molar energy.

Understand and apply the equation

A gas molecule can translate, rotate and vibrate. These motions store energy in different ways. Equipartition counts their average classical contributions: each independent quadratic energy term contributes $\frac12 k_{\mathrm B}T$ . The crucial item is therefore an energy term, not simply a named type of motion. Learn which contributions a model includes and when its predicted heat capacity ceases to apply.

Count energy terms rather than motion names

\begin{aligned}\langle E\rangle&=\frac f2k_{\mathrm B}T\\U&=\frac f2nRT\end{aligned}

f counts active independent quadratic terms, not the speed density f(v) used elsewhere.[1]

Count terms in the energy formula. Translation along one direction contains $\frac{p_x^2}{2m}$ and contributes mean $\frac12 k_{\mathrm B}T$ . A harmonic vibrational mode instead contains two terms: $\frac{p^2}{2m}$ and $\frac{\kappa q^2}{2}$ . Its kinetic and spring energy together contribute $k_{\mathrm B}T$ . Treating one mode as one half contribution would lose a factor of two. Internal energy $U$ includes only the modelled thermal contributions relative to the chosen zero. Common motion of the sample is additional macroscopic energy.[2]

From one term to the whole sample

Three translational terms sum to $\frac32k_{\mathrm B}T$ per particle. For $N$ particles of one species, those averages add. Using $N=nN_{\mathrm A}$ and $R=N_{\mathrm A}k_{\mathrm B}$ gives the molar form. Factor $f$ changes only when further thermally active terms of the chosen model are included. More substance increases total energy, not energy per particle at the same temperature.[1][2]

Explore further: why each quadratic term gets a half contribution

The half contribution follows from a classical equilibrium average, not an instantaneous measurement. Consider an unrestricted coordinate $q$ with energy $aq^2$ , where $a>0$ and the remaining energy does not depend on $q$ . Its equilibrium density is proportional to $e^{-\frac{aq^2}{k_{\mathrm B}T}}$ . The normalised average is a ratio of second-moment and normalisation integrals. Integration by parts gives $\langle q^2\rangle=\frac{k_{\mathrm B}T}{2a}$ . Multiplying by $a$ leaves the half-temperature contribution, independent of that term’s stiffness.[2]

\begin{aligned}\langle aq^2\rangle&=a\frac{\int_{-\infty}^{\infty}q^2e^{-\frac{aq^2}{k_{\mathrm B}T}}\,dq}{\int_{-\infty}^{\infty}e^{-\frac{aq^2}{k_{\mathrm B}T}}\,dq}\\&=\frac12k_{\mathrm B}T\end{aligned}

The boundary term vanishes because the Gaussian decays faster than q grows.[2]

How do you count active quadratic terms rather than just names of motions?

The monatomic ideal model without electronic excitation has three translational terms: $f=3$ . A linear molecule with classically active rotation adds two rotational terms, giving $f=5$ if vibrations are frozen. A rigid nonlinear molecule has three rotational terms, giving $f=6$ in total. Each fully classically active harmonic vibration adds two more terms. Here “linear” describes molecular geometry, not merely small displacement. Mean translational energy of a gas molecule still contains only the three spatial motion contributions, even when the molecule has additional internal contributions.[1][2]

Which contributions are active depends on molecule-specific energy-level spacings and temperature. A quantum-frozen mode does not automatically supply its full classical contribution. Activation is usually gradual. A staircase diagram with universal activation temperatures would therefore not be an appropriate material model.[1]

View diagram: compare contributions using slopes

More active terms give a steeper energy line

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Original model lines with a constant term count in each case over the displayed interval. They compare three models, not a universal activation sequence of a specific substance.[1]

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From U to heat capacity

\begin{aligned}C_{V,m}&=\frac f2R\\C_{p,m}&=\left(\frac f2+1\right)R\\\gamma&=1+\frac2f\end{aligned}

Constant f and fixed composition are required; $C_{p,m}-C_{V,m}=R$ additionally uses the ideal-gas model.[3]

For fixed amount and constant $f$ , the change is $\Delta U=\frac f2nR\Delta T$ , giving constant molar heat capacity at fixed volume. Actual supplied heat still requires a work balance: in a rigid closed vessel without other work or external energy changes, $Q=\Delta U$ . At constant pressure, expansion work is added. If mode activation changes with temperature, actually differentiate $U(T)$ or integrate measured heat capacity. Internal energy of a monatomic ideal gas is the explicitly restricted special case with only three translational terms.[4][1]

Three complete calculations on term count, vibration and heating measurements

The following calculations use explicitly specified classical model regimes. The first two illustrate term counting and vibrational amplitude; the third uses calorimetric data to check an assumed term count.[1][2][3]

A diatomic model regime with $f=5$

Given: $n=0.600\,\mathrm{mol}$ , $T=360\,\mathrm K$ , $f=5$ ; ideal fixed composition, translation and two rotations active, vibration neglected. For the subsequent 80 K warming, the vessel is rigid and closed, with no other work or external energy changes.
Find: Calculate U, Cv,m and rigid-vessel heating by 80 K.

Approach: The five specified quadratic degrees of freedom are classically active. $U=\frac f2nRT$ . Sum contributions.
$U=\frac52(0.600)R(360)=4489.809814\,\mathrm J$
Find capacity and change.
$\begin{aligned}C_{V,m}&=\frac52R\simeq20.78615655\,\mathrm{J\,mol^{-1}\,K^{-1}}\\\Delta U&=nC_{V,m}\Delta T\\&=0.600\cdot20.78615655\cdot80\,\mathrm J\\&\simeq997.7355142\,\mathrm J\end{aligned}$

Result: U≈4.490 kJ; without additional work, 80 K requires about 997.74 J heat.

Check: $\frac{U}{nT}=20.78615655\,\mathrm{J\,mol^{-1}\,K^{-1}}$ agrees with Cv,m. Heating must remain in the same $f=5$ regime; $\frac{\Delta U}{U}=\frac{80}{360}$ .

One classical harmonic mode

Given: $T=500\,\mathrm K$ , $\kappa=0.080\,\mathrm{N\,m^{-1}}$ ; ideal unrestricted harmonic mode in classical equilibrium.
Find: Find both mean energy contributions and RMS displacement.

Approach: The harmonic mode has two quadratic contributions. $\langle E_{\mathrm{kin}}\rangle=\langle E_{\mathrm{pot}}\rangle=\frac12k_{\mathrm B}T$ . One half of kBT for momentum and coordinate.
$\langle E_{kin}\rangle=\langle E_{pot}\rangle=3.4516225\times10^{-21}\,\mathrm J$
Rearrange spring energy for the second moment.
$\begin{aligned}\langle q^2\rangle&=\frac{k_{\mathrm B}T}{\kappa}\\q_{\mathrm{rms}}&=\sqrt{\langle q^2\rangle}\simeq2.937525532\cdot10^{-10}\,\mathrm m\end{aligned}$

Result: Total mean mode energy is 6.903245×10⁻²¹ J; $q_{\mathrm{rms}}\simeq0.294\,\mathrm{nm}$ .

Check: $\frac{\kappa q_{\mathrm{rms}}^2}{2}$ reproduces the potential contribution. Doubling κ halves squared displacement but does not change its mean energy at fixed T.

Testing the model with a heating measurement

Given: 0.400 mol ideal gas in a rigid closed vessel. Between 310 K and 350 K, net heat 332.579 J enters the gas. No vessel contribution, other work or external energy changes. Active classical contributions are constant over this interval.
Find: Infer molar heat capacity and the appropriate term count.

Approach: the specified rigid model has $Q=\Delta U$ . Infer molar heat capacity, then $f$ .
$\begin{aligned}C_{V,m}&=\frac{Q}{n\Delta T}\\f&=\frac{2C_{V,m}}R\end{aligned}$
Insert temperature difference; input data are rounded.
$\begin{aligned}\Delta T&=40\,\mathrm K\\C_{V,m}&=\frac{332.579\,\mathrm J}{0.400\,\mathrm{mol}\cdot40\,\mathrm K}\simeq20.7862\,\mathrm{J\,mol^{-1}\,K^{-1}}\\f&\simeq\frac{2\cdot20.7862}{8.314462618}\simeq5.000\end{aligned}$
Identify a possible molecular model without claiming the measurement alone proves it.

Result: The measurement fits five constant active quadratic contributions. A linear molecule with active rotation and frozen vibration is one possible model.

Check: Using $f=5$ gives $Q=\frac52nR\Delta T\simeq332.578505\,\mathrm J$ , consistent with rounding. Matching heat capacity proves neither molecular geometry nor validity of the same $f$ outside the measured interval.

Rearrange without confusing particles and moles

\begin{aligned}T&=\frac{2U}{fnR}\\f&=\frac{2C_{V,m}}R\\R&=N_{\mathrm A}k_{\mathrm B}\end{aligned}

An f inferred from Cv in a transition regime is only an effective thermal characteristic.[5][1]

For energy per particle use $k_{\mathrm B}$ ; for energy of an amount of substance use $nR$ . One mole of monatomic model gas at 300 K has about 3741.508 J thermal energy, or about $6.2129205\cdot10^{-21}\,\mathrm J$ per particle. Exact SI constants do not remove material-state or model uncertainty. Temperature in the energy formula is absolute. Only a difference in a change calculation has the same numerical value in kelvin and Celsius.[5]

Five exercises with solutions

Calculate · Foundation

A model has $f=6$ , $n=0.20\,\mathrm{mol}$ and $T=250\,\mathrm K$ . Find U.

Show hint

Six half contributions.

Learning goal: Calculate internal energy from active quadratic terms.

Show solution and calculation

Approach: $U=\frac f2nRT$ .
Six contributions give factor three.
Substitution below gives about 1.247 kJ.

Check: twice the amount would have twice the total energy at equal temperature, but unchanged energy per particle.

U=3\cdot0.20\cdot8.31446261815324\cdot250\,\mathrm J\simeq1247.169393\,\mathrm J

Interpret · Application

A harmonic vibrational mode is fully classically active at 400 K. What energy contribution does it supply per mole, and what is its contribution to molar heat capacity?

Show hint

Kinetic and potential terms.

Learning goal: Count the two quadratic terms of a harmonic mode.

Show solution and calculation

Approach: one mode contains a kinetic and a potential quadratic term.
Together they give $U_m=RT$ per mole.
At 400 K this is about 3325.785 J/mol.
The temperature derivative is $R$ , about 8.31446 J/(mol·K).

Check: using only a half contribution would omit either spring or kinetic energy.

\begin{aligned}U_m&=RT=8.31446261815324\cdot400\,\mathrm{J\,mol^{-1}}\\&\simeq3325.785047\,\mathrm{J\,mol^{-1}}\\C_{V,m}^{(\mathrm{mode})}&=R\end{aligned}

Transfer · Application

An ideal constant-f gas has $C_{V,m}=\frac{5R}{2}$ . Find Cp,m and γ.

Show hint

Use Mayer’s relation.

Learning goal: Infer Cp and γ from supplied Cv.

Show solution and calculation

Approach: Mayer’s relation gives $C_{p,m}=C_{V,m}+R$ , then $\gamma=\frac{C_{p,m}}{C_{V,m}}$ .
$C_{p,m}=\frac{7R}{2}$ ≈29.10061916 J/(mol·K), $\gamma=\frac75=1.4$ .
This is not a universal value for every diatomic state.

\gamma=1.4

Interpret · Challenge

In a transition region an effective model gives $U(T)=\frac12nRTf(T)$ with f(T) increasing with T. A graph of U versus T becomes steeper. Can Cv, m still simply equal $\frac12 f(T)R$ ? Differentiate and interpret; f need not be integer here.

Show hint

Apply the product rule to T f(T).

Learning goal: Distinguish temperature-dependent activation from constant f.

Show solution and calculation

Approach: $C_{V, m}=\frac1n\frac{dU}{dT}$ gives $\frac R2[f(T)+Tf^{\prime}(T)]$ .
The extra term represents changing effective mode activation.
It disappears only on a constant-f plateau.

Check: $f'(T)=0$ recovers the classical shortcut. This is differentiation of a supplied effective model, not a claim that fixed integer f fully describes quantum activation.

C_{V,m}=\frac1n\frac{dU}{dT}=\frac R2\left[f(T)+T\frac{df}{dT}\right]

[1][2]

Error analysis · Application

For an ideal gas in a constant-f regime, measured $C_{p, m}=\frac52R$ . A solution sets $f=\frac{2C_{p, m}}R=5$ . Find correct f and γ; the supplied molar heat capacity is explicitly at constant pressure.

Show hint

Use Mayer: $C_{p,m}-C_{V,m}=R$ .

Learning goal: Convert Cp to Cv before inferring active quadratic terms.

Show solution and calculation

Approach: $C_{V, m}=C_{p, m}-R=\frac32R$ , giving $f=3$ and $\gamma=\frac{C_{p, m}}{C_{V, m}}$ = $\frac53$ .
The erroneous inference counts the additional pressure-work R contribution as internal quadratic energy terms.

Check: $f=3$ reconstructs the supplied Cp, m through $\left(\frac f2+1\right)R$ . This inference assumes an ideal gas and constant active classical terms; it does not carry unchanged through an activation transition.

\begin{aligned}C_{V,m}&=\frac52R-R=\frac32R\\f&=\frac{2C_{V,m}}R=3\\\gamma&=\frac{\frac52R}{\frac32R}=\frac53\end{aligned}

[3][1]

The classical result is a limit

Counting blindly does not replace a model check. Nonquadratic energy, restricted coordinates or strong coupling can invalidate the simple independent-term derivation. Freezing of rotation or vibration is a quantum effect; such a mode does not supply its full classical value. Dissociation, ionisation or changing composition require additional energy contributions. Equipartition does not assign equal energy to arbitrary subsystems; it describes appropriate quadratic contributions in classical equilibrium.[2]

Measured heat capacity tests whether the selected contributions match substance and temperature range. A noninteger effective value can indicate a transition and is not automatically an arithmetic mistake. However, a local measurement does not establish a complete temperature-independent molecular model.[1]

In brief

Count independent quadratic energy terms, not merely motion names. Translation has three contributions; a harmonic vibration adds two per mode. Whether they are classically active depends on substance and temperature range. Energy, heat capacity and supplied heat remain distinct quantities. The appropriate work balance connects them.

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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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
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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Frequently asked questions about Equipartition of Energy

What does this relationship describe?+

Equipartition assigns mean $\frac12k_{\mathrm B}T$ to each independent quadratic energy contribution in classical thermal equilibrium. f counts such terms, rather than merely names of motions. Translation has three momentum terms. One harmonic vibrational mode instead has kinetic and potential terms, together kBT. Summation for a fixed ideal sample gives $U=\frac f2nRT$ relative to the chosen energy zero. Bulk sample motion is excluded. A linear molecule with two active rotations and frozen vibration has $f=5$ , for example. This describes a model regime, not full activation of every possible motion at every temperature. Electronic excitation, dissociation and interaction energy require additional contributions.

Where does the formula come from, and what does its structure tell us?+

Consider an unrestricted classical coordinate q with quadratic energy $aq^2$ and $a>0$ . Its equilibrium density is proportional to $e^{-\frac{aq^2}{k_{\mathrm B}T}}$ , provided the remaining energy is independent of q. Dividing the second Gaussian moment by its normalisation gives $\frac{k_{\mathrm B}T}{2a}$ ; multiplying by a supplies $\frac12k_{\mathrm B}T$ . Add this result for f independent quadratic terms. A sample with $N=nN_{\mathrm A}$ and $R=N_{\mathrm A}k_{\mathrm B}$ then has $U=\frac f2nRT$ . The derivation concerns a statistical average, not identical instantaneous energy for every particle. It requires classical equilibrium probabilities and accessible coordinate ranges; it cannot automatically establish full activation of a quantum-frozen vibrational mode.

What does a checked numerical calculation look like?+

For $n=0.600\,\mathrm{mol}$ , $T=360\,\mathrm K$ and constant $f=5$ , $U=\frac52nRT\simeq4489.809814\,\mathrm J$ . Appropriate molar heat capacity is $C_{V,m}=\frac52R\simeq20.78615655\,\mathrm{J\,mol^{-1}\,K^{-1}}$ . Heating by 80 K within the same model regime gives $\Delta U=nC_{V,m}\Delta T\simeq997.7355142\,\mathrm J$ . In a rigid closed vessel without other work or external-energy change this is also heat input. Independently, $\frac U{nT}$ returns the same Cv,m; moreover $\frac{\Delta U}{U}=\frac{80}{360}$ . If thermal activation changes during heating, integrate an appropriate temperature-dependent heat capacity. Correct arithmetic alone does not establish that the substance stays within the assumed $f=5$ regime throughout the process.

Under what conditions may this relationship be used?+

The simple result requires classical thermal equilibrium and the actual independent quadratic terms counted. The gas expression $U=\frac f2nRT$ additionally uses ideal fixed composition and an appropriate energy zero. Low temperatures can quantum-freeze rotation or vibration; transitions are generally gradual and molecule dependent. Nonquadratic energy, restricted coordinate ranges or relevant electronic excitation need another description. $C_{V,m}=\frac f2R$ applies when f is constant. Effective f inferred from measured heat capacity need not be integer in a transition regime. This does not imply half a real motion; it represents partial thermal participation in the system’s ability to change energy.

Which common mistake changes the result?+

A common counting error assigns only $\frac12k_{\mathrm B}T$ to one harmonic vibrational mode. It has two quadratic energy terms and supplies kBT in the fully classical limit. Also distinguish f from speed density f(v), and energy per particle from energy per mole. The former uses kB, the latter R. T must be absolute in kelvin; only temperature differences may use Celsius differences. Equipartition alone does not imply $Q=\Delta U$ : that requires a suitable closed-system process balance without net work or external-energy change. In a transition regime, temperature-dependent f cannot be treated as constant when differentiating U.

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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 Volume 2, §2.3: Heat Capacity and Equipartition of Energy.

More formulas: Heat and thermodynamics

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

Carnot Efficiency

\Delta U = Q - W

First Law of Thermodynamics

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

Stefan-Boltzmann Law

\Delta L = \alpha L_0 \Delta T

Linear thermal expansion

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Physics · Kinetic theory

Equipartition of energy

In classical thermal equilibrium, each quadratic degree of freedom contributes an average $\frac12k_{\mathrm B}T$ per particle.

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Formula

Calculate

⟨ E ⟩ = \frac{f}{2} k_{B} T, U = \frac{f}{2} n RT

E and U in J; kB in J/K; T in K; n in mol; R in J/(mol·K); f is dimensionless.

Variables & units

Variables & units
Symbol	Meaning	Unit
⟨E⟩	mean energy per particle from quadratic degrees of freedom	J
$f$	number of active quadratic degrees of freedom	dimensionless
$kB$	Boltzmann constant	J/K
$T$	absolute temperature	K
$U$	internal energy of the ideal particle system	J
$n$	amount of substance	mol
$R$	molar gas constant	J/(mol·K)

LaTeX

\langle E\rangle=\frac{f}{2}k_BT,\quad U=\frac{f}{2}nRT

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Equipartition energy calculator

Find the mean energy per particle, the thermal energy of a gas sample or the molar heat capacity from the number of degrees of freedom, treated classically.

Calculation model

Find

Degrees of freedom f (quadratic terms)

Number of degrees of freedom, e.g. 3 for a monatomic gas such as helium; a vibration counts twice.

Absolute equilibrium temperature T

Absolute temperature, e.g. 300 K at room temperature; °C is converted to kelvin.

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

Working · rearrange for ⟨E⟩, substitute, evaluate

1. Rearrange for ⟨E⟩
$\langle E\rangle=\frac f2k_{\rm B}T$
2. Substitute values (in SI units)
$\frac{3\cdot 1.38065\times10^{-23}\cdot 300}2=6.21292\times10^{-21}$
3. Result
⟨E⟩ = 6.213 · 10⁻²¹ J

Result: Mean energy per particle ⟨E⟩

⟨E⟩ = 6.213 · 10⁻²¹ J

With f = 3 and T = 300 K, ⟨E⟩ = 6.213 · 10⁻²¹ J.

Quick tips

Each degree of freedom adds ½ kT per particle and ½ RT per mole: monatomic gases have f = 3, diatomic ones without vibration f = 5.
Unit check: kT is an energy per particle, RT an energy per mole. Mixing up k and R puts you off by a factor of 6·10²³.

Assumptions and limits

Derivation & background

Exam blueprint

Validity range

Derivation steps

1Specify classical equilibrium probabilities and independent accessible quadratic energy terms.
2For energy $aq^2$ with $a>0$ , density is proportional to $e^{-\frac{aq^2}{k_{\mathrm B}T}}$ .
3The normalised second Gaussian moment gives $\langle q^2\rangle=\frac{k_{\mathrm B}T}{2a}$ .
4Multiplying by $a$ gives $\langle aq^2\rangle=\frac12k_{\mathrm B}T$ . Add the $f$ active contributions.
5Using $N=nN_{\mathrm A}$ and $R=N_{\mathrm A}k_{\mathrm B}$ gives $U=\frac f2nRT$ relative to the fixed zero.
6Only constant $f$ gives $C_{V,m}=\frac f2R$ . One harmonic mode counts two terms; partial quantum activation needs a different model.

Rearrangements

Energy per particle

\langle E\rangle=\frac f2k_{\mathrm B}T

f counts active quadratic degrees of freedom.

Task variant

One mole of monatomic classical ideal gas at 300 K. Find sample thermal energy and check against energy per particle; electronic excitations are negligible.

One mole of monatomic classical ideal gas at 300 K: approach $U=\frac32nRT$ .
Substitution gives about 3.74 kJ.
Per particle, $\langle E\rangle=\frac32k_{\mathrm B}T\simeq6.21\cdot10^{-21}\,\mathrm J$ .
Multiplication by Avogadro’s constant reconstructs molar energy.

Common mistakes

Counting a harmonic vibrational mode as only one half-energy contribution.

A classical harmonic mode contains one kinetic and one potential quadratic term; together they give mean $k_{\mathrm B}T$ .

Exam context

Problems in Kinetic theory assess formula choice, units, model assumptions and a brief result check.

Formula cluster

Kinetic theory

Connects counted active energy terms with internal energy and heat capacity; translation remains a distinct contribution.

⟨Ekin⟩ = 3/2 kBT

Mean translational kinetic energy per gas molecule

U = 3/2 nRT

Internal energy of a monatomic ideal gas

Q = m·c·ΔT

Heat Capacity: Q = m·c·ΔT

Worked example

Applications

Explain heat capacities and the temperature dependence of molecular degrees of freedom.

How do you calculate with Equipartition of Energy?

Here is how to work through a typical Equipartition of Energy (⟨E⟩=f kBT/2; U=f nRT/2) task step by step:

1
Task
One mole of monatomic classical ideal gas at 300 K. Find sample thermal energy and check against energy per particle; electronic excitations are negligible.
Solution path
1. One mole of monatomic classical ideal gas at 300 K: approach $U=\frac32nRT$ .
2. Substitution gives about 3.74 kJ.
3. Per particle, $\langle E\rangle=\frac32k_{\mathrm B}T\simeq6.21\cdot10^{-21}\,\mathrm J$ .
4. Multiplication by Avogadro’s constant reconstructs molar energy.

Understand and apply the equation

Count energy terms rather than motion names

\begin{aligned}\langle E\rangle&=\frac f2k_{\mathrm B}T\\U&=\frac f2nRT\end{aligned}

f counts active independent quadratic terms, not the speed density f(v) used elsewhere.[1]

From one term to the whole sample

Explore further: why each quadratic term gets a half contribution

\begin{aligned}\langle aq^2\rangle&=a\frac{\int_{-\infty}^{\infty}q^2e^{-\frac{aq^2}{k_{\mathrm B}T}}\,dq}{\int_{-\infty}^{\infty}e^{-\frac{aq^2}{k_{\mathrm B}T}}\,dq}\\&=\frac12k_{\mathrm B}T\end{aligned}

The boundary term vanishes because the Gaussian decays faster than q grows.[2]

How do you count active quadratic terms rather than just names of motions?

View diagram: compare contributions using slopes

More active terms give a steeper energy line

QUANTA STUDYquanta-study.de

Original model lines with a constant term count in each case over the displayed interval. They compare three models, not a universal activation sequence of a specific substance.[1]

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From U to heat capacity

\begin{aligned}C_{V,m}&=\frac f2R\\C_{p,m}&=\left(\frac f2+1\right)R\\\gamma&=1+\frac2f\end{aligned}

Constant f and fixed composition are required; $C_{p,m}-C_{V,m}=R$ additionally uses the ideal-gas model.[3]

Three complete calculations on term count, vibration and heating measurements

A diatomic model regime with $f=5$

Given: $n=0.600\,\mathrm{mol}$ , $T=360\,\mathrm K$ , $f=5$ ; ideal fixed composition, translation and two rotations active, vibration neglected. For the subsequent 80 K warming, the vessel is rigid and closed, with no other work or external energy changes.
Find: Calculate U, Cv,m and rigid-vessel heating by 80 K.

Approach: The five specified quadratic degrees of freedom are classically active. $U=\frac f2nRT$ . Sum contributions.
$U=\frac52(0.600)R(360)=4489.809814\,\mathrm J$
Find capacity and change.
$\begin{aligned}C_{V,m}&=\frac52R\simeq20.78615655\,\mathrm{J\,mol^{-1}\,K^{-1}}\\\Delta U&=nC_{V,m}\Delta T\\&=0.600\cdot20.78615655\cdot80\,\mathrm J\\&\simeq997.7355142\,\mathrm J\end{aligned}$

Result: U≈4.490 kJ; without additional work, 80 K requires about 997.74 J heat.

Check: $\frac{U}{nT}=20.78615655\,\mathrm{J\,mol^{-1}\,K^{-1}}$ agrees with Cv,m. Heating must remain in the same $f=5$ regime; $\frac{\Delta U}{U}=\frac{80}{360}$ .

One classical harmonic mode

Given: $T=500\,\mathrm K$ , $\kappa=0.080\,\mathrm{N\,m^{-1}}$ ; ideal unrestricted harmonic mode in classical equilibrium.
Find: Find both mean energy contributions and RMS displacement.

Approach: The harmonic mode has two quadratic contributions. $\langle E_{\mathrm{kin}}\rangle=\langle E_{\mathrm{pot}}\rangle=\frac12k_{\mathrm B}T$ . One half of kBT for momentum and coordinate.
$\langle E_{kin}\rangle=\langle E_{pot}\rangle=3.4516225\times10^{-21}\,\mathrm J$
Rearrange spring energy for the second moment.
$\begin{aligned}\langle q^2\rangle&=\frac{k_{\mathrm B}T}{\kappa}\\q_{\mathrm{rms}}&=\sqrt{\langle q^2\rangle}\simeq2.937525532\cdot10^{-10}\,\mathrm m\end{aligned}$

Result: Total mean mode energy is 6.903245×10⁻²¹ J; $q_{\mathrm{rms}}\simeq0.294\,\mathrm{nm}$ .

Check: $\frac{\kappa q_{\mathrm{rms}}^2}{2}$ reproduces the potential contribution. Doubling κ halves squared displacement but does not change its mean energy at fixed T.

Testing the model with a heating measurement

Given: 0.400 mol ideal gas in a rigid closed vessel. Between 310 K and 350 K, net heat 332.579 J enters the gas. No vessel contribution, other work or external energy changes. Active classical contributions are constant over this interval.
Find: Infer molar heat capacity and the appropriate term count.

Approach: the specified rigid model has $Q=\Delta U$ . Infer molar heat capacity, then $f$ .
$\begin{aligned}C_{V,m}&=\frac{Q}{n\Delta T}\\f&=\frac{2C_{V,m}}R\end{aligned}$
Insert temperature difference; input data are rounded.
$\begin{aligned}\Delta T&=40\,\mathrm K\\C_{V,m}&=\frac{332.579\,\mathrm J}{0.400\,\mathrm{mol}\cdot40\,\mathrm K}\simeq20.7862\,\mathrm{J\,mol^{-1}\,K^{-1}}\\f&\simeq\frac{2\cdot20.7862}{8.314462618}\simeq5.000\end{aligned}$
Identify a possible molecular model without claiming the measurement alone proves it.

Result: The measurement fits five constant active quadratic contributions. A linear molecule with active rotation and frozen vibration is one possible model.

Rearrange without confusing particles and moles

\begin{aligned}T&=\frac{2U}{fnR}\\f&=\frac{2C_{V,m}}R\\R&=N_{\mathrm A}k_{\mathrm B}\end{aligned}

An f inferred from Cv in a transition regime is only an effective thermal characteristic.[5][1]

Five exercises with solutions

Calculate · Foundation

A model has $f=6$ , $n=0.20\,\mathrm{mol}$ and $T=250\,\mathrm K$ . Find U.

Show hint

Six half contributions.

Learning goal: Calculate internal energy from active quadratic terms.

Show solution and calculation

Approach: $U=\frac f2nRT$ .
Six contributions give factor three.
Substitution below gives about 1.247 kJ.

Check: twice the amount would have twice the total energy at equal temperature, but unchanged energy per particle.

U=3\cdot0.20\cdot8.31446261815324\cdot250\,\mathrm J\simeq1247.169393\,\mathrm J

Interpret · Application

A harmonic vibrational mode is fully classically active at 400 K. What energy contribution does it supply per mole, and what is its contribution to molar heat capacity?

Show hint

Kinetic and potential terms.

Learning goal: Count the two quadratic terms of a harmonic mode.

Show solution and calculation

Approach: one mode contains a kinetic and a potential quadratic term.
Together they give $U_m=RT$ per mole.
At 400 K this is about 3325.785 J/mol.
The temperature derivative is $R$ , about 8.31446 J/(mol·K).

Check: using only a half contribution would omit either spring or kinetic energy.

\begin{aligned}U_m&=RT=8.31446261815324\cdot400\,\mathrm{J\,mol^{-1}}\\&\simeq3325.785047\,\mathrm{J\,mol^{-1}}\\C_{V,m}^{(\mathrm{mode})}&=R\end{aligned}

Transfer · Application

An ideal constant-f gas has $C_{V,m}=\frac{5R}{2}$ . Find Cp,m and γ.

Show hint

Use Mayer’s relation.

Learning goal: Infer Cp and γ from supplied Cv.

Show solution and calculation

Approach: Mayer’s relation gives $C_{p,m}=C_{V,m}+R$ , then $\gamma=\frac{C_{p,m}}{C_{V,m}}$ .
$C_{p,m}=\frac{7R}{2}$ ≈29.10061916 J/(mol·K), $\gamma=\frac75=1.4$ .
This is not a universal value for every diatomic state.

\gamma=1.4

Interpret · Challenge

In a transition region an effective model gives $U(T)=\frac12nRTf(T)$ with f(T) increasing with T. A graph of U versus T becomes steeper. Can Cv, m still simply equal $\frac12 f(T)R$ ? Differentiate and interpret; f need not be integer here.

Show hint

Apply the product rule to T f(T).

Learning goal: Distinguish temperature-dependent activation from constant f.

Show solution and calculation

Approach: $C_{V, m}=\frac1n\frac{dU}{dT}$ gives $\frac R2[f(T)+Tf^{\prime}(T)]$ .
The extra term represents changing effective mode activation.
It disappears only on a constant-f plateau.

Check: $f'(T)=0$ recovers the classical shortcut. This is differentiation of a supplied effective model, not a claim that fixed integer f fully describes quantum activation.

C_{V,m}=\frac1n\frac{dU}{dT}=\frac R2\left[f(T)+T\frac{df}{dT}\right]

[1][2]

Error analysis · Application

For an ideal gas in a constant-f regime, measured $C_{p, m}=\frac52R$ . A solution sets $f=\frac{2C_{p, m}}R=5$ . Find correct f and γ; the supplied molar heat capacity is explicitly at constant pressure.

Show hint

Use Mayer: $C_{p,m}-C_{V,m}=R$ .

Learning goal: Convert Cp to Cv before inferring active quadratic terms.

Show solution and calculation

Approach: $C_{V, m}=C_{p, m}-R=\frac32R$ , giving $f=3$ and $\gamma=\frac{C_{p, m}}{C_{V, m}}$ = $\frac53$ .
The erroneous inference counts the additional pressure-work R contribution as internal quadratic energy terms.

\begin{aligned}C_{V,m}&=\frac52R-R=\frac32R\\f&=\frac{2C_{V,m}}R=3\\\gamma&=\frac{\frac52R}{\frac32R}=\frac53\end{aligned}

[3][1]

The classical result is a limit

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
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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Subject references and further reading

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[1]OpenStax, University Physics Volume 2, §2.3: Heat Capacity and Equipartition of Energy.

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Quanta Study editorial team

Equipartition of energy

Variables & units

Equipartition energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Kinetic theory

Worked example

Applications

How do you calculate with Equipartition of Energy?

Count energy terms rather than motion names

From one term to the whole sample

Explore further: why each quadratic term gets a half contribution

How do you count active quadratic terms rather than just names of motions?

More active terms give a steeper energy line

From U to heat capacity

Three complete calculations on term count, vibration and heating measurements

A diatomic model regime with f=5f=5f=5

One classical harmonic mode

Testing the model with a heating measurement

Rearrange without confusing particles and moles

Five exercises with solutions

A model has f=6f=6f=6, n=0.20 moln=0.20\,\mathrm{mol}n=0.20mol and T=250 KT=250\,\mathrm KT=250K. Find U.

A harmonic vibrational mode is fully classically active at 400 K. What energy contribution does it supply per mole, and what is its contribution to molar heat capacity?

An ideal constant-f gas has CV,m=5R2C_{V,m}=\frac{5R}{2}CV,m​=25R​. Find Cp,m and γ.

In a transition region an effective model gives U(T)=12nRTf(T)U(T)=\frac12nRTf(T)U(T)=21​nRTf(T) with f(T) increasing with T. A graph of U versus T becomes steeper. Can Cv, m still simply equal 12f(T)R\frac12 f(T)R21​f(T)R? Differentiate and interpret; f need not be integer here.

For an ideal gas in a constant-f regime, measured Cp,m=52RC_{p, m}=\frac52RCp,m​=25​R. A solution sets f=2Cp,mR=5f=\frac{2C_{p, m}}R=5f=R2Cp,m​​=5. Find correct f and γ; the supplied molar heat capacity is explicitly at constant pressure.

The classical result is a limit

In brief

Sources for this explanation

Frequently asked questions about Equipartition of Energy

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Equipartition of Energy for exams

Equipartition of energy

Variables & units

Equipartition energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Kinetic theory

Worked example

Applications

How do you calculate with Equipartition of Energy?

Count energy terms rather than motion names

From one term to the whole sample

Explore further: why each quadratic term gets a half contribution

How do you count active quadratic terms rather than just names of motions?

More active terms give a steeper energy line

From U to heat capacity

Three complete calculations on term count, vibration and heating measurements

A diatomic model regime with f=5f=5f=5

One classical harmonic mode

Testing the model with a heating measurement

Rearrange without confusing particles and moles

Five exercises with solutions

A model has f=6f=6f=6, n=0.20 moln=0.20\,\mathrm{mol}n=0.20mol and T=250 KT=250\,\mathrm KT=250K. Find U.

A harmonic vibrational mode is fully classically active at 400 K. What energy contribution does it supply per mole, and what is its contribution to molar heat capacity?

An ideal constant-f gas has CV,m=5R2C_{V,m}=\frac{5R}{2}CV,m​=25R​. Find Cp,m and γ.

In a transition region an effective model gives U(T)=12nRTf(T)U(T)=\frac12nRTf(T)U(T)=21​nRTf(T) with f(T) increasing with T. A graph of U versus T becomes steeper. Can Cv, m still simply equal 12f(T)R\frac12 f(T)R21​f(T)R? Differentiate and interpret; f need not be integer here.

For an ideal gas in a constant-f regime, measured Cp,m=52RC_{p, m}=\frac52RCp,m​=25​R. A solution sets f=2Cp,mR=5f=\frac{2C_{p, m}}R=5f=R2Cp,m​​=5. Find correct f and γ; the supplied molar heat capacity is explicitly at constant pressure.

The classical result is a limit

In brief

Sources for this explanation

Frequently asked questions about Equipartition of Energy

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Equipartition of Energy for exams

A diatomic model regime with $f=5$

A model has $f=6$ , $n=0.20\,\mathrm{mol}$ and $T=250\,\mathrm K$ . Find U.

An ideal constant-f gas has $C_{V,m}=\frac{5R}{2}$ . Find Cp,m and γ.

In a transition region an effective model gives $U(T)=\frac12nRTf(T)$ with f(T) increasing with T. A graph of U versus T becomes steeper. Can Cv, m still simply equal $\frac12 f(T)R$ ? Differentiate and interpret; f need not be integer here.

For an ideal gas in a constant-f regime, measured $C_{p, m}=\frac52R$ . A solution sets $f=\frac{2C_{p, m}}R=5$ . Find correct f and γ; the supplied molar heat capacity is explicitly at constant pressure.

A diatomic model regime with $f=5$

A model has $f=6$ , $n=0.20\,\mathrm{mol}$ and $T=250\,\mathrm K$ . Find U.

An ideal constant-f gas has $C_{V,m}=\frac{5R}{2}$ . Find Cp,m and γ.

In a transition region an effective model gives $U(T)=\frac12nRTf(T)$ with f(T) increasing with T. A graph of U versus T becomes steeper. Can Cv, m still simply equal $\frac12 f(T)R$ ? Differentiate and interpret; f need not be integer here.

For an ideal gas in a constant-f regime, measured $C_{p, m}=\frac52R$ . A solution sets $f=\frac{2C_{p, m}}R=5$ . Find correct f and γ; the supplied molar heat capacity is explicitly at constant pressure.