Law of Equipartition of Energy

PhysicsKinetic TheoryFor NEET aspirants

The law of equipartition of energy states that in thermal equilibrium at absolute temperature $T$ , every independent quadratic term in a molecule's energy (every degree of freedom) gets the same average energy, $\frac{1}{2} k T$ . Counting degrees of freedom (3 for a monatomic gas, 5 for a rigid diatomic, 6 for a rigid non-linear molecule) then gives the internal energy $U = \frac{f}{2} n R T$ of any ideal gas. The law of equipartition of energy is in the JEE Main and NEET syllabus and leads straight to specific heats.

On this page1Degrees of freedom2Translation3Rotation4Vibration5The law6Internal energy7Limits of the law

Key Formulas - Quick Reference

★ Must learnEquipartition: each degree of freedom (quadratic energy term) has average energy $\frac{1}{2} k T$ per molecule, $\frac{1}{2} R T$ per mole
★ Must learnEnergy per molecule $\overline{E} = \frac{f}{2} k T$ ; internal energy of $n$ moles $U = \frac{f}{2} n R T$
★ Must learnRigid molecules: monatomic $f = 3$ , diatomic or linear $f = 5$ , non-linear $f = 6$
Each active vibrational mode adds 2 terms ( $k T$ ): diatomic with vibration $f = 7$
Rigid molecule of $N$ atoms with $c$ independent constraints: $f = 3 N - c$
★ Must learnAll motions: $3 N$ = 3 translations + 2 (linear) or 3 (non-linear) rotations + $3 N - 5$ or $3 N - 6$ vibrational modes
Diatomic (rigid): rotational KE : translational KE $= 2 : 3$ ; per molecule each is $k T$ and $\frac{3}{2} k T$

1. Degrees of Freedom

★ Must learn

The degrees of freedom $f$ of a molecule are the number of independent ways in which it can store energy: the independent coordinates (positions or angles) needed to describe its motion, counted by the quadratic energy terms they produce.

A molecule can store energy in three kinds of motion: translation of its centre of mass, rotation about its centre of mass, and vibration of its atoms along the bonds.

1.1 Translational degrees of freedom

Motion of the centre of mass is described by $x$ , $y$ , $z$ , and the translational kinetic energy has three independent terms:

E_{t r an s} = \frac{1}{2} m v_{x}^{2} + \frac{1}{2} m v_{y}^{2} + \frac{1}{2} m v_{z}^{2}

Every molecule, whatever its shape, has 3 translational degrees of freedom. A monatomic gas (He, Ne, Ar) has only these: the atom is point-like, so it stores no rotational energy.

Figure 1: A monatomic molecule can move independently along

x

y

and

z

: 3 translational degrees of freedom. Rotation of a point-like atom stores no energy (

I \approx 0

), so

f = 3

and

\overline{E} = \frac{3}{2} k T

2. Rotational Degrees of Freedom

A diatomic molecule such as $O_{2}$ or $N_{2}$ is like a rigid dumbbell. Besides translating, it can rotate about the two axes perpendicular to its bond, each with rotational energy $\frac{1}{2} I ω^{2}$ :

E_{r o t} = \frac{1}{2} I_{y} ω_{y}^{2} + \frac{1}{2} I_{z} ω_{z}^{2}

The moment of inertia about the bond axis itself is negligible (the atoms are almost points on that axis), so that rotation stores no energy and is not counted.

Figure 2: A rigid diatomic molecule (

O_{2}

N_{2}

at room temperature) rotates about two axes perpendicular to its bond (

y

and

z

). About the bond axis itself the moment of inertia is almost zero, so that rotation does not count.

f = 3 + 2 = 5

Polyatomic molecules follow the same rule. A linear molecule ( $C O_{2}$ , $C_{2} H_{2}$ ) has 2 rotations, like a diatomic one. A non-linear molecule ( $H_{2} O$ , $N H_{3}$ , $C H_{4}$ ) has a moment of inertia about every axis and has 3.

Figure 3: A linear molecule (

C O_{2}

C_{2} H_{2}

) rotates like a diatomic one: 2 rotations,

f = 5

. A non-linear molecule (

H_{2} O

N H_{3}

C H_{4}

) has a moment of inertia about every axis: 3 rotations,

f = 6

(vibrations ignored).

2.1 Counting rule for rigid molecules

$N$ free atoms need $3 N$ coordinates. Each independent rigid condition (a fixed bond length, a fixed angle) removes one:

f = 3 N - c (c = number of independent constraints)

Figure 4:

N

free atoms need

3 N

coordinates. Each independent rigid constraint removes one:

f = 3 N - c

. A dumbbell gives

5

, a rigid linear triatomic

9 - 4 = 5

(two bonds, two conditions to stay straight), a rigid triangle

9 - 3 = 6

Molecule (rigid)	Translational	Rotational	$f$	Examples
Monatomic	3	0	3	He, Ne, Ar
Diatomic	3	2	5	$H_{2}$ , $O_{2}$ , $N_{2}$ , CO
Linear polyatomic	3	2	5	$C O_{2}$ , $C_{2} H_{2}$
Non-linear polyatomic	3	3	6	$H_{2} O$ , $N H_{3}$ , $C H_{4}$

Quick Recall: tap to check

Why is rotation of a diatomic molecule about its own axis not counted?

Its moment of inertia about that axis is almost zero, so it stores no appreciable energy.

How many degrees of freedom does a rigid

N H_{3}

molecule have?

6: 3 translational and 3 rotational (it is non-linear).

C O_{2}

treated like

H_{2} O

or like

O_{2}

O_{2}

: it is linear, so 2 rotations and

f = 5

when rigid.

3. Vibrational Degrees of Freedom

Real bonds are not perfectly rigid. The atoms of a diatomic molecule can vibrate along the bond like two masses on a spring. The vibrational energy has a kinetic and a potential term, both quadratic:

E_{v ib} = \frac{1}{2} μ (\frac{d r}{d t})^{2} + \frac{1}{2} k_{s} (r - r_{0})^{2}

Here $μ$ is the reduced mass and $k_{s}$ the spring constant of the bond (written $k_{s}$ so it is not confused with Boltzmann's constant $k$ ).

Figure 5: A diatomic molecule that is not rigid vibrates along its bond like two masses on a spring (

μ

is the reduced mass,

k_{s}

the bond's spring constant). The vibrational energy has two quadratic terms, kinetic and potential, so one vibrational mode stores

\frac{1}{2} k T + \frac{1}{2} k T = k T

★ Must learn

One vibrational mode counts as two terms (one kinetic, one potential), so it stores $\frac{1}{2} k T + \frac{1}{2} k T = k T$ . A diatomic molecule with its vibration active has $f = 3 + 2 + 2 = 7$ and energy $\frac{7}{2} k T$ .

Figure 6: In a vibration the energy keeps swapping between kinetic and potential. For a harmonic bond the time averages are equal, and equipartition gives each quadratic term

\frac{1}{2} k T

: one vibrational mode holds

k T

, twice as much as one translational or rotational degree of freedom.

Vibrations need a comparatively large energy to start. For most diatomic gases ( $O_{2}$ , $N_{2}$ , $H_{2}$ ) they are "frozen" at room temperature and become active only at high temperatures (several hundred to a few thousand kelvin). Unless a problem says so, treat molecules as rigid.

In general a molecule of $N$ atoms has $3 N$ independent motions: 3 translations, then 2 rotations and $3 N - 5$ vibrational modes if linear, or 3 rotations and $3 N - 6$ vibrational modes if non-linear. $C O_{2}$ ( $N = 3$ , linear) has $4$ vibrational modes; $H_{2} O$ has $3$ .

Rigid diatomic

3 translational + 2 rotational terms. $f = 5$ , energy $\frac{5}{2} k T$ per molecule. Good for $O_{2}$ , $N_{2}$ at room temperature.

Diatomic with vibration

Adds one vibrational mode = 2 terms (KE + PE). $f = 7$ , energy $\frac{7}{2} k T$ per molecule. Needed at high temperature or when a question says the molecule vibrates.

Key idea

Translation and rotation give one term each; a vibration gives two. Count terms, not motions.

4. The Law of Equipartition of Energy

Kinetic theory already showed that the average translational energy of a molecule is $\frac{3}{2} k T$ . Because the motion is random, no direction is special, so this energy is shared equally among the three components:

\frac{1}{2} m \overline{v_{x}^{2}} = \frac{1}{2} m \overline{v_{y}^{2}} = \frac{1}{2} m \overline{v_{z}^{2}} = \frac{1}{2} k T

Figure 7: Each velocity component has the same bell-shaped distribution, centred on zero (computed for

N_{2}

300 K

). Its rms value is

k T / m = 298 m s^{- 1}

(shaded), so

\frac{1}{2} m \overline{v_{x}^{2}} = \frac{1}{2} m \overline{v_{y}^{2}} = \frac{1}{2} m \overline{v_{z}^{2}} = \frac{1}{2} k T

: the

\frac{3}{2} k T

of translation is shared equally.

Collisions constantly transfer energy between translation, rotation and vibration. Maxwell and Boltzmann showed that in equilibrium every quadratic term receives the same average share. This is the law.

★ Must learn

Law of equipartition of energy: in equilibrium at absolute temperature $T$ , the total energy of a molecule is equally distributed among all its degrees of freedom, each translational and rotational degree of freedom contributing $\frac{1}{2} k T$ and each vibrational mode $k T$ (two quadratic terms).

\overline{E} = \frac{f}{2} k T per molecule, \frac{f}{2} R T per mole

Figure 8: Average energy per molecule

= f \times \frac{1}{2} k T

. Translation always contributes 3, rotation 0, 2 or 3, and each active vibrational mode 2 (one kinetic and one potential term).

Exam Trick

Energy per degree of freedom is $\frac{1}{2} k T$ per molecule, $\frac{1}{2} R T$ per mole, for every gas. So rotational KE of a rigid diatomic gas is $R T$ per mole (2 terms) and translational KE is $\frac{3}{2} R T$ (3 terms): ratio $2 : 3$ , whatever the gas or temperature. For a non-linear molecule the ratio is $3 : 3 = 1 : 1$ .

5. Internal Energy of an Ideal Gas

An ideal gas has no intermolecular forces, so it has no potential energy between molecules. Its internal energy is just the sum of the molecular energies:

★ Must learn

U = N \times \frac{f}{2} k T = \frac{f}{2} n R T = \frac{f}{2} P V

$U$ depends only on temperature (for a given amount of gas), not on pressure or volume separately.

Gas	$f$	Energy per molecule	$U$ for $n$ moles
Monatomic	3	$\frac{3}{2} k T$	$\frac{3}{2} n R T$
Diatomic (rigid)	5	$\frac{5}{2} k T$	$\frac{5}{2} n R T$
Diatomic (vibrating)	7	$\frac{7}{2} k T$	$\frac{7}{2} n R T$
Non-linear (rigid)	6	$3 k T$	$3 n R T$

Figure 9: For an ideal gas

U = \frac{f}{2} n R T

depends only on temperature (no potential energy between molecules). The slope

\frac{d U}{d T} = \frac{f}{2} R

per mole is exactly

C_{V}

, the link to the next concept.

Mixtures. Energy adds: for $n_{1}$ moles with $f_{1}$ and $n_{2}$ moles with $f_{2}$ at the same temperature,

U = (\frac{f _{1}}{2} n_{1} + \frac{f _{2}}{2} n_{2}) R T, f_{mi x} = \frac{n _{1} f _{1} + n _{2} f _{2}}{n _{1} + n _{2}}

Exam Trick

Find $f$ from any energy statement: $f = \frac{2 U}{n R T} = \frac{2 U}{P V}$ . If a gas has $U = 3 P V$ , then $f = 6$ (non-linear). If the energy per molecule is $\frac{5}{2} k T$ , it is a rigid diatomic or linear gas. The same trick with $γ$ is on the next page: $f = \frac{2}{γ - 1}$ .

Quick Recall: tap to check

What is the internal energy of

1 mol

of helium at

300 K

\frac{3}{2} R T \approx 3.74 kJ

Does the internal energy of an ideal gas change in an isothermal expansion?

No.

U = \frac{f}{2} n R T

depends only on

T

What is

f_{mi x}

for 1 mol He and 1 mol

O_{2}

\frac{3 + 5}{2} = 4

6. Limitations of the Law

Equipartition is a result of classical physics. It works well for translation at all ordinary temperatures and for rotation above a few tens of kelvin, but it fails when the energy step needed to start a motion is large compared with $k T$ :

Frozen vibrations: at room temperature $k T \approx 0.026 eV$ is much smaller than the vibrational energy step of $N_{2}$ or $O_{2}$ (a few tenths of an eV), so vibrations take almost no share.
Frozen rotation: light molecules like $H_{2}$ stop rotating below about $80 K$ and behave as monatomic.
Temperature dependence: as a result the measured heat capacities of gases and solids increase with temperature instead of being constant (next concept).
It applies only to energy terms that are quadratic in a coordinate or velocity.

JEE Advanced

Why $\frac{1}{2} k T$ ? In equilibrium the probability of a state with energy $E$ is proportional to $e^{- E / k T}$ (Boltzmann factor). For a term $E = a q^{2}$ :

\overline{a q^{2}} = \frac{\int _{- \infty}^{\infty} a q ^{2} e ^{- a q^{2} / k T} d q}{\int _{- \infty}^{\infty} e ^{- a q^{2} / k T} d q} = \frac{1}{2} k T

The answer does not depend on $a$ : mass, moment of inertia and spring constant all drop out. For a non-quadratic term the share differs, e.g. a potential $E = a ∣ q ∣$ gives $k T$ . Vibrations of $C O_{2}$ : $3 N - 5 = 4$ modes, so with all of them active $f = 3 + 2 + 2 \times 4 = 13$ and $\overline{E} = \frac{13}{2} k T$ .

Key idea

Equipartition is exact for translation, reliable for rotation above a few tens of kelvin, and fails for vibrations at room temperature: quantum effects freeze them out.

7. Solving Problems and Revision Map

Use the flowchart to decide $f$ and the energy, then revise with the mind map.

Figure 10: Finding

f

and the energy. Unless a question says the molecule vibrates (or the temperature is very high), treat it as rigid:

f = 3

5

6

Figure 11: Mind map of this concept. Revise each branch from memory, then check.

8. Solved Examples

Solved Example 1

Find the internal energy of

2 mol

of oxygen at

2 7^{\circ} C

, treating the molecules as rigid. (

R = 8.314 J mol^{- 1} K^{- 1}

)

Solution:

Rigid diatomic: $f = 5$ . $U = \frac{5}{2} n R T = \frac{5}{2} (2) (8.314) (300)$ .

Answer: $U \approx 1.25 \times 1 0^{4} J$ ( $12.5 kJ$ ).

Solved Example 2

For

1 mol

of nitrogen at

300 K

, find the total translational and the total rotational kinetic energy, and their ratio.

Solution:

Translational: 3 terms, $\frac{3}{2} R T = 1.5 (8.314) (300) = 3741 J$ .

Rotational: 2 terms, $R T = 8.314 \times 300 = 2494 J$ .

Answer: $3.74 kJ$ and $2.49 kJ$ ; rotational : translational $= 2 : 3$ .

Solved Example 3

A vessel contains

2 mol

of helium and

3 mol

of oxygen (rigid) at

300 K

. Find the total internal energy.

Solution:

$U = (\frac{3}{2} (2) + \frac{5}{2} (3)) R T = 10.5 R T = 10.5 \times 8.314 \times 300$ .

Answer: $U \approx 2.62 \times 1 0^{4} J$ . (Here $f_{mi x} = \frac{2 ( 3 ) + 3 ( 5 )}{5} = 4.2$ .)

Solved Example 4

The energy associated with each degree of freedom of one mole of an ideal gas at temperature

T

is
(A)

\frac{1}{2} k T

(B)

\frac{1}{2} R T

(C)

\frac{3}{2} R T

(D)

R T

Solution:

Answer: (B). Per molecule it is $\frac{1}{2} k T$ ; for $N_{A}$ molecules it is $\frac{1}{2} N_{A} k T = \frac{1}{2} R T$ . Option (A) is per molecule, not per mole.

Solved Example 5

At a high temperature the molecules of a diatomic gas also vibrate. The average energy of a molecule is
(A)

\frac{3}{2} k T

(B)

\frac{5}{2} k T

(C)

\frac{7}{2} k T

(D)

3 k T

Solution:

Answer: (C). 3 translational + 2 rotational terms give $\frac{5}{2} k T$ ; the vibrational mode adds two terms (KE and PE), $k T$ . Total $\frac{7}{2} k T$ .

Solved Example 6

One mole of an ideal gas at

400 K

has internal energy

9977 J

. Find its degrees of freedom and say what kind of molecule it could be.

Solution:

$f = \frac{2 U}{n R T} = \frac{2 \times 9977}{1 \times 8.314 \times 400} = \frac{19 954}{3325.6}$ .

Answer: $f = 6$ : a rigid non-linear molecule such as $H_{2} O$ vapour or $C H_{4}$ .

Solved Example 7

Find the average energy of a

C O_{2}

molecule at

300 K

(a) treating it as rigid, (b) if all its vibrational modes were active. (

k = 1.38 \times 1 0^{- 23} J K^{- 1}

)

Solution:

(a) Linear, rigid: $f = 5$ , $\overline{E} = \frac{5}{2} k T = 2.5 \times 1.38 \times 1 0^{- 23} \times 300 = 1.04 \times 1 0^{- 20} J$ .

(b) Linear triatomic: $3 N - 5 = 4$ vibrational modes, each $k T$ . $f = 3 + 2 + 8 = 13$ , $\overline{E} = \frac{13}{2} k T = 2.69 \times 1 0^{- 20} J$ .

Answer: (a) $1.04 \times 1 0^{- 20} J$ ; (b) $2.69 \times 1 0^{- 20} J$ . The real value at $300 K$ is only a little above (a), because most vibrations are frozen.

Solved Example 8

A mixture contains equal numbers of moles of helium and hydrogen (rigid) at temperature

T

. The average energy per molecule of the mixture is
(A)

\frac{3}{2} k T

(B)

2 k T

(C)

\frac{5}{2} k T

(D)

4 k T

Solution:

Answer: (B). Half the molecules have $\frac{3}{2} k T$ and half $\frac{5}{2} k T$ ; the average is $\frac{1}{2} (\frac{3}{2} + \frac{5}{2}) k T = 2 k T$ ( $f_{mi x} = 4$ ).

Practice Questions

Find the internal energy of $1 mol$ of argon at $400 K$ .Answer: $\approx 4.99 kJ$
Find the total rotational kinetic energy of $2 mol$ of oxygen at $300 K$ .Answer: $2 R T \approx 4.99 kJ$
How many degrees of freedom does a rigid $N H_{3}$ molecule have?Answer: 6
Find the ratio of the average energy of a helium atom to that of a rigid oxygen molecule at the same temperature.Answer: $3 : 5$
An ideal gas has internal energy $U = 3 P V$ . What is its $f$ ?Answer: 6
How many vibrational modes does a linear triatomic molecule have, and what is $f$ if all are active?Answer: 4 modes; $f = 13$
Find the average total energy of a rigid nitrogen molecule at $2 7^{\circ} C$ .Answer: $\approx 1.04 \times 1 0^{- 20} J$

Common Mistakes to Avoid

Watch out

Counting a vibrational mode as one degree of freedom. It has two quadratic terms (KE and PE) and stores $k T$ , not $\frac{1}{2} k T$ .
Counting rotation of a diatomic or linear molecule about its own axis. Its moment of inertia is negligible; only 2 rotations count.
Treating $C O_{2}$ as non-linear ( $f = 6$ ). It is linear: $f = 5$ when rigid.
Including vibrations when the problem does not mention them. At ordinary temperatures take molecules as rigid.
Mixing up per molecule and per mole: $\frac{1}{2} k T$ per degree of freedom per molecule, $\frac{1}{2} R T$ per mole.
Thinking $\frac{3}{2} k T$ is the total energy of every molecule. It is only the translational part; diatomic molecules have $\frac{5}{2} k T$ in all.
Averaging $f$ of a mixture without weighting by moles: $f_{mi x} = \frac{n _{1} f _{1} + n _{2} f _{2}}{n _{1} + n _{2}}$ .
Applying equipartition at very low temperatures (for example $H_{2}$ below $80 K$ ), where rotations are frozen and the classical law fails.

Frequently Asked Questions

What is the law of equipartition of energy?

In thermal equilibrium at absolute temperature T, the energy of a molecule is shared equally among all its degrees of freedom. Each translational and rotational degree of freedom gets one half k T on average, and each vibrational mode gets k T because it has kinetic and potential energy.

What are degrees of freedom of a gas molecule?

They are the independent ways in which a molecule can store energy: translation along three axes, rotation about two or three axes, and vibration along bonds. A monatomic gas has 3, a rigid diatomic or linear molecule 5, and a rigid non-linear molecule 6.

Why does a diatomic molecule have only two rotational degrees of freedom?

Rotation about the bond axis has an almost zero moment of inertia, because the atoms lie on that axis, so it stores no appreciable energy. Only the two rotations about axes perpendicular to the bond are counted.

Why does a vibrational mode contribute k T instead of one half k T?

A vibration has two quadratic energy terms: the kinetic energy of the moving atoms and the potential energy of the stretched bond. Equipartition gives one half k T to each term, so a full vibrational mode holds k T.

What is the internal energy of an ideal gas according to equipartition?

It is f over 2 times n R T, where f is the number of degrees of freedom and n the number of moles. It depends only on temperature because an ideal gas has no intermolecular potential energy.

When does the law of equipartition fail?

It fails when the energy needed to excite a motion is large compared with k T. Vibrations are frozen at room temperature and rotations of hydrogen freeze below about 80 kelvin, so measured heat capacities are lower than predicted and rise with temperature.

How is equipartition of energy asked in NEET?

NEET asks for the number of degrees of freedom of a molecule, the energy per degree of freedom, the average energy of a diatomic molecule, or the ratio of rotational to translational energy. Remember 3, 5 and 6 for rigid molecules.

How does JEE Main test the law of equipartition?

JEE Main combines it with heat capacities: finding internal energy of gas mixtures, degrees of freedom from the ratio of specific heats, the effect of vibrational modes at high temperature, and energy ratios for diatomic and polyatomic gases.

Previous year questions on Law of Equipartition of Energy

10 questions from past papers, each with a step-by-step solution.

Ready to master Kinetic Theory?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test