Law of equipartition of energy

What is the law of equipartition of energy?

Each quadratic term occurring in the expression for energy is a mode of absorption of energy by the molecule.

NCERT Physics Part II (Class 11), Section 12.5, page 253

In plain words: heat a gas and the energy you put in does not go only into making the molecules fly about faster. It is shared out: some goes into their straight-line motion, some into spinning them, and some, when vibrational levels are thermally populated, into vibration. Each independent quadratic term gets the same classical average share; a complete harmonic vibration has two such terms.

Degrees of freedom

A molecule free to move in space needs three coordinates to fix its position; one confined to a plane needs two; one confined to a line, one. We say it has three, two or one degrees of freedom for motion in space, in a plane or along a line (page 252).

Motion of a body as a whole from one point to another is called translation.

NCERT Physics Part II (Class 11), Section 12.5, page 252

Translation. A molecule free to move in space has three translational degrees of freedom. Its translational kinetic energy is ½mv_x² + ½mv_y² + ½mv_z² (NCERT Eq. 12.22). From kinetic theory, its average is (3/2)k_BT (NCERT Eq. 12.23). Since no direction is special, each term separately averages ½k_BT (NCERT Eq. 12.24).

Rotation. A monatomic molecule such as argon has only translation. A diatomic molecule such as O_₂ or N_₂ can also rotate about its centre of mass, about two independent axes at right angles to the line joining the atoms (NCERT Fig. 12.6, page 252). That adds two rotational degrees of freedom, each with an energy term ½Iω^², the rotational kinetic energy about that axis (NCERT Eq. 12.25). Rotation about the line joining the atoms does not count: its moment of inertia is so small that, for quantum mechanical reasons, it does not come into play (NCERT footnote, page 252).

Vibration. NCERT treats O_₂ at moderate temperatures as a rigid rotator, with no vibration. It adds that this is not always valid and gives CO as a molecule that vibrates even at moderate temperatures, its atoms oscillating along the bond like a one-dimensional oscillator (page 252). That example is misleading: CO’s ideal-gas molar C_ᵥ near room temperature is close to the rigid value (5/2)R (C_ₚ is near (7/2)R). Its vibrational thermal contribution is small, although quantum zero-point vibration remains; “frozen out” does not mean the bond is motionless. Vibration matters at high temperatures and for molecules such as Cl_₂, whose measured specific heat is larger than the rigid prediction (NCERT page 254). Use a full k_B T per vibrational mode when the problem assumes classical, fully excited vibration. Partial excitation does not give the full classical contribution. A vibration adds the energy term

ε_v = ½m(dy/dt)^² + ½ky^² (NCERT Eq. 12.26),

where k is the force constant and y the vibrational coordinate. If y is the bond extension, m denotes the effective reduced mass of the two atoms. This has two squared terms, one for kinetic energy and one for potential energy, like a mass on a spring.

Energy per degree of freedom

That is, in equilibrium, the total energy is equally distributed in all possible energy modes

NCERT Physics Part II (Class 11), Section 12.5, page 253

Each independent quadratic term has classical average energy ½k_B T. A harmonic vibrational mode has two terms and therefore contributes k_B T in this limit. NCERT credits Maxwell with the first proof of this principle of classical statistical mechanics; the proof is beyond the book (page 253). So the average energy of a molecule is ½k_BT times the number of squared terms:

Here f counts each translational and rotational degree of freedom once and each vibrational mode twice (a counting beyond NCERT’s wording, but the same rule). Careful: in NCERT Section 12.6.3, f stands for the number of vibrational modes only, not this total. For a mole, multiply by N_A: U = (f/2)RT for the thermal energy in this model, choosing an energy reference that omits temperature-independent offsets. Use constant f over the temperature interval when computing ΔU from this expression.

The next section of the chapter uses these values to predict the specific heats of gases.

Solved examples

Take k_B = 1.38 × 10^⁻²³ J K^⁻¹ and R = 8.314 J mol^⁻¹ K^⁻¹.

Common mistakes to avoid

Quick revision summary

• Degrees of freedom: 3 translational for any molecule in space; 2 rotational for a diatomic or other linear molecule (such as CO_₂); 3 rotational for a non-linear molecule (such as H_₂O or NH_₃).
• Each squared term in the energy is a mode of absorbing energy.
• Classical equipartition: each active quadratic term has an average energy of ½k_BT; a vibration (kinetic + potential) has k_BT.
• Energy per molecule: (3/2)k_BT (monatomic), (5/2)k_BT (rigid diatomic), (7/2)k_BT (diatomic with vibration).
• For a mole, U = (f/2)RT.

Questions students ask

How do you find the degrees of freedom of a molecule with N atoms?

A molecule of N ≥ 2 atoms has 3N coordinate degrees of freedom in all, since each atom needs three coordinates: 3 are translational, 2 (for a linear molecule) or 3 (for a non-linear one) are rotational, and the remaining 3N − 5 or 3N − 6 are vibrational modes. So a diatomic molecule has 3 + 2 + 1, linear CO_₂ has 3 + 2 + 4 and bent H_₂O has 3 + 3 + 3. For equipartition, each fully excited classical vibrational mode counts twice (kinetic and potential energy); at room temperature most vibrations are only partly excited, if at all, so a rigid triatomic molecule is usually given f = 5 if it is linear and f = 6 if it is not.

Is the kinetic energy of a gas (3/2)RT or (f/2)RT per mole?

Both formulas are right, for different energies: (3/2)RT per mole, or (3/2)k_BT per molecule, is only the translational kinetic energy, while (f/2)RT is the model’s thermal internal energy, including rotation and fully excited classical vibrations, up to a temperature-independent reference offset. Use the translational value for pressure, PV = (2/3)E_trans, and for the rms speed, which depend only on how fast the molecules move from place to place. Use (f/2)RT for heat and internal energy, as in ΔU = μ(f/2)RΔT when f stays constant; the two agree only for a monatomic gas, where f = 3.

Does the law of equipartition of energy hold at all temperatures?

No, the law of equipartition of energy is a result of classical physics and holds for a motion only when k_BT is large compared with the steps between its quantised energy levels. At sufficiently low temperatures, rotational excitation is suppressed and the rotational contribution to heat capacity becomes small; the detailed curve depends on the molecule and its allowed quantum states. Vibrations of N_₂ and O_₂ add noticeably to C_v only above several hundred kelvin and get their full share only at a few thousand kelvin, and the same failure makes the specific heats of solids fall below 3R at low temperatures.

Why does a monatomic gas have only three degrees of freedom?

A monatomic gas such as helium or argon has only three degrees of freedom, all translational, because a single atom has no bond to vibrate and cannot take up rotational energy at ordinary temperatures. In this model an atom is treated as a point particle, with no independent molecular orientation or bond vibration. Electronic and nuclear excitations are ordinarily inaccessible on the thermal energy scale; a nucleus should not be modelled here as a tiny classical rigid rotor. So a monatomic gas has U = (3/2)RT per mole, all of it kinetic energy of translation.

Who gave the law of equipartition of energy?

NCERT credits Maxwell with the first proof of this classical principle. It states that each independent quadratic energy term has average energy ½k_B T in the classical equilibrium limit, and leaves the proof beyond the book.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. State the law of equipartition of energy.

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   In the classical equilibrium limit at temperature T, the energy of a system is shared equally among all its independent energy modes (squared terms in the energy), each having an average energy ½k_BT.

2. Q2. A rigid non-linear molecule, such as water treated as rigid, has 3 rotational degrees of freedom. Find its average energy at 300 K.

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   f = 3 + 3 = 6, so the energy is 3k_BT = 3 × 1.38 × 10^⁻²³ × 300, about 1.24 × 10^⁻²⁰ J.

3. Q3. Why does a vibrational mode contribute k_BT rather than ½k_BT?

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   A vibration has two squared terms in its energy, ½m(dy/dt)^² for kinetic energy and ½ky^² for potential energy. Each gets ½k_BT, giving k_BT in all.