Equipartition of energy and specific heats

Degrees of freedom

Narrated video. Transcript:

A gas molecule can store energy in several ways. The simplest is moving. Its translational kinetic energy is half m v x squared, plus half m v y squared, plus half m v z squared.

Each independent coordinate needed to describe a molecule’s motion is called a degree of freedom. A molecule free to move in space has three translational degrees of freedom: x, y and z.

A monatomic molecule, like helium or argon, has only these three. A single atom spinning on its own stores no energy worth counting.

A diatomic molecule, like oxygen or nitrogen, is shaped like a dumbbell, and it can also rotate. It tumbles about two axes at right angles to its bond, adding half I 1 ω 1 squared plus half I 2 ω 2 squared.

Spinning about the bond itself does not count: the moment of inertia about that axis is almost zero. So a rigid diatomic molecule has 3 translational and 2 rotational degrees of freedom, 5 in all.

At high temperatures the two atoms also vibrate along the bond, like two balls joined by a spring. A vibration has two energy terms: kinetic energy, half m times d y by d t squared, and potential energy, half k y squared.

Look closely: every energy term is a square, of a velocity, an angular velocity, or a displacement. These are called quadratic terms.

The law of equipartition of energy says: in thermal equilibrium at temperature T, every quadratic term has the same average energy, half k B T.

So each translational and each rotational degree of freedom holds half k B T. Each vibrational mode holds k B T: half for its kinetic energy and half for its potential energy.

Specific heats of gases

Narrated video. Transcript:

Equipartition lets us predict how much heat a gas needs. Start with a monatomic gas. Each molecule has 3 × half k B T, so one mole has an internal energy U of three-halves R T.

Heat it at constant volume, and all the heat goes into U. So the molar specific heat at constant volume, C v, is d U by d T: three-halves R, about 12.5 joules per mole per kelvin.

For an ideal gas, C p minus C v equals R, because at constant pressure the gas also does work as it expands. So C p is five-halves R, and γ, which is C p over C v, is 5 over 3: about 1.67.

Measured values for helium and argon: C v is 12.5 and C p is 20.8 joules per mole per kelvin. The prediction is spot on.

A rigid diatomic molecule has 5 degrees of freedom, so U is five-halves R T. Then C v is five-halves R, C p is seven-halves R, and γ is 7 over 5, or 1.4. Nitrogen and oxygen at room temperature agree.

If the molecule also vibrates, add one more k B T per molecule. C v becomes seven-halves R, C p becomes nine-halves R, and γ is 9 over 7, about 1.29.

A polyatomic molecule, like methane, can rotate about all three axes, and it may have f vibrational modes. Then U is three-halves, plus three-halves, plus f, all times R T.

So C v equals 3 plus f, times R. C p equals 4 plus f, times R. And γ equals 4 plus f, over 3 plus f.

The recipe is always the same: count the quadratic terms, give each one half R per mole, and that is C v. Add R to get C p.

Solids and water

Narrated video. Transcript:

Now a solid. Its atoms are held in place by springy bonds, and each one vibrates about a fixed position, in three directions.

Each direction is a vibration with kinetic and potential energy, so it holds k B T. Three directions give 3 k B T per atom, and one mole has U equals 3 R T.

At ordinary pressure a solid hardly expands when heated, so the work it does is negligible, and C p and C v are almost equal. Both are 3 R, about 25 joules per mole per kelvin.

Check the measurements: aluminium 24.4, copper 24.5, silver 25.2, lead 26.4. Most solids agree well at room temperature.

Carbon, as diamond, is a big exception: only 6.1. Its bonds are so stiff that its vibrations are not fully active at room temperature.

Now water. Treat each of its three atoms like an atom in a solid, with 3 R each. That gives C equals 9 R, about 75 joules per mole per kelvin.

Measured: 4186 joules per kilogram per kelvin, times 0.018 kilograms per mole, is 75.3 joules per mole per kelvin. A very good match, and the reason water is so hard to heat up.

Classical equipartition is not the whole story, though. At low temperatures, the specific heats of all substances fall towards zero, as vibrations and rotations freeze out. Explaining this needs quantum mechanics.

Try it: count the energy squares

Interactive simulation. Things to try:

• Each coloured square is one quadratic term holding ½k_BT. Count them to get Cv = (number of squares / 2) R.
• Tick “Count the vibration too” for the diatomic gas: two more squares (KE + PE), and γ drops from 1.40 to 1.29.
• Compare a solid (6 squares per atom, 3R) with water (3 atoms × 6 = 18 squares, 9R).
• Space pauses or restarts the animation.

Which gas has γ = 1.40?

An ideal gas at room temperature has γ = Cp/Cv = 1.40. What are its molecules most likely to be?

1. Single atoms, like argon
2. Rigid diatomic molecules, like nitrogen
3. Vibrating diatomic molecules
4. Rigid non-linear molecules, like methane

Heating helium

A cylinder of fixed volume 44.8 L contains helium at STP. How much heat is needed to raise its temperature by 15 °C? (R = 8.31 J mol⁻¹ K⁻¹)

1. 187 J
2. 249 J
3. 374 J
4. 623 J

Summary

• Equipartition: ½ k_BT for every quadratic term
• Monatomic gas: Cv = (3/2)R, Cp = (5/2)R, γ = 5/3
• Rigid diatomic gas: Cv = (5/2)R, Cp = (7/2)R, γ = 7/5
• Diatomic, vibrating: Cv = (7/2)R, Cp = (9/2)R, γ = 9/7
• Polyatomic (f vibrational modes): Cv = (3 + f)R, γ = (4 + f)/(3 + f)
• Solids and water: C = 3R (solid), C = 9R (water)

• Each translational or rotational degree of freedom holds ½k_BT; each vibrational mode holds k_BT (KE + PE).
• For an ideal gas, Cp − Cv = R.
• A rigid diatomic molecule has only 2 rotational degrees of freedom: spin about the bond does not count.
• Most solids have C ≈ 3R ≈ 25 J mol⁻¹ K⁻¹ at room temperature; water has 9R ≈ 75 J mol⁻¹ K⁻¹.
• At low temperatures, specific heats fall below these values: classical equipartition fails and quantum theory is needed.