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Kinetic Theory

Equipartition of energy and specific heats

Predict C_v, C_p and γ from degrees of freedom.

A degree of freedom is an independent way a molecule can hold energy: each one is a separate squared term in its energy, such as ½mv_x² or ½Iω².

  • A monatomic gas (helium, argon) can only move along x, y and z: 3 translational degrees.
  • A rigid diatomic molecule (O₂, N₂) also spins about 2 axes, so 5. It cannot store energy spinning about its own axis, because its moment of inertia about that axis is almost zero.
  • A non-linear polyatomic molecule (NH₃) spins about all 3 axes: 6.
  • A vibration counts twice, because it has both kinetic and potential energy. Molecules like CO start vibrating at higher temperatures.

Law of equipartition of energy: in thermal equilibrium, the energy is shared equally among all the degrees of freedom, ½k_BT for each one per molecule. So a molecule with f degrees has (f/2)k_BT, and one mole has U = (f/2)RT.

Since C_v = ΔU/ΔT per mole, everything follows from f:

GasfC_vC_p = C_v + Rγ = 1 + 2/f
Monatomic3(3/2)R(5/2)R5/3
Rigid diatomic5(5/2)R(7/2)R7/5
Non-linear polyatomic63R4R4/3

Example: 4 g of oxygen (4/32 = 0.125 mol, diatomic) at 300 K has internal energy U = 0.125 × (5/2) × 8.31 × 300 ≈ 779 J.

Example: 3 mol of nitrogen heated by 10 °C at constant pressure takes Q = μC_pΔT = 3 × (7/2)R × 10 = 105R ≈ 873 J. At constant volume it would take only (5/2)R per mole per degree.

Solids: each atom vibrates in 3 directions, and each vibration has kinetic and potential energy, so an atom holds (3/2)k_BT + (3/2)k_BT = 3k_BT. One mole has U = 3RT, giving C = 3R ≈ 25 J mol⁻¹ K⁻¹ for most solids at room temperature (the Dulong–Petit law). Carbon is an exception. Water, with 3 atoms to a molecule, comes out near 9R ≈ 75 J mol⁻¹ K⁻¹, which matches 4186 J kg⁻¹ K⁻¹ × 0.018 kg.

One warning: the average of a square is not the square of the average. For speeds 1, 2, 3, 4 and 5 m/s, <v²> = 11 but (<v>)² = 9.

Count the squared terms; each brings ½k_BT.

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