What is a degree of freedom?
A degree of freedom is one independent way a molecule can move (and so store energy). We count three kinds:
- Translational: moving as a whole along x, y or z. Every molecule has 3.
- Rotational: spinning about an axis.
- Vibrational: atoms moving in and out like a spring.
Counting degrees of freedom
Monatomic gas (He, Ne, Ar)
Only 3 translational. f = 3. (A point-like atom spinning has almost no rotational energy.)
Diatomic gas (O₂, N₂, H₂) at ordinary temperature
3 translational + 2 rotational (about the two axes at right angles to the bond). Spin about the bond axis has almost no moment of inertia, so it is not counted. f = 5.
Diatomic gas at high temperature
The bond vibrates. One vibration mode brings 2 terms (kinetic + potential). f = 7.
Non-linear polyatomic (NH₃, CH₄, H₂O)
3 translational + 3 rotational. f = 6 without vibration. Linear polyatomic (CO₂) has f = 5 without vibration.
Law of equipartition of energy
In thermal equilibrium at temperature T, the total energy of a molecule is shared equally among its degrees of freedom, and each one holds on average ½kBT.
- Monatomic molecule: 3 × ½kT = (3/2)kT (matches kinetic theory).
- Diatomic (rigid): 5 × ½kT = (5/2)kT.
- Each vibration mode: ½kT (kinetic) + ½kT (potential) = kT.
Energy of one mole: U = (f/2) RT.
Specific heats of gases
Molar specific heat at constant volume: Cv = dU/dT = (f/2) R. Mayer's relation: Cp = Cv + R. Ratio: γ = Cp/Cv = 1 + 2/f.
| Gas | f | Cv | Cp | γ |
|---|---|---|---|---|
| Monatomic | 3 | 3R/2 | 5R/2 | 5/3 ≈ 1.67 |
| Diatomic (rigid) | 5 | 5R/2 | 7R/2 | 7/5 = 1.40 |
| Diatomic (vibrating) | 7 | 7R/2 | 9R/2 | 9/7 ≈ 1.29 |
| Non-linear polyatomic (rigid) | 6 | 3R | 4R | 4/3 ≈ 1.33 |
For a mixture: Cv(mix) = (n₁Cv₁ + n₂Cv₂)/(n₁ + n₂).
Specific heat of solids and water
In a solid each atom vibrates about a fixed spot in 3 directions. Each direction holds kT (kinetic + potential), so one mole has U = 3RT and C = 3R ≈ 25 J mol⁻¹ K⁻¹ (Dulong–Petit law). It works well for many metals at room temperature.
Water: treating each molecule as 3 atoms, C ≈ 9R ≈ 75 J mol⁻¹ K⁻¹, close to the measured 75.3 J mol⁻¹ K⁻¹.
At very low temperatures measured values fall below these, because some motions get "frozen". Classical equipartition cannot explain this; quantum physics does.
Try it: the 3D and at home
- In the 3D: before picking each molecule, write your guess for f and γ. Then pick it and check. Why does γ go down as f goes up?
- At home: hold a pencil by its middle. Count the ways it can move: slide along 3 directions, flip end over end two ways. Twisting about its own length hardly moves anything. That is f = 5, like O₂.
Board exam focus
- State the law of equipartition in one line and use it to find Cv, Cp, γ for monatomic and diatomic gases.
- Know why diatomic f = 5, not 6.
- Specific heat of solids = 3R derivation is a short-answer favourite.
- γ = 1 + 2/f helps check answers quickly.
Key formulas and definitions
- Energy per degree of freedom = ½ k_B T
- Energy per molecule = (f/2) k_B T
- U (1 mole) = (f/2) RT
- Cv = (f/2) R
- Cp = Cv + R
- γ = Cp/Cv = 1 + 2/f
- f: monatomic 3, diatomic 5 (7 with vibration), non-linear polyatomic 6
- Solids: C = 3R
- Mixture: Cv = (n₁Cv₁ + n₂Cv₂)/(n₁ + n₂)
Worked examples
1. Find Cv, Cp and γ for a monatomic gas.
Step 1: f = 3. Step 2: Cv = (3/2)R = 12.5 J/mol K. Step 3: Cp = Cv + R = (5/2)R = 20.8 J/mol K. Step 4: γ = 5/3. Answer: Cv = 1.5R, Cp = 2.5R, γ ≈ 1.67.
2. Find the average energy of one nitrogen molecule at 300 K (rigid diatomic). (k = 1.38 × 10⁻²³ J/K)
Step 1: f = 5. Step 2: E = (5/2)kT = 2.5 × 1.38 × 10⁻²³ × 300. Answer: E ≈ 1.04 × 10⁻²⁰ J.
3. Find the internal energy of 2 mol of oxygen at 27 °C (rigid diatomic). (R = 8.31)
Step 1: T = 300 K, f = 5. Step 2: U = (f/2) n R T = 2.5 × 2 × 8.31 × 300. Answer: U ≈ 1.25 × 10⁴ J.
4. A gas has γ = 1.4. Find its degrees of freedom and say what kind of gas it could be.
Step 1: γ = 1 + 2/f, so 2/f = 0.4. Step 2: f = 5. Answer: f = 5, a rigid diatomic gas like O₂ or N₂.
5. How much heat is needed to raise 3 mol of helium by 10 K at constant volume? (R = 8.31)
Step 1: Cv = (3/2)R = 12.47 J/mol K. Step 2: Q = n Cv ΔT = 3 × 12.47 × 10. Answer: Q ≈ 374 J.
6. Find Cv of a non-linear polyatomic gas whose one vibration mode is active.
Step 1: f = 3 + 3 + 2 = 8. Step 2: Cv = (8/2) R = 4R. Answer: Cv = 4R ≈ 33.2 J/mol K, Cp = 5R, γ = 1.25.
7. A mixture has 1 mol of helium and 1 mol of oxygen (rigid). Find γ of the mixture.
Step 1: Cv = (1 × 1.5R + 1 × 2.5R)/2 = 2R. Step 2: Cp = Cv + R = 3R. Step 3: γ = 3R/2R. Answer: γ = 1.5.
8. Using equipartition, estimate the molar specific heat of copper and its specific heat per kg. (M = 63.5 g/mol)
Step 1: solid, C = 3R = 24.9 J/mol K. Step 2: per kg, c = 24.9 / 0.0635. Answer: c ≈ 392 J/kg K (measured ≈ 385).
Common mistakes
- Counting 3 rotational degrees for a diatomic molecule. Spin about the bond axis is not counted, so f = 5.
- Giving each vibration only ½kT. A vibration has kinetic and potential parts, so it holds kT (counts as 2).
- Writing γ = Cv/Cp. It is Cp/Cv, always greater than 1.
- Adding γ values of a mixture directly. Add the heat capacities first, then find γ.