Equipartition Theorem Calculator
Enter degrees of freedom and temperature to compute the average energy per particle ⟨E⟩ = (f/2)·k·T and molar heat capacity.
Input Data
Results
At a glance:The equipartition theorem states that in thermal equilibrium each independent quadratic term in a system's energy contributes ½kT to the average energy per particle, where k = 1.380649×10⁻²³ J/K is Boltzmann's constant. A molecule with f degrees of freedom (3 translational; 2 or 3 rotational depending on linearity; 2 per vibrational mode) has ⟨E⟩ = (f/2)·kT. Per mole, energy = (f/2)·R·T and constant-volume heat capacity Cv = (f/2)·R. Examples: monatomic gas f=3 → Cv = 3/2 R; linear diatomic (no vibration) f=5 → Cv = 5/2 R. At high enough temperature, vibrational modes add 2 each (kinetic + potential).
Formula
Average energy: ⟨E⟩ = (f/2)·k·T
Molar energy: (f/2)·R·T
Cv = (f/2)·R
$$\langle E \rangle = \frac{f}{2} k_B T, \quad E_{mol} = \frac{f}{2} RT, \quad C_v = \frac{f}{2} R$$How to Use
- Enter the degrees of freedom f.
- Enter the temperature T (K).
- The calculator returns average energy (J and eV), molar energy and Cv.
Case Studies
Diatomic at room temperature
f = 5 (3 trans + 2 rot), T = 300 K.
⟨E⟩ = 2.5×1.38e-23×300 ≈ 1.04×10⁻²⁰ J (≈0.065 eV).
Cv = 2.5×8.314 ≈ 20.8 J/(mol·K).
FAQ
Why doesn't equipartition work at low temperature for vibrations?
Quantum effects freeze out vibrational modes whose spacing exceeds kT; only classical (high-T) behaviour gives the full ½kT per quadratic term. This is why measured Cv of diatomic gases is 5/2 R at room temperature, not 7/2 R.
How many degrees of freedom for a monatomic gas?
Three translational only (no rotation/vibration for a point particle), so f = 3 and Cv = 3/2 R.
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.