Limits of classical equipartition
Explain why classical equipartition fails for some heat capacities and how quantum freezing of modes resolves the discrepancy.
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ContinueClassical counting predicts a contribution from every quadratic energy term. Gas heat capacities show that this rule cannot be applied at every temperature. The missing distinction is between a mode existing and its energy changing appreciably as the sample warms. Here you will use energy gaps to explain freezing without declaring that a frozen mode has no energy.
At a glance
- Prerequisites: equipartition theorem, heat capacity of gases.
- Outcomes: explain the historical mismatch between classical prediction and experiment.
- Key result: low-temperature mode freezing requires quantum energy spacing.
- Common trap: assuming all degrees of freedom are always active at every temperature.
Compare the heat-capacity plateaus
Classical prediction for a diatomic gas with full mode activation:
But experiments at ordinary temperatures often show values closer to
and at lower temperatures can approach (3/2)R.
The missing rotational and vibrational contributions reflect energy spacings that thermal motion cannot readily excite at low T.
Compare the energy gap with thermal energy
Classical equipartition assumes continuous energy and full thermal accessibility. Quantum mechanics gives discrete mode energies.
For a mode with characteristic spacing Δ E:
- If kBT≪Δ E, excited states are rarely occupied.
- That mode contributes little to internal energy change with T.
Physical interpretation:
- Δ E is the typical quantum energy gap between adjacent levels of that mode.
- Heating changes U(T) mainly by populating higher levels. If thermal energy kBT is too small compared with the gap, the population barely changes, so the mode contributes almost nothing to heat capacity.
So heat capacity depends on temperature through mode activation:
- For ordinary molecular gases in their dilute classical regime, translation remains classical over a range in which rotations can be weakly excited.
- Rotational energy spacings are usually smaller than vibrational spacings, so rotational heat capacity becomes appreciable first.
- These are model regimes, not universal activation temperatures. Molecule, density and phase matter; translation cannot be extrapolated as classical all the way to zero temperature.
A frozen mode still exists. A quantum harmonic vibration can retain zero-point energy hν/2; a temperature-independent offset does not contribute to CV = (∂ U/∂ T)V. Freezing means a small temperature-dependent contribution, not zero total energy. Activation is gradual, so a heat capacity between plateaus is not a fractional count of mechanical coordinates. The classical counting rule describes a high-temperature limit for the relevant mode.
Units check:
- Heat capacity remains in J mol⁻¹K⁻¹; what changes is the effective number of active modes.
Physical sanity check:
- In the low-temperature equilibrium limit, heat capacities tend toward zero rather than retaining a classical equipartition value. A real gas may condense before this limit; the dilute classical gas model cannot be extrapolated unchanged to T = 0.
- A smaller heat capacity does not violate energy conservation. It shows the limits of the classical model for thermally accessible energies.
- Distinguish how many modes exist from how strongly they respond to temperature. Use mode counting alongside this lesson.
- Interactions can also change heat capacities; quantum freezing is a distinct effect.
Heat-capacity plateaus from active modes
Worked example 1
Diatomic heat-capacity regimes
Problem
Show full solution
Count thermally active energy terms
Method
Use CV = fR/2 in each plateau regime.Reason
Frozen modes contribute little temperature-dependent energy; an active vibration contributes two quadratic terms.Working
Answer
- Rotation and vibration frozen, translation classical: f = 3 ⇒ CV ≈ (3/2)R.
- Room-temperature range: translation + rotation active ⇒ CV ≈ (5/2)R.
- High T: vibrational mode contributes too ⇒ CV → (7/2)R.
Evaluate the plateau values
Working
Answer
(3/2)R = 12.5, (5/2)R = 20.8, (7/2)R = 29.1 J mol⁻¹K⁻¹.These are limiting plateaus within the gas model, not abrupt transitions or a prediction of 12.5 J mol⁻¹K⁻¹ at absolute zero.
Practice set
Check your understanding 1
Why does a diatomic gas usually gain a vibrational heat-capacity contribution at a higher temperature than its rotational contribution?
Show hint
Compare the characteristic energy gaps with kBT.
Show answer
Vibrational energy spacings are usually much larger than rotational spacings. A higher kBT is needed to populate vibrational excited states appreciably, so their energy changes with temperature later.
Check your understanding 2
What is the molar CV of a diatomic ideal gas in the classical limit with all translational, rotational and vibrational modes active?
Show hint
Show answer
Check your understanding 3
Why does the effective heat capacity decrease as rotational and vibrational modes freeze out on cooling?
Show hint
Heat capacity measures how much energy changes with temperature, not how many modes exist.
Show answer
As kBT falls below a mode’s spacing, its excited-state population changes little with temperature. Its contribution to dU/dT becomes small. Classical translation can still contribute (3/2)R within the dilute gas regime; reaching the absolute-zero limit requires a different description.
Separate stored energy from heat capacity
Optional quantitative model: one harmonic vibration
For a quantum harmonic mode with spacing Δ = hν, the non-degenerate energies are Eⱼ = (j + 1/2)Δ for j = 0,1,…. This spectrum is the harmonic-oscillator model in MIT transport-process lecture notes, section 2.3.4. Summing its Boltzmann weights gives the mean energy ⟨E⟩ = Δ/2 + Δ/(e^(Δ/(kBT))-1). The first term is the temperature-independent zero-point contribution. Differentiating at fixed frequency gives the molar contribution of one mode C_(V,m) = Rx²e^x/((e^x-1)²), x = Δ/kBT. These are supplied model results for optional quantitative practice. For x≫1, thermal excitation and heat capacity are small while mean energy approaches Δ/2. For x≪1, the mean energy approaches kBT and the mode’s molar capacity approaches R, matching the two-term classical vibration. Do not use this single-mode formula as the entire gas heat capacity.
Guided practice 2
Keep zero-point energy separate from heat capacity
Problem
Use the supplied single-mode harmonic model with x = Δ/(kBT) = 5.00. Find ⟨E⟩/Δ and this mode’s molar heat-capacity contribution divided by R. At the same temperature, a second mode has twice the spacing. Find its capacity contribution divided by R. Explain why a larger ground-state energy need not give a larger heat capacity.
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
Which term in the mean energy changes with temperature? How does doubling the spacing change the dimensionless gap?Show solution step by step
Evaluate the two different quantities
Method
Use energy and its temperature derivative separately.Reason
An additive zero-point energy does not contribute to heat capacity.Working
Answer
⟨E⟩/Δ = 1/2 + 1/(e⁵-1) ≈ 0.5068, whereas C_(V,m)/R = 25e⁵/(e⁵-1)² ≈ 0.1707. A low thermal contribution does not mean the oscillator contains no energy.
Change the gap at the same temperature
Method
Use x = 10 for the second mode.Reason
The larger spacing makes thermal excitation harder.Working
Answer
The second capacity contribution is 100e¹⁰/(e¹⁰-1)² ≈ 0.00454. Its ground-state energy is larger, but its temperature response is smaller.
Explain what a frozen vibration retains
Check your understanding 4
For a dilute diatomic gas over a cooling interval, translation and rotation remain classical while the vibrational heat-capacity contribution falls almost to zero. A solution says the molecule has lost its vibrational coordinate and all its vibrational energy, and predicts the total molar capacity must therefore fall to zero. Evaluate the three claims without needing the optional oscillator formula.
Show answer
The vibrational coordinate and energy spectrum still exist. A small excited-state population makes the change of energy with temperature small; a temperature-independent zero-point energy can remain. Within the stipulated classical translation/rotation range, the gas retains approximately 5R/2 molar heat capacity. It is the vibration’s contribution that becomes small, not necessarily the total capacity. Extrapolating this gas model to absolute zero is invalid: other modes, quantum statistics or a phase change can intervene.
What to keep
A mode can exist and retain energy while contributing little heat capacity. Compare its spacing with kBT, and distinguish stored energy from dU/dT. Plateaus are limiting approximations within a stated material and phase model.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.