Skip to lesson content

Lesson 4 of 6

Kinetic Theory · Lesson 4 of 6

Equipartition and Heat Capacities

“Count active quadratic energy terms to explain ideal-gas and solid heat capacities, and recognise the limits of classical predictions.”

Learning Objectives

• Distinguish translational coordinates, rotational modes and vibrational quadratic energy terms. • Apply classical equipartition to thermally active modes without counting one vibration as only one half-kBT term. • Derive molar Cᵥ, Cₚ and γ for monatomic, rigid diatomic and nonlinear polyatomic ideal-gas models. • Interpret predicted and measured heat-capacity tables and explain departures from simple models. • Calculate heat for a rigid helium cylinder and derive the classical solid result C ≈ 3 R.

Why do equal mole samples need different heat inputs?

The previous lesson gave every classical ideal-gas particle the same average translational energy 3 kBT/2 at a fixed temperature. Yet one mole of nitrogen generally needs more heat than one mole of helium for the same temperature rise. The extra energy has places to go: a molecule can rotate and, when thermally excited, vibrate. Heat capacity measures how rapidly total internal energy rises with temperature under the stated constraints.

Degrees of freedom and quadratic energy terms

A free point-like particle requires three coordinates to locate it in space, two if it is constrained to a plane and one if constrained to a line. These are independent translational degrees of freedom. In three dimensions its translational kinetic energy is½m₀vx² +½m₀vy² +½m₀vz². Each term contains the square of one independent velocity component; this is a quadratic energy term.

The three translational energy termsLaTeX
At equilibrium isotropy divides the average translational energy equally. These are averages; instantaneous terms fluctuate.

Classical equipartition generalises this result: each independent thermally active quadratic term in the energy contributes an average½kBT at equilibrium. The theorem itself is stated here rather than proved. “Equal sharing” concerns ensemble averages of energy terms, not identical energy in each molecule at every instant. Its classical use assumes that the relevant modes can actually be thermally excited at the temperature of interest.

A rigid diatomic molecule adds two active rotations

Picture a diatomic molecule as two atoms joined by a bond. It translates along three directions. It can also rotate about two independent axes through its centre of mass perpendicular to the bond. Their kinetic energies are½I₁ω₁² and½I₂ω₂², each adding an average½kBT. In the usual moderate-temperature rigid-diatomic model, rotation about the bond axis is not counted as an active mode: its very small moment of inertia makes the relevant quantum excitation difficult. It is not enough simply to count every imaginable rotation without considering thermal activation.

Rigid-diatomic energyLaTeX
The five active quadratic terms are three translational and two rotational. A rigid molecule has no active bond vibration in this approximation.
A diatomic molecule: two active rotations and one bond vibrationy axisRotation about yRotation about z (out of page)Both rotation axes are perpendicular to the bond.One stretch: kinetic term + potential term = kBT on averageRotation about the bond is not thermally active in this simple model.
Two rotational modes and the vibrational contribution— The illustrated axes are perpendicular to the bond. The lower stretch is one vibrational coordinate, but it stores both kinetic and potential energy. The diagram is schematic.

One vibration stores two half-kBT contributions

A molecule such as CO has a vibrational coordinate, but whether it is thermally active is a separate question. If the bond length can oscillate, let y measure its displacement from equilibrium. Its vibrational energy has a kinetic term½μ(dy/dt)² and a restoring-force potential term½ky². Here μ is the effective mass for that relative vibration and k is a force constant; μ in this formula is not the mole number. The derivative dy/dt means the rate of change of the bond displacement. Both terms are quadratic, so a fully thermally active vibrational mode contributes kBT in total, not½kBT.

Classical energy of one vibrational modeLaTeX
One normal vibration gives two quadratic energy terms. Full classical excitation is an assumption; a physical bond’s ability to vibrate does not imply full equipartition at room temperature.

At sufficiently low temperature, rotational or vibrational energy levels may be too widely spaced for the corresponding modes to absorb classical equipartition amounts of energy. Vibrations often require higher temperatures than rotations. Thus a diatomic molecule can be well approximated as rigid at room temperature even though it has a bond that can vibrate. Adding every possible vibrational mode with its full kBT regardless of temperature would overpredict heat capacity. The detailed quantum calculation is beyond this chapter.

From internal energy to molar heat capacities

For one mole of ideal gas, internal energy Uₘ depends only on temperature for fixed composition. At constant volume and with no other work modes, Q = ΔU, so molar Cᵥ = dUₘ/dT. The derivative describes energy increase per kelvin; if Uₘ = aRT with constant a, its derivative is aR because dT/dT = 1. A temperature-dependent energy Uₘ is not itself a heat capacity.

At constant pressure, heat also supplies expansion work. For one mole, QP = ΔU + PΔV, while PV = RT gives PΔV = RΔT. Dividing by ΔT gives Cₚ = Cᵥ + R. The ratio γ = Cₚ/Cᵥ is dimensionless. Both the difference and the constant-volume definition are molar statements here; mass-specific capacities use different units and a gas-specific constant R/M.

Count active quadratic termsLaTeX
f counts independent thermally active quadratic energy terms per molecule, with the active set approximately unchanged over the temperature interval. Each fully excited vibration counts twice in f.

Monatomic and diatomic predictions

A monatomic gas such as helium or argon has three relevant translational terms, giving Uₘ =3 RT/2. Its Cᵥ is 3 R/2, Cₚ is 5 R/2 and γ is 5/3 ≈1.67. In this simple model no thermally active molecular bond rotations or vibrations are present. The derivative removes T: writing Cᵥ =3 RT/2 would have energy-per-mole units and would be incorrect.

A rigid diatomic gas has five terms: three translation and two rotation. Therefore Uₘ =5 RT/2, Cᵥ =5 R/2, Cₚ =7 R/2 and γ =7/5 =1.40. If one vibrational mode becomes fully active, its two extra quadratic terms raise f to 7. Then Uₘ =7 RT/2, Cᵥ =7 R/2, Cₚ =9 R/2 and γ =9/7. Real transitions between these idealised limits are gradual and depend on the molecule and temperature.

Common ideal-gas heat-capacity limitsLaTeX
All capacities are molar, in J mol⁻¹ K⁻¹. The vibrational row assumes one fully active vibrational mode.

Polyatomic molecules: geometry matters

A nonlinear polyatomic molecule has three active translational and three active rotational terms in the classical rigid model. If s vibrational modes are fully active, each adds kBT, so Uₘ = (3+s) RT, Cᵥ = (3+s) R, Cₚ = (4+s) R and γ = (4+s)/(3+s). Here s counts vibrations while f in the general formula counts quadratic terms; f =6+2 s. Using different symbols avoids confusing a vibration with one energy term.

Nonlinear polyatomic predictionLaTeX
Assumes a nonlinear molecule with three active rotations and s fully active vibrational modes. Linear polyatomic molecules have only two active rigid rotations and must be counted separately.

The distinction between linear and nonlinear is essential. A rigid nonlinear triatomic molecule has Cᵥ ≈3 R and γ≈4/3. A rigid linear triatomic molecule instead has three translational plus two active rotational terms, like the rigid diatomic count before vibrations are added. Therefore “triatomic” alone does not guarantee the 3 R prediction. The prediction below is labelled by its geometry rather than applied indiscriminately to every three-atom molecule.

Model without active vibrationsCᵥ (J mol⁻¹ K⁻¹)Cₚ (J mol⁻¹ K⁻¹)Cₚ−Cᵥγ
Monatomic12.520.8R ≈8.311.67
Rigid diatomic20.829.1R ≈8.311.40
Rigid nonlinear polyatomic24.9333.24R ≈8.311.33

Read heat-capacity measurements as a test of the model

GasCᵥ (J mol⁻¹ K⁻¹)Cₚ (J mol⁻¹ K⁻¹)Reported Cₚ−CᵥReported γ
He12.520.88.301.66
Ne12.720.88.121.64
Ar12.520.88.301.67
H₂20.428.88.451.41
O₂21.029.38.321.40
N₂20.829.18.321.40
H₂O vapour27.035.48.351.31
CH₄27.135.48.361.31

The measured monatomic values cluster around 3 R/2 and 5 R/2, and the common diatomic values around 5 R/2 and 7 R/2. Water vapour and methane are somewhat above the simplest rigid nonlinear prediction, consistent with additional internal energy storage. The capacity entries are rounded, and the difference column is separately reported, so subtracting the displayed capacity entries need not reproduce every final digit. Small real-gas deviations can also occur. Treat the agreement as evidence for an approximate model, not an assertion that every measured row must be exact at all temperatures.

Other gases, including chlorine and ethane, can show larger capacities than the simplest vibration-free model. Additional thermally excited vibrations can account for more energy storage and a lower γ. Both molecule type and temperature matter; one cannot repair every discrepancy merely by counting every possible vibration as fully active. The comparison motivates checking which modes actually participate.

Heating helium in a fixed-capacity cylinder

Problem
A rigid cylinder of volume 44.8 L contains helium at textbook STP. Find the heat required for a 15.0 K rise, using ideal molar volume 22.4 L mol⁻¹ and R =8.31 J mol⁻¹ K⁻¹.

  1. 1.The gas amount isν =44.8/22.4 =2.00 mol.
  2. 2.Helium is monatomic, so molar Cᵥ =3 R/2. The rigid cylinder prevents boundary-volume work; with no other work, Q =ΔU.
  3. 3.Q =νCᵥΔT = (2.00)(1.5)(8.31)(15.0) =373.95 J ≈374 J.
  4. 4.Cₚ would be the wrong heat capacity for this rigid-cylinder process. It would include expansion work that does not occur here.

A solid: three vibrations per atom

An atom in a classical solid vibrates about its equilibrium position in three independent directions. In each direction there is a kinetic and a restoring-force potential term, together averaging kBT. Thus one atom has average vibrational energy 3 kBT, and one mole has Uₘ≈3 RT. In an ordinary solid at atmospheric pressure, volume expansion is small, so PΔV is often negligible compared with ΔU. Heat supplied for a small temperature change is then approximately its internal-energy increase.

Classical solid heat-capacity estimateLaTeX
This is a classical approximation where vibrational modes are thermally active and expansion work is negligible. It is not a universal exact solid heat capacity.
SolidSpecific heat (J kg⁻¹ K⁻¹)Molar heat (J mol⁻¹ K⁻¹)
Aluminium900.024.4
Carbon506.56.1
Copper386.424.5
Lead127.726.5
Silver236.125.5
Tungsten134.424.9

The molar values for most listed solids are near 24.9 J mol⁻¹ K⁻¹, despite large differences per kilogram. Different elements have different numbers of atoms per kilogram, making molar comparisons useful. Carbon is a strong room-temperature exception. At low temperatures the classical prediction also fails broadly because not all vibrational modes absorb classical equipartition energy. The measured table therefore illustrates both the usefulness and the limits of the 3 R estimate.

Quick Quiz

Quiz

Quick check

One fully thermally active harmonic vibrational mode contributes average energy

Explanation

It contains two quadratic terms, kinetic and potential, each contributing½kBT.

Quick check

The monatomic ideal-gas molar Cᵥ is

Explanation

For one mole Um =3 RT/2, so differentiating with respect to T gives 3 R/2, without an extra T.

Quick check

For a rigid diatomic ideal gas with active translation and rotation, γ is

Explanation

Five active quadratic terms give Cᵥ =5 R/2 and Cₚ =7 R/2.

Quick check

Which qualification is needed for the rigid polyatomic Cᵥ =3 R model?

Explanation

A linear molecule has only two active rigid rotations, so geometry and thermal activation must be specified.

Quick check

For 2 mol helium in a rigid cylinder heated by 15 K, Q equals

Explanation

Rigid heating has no boundary work, so Q =νCᵥΔT and helium uses Cᵥ =3 R/2.

Practice Problems

Practice Problems
  1. List the quadratic energy terms for a rigid diatomic molecule in three-dimensional space. Explain why a vibrational mode adds two terms while each active rigid rotation adds one.
  2. Derive Cᵥ, Cₚ and γ for a monatomic ideal gas starting from classical equipartition. Explain why Cᵥ contains R but not RT.
  3. For one mole of rigid diatomic ideal gas warmed from 300 K to 320 K, find ΔU, Q and W at (a) constant volume and (b) constant pressure, using R =8.314 J mol⁻¹ K⁻¹.
  4. A nonlinear ideal-gas molecule has three active translations, three active rotations and two fully active vibrations. Find average energy per molecule, molar Cᵥ, molar Cₚ and γ. State the thermal-activation assumptions.
  5. Explain why the label “triatomic” is insufficient to choose a heat capacity without knowing molecular geometry and which modes are thermally active.
  6. A rigid 44.8 L helium vessel at textbook STP is heated by 30 K. Find heat input using 22.4 L mol⁻¹ and R =8.31 J mol⁻¹ K⁻¹. Explain why using Cₚ gives a different answer.
  7. Estimate the heat required to raise 2.0 mol of a classical solid by 10 K using C≈3 R and R =8.314 J mol⁻¹ K⁻¹. Explain the six quadratic terms per atom and state why this estimate may fail for carbon or at low temperature.
  8. Compare the measured He, N₂ and water-vapour rows with the vibration-free predictions. Explain what the comparison supports, why rounding affects column subtraction and why every possible vibration should not automatically be assigned its full classical energy.

Key Takeaways

Key Takeaways

• Classical equipartition gives½kBT per independent thermally active quadratic energy term. • One vibration has both kinetic and potential terms and contributes kBT when fully excited. • Molar Cᵥ is dUm/dT; for an ideal gas Cₚ−Cᵥ =R. • Molecular geometry and thermal activation determine the appropriate degrees-of-freedom count. • The solid estimate C≈3 R and simple gas capacities are useful models with measurable limits.