Worked examples — Equipartition theorem — ½k_BT per quadratic degree of freedom

Everything below uses only two ideas you already have:

• Each quadratic energy slot (energy of the form $\alpha q^2$) carries $\frac{1}{2}{k}_{B}T$ on average.
• $C_V={\left(\frac{\partial U}{\partial T}\right)}_V$, molar $U$ uses $N_A{k}_B=R$, and $\gamma =C_P/C_V$ with $C_P=C_V+R$ for an ideal gas.

Constants I will reuse: $k_B=1.380649\times 10^{-23}\text{J/K}$, $R=8.314\text{J/(mol⋅K)}$, $N_A=6.022\times 10^{23}$.

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The scenario matrix

Every equipartition problem is one of these cells. The examples below each state which cell(s) they hit.

Cell	What varies	Danger it tests	Example
A — clean count	ordinary gas, all modes active	can you count T-R-V correctly?	Ex 1, Ex 2
B — vibration doubles	high-T diatomic	remembering KE and PE	Ex 3
C — degenerate/zero	a mode with no energy cost (frozen)	contributing nothing	Ex 4
D — solid (Dulong–Petit)	6 slots per atom	3 KE + 3 PE	Ex 5
E — real-world word problem	Brownian speck, actual numbers	plugging into SI, units	Ex 6
F — non-quadratic energy	$E\propto q^4$	the general $\frac{1}{n}{k}_{B}T$ rule	Ex 7
G — exam twist / mixture	gas mixture, derive $\gamma$	averaging over species	Ex 8

We hit all seven cells in eight examples.

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Example 1 — Cell A: monatomic gas energy (clean count)

Step 1 — Count the quadratic slots. A helium atom is a point mass: it can only move in $x,y,z$. Energy $=\frac{1}{2}m{v}_x^2+\frac{1}{2}m{v}_y^2+\frac{1}{2}m{v}_z^2$ = 3 quadratic slots. Why this step? Equipartition is a counting rule — nothing happens until we know how many slots there are.

Step 2 — Energy per atom. $⟨E⟩=3\times \frac{1}{2}{k}_{B}T=\frac{3}{2}{k}_{B}T=\frac{3}{2}(1.380649\times 10^{-23})(300)=6.21\times 10^{-21}\text{J}.$ Why this step? Each slot pays $\frac{1}{2}{k}_{B}T$; three slots pay $\frac{3}{2}{k}_{B}T$.

Step 3 — Total internal energy. $U=n\cdot \frac{3}{2}RT=2.0\times \frac{3}{2}\times 8.314\times 300=7483\text{J}.$ Why this step? Multiply per-mole energy $\frac{3}{2}RT$ by the number of moles; $N_A{k}_B=R$ converts "per atom" to "per mole".

Verify: Sanity — $U=n\cdot \frac{3}{2}RT$ should also equal $N\cdot \frac{3}{2}{k}_{B}T$ where $N=n{N}_A$. Check: $2\times 6.022\times 10^{23}\times 6.21\times 10^{-21}\approx 7.48\times 10^3\text{J}$. ✓ Same number. Forecast answer: $\approx 7500\text{J}$.

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Example 2 — Cell A: diatomic gas at room T (frozen 3rd rotation)

Step 1 — Which modes are active? Translation: 3 slots. Rotation: the molecule is a dumbbell, so it tumbles about 2 axes perpendicular to the bond (rotation about the bond axis has almost no moment of inertia — see Quantum freezing of degrees of freedom — so it is frozen). Vibration: frozen at $300\text{K}$. Total $=3+2=5$ slots.

Step 2 — Heat capacities. $C_V=\frac{5}{2}R=20.79\text{J/(mol⋅K)},C_P=C_V+R=\frac{7}{2}R=29.10\text{J/(mol⋅K)}.$ Why this step? $U=\frac{5}{2}RT$ so $C_V=\partial U/\partial T=\frac{5}{2}R$; for an ideal gas $C_P-C_V=R$ (see Heat capacity of gases).

Step 3 — Adiabatic index. $\gamma =\frac{C_P}{C_V}=\frac{7/2}{5/2}=\frac{7}{5}=\mathrm{1.40.}$

Verify: measured $\gamma$ for air (mostly ${\text{N}}_2$, ${\text{O}}_2$) is $\approx 1.40$. ✓ Forecast: $1.40$.

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Example 3 — Cell B: same gas, high temperature (vibration doubles)

Step 1 — Add the vibrational slots. A vibrating bond is a spring: it stores kinetic energy $\frac{1}{2}\mu {\dot{r}}^2$ and potential energy $\frac{1}{2}\kappa r^2$. That is 2 new quadratic slots, not one. Why this step? This is the classic trap — equipartition counts energy terms, not motions. "Vibes are doubled."

Step 2 — New count and capacities. $3(\text{trans})+2(\text{rot})+2(\text{vib})=7\text{slots}.$ $C_V=\frac{7}{2}R=29.10\text{J/(mol⋅K)},\gamma =\frac{C_P}{C_V}=\frac{9/2}{7/2}=\frac{9}{7}\approx \mathrm{1.286.}$

Verify: $C_V$ rose from $\frac{5}{2}R$ to $\frac{7}{2}R$, i.e. by exactly $R$ (two half-slots), not $\frac{1}{2}R$. ✓ And $\gamma$ dropped from $1.40$ toward $1$, as adding modes always lowers $\gamma$. ✓ Forecast: jump by $R$.

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Example 4 — Cell C: the degenerate mode (a slot that pays nothing)

Step 1 — Look at the moment of inertia about the bond axis. Spinning about the bond axis means spinning the two tiny atomic nuclei on the line itself. The moment of inertia $I_{\parallel }$ is essentially zero (the mass sits on the axis). Why this step? Rotational energy is $\frac{1}{2}I{\omega }^2$; if $I\to 0$ the classical energy of that mode is also tiny — but the honest reason it's frozen is quantum.

Step 2 — Why it truly carries zero. Rotational energy levels are spaced like $\Delta E\sim {\hslash }^2/(2I)$. A vanishing $I$ makes the gap $\Delta E\gg k_{B}T$, so the mode cannot be excited (see Quantum freezing of degrees of freedom). A frozen slot contributes exactly $0$, not $\frac{1}{2}{k}_{B}T$. $⟨E_{\text{3rd rot}}⟩=0.$

Step 3 — Correct count stands. So the room-temperature count is genuinely $3+2=5$ and $\gamma =1.40$, not $1.33$.

Verify: the student's $\gamma =(6/2+1)/(6/2)=4/3\approx 1.333$ is exactly what you'd get IF the mode were active — and experiment says $1.40$, ruling it out. ✓ The degenerate/zero cell gives $0$.

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Example 5 — Cell D: classical solid (Dulong–Petit)

Step 1 — Count slots per atom in a solid. Each atom sits in a 3D harmonic well, bonded to neighbours. In each direction $x,y,z$ it has kinetic energy $\frac{1}{2}m{\dot{x}}^2$ and potential energy $\frac{1}{2}\kappa x^2$. That is $3\times 2=6$ slots per atom. Why this step? A solid atom is like 3 independent springs — each spring, like a vibration, is doubled.

Step 2 — Energy and heat capacity. $⟨E⟩=6\times \frac{1}{2}{k}_{B}T=3k_{B}T⟹U=3RT⟹C_V=3R=24.94\text{J/(mol⋅K)}.$ This is the Dulong–Petit law.

Verify: predicted $3R=24.94$; measured $\approx 24.4$. Agreement within $2\%$. ✓ Forecast: $3R$.

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Example 6 — Cell E: Brownian speck (real-world word problem)

Step 1 — Apply equipartition to the speck's translation in one direction. Even a big object is "one quadratic slot" per direction: $\frac{1}{2}m{v}_x^2$. Equipartition doesn't care how heavy — it still pays $\frac{1}{2}{k}_{B}T$. $\frac{1}{2}m⟨v_x^2⟩=\frac{1}{2}{k}_{B}T.$ Why this step? This is exactly Cell A logic, but applied to a visible particle — the beauty of the mass-independence proved in the parent note.

Step 2 — Solve for the mean-square speed. $⟨v_x^2⟩=\frac{k_{B}T}{m}=\frac{(1.380649\times 10^{-23})(293)}{1.0\times 10^{-15}}=4.045\times 10^{-6}{\text{m}}^2/{\text{s}}^2.$

Step 3 — Take the square root. $v_{\text{rms}}=\sqrt{⟨v_x^2⟩}=2.01\times 10^{-3}\text{m/s}\approx 2\text{mm/s}.$ Why this step? "rms" means root-mean-square — undo the squaring to get a speed.

Verify: units: $\sqrt{\text{J/kg}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$ ✓. Compare with an air molecule ($m\sim 5\times 10^{-26}\text{kg}$) at the same $T$: it would move $\sim 500\text{m/s}$. The speck is $2\times 10^{10}$ times heavier, so slower by $\sqrt{2\times 10^{10}}\approx 1.4\times 10^5$, giving $\sim 3\text{mm/s}$ — same ballpark. ✓ Forecast: millimetres/s.

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Example 7 — Cell F: non-quadratic energy ($E\propto q^4$)

Step 1 — Recall the general theorem. The parent note showed: for $E=\alpha |q{|}^n$, the same $-\frac{d\ln Z}{d\beta }$ machinery gives $⟨E⟩=\frac{1}{n}{k}_{B}T.$ Why this step? The $\frac{1}{2}$ is not universal — it comes from the Gaussian integral, which needs $n=2$. Change the power, change the fraction.

Step 2 — Plug in $n=4$. $⟨E⟩=\frac{1}{4}{k}_{B}T.$

Step 3 — Why smaller than a spring. A quartic well has steeper walls than a parabola, confining $q$ more tightly; the trapped energy per mode is therefore less. The general rule quantifies exactly how much: $1/n$.

Verify: the derivation sketch — sub $u={\beta }^{1/n}q$ into $Z=\int e^{-\beta \alpha q^4}dq$ gives $Z\propto {\beta }^{-1/4}$, so $\ln Z=\text{const}-\frac{1}{4}\ln \beta$ and $-d\ln Z/d\beta =\frac{1}{4\beta }=\frac{1}{4}{k}_{B}T$. ✓ Forecast: $\frac{1}{4}{k}_{B}T$.

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Example 8 — Cell G: gas mixture (exam twist)

Step 1 — Get each species' $C_V$. Helium: $C_{V,\text{He}}=\frac{3}{2}R$. Nitrogen: $C_{V,{\text{N}}_2}=\frac{5}{2}R$ (Ex 2). Why this step? Each gas obeys equipartition independently; they share the same $T$ but count their own slots.

Step 2 — Mole-weighted average $C_V$. Heat capacity is extensive — add total heat capacities, then divide by total moles: $C_{V,\text{mix}}=\frac{n_{\text{He}}{C}_{V,\text{He}}+n_{{\text{N}}_2}{C}_{V,{\text{N}}_2}}{n_{\text{He}}+n_{{\text{N}}_2}}=\frac{1\cdot \frac{3}{2}R+3\cdot \frac{5}{2}R}{4}=\frac{\frac{3}{2}+\frac{15}{2}}{4}R=\frac{9}{4}R.$ So $C_{V,\text{mix}}=2.25R=18.71\text{J/(mol⋅K)}$.

Step 3 — Mixture $\gamma$. For an ideal-gas mixture $C_{P,\text{mix}}=C_{V,\text{mix}}+R=\frac{13}{4}R$, so ${\gamma }_{\text{mix}}=\frac{C_{P,\text{mix}}}{C_{V,\text{mix}}}=\frac{13/4}{9/4}=\frac{13}{9}\approx \mathrm{1.444.}$

Verify: ${\gamma }_{\text{mix}}=1.444$ lies between ${\gamma }_{{\text{N}}_2}=1.40$ and ${\gamma }_{\text{He}}=1.667$, as any weighted mix must. ✓ Forecast: inside the range.

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Recall Which cell was hardest for you?

Non-quadratic ($E=\alpha q^4$) gives $\frac{1}{4}{k}_{B}T$, not $\frac{1}{2}{k}_{B}T$ ::: because the $\frac{1}{2}$ comes only from the Gaussian ($n=2$) integral; general rule $\frac{1}{n}{k}_{B}T$. A vibrating bond adds how much to $C_V$? ::: $R$ (two slots: KE + PE), not $\frac{1}{2}R$. The third rotation of a diatomic contributes ::: $0$ — it is frozen out by the quantum gap. Mixture $C_V$ is combined by ::: mole-weighted average of each species' $C_V$.

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Related: Boltzmann distribution · Partition function Z · Maxwell–Boltzmann speed distribution · Ideal gas law PV=NkT