Boltzmann's entropy S = k_B ln(Ω): Worked examples

Intuition What this page is

The parent note built $S = k_{B} ln Ω$ from scratch. Here we stress-test it: we throw every kind of situation at the formula — tiny discrete systems, huge continuous ones, the weird cases where entropy shrinks, and the exam trick that catches everyone. If a scenario exists, there is a worked cell for it below.

Reminders of symbols we will keep using (all defined in the parent):

$Ω$ = number of microstates consistent with the macrostate (a plain count of "ways").
$S = k_{B} ln Ω$ = entropy; $k_{B} = 1.380649 \times 1 0^{- 23} J/K$ .
$ln$ = natural logarithm — the function that turns products into sums (the whole reason it appears).
$N$ = number of particles; $n$ = number of moles.
$N_{A}$ = Avogadro's number $= 6.022 \times 1 0^{23}$ particles per mole — the count of particles in one mole; it converts a per-particle result into a per-mole one.
$R = N_{A} k_{B} = 8.314 J/(mol\cdotK)$ = the gas constant (Boltzmann's constant scaled up to per mole instead of per particle).
$T$ = absolute temperature in kelvin (K). $Q$ = heat, the energy transferred thermally (in joules, J).
$h$ = Planck's constant $= 6.626 \times 1 0^{- 34} J\cdots$ — the tiny "grain size" of phase space. We meet it in Example 10; it fixes how big one microstate is when position and momentum are continuous.

The scenario matrix

Every problem about $S = k_{B} ln Ω$ falls into one of these cells. The examples below each carry a tag telling you which cell they discharge.

Cell	What makes it distinct	Danger it hides	Example
A. Discrete counting	Count microstates by hand	Confusing micro/macro	Ex 1
B. Ratio of counts	Only $Ω_{f} / Ω_{i}$ matters	Trying to find absolute $Ω$	Ex 2
C. Additivity check	Join two systems	Adding $Ω$ instead of multiplying	Ex 3
D. Degenerate: $Ω = 1$	One single microstate	Thinking $ln 1 \neq = 0$	Ex 4
E. Large- $N$ / Stirling	$N$ astronomically big	Cannot list states	Ex 5
F. Real-world word problem	Physical melting/mixing	Forgetting what counts	Ex 6
G. Sign flip: $S$ decreases	Capped energy, $\partial S / \partial E < 0$	"More energy ⇒ more $S$ "	Ex 7
H. Exam twist: identical particles	$N!$ overcount	Gibbs paradox	Ex 8
I. Limiting behaviour	$E \to 0$ , $N \to \infty$ , $T \to 0$	Ill-defined limits	Ex 9
J. Continuous phase space	Integrals, need $h^{3 N}$	Non-dimensionless $Ω$	Ex 10

Example 1 — Cell A: Discrete counting from scratch

Worked example Four coins, macrostate = "number of heads"

Four fair coins. The macrostate you can see is "how many heads." Find $Ω$ and $S$ for the macrostate "exactly 2 heads." Which macrostate has the largest entropy?

Forecast: guess now — is "2 heads" more or less likely than "4 heads"? By how much?

Steps

List microstates for 2 heads: ${H H T T, H T H T, H T T H, T H H T, T H T H, T T H H}$ → $Ω = 6$ . Why this step? A microstate specifies each coin; the macrostate only counts heads. Counting is the raw meaning of $Ω$ .
Use the combination formula $Ω = (k N) = \frac{N !}{k ! ( N - k )!} = \frac{4 !}{2 ! 2 !} = 6$ . Why this step? Listing works for 4 coins but not 40 — the binomial coefficient is the count of ways to choose which positions are heads.
Entropy of that macrostate: $S = k_{B} ln 6 = 1.792 k_{B}$ . Why this step? Direct substitution into $S = k_{B} ln Ω$ .
Compare "4 heads": $Ω = (4 4) = 1$ , so $S = k_{B} ln 1 = 0$ . Why this step? Shows the extreme, ordered macrostate has zero entropy — it's the tidy-room case.

Verify: the total over all macrostates is $\sum_{k} (k 4) = 1 + 4 + 6 + 4 + 1 = 16 = 2^{4}$ ✓ (every one of the $2^{4}$ raw microstates is counted exactly once). "2 heads" ( $Ω = 6$ ) beats "4 heads" ( $Ω = 1$ ) — the balanced macrostate has the most ways, hence highest $S$ .

Example 2 — Cell B: Only the ratio survives

Worked example Free expansion to triple the volume

A gas of $N = 3.0 \times 1 0^{22}$ molecules expands freely (no heat, no work) so its accessible volume triples. Find $Δ S$ .

Forecast: does $Δ S$ depend on the absolute number of microstates, or only on how they changed?

Steps

Position-choices per molecule scale with volume: tripling $V$ multiplies each molecule's position options by 3, so $Ω_{f} = 3^{N} Ω_{i}$ . Why this step? Each molecule chooses independently → independent choices multiply (the dice logic).
Entropy change: $Δ S = k_{B} ln Ω_{f} - k_{B} ln Ω_{i} = k_{B} ln \frac{Ω _{f}}{Ω _{i}} = k_{B} ln 3^{N} = N k_{B} ln 3$ . Why this step? The unknown absolute $Ω_{i}$ cancels — the log of a ratio only needs the ratio. This is why free expansion is tractable.
Numbers: $Δ S = (3.0 \times 1 0^{22}) (1.380649 \times 1 0^{- 23}) (ln 3) = 0.455 J/K$ . Why this step? Plug in $ln 3 = 1.0986$ .

Verify: thermodynamics gives $Δ S = n R ln (V_{f} / V_{i})$ . Here $n = N / N_{A} = 0.0498$ mol, so $n R ln 3 = 0.0498 \times 8.314 \times 1.0986 = 0.455 J/K$ ✓ — the counting picture reproduces classical thermo exactly.

Example 3 — Cell C: Additivity (multiply counts, add entropies)

Worked example Joining two boxes

Box $A$ has $Ω_{A} = 2^{10}$ microstates; box $B$ has $Ω_{B} = 2^{15}$ . They are independent. Find the combined entropy and confirm it equals $S_{A} + S_{B}$ .

Forecast: is $Ω_{A B}$ equal to $2^{10} + 2^{15}$ or $2^{10} \times 2^{15}$ ?

Steps

Combined count: $Ω_{A B} = Ω_{A} Ω_{B} = 2^{10} \cdot 2^{15} = 2^{25}$ . Why this step? For each of $A$ 's states, $B$ can be in any of its states → multiply, never add.
Combined entropy: $S_{A B} = k_{B} ln 2^{25} = 25 k_{B} ln 2$ . Why this step? Straight substitution.
Separate entropies: $S_{A} = k_{B} ln 2^{10} = 10 k_{B} ln 2$ , $S_{B} = 15 k_{B} ln 2$ ; sum $= 25 k_{B} ln 2$ . Why this step? This is the payoff: $ln$ turned the product $2^{25}$ into the sum $10 + 15$ .

Verify: $S_{A} + S_{B} = 25 k_{B} ln 2 = S_{A B}$ ✓. Additivity holds precisely because of the logarithm — the wrong "add the counts" rule would have given the nonsensical $ln (2^{10} + 2^{15})$ .

Example 4 — Cell D: The degenerate $Ω = 1$ case

Example 5 — Cell E: Large- $N$ , Stirling to the rescue

Worked example Two-state spins, most-probable macrostate

$N = 1 0^{24}$ non-interacting spins, each up or down. Find $S$ of the macrostate "exactly half up" and confirm it is close to the maximum $N k_{B} ln 2$ .

Forecast: with $1 0^{24}$ spins you cannot list states. What tool converts the giant factorial into something usable?

Steps

Count: $Ω = (N /2 N) = \frac{N !}{( N /2 )! ( N /2 )!}$ . Why this step? Same "choose which spins are up" logic as the coins in Ex 1, now enormous.
Take the log and apply Stirling's approximation $ln M! \approx M ln M - M$ (valid for huge $M$ ). Why this step? Factorials of $1 0^{24}$ are uncomputable; Stirling turns $ln (N!)$ into simple arithmetic. This is why statistical mechanics needs it — it's the bridge from counting to smooth thermodynamics.
$ln Ω \approx N ln N - N - 2 [\frac{N}{2} ln \frac{N}{2} - \frac{N}{2}] = N ln N - N ln \frac{N}{2} = N ln 2$ . Why this step? The $- N$ terms cancel; $ln N - ln (N /2) = ln 2$ .
So $S = k_{B} ln Ω \approx N k_{B} ln 2$ . Why this step? This equals the entropy of $N$ free two-state spins — meaning the single "half-up" macrostate contains essentially all the microstates.

Verify (small- $N$ sanity): for $N = 4$ , $(2 4) = 6$ and $4 ln 2 \approx 2.77$ gives $e^{2.77} \approx 16 = 2^{4}$ ; Stirling overshoots slightly for small $N$ (6 vs 16) but the fraction $6/16$ approaches 1's log-scale dominance as $N$ grows ✓. The approximation becomes exact-in-the-log for macroscopic $N$ .

Example 6 — Cell F: Real-world word problem (mixing / melting)

Worked example Melting one mole of ice at

$0^{\circ}$ C Ice melts to water at $T = 273.15$ K with latent heat $L_{f} = 6.01 kJ/mol$ . (a) Find $Δ S$ from thermodynamics. (b) Interpret it as a change in the microstate count $Ω_{water} / Ω_{ice}$ per molecule.

Forecast: liquid molecules wander; crystal molecules are pinned. Should $Ω$ go up or down on melting?

Steps

Reversible melting at constant $T$ : the heat $Q$ (energy absorbed thermally, in joules) equals the latent heat $L_{f}$ , so $Δ S = \frac{Q}{T} = \frac{L _{f}}{T} = \frac{6010}{273.15} = 22.0 J/K$ (per mole). Why this step? At a phase transition the heat enters reversibly at fixed temperature $T$ , so $Δ S = Q / T$ directly (this $Q / T$ is the thermodynamic definition of entropy change).
Connect to counting: $Δ S = k_{B} ln \frac{Ω _{water}}{Ω _{ice}}$ . Per molecule, divide by $N_{A}$ (the number of molecules in one mole): $\frac{Δ S}{N _{A}} = k_{B} ln r$ where $r$ is the per-molecule multiplicity ratio. Why this step? Boltzmann's formula reads the macroscopic $Δ S$ as the log of how many more ways a molecule can be arranged.
Solve for $r$ : $ln r = \frac{Δ S}{N _{A} k _{B}} = \frac{Δ S}{R} = \frac{22.0}{8.314} = 2.646$ , so $r = e^{2.646} = 14.1$ . Why this step? Since $R = N_{A} k_{B}$ , the per-mole $Δ S$ divided by $R$ is the per-molecule $ln r$ . Each molecule gains roughly a factor $\sim 14$ in accessible arrangements on melting — a concrete, feel-able number.

Verify: rebuild $Δ S$ from $r$ : $R ln 14.1 = 8.314 \times 2.646 = 22.0 J/K$ ✓. The abstract count and the measured latent-heat entropy agree.

Example 7 — Cell G: The sign flip (entropy decreases with energy)

Worked example Two-level spins where more energy means fewer ways

$N$ spins, each energy $0$ (down) or $+ ε$ (up). Let $m$ = number of "up" spins, so total energy $E = m ε$ . Show that as $E$ passes half its maximum, $\partial S / \partial E$ turns negative — a negative temperature.

Forecast: we always heat things to raise entropy. Can adding energy ever lower $S$ ?

Steps

Count: $Ω (m) = (m N)$ , so $S = k_{B} ln (m N)$ . Why this step? Fixing $E$ fixes $m$ ; the multiplicity is just "choose which $m$ spins are up."
$Ω$ is largest at $m = N /2$ (half up), and shrinks as $m \to N$ (all up). Why this step? At $m = N$ there is only one way (all up), $Ω = 1$ , $S = 0$ — the energy cap forces order.
Since $E = m ε$ increases with $m$ but $S$ decreases for $m > N /2$ , we get $(\frac{\partial S}{\partial E}) < 0$ . Why this step? $S$ tracks $ln Ω$ , not $E$ . Once past the peak, more energy means fewer arrangements.
Then $\frac{1}{T} = \frac{\partial S}{\partial E} < 0 \Rightarrow T < 0$ — a negative temperature, hotter than any positive $T$ . Why this step? This is exactly the Negative Temperature & Spin Systems regime the parent's fourth mistake-warning flagged.

Verify (numbers, $N = 4$ ): $Ω$ vs $m$ : ${1, 4, 6, 4, 1}$ . Going $m : 2 \to 3$ , $Ω : 6 \to 4$ falls, so $Δ S = k_{B} ln (4/6) = - 0.405 k_{B} < 0$ while $Δ E = + ε > 0$ ✓. Sign of $\partial S / \partial E$ is negative — confirmed.

Example 8 — Cell H: Exam twist — indistinguishable particles (Gibbs)

Worked example Mixing two identical gases — the paradox

Two boxes, each with the same gas at the same $T, P$ , are joined by removing a partition. Naïve counting predicts an "entropy of mixing" $Δ S = 2 N k_{B} ln 2$ . What is the correct answer, and why?

Forecast: nothing physical changes when you mix a gas with an identical gas. Should $Δ S$ be positive or zero?

Steps

Naïve (distinguishable) count treats each half's expansion into the whole as in Ex 2: each of the $2 N$ molecules doubles its volume → $Ω \to 2^{2 N} Ω$ , giving $Δ S = 2 N k_{B} ln 2$ . Why this step? This is the trap: it would say mixing identical gases creates entropy from nothing.
The fix: identical particles are indistinguishable, so swapping two molecules is not a new microstate. We must divide the count by $N!$ per box (the Gibbs correction). Why this step? Overcounting permutations inflates $Ω$ ; the $1/ N!$ removes exactly the fake states. This is the same $N!$ (and the $h^{3 N}$ grain size, see Ex 10) from the parent's third mistake-warning.
With correct indistinguishable counting, $Ω_{final}$ per molecule is unchanged (same density, same $T$ ) → $Δ S = 0$ . Why this step? Physically nothing happened, so a correct formula must give zero — and it does.

Verify: the difference between wrong and right is $Δ S_{wrong} - Δ S_{right} = 2 N k_{B} ln 2 - 0 = 2 N k_{B} ln 2$ , which is precisely the Stirling term $k_{B} ln [(2 N)! / (N! N!)] \approx 2 N k_{B} ln 2$ that the $N!$ factors cancel ✓. See Gibbs Paradox & Indistinguishability.

Example 9 — Cell I: Limiting behaviour (geometry of $S$ vs $E$ )

Worked example Shape of

$S (E)$ for the spin system, all the way across Sketch and interpret $S$ as a function of energy $E = m ε$ for the $N$ -spin system, from $E = 0$ to $E = N ε$ , covering both endpoints and the peak. (Use the figure below as your sketch.)

Forecast: where is $S$ maximal, and what happens at the two ends?

Steps

Endpoint $E = 0$ ( $m = 0$ , all down): $Ω = 1$ , $S = 0$ . Slope $\partial S / \partial E > 0$ → ordinary positive $T$ , in fact $T \to 0^{+}$ . Why this step? The lowest-energy state is unique — degenerate ground, zero entropy (a Third-Law-like edge). Look at the blue dot on the far left of the figure below.
Middle $E = N ε /2$ ( $m = N /2$ ): $Ω$ maximal, $S$ peaks at $N k_{B} ln 2$ , slope $= 0$ → $1/ T = 0$ → $T = \pm \infty$ . Why this step? At the peak, adding energy stops changing $S$ ; that's the infinite-temperature crossover — the pink dot at the top of the dome in the figure.
Right endpoint $E = N ε$ ( $m = N$ , all up): $Ω = 1$ again, so $S = k_{B} ln 1 = 0$ , but the curve reaches this zero coming down with slope $< 0$ → $T \to 0^{-}$ . Why this step? The energy cap makes the top as ordered as the bottom — a single "all up" microstate. There is nowhere else to put the energy, so the count collapses back to one. This is the blue dot on the far right of the figure.
Conclusion: on the whole right half ( $m > N /2$ ) the slope $\partial S / \partial E < 0$ , so $1/ T < 0$ and $T < 0$ . Negative temperature is not "colder than zero" — it is the high-energy half of a capped system, reached only because $S$ turns back down. The endpoint $E = N ε$ is the extreme negative-temperature state ( $T \to 0^{-}$ ), the mirror image of the ordinary $T \to 0^{+}$ ground state. Why this step? This closes the loop with Example 7: the sign flip we proved algebraically is exactly the falling right side of this dome, and its far end is a second $Ω = 1$ , $S = 0$ corner.

Verify (numbers, $N = 4$ , in units of $k_{B}$ ): $S / k_{B} = ln {1, 4, 6, 4, 1} = {0, 1.386, 1.792, 1.386, 0}$ — symmetric, peak at the middle $= ln 6 = 1.792$ , zeros at both ends ✓. Matches the drawn dome exactly.

Example 10 — Cell J: Continuous phase space and the grain size $h$

Worked example Why a classical gas needs

$h$ : counting a continuous system A single classical particle of mass $M$ lives in a 1-D box of length $L$ with momentum magnitude up to $p_{m a x}$ . Its "state" is a point $(x, p)$ in a continuous phase space (position on one axis, momentum on the other). You cannot count points — there are infinitely many. How do we get a finite, dimensionless $Ω$ ?

Forecast: if states form a smooth 2-D region, what turns an area into a pure number of microstates?

Steps

The accessible region is a rectangle in the $(x, p)$ plane: $x \in [0, L]$ , $p \in [- p_{m a x}, p_{m a x}]$ . Its area (a phase-space integral $\int d x d p$ ) is $A = L \cdot (2 p_{m a x})$ . Why this step? In continuous mechanics a microstate is a point, so the "amount of state" is a phase-space area, obtained by integrating over the allowed region (units: position × momentum = $m\cdotkg\cdotm/s = J\cdots$ ), not a plain count.
That area has units of $J\cdots$ — the same units as Planck's constant $h$ . Divide by $h$ to get a pure number: $Ω = \frac{A}{h} = \frac{2 L p _{m a x}}{h}$ . Why this step? $Ω$ inside $ln Ω$ must be dimensionless (you can't take the log of "so many J·s"). $h$ is the natural quantum area of one cell — one microstate occupies phase-space area $h$ per degree of freedom.
For $N$ particles in 3-D the phase space has $3 N$ position and $3 N$ momentum axes, so you divide by $h^{3 N}$ ; and for indistinguishable particles you also divide by $N!$ (Ex 8): $Ω = \frac{1}{N ! h ^{3 N}} \int d^{3 N} x d^{3 N} p$ . Why this step? $h^{3 N}$ makes $Ω$ dimensionless (each of the $3 N$ degrees of freedom contributes one factor of $h$ ); $N!$ removes the permutation overcount. Both together are what the parent's third mistake-warning demanded, and both are what kill Gibbs' paradox.
Numbers (the 1-D single particle): let $L = 1 m$ , $p_{m a x} = 1.0 \times 1 0^{- 24} kg\cdotm/s$ , $h = 6.626 \times 1 0^{- 34} J\cdots$ . Then $Ω = \frac{2 ( 1 ) ( 1 0 ^{- 24} )}{6.626 \times 1 0 ^{- 34}} = 3.02 \times 1 0^{9}$ , and $S = k_{B} ln Ω = k_{B} (21.83) = 3.01 \times 1 0^{- 22} J/K$ . Why this step? A concrete finite number falls out — the continuous system now has a countable entropy, just like the discrete ones above.

Verify (units): $\frac{[ m ] [ kg\cdotm/s ]}{[ J\cdots ]} = \frac{kg\cdotm ^{2} / s}{kg\cdotm ^{2} / s} = 1$ (dimensionless) ✓. Numerically $ln (3.02 \times 1 0^{9}) = 21.83$ and $k_{B} \times 21.83 = 3.01 \times 1 0^{- 22} J/K$ ✓. Links to the Microcanonical Ensemble, where this $h^{3 N} N!$ counting is the standard recipe.

Recall One-line summary of the matrix

Count ways ⇒ log them ⇒ ratios cancel unknowns, products become sums, $Ω = 1$ gives $0$ , huge $N$ needs Stirling, identical particles need $1/ N!$ , continuous states need $/ h^{3 N}$ to become a dimensionless count, and a capped energy makes $S$ come back down — negative temperature.

Active Recall

Free expansion tripling

V

for

N

molecules —

Δ S

N k_{B} ln 3 = n R ln 3

Combined weight of independent systems

2^{10}

and

2^{15}

2^{25}

; entropies add to

25 k_{B} ln 2

Per-molecule multiplicity ratio when ice melts (

Δ S = 22 J/K/mol

r = e^{Δ S / R} = e^{2.646} \approx 14

Where does the spin system's

S (E)

peak?

m = N /2

S = N k_{B} ln 2

, where

\partial S / \partial E = 0

(

T = \pm \infty

What is

S

at the high-energy end

E = N ε

(all spins up)?

Ω = 1

again so

S = 0

; reached with slope

< 0

, i.e.

T \to 0^{-}

(extreme negative temperature).

Why is mixing identical gases

Δ S = 0

Indistinguishability (

1/ N!

) cancels the naïve

2 N k_{B} ln 2

; nothing physical changed.

Why divide continuous phase-space volume by

h^{3 N}

A microstate occupies phase-space area

h

per degree of freedom; dividing makes

Ω

dimensionless so

ln Ω

is well-defined.

Connections

2.4.09 Boltzmann's entropy S = k_B ln(Ω) (Hinglish)
Microcanonical Ensemble
Temperature as $\partial S/\partial E$
Negative Temperature & Spin Systems
Gibbs Paradox & Indistinguishability
Third Law of Thermodynamics
Shannon Information Entropy