Entropy and disorder — Boltzmann S = k·ln(W): Worked examples

You have met the formula $S = k ln W$ in the parent note. Here we drill it into your bones by hitting every kind of situation it can face — big $W$ , tiny $W$ , the degenerate $W = 1$ case, ratios, real chemistry, and an exam trap. Guess before you compute. That is where the learning lives.

The scenario matrix

Every problem the Boltzmann formula throws at you falls into one of these cells. Each worked example below is tagged with the cell it fills.

Cell	What is special about it	Example
A. $W = 1$ (degenerate / ordered)	perfectly unique arrangement, the "zero" of entropy	Ex 1
B. Small finite $W$ (combinatorics)	count by hand with $(n N)$	Ex 2
C. Ratio $W_{f} / W_{i}$ (change, not absolute)	absolute $W$ unknown, only the factor matters	Ex 3
D. Exponential $W \sim a^{N}$ (huge $N$ )	can't list states; use $ln (a^{N}) = N ln a$	Ex 4
E. Real-world numbers (units check)	plug in $k$ , get joules, sanity-check magnitude	Ex 5
F. Limiting behaviour ( $T \to 0$ , $W \to 1$ )	the Third-Law edge case	Ex 6
G. Compare two systems (which is bigger?)	reasoning without full numbers	Ex 7
H. Exam twist (the trap)	" $W$ doubled → $S$ doubled?" style catch	Ex 8

The figure below is your map of this matrix: it plots the master curve $S / k = ln W$ once, and marks on it the three cells whose shape you must feel — the degenerate point (Cell A), the slow doubling nudge (Cell H), and the general slow climb that Cells B–G ride along. Keep glancing back at it: every example lands somewhere on that magenta curve.

Figure — Entropy and disorder — Boltzmann S = k·ln(W)

What the figure shows (alt text): the horizontal axis is the multiplicity $W$ (number of microstates), running from $1$ to $60$ ; the vertical axis is entropy in units of $k$ , i.e. $S / k = ln W$ . A single magenta curve rises steeply near $W = 1$ then flattens — the signature slow growth of a logarithm. A violet dot sits at $(W = 1, S / k = 0)$ : the degenerate zero-entropy point (Cell A). Two orange dots sit at $W = 15$ and $W = 30$ , joined to the vertical axis by dotted horizontal lines; the vertical gap between those two lines is only $ln 2 \approx 0.69$ , showing that doubling $W$ barely raises $S$ (Cell H). Every other example rides somewhere along the same magenta curve — only its $W$ -coordinate differs.

Cell A — the degenerate $W = 1$ state

Worked example Example 1 — A perfect crystal at absolute zero

A block of a pure element is cooled so far that every atom sits frozen in its one lowest-energy lattice site, with no ambiguity anywhere. There is exactly one way to arrange it: $W = 1$ . Find $S$ . (This is the violet dot on the figure.)

Forecast: guess before reading — is $S$ small-but-positive, exactly zero, or negative?

Identify $W$ . Every atom has a single fixed spot, so the whole block has exactly one microstate. $W = 1$ . Why this step? Entropy is a count; you must nail the count first.
Apply the formula: $S = k ln W = k ln 1$ . Why this step? Direct substitution — no combinatorics needed here.
Use $ln 1 = 0$ (because $e^{0} = 1$ ). So $S = k \cdot 0 = 0$ . Why this step? The log of one is zero: " $e$ to the power 0 gives 1."

Answer: $S = 0 J/K$ .

Verify: Only one arrangement means "no missing information" — nothing is uncertain, so entropy should vanish. This is exactly the Third Law of Thermodynamics: as $T \to 0$ , $W \to 1$ , $S \to 0$ . Units: $k$ is J/K times a dimensionless $ln$ , so $S$ is J/K. ✓

Cell B — small finite $W$ by hand

Worked example Example 2 — Six coins, exactly four heads

Toss $N = 6$ distinguishable coins. Consider the macrostate "exactly $n = 4$ heads." Find $W$ and $S$ . (Lands on the magenta curve at $W = 15$ .)

Forecast: more or fewer arrangements than the "all heads" macrostate? (Trust your gut.)

The macrostate fixes how many heads, not which. Count the ways to choose which 4 of the 6 coins show heads: $W = (4 6)$ . Why this step? Choosing a subset is a combination — order of picking doesn't matter, so $(n N)$ .
Compute $(4 6) = \frac{6 !}{4 ! 2 !} = \frac{720}{24 \cdot 2} = 15$ . Why this step? $(n N) = \frac{N !}{n ! ( N - n )!}$ is the number of distinct subsets.
Entropy: $S = k ln 15 = 1.38 \times 1 0^{- 23} \times ln 15$ . Why this step? Now $W$ is a plain number; substitute.
$ln 15 \approx 2.708$ , so $S \approx 1.38 \times 1 0^{- 23} \times 2.708 \approx 3.74 \times 1 0^{- 23} J/K$ . Why this step? This is the final "times $k$ " of count → log → times $k$ : we evaluate $ln 15$ numerically, then multiply by $k$ to convert the pure count into joules-per-kelvin.

Answer: $W = 15$ , $S \approx 3.74 \times 1 0^{- 23} J/K$ .

Verify: "All heads" ( $n = 6$ ) has $(6 6) = 1$ , entropy 0. Fifteen $>$ one, so the mixed macrostate is more probable and has more entropy — matches the intuition that middling ("disordered") outcomes dominate. ✓

Cell C — the ratio $W_{f} / W_{i}$

Worked example Example 3 — Removing a partition (free expansion,

$N = 3$ ) Three distinguishable gas molecules sit in the left half of a box. A partition is pulled so each can now roam the whole box (twice the volume). By what factor does $W$ grow, and what is $Δ S$ ?

Forecast: guess the multiplicity factor before computing.

Each molecule's number of position-choices doubles (twice the accessible volume). Why this step? Twice the space → twice as many position-states per molecule.
Choices of independent molecules multiply: $W_{f} = 2 \times 2 \times 2 \times W_{i} = 2^{3} W_{i} = 8 W_{i}$ . Why this step? Independent choices multiply (this is the multiplication rule of counting).
Entropy change uses only the ratio: $Δ S = k ln W_{f} - k ln W_{i} = k ln \frac{W _{f}}{W _{i}} = k ln 2^{3} = 3 k ln 2.$ Why this step? $ln W_{f} - ln W_{i} = ln (W_{f} / W_{i})$ — the unknown absolute $W_{i}$ cancels, which is why we can solve this without knowing it.
Numerically $3 k ln 2 = 3 (1.38 \times 1 0^{- 23}) (0.6931) \approx 2.87 \times 1 0^{- 23} J/K$ . Why this step? The "times $k$ " finish again: $ln 2 \approx 0.6931$ is the pure-count part, and multiplying by $3 k$ turns it into physical J/K units so we can compare to measured entropies.

Answer: factor $8$ ; $Δ S = 3 k ln 2 \approx 2.87 \times 1 0^{- 23} J/K$ .

Verify: General law $Δ S = N k ln 2$ with $N = 3$ gives $3 k ln 2$ . ✓ See Free Expansion of an Ideal Gas.

Cell D — exponential $W \sim a^{N}$ , huge $N$

Worked example Example 4 — One mole doubling its volume

Now let the same free expansion happen for one mole, $N = N_{A} = 6.022 \times 1 0^{23}$ molecules (recall $N_{A}$ is Avogadro's number, the count of particles in one mole). Find $Δ S$ .

Forecast: will the answer be astronomically large, or an everyday-sized number of J/K?

Same physics as Ex 3 but $N = N_{A}$ : $W_{f} = 2^{N_{A}} W_{i}$ . We cannot list $2^{N_{A}}$ states — that's why we never touch $W$ directly. Why this step? $ln$ tames the exponential: $ln (a^{N}) = N ln a$ .
$Δ S = k ln 2^{N_{A}} = N_{A} k ln 2$ . Why this step? Log pulls the exponent $N_{A}$ out front — the whole reason $ln$ appears in the formula.
Recognise $N_{A} k = R = 8.314 J/(mol\cdotK)$ , the gas constant. So $Δ S = R ln 2$ . Why this step? Boltzmann-per-molecule × Avogadro = universal-per-mole; links stat-mech to classical thermo.
$R ln 2 = 8.314 \times 0.6931 \approx 5.76 J/K$ . Why this step? Final numeric evaluation: $ln 2$ is the count part, $R$ carries the units, so the product is directly J/K for one mole.

Answer: $Δ S = R ln 2 \approx 5.76 J/K$ .

Verify: Despite $W$ ballooning by $2^{6 \times 1 0^{23}}$ , entropy is a humble $5.76 J/K$ — because $ln$ grows painfully slowly. This equals the classical Clausius result for isothermal doubling. Statistical and thermodynamic entropy agree. ✓

Cell E — real-world numbers, units audit

Worked example Example 5 — Two isotopes mixed in a crystal

A crystal has $N = 100$ lattice sites; $50$ are filled by isotope A, $50$ by isotope B, arranged randomly. Even though it looks like a tidy solid, the isotopes give many microstates. Find $S$ from this configurational disorder.

Forecast: a rigid, ordered-looking crystal — will $S$ be zero or clearly positive?

Count ways to choose which 50 of the 100 sites hold isotope A: $W = (50 100)$ . Why this step? Placing A on a subset of sites is again a combination.
This is enormous; use $ln$ . Numerically $(50 100) \approx 1.0089 \times 1 0^{29}$ , so $ln W \approx 66.78$ . Why this step? We need $ln W$ , not $W$ ; the log keeps the number manageable.
$S = k ln W = 1.38 \times 1 0^{- 23} \times 66.78 \approx 9.22 \times 1 0^{- 22} J/K$ . Why this step? The "times $k$ " close: $ln W = 66.78$ is the pure count; multiplying by $k$ delivers the physical entropy in J/K.

Answer: $S \approx 9.2 \times 1 0^{- 22} J/K$ — small but not zero.

Verify: This is the parent note's steel-man: a "perfectly ordered" crystal can still carry entropy from isotopic arrangements. Eyeballing tidiness misleads — you must count. Units: J/K. ✓

Cell F — limiting behaviour, $W \to 1$

Worked example Example 6 — Cooling a two-level spin toward absolute zero

Each spin has just two energy levels separated by the gap $ε$ . We ask only about the two clean limits where the accessible states are equally likely, so Boltzmann's $S = k ln W$ applies exactly. Count $W$ and find $S$ at $T \to 0$ and $T \to \infty$ .

Forecast: at $T \to 0$ , does $S$ head to $0$ , to $k ln 2$ , or blow up?

Cold limit $T \to 0$ : there is nowhere near enough thermal energy to reach the upper level (it costs $ε$ ), so every spin sits in its single ground level. Accessible states per spin: $W = 1$ . Why this step? When only one state is reachable, the count is unambiguous — no partition-function subtlety, just $W = 1$ .
Then $S = k ln 1 = 0$ . Why this step? Degenerate count → zero entropy, the Cell A result again.
Hot limit $T \to \infty$ : thermal energy dwarfs the gap $ε$ , so both levels are reached with equal probability. Accessible, equally-likely states per spin: $W = 2$ . Why this step? Two equally-likely states is exactly the regime where $S = k ln W$ is valid — we count $W = 2$ honestly.
Then $S = k ln 2 \approx 0.693 k \approx 9.57 \times 1 0^{- 24} J/K$ per spin. Why this step? The "times $k$ " close: $ln 2$ is the pure count part, $k$ converts it to J/K for one spin.

Answer: $S \to 0$ as $T \to 0$ ; $S \to k ln 2$ as $T \to \infty$ .

Verify: $T \to 0 \Rightarrow S \to 0$ is precisely the Third Law of Thermodynamics. The high- $T$ ceiling $k ln 2$ says "two equally likely states = 1 bit of uncertainty," echoing Information Entropy (Shannon). (For intermediate $T$ the two states are not equally likely, so one must use the more general $S = - k \sum_{i} p_{i} ln p_{i}$ from the Boltzmann Distribution — not $ln W$ ; that is exactly why we restricted to the two clean limits here.) ✓

Cell G — compare without full numbers

Worked example Example 7 — Which has more entropy: solid or liquid?

Without computing $W$ , argue whether $S_{liquid} > S_{solid}$ for the same substance.

Forecast: predict the sign of $Δ S_{melt}$ .

In the solid each molecule is pinned to a lattice site with limited orientations → few microstates, small $W_{s}$ . Why this step? Fewer accessible positions/orientations ⇒ smaller count.
In the liquid molecules translate and tumble freely → many microstates, $W_{l} ≫ W_{s}$ . Why this step? More accessible arrangements ⇒ larger count.
Since $ln$ is an increasing function, $W_{l} > W_{s} \Rightarrow ln W_{l} > ln W_{s} \Rightarrow S_{l} > S_{s}$ . Why this step? $ln$ preserves order — bigger input, bigger output.
Hence $Δ S_{melt} = k ln (W_{l} / W_{s}) > 0$ .

Answer: liquid has higher entropy; melting increases $S$ .

Verify: Matches everyday fact — heat is absorbed on melting, and by Clausius $Δ S = Q_{rev} / T > 0$ . Two languages, same conclusion. ✓

Cell H — the exam trap

Worked example Example 8 — "

$W$ was doubled, so entropy doubled." True or false? A student claims: "The multiplicity went from $W$ to $2 W$ , therefore the entropy doubled." Correct them, and compute the actual change if $W$ started at $1 0^{20}$ . (This is the orange-dots step on the figure.)

Forecast: does $S$ double, or change by a fixed small amount?

The student confuses $S \propto W$ with $S \propto ln W$ . Write the change honestly: $Δ S = k ln (2 W) - k ln W = k (ln 2 + ln W - ln W) = k ln 2.$ Why this step? $ln (2 W) = ln 2 + ln W$ ; the big $ln W$ cancels, leaving only $ln 2$ .
So doubling $W$ always adds the same tiny amount, independent of the starting $W$ : $k ln 2 \approx 9.57 \times 1 0^{- 24} J/K$ . Why this step? Reveals the trap — the answer doesn't scale with $W$ at all; that is exactly the near-flat orange stretch on the figure.
Compare to the original entropy $S_{i} = k ln (1 0^{20}) = k \times 46.05 \approx 6.36 \times 1 0^{- 22} J/K$ . The change is a $\sim 1.5%$ nudge, not a doubling. Why this step? Puts the additive change in perspective against the base value — a doubling of $W$ moves $S$ by only about one-and-a-half percent.

Answer: False. $Δ S = k ln 2 \approx 9.57 \times 1 0^{- 24} J/K$ , tiny and fixed — not a doubling. The entropy rises by roughly $1.5%$ , not $100%$ .

Verify: $k ln 2$ is the same number that appeared in Ex 6's high- $T$ limit and (per molecule) in free expansion — the "one bit" quantum of entropy. Because $W \sim a^{N}$ , only system size $N$ (via $ln a^{N} = N ln a$ ) can double $S$ ; merely doubling $W$ never can. ✓

Recall Scenario checklist

Which cell needs only a ratio, never the absolute $W$ ? ::: Cell C (and D, H) — the $ln$ turns $W_{f} / W_{i}$ into a difference and unknown $W_{i}$ cancels. What is $S$ whenever $W = 1$ ? ::: $0$ , because $ln 1 = 0$ (Cells A and F cold limit). Doubling $W$ changes $S$ by how much? ::: A fixed $k ln 2$ , regardless of the starting $W$ (Cell H). Why can't we use $ln W$ at intermediate $T$ in Cell F? ::: There the states are not equally likely, so we need $S = - k \sum_{i} p_{i} ln p_{i}$ instead.

Connections

Parent topic — Boltzmann entropy
Microstates and Macrostates — the counting that produces $W$ .
Free Expansion of an Ideal Gas — Cells C & D.
Clausius Entropy dS = dQ_rev / T — verifies Cells D & G.
Third Law of Thermodynamics — Cells A & F.
Boltzmann Distribution — where the equally-vs-unequally-likely question of Cell F is settled, and the source of $S = - k \sum_{i} p_{i} ln p_{i}$ .
Information Entropy (Shannon) — the "one bit = $k ln 2$ " theme.
Second Law of Thermodynamics — why systems drift toward the biggest- $W$ cell.