Entropy and statistical mechanics

Count the ways an outcome can happen.

Count multiplicities, explain a likely macrostate, and identify when equal weighting fails.

Start with: Fractions and probability. A microstate specifies every element; a macrostate groups many such arrangements.

01 · Commit to a prediction

What do you expect?

Four labeled particles are independently in the left or right half of a box, each with probability 1/2. Is exactly two on the left more likely than all four on the left? By what factor?

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02 · Change an assumption

Predict. Change. Explain.

Before changing a control, say what should move and why. Start with the experiments below. Reset restores the starting model; it preserves your written responses.

Experiment 1

Keep N = 4. Predict what happens when k changes from 2 to 0.

Check the prediction

Multiplicity drops from 6 to 1. Probability drops from 6/16 to 1/16; ln W drops from ln 6 to 0.

Experiment 2

Keep k = 2 and change N from 4 to 8. Does multiplicity alone determine probability?

Check the prediction

W increases to 28, but total possibilities rise to 256. Probability is 28/256 = 10.9375%, lower than 37.5%.

Number of labeled particles N

2 to 16 · step 1

Particles on the left k (must be ≤ N)

0 to 16 · step 1

Calculated model output. The table gives the same values. Displayed values are rounded; calculations keep full precision.

Read the plotted data as a table

03 · Connect the mechanism

From the picture to the quantities.

The macrostate hides particle identities. Counting those hidden alternatives explains why a middle-sized split can dominate even though no individual arrangement is privileged. The bars show probability for each possible left count, not a time history. This experiment enumerates possibilities; it does not simulate gas motion.

W(N,k) = N! / [k!(N−k)!]
P(k) = W(N,k) / 2ᴺ
S / k_B = ln W

N and k are dimensionless integers with 2 ≤ N ≤ 16 and 0 ≤ k ≤ N here. W is the number of arrangements. Factorial n! multiplies the integers from 1 to n; 0! = 1. ln is the natural logarithm. k_B = 1.380649 × 10⁻²³ J/K. S is the configurational contribution to entropy for this partition, in J/K.

Worked example

Label the particles A, B, C, D. The two on the left can be AB, AC, AD, BC, BD or CD. That is 6 of the 16 arrangements. P(2) = 0.375 and S/k_B = ln 6 ≈ 1.791759. For all left, W = 1 and this configurational entropy is zero. Zero here does not mean the real gas has no other degrees of freedom.

Assumptions and limits

Equal-volume sides, independent positions and equal weighting are assumptions. Real gas entropy also involves momenta, indistinguishability and physical constraints. This small model allows fluctuations. It does not prove monotonic entropy in every small-system trajectory. The second law concerns total entropy and an appropriate macroscopic description, not an isolated visual impression of disorder.

The annotated sources distinguish established results from this lesson’s original examples.

04 · Follow the structure

Where else does this apply?

Gas expansion

Removing a partition opens more accessible configurations.

Boundary: Counting sides omits velocities and interactions; no temperature or expansion rate is predicted.

Binary records

Many bitstrings share one total count of ones.

Boundary: A count of bitstrings becomes a probability only with a distribution. Bits are not automatically thermodynamic J/K.

05 · Retrieve without hints

Close the explanation. Try a new case.

Write an answer before opening its feedback. Later, return directly here without rereading above. Recognition, explanation and transfer are separate outcomes. No page action or answer reveal measures mastery.

Recognition and explanation

1. Why are six distinct arrangements not six times more likely individually?

Reveal reasoning and rubric

Each has probability 1/16. The macrostate is the union of six mutually exclusive arrangements.

Self-check: Name the level being counted and add mutually exclusive probabilities.

Calculation

2. For three labeled particles, how likely is exactly one on the left? Give W and S/kB.

Reveal reasoning and rubric

The choices are A, B or C. W = 3, probability 3/8 = 37.5%, S/kB = ln 3 ≈ 1.098612.

Self-check: Show the three arrangements and distinguish the probability from the logarithmic count.

Novel transfer

3. A membrane makes each particle independently 90% likely to be left. For N = 4, compare exactly two left with all four left. Can you still divide by 16?

Reveal reasoning and rubric

No. Each two-left arrangement has probability 0.9² × 0.1² = 0.0081, so the macrostate has probability 0.0486. All-left has probability 0.9⁴ = 0.6561. The multiplicities have not changed; their weights have.

Self-check: Reject equal weighting, retain the count of six, and supply the changed weights.

On a later day, try again and record actual evidence in the curriculum. A later unaided explanation and a fresh transfer problem give stronger evidence than immediate familiarity. No reminder is scheduled.

Sources & scope

What supports the lesson?

OpenStax, College Physics 2e §15.7. Microstate counting, equal-probability assumption and Boltzmann entropy.

Original teaching examples. Reference links need a connection; the lesson itself does not. Built 2026-10-11. Learner understanding is not assessed.