Physics · Statistical thermodynamics

Boltzmann entropy

Q: What does this relationship describe?

$S=k_{\mathrm B}\ln\Omega$ assigns entropy to a defined equally probable state set. Ω counts accessible microstates, not particles or volume. A macrostate may fix energy, volume and particle number or a particular occupation count; its microstates are the matching complete realisations. Eight labelled two-state sites with exactly four upper occupations give $\Omega=70$. Taking the logarithm gives $\frac S{k_{\mathrm B}}\approx4.2485$. This is entropy of a small counting model, not one mole. Visual disorder alone does not define multiplicity. Unequal probabilities instead need Gibbs entropy; merely counting possible states then does not suffice to calculate entropy.

Q: Which common mistake changes the result?

Distinguish multiplicity Ω from particle count N. $\Omega=10^{23}$ gives a small entropy because $\ln\Omega$ is only about 53; multiplicity of a macroscopic system generally grows exponentially with size. Doubling Ω adds $k_{\mathrm B}\ln2$, whereas doubling independent identical subsystems squares Ω and doubles S. Use natural logarithms; log10 without a conversion factor changes the result. S has units J/K, not J. With unequal probabilities a count of possible states is insufficient. Occupation-count macrostates in a binomial model are also not equally probable merely because underlying microscopic configurations were assigned equal probabilities.

The Boltzmann relation assigns entropy to a macrostate with Ω accessible, equally probable microstates.

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Formula

Calculate

S = k_{B} ln Ω

S in J/K; kB in J/K; Ω dimensionless.

Variables & units

Variables & units
Symbol	Meaning	Unit
$S$	Boltzmann entropy of the macrostate	J/K
$kB$	Boltzmann constant	J/K
$Ω$	number of accessible microstates	dimensionless

LaTeX

S=k_B\ln\Omega

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Discrete binomial multiplicity bars peak at 70 for four upper occupations. Small-model entropy S/kB=ln70≈4.24850. — Eight labeled binary sites: multiplicities are 1, 8, 28, 56, 70, 56, 28, 8, 1. Equal microstate probabilities do not make macrostates equally probable.

Boltzmann entropy calculator

Find the entropy from the number of equally likely microstates, directly from Ω or from ln Ω for huge counts, plus the entropy change.

Calculation model

Number of microstates Ω

Number of equally likely microstates, a whole number, e.g. 70 for 4 occupied sites out of 8.

Filled with example values. Type your own and the result updates straight away.

Result: Thermodynamic entropy S

S = 5.866 · 10⁻²³ J/K

With Ω = 70, S = 5.866 · 10⁻²³ J/K.

Working · rearrange for S, substitute, evaluate

1. Rearrange for S
$S=k_{\mathrm B}\ln\Omega$
2. Substitute values (in SI units)
$1.38065\times10^{-23}\cdot\ln\left(70\right)$
3. Result
S = 5.866 · 10⁻²³ J/K

Quick tips

Whenever Ω doubles, S grows by the same amount k · ln 2 ≈ 9.57·10⁻²⁴ J/K, however large Ω already is.
Use the natural logarithm, not log₁₀: Ω = 10²³ gives ln Ω = 23 · ln 10 ≈ 53. That keeps even huge state counts manageable.

Assumptions and limits

Discrete equally probable accessible microstates of an explicitly defined model. Not a material, visual-order or total-particle counter. Exact small multiplicity must be representable as a safe integer; use ln Ω separately for large counts. Entries with more than 15 significant digits are rounded to the nearest floating-point value before the integer check, so 1.0000000000000001 is read as 1.

Derivation & background

Ω counts accessible microstates under explicit macroconstraints, not particles. Equal probabilities reduce $S_{\mathrm G}=-k_{\mathrm B}\sum_i p_i\ln p_i$ to $S=k_{\mathrm B}\ln\Omega$ . Independent subsystems multiply multiplicities and add entropies; correlations require joint description. Unequal weights need Gibbs summation. Continuous classical states and identical particles need consistent dimensionless counting conventions.

Exam blueprint

Validity range

Defined dimensionless counting of accessible equally probable microstates under fixed macroconstraints. Independence for product counting; unequal probabilities require Gibbs summation. Continuous phase space and identical particles need consistent counting conventions.

Derivation steps

A macrostate can have many microscopic realisations. Boltzmann’s logarithm turns that multiplicity into additive entropy. The crucial questions are which states are counted under fixed constraints and when their probabilities really are equal.

1Specify the macroconstraints and count accessible microstates. Multiplicity $\Omega$ is dimensionless.
2For equally probable states, $p_i=\frac1\Omega$ and $\sum_{i=1}^{\Omega}p_i=1$ .
3Gibbs entropy is $S_{\mathrm G}=-k_{\mathrm B}\sum_i p_i\ln p_i$ .
4Substitution gives $S=-k_{\mathrm B}\Omega\frac1\Omega\ln\frac1\Omega=k_{\mathrm B}\ln\Omega$ .
5Independent subsystems have $\Omega_{AB}=\Omega_A\Omega_B$ . Taking the logarithm gives $S_{AB}=S_A+S_B$ .
6Two appropriately defined state sets give $\Delta S=k_{\mathrm B}\ln\frac{\Omega_2}{\Omega_1}$ . Unequal probabilities require the Gibbs sum.

Rearrangements

Number of microstates

\Omega=\exp\!\left(\frac{S}{k_{\mathrm B}}\right)

Assumes equally probable microstates.

Task variant

An equally probable state set has $\Omega=10^{23}$ . Calculate its entropy, then the entropy change when accessible multiplicity doubles. Is this multiplicity the particle count?

Approach: $S=k_{\mathrm B}\ln\Omega$ .
Since $\ln10^{23}=23\ln10$ , $S\approx7.31184\cdot10^{-22}\,\mathrm{J\,K^{-1}}$ .
Doubling adds $\Delta S=k_{\mathrm B}\ln2\approx9.56993\cdot10^{-24}\,\mathrm{J\,K^{-1}}$ , rather than doubling entropy.

Check: the logarithmic difference gives the same increment. $\Omega$ counts states, not particles.

Common mistakes

Equating multiplicity with particle count.

Count complete allowed configurations under the specified macroconstraints. Eight sites with exactly four upper occupations give 70 states, not eight.

Using the Boltzmann shortcut for unequal probabilities.

Check the probabilities; otherwise use $S_{\mathrm G}=-k_{\mathrm B}\sum_i p_i\ln p_i$ .

Multiplying state counts despite correlations.

Count only allowed joint combinations. Product counting and additive entropy here require independent subsystems.

Exam context

Count a constrained state set combinatorially and check by exchanging occupations.
Distinguish multiplicity, unequal probabilities and correlations in small models.

Formula cluster

Statistical thermodynamics

Statistical counting explains entropy of a defined macrostate. Reversible heat balance provides a complementary thermodynamic approach.

ΔS = ∫δQrev/T; bei konstantem T: ΔS = Qrev/T

Entropy change along a reversible path

⟨Ekin⟩ = 3/2 kBT

Mean translational kinetic energy per gas molecule

ΔU = Q − W

First Law of Thermodynamics

Worked example

For $\Omega=10^{23}$ , $S=k_{\mathrm B}\ln\Omega=23k_{\mathrm B}\ln10\approx7.31\cdot10^{-22}\,\mathrm{J\,K^{-1}}$ . Check: doubling the equally probable multiplicity adds $k_{\mathrm B}\ln2\approx9.57\cdot10^{-24}\,\mathrm{J\,K^{-1}}$ ; it does not double entropy. Ω is not particle count.

Applications

Connect macroscopic entropy with statistical state counts and interpret entropy changes.

How do you calculate with Boltzmann Entropy and Microstates?

Here is how to work through a typical Boltzmann Entropy and Microstates (S=kB ln Ω) task step by step:

1
Task
An equally probable state set has $\Omega=10^{23}$ . Calculate its entropy, then the entropy change when accessible multiplicity doubles. Is this multiplicity the particle count?
Solution path
1. Approach: $S=k_{\mathrm B}\ln\Omega$ .
2. Since $\ln10^{23}=23\ln10$ , $S\approx7.31184\cdot10^{-22}\,\mathrm{J\,K^{-1}}$ .
3. Doubling adds $\Delta S=k_{\mathrm B}\ln2\approx9.56993\cdot10^{-24}\,\mathrm{J\,K^{-1}}$ , rather than doubling entropy.
Check: the logarithmic difference gives the same increment. $\Omega$ counts states, not particles.

Understand and apply the equation

What Ω counts

\begin{aligned}S&=k_{\mathrm B}\ln\Omega\\\Delta S&=k_{\mathrm B}\ln\frac{\Omega_2}{\Omega_1}\end{aligned}

Ω is a dimensionless state count, not particle count or gas volume.[1][2]

A microstate specifies a complete allowed microscopic realisation of the model. A macrostate fixes only a few constraints, such as energy, volume, particle count or number occupying higher levels. Many microstates can realise the same macrostate. Define that mapping before calculation; an arbitrary visual impression of disorder does not define Ω.[3]

Why the logarithm adds

\Omega_{AB}=\Omega_A\Omega_B\quad\Longrightarrow\quad S_{AB}=k_{\mathrm B}\ln\Omega_A+k_{\mathrm B}\ln\Omega_B

Product counting requires independent accessible states of the two subsystems.[2]

Each allowed state of A can then be paired with every state of B. An additive macroscopic quantity must convert that product into a sum. The logarithm does so; kB connects it to thermodynamic units J/K. Additivity alone, without further statistical and thermodynamic assumptions, is not a complete proof of every entropy law.[2]

Explicitly check equal probabilities

\begin{aligned}S_G&=-k_{\mathrm B}\sum_i p_i\ln p_i\\{} p_i&=\frac{1}{\Omega}\Rightarrow S_G=k_{\mathrm B}\ln\Omega\end{aligned}

For $p=0$ use the limit $\lim_{p\to0^+}p\ln p=0$ ; probabilities must sum to 1.[2]

The Boltzmann form arises as the equally probable special case of the Gibbs form. In an isolated equilibrium model count states under the appropriate energy constraints; thermal-bath coupling usually weights different energies differently. A count of possible states alone is then insufficient. Macrostates likewise need not be equally probable even when microstates are, if their multiplicities differ.[2][3]

Shannon entropy uses the same probability weighting as an information measure. Logarithm base determines its unit: base 2 gives bits and the natural logarithm gives nats. Thermodynamic Gibbs entropy multiplies the natural logarithm by Boltzmann’s constant. An information value in bits therefore cannot be used unchanged as entropy in joules per kelvin.[2][7]

Count states and weight probabilities

Eight distinguishable fixed sites

Given: Eight fixed labelled sites each have two possible states. The macrostate fixes exactly four upper occupations; all matching configurations are equally probable.
Find: Count Ω and calculate this macrostate’s entropy.

Approach: Labeled sites are distinguishable and all microstates are equiprobable. $\Omega=\binom Nk,\quad S=k_{\mathrm B}\ln\Omega$ . Choose four sites from eight.
$\Omega={8\choose4}=\frac{8!}{4!4!}=70$
Use the natural logarithm.
$S=k_{\mathrm B}\ln70=5.865680707\times10^{-23}\,\mathrm{J/K}$

Result: $\Omega=70$ , $\frac{S}{k_{\mathrm B}}\approx4.248495242$ . This small entropy belongs to the eight-site model, not one mole.

Check: Binomial counting is symmetric under k↔8−k. If instead all 256 binary configurations are equally weighted, the four-upper macrostate has probability $\frac{70}{256}=0.2734375$ . That is a different, explicitly broader total state set.

Two states with unequal probabilities

Given: A precisely defined two-state model has $p_1=0.75,\quad p_2=0.25$ . Probabilities are unequal.
Find: Calculate SG and compare with maximum two-state entropy.

Approach: For unequal probabilities use Gibbs entropy. $S_G=-k_{\mathrm B}\sum_i p_i\ln p_i$ . Add weighted logarithms.
$\frac{S_G}{k_{\mathrm B}}=-(0.75\ln0.75+0.25\ln0.25)=0.5623351446$
Compare with the equally probable case.
$\frac{S_{\mathrm{max}}}{k_{\mathrm B}}=\ln2=0.6931471806$

Result: SG≈7.763874551×10⁻²⁴ J/K, below $k_{\mathrm B}\ln2$ ≈9.569929617×10⁻²⁴ J/K.

Check: $p_1+p_2=1$ ; exchanging labels leaves SG unchanged. Inserting $\Omega=2$ into Boltzmann’s formula without checking probabilities would overestimate entropy.

Release an occupation-count constraint

Given: Eight labelled two-state sites are divided into two groups of four. Initially each group must contain exactly two upper occupations. Afterwards only the total of four upper occupations remains fixed. Microstates in each allowed set are equiprobable; all upper levels have the same energy.
Find: Compare multiplicities and entropy difference between the two constrained equilibrium models.

Approach: with separately fixed occupation counts, combine the two state lists independently. With only the common constraint, count every arrangement of four upper sites.
$\begin{aligned}\Omega_i&=\binom42\binom42\\\Omega_f&=\binom84\\\Delta S&=k_{\mathrm B}\ln\frac{\Omega_f}{\Omega_i}\end{aligned}$
The separate constraint allows 36 configurations and the common one 70. The larger set contains the earlier set as a subset.
$\Omega_i=6\cdot6=36,\quad\Omega_f=70,\quad\frac{\Delta S}{k_{\mathrm B}}=\ln\frac{70}{36}\approx0.6649763$
Within the larger model, check the probability of the former subset.
$p(2\text{ obere Plätze je Gruppe})=\frac{36}{70}\approx0.514286$

Result: Removing the extra constraint enlarges the state set. The models differ in entropy by about 0.664976 kB without changing total occupation energy.

Check: Vandermonde’s identity counts the same 70 states by summing over one group’s occupation. The product 6×6 belongs to the original separate constraint; afterwards the fixed total correlates the groups. Counting the model states does not determine equilibration time.

Equal microstate chances give unequal macrostate chances

QUANTA STUDYquanta-study.de

The count is

\Omega(k)=\binom8k

. Only the nine integer points are allowed occupation counts; the connecting line guides the eye. Under 256 equiprobable microstates, each macrostate has probability

\frac{\Omega(k)}{256}

.[1][3]

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Application: entropy change of an ideal gas

\Omega_{pos}\propto V^N\quad\Longrightarrow\quad\Delta S=Nk_{\mathrm B}\ln\frac{V_2}{V_1}=nR\ln\frac{V_2}{V_1}

Ideal gas at fixed N and equal endpoint temperatures; common counting constants cancel from the change.[1][4]

The volume factor illustrates additional position possibilities. Absolute classical gas entropy requires a consistently dimensionless phase-space measure; identical particles need proper indistinguishability counting. That is more than freely labelling every particle. The volume relation used here for the same fixed sample avoids that additional absolute-entropy calculation because unchanged factors cancel in the ratio.[5][1]

For 0.20 mol and volume factor 1.8, $\Delta S=0.977426051\,\mathrm{J\,K^{-1}}$ . Free isolated expansion gives this state change despite $Q=0$ ; entropy increase does not always require actual heat absorption.[4]

Entropy change along a reversible path connects this state counting to thermodynamic calculation. The path may be an imagined reversible comparison between the same endpoints. For an actual irreversible expansion, it does not specify the heat exchanged in that process.[4][1]

Exponential counts without overflow

\begin{aligned}\ln\Omega&=\frac S{k_{\mathrm B}}\\{} \frac{\Omega_2}{\Omega_1}&=\exp(\frac{\Delta S}{k_{\mathrm B}})\end{aligned}

Large systems are computed through $\ln\Omega$ rather than writing out enormous multiplicities.[1][2]

$\Omega=10^{23}$ does not mean 10²³ particles. That state-count case gives $S=23k_{\mathrm B}\ln10\approx7.311842154\cdot10^{-22}\,\mathrm{J\,K^{-1}}$ . A macroscopic system typically has a multiplicity whose logarithm grows with particle count. Doubling Ω adds $k_{\mathrm B}\ln2$ ; doubling independent identical subsystems instead squares Ω and doubles S.[1][2]

Five exercises with solutions

Calculate · Foundation

Two independent equally probable state sets have $\Omega_A=6$ and $\Omega_B=5$ . Find ΩAB and $\frac{S_{AB}}{k_{\mathrm B}}$ .

Show hint

Count products, add logarithms.

Learning goal: Multiply independent state counts.

Show solution and calculation

Approach: each state of A can be paired with every state of B.
$\Omega_{AB}=6\cdot5=30$ .
$\frac{S_{AB}}{k_{\mathrm B}}=\ln30\approx3.401197382$ .
$\ln6+\ln5=\ln30$ gives the same result.

S_{AB}=k_{\mathrm B}\ln30

Calculate · Application

An equally probable multiplicity increases by factor 8. Find ΔS.

Show hint

Use the ratio instead of individual counts.

Learning goal: Calculate entropy change from a multiplicity ratio.

Show solution and calculation

Approach: $\Delta S=k_{\mathrm B}\ln\frac{\Omega_2}{\Omega_1}=k_{\mathrm B}\ln8$ .
$\Delta S=k_{\mathrm B}\ln8\approx2.870978885\cdot10^{-23}\,\mathrm{J\,K^{-1}}$ .
It is three times $k_{\mathrm B}\ln2$ , not eight times the old entropy.

\Delta S=3k_{\mathrm B}\ln2

Interpret · Application

Four fixed labelled two-state sites have exactly one upper occupation. Find Ω; may the five possible occupation-count macrostates be equally weighted?

Show hint

Choose one upper site from four; compare multiplicities.

Learning goal: Distinguish microstate and macrostate probabilities.

Show solution and calculation

Approach: choose one upper-occupied site from four: $\Omega=\binom41$ .
$\Omega=4$ .
The five macrostates have multiplicities 1,4,6,4,1; under 16 equally probable microstates their probabilities therefore differ.

Check: the five multiplicities sum to 16.

\Omega={4\choose1}=4

Interpret · Challenge

Two labelled two-state sites A and B may occupy only joint microstates 00 and 11, equally likely. Each site separately can be 0 or 1. Find ΩAB and $\frac{S_{AB}}{k_{\mathrm B}}$ . Why is ΩAΩB not the joint multiplicity?

Show hint

Count allowed joint states, not every formal combination.

Learning goal: Identify correlations that prevent multiplicative state counting.

Show solution and calculation

Approach: the joint list contains two states, so $S_{AB}=k_B\ln2$ .
Multiplying $2\cdot2=4$ also counts forbidden states 01 and 10.
The sites are perfectly correlated rather than independent.

Check: probabilities $\frac12$ for both allowed joint states give ln2 from the Gibbs sum as well. Summing the two marginal entropies would overestimate joint entropy.

\begin{aligned}\Omega_{AB}&=2\ne\Omega_A\Omega_B=4\\\frac{S_{AB}}{k_{\mathrm B}}&=\ln2\approx0.693147\end{aligned}

[3][2]

Error analysis · Application

A model has two allowed microstates with probabilities 0.90 and 0.10. A solution sets $S=k_B\ln2$ because there are two states. Instead find $\frac S{k_{\mathrm B}}$ using the article’s Gibbs formula and compare.

Show hint

The two states are not equally probable.

Learning goal: Check equiprobability before using the Boltzmann shortcut.

Show solution and calculation

Approach: the probability-weighted logarithm sum gives 0.325083 rather than ln2≈0.693147.
State count alone does not justify the shortcut for unequal probabilities.

Check: replacing the probabilities by 0.50 and 0.50 would give ln2. Entropy is positive here but smaller than for the uniform model; multiplying by kB supplies units J/K.

\frac S{k_{\mathrm B}}=-0.90\ln0.90-0.10\ln0.10\approx0.325083

[2]

Statistical likelihood is not a mechanical constraint

Rare configurations and fluctuations are possible in a small model. Large multiplicity differences make some macrostates very likely, but do not prove every reverse fluctuation microscopically impossible. Macroscopic thermodynamics also involves system size, observation time and chosen constraints.[1]

Continuous measurements, correlations or unequal probabilities need an appropriate state description. Coupled subsystems do not automatically satisfy $\Omega_{AB}=\Omega_A\Omega_B$ . $S=k_{\mathrm B}\ln\Omega$ is a precise statistical model, not a scoring system for a room’s visual tidiness.[2][3]

Define allowed states and their probabilities before taking a logarithm. Multiply independent lists, but count common constraints first. Entropy change uses the ratio of two consistently defined state sets; a larger particle count alone is not a sufficient explanation.

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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Change history

October 2, 2026Added connections to Shannon entropy and to entropy change along a reversible path, citing Shannon’s original paper as an additional source; preserved existing derivations, exercises and original figures.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Refined the learning sequence, symbolic exercise approaches and model limits; added a third original example and quantitative figure, with suitable further explanations optional.

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Quanta. Boltzmann Entropy and Microstates. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/boltzmann-entropie-mikrozustaende

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Frequently asked questions about Boltzmann Entropy and Microstates

What does this relationship describe?+

$S=k_{\mathrm B}\ln\Omega$ assigns entropy to a defined equally probable state set. Ω counts accessible microstates, not particles or volume. A macrostate may fix energy, volume and particle number or a particular occupation count; its microstates are the matching complete realisations. Eight labelled two-state sites with exactly four upper occupations give $\Omega=70$ . Taking the logarithm gives $\frac S{k_{\mathrm B}}\approx4.2485$ . This is entropy of a small counting model, not one mole. Visual disorder alone does not define multiplicity. Unequal probabilities instead need Gibbs entropy; merely counting possible states then does not suffice to calculate entropy.

Where does the formula come from, and what does its structure tell us?+

For independent subsystems, every allowed microstate of A can be paired with every state of B. Multiplicities therefore multiply: $\Omega_{AB}=\Omega_A\Omega_B$ . The logarithmic form converts that product into $S_A+S_B$ . kB supplies thermodynamic units J/K. Another route starts from $S_{\mathrm G}=-k_{\mathrm B}\sum_i p_i\ln p_i$ . Substituting $p_i=\frac1\Omega$ for Ω equally probable states makes the sum exactly $k_{\mathrm B}\ln\Omega$ . Additivity alone, without further assumptions, does not prove every entropy law. With correlations, do not automatically count joint states as an independent product; their joint probabilities likewise may not factorise into separate subsystem probabilities.

What does a checked numerical calculation look like?+

Eight labelled sites have two possible states, with exactly four upper occupations required. There are $\frac{8!}{4!4!}=70$ matching configurations. Thus $S=k_{\mathrm B}\ln70\approx5.865680707\cdot10^{-23}\,\mathrm{J\,K^{-1}}$ . A combinatorial check gives the same count when upper and lower occupations are exchanged. If instead all 256 binary configurations are equally weighted, the four-upper macrostate has probability $\frac{70}{256}=0.2734375$ . That is a broader total state set and must not be confused with the set constrained to exactly four. The separate existing case $\Omega=10^{23}$ gives 7.311842154×10⁻²² J/K; $\Omega=10^{23}$ does not mean 10²³ particles. Both calculations use their explicitly defined state sets.

Under what conditions may this relationship be used?+

The simple Boltzmann form requires accessible equally probable microstates under defined macroscopic constraints. Continuous classical phase space needs consistent dimensionless counting; identical particles additionally require proper indistinguishability treatment. With unequal probabilities use $S_{\mathrm G}=-k_{\mathrm B}\sum_i p_i\ln p_i$ and $\sum_i p_i=1$ . For example, a two-state model with 0.75 and 0.25 gives $\frac{S_{\mathrm G}}{k_{\mathrm B}}\approx0.5623351$ , below ln2. Coupled subsystems do not automatically have product counting. Small models allow fluctuations and rare macrostates; high statistical likelihood does not mechanically forbid every reverse fluctuation. A gas-volume relation additionally requires the same fixed sample and suitable thermodynamic endpoint states.

Which common mistake changes the result?+

Distinguish multiplicity Ω from particle count N. $\Omega=10^{23}$ gives a small entropy because $\ln\Omega$ is only about 53; multiplicity of a macroscopic system generally grows exponentially with size. Doubling Ω adds $k_{\mathrm B}\ln2$ , whereas doubling independent identical subsystems squares Ω and doubles S. Use natural logarithms; log10 without a conversion factor changes the result. S has units J/K, not J. With unequal probabilities a count of possible states is insufficient. Occupation-count macrostates in a binomial model are also not equally probable merely because underlying microscopic configurations were assigned equal probabilities.

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Linear thermal expansion

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Symbol

Meaning

Unit

S

Boltzmann entropy of the macrostate

J/K

kB

Boltzmann constant

J/K

Ω

number of accessible microstates

dimensionless

Derivation & background

Exam blueprint

Validity range

Derivation steps

1Specify the macroconstraints and count accessible microstates. Multiplicity $\Omega$ is dimensionless.
2For equally probable states, $p_i=\frac1\Omega$ and $\sum_{i=1}^{\Omega}p_i=1$ .
3Gibbs entropy is $S_{\mathrm G}=-k_{\mathrm B}\sum_i p_i\ln p_i$ .
4Substitution gives $S=-k_{\mathrm B}\Omega\frac1\Omega\ln\frac1\Omega=k_{\mathrm B}\ln\Omega$ .
5Independent subsystems have $\Omega_{AB}=\Omega_A\Omega_B$ . Taking the logarithm gives $S_{AB}=S_A+S_B$ .
6Two appropriately defined state sets give $\Delta S=k_{\mathrm B}\ln\frac{\Omega_2}{\Omega_1}$ . Unequal probabilities require the Gibbs sum.

How do you calculate with Boltzmann Entropy and Microstates?

Here is how to work through a typical Boltzmann Entropy and Microstates (S=kB ln Ω) task step by step:

Task

An equally probable state set has $\Omega=10^{23}$ . Calculate its entropy, then the entropy change when accessible multiplicity doubles. Is this multiplicity the particle count?

Solution path

Approach: $S=k_{\mathrm B}\ln\Omega$ .
Since $\ln10^{23}=23\ln10$ , $S\approx7.31184\cdot10^{-22}\,\mathrm{J\,K^{-1}}$ .
Doubling adds $\Delta S=k_{\mathrm B}\ln2\approx9.56993\cdot10^{-24}\,\mathrm{J\,K^{-1}}$ , rather than doubling entropy.

Check: the logarithmic difference gives the same increment. $\Omega$ counts states, not particles.

What Ω counts

\begin{aligned}S&=k_{\mathrm B}\ln\Omega\\\Delta S&=k_{\mathrm B}\ln\frac{\Omega_2}{\Omega_1}\end{aligned}

Ω is a dimensionless state count, not particle count or gas volume.[1][2]

Why the logarithm adds

\Omega_{AB}=\Omega_A\Omega_B\quad\Longrightarrow\quad S_{AB}=k_{\mathrm B}\ln\Omega_A+k_{\mathrm B}\ln\Omega_B

Product counting requires independent accessible states of the two subsystems.[2]

Explicitly check equal probabilities

\begin{aligned}S_G&=-k_{\mathrm B}\sum_i p_i\ln p_i\\{} p_i&=\frac{1}{\Omega}\Rightarrow S_G=k_{\mathrm B}\ln\Omega\end{aligned}

For $p=0$ use the limit $\lim_{p\to0^+}p\ln p=0$ ; probabilities must sum to 1.[2]

Count states and weight probabilities

Eight distinguishable fixed sites

Given: Eight fixed labelled sites each have two possible states. The macrostate fixes exactly four upper occupations; all matching configurations are equally probable.
Find: Count Ω and calculate this macrostate’s entropy.

Approach: Labeled sites are distinguishable and all microstates are equiprobable. $\Omega=\binom Nk,\quad S=k_{\mathrm B}\ln\Omega$ . Choose four sites from eight.
$\Omega={8\choose4}=\frac{8!}{4!4!}=70$
Use the natural logarithm.
$S=k_{\mathrm B}\ln70=5.865680707\times10^{-23}\,\mathrm{J/K}$

Result: $\Omega=70$ , $\frac{S}{k_{\mathrm B}}\approx4.248495242$ . This small entropy belongs to the eight-site model, not one mole.

Two states with unequal probabilities

Given: A precisely defined two-state model has $p_1=0.75,\quad p_2=0.25$ . Probabilities are unequal.
Find: Calculate SG and compare with maximum two-state entropy.

Approach: For unequal probabilities use Gibbs entropy. $S_G=-k_{\mathrm B}\sum_i p_i\ln p_i$ . Add weighted logarithms.
$\frac{S_G}{k_{\mathrm B}}=-(0.75\ln0.75+0.25\ln0.25)=0.5623351446$
Compare with the equally probable case.
$\frac{S_{\mathrm{max}}}{k_{\mathrm B}}=\ln2=0.6931471806$

Result: SG≈7.763874551×10⁻²⁴ J/K, below $k_{\mathrm B}\ln2$ ≈9.569929617×10⁻²⁴ J/K.

Check: $p_1+p_2=1$ ; exchanging labels leaves SG unchanged. Inserting $\Omega=2$ into Boltzmann’s formula without checking probabilities would overestimate entropy.

Release an occupation-count constraint

Given: Eight labelled two-state sites are divided into two groups of four. Initially each group must contain exactly two upper occupations. Afterwards only the total of four upper occupations remains fixed. Microstates in each allowed set are equiprobable; all upper levels have the same energy.
Find: Compare multiplicities and entropy difference between the two constrained equilibrium models.

Approach: with separately fixed occupation counts, combine the two state lists independently. With only the common constraint, count every arrangement of four upper sites.
$\begin{aligned}\Omega_i&=\binom42\binom42\\\Omega_f&=\binom84\\\Delta S&=k_{\mathrm B}\ln\frac{\Omega_f}{\Omega_i}\end{aligned}$
The separate constraint allows 36 configurations and the common one 70. The larger set contains the earlier set as a subset.
$\Omega_i=6\cdot6=36,\quad\Omega_f=70,\quad\frac{\Delta S}{k_{\mathrm B}}=\ln\frac{70}{36}\approx0.6649763$
Within the larger model, check the probability of the former subset.
$p(2\text{ obere Plätze je Gruppe})=\frac{36}{70}\approx0.514286$

Result: Removing the extra constraint enlarges the state set. The models differ in entropy by about 0.664976 kB without changing total occupation energy.

Equal microstate chances give unequal macrostate chances

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The count is

\Omega(k)=\binom8k

. Only the nine integer points are allowed occupation counts; the connecting line guides the eye. Under 256 equiprobable microstates, each macrostate has probability

\frac{\Omega(k)}{256}

.[1][3]

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Application: entropy change of an ideal gas

\Omega_{pos}\propto V^N\quad\Longrightarrow\quad\Delta S=Nk_{\mathrm B}\ln\frac{V_2}{V_1}=nR\ln\frac{V_2}{V_1}

Ideal gas at fixed N and equal endpoint temperatures; common counting constants cancel from the change.[1][4]

Exponential counts without overflow

\begin{aligned}\ln\Omega&=\frac S{k_{\mathrm B}}\\{} \frac{\Omega_2}{\Omega_1}&=\exp(\frac{\Delta S}{k_{\mathrm B}})\end{aligned}

Large systems are computed through $\ln\Omega$ rather than writing out enormous multiplicities.[1][2]

Five exercises with solutions

Calculate · Foundation

Two independent equally probable state sets have $\Omega_A=6$ and $\Omega_B=5$ . Find ΩAB and $\frac{S_{AB}}{k_{\mathrm B}}$ .

Show hint

Count products, add logarithms.

Learning goal: Multiply independent state counts.

Show solution and calculation

Approach: each state of A can be paired with every state of B.
$\Omega_{AB}=6\cdot5=30$ .
$\frac{S_{AB}}{k_{\mathrm B}}=\ln30\approx3.401197382$ .
$\ln6+\ln5=\ln30$ gives the same result.

S_{AB}=k_{\mathrm B}\ln30

Calculate · Application

An equally probable multiplicity increases by factor 8. Find ΔS.

Show hint

Use the ratio instead of individual counts.

Learning goal: Calculate entropy change from a multiplicity ratio.

Show solution and calculation

Approach: $\Delta S=k_{\mathrm B}\ln\frac{\Omega_2}{\Omega_1}=k_{\mathrm B}\ln8$ .
$\Delta S=k_{\mathrm B}\ln8\approx2.870978885\cdot10^{-23}\,\mathrm{J\,K^{-1}}$ .
It is three times $k_{\mathrm B}\ln2$ , not eight times the old entropy.

\Delta S=3k_{\mathrm B}\ln2

Interpret · Application

Four fixed labelled two-state sites have exactly one upper occupation. Find Ω; may the five possible occupation-count macrostates be equally weighted?

Show hint

Choose one upper site from four; compare multiplicities.

Learning goal: Distinguish microstate and macrostate probabilities.

Show solution and calculation

Approach: choose one upper-occupied site from four: $\Omega=\binom41$ .
$\Omega=4$ .
The five macrostates have multiplicities 1,4,6,4,1; under 16 equally probable microstates their probabilities therefore differ.

Check: the five multiplicities sum to 16.

\Omega={4\choose1}=4

Interpret · Challenge

Two labelled two-state sites A and B may occupy only joint microstates 00 and 11, equally likely. Each site separately can be 0 or 1. Find ΩAB and $\frac{S_{AB}}{k_{\mathrm B}}$ . Why is ΩAΩB not the joint multiplicity?

Show hint

Count allowed joint states, not every formal combination.

Learning goal: Identify correlations that prevent multiplicative state counting.

Show solution and calculation

Approach: the joint list contains two states, so $S_{AB}=k_B\ln2$ .
Multiplying $2\cdot2=4$ also counts forbidden states 01 and 10.
The sites are perfectly correlated rather than independent.

Check: probabilities $\frac12$ for both allowed joint states give ln2 from the Gibbs sum as well. Summing the two marginal entropies would overestimate joint entropy.

\begin{aligned}\Omega_{AB}&=2\ne\Omega_A\Omega_B=4\\\frac{S_{AB}}{k_{\mathrm B}}&=\ln2\approx0.693147\end{aligned}

[3][2]

Error analysis · Application

A model has two allowed microstates with probabilities 0.90 and 0.10. A solution sets $S=k_B\ln2$ because there are two states. Instead find $\frac S{k_{\mathrm B}}$ using the article’s Gibbs formula and compare.

Show hint

The two states are not equally probable.

Learning goal: Check equiprobability before using the Boltzmann shortcut.

Show solution and calculation

Approach: the probability-weighted logarithm sum gives 0.325083 rather than ln2≈0.693147.
State count alone does not justify the shortcut for unequal probabilities.

Check: replacing the probabilities by 0.50 and 0.50 would give ln2. Entropy is positive here but smaller than for the uniform model; multiplying by kB supplies units J/K.

\frac S{k_{\mathrm B}}=-0.90\ln0.90-0.10\ln0.10\approx0.325083

[2]

Statistical likelihood is not a mechanical constraint

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Added connections to Shannon entropy and to entropy change along a reversible path, citing Shannon’s original paper as an additional source; preserved existing derivations, exercises and original figures.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Refined the learning sequence, symbolic exercise approaches and model limits; added a third original example and quantitative figure, with suitable further explanations optional.

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Quanta. Boltzmann Entropy and Microstates. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/boltzmann-entropie-mikrozustaende

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Boltzmann entropy

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Applications

How do you calculate with Boltzmann Entropy and Microstates?

What Ω counts

Why the logarithm adds

Explicitly check equal probabilities

Count states and weight probabilities

Eight distinguishable fixed sites

Two states with unequal probabilities

Release an occupation-count constraint

Equal microstate chances give unequal macrostate chances

Application: entropy change of an ideal gas

Exponential counts without overflow

Five exercises with solutions

Two independent equally probable state sets have ΩA=6\Omega_A=6ΩA​=6 and ΩB=5\Omega_B=5ΩB​=5. Find ΩAB and SABkB\frac{S_{AB}}{k_{\mathrm B}}kB​SAB​​.

An equally probable multiplicity increases by factor 8. Find ΔS.

Four fixed labelled two-state sites have exactly one upper occupation. Find Ω; may the five possible occupation-count macrostates be equally weighted?

Two labelled two-state sites A and B may occupy only joint microstates 00 and 11, equally likely. Each site separately can be 0 or 1. Find ΩAB and SABkB\frac{S_{AB}}{k_{\mathrm B}}kB​SAB​​. Why is ΩAΩB not the joint multiplicity?

A model has two allowed microstates with probabilities 0.90 and 0.10. A solution sets S=kBln⁡2S=k_B\ln2S=kB​ln2 because there are two states. Instead find SkB\frac S{k_{\mathrm B}}kB​S​ using the article’s Gibbs formula and compare.

Statistical likelihood is not a mechanical constraint

Sources for this explanation

Frequently asked questions about Boltzmann Entropy and Microstates

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Boltzmann Entropy and Microstates for exams

Boltzmann entropy

Variables & units

Boltzmann entropy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Statistical thermodynamics

Worked example

Applications

How do you calculate with Boltzmann Entropy and Microstates?

What Ω counts

Why the logarithm adds

Explicitly check equal probabilities

Count states and weight probabilities

Eight distinguishable fixed sites

Two states with unequal probabilities

Release an occupation-count constraint

Equal microstate chances give unequal macrostate chances

Application: entropy change of an ideal gas

Exponential counts without overflow

Five exercises with solutions

Two independent equally probable state sets have ΩA=6\Omega_A=6ΩA​=6 and ΩB=5\Omega_B=5ΩB​=5. Find ΩAB and SABkB\frac{S_{AB}}{k_{\mathrm B}}kB​SAB​​.

An equally probable multiplicity increases by factor 8. Find ΔS.

Four fixed labelled two-state sites have exactly one upper occupation. Find Ω; may the five possible occupation-count macrostates be equally weighted?

Two labelled two-state sites A and B may occupy only joint microstates 00 and 11, equally likely. Each site separately can be 0 or 1. Find ΩAB and SABkB\frac{S_{AB}}{k_{\mathrm B}}kB​SAB​​. Why is ΩAΩB not the joint multiplicity?

A model has two allowed microstates with probabilities 0.90 and 0.10. A solution sets S=kBln⁡2S=k_B\ln2S=kB​ln2 because there are two states. Instead find SkB\frac S{k_{\mathrm B}}kB​S​ using the article’s Gibbs formula and compare.

Statistical likelihood is not a mechanical constraint

Sources for this explanation

Frequently asked questions about Boltzmann Entropy and Microstates

Subject references and further reading

More formulas: Heat and thermodynamics

Retain Boltzmann Entropy and Microstates for exams

Two independent equally probable state sets have $\Omega_A=6$ and $\Omega_B=5$ . Find ΩAB and $\frac{S_{AB}}{k_{\mathrm B}}$ .

Two labelled two-state sites A and B may occupy only joint microstates 00 and 11, equally likely. Each site separately can be 0 or 1. Find ΩAB and $\frac{S_{AB}}{k_{\mathrm B}}$ . Why is ΩAΩB not the joint multiplicity?

A model has two allowed microstates with probabilities 0.90 and 0.10. A solution sets $S=k_B\ln2$ because there are two states. Instead find $\frac S{k_{\mathrm B}}$ using the article’s Gibbs formula and compare.

Two independent equally probable state sets have $\Omega_A=6$ and $\Omega_B=5$ . Find ΩAB and $\frac{S_{AB}}{k_{\mathrm B}}$ .

Two labelled two-state sites A and B may occupy only joint microstates 00 and 11, equally likely. Each site separately can be 0 or 1. Find ΩAB and $\frac{S_{AB}}{k_{\mathrm B}}$ . Why is ΩAΩB not the joint multiplicity?

A model has two allowed microstates with probabilities 0.90 and 0.10. A solution sets $S=k_B\ln2$ because there are two states. Instead find $\frac S{k_{\mathrm B}}$ using the article’s Gibbs formula and compare.