Second Law of Thermodynamics and Entropy

Drop a hot stone into cold water and the water warms while the stone cools; you will wait forever to see the stone spontaneously reheat itself by chilling the water further. Shuffle a sorted deck and it disorders; keep shuffling and it never sorts. Let a gas fill a room and it never crowds back into one corner. Every one of these processes conserves energy perfectly — the first law is satisfied whichever direction they run. Yet only one direction ever happens. The second law of thermodynamics is the principle that supplies this arrow: it introduces an extensive state function, the entropy $S$ , and asserts that for any process in a thermally isolated system $S$ can only increase or stay the same. Historically the law was discovered through engines. Carnot's 1824 analysis of idealized reversible cycles, sharpened by Kelvin and Clausius, showed that the efficiency of a reversible engine depends only on the two reservoir temperatures — never on the working substance. Clausius distilled this into an inequality, $\oint\delta Q/T \leq 0$ , whose equality case for reversible cycles guarantees that $\delta Q_{\mathrm{rev}}/T$ is an exact differential. That defines entropy and hands the absolute thermodynamic temperature $T$ the role of integrating factor, converting the ragged, inexact heat differential $\delta Q$ into the perfectly exact $\mathrm{d}S$ . From entropy as a state function the entire structure of equilibrium thermodynamics unfolds. The fundamental relation $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V + \mu\,\mathrm{d}N$ identifies $T,p,\mu$ as derivatives of energy — or, inverted, of entropy. Maximizing total entropy at fixed energy, volume, and particle number yields the equilibrium conditions of equal temperatures, pressures, and chemical potentials between subsystems: the thermodynamic content of "reaching equilibrium." The free energies follow as Legendre transforms recasting the entropy-maximum principle into energy-minimum principles for systems coupled to reservoirs. Boltzmann gave the law its microscopic meaning: $S = k_{\mathrm B}\ln\Omega$ , where $\Omega$ counts the microstates compatible with a macrostate. Because the number of microstates grows astronomically with available phase-space volume, an isolated system overwhelmingly likely drifts toward macrostates of larger $\Omega$ . Entropy increase is then not an ironclad law of mechanics but a statement of overwhelming statistical probability for systems with $\sim10^{23}$ degrees of freedom — reconciling time-symmetric microscopic dynamics with the macroscopic arrow of time.

Key takeaways

The second law states that the total entropy of an isolated system never decreases: $\Delta S_{\text{tot}} \geq 0$ , with equality only for reversible processes.
Equivalent formulations — Clausius (heat does not spontaneously flow cold→hot), Kelvin–Planck (no cyclic engine fully converts heat to work), Carathéodory (adiabatic inaccessibility) — all imply the same entropy function.
The Clausius inequality $\oint\delta Q/T \leq 0$ defines entropy $\mathrm{d}S = \delta Q_{\mathrm{rev}}/T$ , integrable because $1/T$ is an integrating factor for heat.
Entropy distinguishes reversible from irreversible processes and sets the arrow of time: spontaneous changes increase total entropy and dissipate available work.
Equilibrium maximizes entropy at fixed energy; the conjugate variables $T,p,\mu$ emerge as partial derivatives of $S$ .
Statistically $S = k_{\mathrm B}\ln\Omega$ : entropy counts accessible microstates, and the law reflects the overwhelming probability of evolving toward larger phase-space volume.
Key idea: There exists an extensive state function $S$ , the entropy, such that for any adiabatically isolated system $\Delta S \geq 0$ , with equality iff the process is reversible. Equivalently $\mathrm{d}S = \delta Q_{\mathrm{rev}}/T$ , with $T$ acting as integrating factor for heat.

The core insight

There exists an extensive state function $S$ , the entropy, such that for any adiabatically isolated system $\Delta S \geq 0$ , with equality iff the process is reversible. Equivalently $\mathrm{d}S = \delta Q_{\mathrm{rev}}/T$ , with $T$ the absolute temperature acting as integrating factor for heat.

Where this lives today

The second law shapes the entire enterprise of energy conversion. A power plant with $T_h \approx 800\ \mathrm{K}$ and $T_c \approx 300\ \mathrm{K}$ cannot exceed the Carnot ceiling $\eta_{\max} = 1 - T_c/T_h = 0.625$ no matter the engineering; real plants reach $\sim 0.40$ because irreversibilities generate entropy. Refrigerators and heat pumps run the logic backward, and their ideal coefficient of performance $T_c/(T_h - T_c)$ explains why cooling near ambient is cheap and deep cryogenics is expensive. At the research frontier the law is being tested and extended at the smallest scales. In stochastic thermodynamics the entropy production of a single fluctuating trajectory — a colloidal bead, a stretched biomolecule — is a random quantity, and the Jarzynski equality and Crooks fluctuation theorem sharpen " $\Delta S \geq 0$ " into exact statements about the probability of transient entropy decreases, which vanish exponentially in system size. Meanwhile the thermodynamic arrow of time is tied to cosmology: because the microscopic laws are time-symmetric, the observed arrow requires a low-entropy initial condition, and the smooth, low-gravitational-entropy early universe is thought to supply exactly that boundary condition.

The physical picture

Think of entropy as a count of how many microscopic arrangements look the same macroscopically. A shuffled deck has vastly more disordered orderings than sorted ones, so random shuffling overwhelmingly lands you in disorder. The second law is the macroscopic shadow of this counting: with $10^{23}$ particles the higher-entropy macrostates are so overwhelmingly more numerous that the reverse is simply never seen. The arrow of time is a probability statement wearing the costume of a law. The absolute temperature $T$ is the price, in energy, of one unit of entropy: at fixed volume $\mathrm{d}U = T\,\mathrm{d}S$ . This single idea explains why heat flows hot-to-cold, and the figure below turns it into a number. Move an amount of heat $Q$ from a hot body to a cold one: the hot body's entropy drops by $Q/T_h$ , the cold body's rises by $Q/T_c$ , and because $T_c < T_h$ the rise beats the drop — the universe's entropy goes up.

Figure

The arrow of time as a number: heat Q from hot to cold raises total entropy by Q(1/Tc − 1/Th)

Reversibility is the boundary of the law. A process that proceeds through a continuous sequence of equilibrium states leaves the total entropy exactly unchanged and can be run backward. Real processes — free expansion, friction, finite-temperature heat flow — generate entropy and thereby pick out a preferred direction of time. The figure below contrasts the two: a reversible loop returns to zero, an irreversible one ratchets upward and never comes back down.

Figure

Reversible vs irreversible: total entropy returns to zero or ratchets up forever

Entropy maximization is an optimization principle. At fixed total energy an isolated system partitions energy, volume, and particles among its parts so as to maximize $S$ ; the stationarity conditions are the familiar equalities of intensive variables, and the second-order conditions guarantee thermodynamic stability.

The core equations

The engine-based origin of the law is the Clausius inequality: sum the ratio heat-over-temperature around any cycle and it cannot be positive.

\oint\frac{\delta Q}{T} \leq 0 \qquad (1)

where: $\delta Q$ — heat absorbed by the system at a step (inexact); $T$ — temperature of the reservoir supplying it; $\oint$ — around a complete cycle. Equality holds for reversible cycles; strict inequality signals irreversibility. This single line contains the entire second law. For a reversible cycle the equality forces path-independence, so $\delta Q_{\mathrm{rev}}/T$ is the differential of a state function — the entropy.

\mathrm{d}S = \frac{\delta Q_{\mathrm{rev}}}{T} \qquad (2)

where: $\mathrm{d}S$ — exact differential of entropy; $\delta Q_{\mathrm{rev}}$ — heat along a reversible path connecting the same endpoints. Because it is exact, $S(B) - S(A) = \int_A^B \delta Q_{\mathrm{rev}}/T$ is path-independent — the defining property of a state function. Comparing an irreversible path to a reversible return path yields the general inequality relating entropy change to actual heat.

\mathrm{d}S \geq \frac{\delta Q}{T} \qquad (3)

where: $\delta Q$ — actual (possibly irreversible) heat absorbed; $T$ — temperature of the surroundings. Equality for reversible steps; for an adiabatic ( $\delta Q = 0$ ) process this reads $\mathrm{d}S \geq 0$ . Applied to the isolated system-plus-surroundings, this becomes the law's headline statement.

\mathrm{d}S_{\text{tot}} = \mathrm{d}S_{\text{sys}} + \mathrm{d}S_{\text{surr}} \geq 0 \qquad (4)

where: $S_{\text{tot}}$ — entropy of the whole isolated universe; the subsystem $S_{\text{sys}}$ may freely decrease provided $S_{\text{surr}}$ rises by more. With entropy in hand, the first and second laws combine into the fundamental relation of thermodynamics.

\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V + \mu\,\mathrm{d}N \qquad (5)

where: $U$ — internal energy; $p$ — pressure conjugate to $V$ ; $\mu$ — chemical potential conjugate to $N$ . Every equilibrium condition and Maxwell relation descends from this line. Inverting to the entropy representation exhibits $T,p,\mu$ as slopes of $S$ .

\frac{1}{T} = \left(\frac{\partial S}{\partial U}\right)_{V,N}, \quad \frac{p}{T} = \left(\frac{\partial S}{\partial V}\right)_{U,N}, \quad \frac{\mu}{T} = -\left(\frac{\partial S}{\partial N}\right)_{U,V} \qquad (6)

where: each intensive variable is an entropy derivative. That $1/T = (\partial S/\partial U)$ is positive is why heat flows toward lower temperature — energy migrates to where it buys the most entropy. Boltzmann's bridge makes entropy a microstate count, connecting thermodynamics to combinatorics.

S = k_{\mathrm B}\ln\Omega \qquad (7)

where: $k_{\mathrm B} = 1.380649\times10^{-23}\ \mathrm{J/K}$ (exact, SI 2019); $\Omega$ — number of microstates consistent with the macrostate. The logarithm makes entropy additive: independent systems multiply their counts, and $\ln$ turns the product into a sum.

Derivation AThe universe gains entropy in irreversible heat flow, but exactly zero in a reversible loop

This is the quantitative heart of the second law, and the owned structural result of this article: we show that spontaneous heat flow across a finite temperature difference strictly increases total entropy, while an idealized reversible cycle changes it by exactly zero — pinning down where irreversibility lives. The strategy is to compute $\Delta S_{\text{tot}}$ in two clean cases. First, direct heat conduction between two reservoirs, where the sign follows from $T_c < T_h$ . Second, a reversible engine cycle, where the reversible heats satisfy $Q_c/Q_h = T_c/T_h$ (Carnot's theorem) and the two entropy fluxes cancel to zero. Together they localize the entropy production to the irreversible step.

Step 1. Let heat $Q > 0$ pass from a hot reservoir at $T_h$ to a cold reservoir at $T_c$ (finite gap, $T_h > T_c$ ). The hot reservoir loses $Q$ : its entropy change is $-Q/T_h$ . The cold reservoir gains $Q$ : its change is $+Q/T_c$ .

Step 2. Add them for the isolated pair:

\Delta S_{\text{tot}} = \frac{Q}{T_c} - \frac{Q}{T_h} = Q\left(\frac{1}{T_c} - \frac{1}{T_h}\right).

Why this is positive: since $T_c < T_h$ , we have $1/T_c > 1/T_h$ , so the bracket is strictly positive — the process generates entropy. It is reversible only in the limit $T_h \to T_c$ (infinitesimal gap).

Step 3. Now the reversible cycle. A reversible engine absorbs $Q_h$ at $T_h$ and rejects $Q_c$ at $T_c$ ; Carnot's theorem gives the signed relation $Q_c/T_c = -Q_h/T_h$ (heat rejected, $Q_c < 0$ ). Sum the entropy exchanged with the reservoirs:

\oint\frac{\delta Q_{\mathrm{rev}}}{T} = \frac{Q_h}{T_h} + \frac{Q_c}{T_c} = 0.

Why exactly zero: substitute $Q_c = -Q_h T_c/T_h$ and the terms cancel identically — no entropy is produced in a reversible loop.

Step 4. The contrast is the whole content of the law: the irreversible heat flow of Step 2 has $\Delta S_{\text{tot}} > 0$ , while the reversible loop of Step 3 has $\Delta S_{\text{tot}} = 0$ . Entropy production is the fingerprint of irreversibility, and its magnitude measures the lost opportunity to do work. Verified: $\Delta S_{\text{tot}} = Q(1/T_c - 1/T_h) = 0.833\ \mathrm{J/K}$ for $Q=1000\ \mathrm{J}$ , $T_h=400$ , $T_c=300$ ; and $Q_h/T_h + Q_c/T_c = 0$ symbolically — both in verify_aq-sm-second-law-entropy.py. Cf. Reif §5.5, Callen Ch. 4.

Go deeper · Derivation B · 1 minEntropy maximization gives thermal equilibrium: equal temperatures

Derivation BEntropy maximization gives thermal equilibrium: equal temperatures

We show that the abstract "maximize total entropy" principle produces the concrete, familiar equilibrium condition $T_1 = T_2$ , and that the approach to it obeys the Clausius statement (heat flows hot→cold). Two subsystems exchange energy through a fixed, impermeable wall inside an isolated composite. Because the total energy is conserved, only one energy variable is free; maximizing the additive entropy over it and inserting the thermodynamic definition of temperature delivers the result, while the sign of the first-order change gives the direction of spontaneous heat flow.

Step 1. The composite is isolated: $U = U_1 + U_2$ fixed, volumes and particle numbers fixed. Entropy is additive over weakly coupled parts:

S = S_1(U_1) + S_2(U_2), \qquad U_2 = U - U_1.

Step 2. At the entropy maximum the derivative with respect to the free variable $U_1$ vanishes:

\frac{\mathrm{d}S}{\mathrm{d}U_1} = \frac{\partial S_1}{\partial U_1} - \frac{\partial S_2}{\partial U_2} = 0.

Why the minus sign: $\mathrm{d}U_2 = -\mathrm{d}U_1$ because total energy is fixed.

Step 3. Insert the definition $1/T = (\partial S/\partial U)_{V,N}$ :

\frac{1}{T_1} = \frac{1}{T_2} \ \Rightarrow\ T_1 = T_2.

Equal temperatures is the equilibrium condition — the thermodynamic meaning of "thermal contact settling down."

Step 4. Away from equilibrium, the entropy change for a small energy transfer is

\mathrm{d}S = \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\mathrm{d}U_1 \geq 0.

If $T_1 < T_2$ then $1/T_1 > 1/T_2$ , so $\mathrm{d}U_1 > 0$ : energy flows into the colder body. Heat flows hot→cold — the Clausius statement, derived. Verified: the second-order condition $\partial^2 S/\partial U_1^2 < 0$ requires $C_V > 0$ (thermal stability); the two-body mixing $\Delta S = C\ln[T_f^2/(T_1 T_2)] > 0$ (AM > GM) is checked in the verify script. Cf. Callen Ch. 2, 8.

Limiting cases

The behavior of entropy under representative idealized processes:

Case	Result	Meaning
Reversible process	$\Delta S_{\text{tot}} = 0,\ \ \mathrm{d}S = \delta Q_{\mathrm{rev}}/T$	Quasi-static, no dissipation; entropy merely transferred, total conserved, invertible.
Irreversible process	$\Delta S_{\text{tot}} > 0,\ \ \mathrm{d}S > \delta Q/T$	Free expansion, friction, mixing, finite- $\Delta T$ heat flow; entropy generated internally.
Isolated at equilibrium	$S = S_{\max}\big\|_{U,V,N},\ \ \delta S = 0,\ \delta^2 S < 0$	Entropy stationary and maximal; perturbations only lower it — stability.
Approach to $T\to 0$	$\lim_{T\to0}S = S_0$ (const, $= 0$ for a perfect crystal)	Third law: entropy tends to a universal constant; $T = 0$ is unattainable in finite steps.

Worked examples

Example 1 — The entropy generated when hot and cold water mix (a real number).

Step 1. Bring $Q = 1000\ \mathrm{J}$ of heat directly from a hot reservoir at $T_h = 400\ \mathrm{K}$ to a cold one at $T_c = 300\ \mathrm{K}$ (large reservoirs, so temperatures stay fixed).

Step 2. Hot reservoir entropy change: $\Delta S_h = -Q/T_h = -1000/400 = -2.50\ \mathrm{J/K}$ . Cold reservoir: $\Delta S_c = +Q/T_c = 1000/300 = +3.33\ \mathrm{J/K}$ .

Step 3. Total: $\Delta S_{\text{tot}} = -2.50 + 3.33 = \mathbf{+0.83\ J/K}$ . Positive, so the process is irreversible — exactly as observed; you never see heat crawl back uphill.

Step 4. The lost work is $T_c\,\Delta S_{\text{tot}} = 300 \times 0.833 = \mathbf{250\ J}$ . Had a reversible engine handled the same $1000\ \mathrm{J}$ , it could have delivered $Q(1 - T_c/T_h) = 1000(1 - 0.75) = 250\ \mathrm{J}$ of work; direct conduction threw all of it away as generated entropy.

Example 2 — Free expansion of one mole of ideal gas.

Step 1. One mole of ideal gas doubles its volume by rushing into an evacuated chamber inside an insulated box: $V_f = 2V_i$ , with $\delta Q = 0$ and $\delta W = 0$ , so $\Delta U = 0$ and (ideal gas) $T_f = T_i$ .

Step 2. Entropy is a state function, so compute $\Delta S$ along a reversible isotherm between the same endpoints: $\Delta S = \int_{V_i}^{V_f}(nR/V)\,\mathrm{d}V = nR\ln(V_f/V_i)$ .

Step 3. For $V_f/V_i = 2$ and $n = 1$ mol: $\Delta S = (1)(8.314)\ln 2 = 8.314 \times 0.6931 = \mathbf{5.76\ J/K}$ . The surroundings are untouched ( $\Delta S_{\text{surr}} = 0$ ), so $\Delta S_{\text{tot}} = +5.76\ \mathrm{J/K} > 0$ — irreversible.

Step 4. Cross-check microscopically: doubling the volume gives each of $N = N_{\mathrm A}$ molecules twice as many accessible positions, so $\Omega_f/\Omega_i = 2^N$ and $\Delta S = k_{\mathrm B}\ln 2^N = N k_{\mathrm B}\ln 2 = nR\ln 2 = \mathbf{5.76\ J/K}$ — the counting picture and the thermodynamic integral agree exactly.

Going further

References. Fermi, Thermodynamics, Ch. 3–5 · Pathria & Beale, Statistical Mechanics, 3rd ed., Ch. 1 · Kardar, Statistical Physics of Particles, Ch. 1 · Reif, Fundamentals of Statistical and Thermal Physics, Ch. 3, 5 · Sethna, Statistical Mechanics, Ch. 5 · Callen, Thermodynamics and an Introduction to Thermostatistics, Ch. 1–4

Provenance. Equations and inequalities follow Pathria & Beale Ch. 1, Reif Ch. 3 & 5, Kardar Ch. 1, and Callen Ch. 1–4. A verifier should: (1) confirm the Carnot efficiency $\eta = 1 - T_c/T_h$ and the Clausius inequality; (2) recompute the irreversible heat-flow $\Delta S_{\text{tot}} = Q(1/T_c - 1/T_h) = 0.833\ \mathrm{J/K}$ and the reversible-loop $\oint\delta Q/T = 0$ ; (3) recompute $\Delta S = nR\ln 2 = 5.76\ \mathrm{J/K}$ for one mole using $R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ , and $\Delta S = k_{\mathrm B}\ln 2^N$ ; (4) verify the entropy-maximum equilibrium condition $T_1 = T_2$ and the stability requirement $C_V > 0$ . Constants: $k_{\mathrm B} = 1.380649\times10^{-23}\ \mathrm{J/K}$ (SI 2019 exact).

Builds on: First Law of Thermodynamics and Internal Energy · Work, Heat, and Inexact Differentials · Carnot Cycle and Thermodynamic Temperature · Reversible and Irreversible Processes · Zeroth Law of Thermodynamics and Empirical Temperature Leads to: Thermodynamic Potentials and the Fundamental Relation · Temperature, Pressure, and Chemical Potential as Entropy Derivatives · Helmholtz Free Energy, Enthalpy, and Gibbs Free Energy · Third Law of Thermodynamics · Maxwell Relations

Origins · 1 minWilliam Thomson and how this was discovered

William Thomson

1824–1907 · Glasgow, Scotland

William Thomson — ennobled as Lord Kelvin in 1892 — gave the Second Law one of its two canonical statements: no cyclic process can convert heat entirely into work with no other effect (the Kelvin–Planck statement). In 1848 he used Carnot's substance-independent efficiency to define an absolute thermodynamic temperature scale, the Kelvin scale, freeing temperature from any particular thermometric fluid. His 1852 paper on the "universal tendency in nature to the dissipation of mechanical energy" was among the first to state the arrow of time physically: usable energy degrades irreversibly even though the total is conserved. Clausius then named the degrading quantity entropy; Thomson supplied the engine-and-dissipation picture behind $\Delta S_{\text{univ}}\ge 0$ .

History

1824

— Carnot analyzes the ideal reversible engine and the temperature-only efficiency limit

1848

— Kelvin (William Thomson) proposes an absolute thermodynamic temperature scale from Carnot's result

1850

— Clausius states the second law: heat cannot spontaneously flow from cold to hot

1851

— Kelvin gives the Kelvin–Planck statement: no cyclic engine fully converts heat to work

1854

— Clausius formulates the inequality and the equivalence value of a transformation

1865

— Clausius names the state function entropy; states "the entropy of the universe tends to a maximum"

1877

— Boltzmann connects entropy to microstate counting, S = k ln W

Does the second law mean everything must become more disordered?

Only the total entropy of an isolated system is forced to increase; local order can and does grow constantly. Snowflakes crystallize, planets and stars form, life organizes itself — all while raising the entropy of their surroundings by more than they lower their own. "Disorder" is also a loose word; entropy precisely counts accessible microstates, and a system can look more "ordered" while occupying more phase space (a stretched polymer, for instance). The rigorous statement is $\Delta S_{\text{tot}} \geq 0$ , not "everything decays."

If microscopic physics is time-reversible, where does the arrow of time come from?

This is the deep puzzle the second law forces on us. The equations of mechanics run equally well forwards and backwards, so nothing in them singles out a direction. The arrow is statistical: there are astronomically more microstates corresponding to "gas spread out" than to "gas in one corner," so an isolated system drifts overwhelmingly toward the more numerous macrostates. But this requires the system to start in an improbable low-entropy state — and explaining why the universe began that way is a genuinely open question, pushed back to cosmological initial conditions. The law describes the arrow; it does not fully explain why there is one.

Is the entropy increase absolutely certain, or just very likely?

For a macroscopic system it is so overwhelmingly likely that "certain" is a fair everyday description, but strictly it is statistical. The probability of a spontaneous entropy decrease is roughly $e^{-\Delta S/k_{\mathrm B}}$ , and for any macroscopic $\Delta S$ this is a number like $e^{-10^{20}}$ — you would wait far longer than the age of the universe to see a mole of gas fluctuate back into one corner. For tiny systems (a few molecules, a colloidal bead) transient decreases are observed, and the Crooks and Jarzynski relations quantify exactly how often — turning " $\Delta S \geq 0$ " into a precise probabilistic statement.

Why is the absolute temperature the integrating factor, and what fixes its scale?

The Clausius inequality becomes an equality for reversible cycles, which means $\delta Q_{\mathrm{rev}}/T$ has vanishing loop integral — it is exact, so $1/T$ is the integrating factor that turns inexact heat into the exact differential of entropy. What fixes $T$ is Carnot's theorem: the ratio of reversible heats between two reservoirs is the same for every working substance, and that universal ratio defines the temperature ratio $T_h/T_c$ . Setting the triple point of water (or, since 2019, fixing $k_{\mathrm B}$ ) pins the scale. The absolute temperature is not just a thermometer reading; it is the substance-independent quantity that makes entropy exist.

What is the difference between the Clausius and Kelvin–Planck statements?

They sound different but are logically equivalent. Clausius: heat cannot spontaneously flow from a colder to a hotter body (no free refrigeration). Kelvin–Planck: no cyclic process can take heat from a single reservoir and convert it entirely into work (no perfect heat-to-work engine). You prove their equivalence by contradiction: a device violating one can be combined with an ordinary engine or refrigerator to violate the other. Both are subsumed by the entropy formulation $\Delta S_{\text{tot}} \geq 0$ , which is why that single inequality is taken as the master statement.

How can entropy be both "heat over temperature" and "log of microstate count"?

Because they are the same quantity seen from two levels. Clausius defined entropy macroscopically as the state function whose differential is $\delta Q_{\mathrm{rev}}/T$ — pure thermodynamics, no atoms. Boltzmann later showed this thermodynamic entropy equals $k_{\mathrm B}\ln\Omega$ , the (log of the) number of microstates realizing the macrostate. The bridge is checkable: for the free expansion of an ideal gas, the thermodynamic integral gives $\Delta S = nR\ln 2$ , and the microscopic count gives $k_{\mathrm B}\ln 2^N = nR\ln 2$ — identical. The macroscopic $\delta Q/T$ and the microscopic $k_{\mathrm B}\ln\Omega$ are two faces of one function.