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The entropy contribution from populated molecular rotational states. A nonlinear rigid rotor has $q_{\mathrm{rot}}\propto T^{3/2}$ at high temperature, giving $S_{\mathrm{rot},\mathrm{m}}=R(\ln q_{\mathrm{rot}}+3/2)$. Its rotational symmetry number removes redundant orientations.

The rotational symmetry number corrects a rotational phase-space count for indistinguishable molecular orientations. A linear molecule with identical ends, such as carbon dioxide with identical oxygen isotopes, has $\sigma =2$; making its ends isotopically distinct gives $\sigma =1$. In the high-temperature linear rigid rotor approximation, $q_{\mathrm{rot}}=8{\pi }^{2}IkT/(\sigma h^2)$. Holding the spectrum otherwise fixed, doubling $\sigma$ lowers molar rotational entropy by $R\log 2$.

In a flexible crosslinked polymer network, deformation restricts the configurations of chain segments between cross-links. The resulting decrease of entropy produces a restoring entropic force. For an ideal affine Gaussian network, the shear modulus is $G={\nu }_s{k}_{B}T$, where ${\nu }_s$ is the number density of elastically active strands. A nearly incompressible isotropic rubber has Young's modulus $E\simeq 3G$. At fixed network density within the rubbery regime, both increase approximately linearly with absolute temperature. This statement does not describe crossing the glass transition or chemical degradation.

The strings in Boltzmann entropy of a two-box occupancy are complete microstates only within the binary toy model. They retain a coarse-grained variable for each particle and omit its exact position within a box and its momentum. An actual classical gas instead occupies continuous phase space; obtaining a dimensionless count from a phase space volume requires a specified cell convention and the appropriate treatment of particle identity. A quantum state count must respect the allowed quantum states and quantum statistics. Thus $2^N$ is not a general count of microstates for a gas obeying quantum ideal-gas statistics. ^[ref]

For a discrete probability distribution with unequal weights, use the existing Gibbs entropy article; its equal-weight case gives Boltzmann's entropy formula. Shannon entropy uses the corresponding information measure, with the logarithm base setting its units. A count alone cannot replace the specified probability distribution, as the Boltzmann distribution examples demonstrate. ^[ref]

Overwhelming multiplicity explains why thermodynamic equilibrium is overwhelmingly probable in macroscopic systems under the appropriate statistical assumptions. It does not make low-entropy thermal fluctuations logically impossible. Nor does counting alone derive irreversible macroscopic evolution: one also needs dynamics, an initial preparation, and a reason the system samples the allowed microstates on the timescale of interest. ^[ref]

Historically, Boltzmann's 1877 paper counted microscopic arrangements, called complexions, and related the natural logarithm of their number to entropy. The Sharp and Matschinsky translation appeared in Entropy 17 (2015), pages 1971–2009, DOI 10.3390/e17041971. Its notation differs from ours: Boltzmann's permutability measure $\Omega$ is already logarithmic, whereas here $\Omega$ is the count. ^[ref] Josiah Willard Gibbs's Elementary Principles in Statistical Mechanics (1902) developed the statistical ensemble framework; Chapter XIV compares several statistical definitions of entropy with thermodynamics. ^{[ref] [ref]}

For an isolated system, the second law requires $\Delta S\ge 0$. More generally, include all affected surroundings to obtainEquality describes the reversible limit; irreversible macroscopic processes produce entropy. A closed system can lose entropy by exporting heat, so nondecrease does not apply to that system alone. The Clausius statement of the second law forbids a cyclic device whose sole effect is transferring heat from cold to hot. The Kelvin-Planck statement of the second law forbids a cyclic device whose sole effect is converting heat from one thermal reservoir completely into work. "Sole effect" matters: a refrigerator moves heat from cold to hot by consuming work. ^[ref]

For direct transfer of $Q=1000\mathrm{J}$ from a thermal reservoir at $600\mathrm{K}$ to one at $300\mathrm{K}$, with no work extracted,The reverse transfer alone would have negative total entropy change. Both directions conserve energy, but only the hot-to-cold direction is spontaneous. The finite-body example under entropy replaces these constant temperatures by changing ones.

Friction, free expansion and heat transfer across a finite temperature difference all introduce irreversibility. Moving arbitrarily slowly is insufficient if friction remains. A reversible thermodynamic process must permit restoration of both the system and surroundings. This gives engineers a useful ideal benchmark: irreversibility spends an opportunity to obtain work, even though energy itself is conserved. ^[ref]

Second sound is an entropy and temperature wave. At negligible thermal expansion, it has almost zero net mass flux. Lev Davidovich Landau predicted two sound modes in 1941.^[ref]

After thermal decoupling in cosmology, two homogeneous sectors that evolve adiabatically without entropy injection satisfy $s_{\mathrm{SM}}{a}^3=\mathrm{constant}$ and $s_{\varphi }{a}^3=\mathrm{constant}$, hence their entropy density ratio $\xi$ is fixed. At a common relativistic temperature before decoupling, $\xi =g_{\ast s}^{\mathrm{SM}}/g_{\ast s}^{\varphi }$. Internal annihilations can change each sector's temperature and effective entropy degrees of freedom without changing its total comoving entropy.

The Sherrington-Kirkpatrick model couples every distinct pair of Ising spins using independent standard normal random variables $z_{ij}$ for $i<j$, with $J_{ji}=J_{ij}$. Here $J_0$ and $\mathcal{J}$ have units of energy. David Sherrington and Scott Kirkpatrick introduced this scaling in 1975; their original replica-symmetric calculation gave an unphysical negative low-temperature entropy. The model and that initial proposed solution must be distinguished.^[ref]

A shock polar is the locus of possible downstream velocity vectors, or an equivalent pressure/deflection relation, as the orientation of an oblique shock varies for fixed upstream conditions. Entropy admissibility selects physical compressive shocks. The strong-shock perfect-gas velocity polar is a circle, while finite-Mach polars have a different shape.

A thermodynamic system for which the reversible mechanical work is entirely pressure-volume work. At fixed composition its internal energy is specified by entropy and volume; additional work modes require additional state variables.

Consider an original counting example: a square two-dimensional XY model of side $L$, a microscopic core radius $a$, and a fixed phase stiffness ${\rho }_s$. This deliberately neglects screening by other vortices. Away from a unit phase vortex, $\theta =\arg (x+iy)$ and $|\nabla \theta |=1/r$. Integrating the Goldstone-mode effective free energy over an annulus givesBoundary shape changes finite terms. There are order $(L/a{)}^2$ core positions, so the positional entropy is $S_{\mathrm{v}}=2k_B\log (L/a)+O(k_B)$, where $k_B$ is the Boltzmann constant. With $K={\rho }_s/(k_{B}T)$, subtracting $T{S}_{\mathrm{v}}$ gives the displayed free energy.

For $L/a=1000$ and $E_c/(k_{B}T)=2$, the dimensionless free energy cost is about $5.546$ at $K=0.8$ and $-0.965$ at $K=0.5$. The same logarithm appears in the cost and the positional reward: increasing the system size strengthens whichever has the larger coefficient. The corresponding one-defect Boltzmann factors, about $0.00390$ and $2.62$, are relative statistical weights, not normalized probabilities. The latter signals failure of a dilute single-defect estimate.

A single net phase vortex requires compatible boundaries: on a periodic sample use neutral pairs, as required by vortex charge neutrality. This argument identifies a possible threshold, but predicting the universal stiffness jump requires the renormalized phase stiffness, after smaller pairs have responded. It does not justify setting the transition temperature equal to a bare lattice coupling divided by a universal constant.

In ideal reduced magnetohydrodynamics with constant background sound and Alfvén speeds, perpendicular total-pressure balance gives $\delta p+{\rho }_0{v}_A^2\delta B_{\parallel }/B_0=0$. Put $s=\delta p/p_0-\gamma \delta \rho /{\rho }_0$, $r=\delta \rho /{\rho }_0+s/\gamma$, $c_T=v_A{c}_s/(v_A^2+c_s^2{)}^{1/2}$ and $h=c_s^{2}r/c_T$. Then the parallel momentum, continuity and induction equations yieldThus $f^{\pm }=u_{\parallel }\pm h$ satisfy $D{f}^{\pm }=\mp c_T{\nabla }_{\parallel }{f}^{\pm }$. The two slow-wave populations propagate along bent magnetic field lines and are mixed by the perpendicular Alfvénic turbulence, while the entropy fluctuation is simply advected. They do not force the perpendicular Alfvénic system at this leading order.

Molar crystal volume follows from $V_m=N_A{V}_{\mathrm{cell}}/Z$ with $Z$ atoms per cell for an elemental solid. Combining two such values with their molar entropy difference gives $\mathrm{d}P/\mathrm{d}T=\Delta S_m/\Delta V_m$. Ambient values supply only an approximation to a high-pressure coexistence slope, not a phase-boundary intercept.

Solubility specifies the amount that can remain dissolved at chemical equilibrium under stated conditions; it is different from how quickly a sample dissolves. Separating ions from an ionic crystal costs energy, but solvation establishes new interactions with the solvent. In water, hydration of ions couples charged ions to polar water molecules. The net enthalpy of dissolution can have either sign; entropy and concentration also enter the Gibbs free energy balance. Consequently, being a salt does not guarantee high solubility in water or any other solvent.^[ref]

Low solubility must also be distinguished from weak electrolyte behavior. A small amount of an ionic compound can dissolve essentially completely as ions, giving few charge carriers because little material dissolved. Calling that a weak electrolyte solely because its chemical solution conducts poorly would conflate two different questions.^[ref]

At fixed temperature and pressure, compare fully specified initial and final states usingThe entropy change is for the whole selected system, including both solute and solvent. Dispersing a solute can increase accessible arrangements while orienting nearby solvent molecules restricts others. One picture of an ordered hydration shell cannot determine the sign of the total entropy change. A negative Gibbs free energy change favors the specified process; it does not determine its reaction rate.^[ref]

An illustrative enthalpy accounting path separates solute constituents, reorganizes the solvent, and establishes their interactions:This is a chosen decomposition between compatible states, not a measured sequence. If a gaseous-ion-to-chemical solution solvation term already includes solvent reorganization, do not add that reorganization a second time. Lattice enthalpy from solution calorimetry gives the alternative Hess's law route that combines lattice separation with the complete hydration of ions term.^[ref]

The invented inputs are $+100$ for separation and $+20$ for solvent reorganization, in $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Interaction terms of $-150$ in solvent A and $-90$ in solvent B give $\Delta H_A=-30$ and $\Delta H_B=+30$. Stipulated total entropy changes of $-50$ and $+150\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ give $-T\Delta S_A=+15$ and $-T\Delta S_B=-45\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at $300\mathrm{K}$. Both selected processes therefore have $\Delta G=-15\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. This deliberately constructed comparison shows why the signs of the enthalpy and Gibbs free energy changes need not agree; it does not predict either solvent's solubility.

For incremental dissolution at a particular composition, the relevant chemical potential difference also depends on thermodynamic activities. At chemical equilibrium that driving difference vanishes. A quoted standard Gibbs free energy change and the finite change between arbitrary mixtures answer different questions; use compatible standard states and the reaction quotient when discussing solubility.

The motivating calorimetry experiment is especially instructive. A weighed potassium chloride sample dissolves in a known amount of water in a platinum-lined, shielded calorimeter; a quartz-oscillator thermometer and measured electrical heating provide the readouts. Electrical calibration before and after the reaction, comparison with a resistance thermometer, and checks on residual water and particle size constrain errors. The inspected authored article reports positive enthalpy of solution at $298.15\mathrm{K}$ for the specified dilution, despite complete dissolution. This is a combined enthalpy balance, not an isolated solvation free energy or direct single-ion hydration of ions measurement. Shield tracking and final concentration matter; the existing lattice enthalpy from solution calorimetry discussion gives the numerical result and the conversion limits.^[ref]

Specific entropy is entropy per unit mass. For a perfect gas with constant specific heat capacity $c_v$ at constant volume and specific-heat ratio $\gamma$, differences obeywhere $p$ is pressure and $\rho$ is mass density. Thus a homentropic flow has constant $p/{\rho }^{\gamma }$.

The sum of specific enthalpy and macroscopic kinetic energy per unit mass. In a steady adiabatic flow without shaft work, its change is determined by the change in external potential energy. At essentially constant external potential energy, it is conserved through a stationary nonradiating shock wave even though entropy increases.

The temperature reached by bringing a flow to rest adiabatically while converting its kinetic energy to enthalpy. For a calorically perfect gas, $T_{\mathrm{stag}}=T+u^2/(2c_p)=T[1+(\gamma -1)M^2/2]$. It is conserved across an adiabatic normal shock even though entropy increases and stagnation pressure decreases.

A state function assigns a value to each equilibrium thermodynamic state of a specified thermodynamic system. Its change depends only on the initial and final states. Internal energy $U$, entropy $S$ and enthalpy $H=U+pV$ are state functions; here $p$ is pressure and $V$ is volume. Transferred heat $Q$ and work $W$ belong to a thermodynamic process, rather than to a state. A system stores internal energy; it does not store separate inventories of “heat received” and “work received”.^[ref]

Use the explicit convention $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work done by the thermodynamic system on its surroundings. For a closed system with unchanged endpoint bulk kinetic energy and external potential energy, the first law of thermodynamics readsThe notation $dU$ denotes an exact differential of a state function; $\delta Q$ and $\delta W_{\mathrm{by}}$ denote incremental transfers. Along a reversible thermodynamic process of a simple compressible system,where $T$ is absolute temperature. These are rules for particular thermodynamic processes, not definitions of stored heat or stored work. In particular, a quasistatic process with friction can remain near thermodynamic equilibrium while producing entropy.^[ref]

For any state function $F$ and two equilibrium states $A,B$, an integral of its exact differential givesThe proof is telescoping: add the changes $F(x_{i+1})-F(x_i)$ along a subdivided path, and all intermediate values cancel. The smooth limit gives the integral. Conversely, if an integrable differential form has zero integral around every accessible closed path in a connected region, define $F(B)$ by integrating from a reference state. Two routes differ by a closed path, so the definition is unambiguous. This argument needs every accessible loop, rather than a local test that might miss a hole in the domain; see path independence.

The same cancellation does not hold separately for heat and work. A thermodynamic cycle has $\Delta U=0$ but can have nonzero net transfers, with $\oint \delta Q=\oint \delta W_{\mathrm{by}}$. Nor does the fact that entropy is a state function make $Q/T$ its general change: the formula $dS=\delta Q_{\mathrm{rev}}/T$ uses a reversible thermodynamic process and its varying temperature.

This distinction makes endpoint calculations powerful. We can compute an internal energy, entropy or enthalpy difference along a convenient reversible reference path even when the actual thermodynamic process is irreversible. Its actual heat and work must still be obtained from its own boundary interactions. The appealing cancellation is visible in two reversible ideal-gas paths between the same states: a different mechanical output requires a matching difference in thermal input, while stored internal energy arrives at the same value.

Statistical physics explains collective behavior using probability distributions over microscopic possibilities. Its central surprise is that incomplete microscopic knowledge can still produce precise predictions for internal energy, entropy and heat capacity. The bridge is statistical mechanics: specify the available microstates, assign their probabilities, then compute the expected values of measured quantities.

A two-level system already displays the idea. Warming it makes its higher energy more likely; its heat capacity first rises, then falls as the two probabilities approach equality. With two distinguishable units, merely counting microstates explains why the middle energy has twice the state degeneracy. These small examples expose the machinery before the much larger sums needed for a photon gas or an Ising model.

The appeal is economical: the same microscopic energies and probabilities explain both an average reading and its variance. A measurement's spread therefore becomes useful physical information rather than an inconvenience to discard.