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A magnetic field favors alignment of the magnetic moments of a paramagnet, reducing the number of substantially occupied magnetic configurations at fixed temperature. Its magnetic entropy therefore decreases as the field magnitude increases, explaining the negative sign of $(\partial S/\partial B{)}_T=-\alpha B/T^2$.

Choose the magnetic Helmholtz free energy convention $dF=-SdT-mdB$. Equality of its mixed partial derivatives gives the magnetic Maxwell relationThe magnetic moment inferred from an entropy derivative, integrated at fixed $B$, isThe given entropy derivative alone cannot determine the temperature-independent function $m_0(B)$. With the usual independent-paramagnet condition $m\to 0$ as $T\to \infty$ at fixed $B$, $m_0(B)=0$, giving Curie's law, $m=\alpha B/T$. The sketch uses that additional condition. The inverse-$T$ expression applies in the weak-field regime; its divergence cannot describe the actual low-temperature saturation of a finite set of magnetic moments.

For reversible quasistatic thermal isolation, $dS=0$. With positive field-dependent heat capacity $C_B=T(\partial S/\partial T{)}_B$,Reducing a positive $B$ therefore lowers $T$: this is adiabatic demagnetization. Equivalently, reducing $B^2$ cools for either field sign. Reduced alignment would raise magnetic entropy at fixed $T$; cooling compensates so total entropy remains constant. Insulation without reversibility would allow additional entropy production and need not follow this adiabat exactly.

For pure phases at the $1$ bar reference pressure, the equilibrium condition is ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }=0$. ThusThe calculation uses the stipulated constant reaction enthalpy and entropy, with the fluid at its reference chemical fugacity.

At the one-bar standard state, pure phases have unit thermodynamic activities. The Gibbs energy of reaction balance is ${\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }=0$, givingThis uses the stipulated temperature-independent reaction enthalpy and entropy differences, taken per mole of the written reaction.

At the one-bar standard state, pure mineral endmembers and water vapour treated as an ideal gas all have unit thermodynamic activity. The Gibbs energy of reaction is ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$. Setting it to zero givesThe quantities are taken per mole of the reaction as written. The simplified result uses the stipulated temperature-independent enthalpy and entropy differences.

The hypothesis of a special low-entropy boundary condition in the past, used in statistical accounts of the thermodynamic arrow of time. A complete account must also state its probability measure over compatible microstates and the macroscopic description used to define entropy.

The reciprocal of the baryon-to-photon ratio, not a synonym for it. In a thermal photon gas this number ratio is proportional to, but differs from, the thermodynamic entropy per baryon. The exact photon entropy per photon is $4u_{\gamma }/(3k_{B}T{n}_{\gamma })\simeq 3.60$, so the photon-sector entropy per baryon in units of $k_B$ is about $3.60s_{\gamma }$. The entropy sector must be specified when other relativistic species contribute.

Many ionic crystals have high melting temperatures compared with small molecular solids, because reorganizing their strongly interacting ions costs energy. But melting leaves a dense liquid with interactions; it does not perform the complete separation defining a lattice energy. At fixed pressure, the transition occurs when the Gibbs free energies of the solid and liquid match. The relevant enthalpy and entropy differences both matter, so lattice energy alone is not a universal ordering of melting temperatures. Strong covalent bonds in extended networks can also give high melting temperatures.^[ref]

Polymer adsorption binds a polymer near an attractive interface through contacts of some of its monomers. The binding energy competes with the loss of conformational entropy. A chain need not lie flat: noncontacting loops and tails can extend into the solvent. In a weak-adsorption blob model, the layer thickness exceeds the Kuhn length and local chain statistics remain approximately those of the free chain below that thickness.

A layer of sufficiently crowded tethered polymer chains stretched away from their attachment surface. In a good solvent, overlap increases the interaction contribution to free energy; extension relieves this crowding at a cost of chain entropy. The mushroom-to-brush crossover occurs at ${\sigma }_g{R}^2\sim 1$, where $R$ is an isolated coil size. In a simple good-solvent scaling model, the brush height is $H\sim Nb(v{\sigma }_g/b{)}^{1/3}$ for effective segment excluded volume $v$ and statistical segment length $b$; if $v\sim b^3$, this becomes $H\sim Nb({\sigma }_g{b}^2{)}^{1/3}$. The scaling requires strong stretching relative to the isolated coil but extension well below the contour length $Nb$.

Low-temperature solid helium-3 can have greater entropy than the liquid through disordered nuclear spins, enabling compression cooling. ^[ref]

The sign follows from the Clapeyron equation. Let $S_l,S_s$ be molar entropies and $V_l,V_s$ molar volumes of liquid and solid, respectively, on their phase coexistence curve. Equality of their chemical potentials, with $d\mu =-SdT+VdP$, givesWith $S_s>S_l$ and $V_l>V_s$, the slope is negative. Reversibly freezing one mole at fixed temperature requires heat into the sample,Here $L_{s\to l}$ uses the melting convention for latent heat of fusion. This is the opposite sign from ordinary ice melting. The inequalities specify the regime; freezing does not universally cool a substance. In an approximately isolated sample, the growing higher-entropy phase draws on the sample's thermal energy instead of an external heater. Irreversible losses and incoming heat oppose the cooling.

An original deliberately local model makes the cooling quantitative. Fix a total amount $n$ of material, let $x$ be the mole fraction of solid, and stipulate molar entropiesThe constants $a$ and $b$ have units of molar entropy per temperature and molar entropy, respectively. Ignore their pressure dependence and the cell's own heat capacity. This is a local thermodynamics model, not an extrapolation to zero temperature. A constant positive $b$ cannot remain valid once the solid's spins order.

For a reversible thermodynamic process with no heat exchange, total entropy is constant. Differentiating giveswhen $b>aT$ and $x<1$. Define $T_0=b/a$; solving between two states givesIn particular, stipulate $x_0=0$, $T_i=T_0/4$ and $x=1/8$. ThenThe last expression is the local reversible heat requirement per mole frozen at the final state, divided by $b{T}_0$; it is not the heat exchanged during the isolated compression. That compression exchanges no heat. This distinction separates the latent-heat sign from the actual boundary heat budget.

The utility of the model is its explicit condition: $S_s>S_l$ makes $dT/dx<0$, while positive total heat capacity supplies a stable thermal response. More generally, for molar entropies depending only on temperature, let $C_l,C_s$ be the molar heat capacities. Then $dT/dx=-T(S_s-S_l)/[(1-x)C_l+x{C}_s]$. Incoming heat or entropy production adds terms to the balance and can overwhelm the ideal cooling. Neither the numerical ratios above nor the straight entropy lines are measurements of helium-3.

If otherwise comparable isolated planets of one mass follow $L_{\mathrm{int}}=C{t}^{-\alpha }$, their intrinsic luminosity ratio is $L_{\mathrm{int}}(t_1)/L_{\mathrm{int}}(t_2)=(t_2/t_1{)}^{\alpha }$. The normalization depends on composition and initial entropy. Strong stellar irradiation, deep heating, or accretion can invalidate extrapolating the same cooling law.

Pressure measures the compressive normal force per unit surface area. For a uniformly loaded surface of surface area $A$, the magnitude of the perpendicular force is $F_{\perp }=PA$; locally, $P=d{F}_{\perp }/dA$. Its SI unit is the pascal, $1\mathrm{Pa}=1\mathrm{N}{\mathrm{m}}^{-2}$. A force parallel to the surface contributes shear stress, rather than pressure. In a gas, the steady macroscopic force comes from many microscopic transfers of momentum; ideal gas pressure from wall collisions makes this connection explicit. ^[ref]

For a computed example, a uniform pressure of $100\mathrm{kPa}$ on a piston of surface area $10\mathrm{c}{\mathrm{m}}^2=10^{-3}{\mathrm{m}}^2$ produces a perpendicular force of $100\mathrm{N}$. If the opposite face is exposed to $80\mathrm{kPa}$, the net force due to the two pressures is $(100-80)\times 10^3\times 10^{-3}=20\mathrm{N}$, directed toward the lower-pressure side. The pressure on either face and their difference answer different questions; gauge pressure records a difference from a specified reference, often atmospheric pressure.

A scalar pressure describes a gas in local thermodynamic equilibrium with isotropic random particle velocities: every wall orientation has the same normal force per unit surface area. Without this isotropy, normal stresses can depend on direction and a pressure tensor is needed. Thus random motion alone is insufficient; a preferred direction matters.

In thermodynamics, pressure and volume are thermodynamic conjugate variables. For a simple compressible system, reversible expansion delivers work $\delta W=PdV$, and at fixed entropy $S$ and particle number $N$,where $U$ is internal energy. A large pressure therefore means both a large wall force per unit surface area and a large amount of expansion work per small increase in volume.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

A shock wave with significant radiation losses in its downstream evolution. Often a narrow dissipative jump initially heats the gas according to adiabatic Rankine-Hugoniot conditions, followed by a much longer radiative cooling layer. The gas's specific entropy increases at the jump and may then fall as radiation carries entropy away. If both distant states have equal temperature, the combined mass and momentum balances give an isothermal shock; the gas's energy flux is not conserved across the whole radiating region.

The entropy change per specified extent of a reaction, equal to the stoichiometric product-minus-reactant sum of partial molar entropies.

Residual entropy is entropy remaining in the zero-temperature limit. An equilibrium ground-state degeneracy $g_0$ gives $S_0=k_B\ln g_0$ for the complete system; an extensive residual entropy requires a correspondingly large state count. Frozen configurational disorder can also leave an experimentally inferred residual entropy when a sample fails to equilibrate. The zero-entropy perfect-crystal form of the third law of thermodynamics assumes a unique, fully ordered equilibrium ground state.

An ideal cyclic heat engine producing no entropy overall. Between two equilibrium reservoirs its efficiency reaches the Carnot bound; finite reservoir temperatures can be treated as varying through successive reversible cycles.

A thermodynamically reversible process can be reversed without a net change of the system and surroundings. Its entropy change equals reversible heat transfer divided by temperature. A reversible adiabatic process conserves entropy; adiabaticity alone does not forbid internal entropy production in an irreversible evolution.

Chain growth by successive olefin metathesis steps that open cyclic alkenes. Relief of ring strain can offset the loss of monomer translational entropy.