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A bubble chamber photographs vapor bubbles along charged particle paths in a transparent superheated liquid. The liquid serves as target and recording medium.^[ref]

Compare the two phase changes: a cloud chamber forms liquid drops inside supersaturated vapor, whereas a bubble chamber forms vapor cavities inside a superheated liquid. Both use metastability to amplify a microscopic disturbance. The photons reaching the camera reveal the drops or bubbles. The photographed trail's width is not the particle's width: it includes bubble growth, illumination and spatial resolution.

The Gibbs free energy iswhere $H$ is enthalpy. For a closed system of fixed composition, with only pressure-volume work, spontaneous relaxation at fixed temperature and pressure cannot increase $G$; stable thermodynamic equilibrium minimizes it over the allowed states. A loaded piston can move while maintaining the imposed pressure, so the enthalpy term includes the displacement cost. Along a smooth equilibrium branch of fixed composition,Thus its temperature slope at fixed pressure is $-S$, while its pressure slope at fixed temperature is $V$. See MIT 5.60, Lecture 13, Gibbs free energy and its derivatives.

Consider two illustrative thermodynamic phases, A and B, of the same substance at one fixed pressure. Write B minus A, using quantities per mole:Treat these differences as approximately constant over a limited temperature window. ThenAt $280\mathrm{K}$, $\Delta G=+0.8\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, so A is preferred. At $320\mathrm{K}$, $\Delta G=-0.8\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, so B is preferred. At $300\mathrm{K}$, both have equal molar Gibbs free energy, permitting phase coexistence in this bulk model. B's greater entropy outweighs its greater enthalpy above the crossing.

Each plotted point compares alternatives at the same temperature; moving along the curve changes the imposed temperature. The model predicts preference, not instantaneous conversion: nucleation barriers can sustain metastability. Real enthalpy and entropy differences generally vary with temperature through heat capacities, bending this line outside the approximation.

At fixed pressure, the first law of thermodynamics gives $Q=\Delta H$ under the stated work restriction, so the same thermal reservoir accounting yields $\Delta S_{\mathrm{total}}=-\Delta G/T\ge 0$. Variable composition restores ${\sum }_i{\mu }_{i}d{n}_i$ to $dG$; constrained Gibbs minimization for chemical equilibrium handles those permitted changes.

The historical foundation is Gibbs's ^[ref]. Its equations (87) and (91) introduce the combinations now called Helmholtz free energy and Gibbs free energy, connecting entropy criteria, chemical potentials and phase coexistence within one framework.

Nucleation forms a sufficiently large region of a new phase within an old metastable phase. In classical nucleation theory, positive surface energy competes with bulk free-energy gain, producing a critical nucleus and a barrier to conversion. The Arrhenius nucleation time depends exponentially on this barrier; boundaries and impurities can alter it. This explains how metastability and hysteresis can persist despite a different globally stable phase.

A first-order phase transition has a discontinuity in a first derivative of the equilibrium free energy, such as entropy or the order parameter; a thermal first-order transition normally has latent heat. At a continuous phase transition the order parameter goes continuously to zero and there is no latent heat, although response functions and higher derivatives of the free energy can be singular.

For the symmetric scalar Landau free energy at zero conjugate field, minimizeIf $u>0$, the sixth-order term is unnecessary near $r=0$. For $r>0$ the minimum is $M=0$, whereas for $r<0$ there are two symmetry-related minima with $M^2=-r/u$ to leading order. The order parameter vanishes continuously as $r\uparrow 0$. The equilibrium free-energy density changes from zero to $-r^2/(4u)$, so its first temperature derivative is continuous and its second derivative has a jump. This is the ordinary mean-field continuous phase transition.

If $u<0$, keep $v>0$ for stability. A nonzero stationary order parameter, with $s=M^2$, obeys $r+us+v{s}^2=0$. For phase coexistence with $M=0$ its free energy must also vanish. Substituting $r=-us-v{s}^2$ into $V$ givesThe nonzero solution of $V=0$ is thereforeThe competing minima exchange stability at positive $r_t$, with a finite jump of the order parameter. This is a first-order phase transition. The metastability region is also visible: nonzero stationary points first appear at the spinodal $r=u^2/(4v)$, and the disordered local minimum loses stability at $r=0$.

For example, holding $u,v$ fixed and taking $r=a(T-T_0)$, the ordered phase has entropy lower than the disordered phase by $a{M}_t^2/2$ at coexistence. Heating through the transition absorbs latent heat per volume $T_{t}a{M}_t^2/2$. More generally the temperature dependence of all coefficients contributes to this entropy difference. With no inversion symmetry, a cubic term can also produce a first-order phase transition, as comparison of the competing minima of the Landau free energy shows.

The uniform stationarity equation and local stability condition areFor $u>0$, $h=0$, and small negative $r$, the stable nonzero solution satisfies $M^2=-r/u+O(r^2)$. Thus the order parameter vanishes continuously as $r\uparrow 0$. The equilibrium Landau free energy has leading singular contribution $f_{\min }=-r^2/(4u)$ below the transition and zero above it. Its first thermal derivative is continuous, while its second derivative has a finite jump: the mean-field approximation predicts a continuous phase transition with no latent heat. The magnetic susceptibility follows by differentiating the stationarity equation:It diverges on approaching the ordinary thermodynamic critical point, and the quadratic Landau scalar correlation length behaves as $\xi \sim \sqrt{\kappa /|r|}$, with a different amplitude on the two sides.

For $u<0$, a nonzero stationary point at $h=0$ has $r+u{M}^2+v{M}^4=0$. Substitution into $V(M)-V(0)$ gives $-u{M}^4/4-v{M}^6/3$. Setting this to zero yieldsThe finite jump of $M$ makes this a first-order phase transition. For a thermal path $r=r(T)$ with $r^{\prime }(T)\ne 0$ and the other coefficients fixed, $\partial f_{\min }/\partial T=r^{\prime }(T)M^2/2$ jumps, giving an entropy discontinuity and nonzero latent heat. The Landau free energy itself remains continuous. The disordered minimum loses stability at $r=0$; the two nonzero stationary solutions merge at $r=u^2/(4v)$. These spinodal points bound metastability, but neither replaces the equilibrium phase coexistence condition.

Writing $r=at$ with $a>0$ near an ordinary thermodynamic critical point, the zero-field equation gives $M\sim (-t{)}^{1/2}$. At $r=0$, small $h$ obeys $h=u{M}^3+O(M^5)$, hence $M\sim \mathrm{sgn}(h)|h{|}^{1/3}$. The ordinary mean-field critical exponents are therefore $\beta =1/2$ and $\delta =3$. These are Landau theory predictions, not dimension-independent values for an interacting theory below its upper critical dimension.

A continuous phase transition has a continuously vanishing order parameter and no latent heat. A first-order phase transition exchanges globally stable phases with discontinuous first derivatives of a thermodynamic potential; the order parameter generally jumps, and an entropy jump gives latent heat. In Landau theory the distinction is determined by global minima, not just by the appearance of additional stationary points.

For the uniform scalar potential $U(M)=r{M}^2/2+u{M}^4/4$ with $u>0$ and $h=0$, its stationary equation is $M(r+u{M}^2)=0$. For $r>0$, $M=0$ is the minimum; for $r<0$, the minima are $M=\pm \sqrt{-r/u}$. They approach zero continuously as $r\uparrow 0$, giving an ordinary continuous transition.

For $u<0$ and $v>0$, use the sextic potential. Write $y=M^2$. Stationarity of a nonzero phase requires $r+uy+v{y}^2=0$, while equality of its potential with $U(0)=0$ requires $ry/2+u{y}^2/4+v{y}^3/6=0$. Eliminating $r$ yieldsAt this value the potential factors as $U(M)=\frac{v}{6}{M}^2(M^2+3u/(4v){)}^2\ge 0$, proving that the three stationary zeros really are degenerate global minima. Crossing this line makes the order parameter jump: the transition is first order. The disordered and ordered spinodal points instead occur at $r=0$ and $r=u^2/(4v)$ respectively, delimiting metastability; these are not the equilibrium transition. If the symmetry allows a cubic term, that term can also produce a first-order transition.

At a continuous phase transition the equilibrium order parameter tends continuously to its critical value, there is no latent heat, and the correlation length diverges. For $u>0$ at zero field, minimizing $V(M)=r{M}^2/2+u{M}^4/4$ gives $M=0$ when $r\ge 0$ and $M=\pm \sqrt{-r/u}$ when $r<0$. These minima join continuously at $r=0$. The minimized singular free-energy density is zero above and $-r^2/(4u)$ below, so its first thermal derivative is continuous, while its second derivative has a finite jump in mean-field approximation.

A first-order phase transition instead has competing minima whose free energies cross while their order parameters remain distinct. The first derivatives of the equilibrium free energy generally jump; along a temperature-driven crossing this gives latent heat. For the symmetric sextic potential with $u<0,v>0$, let $y=M^2>0$. At phase coexistence, the stationary points and equality of the two free energies requireEliminating $r$ gives $-u{y}^2/4-v{y}^3/3=0$, henceAt coexistence $V(M)=v{M}^2(M^2+3u/(4v){)}^2/6$, proving that all three degenerate minima are global. The equilibrium order parameter jumps from zero to one of the two nonzero values even though $r_{\mathrm{coex}}>0$. The disordered minimum loses local stability at $r=0$, and the nonzero stationary points first appear at $r=u^2/(4v)$. These spinodal points delimit metastability and are not the equilibrium transition line. Cubic terms, when symmetry permits them, provide another common route to a first-order transition.

A first-order phase transition exchanges two distinct global minima of the Landau free energy. The equilibrium free energy remains continuous, but a first derivative such as the magnetization or entropy jumps. A temperature-driven transition generically has latent heat; crossing $h=0$ between symmetry-related ordered phases instead has a magnetization jump without requiring latent heat. A continuous phase transition has no such order-parameter jump and usually has a divergent correlation length and nonanalytic higher responses.

For $u>0$, $h=0$ and $r$ crossing zero, minimizing $V=r{M}^2/2+u{M}^4/4$ gives $M=0$ for $r>0$ and $M=\pm \sqrt{-r/u}$ for $r<0$. The minima merge continuously at zero. The curvature vanishes there, so the long-distance response becomes singular.

For $u<0<v$, put $z=M^2$. A nonzero stationary minimum satisfies $r+uz+v{z}^2=0$. Equality with the $M=0$ free energy givesThe finite separation demonstrates the first-order phase transition. Metastability occurs where a nonglobal local minimum persists; its disappearance at a spinodal point is not the equilibrium coexistence criterion.

An order parameter distinguishes thermodynamic phases and transforms under the symmetry that may be broken. For a scalar ferromagnet it is the magnetization per site; a disordered zero-field phase has $m=0$, while ordered pure thermodynamic phases have $m=\pm m_0$. In a fluid one can instead use the density measured relative to its critical value; the field conjugate to it is then related to the chemical potential rather than literally a magnetic field. A nonzero value in the presence of an explicit conjugate field is not by itself evidence of a spontaneous transition.

The LG theory describes a slowly varying local order parameter by a symmetry-constrained free-energy functional, for exampleAt $h=0$, a ${\mathbb{Z}}_2$ symmetry excludes odd powers. The coefficients are assumed analytic functions of the controls near the transition, and the expansion is stabilized by a positive highest retained even coefficient. The Landau approximation obtains equilibrium by minimizing this functional, neglecting long-wavelength fluctuation corrections. Its nonconvex local potential describes distinct candidate phases and mean-field metastability; the exact thermodynamic potential is convexified when macroscopic mixtures are admitted.

In a finite symmetric system, the zero-field expectation of $m$ can vanish even below the transition because both ordered orientations are sampled. Spontaneous order is defined by taking the thermodynamic limit before removing a selecting field, for example $m_0={\lim }_{h\downarrow 0}{\lim }_{N\to \infty }⟨m⟩$. This distinction also matters for the connected correlations used below.

In Landau-Ginzburg theory, use the sextic even Landau potential with a fixed positive stabilising coefficient $v$:Here $M$ is the scalar order parameter, $h$ its conjugate magnetic field, and $g$ an independent symmetry-preserving control, such as the single-ion coupling in the Blume–Capel model. Thus $g$ need not break $M\mapsto -M$, whereas $h$ does. The Blume–Capel mean-field tricritical parameters are one realization; generally, a tricritical point occurs whenTwo controls tune away the quadratic and quartic terms. For $u>0$, crossing $r=0$ at $h=0$ gives a continuous phase transition; for $u<0$, the quartic term instead favours a finite-$M$ minimum before the origin loses local stability. The sextic term bounds the free energy below. The tricritical point joins the continuous and discontinuous transition loci.

For the symmetric phase diagram, write $s=M^2$. A nonzero stationary point obeys $r+us+v{s}^2=0$. Its free-energy density relative to the origin is $rs/2+u{s}^2/4+v{s}^3/6$. Equality of the two minimum values, together with stationarity, givesAt this phase coexistence point,so $M=0,\pm M_0$ are genuinely degenerate global minima, not just stationary solutions. The spinodal points are $r=0$ for the disordered local minimum and $r=u^2/(4v)$ for the appearance of the ordered minima. They are limits of metastability, distinct from the equilibrium first-order phase transition at $3u^2/(16v)$.

Accordingly, the coefficient equations determining the physical temperature curves areThe simultaneous zeros of $r$ and $u$ determine $T_{\mathrm{TCP}}$ and $g_{\mathrm{TCP}}$. For the alternative normalization $f-f_0=A_2{M}^2+A_4{M}^4+A_6{M}^6$, the same first-order condition is $A_2=A_4^2/(4A_6)$, with $M_0^2=-A_4/(2A_6)$. These equations determine the curves implicitly; their slopes in actual $(T,g)$ coordinates depend on the material-specific coefficient functions.

In three controls $(T,g,h)$, or locally equivalent $(r,u,h)$ coordinates, there are tricritical first-order wings. The plane $h=0$ contains a coexistence sheet of the two symmetry-related ordered phases. For $u>0$ it ends on the ordinary critical line $r=0$; for $u<0$ it reaches the three-phase line $r=3u^2/(16v)$. Two further first-order surfaces extend from that line into $h>0$ and $h<0$. Across a wing, two minima of the same sign but different magnitudes exchange global stability; the field favours that sign. Each wing ends on an ordinary critical edge where those minima and their intervening maximum merge. The three surfaces and their critical boundaries meet at the tricritical point.

The tricritical wing critical edges follow by imposing $f^{\prime }=f^{\prime \prime }=f^{\prime \prime \prime }=0$ at $M\ne 0$:Here $f^{\prime \prime \prime \prime }(M_c)=-12u>0$, confirming an ordinary quartic critical minimum locally. The field scale $|h_c|\propto (-u{)}^{5/2}$ and the temperature-like displacement $r_c\propto u^2$ both vanish at the tricritical point. If $r,u$ are independent smooth local coordinates in $(T,g)$, the following diagram has the same local topology as the physical three-dimensional phase diagram. Its surfaces use exact equal-minimum conditions, as described by tricritical wing coexistence factorization.

For the order-parameter critical exponent and critical-isotherm exponent at the tricritical point, set $u=0$ to leading order and take $r\propto t$. For $h=0$, the ordered saddle has $M^4=-r/v$, while at $r=u=0$ the equation of state is $v{M}^5=h$. Hence the tricritical mean-field critical exponents areThese are mean-field critical exponents. Below the tricritical upper critical dimension they need not be the interacting exponents, and at that dimension logarithmic corrections can accompany the powers. Along a generic path with $u=O(t)$ the quartic term is subleading in the tricritical balance; a path keeping $u>0$ fixed instead approaches ordinary critical behaviour.

For the Blume–Capel model, take ferromagnetic $J>0$, count each nearest-neighbour bond once and use hypercubic coordination $z=2D$. The nearest-neighbour direction sum in the printed Hamiltonian is understood implicitly. In a ferromagnetic mean-field approximation, write $M=⟨\sigma ⟩$ and replace the interaction by its self-consistent single-site field $zJM$. The single-site partition function and variational free-energy density areThe first term prevents double counting the interaction energy. Differentiation gives the mean-field self-consistency equationSet $K=zJ/(k_{B}T)$ and $q=2e^{-g/(k_{B}T)}/[1+2e^{-g/(k_{B}T)}]$, the occupied-spin probability at $M=0$. Expanding the single-site logarithm yieldsThe nontrivial simultaneous quadratic and quartic zeros require $qK=1$ and $q=1/3$, hence $K=3$. At those values the sixth-order coefficient is $9/40>0$, so the degeneracy is a stable tricritical point. The mean-field tricritical parameters areThis is a mean-field result, not a dimension-independent exact lattice transition temperature. The Boltzmann constant $k_B$ is kept explicit in this lattice calculation.

For the sextic even Landau potentialput $q={\varphi }^2$. The stationary-point equation isso the nonzero stationary points haveAt such a point, $\epsilon =-q^2-gq$, and henceA first-order transition occurs when a nonzero minimum has the same free energy as the disordered minimum $G(0)=0$. ThereforeAt $g=0,\epsilon =0$, the first-order line meets the continuous line and forms the tricritical point. For $g>0$, the transition at $\epsilon =0$ is continuous.

For $g<0$, the disordered state $\varphi =0$ is a local minimum when $\epsilon >0$ and loses stability at the disordered spinodal point $\epsilon =0$. The ordered minima exist whileand merge with the intervening maxima at the ordered spinodal $\epsilon =g^2/4$. Consequently the two spinodals enclose a region of metastability. If increasing $\epsilon$ corresponds to heating, the ordered phase can persist above the coexistence curve throughoutwhile the disordered phase can persist below it throughout

At zero applied field, time-reversal symmetry interchanges $M$ and $-M$ without changing the free energy, so the analytic expansion contains only even powers. The term $-{\mu }_{0}HM$ breaks this symmetry at fixed nonzero H. Stationarity and local stability requireFor $T>T_c$ at small field, the cubic term is higher order and $M\simeq {\mu }_{0}H/a(T)$. Thus the susceptibility as defined in the question isThe three requested free-energy sketches and their minima are given separately above.

For the plotted isotherms, divide the equation of state by nonzero M and rearrange:This is an Arrott plot: its slope determines b, and the isotherm through the origin identifies the mean-field Curie temperature.^[ref] Under the paper's stated uniform-field Landau model, use H as supplied on its axis.

The ordinate unit is $(10^4\mathrm{A}{\mathrm{m}}^{-1}{)}^2=10^8{\mathrm{A}}^2{\mathrm{m}}^{-2}$. Reading the original graph gives slopes about $0.076$–$0.080$ ordinate units per unit H/M, so the physical slope is approximately $7.8\times 10^6{\mathrm{A}}^2{\mathrm{m}}^{-2}$. ConsequentlyApproximate horizontal intercepts and the corresponding $a={\mu }_0(H/M{)}_{M^2=0}$ are:

The sign change brackets the transition between 19 and 25 K. A linear fit of $a(T)=\alpha (T-T_c)$ across the four read-offs gives approximately $\alpha =4.8\times 10^{-6}\mathrm{H}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ andInterpolating only the two nearest isotherms also gives about 22 K. These are graph estimates, not exact recovered measurements.

Below $T_c$ at zero field, the stable branches have $M_0^2=-a/b$. Substitution gives the free-energy density relative to the omitted smooth background:The equilibrium excess heat-capacity density is $c_H=-T{\partial }_T^2{F}_{\min }=T{\alpha }^2/(2b)$ below $T_c$ and zero above it. ThereforeThe all-isotherm linear fit gives about $1.5\times 10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. Estimating the slope only between the 19 K and 25 K curves instead gives $\alpha \simeq 6.9\times 10^{-6}\mathrm{H}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and a jump near $3\times 10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. The lines are not exactly parallel and the intercepts are not perfectly linear in T, so a jump of order a few $10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$ is supported; finer precision is not. This is per volume because F is a density; no sample volume is given to convert it to a total heat capacity.

If $b<0$, the quartic truncation is unbounded below. Add a positive sixth-order term $c{M}^6/6$ with $c>0$. At zero field the coexistence conditions giveThus the zero-field transition is generically first order, with a discontinuous spontaneous magnetisation and possible metastability. It occurs before the quadratic coefficient reaches zero. The case $b=0$ is the tricritical boundary of this mean-field model; no value of c is invented.

A phase diagram partitions a space of thermodynamic variables into regions occupied by different equilibrium thermodynamic phases and records the phase coexistence curves between them. For a pure, one-component substance, a pressure–temperature phase diagram gives one point for each pair $(T,p)$. In a single-thermodynamic phase region both controls can vary independently. On a two-thermodynamic phase phase coexistence curve, they obey one relation; at a triple point, both are fixed. These statements assume bulk thermodynamic equilibrium, fixed composition and no additional applied fields.^[ref]

At fixed temperature and pressure, the stable bulk thermodynamic phase minimizes Gibbs free energy for a fixed amount of substance. For a pure substance this is a comparison of chemical potentials on a common normalization. Two candidate thermodynamic phases coexist where ${\mu }_{\alpha }(T,p)={\mu }_{\beta }(T,p)$ and neither is exceeded in stability by a third thermodynamic phase. Differentiating their equality gives the Clapeyron equation. Thus a phase coexistence curve is a locus of equal stability, rather than a trajectory that every heated sample must follow.^[ref]

The following original cartoon uses dimensionless positive coordinates $\theta =T/T_{\ast }$ and $\pi =p/p_{\ast }$, with arbitrary reference scales $T_{\ast },p_{\ast }$. It represents no particular substance and contains no measured data. Stipulate a triple point at $(1,1)$, a liquid-gas critical point at $(3,5)$, and these drawing rules:They specify sublimation, liquid–gas phase coexistence and melting boundaries for the picture. They are not an equation of state, do not predict latent heat or mass density, and do not supply a quantitative critical theory. In particular, the melting slope is merely an illustrative choice.

For example, at fixed $\pi =3$ the melting boundary is at $\theta =1.2$ and the liquid–gas boundary is at $\theta =2$. Heating from $(\theta ,\pi )=(1.6,3)$ to $(3.6,3)$ therefore starts in liquid, crosses phase coexistence at $(2,3)$, and ends in gas. In an equilibrium conversion at that crossing, added heat can change the phase fraction at fixed temperature through latent heat. The pressure–temperature point alone does not specify how much of each thermodynamic phase is present: for that one needs an additional amount or volume constraint.

Now connect exactly the same endpoints by raising $\pi$ from $3$ to $6$ at $\theta =1.6$, moving to $\theta =3.6$ at $\pi =6$, then lowering $\pi$ to $3$. The first leg stays on the liquid side because ${\theta }_{\mathrm{melt}}$ ranges only from $1.2$ to $1.5$ and ${\pi }_{\mathrm{vap}}(1.6)=2.2<3$. The second leg lies above the entire liquid–gas boundary, whose largest drawn pressure coordinate is $5$; the third lies beyond its largest temperature coordinate, $3$. No leg crosses a first-order phase transition boundary. The middle leg passes through a supercritical fluid region, with $\theta >3$ and $\pi >5$, and the final leg returns to a gaslike state continuously.

Avoiding a first-order phase transition does not mean that no heat or work is exchanged. Along the detour, mass density, enthalpy and heat capacity may change substantially but continuously. The liquid-gas critical point itself has singular behavior, so the detour goes around it rather than through it. This connected fluid region is the striking lesson of a phase diagram: a familiar change from liquid to gas need not involve boiling across a sharp boundary.

An equilibrium phase diagram also does not predict how rapidly a conversion occurs. Metastability and a barrier described by classical nucleation theory can delay the observed phase transition. Nor should a binary-composition phase diagram be read as the same picture: its coexistence regions and lever rule encode composition and phase fractions, and can occupy an area even though pure-substance two-thermodynamic phase coexistence occupies a curve in the pressure–temperature plane.

Crystal polymorphism is the existence of different crystal structures for the same chemical substance. Each crystal polymorph can have its own mass density, melting behavior and range of stability on a phase diagram. A different external shape of the same crystal structure alone does not establish a different crystal polymorph.^[ref]

At fixed temperature and pressure, compare the chemical potentials of the possible crystal polymorphs. The lower value determines the stable bulk thermodynamic phase; equality permits phase coexistence when neither form is beaten by another thermodynamic phase. A higher-chemical potential form can persist through metastability, so finding it in a sample does not by itself locate an equilibrium boundary.

A supersaturated solution has an ion product above the relevant solubility product. It can persist through metastability: nucleation must first create a sufficiently large new solid region. Added surfaces, impurities and mixing history can affect that barrier. Therefore a clear sample during a finite observation does not prove undersaturation, and visible cloudiness does not measure the ion product.

A free energy is a thermodynamic potential that compares possible states under specified environmental constraints. For a closed system of fixed composition, treated as a simple compressible system with only pressure-volume work, stable thermodynamic equilibrium minimizes Helmholtz free energy at fixed temperature and volume, or Gibbs free energy at fixed temperature and pressure. The choice follows the experiment: a rigid sealed vessel and a sealed cylinder with a freely moving loaded piston impose different constraints. Both may exchange heat with a thermal reservoir. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19.

The useful insight is that internal energy alone cannot rank these alternatives: exchanging heat also changes the surroundings' entropy. Subtracting $TS$ packages that accounting into a property of the chosen system. Adding $pV$ additionally accounts for displacing the surroundings at fixed pressure. A free energy is a state function; supplied heat describes a transfer during a process. Under the stated constraints, supplied heat is $Q=\Delta U$ at fixed volume or $Q=\Delta H$ at fixed pressure, whereas neither $\Delta F$ nor $\Delta G$ generally equals $Q$.

A lower free energy identifies a thermodynamic preference, without specifying a timescale. A barrier can trap a system in a local minimum, producing metastability although a lower global minimum exists. The enzyme example illustrates the corresponding distinction for chemical reactions: changing the route can change the reaction rate while preserving the endpoint Gibbs free energy difference and chemical equilibrium.

A Van der Waals isotherm is the pressure–particle volume curve at a fixed temperature. Put $x=v/v_c$, $t=T/T_c$, and $q=p/p_c$, using the Van der Waals critical point values below. Then the existing Van der Waals law of corresponding states becomesAll curves in the figure are calculations in this reduced model.

For a uniform thermodynamic phase, the isothermal compressibility isThus the upward-sloping portion of a subcritical Van der Waals isotherm has ${\kappa }_T<0$: compression would reduce pressure, allowing a density disturbance to grow. It is mechanically unstable. The spinodal points bound that portion; setting the derivative $\partial q/\partial x$ to zero gives their curve $t=(3x-1{)}^2/(4x^3)$. Negative slope alone guarantees only local mechanical stability: portions between the spinodal and the phase coexistence endpoints have metastability. Equilibrium replaces the entire interval between those endpoints, including these metastable portions, by a Maxwell construction plateau.^[ref]

Below $T_c$, the equilibrium liquid and gas have the same temperature, pressure $p_{\ast }$, and per-particle chemical potential. For macroscopic bulk thermodynamic phases, neglect interface contributions. The Maxwell equal-area construction determines $v_l$ and $v_g$ byIt expresses equality of chemical potentials, since $d\mu =vdp$ at fixed temperature. Equivalently, the equilibrium Helmholtz free energy per particle follows a common tangent line between the two endpoint states, lying below the homogeneous continuation between them. The construction uses that mathematical continuation to compare endpoints; it is not a physical compression path through negative isothermal compressibility states.^[ref]

Here is a reproducible calculated example at $t=0.9$. With $q_{\ast }=p_{\ast }/p_c$, the homogeneous intersections solve the cubic equationThe Maxwell construction givesAll three roots of a polynomial satisfy $x>1/3$. Only the smallest and largest are the phase coexistence endpoints; the middle one has negative isothermal compressibility. Equal pressures at three admissible values of volume therefore do not imply three stable thermodynamic phases. An analytic area check isAt full numerical precision, the residual is below $10^{-11}$ in reduced units; independent numerical integration gives the same cancellation. At $t=0.9$, the spinodal points are instead $x\simeq 0.718597189$ and $1.528504964$.

The equilibrium plateau changes the phase fraction rather than the endpoint densities. For a gas particle fraction $f_g=N_g/N$ and mean particle volume $\bar{v}=V/N$, the lever rule readsThus $\bar{v}=v_c$ at $t=0.9$ gives $f_g\simeq 0.227219$: this is a particle fraction, not the volume fraction occupied by gas. Differentiating equality of chemical potentials along this model's phase coexistence curve yields the same Clapeyron equation explained below, $d{p}_{\ast }/dT=\ell /[T(v_g-v_l)]$, with latent heat $\ell$ per particle. Both endpoint differences vanish as the Van der Waals critical point is approached; the equation's critical limit must be taken as a ratio.

The Maxwell construction presumes a pure bulk fluid at thermodynamic equilibrium. Its endpoint equality does not specify how rapidly a real compression reaches equilibrium. A homogeneous metastable sample can persist while nucleation is delayed; the spinodal marks loss of local mechanical stability in this mean-field theory, rather than the equilibrium onset of condensation. Surface tension, finite sample size and impurities can affect the observed transition. These qualifications matter when comparing a mathematical Van der Waals isotherm with a visible liquid meniscus.

A hypothetical Van der Waals equation model illustrates positive and negative Joule-Thomson coefficient regions and a finite throttle that first warms, then cools. Use the molar convention already distinguished in Van der Waals equation:Here $v_{\mathrm{m}}$ is molar volume, $R$ the molar gas constant, $u_{\mathrm{m}}$ internal energy per mole, and $C_{V,\mathrm{m}}$ the stipulated constant molar heat capacity at fixed volume. The second equation is an additional caloric assumption: an equation of state alone does not specify the temperature dependence of internal energy. It satisfies the equilibrium identity $(\partial u_{\mathrm{m}}/\partial v_{\mathrm{m}}{)}_T=T(\partial p/\partial T{)}_{v_{\mathrm{m}}}-p=a_{\mathrm{m}}/v_{\mathrm{m}}^2$. The arbitrary energy reference is set to zero.

Define dimensionless variables $r=v_{\mathrm{m}}/b_{\mathrm{m}}$, $\theta =R{b}_{\mathrm{m}}T/a_{\mathrm{m}}$, $\pi =b_{\mathrm{m}}^{2}p/a_{\mathrm{m}}$ and $\eta =b_{\mathrm{m}}{h}_{\mathrm{m}}/a_{\mathrm{m}}$, where $h_{\mathrm{m}}$ is molar enthalpy. These differ from the critical-value scaling in Van der Waals law of corresponding states. The throttle calculation reduces toThus a prescribed inlet fixes $\eta$, and solving the first and last equations at the outlet pressure gives the final temperature. For an inlet $({\theta }_1,{\pi }_1)=(0.8,0.12)$ and outlet ${\pi }_2=0.025$, the model gives ${\theta }_2\simeq 0.752287$, a decrease of $0.047713$. The inlet $(2.2,0.12)$ instead gives ${\theta }_2\simeq 2.205786$, an increase of $0.005786$. The purple inlet $(1.45,0.30)$ initially warms but ends at ${\theta }_2\simeq 1.437585$ for the same outlet pressure. These are reduced toy values, not kelvin readings or experimental evidence. No physical gas parameters were fitted.

For a short inversion derivation, implicit differentiation of the Van der Waals equation at fixed pressure givesInsert this into $T(\partial v_{\mathrm{m}}/\partial T{)}_p=v_{\mathrm{m}}$ and substitute back into the equation of state. The Joule-Thomson inversion curve isPositive inversion pressure requires $r>3/2$. Its maximum pressure is $\pi =1/3$ at $r=3$, while the upper dilute limit is ${\theta }_{\mathrm{inv}}\to 2$. Using Van der Waals critical point values, the latter is $T_{\mathrm{inv},\max }/T_c=27/4$. The inversion locus depends on the equation of state; the chosen molar heat capacity additionally determines how much temperature changes along each finite throttle.^[ref]

The displayed region has $\theta >8/27$, above the model's critical temperature, and uses only $r>1$, $p>0$, $(\partial p/\partial v_{\mathrm{m}}{)}_T<0$ and positive $C_{p,\mathrm{m}}$. Subcritical calculations require attention to phase coexistence, metastability and the Maxwell construction; an unstable homogeneous continuation is not an equilibrium throttle endpoint. Even on a stable branch, this mean-field theory with constant $C_{V,\mathrm{m}}$ is a qualitative model, not an accurate inversion map for a particular real gas. Phase transitions, changing composition, appreciable heat leakage or bulk kinetic energy changes require a different property calculation or steady-flow energy balance. At phase coexistence, the smooth single-thermodynamic phase coefficient formula cannot replace a balance involving the phase fraction.