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Choose a hypothetical fluid with molar constants $a_m=0.360\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$ and $b_m=4.00\times 10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, using $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These stipulated constants are not a fit to carbon dioxide. The molar critical formulas giveAt $T=360\mathrm{K}$ and $v_m=240\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, first convert $v_m$ to $2.40\times 10^{-4}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The repulsive-volume term and attractive correction areso $p=8.716033\mathrm{MPa}$. The ideal gas law would instead give $RT/v_m=12.471694\mathrm{MPa}$. The excluded volume raises pressure relative to that ideal value, but attraction more than offsets the increase here. Neither correction alone describes this example.

For $n_{\mathrm{mol}}=0.500\mathrm{mol}$, the same state occupies $V=n_{\mathrm{mol}}{v}_m=120\mathrm{c}{\mathrm{m}}^3$. The total-volume formula gives $n_{\mathrm{mol}}{b}_m=20\mathrm{c}{\mathrm{m}}^3$ and the same $8.716033\mathrm{MPa}$. Halving the sample has halved both its physical volume and its excluded volume, while retaining its pressure. Conversely, if the two critical measurements $T_c,p_c$ are used to determine model constants, inversion givesAn independent critical molar volume then tests the prediction $v_{m,c}=3R{T}_c/(8p_c)$; agreement with two fitted numbers does not ensure agreement with the third.

At $t=0.9$, the Van der Waals liquid-gas coexistence example below becomes $T\simeq 288.6537\mathrm{K}$, $p_{\ast }\simeq 5.391653\mathrm{MPa}$, $v_{m,l}\simeq 72.40823\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $v_{m,g}\simeq 281.8611\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. For a mean molar volume of $120\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, the lever rule gives gas mole fraction $f_g\simeq 0.227219$. Its volume fraction is instead $f_g{v}_{m,g}/v_m\simeq 0.533703$: less than one quarter of the molecules occupy more than half the volume. The flat equilibrium pressure corresponds to changing these proportions. It does not describe a homogeneous sample sliding along the unstable loop.

The saturation concentration $c_{\mathrm{sat}}(T)$ is the dilute-phase coexistence concentration at the specified temperature and solution conditions. On increasing the overall concentration through this boundary, a macroscopic condensed phase can first coexist with the dilute phase. The other branch gives the dense-phase concentration $c_{\mathrm{dense}}(T)$.

At the thermodynamic critical point, the two branches meet: their concentration difference vanishes. Its coordinates define the critical concentration $c_c$ and critical temperature $T_c$. In the upper-critical model, $T>T_c$ admits no two-liquid coexistence. Thus $c_{\mathrm{sat}}$ is a temperature-dependent branch, whereas $c_c$ is the concentration at the endpoint.

Between coexistence concentrations, the lever rule gives the dense-phase volume fraction $v=(\bar{c}-c_{\mathrm{sat}})/(c_{\mathrm{dense}}-c_{\mathrm{sat}})$, assuming consistent concentration and volume definitions. Increasing total chromatin then primarily changes phase amounts rather than allowing the dilute phase to increase without bound.

Apply the lever rule on the $779^{\circ }\mathrm{C}$ solid tie line. All compositions are copper mass percentages:Thus the equilibrium material is 77.1% $\alpha$ and 22.9% $\beta$ by mass just below the eutectic temperature. These are phase fractions; converting them to volume fractions would require densities.

The profile estimate and equilibrium lever-rule estimate need not agree exactly. The height trace contains only four major ridges, making it a small sample of the microstructure. Pixel size, ridge-threshold choice, tip geometry, contamination, polishing damage and unequal etching change apparent widths. A finite-radius AFM tip can broaden the narrow raised cementite regions, biasing their inferred fraction upward.

The equilibrium calculation assumes eutectoid pearlite with the chosen carbon composition and tie-line values. Off-eutectoid steel can include proeutectoid ferrite or cementite outside the pearlite; alloying, carbon segregation and non-equilibrium cooling alter the simple Fe–C estimate. Carbon solubility also varies with temperature. The mass-to-volume conversion requires phase densities.

The observed difference between roughly $17\%$ ridge fraction and $11\%$ equilibrium phase volume is therefore a reason to examine sampling and the etched-surface measurement, not to force the image widths to match the lever rule. Multiple colonies, repeated calibrated traces and a phase-sensitive measurement would improve the comparison.

For eutectoid pearlite, apply the lever rule between ferrite and cementite at the transformation temperature. Using standard approximate Fe–C values $C_e=0.76$, $C_{\alpha }=0.022$ and $C_{\mathrm{cem}}=6.67$ in weight percent carbon,These are mass fractions. To compare with an unbiased geometric line fraction, convert using the phase densities:With representative densities ${\rho }_{\mathrm{cem}}\simeq 7.69$ and ${\rho }_{\alpha }\simeq 7.87\mathrm{Mg}{\mathrm{m}}^{-3}$, this gives about $11.3vol\%$ cementite and $88.7vol\%$ ferrite. Because the densities are similar, the mass and volume fractions differ only modestly. The phase-diagram values here are conventional approximations, not a claim that the exact 2018 Data Book has been recovered.

For a mixture with phase compositions $x_1,x_2$, its molar Gibbs free energy is the straight-line combination $g=(1-f)g_1(x_1)+f{g}_2(x_2)$ at overall $x=(1-f)x_1+f{x}_2$. Minimizing over phase compositions gives equal slopes and equal tangent intercepts: the common-tangent construction expresses equality of both component chemical potentials. The equilibrium curve is the lower convex envelope of the candidate phase curves. Phase fractions follow the lever rule.

Convert atomic fractions to mass fractions before using the diagram's weight-percent scale:With $M_{\mathrm{Sn}}=118.71$ and $M_{\mathrm{Bi}}=208.98\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, the overall $x=0.30$ is $19.58wt\%$ Sn. At $150{}^{\circ }\mathrm{C}$, the diagram gives approximately $x_B=0.028$ and $x_L=0.555$. These are about $1.61$ and $41.47wt\%$ Sn. The lever rule givesThe microstructure has primary Bi-rich crystals surrounded by remaining liquid. The phase compositions and fractions are approximate readings of a printed phase diagram, not exact thermodynamic values. Faster cooling generally gives finer crystals, solute segregation and departure from these equilibrium compositions.

The Maxwell construction replaces a nonconvex homogeneous free-energy density by its coexistence envelope. If $U_0(M)$ excludes the source, the coexisting values $M_1,M_2$ satisfyThus one straight line is tangent at both endpoints. Equivalently,the equal-area form of the common-tangent construction. At fixed average $\overline{M}$ between the endpoints, the lever rule gives $\overline{M}=p{M}_1+(1-p)M_2$. Such a mixture has the tangent-line bulk energy, below the energy of an unstable homogeneous state.

The two phases occupy spatial domains separated by domain walls. The gradient term gives a positive wall energy, but a macroscopic interface has area of order $L^{D-1}$ rather than bulk volume $L^D$, so its cost per volume vanishes in the thermodynamic limit. For the symmetric quartic example a scalar quartic domain wall obeys $\kappa {\varphi }^{\prime \prime }=r\varphi +u{\varphi }^3$ and interpolates between $\pm M_0$ as $\varphi =M_0\tanh (x/\ell )$, with $M_0^2=|r|/u$ and $\ell =\sqrt{2\kappa /|r|}$. Multiplication of the differential equation by ${\varphi }^{\prime }$ gives the first integral $\kappa ({\varphi }^{\prime }{)}^2/2=u({\varphi }^2-M_0^2{)}^2/4$, explaining the finite positive interface cost. A uniform unconstrained system need not form both domains: phase separation is particularly relevant when the average order parameter is fixed or boundaries enforce competing phases.

Hysteresis occurs when the system follows a metastable local minimum after another minimum has become the equilibrium one. For $r<0,u>0$, the equation of state is $h=rM+u{M}^3$. The hysteresis spinodals of a scalar quartic potential follow from $r+3u{M}^2=0$:Following local minima until these limits gives different switching fields on increasing and decreasing $h$. Equilibrium instead switches at $h=0$. Actual switching can occur before a mean-field spinodal through nucleation, and depends on time scales; hysteresis is not a property of a globally minimized equilibrium potential. The second panel below distinguishes the local branches and their limiting switches.

Take $z$ positive downwards from the roof. Write $T_0=T_m-m{C}_0$, $C_B=(T_m-T_B)/m$, $\Delta T=T_0-T_B=m(C_B-C_0)>0$ and $\Delta C=C_B-C_0$. The cooled magma follows the decreasing liquidus from $(C_0,T_0)$ towards $(C_B,T_B)$; the initially superheated liquid starts at $(C_0,T_{\infty })$ above that liquidus. The solid has zero solute concentration. At each temperature in the mushy layer, a horizontal tie line joins the pure solid to the residual liquid on the liquidus. The lever rule gives the liquid fraction from the bulk concentration. Since $T_B>T_E$, this path stops before the eutectic temperature and does not cross the eutectic composition.

The porosity $\varphi$ denotes the liquid fraction, so $1-\varphi$ is the solid fraction. Equal phase mass densities remove volume-change flow; the stable density stratification suppresses convection. In the stagnant mushy-layer model, define bulk specific enthalpy and bulk solute concentration byThe arbitrary reference of specific enthalpy has been omitted. The two conservation laws in $0<z<h(t)$ areHere $\kappa =k_T/(\rho c_p)$ is the thermal diffusivity, with $k_T$ the thermal conductivity. The positive $L{\varphi }_t$ term records increased specific enthalpy on melting; during freezing it supplies latent heat. In the liquid $z>h(t)$, the heat equation and solute diffusion equation arewhere $D$ is the solutal diffusivity. Neglecting solute diffusion inside the mushy layer does not permit dropping $D{C}_{zz}$ in the liquid at this stage.

Use $T(0,t)=T_B$, liquid $T\to T_{\infty }$ and $C\to C_0$ as $z\to \infty$, and initially $h=0$, $T=T_{\infty }$, $C=C_0$ for $z>0$. These are the half-space boundary conditions appropriate before cooling reaches the bottom of the chamber. For a closed chamber of finite depth $H$, replace the far-field conditions by the specified bottom conditions, for example $T_z=C_z=0$ at an insulated impermeable floor; the half-space similarity solution then ceases to apply when the thermal penetration depth approaches $H$.

At $z=h(t)$, let $C_i$ and $T_i=T_m-m{C}_i$ be the common liquid concentration and temperature, and let ${\varphi }_i$ be the limiting porosity on the mush side. Continuity of $T,C$, the Rankine-Hugoniot condition for solute, and the interfacial conservation of energy giveThe first condition has the sign of solute rejection into the liquid. For the sharp-edge stagnant mushy-layer model with nonzero liquid solutal diffusivity, close the free-boundary problem by marginal equilibrium at a mush-liquid boundary:This prevents constitutional supercooling immediately ahead of the mushy layer. A small solid fraction jump at its edge is allowed: imposing ${\varphi }_i=1$ as well as these finite-$D$ conditions would generally overdetermine the reduced model. At the roof, $C=C_B$ follows from the liquidus, and no additional solute-flux boundary condition is needed in the diffusion-free mush.

To obtain the large-concentration similarity solution, putThe positive superheat specifies the physically intended liquid initial state. On a similarity solution, $C_i,{\varphi }_i$ are constant, so every newly incorporated level leaves the same constant $B={\varphi }_i{C}_i$. Write $C_i=C_0+\Delta C\delta$ and $B=\Delta C(\mathcal{C}+a)$. Then throughout the mushy layerFor $a,\theta =O(1)$, $\mathcal{C}\gg 1$, and $\mathcal{S}/\mathcal{C}=O(1)$, the effective heat capacity is constant to relative error $O({\mathcal{C}}^{-1})$. Thus setIntegration of the similarity solution ordinary differential equation gives the leading analytical profilesThe temperature is $T=T_0+\Delta T\theta$. In the mushy layer, the concentration and porosity are the displayed $C,\varphi$; in the liquid,These retain the liquid solutal boundary layer. For completeness, finite-$r$ values of $\delta ,a,\lambda$ are determined analytically, to the same large-$\mathcal{C}$ order, by settingand solvingThe solute jump gives ${\varphi }_i=1-b/(\mathcal{C}+\delta )$ exactly; replacing $\mathcal{C}$ by $\mathcal{C}+\delta$ on the right of the last equation retains that jump's next-order factor, but does not make the linearized thermal profiles exact at finite $\mathcal{C}$. The branch has $0\le \delta <1$, $0<\varphi \le 1$. All these profiles are leading asymptotic expressions, not exact solutions of the nonlinear finite-$\mathcal{C}$ heat equation.

For $r\to 0$ at fixed positive $\lambda$, the complementary error function asymptotic gives $K(\lambda /\sqrt{r})\sim \sqrt{\pi }\lambda /\sqrt{r}$, hence $\delta =O(r)$, $b=O(r)$, $a\to 0$, and ${\varphi }_i\to 1$. The result simplifies toIn particular, $\varphi (0,t)=C_0/C_B=\mathcal{C}/(\mathcal{C}+1)$ and $\varphi (h^{-},t)=1$. There is no finite interfacial latent heat jump in this limit: the latent heat is released continuously throughout the mushy layer. Matching the two temperature gradients givesThe printed final equation is missing the square on $\Omega$ in the exponential. The corrected exponent follows directly from differentiating $\mathrm{erf}(\Omega \eta )$ and is consistent with ${\Omega }^2=1+\mathcal{S}/\mathcal{C}$. The positive root is unique: the mush-side gradient decreases from infinity to zero with $\lambda$, whereas ${\theta }_{\infty }K(\lambda )$ increases. The zero-superheat limit is singular and is not described by a finite positive $\lambda$ in this two-region construction.

Interpret the stated transition as an upper critical miscibility temperature: cooling increases the effective $\chi$ through ${\chi }_c$. Above $T_c$, the homogeneous mixing free energy is convex throughout $0<\varphi <1$, with no two-phase common tangent. Below $T_c$, it acquires a region with $f^{\prime \prime }<0$, bordered by the spinodal. A lower straight line touches the curve at two binodal compositions ${\varphi }_{-}$ and ${\varphi }_{+}$; this common-tangent construction gives the equilibrium polymer-poor and polymer-rich phases.

The precise raw curve need not have two equal minima: subtracting an affine function of $\varphi$ changes reference chemical potentials without changing curvature or coexistence compositions. The free-energy panels above subtract the tangent at ${\varphi }_c$ from the upper-temperature curve and the common tangent from the lower-temperature curve. This makes the common tangent and the convex-versus-nonconvex distinction visible.

If the mean initial composition ${\varphi }_0$ lies inside the low-temperature spinodal region, an arbitrarily small separation into equal-volume regions with compositions ${\varphi }_0\pm \delta \varphi$ changes the mean free energy bySmall fluctuations therefore grow spontaneously: this is spinodal decomposition. Interfacial costs limit very short-wavelength fluctuations, and the polymer-poor and polymer-rich domains subsequently coarsen towards their binodal compositions.

If ${\varphi }_0$ lies just outside a spinodal branch but still inside the binodal, then $f^{\prime \prime }({\varphi }_0)>0$. Small fluctuations increase free energy, so the uniform mixture is metastable. Separation requires a sufficiently large nucleus to overcome the interfacial barrier, followed by nucleation and growth. There is no guarantee that this happens rapidly. If the initial composition is outside the binodal as well, it remains a stable uniform phase after cooling.

For a mean composition in the coexistence interval, conservation of polymer fixes the volume fraction $\alpha$ of the rich phase by the lever rule:The temperatures alone do not determine which mechanism occurs without the initial composition. Numerical free-energy curves additionally require $\chi (T)$; the plotted example uses $\chi (T)={\chi }_c{T}_c/T$ with absolute temperatures $T_c=298.15\mathrm{K}$, $T_1=303.15\mathrm{K}$ and $T_2=293.15\mathrm{K}$.

The proportions of ferrite and cementite within the two-phase pearlite microstructure. A lever rule on a weight-percent-carbon diagram gives mass fractions; converting those to volume fractions requires the phase densities. Proeutectoid phases outside pearlite must be treated separately.

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.

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.