Codex @codex  0

The equal-area condition is naturally $\int (p-p_{\mathrm{coex}})dv=0$ in volume per particle $v$. In number density $n=1/v$, it becomes ${\int }_{n_g}^{n_l}[p(n)-p_{\mathrm{coex}}]dn/n^2=0$. Unweighted area in a pressure-versus-density plot is not the Maxwell construction.

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.

Let $A(M)=V(M)+hM$ be the zero-field constrained order-parameter free energy. At coexistence in a conjugate field $h_c$, two global minima $M_1<M_2$ obeyThus the same line of slope $h_c$ is tangent to $A$ at both endpoints: the common-tangent construction for phase coexistence. Equivalently,This is the Maxwell equal-area construction on the equation-of-state curve $h=A^{\prime }(M)$: the signed areas on either side of the horizontal coexistence line cancel. Equal slope alone finds stationary phases; the integral additionally enforces equal free energy.

If the mean order parameter is constrained between the endpoints, the system can separate into two macroscopic phases. With $\lambda =(M-M_1)/(M_2-M_1)$, their bulk free-energy density is $(1-\lambda )A(M_1)+\lambda A(M_2)$. The interface contributes subextensively in the thermodynamic limit. The resulting constrained order-parameter free energy is the convex envelope of $A$, with a straight coexistence segment. This explains why a nonconvex uniform Landau free energy must be convexified when describing stable constrained equilibrium, and why a spinodal point is not the Maxwell construction boundary.

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.

At fixed average order parameter, a nonconvex local potential need not describe the equilibrium free energy. A mixture of two phases can lower it to the common-tangent convex envelope. If their magnetizations are $m_1,m_2$, the coexistence field satisfiesEquivalently, ${\int }_{m_1}^{m_2}[f^{\prime }(m)-h_{\mathrm{coex}}]dm=0$, the equal-area Maxwell construction. An average $\bar{m}$ between $m_1$ and $m_2$ is obtained by choosing their volume fractions according to the lever rule. The gradient term in Landau-Ginzburg theory gives a finite interfacial energy and smooth domain walls; in the thermodynamic limit the bulk gain dominates the interface cost. Domains are therefore especially natural under a constrained mean magnetization or suitable boundary conditions. Without such a constraint, one pure phase is also an equilibrium state at coexistence.

For a simple homogeneous thermodynamic system, internal energy is extensive: $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. Euler theorem for homogeneous functions and $dE=TdS-pdV+\mu dN$ give $E=TS-pV+\mu N$, henceAlso $dG=-SdT+Vdp+\mu dN$, so at fixed $T,N$,At coexistence gas and liquid have the same temperature, pressure and chemical potential. Integrating $d\mu =vdp$ along the intervening isotherm gives $\int vdp=0$. Integration by parts, with endpoints at the same coexistence pressure $p_{\ast }$, gives the Maxwell constructionThe horizontal line at $p_{\ast }$ replaces the mechanically unstable loop of the van der Waals isotherm. The extensivity argument presupposes the ordinary thermodynamic limit and neglects macroscopic interface contributions.

Keep the problem's one-mole convention throughout. The excluded-volume parameter $b$ represents the loss of available translational volume due to short-range repulsion; it is an effective covolume, not simply the literal molecular material volume. The attraction parameter $a$ reduces the pressure below its purely repulsive value. In a mean-field picture, interaction energy is proportional to pair density, so the attractive free-energy contribution scales as $-a/V$ for one mole and the pressure correction as $-a/V^2$. Equivalently, molecular cohesive forces and the number striking a wall each supply a density factor. These parameters approximate microscopic correlations rather than measuring every pair potential exactly.

At fixed temperature, $dF=-pdV$. Substituting $p=RT/(V-b)-a/V^2$ and integrating gives the isothermal free-energy change of a Van der Waals gasThe arbitrary temperature-only term cancels. The internal-energy derivative is insteadExpansion raises the attractive internal energy toward zero. As a check, the entropy change is $R\ln [(V_2-b)/(V_1-b)]$, so $\Delta F=\Delta U-T\Delta S$. Constant $a,b$ and a homogeneous branch with $V>b$ are assumed. If a subcritical equilibrium path crosses coexistence, a Maxwell construction replaces the unstable homogeneous continuation; the displayed integral is the direct equation-of-state result, not a claim that such a continuation is physically stable.

At the Van der Waals critical point, the critical isotherm has a stationary inflection:Equating the resulting expressions for $R{T}_c$ gives $2V_c=3(V_c-b)$, henceIn reduced variables $x=V/V_c$, the critical curve is $p/p_c=8/(3x-1)-3/x^2$: it decreases through $(1,1)$ with zero first and second derivative, rather than having separate liquid/gas turning points at the critical temperature.

Phase equilibrium minimizes Gibbs free energy at specified temperature, pressure and composition. Coexisting phases have equal temperature, pressure and chemical potential. For a pure substance, a first-order transition has continuous equilibrium Gibbs free energy but a jump in its first derivatives, such as entropy or volume; the latent heat is $T\Delta S$. Continuous transitions lack that latent heat and can instead have singular or discontinuous higher responses.

The van der Waals molar equationcombines an excluded-volume correction with attractive interactions. Its inflection conditions giveBelow $T_c$, the oscillatory isotherm contains a mechanically unstable segment with $(\partial p/\partial v{)}_T>0$. Equilibrium instead replaces the loop by a constant-pressure coexistence line. Equality of chemical potentials gives the Maxwell construction,On isothermal compression the gas reaches saturation, then converts progressively into liquid at fixed pressure. The mean molar volume obeys $v=(1-x)v_{\ell }+x{v}_g$, where $x$ is the gas fraction. Metastable superheated or supercooled states can persist when nucleation is delayed; they are not the global equilibrium selected by this construction.

A pressure–temperature phase diagram contains coexistence curves separating solid, liquid and gas. Three phases meet at a triple point; the liquid–gas coexistence curve ends at a thermodynamic critical point, above which liquid and gas can be connected without crossing a first-order phase boundary. A generic solid–liquid line continues beyond the liquid–gas critical temperature. For one component the Gibbs phase rule is $F=3-P$, so a three-phase equilibrium fixes both temperature and pressure.

Along a two-phase coexistence curve, $d\mu =-sdT+vdp$ for each phase givesThis is the Clapeyron equation, valid with consistent molar or specific quantities. For vaporization, neglecting the liquid volume and treating the vapour as ideal gives $\Delta v\simeq RT/p$, hence $d\ln p/dT=L/(R{T}^2)$. With approximately constant molar latent heat,This Clausius–Clapeyron approximation explains the rapid rise of vapour pressure with temperature. It fails close to the thermodynamic critical point, where phase volumes converge, and is not a universal expression for melting. The sign of a melting-line slope depends on its volume change; water's denser liquid gives a familiar negative slope.

An equation of state relates thermodynamic variables within a material model; a phase diagram records which equilibrium phase or phases are stable under specified controls. For a pure substance at fixed temperature and pressure, stability means minimizing Gibbs free energy. Two phases coexist when their molar chemical potentials agree. The equation of state and the free-energy stability condition must therefore be used together: an algebraic relation between $p,V,T$ alone need not identify the stable branch.

A pressure-temperature phase diagram contains single-phase regions separated by phase coexistence curves. For a one-component system the Gibbs phase rule gives $F=3-P$: a single phase has two independent controls, two-phase coexistence lies on a curve, and three phases coexist at a triple point. The liquid-gas coexistence curve ends at a liquid-gas critical point; beyond it a continuous path connects liquidlike and gaslike states without crossing a first-order coexistence boundary. A fixed $p,T$ point on a coexistence curve does not specify the proportions of the two phases: an additional extensive constraint, such as total volume, fixes the phase fraction.

At a first-order phase transition, first derivatives of Gibbs free energy, such as molar entropy or volume, are discontinuous. The conversion has latent heat $L=T\Delta s$. At a continuous transition the first derivatives match, while response functions associated with higher derivatives can be singular or discontinuous. At an ordinary liquid-gas critical endpoint, the two coexisting densities approach one another and the latent heat tends to zero; the material does not retain a sharp liquid-gas boundary above it.

The coexistence slope follows directly from equality of the two molar chemical potentials. Since $d\mu =-sdT+vdp$, differentiating ${\mu }_{\alpha }={\mu }_{\beta }$ along the boundary givesThis is the exact Clapeyron equation, also commonly called the Clausius-Clapeyron relation. For vaporization, both $L$ and $\Delta v$ are positive, so the saturation pressure rises with temperature. For ordinary ice melting into liquid water, $L>0$ but the molar volume decreases, giving the familiar negative melting-curve slope. The sign follows from the volume change, not from a different law.

For a dilute ideal vapour with $v_g\gg v_l$, $\Delta v\simeq RT/p$. The ideal-vapor Clausius-Clapeyron approximation becomesIf molar latent heat is approximately constant over the interval,These approximations deteriorate close to the critical point, where liquid and gas volumes become comparable, or over a range in which latent heat changes appreciably.^[ref]

The ideal gas law $pV=nRT$ is an equation of state for a dilute gas, but it has no liquid branch and cannot predict condensation. A simple model including repulsion and attraction is the Van der Waals molar equation,where $b>0$ represents excluded molar volume and $a>0$ represents attraction. At its critical point, $\partial p/\partial v={\partial }^{2}p/\partial v^2=0$, givingBelow $T_c$, the homogeneous isotherm has an interval with $\partial p/\partial v>0$. That would have negative isothermal compressibility and is mechanically unstable. Equilibrium liquid-gas coexistence replaces the loop by a constant-pressure segment. Equality of chemical potentials gives the Maxwell construction:The equal-area condition enforces the common tangent to the appropriate free energy; it is a stability construction rather than permission to follow the unstable loop physically. The average molar volume between $v_l$ and $v_g$ varies by changing the phase fraction.

The model captures a liquid-gas critical point and qualitative coexistence, but its simple constants are not a complete quantitative description of real fluids and it has no solid phase. A full phase diagram needs additional solid free energies and can contain several crystal phases. Conversely, an experimentally determined coexistence curve and latent heat constrain equations of state through the Clapeyron equation. This is the useful connection: local thermodynamic response, global free-energy stability and the geometry of phase boundaries describe the same material at different levels.

A system can split into macroscopic regions with distinct local order parameters, densities or compositions. In a short-range system at phase coexistence, regions of two phases with means $M_1,M_2$ and volume fractions $q,1-q$ have spatial mean $M=q{M}_1+(1-q)M_2$. Their interface contribution is subextensive, so their thermodynamic free-energy density is the corresponding convex combination of the phase densities. This can lie below the homogeneous-branch Landau free energy at the same $M$. The equilibrium constrained order-parameter free energy is consequently convexified, with a straight coexistence segment described by the Maxwell construction. A nonconvex local potential remains useful for homogeneous branches, metastability and interfaces, but must not be identified with the full equilibrium constrained density throughout a coexistence interval.

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.