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The 1869 The Bakerian Lecture: On the Continuity of the Gaseous and Liquid States of Matter by Thomas Andrews describes dried carbon dioxide trapped above mercury in a sealed capillary tube. A steel screw pressurized water in a strong copper tube; movable mercury transmitted the load. A second capillary tube held air as a pressure gauge. Flowing water around the visible capillary tubes controlled temperature; a thermometer read the bath. A cathetometer read column positions, converted to volume readings using weighed-mercury calibration.^[ref]

Below the critical temperature, compression produced a visible liquid–gas boundary; above it, large density changes occurred without that boundary. Three trials gave $30.92{}^{\circ }\mathrm{C}$ as his critical temperature. Purity checks detected residual air; mercury height corrections accounted for unequal hydrostatic pressures. He explicitly left air nonideality and pressure-induced tube-capacity corrections unresolved, so his air-gauge pressures were approximate. These readouts support a liquid-gas critical point; they do not measure a homogeneous unstable loop or establish universal quantitative accuracy of the later Van der Waals equation.^[ref]

A disappearing liquid meniscus by itself has more than one possible explanation: in ordinary subcritical condensation or evaporation, one thermodynamic phase can simply run out. The stronger critical-point inference combines temperature control, compression readings and the existence of a continuous route around the endpoint of the phase coexistence curve. On that route a dilute gas can become dense without crossing a two-phase boundary. This is what continuity of liquid and gas means; it does not remove their distinct equilibrium number densities below $T_c$.

In his 12 December 1910 Nobel lecture, Johannes Diderik van der Waals explained gases and liquids through the same moving molecules, with attraction and finite size modifying their pressure. He also rejected the idea that constant $a,b$ should give universal numerical agreement. His critical-volume discussion explicitly retained derivatives of a volume-dependent $b$: the familiar $v_c=3b$ follows only when those derivatives are neglected. Thus the elegant constant-parameter Van der Waals critical point is a model result whose limitations were already part of its author's account.^[ref]

The Antoine equation is an empirical vapor pressure fit whose coefficients depend on the pressure and temperature units and on a stated validity interval. Its slope is $d\ln p_{\mathrm{sat}}/dT=\ln (10)B/(T+C{)}^2$. In the ideal-vapor Clausius-Clapeyron approximation, this implies latent heat $L=R_v\ln (10)B[T/(T+C){]}^2$. An accurate pressure fit need not give an equally accurate derivative; extrapolation to the liquid-gas critical point cannot replace the full Clausius-Clapeyron relation.

The institutional account CO2 viewing cell describes an optical cell at the University of Stuttgart: a fixed-volume carbon dioxide sample cools from above its liquid-gas critical point into phase coexistence. The apparatus supplies a view through the cell and displayed temperature and pressure readings. Near roughly $31{}^{\circ }\mathrm{C}$, the account reports critical opalescence: enhanced scattering of light makes reflected light appear whitish and transmitted light dark. This readout reveals growing density thermal fluctuations; it is not a direct count of molecules or a measurement of an infinite correlation length. ^[ref]

Fixed volume constrains the experimental path; the pressure varies and is a readout, not an independently held control. The account explicitly identifies the clip as part of a longer cooldown and says its endpoints are not steady states. To test an equilibrium interpretation more closely, one would hold temperature until the readout settles, compare cooling and warming, and check optical background away from the liquid-gas critical point; these are proposed controls, not measurements reported on that page. The account does not provide cell dimensions, filling number density, calibration uncertainties or such control runs. The sample is finite: its striking appearance motivates the thermodynamic limit as a model of macroscopic sharpness, rather than demonstrating a literal equilibrium singularity.

A physicist.^[ref]

The Van der Waals equation connects a dilute gas model to the possibility of a liquid-gas critical point. Its usefulness includes showing how attractions and excluded volume can change an equation of state; its simple constants do not make it an exact model for every fluid.

An order parameter distinguishes thermodynamic phases. For a ferromagnet it is the magnetization; for a fluid near its liquid-gas critical point it can be the density relative to its critical value. In a system with a scalar order parameter, introduce a slowly varying real scalar field $\varphi (x)$ whose spatial average represents that order parameter. The Landau-Ginzburg theory organizes the effective free energy as a local expansion consistent with the symmetry:Here $K>0$ is the stiffness in the gradient energy, and $H$ is the field conjugate to an order parameter. The even powers assume a symmetry $\varphi \mapsto -\varphi$ at $H=0$. Without that symmetry further terms, including a cubic term, may be allowed. The coefficients are smooth functions of temperature and other controls; near a mean-field transition one usually writes $r=a(T-T_0)$ with $a>0$. A positive highest retained even coefficient makes the local potential bounded below.

The partition function is obtained by weighting configurations with $\exp [-\mathcal{F}/(k_{B}T)]$. The Landau approximation replaces this functional average by minimization, taking a uniform $\varphi =M$ in a homogeneous phase. In a symmetric finite system the zero-field average can vanish even below the transition because it averages both signs of $M$. The spontaneous order parameter selects a pure thermodynamic phase by taking the thermodynamic limit before removing the conjugate field:Thus the order parameter measures the ordering of a selected thermodynamic phase, rather than the average over all symmetry-related phases.

The liquid-gas critical point is the endpoint of the gas-liquid coexistence curve where the distinct phases merge. For this equation of state it is the horizontal inflection of an isotherm, so $p_v=p_{vv}=0$:Dividing these conditions gives $v_c=3b$, then substitution givesAll lie in the physical domain $v>b$.

The expansion about the liquid-gas critical point isDifferentiating at fixed $t$ gives the requested $-9{\varphi }^2/2+O({\varphi }^3)+t[-6+O(\varphi )]$. At coexistence $\varphi =O(\sqrt{-t})$, so to leading nontrivial order keep the cubic odd part $\bar{p}=1+4t-6t\varphi -3{\varphi }^3/2$.

Let $a={\varphi }_l$, $b={\varphi }_g$ for this leading cubic model. Equality of endpoint pressures gives $a^2+ab+b^2=-4t$. The Maxwell condition in pressure coordinates is ${\int }_a^b(1+\varphi ){\bar{p}}^{\prime }(\varphi )d\varphi =0$. Since the endpoint pressures agree, its first term vanishes, leavingIf $a+b\ne 0$, these two equations would imply $(a-b{)}^2=0$, contrary to distinct phases. Therefore $a=-b$, and the pressure equality gives $a^2=b^2=-4t$. HenceThus the coexistence gap has exponent one half. The printed equality ${\varphi }_l(t)=-{\varphi }_g(t)$ is a leading-order statement, not an exact identity for the full Van der Waals equation. Discarded asymmetric terms shift both phase volumes at higher order. This distinction is needed even though the leading cubic calculation has exact antisymmetry within its truncation.

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.

An equation of state relates equilibrium state variables such as pressure, volume, temperature and material amount. At fixed composition, it reduces the number of independent thermodynamic variables. It does not by itself specify all caloric properties: information such as heat capacity or internal energy is also needed.

At a liquid-gas first-order transition, an equilibrium isotherm can contain a constant-pressure interval in which changing total volume changes the phase fractions. A metastable single-phase extrapolation can instead show a region with $(\partial p/\partial V{)}_T>0$; that region has negative isothermal compressibility and is mechanically unstable. Thermodynamic phase coexistence requires equal temperature, pressure and chemical potential, and a consistent free-energy description replaces the unstable region by coexistence. For an appropriate analytic fluid equation of state, a liquid-gas critical point occurs where $(\partial p/\partial V{)}_T=({\partial }^{2}p/\partial V^2{)}_T=0$.

These tests diagnose the phase behaviour represented by that equation of state. They do not guarantee detection of every possible transition involving an additional order parameter which the mechanical equation omits.

Entropy at absolute zero. The third law of thermodynamics concerns the low-temperature limit of equilibrium states. In the zero-entropy formulation, the entropy of a perfect crystal with a nondegenerate ground state tends to zero as $T\to 0$. The statistical interpretation is $S=k_B\ln \Omega$: a unique accessible ground state has $\Omega =1$. The entropy-difference formulation of the third law of thermodynamics states that entropy differences associated with regular changes of mechanical parameters tend to zero as absolute zero is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate ground state can carry residual entropy.

For a phase of fixed composition and pressure,A finite limiting entropy constrains this integral. Ordinary equilibrium heat capacities therefore vanish as temperature approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic phonons and $C\propto T$ for a normal degenerate Fermi liquid. A constant positive heat capacity extrapolated to zero would give a logarithmically divergent entropy difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.

Gibbs free energy. For fixed particle numbers,Thus, when $S\to 0$, the Gibbs free energy approaches its zero-temperature value with a horizontal tangent:The equality $G(0,p)=H(0,p)$ uses the enthalpy $H=U+pV$; neither limiting value needs to be zero. The low-temperature Gibbs free-energy expansion makes the approach quantitative. For $C_p=a{T}^r$ with $r>0$ and $S(0)=0$,The second derivative is ${\partial }_T^{2}G=-C_p/T\le 0$, although it need not tend to zero. For a reaction between such equilibrium phases, ${\partial }_T\Delta G=-\Delta S\to 0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine residual entropy $S_0$ is retained instead, $G=G(0,p)-S_{0}T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.

Why helium remains a liquid. Helium atoms attract each other weakly and are exceptionally light. Their zero-point energy is large compared with the binding that would localize them in a solid. Both helium-4 and helium-3 can therefore remain liquid down to the lowest temperatures at low pressure; substantial applied pressure is needed to solidify them. This is not a violation of the third law. A quantum liquid can have a unique low-entropy ground state, and zero-point motion does not itself imply thermal entropy. The third law does not require every substance to crystallize at atmospheric pressure.

A helium-4 atom has integer spin zero and is a boson. Below the lambda point near $2.17\mathrm{K}$ at saturated vapor pressure, its liquid is superfluid. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the condensate fraction need not equal the superfluid fraction. The two-fluid model of superfluidity separates an entropy-carrying normal component of a superfluid from the superfluid component, whose entropy is zero. At sufficiently low temperature the dominant thermal excitations are acoustic phonons: their linear dispersion in three dimensions gives $U-U_0\propto V{T}^4$, $C_V\propto V{T}^3$ and $S\propto V{T}^3$. Higher-energy rotons are exponentially suppressed. Thus the normal component of a superfluid and entropy disappear as $T\to 0$, while superfluid flow and quantum motion can persist.

A helium-3 atom has spin $1/2$ and is a fermion. Well below its Fermi temperature, but above its superfluid transition, liquid helium-3 is a degenerate interacting Fermi liquid. Only quasiparticles close to the Fermi surface are thermally excited, giving $C_V\propto T$ and $S\propto T$. At temperatures of order a millikelvin, helium-3 also becomes superfluid through formation of Cooper pairs of fermions. This is not Bose-Einstein condensation of individual helium-3 atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic quasiparticle contribution to heat capacity is exponentially suppressed at temperatures well below the pairing gap. Gapless collective modes can still give thermal terms following a power law. The entropy nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.^[ref]

Helium-3 also illustrates the importance of nuclear spin entropy. Over an intermediate low-temperature range, localized spins in the solid can retain approximately $k_B\ln 2$ of entropy per atom, while the degenerate liquid has much less. Reversible conversion of liquid to the higher-entropy solid then absorbs heat; adiabatic compression along the melting curve can cool the sample. This is Pomeranchuk cooling. At still lower temperature, interactions between the nuclear spins remove this nearly free-spin entropy, so a constant $k_B\ln 2$ cannot be extrapolated to absolute zero.^[ref]

Liquefaction and cooling methods. A gas must be below its liquid-gas critical point to form a distinct liquid phase by compression. Compression above the critical temperature alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression heat, cools the compressed gas in a heat exchanger and uses expansion to lower its temperature further.

A throttling valve operates approximately at constant specific enthalpy when kinetic-energy changes and heat leakage are negligible. The Joule-Thomson coefficient isusing consistent molar quantities, or consistent total quantities for a fixed amount. A pressure drop cools the gas if ${\mu }_{\mathrm{JT}}>0$. An ideal gas has zero Joule-Thomson coefficient; real intermolecular interactions are essential. In particular, room-temperature helium warms on throttling, so it must first be brought into a region of positive Joule-Thomson coefficient. Expansion through an engine or turbine can instead remove work and produce approximately isentropic cooling.

A regenerative liquefier sends the cold returning gas through a counterflow heat exchanger to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces liquid. Modern helium liquefaction commonly combines work-producing expansion with a final throttle. ^[ref]

Once liquid exists, pumping away vapour lowers its saturation pressure and boiling temperature. The escaping vapour carries latent heat away, cooling the remaining liquid. At still lower temperatures, a dilution refrigerator exploits the different entropy carried by helium-3 in concentrated and dilute helium mixtures; adiabatic demagnetization and Pomeranchuk cooling provide other routes.

Approaching zero. The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach absolute zero in finitely many stages. As the entropy available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero entropy cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. Quantum liquids and refrigeration methods can approach $0\mathrm{K}$ while remaining consistent with vanishing equilibrium entropy; they do not provide a finite-operation route to absolute zero.

Three phase coexistence curves meet at the triple point, where ice, liquid water and water vapour have equal chemical potential. In the ordinary-pressure region, the low-temperature phase is ice, the high-pressure moderate-temperature phase is liquid, and the low-pressure high-temperature phase is vapour. The triple point is approximatelyBelow $p_t$, warming stable ordinary ice crosses the sublimation boundary directly into vapour; there is no intervening stable liquid. Above $p_t$, melting and then boiling can occur. The liquid–vapour curve terminates at the liquid-gas critical point, approximately $647.1\mathrm{K}$ and $22.06\mathrm{MPa}$, beyond which there is no liquid–vapour phase boundary.

The distinctive feature is the negative slope of the ordinary ice–water melting line. The Clapeyron equation givesMelting increases entropy but reduces volume because ordinary ice is less dense than liquid water, so $\Delta s>0$ and $\Delta v<0$. Increasing pressure consequently lowers the melting temperature locally. Sublimation and vaporization both increase volume and entropy, giving positive slopes. The following original schematic shows these signs; it is not a quantitative fit and omits the additional high-pressure ice polymorphs.

The phase diagram must contain three coexistence curves meeting at the triple point: a solid-gas sublimation curve, a solid-liquid melting curve and a liquid-gas vaporization curve. The vaporization curve ends at a liquid-gas critical point; beyond it liquid and gas are continuously connected as a supercritical fluid.

The imposed contraction on freezing means $v_{m,s}<v_{m,l}$. Melting therefore has $\Delta v_m=v_{m,l}-v_{m,s}>0$ and $L_{\mathrm{fusion}}>0$. The Clapeyron equation then requiresThus raising the pressure raises the melting temperature. The liquid-gas and solid-gas slopes are also positive because the transformations into gas increase both entropy and volume. The liquid region lies between the positively sloping melting line and the vaporization curve, while gas occupies the low-pressure side. The figure is a qualitative solution sketch, not a reproduction or fit of the paper's vapor-pressure measurements; its axes are schematic.

Rename the Antoine equation coefficients $A_A,B_A,C_A$ to distinguish them from the heat-capacity constants. The logarithmic slope is independent of the pressure unit:Using an ideal vapour and neglecting liquid specific volume, Clausius–Clapeyron gives $d\ln p_{\mathrm{sat}}/dT=L_v/(R_v{T}^2)$. Therefore the positive magnitude of the latent heat released on condensation isFor $C_A=-198.043\mathrm{K}$, $d\ln L_v/dT=2C_A/[T(T+C_A)]<0$ over the quoted range. With $R_v\simeq 461.5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, this fit implies approximately $2.984\mathrm{MJ}\mathrm{k}{\mathrm{g}}^{-1}$ at $380\mathrm{K}$ and $1.598\mathrm{MJ}\mathrm{k}{\mathrm{g}}^{-1}$ at $573\mathrm{K}$. The required ratio does not depend on $R_v$:The decreasing trend is qualitatively consistent with the liquid and vapour phases becoming less distinct towards the liquid-gas critical point. However, the full Clapeyron relation is $L_v=T(v_v-v_l)d{p}_{\mathrm{sat}}/dT$: at criticality the two specific volumes merge and the latent heat vanishes. Neglecting $v_l$ and treating the vapour as ideal then fail. The Antoine coefficients are not licensed for extrapolation beyond $573\mathrm{K}$, and derivatives of an approximate pressure fit can be less accurate than the fitted pressures themselves.

A concrete check exposes the extrapolation failure: at the supplied $647\mathrm{K}$ critical temperature the formula would still give $L_v\simeq 1.42\mathrm{MJ}\mathrm{k}{\mathrm{g}}^{-1}$, rather than zero, and $p_{\mathrm{sat}}\simeq 134\mathrm{bar}$ rather than $221\mathrm{bar}$. Above the true liquid-gas critical point there is no liquid–vapour coexistence curve or condensation latent heat in the same sense. The last panel of the original figure in solution makes this limitation visible.

For the variable-latent-heat saturation curve of a pure saturated water atmosphere, the total pressure equals the saturation pressure. The exact Clausius-Clapeyron relation along the phase coexistence curve iswhere $v_v,v_c$ are the specific volumes of the vapor and condensed thermodynamic phase. Neglecting $v_c$ and using the ideal gas law, $v_v=R_{\mathrm{wv}}T/p$, givesThe triple point provides the integration constant. Below it, the prescribed latent heat is constant, soAbove it, write $K=L_{\ast }/[R_{\mathrm{wv}}(T_c-T_{\ast })]$. Integrating $K(T_c/T^2-1/T)$ givesThese are the requested implicit temperature profiles $T(p)$. Both branches meet at $(T_{\ast },p_{\ast })$, and each is monotone before the limiting liquid-gas critical point. The lower branch describes equilibrium with ice, rather than a stable liquid surface. A stable liquid ocean surface belongs to the upper branch; the colder upper atmosphere may use the lower branch. Hydrostatic balance is needed to convert this relation into a height profile, but not to obtain $T(p)$.

Let $Y=-\ln (p/p_{\ast })$. For the temperature-dependent latent heat,Keeping the latent heat of vaporization fixed changes only the upper branch toBoth curves fall from $+\infty$ as $T\to 0^{+}$, pass through $Y=0$ at the triple point, and are convex. At the triple point there is a corner because the latent heat changes from $L_s$ to $L_{\ast }$. Their right-hand slopes there agree. Above the triple point,The variable-latent heat curve lies above the fixed-latent heat curve: its downward slope has smaller magnitude at every $T>T_{\ast }$. It reaches a horizontal tangent at $T_c$, whereas the fixed-latent heat curve is still descending. Both limiting ordinates are finite; this model gives $Y_{\mathrm{variable}}(T_c)\simeq -7.30$ and $Y_{\mathrm{fixed}}(T_c)\simeq -11.40$. These are extrapolations of the stipulated model, not accurate water liquid-gas critical point data. Extending the variable branch beyond $T_c$ would give negative latent heat and is physically inadmissible.

No. Near the liquid-gas critical point, the mass densities of liquid and vapor approach each other, so the neglected condensed-phase specific volume becomes comparable with the vapor specific volume. The vapor also ceases to obey the dilute ideal gas law. Although the latent heat tends to zero, the specific-volume difference tends to zero too: the exact Clausius-Clapeyron relation contains their ratio. Setting only the numerator to zero therefore does not justify a vanishing physical saturation pressure slope. A realistic equation of state and coexistence data are required near the surface; the dilute colder upper atmosphere can still be described approximately by the earlier expression.

A phase transition is a qualitative change of thermodynamic phase as control parameters such as temperature, pressure or chemical potential are varied. A genuine equilibrium singularity is a nonanalyticity of the appropriate thermodynamic potential in the thermodynamic limit; finite systems generally show rounded crossovers instead.

An order parameter is a macroscopic quantity that distinguishes the phases. It often vanishes in a symmetric disordered phase and becomes nonzero after spontaneous symmetry breaking, although density-driven transitions need not have that simple symmetry interpretation. It can be a scalar, vector or tensor. At a first-order phase transition an appropriate first derivative of the free energy is discontinuous, and the order parameter commonly jumps; at a continuous phase transition the order parameter can approach zero continuously while a susceptibility or a higher derivative becomes singular.

Three examples are:

These examples also illustrate why an order parameter must reflect the physical distinction between phases rather than merely be any quantity that changes with temperature.

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.

A renormalization group compares models at different observation lengths. A Wilsonian coarse-grained statistical Hamiltonian sums over short-distance fluctuations; a renormalization-group transformation rescales coordinates and fields for comparison. Kenneth Geddes Wilson's 1982 lecture treats fluctuations across successive length scales to explain critical phenomena.^[ref]

The experimental motivation is Andrews's carbon dioxide experiment. Dried carbon dioxide occupied a capillary tube above mercury; a screw compressed the transmitting water. A second air-filled tube gauged pressure. Flowing water controlled temperature, and a cathetometer read column positions for volume. The liquid meniscus vanished near the liquid-gas critical point. Residual air and imperfect air-gauge corrections limited accuracy.^[ref] These observations motivate critical-point theory without determining a correlation-length critical exponent. The renormalization group explains how critical measurements can become insensitive to many molecular details.

In 1972, Kenneth Geddes Wilson and Michael Ellis Fisher used an epsilon expansion about $d=4$ to compute nonclassical critical exponents. Their paper distinguishes a Gaussian fixed point from an interacting renormalization-group fixed point, tunes the thermal parameter, and extracts powers from a linearized recursion. Its Ising-like result is $\gamma =1+ϵ/6+O({ϵ}^2)$ for $d=4-ϵ$.^[ref] The worked model below isolates this mechanism with chosen coefficients; it is not a precision calculation for the three-dimensional Ising model.

A state of matter is a broad category of material behavior. A solid resists a sustained change of shape, a liquid flows while remaining difficult to compress, and a gas expands to fill its container. These distinctions are useful first descriptions, but they do not enumerate all thermodynamic phases. Plasma adds collective behavior of charged particles, while liquid crystals show that flow and structural order need not disappear together.

Distinguish three questions: what broad state of matter does the material resemble, which thermodynamic phase is present, and what are its temperature, pressure and other state variables? Two samples at different temperatures can be in the same liquid thermodynamic phase. Conversely, crystal polymorphism gives different solid thermodynamic phases of the same substance, distinguished by their crystal structures. A phase transition between them need not change the broad state-of-matter category.

For an imagined substance with crystalline forms A and B, melting A produces a solid–liquid phase transition, while converting A to B produces a solid–solid phase transition. A sample containing A and B has two thermodynamic phases even though all its material is solid. Cutting one uniform crystal into two pieces does not create a second thermodynamic phase: count distinct equilibrium forms, rather than pieces or grains. This distinction matters when applying the phase rule.

A phase diagram locates stable thermodynamic phases using temperature and pressure. Its liquid–gas phase coexistence curve ends at a liquid-gas critical point, around which a path connects liquid and gas without a first-order phase transition. Broad categories can therefore remain useful without a sharp boundary.^[ref]