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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]

Johannes Diderik van der Waals received the award for an equation of state describing gases and liquids.^[ref] The Van der Waals equation connects that description to a Van der Waals critical point and Van der Waals liquid-gas coexistence.

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.

Pressure and its physical terms. Solving the molar Van der Waals equation for pressure givesThe excluded-volume parameter $b$ reduces the volume available for translational motion, raising the kinetic pressure relative to $RT/V$. The attraction parameter $a$ lowers the pressure because attractive interactions reduce the momentum transferred to the boundary. This mean-field model treats $a,b$ as temperature-independent material constants. For one mole, $V$ is the sample volume with the molar amount already fixed; $b$ has units of that volume and $a$ of pressure times volume squared.

volume dependence of entropy. The Helmholtz free energy has differential $dF=-SdT-PdV$. Equality of mixed derivatives yields the Maxwell relationFor $a=0$ this derivative is $R/(V-b)$, givingEquivalently, $S=R\ln (V-b)+f(T)$ with a common reference volume understood inside the logarithm. The equation of state fixes this volume dependence, not the otherwise unspecified function of temperature.

Excess entropy at fixed volume and temperature. The ideal gas has $(\partial S_{\mathrm{id}}/\partial V{)}_T=R/V$. ThereforeIntegrate from infinite volume, where the entropy difference is specified to vanish:It is negative because excluding configurations reduces the accessible state count. It tends to zero as $V\to \infty$ and has leading dilute limit $-Rb/V$. This is the entropy excess of a van der Waals gas.

Effect of attraction. Restoring a temperature-independent $a$ changes the pressure by $-a/V^2$, but not its derivative with respect to temperature. With the same infinite-volume entropy reference, it therefore does not change $\Delta S^{\mathrm{ex}}$ at fixed $V,T$. This can also be seen fromThe attraction affects internal energy and the equilibrium volume, so it can affect entropy indirectly when different state points are compared. If $a$ were temperature dependent, its derivative would contribute directly to the entropy; that is outside the stipulated constant-parameter model.

Coexistence isotherm. Below the Van der Waals critical point, the formal homogeneous isotherm has a local minimum and maximum. Its middle branch has $\partial P/\partial V>0$, hence negative isothermal compressibility, and is mechanically unstable. Portions of the negative-slope branches can be metastable rather than globally stable. The equilibrium isotherm replaces the loop between liquid volume $V_l$ and vapour volume $V_g$ by a constant saturation pressure $P_s$. The two volume are fixed byThe Maxwell equal-area construction expresses equality of the two phases' chemical potentials. Intermediate total volume are mixtures: changing the vapour fraction changes total volume at the same pressure, rather than following the unstable homogeneous curve. The spinodals are the turning points of the formal isotherm; they are distinct from the coexistence endpoints.

Dense-phase volume at low temperature. At a general subcritical temperature, the two coexistence equations above must be solved together. For this equation of state the area condition is explicitlyThis supplies an exact implicit determination of the liquid volume even when a simple closed form is unavailable.

For the asymptotically low-temperature limit, put $\tau =RTb/a\ll 1$. The liquid is dense, $V_l\simeq b$, while the vapour is dilute, $P_s{V}_g\simeq RT$ and $V_g\gg V_l$. The area condition then givesThus the vapour volume grows exponentially in $1/\tau$, and the saturation pressure is exponentially small. It is consequently consistent to neglect $P_s$ in the liquid equation to algebraic order in $\tau$:Choose the root near $b$, not the large-volume zero-pressure root:Expanding gives the leading requested dense-phase behavior,This is the low-temperature liquid volume of a van der Waals gas. Setting $P_s=0$ is an asymptotic approximation, not an exact coexistence condition at every temperature below $T_c$. The figure illustrates generic subcritical coexistence; its finite-temperature liquid volume is determined by the full equal-area equations. The van der Waals liquid also ceases to be a realistic description at sufficiently low physical temperature, where other phases or quantum effects can matter.

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.

The molar volume $v_m=V/n_{\mathrm{mol}}$ is the volume per amount of substance, not the total container volume or the volume per particle. In the molar convention of the Van der Waals equation, take constant positive parametersThenor, for any fixed amount of substance,Both terms in the first equation are pressures. The total excluded volume is $n_{\mathrm{mol}}{b}_m$: using $V-b_m$ would subtract quantities with different units. Changing the sample size at fixed molar volume and temperature changes $V$ and $n_{\mathrm{mol}}$ proportionally, while leaving pressure unchanged. This is a useful unit and extensivity check before fitting a real gas.

For the molar Van der Waals critical point, differentiate with respect to molar volume at fixed temperature:Setting both to zero gives $R{T}_c=2a_m(v_{m,c}-b_m{)}^2/v_{m,c}^3$ and $R{T}_c=3a_m(v_{m,c}-b_m{)}^3/v_{m,c}^4$. Equating them and dividing by the nonzero factors yields $2v_{m,c}=3(v_{m,c}-b_m)$, soHere $v_{m,c}$ has units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, $T_c$ has units $\mathrm{K}$, and $p_c$ has units $\mathrm{Pa}$. This is the molar version of the particle derivation below: $v_{m,c}=N_A{v}_c$, with the same physical $T_c$ and $p_c$. If one writes $v_c=3b$, $T_c=8a/(27Rb)$ and $p_c=a/(27b^2)$ without subscripts, then all of $v_c,b,a$ in those formulas must be molar quantities.

Now define the dimensionless reduced variables explicitly as $x=v_m/v_{m,c}$, $t=T/T_c$ and $q=p/p_c$. Substituting the three critical values cancels $a_m,b_m,R$ and givesBecause $v_m/v_{m,c}=v/v_c$, these are exactly the reduced variables in the Van der Waals isotherm discussion. The Van der Waals law of corresponding states is therefore independent of whether one starts with particle or molar units. It predicts a critical ratio $p_c{v}_{m,c}/(R{T}_c)=3/8$, providing a third check on any proposed parameter fit.

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.