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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.

Vapor-compression refrigeration circulates a working fluid through cold and hot coils. In the cold coil, liquid becomes gas while absorbing heat from the refrigerated region. Compression supplies work and raises the pressure and temperature; the hot coil releases heat as the gas returns to liquid. A restriction reduces the pressure before the next cold-side passage. Changing the pressure changes the temperature of the phase transition, letting one circulating fluid receive and release heat at different temperatures. This physical implementation serves both refrigerators and heat pumps; its ordinary dissipative operation differs from the reversible Carnot cycle used to bound performance. ^{[ref] [ref]}

An ideal gas law measurement obtains amount of substance without knowing the chemical formula in advance:Here $p$ is vapor pressure, $V$ is its occupied volume, $T$ is absolute temperature, $R$ is the molar gas constant, and $m$ is the mass of vapor occupying that volume at those conditions.

Santa Monica College's teaching experiment uses a $125\mathrm{mL}$ Erlenmeyer flask with a pinhole foil cap, a $600\mathrm{mL}$ water bath on a hot plate, a thermometer, and a barometer. Excess volatile liquid displaces air. A dissolved dye loses its colour and deposits when the liquid evaporates; three more minutes of heating follow. Cooling condenses retained vapor. Subtracting the dry assembly mass from the cooled assembly mass supplies $m$. The flask volume comes from brim-full water mass and mass density at its measured temperature. Controls include a dry exterior, excluding bath water, and three trials.^[ref]

The inference assumes vapor pressure equals atmospheric pressure, vapor and bath reach thermal equilibrium, the retained vapor mass is recovered on cooling, and the vapor obeys the ideal gas law. Residual air, thermal expansion of the flask, and vapor temperature below bath temperature are explicit error questions in the teaching experiment. Repeating trials assesses repeatability; it does not remove a shared bias.^[ref]

The equilibrium pressure of a vapor in contact with its liquid or solid at a specified temperature. For pure liquid–gas phase coexistence with negligible interface effects, it is the pressure on their phase coexistence curve. In a gas mixture, distinguish the component's partial pressure from the total pressure; the pure-substance Integrated Clausius-Clapeyron relation does not describe the total pressure of arbitrary humid air.

The time for a vertical pressure or gravity response. A sound-crossing time and a small vertical oscillation period have this scale in a thin disk. It need not equal a damping time.