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Substitute $v=3b\bar{v}$, $T=(8a/(27bk))\bar{T}$, and $p=(a/(27b^2))\bar{p}$ into the equation of state. After cancellation of the material parameters,This is the Van der Waals law of corresponding states: fluids described by this model share the same reduced equation, although real fluids need not obey it exactly.

The liquid-gas critical point of the Van der Waals equation occurs where its two subcritical stationary points merge into a horizontal inflection point. The conditions are $(\partial p/\partial v{)}_T=({\partial }^{2}p/\partial v^2{)}_T=0$. The short calculation isDividing these relations gives $2/(v_c-b)=3/v_c$, henceSubstitution gives the Van der Waals law of corresponding states and the parameter-independent ratio $p_c{v}_c/(k_B{T}_c)=3/8$. That ratio belongs to this equation of state; reducing a real fluid's variables does not enforce it.^[ref]

An exact rearrangement of the reduced critical Van der Waals isotherm makes its flatness visible:The first nonzero departure near $x=1$ is $-3(x-1{)}^3/2$. On the critical particle volume approached from $t>1$, $(\partial q/\partial x{)}_{x=1}=-6(t-1)$ andThe divergent isothermal compressibility and cubic flattening explain why the critical region is mechanically soft. These are mean-field theory predictions; thermal fluctuations limit their use for real fluids very close to a thermodynamic critical point.

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]

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.