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For inviscid planar incompressible flow, write velocity as $(-{\psi }_y,{\psi }_x)$ and vorticity as $\zeta ={\nabla }^2\psi$. Taking curl of the Euler equations for an inviscid fluid removes pressure and gives the displayed transport equation, with $J(\psi ,\zeta )={\psi }_x{\zeta }_y-{\psi }_y{\zeta }_x$. The Poisson equation then recovers the streamfunction at each instant. A velocity tangent to the plane cannot stretch its perpendicular vorticity vector.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.

In a connected inviscid uniformly rotating barotropic fluid, momentum balance makes the displayed sum constant. The pressure term is the barotropic enthalpy function. Perturbations experience Coriolis acceleration, and their rotating-frame frequency depends on the chosen exponential sign convention.

Put $Q=2|\Omega |c/(\pi G{\Sigma }_0)$. The uniformly rotating gas-sheet dispersion relation has growing waves precisely for $Q<1$, at $q_{-}<|k|<q_{+}$ with $q_{\pm }=(\pi G{\Sigma }_0/c^2)(1\pm \sqrt{1-Q^2})$. Pressure stabilizes short waves and rotation stabilizes long waves. Equality $Q=1$ is marginal; $Q\ge 1$ is the gas Toomre stability criterion.

A homogeneous canonical scalar field with positive scalar potential has energy density ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and pressure $P_{\varphi }={\dot{\varphi }}^2/2-V$. When its potential energy dominates its kinetic energy, the equation-of-state parameter approaches $-1$. If other components and spatial curvature are negligible, the Friedmann acceleration equation gives accelerating expansion. Small initial kinetic energy is not enough to ensure long cosmic inflation; the field must also remain on a sufficiently flat part of the scalar potential, as measured by the potential slow-roll parameters.

The Van der Waals equation is a simple real gas equation of state that can also describe a liquid-gas critical point. With particle count $N$ fixed, write the volume per particle as $v=V/N$:or equivalently $(p+a{N}^2/V^2)(V-Nb)=N{k}_{B}T$. Here $p$ is pressure, $T$ is absolute temperature, and $k_B$ is the Boltzmann constant. In SI, $v$ and $b$ have units ${\mathrm{m}}^3$ per particle, $a$ has units $\mathrm{Pa}{\mathrm{m}}^6=\mathrm{J}{\mathrm{m}}^3$, and $k_B$ has units $\mathrm{J}{\mathrm{K}}^{-1}$; the particle count is dimensionless. The pole at $v=b$ excludes smaller volume values from this model.

The $b$ correction represents excluded volume; the attraction term lowers pressure at fixed temperature and particle number density $1/v$. This is a mean-field theory, replacing detailed intermolecular structure by two parameters. At $v\gg b$ with $a/v^2\ll k_{B}T/v$, it approaches the ideal gas law $pv=k_{B}T$. It supplies a useful qualitative route from a dilute gas to phase coexistence, without a claim of quantitative accuracy for every fluid.^[ref]

Do not mix this particle convention with the molar volume convention elsewhere in this article. With amount of substance $n_{\mathrm{mol}}=N/N_A$, Avogadro constant $N_A$, and molar gas constant $R=N_A{k}_B$, one has $v_m=N_{A}v$, $b_m=N_{A}b$, and $a_m=N_A^{2}a$, so $p=RT/(v_m-b_m)-a_m/v_m^2$. The molar parameters have units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$, respectively.

Johannes Diderik van der Waals introduced this equation of state in 1873, connecting molecular attraction with departures from the dilute ideal gas law. Its historical significance is the unified treatment of gases and liquids, rather than a universally exact fit.^[ref]

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 Van der Waals equation becomes independent of its material parameters when pressure, temperature and molecular volume are divided by their critical values: $\bar{p}=8\bar{T}/(3\bar{v}-1)-3/{\bar{v}}^2$. This explains the common reduced isotherms within this model, without claiming exact agreement for all real fluids.

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.

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.