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A flat circular contact of radius $a$ carrying total normal force $N$ has uniform pressure $p=N/(\pi a^2)$. In steady relative rotation, a ring of radius $s$ and thickness $ds$ experiences tangential kinetic friction $dF=\mu p2\pi sds$. Its resisting torque is $d\tau =sdF$. Integration from zero to $a$ gives ${\tau }_f=2\mu Na/3$. A different pressure distribution changes the result; for an axisymmetric distribution use ${\tau }_f=2\pi \mu {\int }_0^{a}p(s)s^{2}ds$.

For a uniform ideal gas sphere, set $\rho =mn$, $p=n{k}_{B}T$ and $M=4\pi \rho R^3/3$. Building the sphere by concentric shells gives gravitational potential energy $W=-3G{M}^2/(5R)$. During a small homologous contraction, released gravitational energy balances the work against gas pressure when $3G{M}^2|dR|/(5R^2)=4\pi R^{2}p|dR|$, or $GMm/R=5k_{B}T$. Eliminating $R$ gives the displayed mass. At the same initial mass density and temperature, larger masses have stronger gravitational contraction than pressure support in this estimate. Its numerical coefficient belongs to the uniform-sphere pressure-work model; the infinite-medium Jeans instability and a pressure-confined isothermal sphere use different equilibrium and boundary assumptions.

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 U-tube differential manometer contains a connected column of liquid with one gas connection above each liquid meniscus. Write $p_L,p_R$ for the left and right pressures, and $z_L,z_R$ for the meniscus heights on an upward vertical coordinate. Assume constant mass density $\rho$ and downward gravitational acceleration of magnitude $g$; neglect the gas column's mass density and differences in capillary pressure. The signed vertical length difference $\Delta h=z_R-z_L$ then givesLarger left pressure means a lower left liquid meniscus: $p_L>p_R$ implies $z_L<z_R$. Equal pressures give equal meniscus heights.

The balance follows directly from hydrostatic pressure. At a common height $z_0$ below both surfaces, the two points in the connected homogeneous liquid have equal pressure:Subtracting cancels the common bottom depth and leaves $p_L-p_R=\rho g(z_R-z_L)$. The same-height rule follows from hydrostatic equilibrium, $\nabla p=-\rho g{\mathbf{e}}_z$: a horizontal displacement changes no pressure, and for constant $\rho$ the quantity $p+\rho gz$ is constant throughout the connected liquid. Equal height alone is insufficient for unrelated vessels or for fluid flow with significant acceleration.

For an original constructed example, choose water of stipulated mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and vertical length difference $\Delta h=0.060\mathrm{m}$. ThenThe SI pressure unit is the pascal: the product has units $\mathrm{kg}{\mathrm{m}}^{-3}\times \mathrm{m}{\mathrm{s}}^{-2}\times \mathrm{m}=\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}=\mathrm{Pa}$. Swapping the two connections gives $-588.6\mathrm{Pa}$ and reverses which liquid meniscus is lower. In equal-bore arms, volume conservation makes each surface move $30\mathrm{mm}$ from its initial equal-height position; using only one surface's displacement would halve the inferred pressure. With unequal arm areas $A_L,A_R$, a left fall $a$ and right rise $b$ instead obey $A_{L}a=A_{R}b$, while the vertical length difference is still $a+b$.

If the right port is open to the local atmosphere, $p_L-p_R$ is the left gauge pressure. Only with that reference and known atmospheric pressure can one obtain absolute pressure fromFor an explicitly stipulated atmospheric pressure of $100000\mathrm{Pa}$, the example gives $100588.6\mathrm{Pa}$ absolute pressure. If the right port is connected to another vessel or a pressure tapping, the same reading is a differential pressure and supplies no absolute pressure without knowing that reference.^[ref]

Vacuum distillation carries out distillation at reduced pressure.

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.

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.