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The virial expansion describes the first departures from the ideal gas law as a power series in number density. For a homogeneous, single-species gas in three dimensions at thermal equilibrium, write $n=N/V$, with $V$ the volume, and define the dimensionless compressibility factor $Z$ byHere $p$ is absolute pressure, $T$ is absolute thermodynamic temperature, and $k_B$ is the Boltzmann constant. The second virial coefficient determines the initial slope away from $Z=1$; higher coefficients describe further corrections. The symbol $Z$ here is an equation of state ratio, not a canonical partition function. ^[ref]

The left panel sketches the measurement principle; the right panel uses deliberately chosen polynomials, $Z=1+bx+6x^2$ with $x=n{\ell }^3$ and $b=\pm 2$, where $\ell$ is a reference length. These curves are not measured data. Their initial slopes have opposite signs, but both bend because of the quadratic term. Using a finite secant from $x=0$ to $x=h$ gives $(Z(h)-1)/h=b+6h$, rather than $b$: reducing the fitting range reduces this particular truncation bias. This algebra explains why measuring one nonideal pressure is insufficient to extract a reliable second virial coefficient.

Waxman and Hastings's 1971 Burnett experiment used sample and expansion chambers in a thermostatic bath, with a valve between them and a pressure transducer linked to a piston gauge. They measured successive isothermal expansions: evacuate the expansion chamber, open the valve, record the equilibrated pressure, isolate and evacuate that chamber again, then repeat. This avoids separately measuring the absolute gas amount and chamber volumes. For carbon dioxide at $50{}^{\circ }\mathrm{C}$, their measurements below $35$ bar gave a molar second virial coefficient of approximately $-102.2\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. ^[ref]

An idealized reduction makes the signal explicit. Let $q=(V_A+V_B)/V_A$ be the calibrated chamber volume ratio, $p_i$ the pre-expansion pressure, and $p_f$ the post-expansion pressure. Particle conservation during one valve opening gives $n_f=n_i/q$. At the same temperature, division of the two equations of state yieldsThus the intercept determines $q$, while the dilute-limit slope determines the second virial coefficient. In the actual experiment, helium runs calibrated the cell constant and chamber deformation corrections were included. Agreement of the carbon dioxide low-pressure intercept with the helium calibration, and the absence of an anomalous low-pressure upturn, checked for gas adsorption on chamber walls. ^[ref]

For a limited dilute-gas fit, hold temperature fixed, allow thermal equilibrium after every expansion, calibrate absolute pressure and the chamber ratio, and check leaks and sample composition. Repeat the analysis with a smaller upper pressure cutoff and an added curvature term; compare the resulting second virial coefficients with their uncertainties. Too wide a range mixes in higher coefficients; too narrow a range makes the nonideal signal hard to distinguish from instrument error. The displayed chamber equation neglects deformation, dead spaces and adsorption, so it is a teaching reduction rather than the complete experimental analysis.

Units matter. With number density in ${\mathrm{m}}^{-3}$, $B_2$ has units ${\mathrm{m}}^3$ and $B_3$ has units ${\mathrm{m}}^6$. For molar density $c=n/N_{\mathrm{A}}$ in $\mathrm{mol}{\mathrm{m}}^{-3}$, the same equation of state isHere $N_{\mathrm{A}}$ is the Avogadro constant and $R=N_{\mathrm{A}}{k}_B$ is the molar gas constant; the molar coefficients have units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and ${\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$. A coefficient tabulated per mole must therefore multiply molar density, not number density.

For the classical formulas below, use a nondegenerate gas with thermal de Broglie wavelength ${\lambda }_{\mathrm{th}}$ satisfying $n{\lambda }_{\mathrm{th}}^3\ll 1$, negligible quantum corrections to relative motion, and a stable, sufficiently short-ranged pair potential. Diluteness must also control the omitted interaction terms: $|B_{2}n|\ll 1$ alone is not enough when positive and negative contributions cancel. A finite truncation is a local low-number density approximation, not an equation of state for a dense liquid or across a phase transition. ^{[ref] [ref]}

The proof is a single time derivative. Apply the product rule to each dot product and use Newton's second law:Integrating from $0$ to $T$ gives the endpoint identity; the stated sublinear condition removes the endpoint term only in the infinite-time limit. Equivalently, for the scalar inertia $I=\frac{1}{2}{\sum }_i{m}_i|{\mathbf{r}}_i{|}^2$ used in the pairwise moment-of-inertia identity, $\dot{I}=G$ and $\ddot{I}=2K+{\sum }_i{\mathbf{r}}_i\cdot {\mathbf{F}}_i$. Specifying this factor of one half avoids confusion with another common inertia convention.

In On a Mechanical Theorem applicable to Heat by R. Clausius (1870, pp. 123-124), the author defines stationary motion through bounded positions and velocities, includes planetary orbits and elastic vibrations, and names the average negative half of the force-position sum the “virial”. That historical convention differs from the auxiliary $G(t)$ defined above. His proof on pp. 126-127 keeps an endpoint term, then explains why it vanishes over a complete period or becomes small in a long bounded irregular motion. The same paper separates internal and external forces and includes uniform boundary pressure on pp. 125-126: confinement was already part of the formulation, rather than a reason to omit external stresses.^[ref]

A viscometric flow has the local relative deformation history of steady simple shear flow, up to rigid changes of frame. For an isotropic simple fluid, symmetry leaves one shear stress and two independent normal-stress differences. The corresponding viscometric functions depend on the magnitude of the shear rate, while pressure supplies the isotropic part.

A slender low-viscosity fluid column inside a much more viscous fluid rises under a density contrast. At negligible inertia, its axial flux is approximately $q=\pi a^4(\Delta \rho g-P_{i,z})/(8{\mu }_i)$. Radial expansion strains the outer fluid and gives the normal-stress pressure difference $P_i-P_o=2{\mu }_o{a}_t/a$ at leading order when inner normal viscous stress and interfacial tension are negligible. Coupling that pressure to volume conservation yields the conduit equation. This two-fluid viscous model differs from a turbulent entraining thermal plume.

A convex scalar function $\Omega (D)$ on symmetric rate-of-strain tensors generates stress by $S=\partial \Omega /\partial D$. For incompressible flow, its argument is trace free and a pressure reaction must be added to the stress. At nondifferentiable points the law is $S\in \partial \Omega (D)$, where the subdifferential is defined by $\Omega (E)-\Omega (D)\ge S:(E-D)$ for every admissible $E$. This is a strain-rate potential, in contrast with potentials parameterized by thermodynamic forces.

The viscous stress tensor is the part of the Cauchy stress tensor associated with deformation rates. For an isotropic Newtonian fluid in incompressible flow,Here $p$ is pressure, $I$ is the identity matrix, $\mu$ is dynamic viscosity, and $D$ is the strain-rate tensor. The convention is $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$; $T$ denotes transpose and $\mathrm{tr}$ denotes trace. Across a plane with unit normal vector $\mathbf{n}$, the viscous traction is ${\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}$. The tensor makes this rule work for any plane orientation. ^[ref]

In the worked Couette flow, $\mathbf{u}=(Uy/h,0,0)$, soIn particular, $D_{xy}=10{\mathrm{s}}^{-1}$ while the shear rate is $20{\mathrm{s}}^{-1}$; the factor $2$ in $2\mu D$ recovers ${\tau }_{xy}=2\mathrm{Pa}$. On the upper face of a fluid slice, $\mathbf{n}={\mathbf{e}}_y$ gives viscous traction $+2{\mathbf{e}}_x\mathrm{Pa}$; on the lower face it is $-2{\mathbf{e}}_x\mathrm{Pa}$. Constant stress can therefore transmit a force between the plates even though its divergence, and the net viscous force on each slice, vanish.

Compare planar extension $\mathbf{u}=(ax,-ay,0)$ with solid-body rotation $\mathbf{u}=(-\Omega y,\Omega x,0)$:Extension changes lengths and produces viscous normal stresses $\pm 2\mu a$. Rotation preserves distances and gives zero viscous stress, despite its nonzero velocity gradient and vorticity $2\Omega {\mathbf{e}}_z$. For $a=\Omega =1{\mathrm{s}}^{-1}$ and $\mu =0.10\mathrm{Pa}\mathrm{s}$, extension gives ${\sigma }_{\mathrm{visc},xx}=0.20\mathrm{Pa}$ and ${\sigma }_{\mathrm{visc},yy}=-0.20\mathrm{Pa}$; rotation gives zero. A rotating fluid may still have a pressure gradient supplying its centripetal acceleration.

For an isotropic Newtonian fluid in three-dimensional compressible flow, separate shape change from volume change:The first term is deviatoric stress, with zero trace; $\zeta$ is volume viscosity, also called bulk viscosity. For uniform expansion $\mathbf{u}=b\mathbf{x}$, $D=bI$ and $\nabla \cdot \mathbf{u}=3b$, so the shear contribution vanishes and the viscous stress is $3\zeta bI$. Thus dynamic viscosity and volume viscosity describe distinct responses, and incompressible flow removes only the volume-change term. ^[ref]

For an ideal gas mixture, the volume mixing ratio $x_i$ equals the mole fraction, number fraction, and ratio of partial to total pressure: $x_i=N_i/N=n_i/n=P_i/P$, with ${\sum }_i{x}_i=1$. This equivalence fails for general nonideal mixtures.

For a Newtonian fluid and a flat wall with $0<b<\infty$, the Navier slip boundary condition is equivalent to a tangential wall-on-fluid tractionThe sign expresses resisting friction: ${\partial }_{n}u$ uses the inward unit normal, whereas the wall-on-fluid traction uses the outward unit normal of the fluid domain. The irreversible loss of mechanical energy per unit wall area and time isFor two tangential directions replace $u_{\mathrm{rel}}^2$ by the squared Euclidean norm of relative tangential velocity. At a moving wall this relative-velocity loss must be distinguished from the power supplied by the wall's motion. Bulk viscous dissipation also remains present. A finite slip length models reduced interfacial resistance, without declaring every boundary ideal or removing dynamic viscosity.

The limiting loss depends on how the flow is driven. In Couette flow with a slipping lower wall,It tends to zero for both $b\to 0$ and $b\to \infty$ at fixed $H$ and $U$. At $b=0$ impose the no-slip boundary condition directly, rather than dividing by zero. In planar Poiseuille flow with equal wall slip at fixed $G$, each wall instead dissipates $G^{2}b{H}^2/(4\mu )$, which grows with $b$ because relative velocity grows. The bulk loss per unit width is $G^2{H}^3/(12\mu )$; adding both wall losses gives $Gq$, the pressure work per unit channel length and width. This energy balance shows why small friction coefficients do not guarantee small losses under every driving constraint.

Water potential expresses the chemical potential of water relative to a stated pure-water reference in units of pressure. For a nearly incompressible liquid at fixed temperature, use a constant reference molar volume ${\bar{V}}_w$ and writeThe reference is pure water at the same temperature and a chosen pressure $P_0$. Both sides must use the same reference. Neglecting height differences and interactions with solid surfaces, the approximation isHere $a_w$ is the dimensionless thermodynamic activity of water, $R$ the molar gas constant, and $T$ the absolute temperature. The natural logarithm is therefore applied to a dimensionless quantity. The pressure contribution ${\Psi }_p=P-P_0$ increases with pressure; the solute potential ${\Psi }_s$ is negative when $a_w<1$. Across an available passive water route, net water transfer favors higher to lower water potential. The familiar plant description adds effects of height and interactions with solid surfaces when they matter; they are omitted in the two liquid compartments here.^[ref]

Adding solute and raising pressure can thus cancel in water potential while each contribution remains nonzero. Equality of water potential does not imply equality of pressure or of every solute concentration. In a supported plant cell, the cell wall permits a positive turgor pressure to balance a negative solute potential; plasmolysis illustrates the opposite loss of support after water leaves.