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For any fixed real $n$-by-$n$ matrix $A$, as the real parameter $t$ tends to zero,The big O notation means the remainder is bounded in absolute value by $C{t}^2$ for some constant $C$ near zero. The determinant stays positive for sufficiently small $t$, so its oriented volume factor is also its ordinary volume factor. Thus the trace is the first-order fractional volume change, measured from the identity matrix.

A proof follows directly from multilinearity of the determinant. Each column of $I+tA$ is $e_j+tA{e}_j$. The constant term chooses all the standard basis columns and equals one. A term containing exactly one factor $t$ replaces just column $j$ by $A{e}_j$; expanding that vector in the standard basis leaves only its $e_j$ component, $a_{jj}$, because every other choice repeats a column. Summing gives $\mathrm{tr}A$. Terms with two or more replacements have order $t^2$ or higher. This also proves the derivative of the determinant at the identity matrix.

This coefficient has a direct visual meaning. At the point $e_j$, the linear map $x\mapsto Ax$ has component $a_{jj}$ along direction $e_j$. In a small step $x\mapsto x+tAx$, these are the first-order changes of the square's or box's sides in their own directions. Summing them gives its fractional area or volume change; the off-diagonal entries initially change shape rather than contributing to this first-order total.

For the constant-coefficient linear system of differential equations $x^{\prime }=Ax$, the exact linear flow is $x(t)=e^{tA}x(0)$, using the matrix exponential. Its volume factor isThis is the matrix exponential determinant identity. One can derive it without diagonalization: set $E(t)=e^{tA}$, whose matrix inverse is $E(-t)$, and use $E^{\prime }=AE$ in Jacobi's formula to get $V^{\prime }=\det E\mathrm{tr}(E^{-1}AE)=(\mathrm{tr}A)V$. The scalar differential equation with initial value one gives the displayed exponential function. Thus positive trace causes volume expansion, negative trace causes contraction, and zero trace preserves volume, even when individual directions behave differently.

In two dimensions, compare four generators at $t=0.4$:The first and third diagonal matrices expand or contract both directions. The second expands horizontally and contracts vertically by reciprocal factors, making a rectangle of unchanged area. For the last nilpotent matrix, $A^2=0$, so its matrix exponential is exactly $I+tA$; it sends $(x,y)$ to $(x+ty,y)$, a shear mapping. Zero trace allows substantial shape change and does not mean the flow map is stationary.

The first-order statement and the exact flow map should be distinguished. For $A=\mathrm{diag}(1,-1)$, $\det (I+tA)=1-t^2$, whereas $\det (e^{tA})=1$ for all real $t$. Zero trace eliminates the first-order change in the approximate step, and it preserves volume exactly in the matrix exponential flow. The examples use an abstract parameter $t$ and coordinate vectors; no physical units are assumed.

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.

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 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]

Volume magnetic susceptibility is the dimensionless ratio of longitudinal magnetization to internal magnetic field intensity in a linear regime through the origin. Positive response gives paramagnetism. Always distinguish magnetization $M=\mathcal{M}/V$ from total magnetic moment $\mathcal{M}$: dividing by volume changes the normalization. Mass-normalized and mole-normalized values likewise require the mass density or the volume per amount of substance before comparison with ${\chi }_v$.

In the existing Pauli paramagnetism formula, $M$ denotes total magnetic moment, and $g(E_F)$ is the total spin-summed density of states. Relabel that response asHere ${\mu }_B$ is the Bohr magneton. If its $B$ is a prescribed vacuum/applied magnetic flux density $B_0={\mu }_0{H}_{\mathrm{app}}$, the dimensionless apparent slope is ${\mu }_0{\chi }_B^{\mathrm{vol}}$. It agrees approximately with intrinsic ${\chi }_v$ only when the demagnetizing field correction and magnetic feedback are negligible. The existing Curie's law formula also uses a total-magnetic moment, $B$-derivative convention; its $N$ counts moments rather than giving a number density.

The $(n-1)$-dimensional volume of the Euclidean unit sphere equals $n$ times the $n$-dimensional volume of its unit ball. If $R={\sum }_i{x}^i{\partial }_i$ is the radial vector field and $\omega =d{x}^1\wedge \cdots \wedge d{x}^n$, then $d({\iota }_R\omega )=n\omega$, and the Generalized Stokes theorem proves the formula.

Thermodynamic volume is the volume occupied by a thermodynamic system, treated as an extensive variable.

The Young–Laplace equation relates the pressure jump across a stationary interface to its curvature and surface tension. Choose a unit normal vector $\mathbf{n}$ pointing from the inside region to the outside region, and define $\kappa ={\nabla }_s\cdot \mathbf{n}$ to be positive for a convex sphere. In mechanical equilibrium, with constant surface tension and no additional normal viscous stresses,Here $k_1,k_2$ are the signed principal curvatures in this convention. Their sum is twice the corresponding signed mean curvature; stating the convention prevents a factor-of-two ambiguity. The Young–Laplace equation is a local normal stress balance for one interface. ^[ref]

A spherical liquid droplet of radius $R$ in a gas has one interface, with $k_1=k_2=1/R$. An ideal thin soap bubble has two liquid–gas interfaces. Adding their pressure jumps givesFor the soap bubble, $\Delta p$ now means the inner gas pressure minus the outer gas pressure. The second formula assumes equal surface tensions on both faces and film thickness negligible relative to $R$, so their radii can be identified. Retaining distinct radii gives $2\gamma /R_{\mathrm{inner}}+2\gamma /R_{\mathrm{outer}}$. A gas bubble inside bulk liquid has only one interface and therefore uses $2\gamma /R$. ^[ref]

With the assumed constant toy surface tension $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, a droplet with $R=1\mathrm{mm}$ has excess pressure $144\mathrm{Pa}$; at $R=0.5\mathrm{mm}$ it has $288\mathrm{Pa}$. A thin soap bubble assigned the same $\gamma$ and $R=1\mathrm{mm}$ has excess pressure $288\mathrm{Pa}$. These are differences from the surrounding pressure, rather than absolute pressures. Halving the radius doubles the capillary pressure; doubling the interface count also doubles it at fixed radius and surface tension.

The spherical case has a short work argument. A virtual increase in radius gives $dV=4\pi R^{2}dR$ and $dA=8\pi RdR$. At mechanical equilibrium, the net pressure work balances the increase in surface energy:For a thin soap bubble, both interfaces gain approximately the same surface area, doubling the right-hand side and giving $4\gamma /R$. This is a virtual work balance between neighboring spherical configurations, rather than an assumption that a fixed-volume droplet actually expands. The general Young–Laplace equation replaces the spherical area-to-volume ratio by local curvature, making geometry directly measurable through pressure.