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Consider a particle of mass $m$ in a rigid rectangular box of length $L$ along $x$ and wall surface area $A$, so $V=AL$ is the volume. Choose $+x$ toward the right wall. For an incoming normal velocity $v_x=v>0$, an elastic collision reverses the normal momentum:The impulse delivered to the wall points outward, toward $+x$, by Newton's third law. The particle's opposite momentum change is not the wall's impulse. A tangential velocity component is unchanged by the ideal smooth-wall collision.

Ignoring intervening particle-particle collisions in this counting model, the particle travels a normal distance $2L$ before hitting the same wall again. Its hit rate is $|v_x|/(2L)$; multiplying by the impulse per hit gives the time-averaged normal force and its contribution to pressure:For example, stipulate $m=4\times 10^{-26}\mathrm{kg}$, $|v_x|=500\mathrm{m}{\mathrm{s}}^{-1}$, $L=0.10\mathrm{m}$, and $A=0.010{\mathrm{m}}^2$. Then $V=0.0010{\mathrm{m}}^3$, the impulse per hit is $4\times 10^{-23}\mathrm{N}\mathrm{s}$, the hit rate is $2500{\mathrm{s}}^{-1}$, and $P_1=10^{-17}\mathrm{Pa}$. This is one particle's tiny time-averaged contribution to one wall's pressure, not an isotropic equilibrium gas by itself.

For $N$ particles of equal mass, add their force contributions and define $⟨v_x^2⟩=N^{-1}{\sum }_{i=1}^N{v}_{x,i}^2$. In the gas's rest frame, isotropy gives $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=⟨v^2⟩/3$, where $v^2=v_x^2+v_y^2+v_z^2$. HenceThe final equality uses the classical equipartition theorem. The kinetic pressure of an isotropic gas therefore recovers the ideal gas law from momentum transfers. Doubling the normal speed would double both the impulse per hit and the hit rate, giving four times the pressure contribution; pressure measures a mean squared speed. Interparticle elastic collisions redistribute velocities and help establish isotropy; the individual round-trip times need not remain $2L/|v_x|$, but the equilibrium momentum flux result survives. ^[ref]

For a fixed Newtonian gravitational potential, total energy density is $\mathcal{E}=\rho (u^2/2+\Phi )+p/(\gamma -1)+B^2/(2{\mu }_0)$. Its flux combines advected kinetic/potential energy, specific enthalpy and the Poynting vector. Dotting the ideal magnetohydrodynamic momentum equation with velocity, adding the adiabatic internal energy equation and the magnetic energy equation cancels the magnetic work. The pressure terms combine into $-\nabla \cdot (p\mathbf{u})$, proving the conservation law. A time-dependent imposed potential instead contributes $\rho {\partial }_t\Phi$.

The smooth ideal-MHD system couples mass conservation, momentum, adiabatic pressure transport and the ideal magnetohydrodynamic induction equation, subject to the solenoidal magnetic-field constraint. Without gravity it is a quasilinear partial differential equation system. Across shocks, conservative total energy replaces smooth adiabatic pressure transport.

The inviscid magnetohydrodynamic momentum equation balances inertia, pressure, gravitational acceleration and Lorentz force density. The magnetic force separates into magnetic tension and magnetic pressure. It is coupled to the continuity equation and the ideal magnetohydrodynamic induction equation.

For liquid–gas phase coexistence, make two distinct approximations: the gas obeys the ideal gas law, $v_{m,g}=RT/p$, and its molar volume greatly exceeds the liquid molar volume, $v_{m,g}\gg v_{m,l}$. Here $R$ is the molar gas constant and $p$ is the saturation pressure of the pure substance. Substitution into the Clapeyron equation giveswhere $p_{\ast }>0$ is any fixed reference pressure, making the logarithm dimensionless. Constant latent heat is a further approximation, needed for the elementary Integrated Clausius-Clapeyron relation below.^[ref]

The neglected liquid molar volume has a transparent error estimate. If the gas is ideal but $ϵ=v_{m,l}/v_{m,g}$ is retained, thenThus ignoring the liquid molar volume underestimates this slope by a relative amount $ϵ$. This estimate does not bound errors from treating a real gas as ideal or from freezing $L_m(T)$. Near a thermodynamic critical point, the gas and liquid molar volumes approach each other and these dilute-vapor assumptions fail.

An incompressible flow preserves the volume of each moving fluid element. Its local condition is zero velocity divergence, $\nabla \cdot \mathbf{u}=0$. Combined with mass conservation, it says that each parcel's mass density stays unchanged as it moves. This does not require every parcel to have the same mass density: an initially nonuniform distribution can be transported without compression. Nor does incompressibility require a constant velocity or constant pressure.

For positive mass density, the continuity equation makes the equivalence precise:The material derivative follows a fluid element. Thus $D\rho /Dt=0$ if and only if the velocity field is a divergence-free vector field. The distinction between following a parcel and observing a fixed position prevents confusing incompressibility with an unchanging density field. ^[ref]

For a rigid pipe with impermeable walls, no leaks and steady flow, mass conservation at constant mass density implies a constant volumetric flow rate:Here $\mathbf{n}$ points downstream, $A$ is the section's area and $\overline{U}$ is the mean normal velocity, not necessarily the centreline speed. No assumption of inviscid flow or a uniform velocity profile is needed. If unequal amounts entered and left a fixed segment, its mass would accumulate, contradicting steady flow.

For a worked example, take horizontal water flow with stipulated mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$, inlet radius $R_1=0.010\mathrm{m}$, outlet radius $R_2=0.005\mathrm{m}$ and inlet mean velocity ${\overline{U}}_1=0.20\mathrm{m}{\mathrm{s}}^{-1}$. The circular areas giveThe radius halves, the area falls to one quarter, and the mean velocity quadruples. The mass flow rate is $\rho Q=0.02\pi \mathrm{kg}{\mathrm{s}}^{-1}$. The same amount of water passes both sections each second; narrowing changes its speed rather than compressing it. The Reynolds number and Bernoulli's principle discussions use this same example to examine viscous effects and the additional assumptions needed to calculate pressure.

In a uniform-density incompressible flow inside a smooth rigid sphere rotating at angular speed $\Omega$, a normal mode with phase $e^{i\sigma t+im\varphi }$ has intrinsic frequency $s=\sigma +m\Omega$. For $s\ne 0,\pm 2\Omega$, eliminating velocity from the linearized Euler equations gives the displayed pressure equation, with $W=p^{\prime }/\rho$. Impermeability gives $R{W}_R+(2m\Omega /s)W+(1-4{\Omega }^2/s^2)z{W}_z=0$ on the sphere. There is no condition $W=0$: the rigid wall supplies variable normal pressure. These inertial waves are restored by Coriolis acceleration, and their pressure equation is hyperbolic for $|s|<2|\Omega |$.

For $0<{\omega }^2<{\kappa }^2$, set $S=kH\omega /\sqrt{{\kappa }^2-{\omega }^2}$. Even-pressure modes satisfy ${\omega }^2=-{\Omega }_z^{2}S\tan S$, while odd-pressure modes satisfy ${\omega }^2={\Omega }_z^{2}S\cot S$. In the case ${\Omega }_z>\kappa >0$, the branches approach $\kappa$ as $kH\to 0$. For each fixed vertical order $n\ge 1$ and large $kH$, $S\to n\pi /2$ and $\omega \sim \kappa n\pi /(2kH)$. Even orders have even pressure and odd orders odd pressure. The limiting boundary condition is zero normal derivative of pressure at both interfaces.

// Target: astrophysics.bigb

For reference radii $a_0,b_0$, current inner radius $a$ and uniform axial stretch $z>0$, incompressibility integrates to $r^2=a^2+(R^2-a_0^2)/z$. The principal stretches are $r_{,R}=1/(z\lambda )$, $\lambda =r/R$ and $z$. Write $\hat{W}(\lambda ,z)=W((z\lambda {)}^{-1},\lambda ,z)$ and $\mathcal{E}=2\pi {\int }_{a_0}^{b_0}\hat{W}RdR$ per reference length. Virtual work gives $p={\mathcal{E}}_{,a}/(2\pi az)$ and the total axial wall load $N_{\mathrm{wall}}={\mathcal{E}}_{,z}$, both derivatives holding the other displayed parameter fixed. For pressurized closed ends the independently applied axial load is $N_{\mathrm{ext}}=N_{\mathrm{wall}}-\pi a^{2}p$. The distinction follows from the axial pressure power $\pi a^{2}p\dot{z}$.

An extensive quantity measures how much of a thermodynamic system there is; an intensive quantity describes its state independently of that amount. The clean comparison is replication: take independent identical systems at the same thermodynamic equilibrium, with the same substance, composition and thermodynamic phase. Combining their accounting boundaries multiplies internal energy $U$, entropy $S$, volume $V$, mass $m$ and amount of substance $n$, while leaving temperature $T$, pressure $p$ and chemical potential $\mu$ unchanged.^[ref]

For $\lambda$ copies the ledger isAn integer $\lambda$ counts actual copies; a positive real $\lambda$ describes the corresponding bulk scaling law. Copies can remain separated: replication need not involve removing a partition, changing a container or performing any work. Distinguish $n$, the amount of substance measured in moles, from $N=n{N}_A$, the dimensionless count of specified entities, where $N_A$ is the Avogadro constant in ${\mathrm{mol}}^{-1}$. For example, $2\mathrm{mol}$ contains $1.204428152\times 10^{24}$ specified entities; tripling it gives $6\mathrm{mol}$ and $3.613284456\times 10^{24}$ entities. This uses the exact Avogadro constant $N_A=6.02214076\times 10^{23}{\mathrm{mol}}^{-1}$, which does not triple. In the differential $dU=TdS-pdV+\mu dn$, $\mu$ is a molar chemical potential in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$; the coefficient of $dN$ instead has units of energy per entity.

As a direct check, the ideal gas law $pV=nRT$ givesFor a monatomic ideal gas with only translational degrees of freedom active, internal energy of an ideal gas gives $U=3nRT/2$, so $U^{\prime }=\lambda U$ at unchanged temperature. Here $R$ is the molar gas constant. The scaling distinction lets a small sample supply an equation of state for a much larger amount: material state and inventory become separate questions.

Replication does not mean multiplying every linear dimension by $\lambda$: a geometrically similar container then has ${\lambda }^3$ times the volume. Nor does it mean combining samples at different states. Equal temperature establishes thermal equilibrium but does not imply equal pressure, composition or mass density. Density of a composite system makes this distinction concrete.

An intensive quantity is unchanged under replication at the same thermodynamic equilibrium. Temperature, pressure, chemical potential and the mass density of a uniform substance are examples. Independence from sample size does not mean independence from state: a material's mass density can change with temperature or pressure.

The relation to extensive quantities is particularly elegant for a differentiable bulk fundamental relation. Put $\mathbf{x}=(S,V,n)$ and writeTaking a partial derivative with respect to $x_i$ and cancelling $\lambda$ givesThe equilibrium differential $dU=TdS-pdV+\mu dn$ identifies these derivatives with temperature, minus pressure and molar chemical potential. They are degree-zero homogeneous functions. The same degree-one assumption yields the Euler relation in thermodynamics, but a surface energy contribution requires extra care.

An intensive quantity generally has no meaningful sum across samples: two equal temperatures do not add to a doubled temperature. Neither is the final temperature after thermal contact generally an unweighted average; heat capacities and the first law of thermodynamics determine the balance. Ratios of additive inventories lead to the useful weighted rules in normalizing extensive quantities.

Constant temperature does not generally mean constant internal energy. At fixed amount and composition, within one smooth thermodynamic phase of a simple compressible system, an equilibrium identity isThe bracket vanishes for the ideal gas law $pV=nRT$. It need not vanish for a real gas: changing the volume changes the average interaction potential energy, even when the random-motion contribution stays at the same temperature. The identity follows by writing $dU=TdS-pdV$ and using a Maxwell relation; it does not require the actual process to be reversible. ^[ref]

For a Van der Waals equation model with volume per mole $v=V/n$, $p=RT/(v-b)-a/v^2$ gives $(\partial U/\partial V{)}_{T,n}=a/v^2$. Thus an isothermal expansion from $v_1$ to $v_2>v_1$, at fixed amount of substance $n$, increases internal energy bySeparating attractive particles makes their interaction potential energy less negative. For stipulated $n=1\mathrm{mol}$, $a=0.10\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$, $b=10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, $T=300\mathrm{K}$, $v_1=0.010$ and $v_2=0.020{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, the calculated increase for this one-mole sample is $5\mathrm{J}$, despite $\Delta T=0$. This dilute stable-branch model example is not a measurement of a named gas. ^[ref]

A phase transition gives another reason. During equilibrium vaporization of a pure substance at fixed pressure, the temperature can stay at its phase coexistence value while the phase fraction changes. Molecular separation changes interaction potential energy.^[ref] If the closed system does only pressure-volume work, its latent heat input obeysHere $H=U+pV$ is enthalpy. A vaporization latent heat therefore is not automatically the internal energy increment: some input supplies expansion work. A fixed temperature reading alone cannot tell how much internal energy changed.