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Kinetic theory describes many-particle systems through a distribution on phase space, retaining both position and velocity. In the number-density convention, $f(t,\mathbf{x},\mathbf{v})d^{3}x{d}^{3}v$ is the expected number of particles in a small position–velocity region. For identical particles of mass $m$, its integrals give the number density $n$ and bulk velocity $\mathbf{u}$:This $f$ is not normalized to one: its integral over the whole phase space gives the total expected particle number. Transport moves particles along trajectories, while collision or diffusion terms change the distribution. The Boltzmann equation, Hamiltonian Liouville equation and Fokker-Planck equation describe different physical assumptions, rather than interchangeable collision laws.

An ideal gas is a motivating example: the pressure on a wall comes from microscopic momentum transfers, even when the bulk velocity is zero. The Maxwell-Boltzmann velocity distribution separates this random molecular motion from bulk motion; see ideal gas pressure from wall collisions. The attractive feature of kinetic theory is that a distribution can describe two opposing populations where one average velocity hides almost everything. It also tells us when averaging into a continuum approximation loses essential information: compare the intermolecular mean free path with the apparatus scale using the Knudsen number. ^[ref]

The Knudsen number is the dimensionless ratio of an intermolecular mean free path ${\ell }_{\mathrm{mfp}}$ to a specified characteristic length $L$ of the apparatus or flow. The mean free path measures the distance travelled between gas–gas collisions; $L$ measures the distance over which boundaries or macroscopic conditions matter. Wall encounters must not be counted as gas–gas collisions in this comparison: doing so would hide the very wall-dominated behavior we want to detect. The same gas can therefore have a small Knudsen number in a large vessel and a large Knudsen number in a tiny opening.

For a dilute hard-sphere gas of identical molecular diameter $d$ in thermal equilibrium, use the conventionHere $n$ is the number density, $p$ the pressure, $T$ the absolute temperature, and $k_B$ the Boltzmann constant; the second equality uses the ideal gas law. The collision cross-section is $\pi d^2$, and the $\sqrt{2}$ accounts for both partners moving with independent equilibrium velocities. Mixtures and velocity-dependent collision cross-sections require a stated molecular model and mean free path convention; the displayed expression is not an exact formula for every real gas. ^{[ref] [ref]}

As a numerical toy example, choose $T=300\mathrm{K}$ and $d=0.37\mathrm{nm}$, without claiming that this is a measured molecular diameter for a particular gas. The formula gives ${\ell }_{\mathrm{mfp}}\simeq 68.1\mathrm{nm}$ at $p=10^5\mathrm{Pa}$. It yields $\mathrm{Kn}\simeq 6.81\times 10^{-6}$ for a $1\mathrm{cm}$ vessel but $\mathrm{Kn}\simeq 0.681$ for a $100\mathrm{nm}$ gap. At $p=1\mathrm{Pa}$ the same toy gas has ${\ell }_{\mathrm{mfp}}\simeq 6.81\mathrm{mm}$, comparable to the centimetre vessel. Reducing either pressure or apparatus size can thus expose molecular transport.

Choose $L$ for the question being asked: a tube diameter compares the mean free path with wall separation, while tube length describes a different geometric scale. For a positive macroscopic field $q$, a local gradient scale can instead be $L_q=|q|/|\nabla q|$, where its gradient is nonzero. A large vessel can contain a thin region with a large local Knudsen number. Small geometric Knudsen number alone does not guarantee a slowly varying near-equilibrium flow. The time, velocity and space scalings must also be stated when the Knudsen number appears as a coefficient in a Boltzmann equation. ^[ref]

For a negligible-mass ideal-gas envelope above a core of mass $M_c$, with constant luminosity $L$ and Kramers' opacity law, division of hydrostatic equilibrium by radiative diffusion gives $PdP/dT=16\pi acG{M}_c\mathcal{R}{T}^{15/2}/(3{\kappa }_{0}L\mu )$. Taking the outer pressure constant to be negligible yields $P=C{T}^{17/4}$, where $C^2=64\pi acG{M}_c\mathcal{R}/(51{\kappa }_{0}L\mu )$, and $\rho =(\mu C/\mathcal{R})T^{13/4}$. Hydrostatic equilibrium then gives $dT/dr=-4\mu G{M}_c/(17\mathcal{R}{r}^2)$, so $T=B/r+T_0$ with $B=4\mu G{M}_c/(17\mathcal{R})$. The radiative-zero choice $T_0=0$ is exact for a vanishing-temperature boundary at infinity and a deep-envelope approximation for a very extended finite star. It is not implied by small surface temperature alone at a specified finite radius.

The Lagrangian pressure perturbation is the change in pressure experienced by a displaced fluid element: $\Delta p=\delta p+\mathbf{\xi }\cdot \nabla p_0$. For a free surface against fixed external pressure, $\Delta p=0$, rather than $\delta p=0$ at the original surface.

Laminar flow is a regime of fluid flow in which the motion remains organized, with neighboring fluid parcels following smooth, coherent paths rather than the irregular interacting fluctuations characteristic of turbulence. In a straight pipe, the simplest example consists of coaxial layers sliding past one another. This describes the character of the motion, not a requirement that every parcel have the same velocity. Nor does it require steady flow: a smooth profile can evolve in time as viscosity transmits motion from a newly moving wall into a resting fluid. The unsteady wall-driven solution provides an explicit example. ^[ref]

The Reynolds pipe-and-dye experiment makes this distinction visible. A thin injected dye band can remain coherent in laminar flow, whereas irregular motion can fold it and carry dye across the pipe. However, molecular diffusion can spread dye even in laminar flow, and an inlet disturbance can move a coherent band without producing turbulence. A dye band is a streakline, not a measurement of the entire velocity field; in steady flow an ideal passive tracer's streakline coincides with a streamline. Reynolds explicitly distinguished a wandering band from one broken into eddies. ^[ref]

For a circular pipe of internal diameter $D$, define the mean axial velocity $U$ using the volumetric flow rate $Q$ and cross-sectional area $A$:Here $\rho$ is mass density, $\mu$ is dynamic viscosity and $\nu =\mu /\rho$ is kinematic viscosity. This particular Reynolds number uses the diameter and mean velocity, not the radius or centreline velocity. Small Reynolds number generally favors laminar flow, but a single universal critical Reynolds number does not distinguish it from turbulence in all geometries and experiments. Even in a given pipe, inlet conditions and the magnitude of disturbances affect the observed onset. Reynolds demonstrated this dependence by deliberately inserting a wire coil. ^[ref]

The quantitative example above assumes steady flow, incompressible flow, a Newtonian fluid with constant dynamic viscosity, a straight horizontal circular pipe of radius $R=D/2$, stationary walls satisfying the no-slip boundary condition, and fully developed flow. The last assumption means that the axial velocity profile no longer changes downstream. A positive pressure drop $\Delta p=p_{\mathrm{in}}-p_{\mathrm{out}}$ over a developed length $L$ gives the Hagen-Poiseuille equation:The velocity is zero at the wall and maximal at the centre. Averaging over the circular area, whose annuli have area $2\pi rdr$, gives $U=u(0)/2$; averaging along one diameter would give a different answer. These assumptions produce a parabola, rather than every laminar flow having a parabolic profile. ^[ref]

As a stipulated calculation, take $D=0.010\mathrm{m}$, $U=0.10\mathrm{m}{\mathrm{s}}^{-1}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, $\mu =0.0010\mathrm{Pa}\mathrm{s}$ and $L=1.0\mathrm{m}$, and assume the developed laminar flow above. Then $\nu =10^{-6}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, ${\mathrm{Re}}_D=1000$, $u(0)=0.20\mathrm{m}{\mathrm{s}}^{-1}$, $Q=7.85\times 10^{-6}{\mathrm{m}}^3{\mathrm{s}}^{-1}$ (about $0.471\mathrm{L}\mathrm{mi}{\mathrm{n}}^{-1}$), and $\Delta p=32\mu LU/D^2=32\mathrm{Pa}$. These are predictions from chosen parameters, not measurements or a claim that ${\mathrm{Re}}_D=1000$ guarantees laminar flow in every apparatus.

The beauty of this example is that a spatially varying velocity field still gives an exact relation between two accessible measurements, pressure drop and volumetric flow rate. It predicts the hydraulic resistance $\Delta p/Q=8\mu L/(\pi R^4)$. Doubling the radius therefore multiplies $Q$ by sixteen at fixed pressure drop, length and dynamic viscosity, provided the model remains applicable. This makes the laminar flow model useful for sizing narrow tubes or determining dynamic viscosity. It also explains why a persistent narrow dye band can be informative without implying that the fluid is moving as a uniform plug.

Latent heat is the heat absorbed or released in a first-order phase transition through its entropy jump. For an equilibrium conversion from thermodynamic phase $\alpha$ to thermodynamic phase $\beta$ at fixed pressure, evaluate both thermodynamic phases at the same temperature and pressure on their phase coexistence curve:Use one normalization throughout: $L$, $S$ and $H$ may all be per mole, giving $L$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or all per unit mass, giving $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Here $S$ is entropy and $H$ is enthalpy on that basis. Equal Gibbs free energies of the coexisting thermodynamic phases give $\Delta H=T_{\mathrm{tr}}\Delta S$. Identifying this enthalpy difference with supplied heat assumes a closed system and an equilibrium conversion with only pressure-volume work. ^[ref]

For a pure substance with positive latent heat per unit mass $L$, let $x$ be the phase fraction of the higher-enthalpy thermodynamic phase, measured by mass. While both thermodynamic phases coexist at the fixed pressure,Adding heat changes $x$ rather than the temperature; this creates the horizontal portion of a heating curve. Once $x=1$, additional heat raises the temperature according to specific heat capacity at constant pressure. Reversing the conversion releases heat. The plateau represents changing enthalpy, rather than a finite single-thermodynamic phase heat capacity satisfying $Q=C_p\Delta T$. ^[ref]

For a constructed fixed-pressure model, take $m=1\mathrm{kg}$, initial temperature $280\mathrm{K}$, transition temperature $T_{\mathrm{tr}}=300\mathrm{K}$, solid specific heat capacity at constant pressure $c_s=2\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, latent heat $L=100\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, and liquid specific heat capacity at constant pressure $c_l=3\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. These are stipulated model parameters, not empirical properties of a named material. Heating the solid to $300\mathrm{K}$ takes $40\mathrm{kJ}$ of sensible heat; converting it fully to liquid takes another $100\mathrm{kJ}$ of heat; heating the liquid to $320\mathrm{K}$ takes $60\mathrm{kJ}$. The total is $200\mathrm{kJ}$. After a total heat input of $90\mathrm{kJ}$, $x=(90-40)/100=1/2$ and the temperature is still $300\mathrm{K}$. The corresponding per-unit-mass entropy jump is $L/T_{\mathrm{tr}}=\frac{1}{3}\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$.

A continuous phase transition has no latent heat. Its heat capacity can have a jump or a singularity without a finite interval in which added heat converts coexisting thermodynamic phases at a constant temperature. A heating plateau therefore requires the stated phase coexistence conditions; it is not a feature of every phase transition. ^[ref]

Within the Ideal-vapor Clausius-Clapeyron approximation, a local slope of the natural logarithm of saturation pressure against inverse temperature estimates the molar latent heat:In the stipulated vapor-pressure calculation, the slope is $-4811.16\mathrm{K}$, which recovers $40.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. For variable $L_m(T)$, a two-point estimate gives an average weighted by $T^{-2}$, rather than the latent heat at an arbitrary midpoint. Differentiating noisy pressure measurements amplifies their errors. Outside the ideal-vapor regime, use independently determined molar volumes in $L_m=T\Delta v_{m}dp/dT$; a straight-looking graph alone cannot establish that the gas is ideal.

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]