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Hydraulic resistance relates the pressure drop across a resistive fluid element to its volumetric flow rate. In a linear model, $\Delta p=RQ$ with $R>0$, so dissipated mechanical power is $\Delta pQ=R{Q}^2$. A peripheral circulatory bed can be represented by such a resistance when its detailed dynamics are not retained. This terminal pressure-to-flow resistance differs from the area-dependent friction rate multiplying velocity in a distributed tube momentum equation.

An incompressible point source of volume flux $Q$ at height $a$ above an impermeable plane has velocity potential $\varphi =-Q(1/r_{+}+1/r_{-})/(4\pi )$, with an equal image source below the plane. On the wall, $u_s=Qs/[2\pi (s^2+a^2{)}^{3/2}]$. The Bernoulli equation gives $p-p_{\infty }=-\rho Q^2{s}^2/[8{\pi }^2(s^2+a^2{)}^3]$. Integrating the pressure deficit over the plane gives the displayed attraction force relative to ambient pressure. If source strength instead denotes the coefficient $Q/(4\pi )$ of radial inverse-square velocity, the force is $\pi \rho m^2/a^2$.

The hydrophobic effect is the solvent-mediated tendency of nonpolar regions to associate in water. Exposing a hydrocarbon region changes how surrounding water molecules can arrange and interact; reducing that exposure can lower the total free energy. Both water and lipids enter the accounting. The effect is not an additional covalent bond between chains. See IUPAC Gold Book: hydrophobic interaction.

At fixed temperature and pressure, compare two candidate organizations by the change in Gibbs free energy, $\Delta G=\Delta H-T\Delta S$, where $H$ is enthalpy, $T$ is absolute temperature, and $S$ is entropy. An organization with lower $G$ is favored in this comparison. Reducing exposed hydrocarbon area is one contribution; changes in chain packing and the entropy of the lipids also matter. This explains why hydrophobicity alone cannot predict every aggregate shape. See 16.4 Free Energy.

For metric signature $(+---)$, a stationary perfect fluid follows a timelike Killing vector field $K$ with four-velocity $U=K/N$. Its Killing-flow four-acceleration is $U^b{\nabla }_b{U}_a=-{\nabla }_a\log N$. The relativistic Euler equation and stationarity of the pressure give the displayed hydrostatic equilibrium equation. This uses stationarity along $K$ and does not require hypersurface orthogonality.

The hydrostatic pressure support equation in enclosed-mass coordinate is $dP/dm=-Gm/(4\pi r^4)$. Since $r(m)\le R$, integration from surface pressure zero to the centre givesIf mass density decreases outward, the mean density within $r$ is at least the global mean, so $r(m)\le R(m/M{)}^{1/3}$. This improves the bound towith equality for uniform mass density. Centrally concentrated planets need larger central pressure to support their inner mass.

Hydrostatic pressure is the pressure distribution in a resting fluid, balancing its body force from gravity. Take $z$ upward and a uniform downward gravitational field $\mathbf{g}=-g{\mathbf{e}}_z$, where $g>0$. The vertical pressure gradient satisfiesA short derivation balances the vertical forces on a thin horizontal layer of area $A$ and thickness $\Delta z$: $p(z)A-p(z+\Delta z)A-{\rho }_{f}gA\Delta z=0$. Dividing by $A\Delta z$ and taking the derivative gives the equation above. For constant mass density ${\rho }_f$, integration from the surface at $z_s$ givesThus pressure increases with depth $h$. For the stipulated water with mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$ and gravitational acceleration $9.81\mathrm{m}{\mathrm{s}}^{-2}$, depths $0.20\mathrm{m}$ and $0.30\mathrm{m}$ give gauge pressures $1962\mathrm{Pa}$ and $2943\mathrm{Pa}$. Their difference produces the buoyancy of the example cube. ^[ref]

Ice-melting calorimetry measures the fusion enthalpy through conservation of energy, rather than inferring it from the small temperature change of the melting ice itself. In a mixing experiment, warm water and the container cool while added ice melts and the resulting water warms. Their combined energy balance separates fusion from sensible heat.

Weigh the container and stirrer, then weigh the added warm water to obtain $m_w$. Let the container and water reach a common initial temperature $T_i$. Prepare ice in a clean ice-water mixture near $T_m$, rather than assuming freezer ice is already at $T_m$; drain and blot its surface immediately before transfer. Add it without splashing, stir gently, and record temperature before addition and until after the last ice disappears. Reweigh to find the added mass $m_i$. Wisconsin's laboratory guide specifies drying each piece, weighing before and after, and recording the temperature history; it places the initial and final temperatures on opposite sides of the jacket temperature to reduce net environmental exchange. ^[ref]

For fully melted, initially dry ice at $T_m$, a common final temperature $T_f>T_m$, constant water specific heat capacity at constant pressure $c_w$, and apparatus heat capacity $C_{\mathrm{cal}}$ initially at $T_i$,Here $Q_{\mathrm{env}}>0$ denotes net heat entering the apparatus from its surroundings. The left side supplies the fusion enthalpy and the sensible heat on the right; the water produced by melting must be warmed too. The balance extends Case Western's mixing demonstration by including the apparatus and environmental terms. If the specific heat capacities vary appreciably, replace their products with the corresponding temperature integrals. ^[ref]

Determine $C_{\mathrm{cal}}$ independently, for example by a warm/cold water mixing balance with no phase transition, including the wetted stirrer and thermometer in the apparatus. Check the thermometer calibration, keep the lid in place between transfers, and monitor the surroundings. Before/after drift and a comparable run without ice help bound $Q_{\mathrm{env}}$; such corrections assume comparable heat transfer conditions and are not automatically exact. Choosing a jacket temperature between $T_i$ and $T_f$ reduces exchange but does not prove $Q_{\mathrm{env}}=0$. Ensure all the ice has melted and the contents are well mixed before using the measured $T_f$: with residual ice, the weighed addition is not the mass that has undergone fusion. Repeat the experiment and report measurement errors in mass, temperature, $C_{\mathrm{cal}}$ and the leak correction.

An independent way to see the plateau is David Roundy's Oregon State experiment: heat an ice-water mixture slowly with a waterproof resistor in an insulated cup, stir, and record temperature, heater electric current $I(t)$, electric potential difference $V(t)$ and elapsed time. Plot temperature against $E_{\mathrm{el}}=\int I(t)V(t)dt$, recording changes in $I$ or $V$. The initial solid mass must exclude accompanying water. For a boundary enclosing the heater as well as the sample, this input is electrical work, so $\Delta H=E_{\mathrm{el}}+Q_{\mathrm{env}}$ under the constant-pressure assumptions above. Across an ideal isothermal plateau the apparatus has no sensible heat change, giving $m_{s}L=E_{\mathrm{el}}+Q_{\mathrm{env}}$; departures from uniform temperature require accounting for the heater and container response. ^[ref]

An ideal gas is a model of a dilute gas whose particle centres move according to classical mechanics and occupy negligible volume compared with their container. Treating the particles as point masses removes the molecular-size correction; neglecting interactions between separated particles removes the intermolecular contribution to potential energy. The usual kinetic theory model retains brief elastic collisions with the walls and, optionally, with other particles: these transfer momentum while conserving the total kinetic energy of the colliding bodies. There is no appreciable interaction energy stored between collisions. These assumptions lead to the ideal gas law, $PV=N{k}_{B}T$. ^{[ref] [ref]}

The model's attraction is its universality: at the same temperature and number density, different classical ideal gases have the same pressure, even when their particle masses differ. In ideal gas pressure from wall collisions, heavier particles move more slowly at a given temperature, and the two effects cancel. For example, one mole of any ideal gas at $300\mathrm{K}$ in $0.025{\mathrm{m}}^3$ has pressure approximately $99.774\mathrm{kPa}$, as computed under the ideal gas law.

The ideal gas law describes the translational contribution to pressure; it does not specify all of the internal energy or the heat capacity. A monatomic gas has three classical translational degrees of freedom, whereas a gas of molecules can also store rotational kinetic energy and energy in vibrational normal modes. An equation of state alone therefore does not determine how much heat a chosen temperature increase requires. It also does not select a process: isothermal expansion and adiabatic expansion need different additional constraints. ^[ref]

A real gas departs from this approximation when particle size or interactions become important, especially at high number density or low temperature near a phase transition. The Van der Waals equation adds such corrections. Independently, at sufficiently low temperature or high number density, quantum statistics can become essential: even particles with negligible interactions need not obey the classical ideal gas law. The classical statistical assumption is the nondegenerate gas regime, which is distinct from neglecting interactions. ^{[ref] [ref]}

The ideal gas law is the equation of state relating the absolute pressure $P$, volume $V$, particle count $N$, and absolute thermodynamic temperature $T$ of a classical ideal gas:Here $k_B$ is the Boltzmann constant, $N_{\mathrm{A}}$ is the Avogadro constant, $R$ is the molar gas constant, and $n_{\mathrm{mol}}$ is the amount of substance in moles. The distinct symbol $n_{\mathrm{mol}}$ avoids confusing amount of substance with the number density $n=N/V$, measured in ${\mathrm{m}}^{-3}$: the same law can be written $P=n{k}_{B}T$. Use absolute pressure, relative to a vacuum, rather than an unconverted gauge pressure. ^[ref]

The exact SI definitions of the Avogadro constant and Boltzmann constant giveThis numerical value is exact; the applicability of the ideal gas model to a real gas is an approximation. ^{[ref] [ref]}

For $n_{\mathrm{mol}}=1\mathrm{mol}$, $T=300\mathrm{K}$, and $V=0.025{\mathrm{m}}^3$, direct substitution givesAt fixed temperature and particle count, doubling the volume to $0.050{\mathrm{m}}^3$ halves the pressure to approximately $49.887\mathrm{kPa}$: this is Boyle's law. At fixed volume and particle count, doubling the temperature from $300\mathrm{K}$ to $600\mathrm{K}$ doubles the pressure to approximately $199.547\mathrm{kPa}$. These ratios use kelvins, not readings in the degree Celsius scale: $300\mathrm{K}$ is $26.85{}^{\circ }\mathrm{C}$, and $600\mathrm{K}$ is $326.85{}^{\circ }\mathrm{C}$.

The curved slices show Boyle's law at three temperatures; the straight slice shows proportionality to absolute temperature at one fixed volume. They are relations among equilibrium states, rather than a prescribed process history. Compression could hold the temperature fixed through heat transfer, or raise it during adiabatic compression. To calculate a process's heat or work, supplement the ideal gas law with the process constraint, the first law of thermodynamics, and the relevant internal energy or heat capacity information. The graph's extrapolation toward $T=0$ is mathematical; it does not assert that a real gas remains classical or gaseous there.

For a stellar mixture with fixed mean molecular weight, the ideal gas law can be written $P=\mathcal{R}\rho T/\mu$, where $\rho$ is mass density and $\mathcal{R}$ is the stellar gas constant. It assumes negligible interaction corrections and a nondegenerate gas; excitation or ionization can change the particle count and thermodynamic response.

In his 1834 memoir on the motive power of heat, Émile Clapeyron combined gas laws into a relation between pressure, volume, and temperature while analysing heat engines (printed page 164). The equation above uses modern SI notation and absolute thermodynamic temperature; it is not a transcription of his historical temperature convention. ^[ref]

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]