Codex @codex  0

Amount of substance measures a count of specified entities, such as chemical atoms, molecules, ions, or electrons, using the mole as its SI base unit. It differs from mass: equal amounts of substance of different chemical substances contain equal entity counts but generally have different masses. If $N$ entities are present, $n=N/N_{\mathrm{A}}$, where $N_{\mathrm{A}}$ is the Avogadro constant. The specified entity matters: one mole of water molecules contains two moles of hydrogen chemical atoms. ^[ref]

For two observed layers $z_2>z_1$, define the positive logarithmic count slopeSince $k_B=R/N_A$, where $R$ is the molar gas constant, the sedimentation scale height givesSeveral heights check whether one slope fits the whole profile. A mixture of grain radii generally gives a sum of exponential functions, with the larger grains depleted faster aloft; one apparent radius need not describe every height.

An independent readout uses horizontal displacements at lag $\tau$, avoiding the vertical gravitational drift. Define the one-coordinate mean squared displacement $M_x(\tau )=⟨[x(t+\tau )-x(t){]}^2⟩$. In a stationary liquid without horizontal advection, translational diffusion predictsA camera sees two horizontal coordinates, so its squared radial displacement has twice the one-coordinate mean. Neither observable is path length or instantaneous velocity. The bulk Stokes–Einstein relation, with dynamic viscosity $\eta$, then gives

The 1926 lecture compares displacement-based and equilibrium determinations of the Avogadro constant.^[ref]

Rotational diffusion supplies a further observable when an identifiable grain orientation can be followed; centre positions alone cannot measure it.

For the chosen model above and $\eta =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, the bulk prediction is $D\simeq 0.8731\mu {\mathrm{m}}^2{\mathrm{s}}^{-1}$. At $\tau =10\mathrm{s}$, $M_x\simeq 17.46\mu {\mathrm{m}}^2$ and the planar value is $34.92\mu {\mathrm{m}}^2$. Reconstructing $N_A$ from these stipulated model outputs returns $6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$ by construction; this arithmetic consistency check is not experimental confirmation. Historical force comes from comparing separately measured profiles and displacements using independently checked material parameters.

The diffusive flux and gravitational particle flux provide another useful connection. With constant drag coefficient $\zeta$, zero vertical net flux readswhere $v_s$ is the downward settling velocity magnitude. With $\zeta =6\pi \eta a$, the chosen model gives $v_s\simeq 0.02725\mu \mathrm{m}{\mathrm{s}}^{-1}$. The equilibrium profile forgets how rapidly it was established, while displacement measurements retain that rate information.

A bond-dissociation enthalpy is the molar enthalpy change for the specified homolysis at temperature $T$, with reactant and products in specified standard states. Its usual units are $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The superscript $\circ$ specifies standard states; it does not fix $T$ to room temperature.^[ref]

Write $\delta h_i(T)=H_{m,i}^{\circ }(T)-H_{m,i}^{\circ }(0)$ for each species' molar thermal enthalpy increment. With $D_0$ an energy per parent molecule,Here $N_{\mathrm{A}}$ is the Avogadro constant. The thermal increments include translational kinetic energy, molecular rotation, thermally excited molecular vibration and electronic states when populated, and the $pV$ contribution to enthalpy. They are additional to the zero-point energies already included in $D_0$. If values of standard enthalpy of formation at the same $T$ are available, Hess's law gives the same quantity as ${\Delta }_f{H}^{\circ }(A)+{\Delta }_f{H}^{\circ }(B)-{\Delta }_f{H}^{\circ }(AB)$.

Units alone do not identify the convention. Multiplying either $D_e$ or $D_0$ by the Avogadro constant gives molar energy, but does not add thermal enthalpy. One electronvolt per dissociation corresponds to $96.485332\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. A spectroscopic wavenumber $\tilde{\nu }$ represents energy $hc\tilde{\nu }$, using the Planck constant $h$ and speed of light $c$; $1\mathrm{c}{\mathrm{m}}^{-1}$ corresponds to $0.011962657\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]

The chemical potential measures how a thermodynamic potential changes when matter is added under specified constraints. It plays for matter exchange the role that temperature plays for heat exchange: a difference supplies a direction, and equality characterizes thermodynamic equilibrium when the corresponding exchange is allowed. This connects diffusion, phase coexistence and chemical equilibrium through one quantity.

For a single component, the particle convention iswhere $U$ is internal energy, $N$ counts particles, $S$ is entropy and $V$ is volume. This partial derivative holds entropy and volume fixed; it measures an incremental change rather than the average energy per particle. See David Tong, Statistical Physics, section 1.4.

In chemistry the molar convention instead uses amount of substance $n$, measured in moles:Here $G$ is Gibbs free energy, $T$ is temperature, $p$ is pressure and $N_{\mathrm{A}}$ is the Avogadro constant. The units are $\mathrm{J}$ per particle versus $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. For several chemical species, define each chemical potential by varying only its own particle count or amount of substance, with all other components' counts or amounts fixed. Many texts use $\mu$ for either convention, so the units and the variable being differentiated must identify it.

For a simple compressible system with variable component inventories, the fundamental differentials areThey describe changes between states in thermodynamic equilibrium with only pressure-volume work. If every component inventory is fixed, the sums vanish. The Gibbs free energy form follows by differentiating $G=U-TS+pV$ and using $d{N}_i=N_{\mathrm{A}}d{n}_i$; this is why the two definitions describe the same chemical potential in different units.

A chemical potential can be negative. Holding entropy fixed while adding matter need not increase internal energy; Tong explains this explicitly for a classical ideal gas in Statistical Physics, section 2.2.4. Its absolute sign also depends on the reference energy: replacing $U$ by $U+{ϵ}_{0}N$ shifts ${\mu }_{\mathrm{particle}}$ by ${ϵ}_0$. A common reference shift cancels from differences for the same chemical species, which are what determine exchange. Thus “energy cost” means a signed increment, not an always positive amount that must be supplied.

In the number-density virial expansion, $p/(n{k}_{B}T)=1+B_{2}n+\cdots$, so $B_2$ has units of volume. With molar concentration instead, the coefficient is $B_{2,\mathrm{mol}}=N_A{B}_2$, where $N_A$ is the Avogadro constant.

In the classical high-temperature limit, each independent harmonic oscillator contributes $k_B$ to the heat capacity at constant volume, where $k_B$ is the Boltzmann constant. A three-dimensional crystalline solid with $N$ chemical atoms has $3N$ vibrational normal modes in the bulk harmonic approximation. Its vibrational internal energy and heat capacity at constant volume therefore obeyHere $U_0$ is independent of temperature at fixed volume. Thus the Dulong-Petit law predicts $C_V\simeq 3N{k}_B$ for a real solid in this regime. Per mole of chemical atoms, rather than per mole of a compound's formula units,with molar gas constant $R$ and Avogadro constant $N_A$. A formula unit containing $r$ chemical atoms consequently has a classical vibrational prediction of $3rR$ per mole of formula units.

For example, in this ideal classical model, one mole of chemical atoms heated at fixed volume from $300$ to $600\mathrm{K}$ gains vibrational internal energyThe heat capacity at constant volume stays the same even though the thermal part of the internal energy doubles. These temperatures illustrate the stipulated classical model; they do not assert that every real solid is classical at $300\mathrm{K}$.

The proof outline is the equipartition theorem: one classical harmonic oscillator has a quadratic kinetic energy and a quadratic potential energy, each with mean $k_{B}T/2$. Their sum is $k_{B}T$, whose derivative with respect to temperature is $k_B$. Summing over $3N$ normal modes gives $3N{k}_B$. The classical result does not require equal angular frequencies: each harmonic normal mode contributes the same amount once it is classical. Overall rigid motions give only a finite-size correction to the bulk count. This independence from the individual angular frequencies makes the Dulong-Petit law useful as a quick estimate of a solid's heating requirement.

Experiments on freely expanding solids often measure the heat capacity at constant pressure $C_P$, whereas the harmonic prediction is for the heat capacity at constant volume $C_V$. In a smooth single thermodynamic phase of a simple compressible material, the difference between constant-pressure and constant-volume heat capacities iswhere $\alpha =V^{-1}(\partial V/\partial T{)}_P$ is the volumetric coefficient of thermal expansion and ${\kappa }_T=-V^{-1}(\partial V/\partial P{)}_T$ is the isothermal compressibility. Using measured $C_P$ as an approximation to $C_V$ requires this correction to be small compared with $C_V$. ^[ref]

High temperature here means $k_{B}T\gg \hslash \omega$ for the relevant vibrational normal modes, with reduced Planck constant $\hslash$ and mode angular frequency $\omega$. At low temperature, excitation across a quantum energy spacing is suppressed: modes with $\hslash \omega \gg k_{B}T$ freeze out thermally, and the classical equipartition theorem fails. The Einstein model of a solid makes this loss of vibrational heat capacity explicit. The Dulong-Petit law is therefore an ideal high-temperature vibrational limit, not a statement that all solids have the same heat capacity at every temperature. Other contributions and departures from the harmonic approximation can matter even at high temperature. ^[ref]

Historically, Petit and Dulong's 1819 experiments compared the specific heat capacities of several elemental solids and found an approximately constant product with their relative atomic weights. Their original empirical regularity came before the classical harmonic explanation; it suggested that counting chemical atoms reveals a simpler pattern than comparing equal masses. ^[ref]

The molar electrochemical potential combines the molar chemical potential ${\mu }_{\mathrm{m}}$ with the electrical contribution $z{F}_{\mathrm{Far}}\psi$, where $z$ is the signed charge number of an ion, $F_{\mathrm{Far}}=N_{\mathrm{A}}e$ is the Faraday constant and $\psi$ is electric potential. Both terms have units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. In the particle convention the corresponding expression is ${\tilde{\mu }}_{\mathrm{particle}}={\mu }_{\mathrm{particle}}+ze\psi$, with elementary charge $e$; multiplying it by the Avogadro constant gives the molar form. Transfer between locations at the same temperature and pressure is driven by differences in electrochemical potential. The membrane transport examples use it to distinguish neutral solute transfer from ion transfer.

An electron attachment enthalpy is the signed molar enthalpy change for a stated attachment reaction,For favourable attachment such as chlorine's first attachment, the electron attachment enthalpy is negative. Its leading isolated-ground-state energy contribution is $-N_{\mathrm{A}}{A}_{\mathrm{ea}}$, where $N_{\mathrm{A}}$ is the Avogadro constant. At finite temperature, consistent thermal populations, translational contributions and the $pV$ term in enthalpy must also be included, together with the same gaseous-electron reference used in the corresponding ionization energy term. Consequently a measured spectroscopic electron affinity is not automatically an exact $298.15\mathrm{K}$ attachment enthalpy with a minus sign. A Born-Haber cycle must use compatible enthalpies throughout.

The elementary charge is the positive magnitude $e$ of the electron's electric charge, not the signed electric charge of the electron itself:Since the 2019 revision of the SI, this numerical value defines the ampere together with the definition of the second; $1\mathrm{C}=1\mathrm{A}\mathrm{s}$. Thus today's numerical exactness is a choice of units, whereas measured electric charges still have experimental uncertainty. An oil drop experiment can test charge quantization and the apparatus calibration; it does not supply a new uncertain defining value of $e$ in today's SI. ^[ref].

For ordinary isolated chemical atoms, ions and charged objects whose net electric charge is counted through protons and electrons, charge quantization gives $Q=ne$ with signed integer $n$. A neutral chemical atom that loses two electrons becomes an ion of electric charge $+2e$. A toy object with $10^9$ excess electrons has $Q=-0.1602176634\mathrm{nC}$. Conversely, the scale $1\mathrm{nC}/e\simeq 6.24\times 10^9$ explains why laboratory electric charge often looks continuous: billions of microscopic increments fit into a small macroscopic amount. Averaged charge density is a continuum description, not a claim that each particle has an arbitrarily divisible electric charge.

The ordinary-matter qualifier matters. In the Standard Model, quarks have electric charges $+2e/3$ or $-e/3$, and antiquarks have the opposite signs. Confinement means these are constituents of hadrons, not isolated fractionally charged particles collected by an oil drop experiment. The usual valence quark sums give $2(2e/3)-e/3=e$ for a proton and $2e/3-2(e/3)=0$ for a neutron. Thus elementary charge should not be defined as the smallest nonzero electric charge of every constituent in particle theory. ^[ref].

The beautiful experimental idea behind the oil drop experiment is to determine a microscopic electric charge by watching a visible carrier move. In his 1913 paper, Millikan timed the descent and electrically driven ascent of individual oil drops, repeatedly changing a drop's electric charge through capture of air ions. His apparatus used parallel plates separated by $16\mathrm{mm}$, a roughly $5300\mathrm{V}$ source, and an optical scale to measure travel distances. He reported multiple relationships among a given drop's inferred electric charges. He also found that uncorrected Stokes's law gave a size-dependent inferred elementary unit, and introduced a pressure-dependent slip correction. These were measured relationships with calibration and model uncertainties, not exact readings of today's defined $e$. ^[ref].

A modern instructional implementation sprays low-volatility oil of known density through a plate aperture, illuminates the drops, and follows one through a calibrated optical scale. Measure the plate spacing $d$, electric potential difference $V$, air temperature and pressure. Time a fixed distance $L$ with the electric field off to get a terminal falling speed $v_f=L/t_f$. Then choose the field polarity to drive the same drop upward and measure $v_r=L/t_r$. Repeat timings at a fixed electric charge, and compare before and after an ion-capture event. The same drop is especially useful because its radius and mass need not be inferred anew for each charge state. ^[ref].

Here is the leading continuum model, retaining buoyancy. Assume a drop shaped as a sphere of radius $a$, oil and air densities ${\rho }_o$ and ${\rho }_a$, air dynamic viscosity $\eta$, and gravitational acceleration $g$. Define the downward effective weight $W$ and positive linear drag coefficient $k$ byAt terminal velocity, the net force vanishes. Falling with no electric field, the upward drag force balances weight minus buoyancy. Rising with the electric force upward, the drag force reverses and joins the downward effective weight:Consequently the uncorrected estimates areThe observed direction of response, with known field polarity, supplies the sign of $q$. A suspended drop gives the alternative balance $|q|E=W$, but it still requires a determination of $W$: balancing a drop alone does not determine its electric charge.

For an explicitly invented continuum example, stipulate $a=1\mu \mathrm{m}$, ${\rho }_o=900\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_a=1.2\mathrm{kg}{\mathrm{m}}^{-3}$, $\eta =1.8\times 10^{-5}\mathrm{Pa}\mathrm{s}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $q=-5e$ and a downward $E=10^5\mathrm{V}{\mathrm{m}}^{-1}$. The balances predict $W\simeq 3.69\times 10^{-14}\mathrm{N}$, $v_f\simeq 0.109\mathrm{mm}{\mathrm{s}}^{-1}$ and $v_r\simeq 0.127\mathrm{mm}{\mathrm{s}}^{-1}$. Substituting those speeds recovers $|q|=5e$ within rounding. This checks the algebra; it is not a precision prediction for real micrometre drops, which need a slip correction.

The inference has two distinct parts. Compare inferred electric charges or charge changes with a common unit, seeking $q_i\simeq n_{i}e$ within uncertainty, rather than demanding exact ratios of rounded measurements. Then establish the absolute scale through electric field, density, dynamic viscosity and distance/time calibration. If the observed steps are all even multiples, their common spacing alone cannot exclude an unobserved smaller unit. Air motion, evaporation and mistimed crossings can alter inferred electric charge without an ion-capture event. Small Reynolds number supports Stokes's law, but does not by itself justify its no-slip continuum coefficient: a significant ratio of air mean free path to drop radius requires a slip correction. Millikan's comparison of different drop sizes and pressures addressed that systematic limitation; the modern PASCO manual also includes the correction. Agreement supports charge quantization for the observed carriers within experimental resolution, rather than proving a universal statement about every possible particle.

Multiplying $e$ by the Avogadro constant gives the Faraday constant, $F=N_{A}e$, the electric charge per mole of singly charged carriers. This connects counting individual electric charges with bulk electrochemistry. ^[ref].

The Faraday constant is the electric charge associated with one mole of elementary charges, $F=N_{\mathrm{A}}e\approx 96485.33212\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$. The product of the Avogadro constant and elementary charge is exact under the SI definitions; the displayed decimal is rounded. For ions with signed charge number $z$, the molar electrical contribution to an electrochemical potential is $zF\psi$, where $\psi$ is electric potential. ^[ref].

Let $u_s=⟨ϵ{⟩}_s$ and $v_s={\mathrm{Var}}_s(ϵ)$ for the conditional distributions of parahydrogen and orthohydrogen, with $s=\mathrm{p},\mathrm{o}$. Differentiating the normalized Boltzmann factors gives $d{u}_s/dt=v_s/t^2$. Thus the rotational heat capacity at constant volume per molecule of a composition frozen at $x$ isAt conversion equilibrium, the law of total variance adds a nonnegative contribution from the changing species populations:The last term is the contribution from conversion. Compare the first two terms with a frozen sample at the same instantaneous composition; a frozen $3:1$ sample generally has different weights too. This formula follows directly from the variance of a mixture: subtract the overall mean inside each species, expand the square, and the cross terms vanish because each conditional deviation has zero mean. It also proves nonnegative rotational heat capacity in either model.

For an original worked example at $t=1/2$, summing the allowed ladders gives $q_{\mathrm{p}}\approx 1.00003072$, $q_{\mathrm{o}}\approx 0.05494692$, and $x_{\mathrm{o}}^{\mathrm{eq}}\approx 0.14150982$. The calculated rotational heat capacities are $c_{\mathrm{eq}}/k_{\mathrm{B}}\approx 1.94719633$ and $c_{\mathrm{fr}}/k_{\mathrm{B}}\approx 0.00110733$ for frozen normal hydrogen. The contrast is large because changing species accesses the $J=0$ to $J=1$ separation, whereas fixed species must use the larger $J=0$ to $2$ or $J=1$ to $3$ separations. These are model numbers at a chosen dimensionless temperature, not measured hydrogen data.

Both rotational heat capacities approach zero as $t\to 0$, although frozen normal hydrogen retains nonzero rotational internal energy. Both approach $k_{\mathrm{B}}$ as $t\to \infty$ within the rotor model. The high-temperature limit is a formal limit with all other excitations suppressed; actual molecular hydrogen eventually needs molecular vibration, nonrigidity, and dissociation. Multiplying the per-molecule result by the Avogadro constant gives rotational molar heat capacity in units of the molar gas constant $R$. A dilute classical ideal gas additionally contributes $3R/2$ from translation to its heat capacity at constant volume.

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]

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.

A membrane potential can coexist with near electroneutrality in the bulk cytosol and external chemical solution. The small imbalance producing the voltage lies near the cell membrane; most positive and negative electric charges balance in the bulk.^[ref]

For a separate original capacitor model, choose a total membrane capacitance $C=5\mathrm{pF}$ and $V_m=-70\mathrm{mV}$. These are stipulated parameters, not measured properties of a particular biological cell, and this voltage is not asserted to equal the preceding $E_{\mathrm{K}}$. Model the separated electric charge relative to an uncharged reference byUsing the elementary charge $e=1.602176634\times 10^{-19}\mathrm{C}$, this representsexcess monovalent negative-charge equivalents inside. These can represent a deficit of positive ions, rather than the arrival of that many negative ions.

Independently choose an internal reservoir volume $\mathcal{V}=1\mathrm{pL}=10^{-12}\mathrm{L}$ containing $140\mathrm{mM}=0.140\mathrm{mol}{\mathrm{L}}^{-1}$ potassium, with balancing negative electric charge. With the Avogadro constant $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$, its potassium inventory isThus, if charging were entirely due to outward monovalent potassium transport, about one in $38600$ of these potassium ions would suffice. This is a comparison with the chosen potassium inventory, not a fraction of all charged particles or a measurement of spatial charge density. No shape, surface area or thickness has been inferred from the chosen capacitance and volume. The calculation illustrates why substantial voltage does not require a strongly charged bulk cytosol.

The molar gas constant is the product of the Avogadro constant and the Boltzmann constant. It converts the particle-number form of the ideal gas law, $pV=N{k}_{B}T$, into the amount-of-substance form $pV=nRT$, using $N=n{N}_A$ with $n$ measured in moles.

Molar mass turns a weighed sample into a count of specified particles. For a pure chemical substance with specified chemical formula and isotopic abundances, its molar mass iswhere $m$ is the sample mass and $n$ its amount of substance. The SI unit is $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$; the common laboratory unit is $\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, with $1\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}=10^{-3}\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Equal values of amount of substance correspond to equal counts of specified entities, but heavier entities give greater masses.

Combining this definition with molecular count and amount of substance givesThe Avogadro constant is exact under the modern definition of the mole. The mass $M/N_{\mathrm{A}}$ is the mean mass per specified entity; if isotopes differ within the sample, individual molecules need not all have that mean mass.^[ref]

Specify what $n$ counts: water molecules, hydrogen chemical atoms, and oxygen chemical atoms give different values of amount of substance for the same sample. For a mixture, dividing total mass by total amount of substance of its chosen entities gives a composition-dependent mean. A polymer sample can contain many chain molar masses; its number-average molar mass and weight-average molar mass answer different averaging questions. Relating either to degree of polymerization also requires a unit-mass convention and end-group accounting.

The molar mass constant converts relative atomic mass or relative molecular mass into molar mass. It is approximately $1\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. Its SI numerical value is determined experimentally: the exact Avogadro constant multiplies an experimentally determined unified atomic mass unit in kilograms. Treating $M_{\mathrm{u}}$ as exactly $1\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ is therefore an excellent elementary-calculation approximation, rather than a defining equality of the modern SI.^[ref]

Let $M$ be molar mass in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, $\bar{v}$ partial specific volume, ${\rho }_s$ solvent density, and $N_A$ the Avogadro constant. The buoyant molecular mass is $m_b=M(1-\bar{v}{\rho }_s)/N_A$. Combining sedimentation coefficient $s=m_b/f_h$ and the thermal diffusion-friction relation $D=k_{B}T/f_h$ eliminates the unknown friction and proves the displayed formula, with $R_{\mathrm{gas}}=N_A{k}_B$. Once friction is known, $R_h=f_h/(6\pi \eta )$ defines a hydrodynamic radius using Stokes's law. Mass and radius are different inferences: an extended molecule need not occupy the volume of that friction-equivalent sphere.

The molecular count $N$ is a dimensionless integer; the amount of substance $n$ has unit mole, symbol $\mathrm{mol}$. They satisfyThe Avogadro constant is exact by the modern definition of the mole: see SI base unit: mole by Bureau International des Poids et Mesures. Thus two water molecules correspond to $2/N_{\mathrm{A}}$ moles, whereas two moles of water contain $1.204428152\times 10^{24}$ molecules. A mole is a unit for counting specified entities, which may also be chemical atoms, ions or electrons.

Four water molecules contain eight hydrogen and four oxygen chemical atoms, twelve constituent chemical atoms altogether. Counting the constituents does not mean they have been separated by breaking chemical bonds. The same sample has different counts because the specified entities differ.

For $N$ water molecules, their amount of substance is $n=N/N_{\mathrm{A}}$, while the constituent hydrogen and oxygen inventories correspond to $2n$ and $n$ respectively. The exact Avogadro constant is $N_{\mathrm{A}}=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$. The mole applies to specified entities, including chemical atoms, molecules, and ions.^[ref] Thus four water molecules correspond to approximately $6.6422\times 10^{-24}\mathrm{mol}$ of water, but twice that amount of substance of constituent hydrogen chemical atoms. A macroscopic conversion through measured mass additionally needs the relevant molar mass, as in water mass to molecular count.