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Mayer's relation states that the difference between heat capacity at constant pressure and heat capacity at constant volume for a fixed-composition ideal gas isHere both capacities are total heat capacities in $\mathrm{J}{\mathrm{K}}^{-1}$, $n$ is amount of substance, $N$ is the number of particles, $R$ is the molar gas constant and $k_B$ is the Boltzmann constant. In other normalizations the same statement isThe first is a difference of molar heat capacities, in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$; the second uses specific heat capacities, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Writing $C_p-C_V=R$ with total capacities would omit the sample's amount of substance. The relation holds pointwise even when the capacities depend on temperature. ^[ref]

The proof outline is short: enthalpy is $H=U+pV=U+nRT$, and internal energy of an ideal gas is independent of volume at fixed temperature. Differentiating at fixed composition and amount of substance therefore gives $C_p=dH/dT=dU/dT+nR=C_V+nR$. It uses the ideal gas law, rather than assuming any particular number of molecular degrees of freedom. At fixed pressure, the additional heat supplies expansion work, $pdV=nRdT$. The two processes can have the same temperature endpoints and equal $\Delta U$ while requiring different heat inputs.

For one mole of the rigid-rotation diatomic ideal gas model, $C_V=5R/2$ and $C_p=7R/2$. Heating by $10\mathrm{K}$ with $R\simeq 8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ gives $Q_V\simeq 207.86\mathrm{J}$ in a rigid sealed vessel and $Q_p\simeq 291.01\mathrm{J}$ under a frictionless piston at fixed pressure. Both gain $207.86\mathrm{J}$ of internal energy, while the piston delivers approximately $83.14\mathrm{J}$ of work. This is a calculated model budget. If the capacities vary with temperature, the corresponding finite statement is stillunder the fixed-pressure and fixed-volume heating assumptions in heat capacity. Their final volume values generally differ. Combining Mayer's relation with heat capacity ratio also gives $C_V=nR/(\gamma -1)$ and $C_p=\gamma nR/(\gamma -1)$ locally; constant values cannot be extrapolated across intervals in which $\gamma (T)$ changes.

For a smooth stable single thermodynamic phase of a general simple compressible system, the broader Maxwell relation identity iswhere $\alpha$ is the coefficient of thermal expansion and ${\kappa }_T$ the isothermal compressibility, with composition fixed throughout. Substituting the ideal gas law gives $nR$; a real gas need not do so. This distinction matters near a phase transition or when interactions appreciably change the equation of state. It explains why Mayer's relation is useful: thermal measurements under different constraints become linked to a mechanical response, while the difference retains a particularly simple value for an ideal gas.^[ref]

Molar concentration is amount of substance of a named chemical species per final chemical solution volume: $c=n/V$. State whether the count describes dissolved species or a conserved analytical concentration.

Molar enthalpy is enthalpy per amount of substance $n$, measured in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. With $u_{\mathrm{m}}=U/n$ and molar volume $v_{\mathrm{m}}=V/n$, $h_{\mathrm{m}}=u_{\mathrm{m}}+p{v}_{\mathrm{m}}$. Its fixed-pressure temperature derivative is the corresponding molar heat capacity $C_{p,\mathrm{m}}$, at fixed composition.

Molar heat capacity is heat capacity divided by amount of substance $n$, with units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. Its constraint must still be specified: $C_{V,\mathrm{m}}=C_V/n$ and $C_{p,\mathrm{m}}=C_p/n$. For the classical monatomic gas example in heat capacity, these are $3R/2$ and $5R/2$, where $R$ is the molar gas constant. Some textbooks use $C_V$ and $C_p$ for these per-mole quantities; the convention here reserves them for the whole sample.

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 solubility $s$ is the amount of substance of a specified solid dissolved per volume of saturated chemical solution, under stated conditions. It counts dissolved formula units, not the sum of all dissolved ion concentrations. For AB with no initial A or B and no competing chemical reactions, one-to-one stoichiometry gives $c_{\mathrm{A}}=c_{\mathrm{B}}=s$. For an AB${}_2$ solid, the corresponding concentrations are $s$ and $2s$.^[ref]

The volume per amount of substance, with $n$ measured in moles. For a uniform pure substance, $v_m=M/\rho$, where $M$ is its molar mass and $\rho$ its mass density. The Clapeyron equation pairs a difference of molar volumes with latent heat per mole.

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.

Molecular diffusion is the component of material transport caused by the random thermal motion of molecules, measured relative to bulk motion of their surroundings. A diffusion coefficient is a property of a particular substance in a particular medium under specified conditions, not a universal speed of spreading. Its dependence on the solute, solvent and temperature is precisely what makes a spreading experiment informative. A random walk idealizes this motion; the Brownian motion description provides a continuous approximation over sufficiently long times. ^[ref]

A motivating experiment: watch dye enter a gel. Samprovalaki, Robbins and Fryer studied rhodamine 6G and methylene blue entering agar gel from two side reservoirs. A covered slab, $10\times 5\times 1\mathrm{mm}$, sat on a controlled thermal stage; the dye was brought to the same temperature. They photographed the gel every ten minutes, averaged image intensity across a strip, and fitted the changing concentration to a diffusion equation model. The cover limited evaporation and stopped dye travelling over the gel surface. Runs at $30$, $50$ and $70{}^{\circ }\mathrm{C}$ were repeated in triplicate. The measured spatial colour change therefore supplied a quantitative transport test, rather than merely showing that dye eventually spreads. Changing lighting also changes image intensity, so the optical response had to be checked. Importantly for interpretation, their constant-diffusion coefficient fit was imperfect and the gel coefficient is an effective transport coefficient, not automatically the value in free water. ^[ref]

A finite-pulse version to test width growth. The following is a proposed teaching variant, not that paper's apparatus or a report of measurements. Make a long, narrow, transparent gel channel of uniform cross-section. Join a thin dyed gel band between two undyed lengths of the same gel, with good contact and no air gaps; the dye should initially fill the cross-section so the dominant concentration variation is along the channel coordinate $x$. After assembly, keep the covered channel horizontal at uniform temperature, away from vibration. The dye is a finite initial pulse: no reservoir supplies more dye. Choose a dye whose image response can be calibrated and whose interaction with the gel is small enough for an approximately constant diffusion coefficient to be plausible.

Photograph the whole pulse at recorded times with fixed lighting, focus and exposure, including a length scale. Use uniformly dyed standards of the same gel thickness and a blank gel to convert image intensity to concentration $C(x,t)$; do not assume raw brightness is proportional to concentration. Average across the channel width and examine several parallel strips to check the one-dimensional approximation. Measure the initial profile instead of treating the band as infinitely thin. For a finite pulse with negligible end effects, define its spatial integral $Q$, centre $\bar{x}$ and width $\sigma$ byHere ${\sigma }^2$ is the variance of the normalized profile $C/Q$, and $\sigma$ is its standard deviation. If $C$ is measured in $\mathrm{mol}{\mathrm{m}}^{-3}$, then $Q$ has units $\mathrm{mol}{\mathrm{m}}^{-2}$; multiplying by the channel cross-section gives the total amount of substance. For the unbounded one-dimensional diffusion equation with constant $D$, conserved $Q$, finite variance and sufficiently decaying tails,Thus plot width squared against time: its slope determines the diffusion coefficient, while its intercept includes the finite starting width. This moment law holds even for a non-Gaussian distribution initially; it follows by applying integration by parts twice to the diffusion equation. ^[ref]

For the particularly readable case of an initially Gaussian distribution centred at zero, uniform advection at speed $u$ and constant diffusion coefficient $D$ give the one-dimensional concentration profileIts centre moves as $\bar{x}=ut$ and its variance grows as ${\sigma }_0^2+2Dt$. Our illustrative choice ${\sigma }_0=0.50\mathrm{mm}$ and $D=0.20\mathrm{m}{\mathrm{m}}^2{\mathrm{h}}^{-1}$ gives widths $0.50$, $0.81$ and $1.36\mathrm{mm}$ at zero, one and four hours. These are calculated values, not a measured dye coefficient. Once $2Dt$ dominates the initial variance, width grows as $\sqrt{2Dt}$: four times the elapsed time gives approximately twice the width. A rectangular starting band will not immediately have a Gaussian distribution, although the same width-squared law still applies under the stated assumptions.

Controls and limits of the interpretation. In the proposed experiment, check that $Q$ stays constant and that the centre remains stationary while the width increases. Include a blank and a uniformly dyed gel photographed over the same period: an apparent signal change there exposes lighting drift, dye fading or evaporation. Repeat with a lower starting dye concentration and a second initial width; agreement of the slopes is a check on constant $D$ and on preparation effects. Keep the pulse far from the channel ends and record its faint tails: truncation and background errors can strongly bias a variance. Poor contact, dye binding, chemical reactions or transport in a liquid film can invalidate the simple model. A gel suppresses bulk liquid motion, but its inferred diffusivity need not equal that in the pure solvent. A stationary centre alone does not exclude symmetric flows, and linear width-squared growth alone does not prove a molecular mechanism: turbulence can also produce an effective diffusive description. These are reasons to combine the profile test with the controlled setup.

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.

The mole is the SI base unit of amount of substance. One mole contains exactly $6.02214076\times 10^{23}$ specified entities, as fixed by the Avogadro constant. It counts entities rather than a fixed mass or volume. ^[ref]

A net ionic equation retains the chemical species changed by a chemical reaction and cancels spectator ions present unchanged on both sides. Dissolved strong electrolytes are represented by their separated ions, while a solid product retains its chemical formula. Cancellation preserves both elemental inventories and electric charge conservation.^[ref]

For the calcium carbonate example, the complete ionic account isCanceling the sodium cations and chloride anions yields the net ionic equation above. Neither cancellation nor the conventional sodium chloride formula implies that the remaining dissolved ions have become paired molecules.

For this schematic, let $\xi$ be the amount of substance of AB formed. With no competing chemical reactions or losses,Adding each dissolved inventory to its contribution in the solid recovers its initial total. Each dissolved pair has net electric charge zero, so transferring a pair into neutral AB also preserves electric charge conservation. This is a short stoichiometry argument; choosing the actual $\xi$ additionally needs chemical equilibrium or chemical kinetics information.

A nonelectrolyte supplies neutral molecules rather than ions on dissolving in the specified solvent, as illustrated by sucrose in water.

For example, dissolving an idealized $1\mathrm{mmol}$ sucrose inventory gives $1\mathrm{mmol}$ dissolved sucrose molecules, rather than separate cations and anions. Dissolving the same amount of substance of a fully dissociated one-to-one strong electrolyte instead gives $1\mathrm{mmol}$ of each ion. Both can disappear visibly into the solvent; the resulting chemical species distinguish them.^[ref]

A nonelectrolyte is not necessarily insoluble: sucrose provides a soluble counterexample. Nor does the label assert exactly zero measured electrical conductivity. Self-ionization of water, impurities or residual electrolyte from a probe can supply background ions. Comparing with a solvent blank and rinsing between samples, as in the strong electrolyte experiment, tests whether the dissolved substance itself produces appreciable additional mobile ions. This separates visible dissolution from ion production without treating lamp darkness as an absolute absence of electric charge.

Dividing two extensive quantities with the same replication factor produces an intensive quantity:Division by volume gives a density, such as mass density $\rho =m/V$ or energy density $U/V$. Division by mass gives a specific quantity, such as specific internal energy $u=U/m$ or specific volume $v=V/m$. Division by amount of substance gives a molar quantity, such as molar volume $V_m=V/n$. Their units are respectively $\mathrm{kg}{\mathrm{m}}^{-3}$, $\mathrm{J}{\mathrm{m}}^{-3}$, $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, ${\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$ and ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The denominator must be stated: “per particle”, “per mole” and “per unit mass” are different normalizations.

For a hypothetical uniform parcel with mass $2\mathrm{kg}$, volume $0.002{\mathrm{m}}^3$ and internal energy $40\mathrm{kJ}$ relative to a specified reference, $u=20\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$ and $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. Three independent copies have $6\mathrm{kg}$, $0.006{\mathrm{m}}^3$ and $120\mathrm{kJ}$; both ratios remain unchanged. This permits comparison of materials without confusing the amount tested with the state tested.

If $X$ and $Y$ are additive but the parcels have different $r_i=X_i/Y_i$, their composite ratio is a weighted arithmetic mean:The denominator supplies the weights: specific internal energy uses mass weights, while mass density uses volume weights. This identity is accounting for the specified subsystems; predicting a new state after mixing also needs the relevant equation of state.

In the dilute ideal particle-count description, osmolarity is the total amount of substance of independently dissolved solute particles per volume of chemical solution: $c_{\mathrm{osm}}={\sum }_j{n}_j/V$. It includes both membrane-permeating and retained solutes; specifying a particular membrane is unnecessary to count particles. A neutral undissociated solute contributes one particle per molecule, whereas separated ions are counted individually. Common units are $\mathrm{osmol}{\mathrm{L}}^{-1}$ or $\mathrm{mOsm}{\mathrm{L}}^{-1}$, with one osmole denoting one mole of the counted particles. Osmolality uses mass of solvent instead of volume of chemical solution; the two numerical values should not be identified without the appropriate mass-to-volume conversion.^[ref]

This ideal counting definition states its scope: interactions can require a thermodynamic activity description of osmotic pressure. Tonicity asks a different question, namely how that surrounding chemical solution changes a specified biological cell's volume through its particular cell membrane.

Osmotic pressure is the excess pressure needed on a chemical solution to maintain osmotic thermodynamic equilibrium with pure solvent across a semipermeable membrane. For a nearly incompressible solvent, equality of chemical potential gives $\Pi \simeq -(RT/{\bar{V}}_w)\ln a_w$. It is an equilibrium property; an observed initial swelling rate is a different measurement.^[ref]

For a dilute ideal chemical solution of retained particles,Here $n$ is the amount of substance of independently dissolved solute particles, measured in moles; $V$ is the final volume of chemical solution, not the starting volume of pure solvent. Count the resulting ions separately if a solute dissociates; do not multiply twice if $n$ already counts them. With $c_{\mathrm{osm}}$ in $\mathrm{mol}{\mathrm{m}}^{-3}$ and $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, $RT{c}_{\mathrm{osm}}$ has units $\mathrm{J}{\mathrm{m}}^{-3}=\mathrm{Pa}$. This resembles the ideal gas law, although the particles here are dissolved in a liquid.

The assumptions are uniform temperature, a dilute ideal chemical solution, a nearly incompressible solvent, and a semipermeable membrane that retains the counted solutes while passing solvent. Association, incomplete dissociation and interactions change the relation between added material and independent particles; nonideal thermodynamic activities invalidate simple particle counting. The formula does not describe a concentrated brine merely because the apparatus contains a membrane.

A short derivation explains why particles are counted. In an ideal mixture, $a_w=x_w=n_w/(n_w+n)$, with $n_w$ the amount of substance of water. If $n\ll n_w$, then $-\ln x_w=\ln (1+n/n_w)\simeq n/n_w$ and $V\simeq n_w{\bar{V}}_w$. Substitution gives $\Pi \simeq RTn/V$. This is a dilute-limit approximation, not a claim that water molecules stop moving or that dissolved particles are a gas.

Limited Papain digestion of IgG cuts each immunoglobulin heavy chain on the N-terminus side of the hinge crosslinks, releasing two Fab fragments and one Fc fragment.^[ref]

A connectivity argument explains the three products. Treat each continuous backbone segment as a graph vertex, and retained interchain disulfide bonds as edges. After the two stipulated cuts, six backbone segments remain: two complete immunoglobulin light chains, two immunoglobulin heavy chain arm segments and two immunoglobulin heavy chain stem segments. Each light chain remains attached to its own arm segment, producing two separate Fab fragments. The two stem segments remain attached to each other, producing one Fc fragment. Thus six backbone segments form three connected components of a graph, not six independently released products. Removing crosslinks instead, while keeping each backbone intact, would leave four chain components in this simplified covalent-connectivity model.

For a constructed mixture, start with $5\mathrm{pmol}$ of intact IgG and stipulate that $60\%$ undergoes the complete two-cut event while the rest remains intact. Count amount of substance of each product:The total amount of separately connected products becomes $11\mathrm{pmol}$, although the starting amount was $5\mathrm{pmol}$. Each digested IgG replaces one connected assembly with three. The antibody binding site inventory is conserved in this model:The factor two counts sites per intact IgG, not extra immunoglobulin light chains per Fab fragment. Fragment counts, backbone-segment counts and antibody binding site counts must be kept distinct. A real digest's retained binding would additionally depend on preservation of the paratopes.

The historical functional test was more subtle than asking whether a fragment still formed a visible precipitate. Rodney Robert Porter's earlier rabbit anti-ovalbumin preparation was digested with papain powder. The resulting fragment did not precipitate ovalbumin itself, but specifically inhibited ovalbumin precipitation with undigested antiserum. The intact-antiserum reaction supplied the positive comparison: failure to precipitate alone did not mean failure to recognize the antigen.^[ref]

The inference is that binding and assembly of a precipitating network are different abilities. A one-site Fab fragment can occupy an epitope without providing the two-arm bridge available to intact IgG. The inhibition readout supports retained recognition under the tested conditions; it does not locate individual contacts or show that every papain-generated product binds. This is the conceptual benefit of fragmenting an antibody: one can ask which functions survive loss of a structural connection.

Treat the membrane as a capacitor, and count the additional positive charge on its inner surface. With radius $r=10^{-6}$ m and length $\ell =10^{-4}$ m, including both end caps givesThe membrane capacitance is $C=C_{m}A=6.3460$ pF. Thus $\Delta Q=C\Delta V=5.3941\times 10^{-13}$ C, corresponding to $\Delta N=\Delta Q/e=3.3667\times 10^6$ monovalent sodium ions. Converting charge to amount of substance using the Faraday constant,Omitting the ends changes the answer to $0.01762$ mM, a one-percent correction. A large membrane potential change therefore requires only about $0.12\%$ of the intracellular sodium population to enter. The bulk liquid remains nearly electrically neutral; excess charge resides chiefly near the cell membrane. This explains why an action potential need not exhaust the concentration difference that drives it, and why fixed reversal potentials are a useful approximation during one spike. This membrane capacitive ion budget assumes sodium supplies the minimum net charging requirement. Simultaneous opposing electric currents can require more sodium entry; repeated spikes eventually require the sodium-potassium ATPase to maintain the gradients.

Here $p$ is absolute pressure, $V$ is volume, $n$ is amount of substance in moles, $R$ is the molar gas constant and $T$ is absolute thermodynamic temperature in kelvin. This is the ideal gas law; consistent SI units give pressure in pascals, volume in cubic metres and $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

The relation describes the slope of the phase coexistence curve of a pure substance in the pressure-temperature plane:Here $p(T)$ is the common equilibrium pressure of the two phases, $T$ is their common absolute temperature in kelvins, and the derivative is taken along their coexistence curve. Choose a direction of transformation from phase 1 to phase 2. Then $L$ is the reversible latent heat for that transformation and $\Delta V=V_2-V_1$ is the corresponding volume change, using the same amount of substance for both quantities.

In the molar convention used below, $L=h_{m,2}-h_{m,1}$ has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and $\Delta V=v_{m,2}-v_{m,1}$ has units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, so the right-hand side has units $\mathrm{Pa}{\mathrm{K}}^{-1}$. For vaporization, both $L$ and $\Delta V$ are positive. The displayed general relation is also called the Clapeyron equation; the ideal-vapor approximation used in (c) gives its usual Clausius form.