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Molar mass is mass per amount of substance, commonly expressed in $\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. For chain counts $n_i$ and chain molar masses $M_i$, the number average is $M_n={\sum }_i{n}_i{M}_i/{\sum }_i{n}_i$. Relating it to a degree of polymerization requires a defined unit-mass convention and end-group accounting.

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 BIPM, SI base unit: mole. 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.

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 proposed modern teaching measurement puts a pipette-measured portion of an iron ${\mathrm{Fe}}^{2+}$ chemical solution in a conical flask, acidifies with excess dilute sulfuric acid, and adds standardized potassium permanganate from a clamped burette while swirling. The documented endpoint principle is that added purple permanganate is consumed until the first small excess leaves a pale purple colour; no separate colour indicator is required. Insufficient acidity can produce brown manganese dioxide, invalidating the assumed five-Electron half-reaction.^[ref]

Record initial and final burette readings; their difference is the delivered volume $V$. The following controls are proposed additions to test the inference. Run a reagent blank with the same water, sulfuric acid and endpoint criterion but no sample, obtaining $V_b$. Use a separately established potassium permanganate concentration $c$, clear the burette tip of air, and repeat sample portions. A known added amount of substance of ${\mathrm{Fe}}^{2+}$ should increase consumption by one fifth of that amount. A blank estimates reagent demand and the small endpoint excess; it cannot remove an interfering sample reactant.

Assuming complete, selective reaction according to the acidic permanganate-iron balance, the corrected inference isWith $c$ in $\mathrm{mol}{\mathrm{L}}^{-1}$, use $V$ and $V_b$ in litres. As an original stipulated calculation, not an experimental result, take $c=0.02000\mathrm{mol}{\mathrm{L}}^{-1}$, $V=12.60\mathrm{mL}$ and $V_b=0.10\mathrm{mL}$. Then the permanganate consumption is $0.2500\mathrm{mmol}$ and the inferred ${\mathrm{Fe}}^{2+}$ amount is $1.250\mathrm{mmol}$. In a $25.00\mathrm{mL}$ sample portion this corresponds to $0.05000\mathrm{mol}{\mathrm{L}}^{-1}$.

The endpoint detects residual permanganate, not the iron identity directly. Other sample reactants that consume permanganate raise the apparent ${\mathrm{Fe}}^{2+}$ amount; prior oxidation of sample ${\mathrm{Fe}}^{2+}$ to ${\mathrm{Fe}}^{3+}$ lowers it. A recovery check probes response to the added iron but does not prove absence of every interference. The readout determines the reacting ${\mathrm{Fe}}^{2+}$ inventory under the stated conditions, rather than total iron in an arbitrary mixture. No measured titres are asserted here.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

For one chemical reaction in a closed system of fixed volume $V$, with uniform concentrations and no competing reactions or appreciable accumulation of unlisted species, choose the chemical equationBrackets denote amount of substance per volume, for example $[\mathrm{A}]=n_{\mathrm{A}}/V$. The stoichiometry gives one instantaneous reaction ratewhere $\xi$ is the existing extent of reaction. The minus signs make forward consumption positive; dividing by the coefficients makes measurements of different chemical species agree. Units can be $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. See IUPAC Gold Book: rate of reaction and Flowers et al., Chemistry 2e, §12.1: chemical reaction rates.

For the illustrative chemical equation $2\mathrm{A}+\mathrm{B}⟶3\mathrm{C}$, a measured $d[\mathrm{C}]/dt=0.060\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ implies $v=0.020\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, $d[\mathrm{A}]/dt=-0.040\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, and $d[\mathrm{B}]/dt=-0.020\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. These numbers express stoichiometry, without supplying a rate law.

Each derivative is the slope of a concentration curve at one instant. Over a finite interval, $-\Delta [\mathrm{A}]/(a\Delta t)$ instead gives an average reaction rate. For A converting reversibly to B, the net reaction rate is $v=d[\mathrm{B}]/dt=J_f-J_r$; both directional reaction rates can remain positive when $v=0$ at chemical equilibrium.

A redox titration adds a measured amount of substance of one reactant to determine another through a known redox chemical equation. The burette reading becomes an amount through concentration; the balanced chemical equation supplies the amount ratio. A colour endpoint additionally needs a chemical explanation, as in the permanganate-iron titration.

Specific heat capacity is heat capacity per unit mass:Its units are $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and the constraint remains part of the definition. In particular, specific heat capacity at constant volume and specific heat capacity at constant pressure are different responses. For homogeneous material, doubling the mass doubles its heat capacity while leaving its specific heat capacity unchanged. ^[ref]

As an illustrative material model, give a fixed-volume body mass $m=2\mathrm{kg}$ and constant specific heat capacity at constant volume $c_V=400\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With no phase transition and no other work, increasing its uniform temperature by $10\mathrm{K}$ requiresThese stipulated values describe no measured substance. The right panel of the diagram in heat capacity shows the straight line $Q=m{c}_V\Delta T$: a temperature increase of $5\mathrm{K}$ takes $4000\mathrm{J}$ of heat. If the specific heat capacity at constant volume depends on temperature, use $Q=m{\int }_{T_1}^{T_2}{c}_V(T)dT$ instead. The corresponding fixed-pressure calculation uses specific heat capacity at constant pressure and gives $Q=\Delta H$ under the assumptions in heat capacity.

Choosing between specific heat capacity, molar heat capacity and volumetric heat capacity is choosing whether to compare equal mass, equal amount of substance or equal volume. A small sample with large specific heat capacity can still have less total heat capacity than a much larger sample.

Stoichiometry relates the quantities consumed and formed in a chemical reaction. Its coefficients are ratios of entity counts, or equivalently of amount of substance, rather than ratios of masses. For the water chemical equation, consuming two molecular hydrogen molecules per molecular oxygen molecule gives the same $2:1$ ratio as consuming two moles of molecular hydrogen per mole of molecular oxygen. Multiplying every entity count by the same factor preserves the ratio. See Flowers et al., Chemistry 2e, reaction stoichiometry.

Coefficients and subscripts do different jobs. The coefficient in $2{\mathrm{H}}_2\mathrm{O}$ counts two water molecules; the subscript says that each contains two hydrogen chemical atoms. Changing coefficients scales quantities. Changing a subscript changes the specified chemical substance.

This distinction catches a common counting error: one mole of water molecules contains two moles of hydrogen chemical atoms and one mole of oxygen chemical atoms. Always specify what is counted.

Stoichiometry records reactant and product multiplicities in a chemical reaction or a chemical reaction network. The change caused by one reaction is its stoichiometric vector; multiplying by the reaction rate gives its contribution to the concentration equations. For two identical reactants the rate-constant convention absorbs any combinatorial factor.

The Van der Waals equation is a simple real gas equation of state that can also describe a liquid-gas critical point. With particle count $N$ fixed, write the volume per particle as $v=V/N$:or equivalently $(p+a{N}^2/V^2)(V-Nb)=N{k}_{B}T$. Here $p$ is pressure, $T$ is absolute temperature, and $k_B$ is the Boltzmann constant. In SI, $v$ and $b$ have units ${\mathrm{m}}^3$ per particle, $a$ has units $\mathrm{Pa}{\mathrm{m}}^6=\mathrm{J}{\mathrm{m}}^3$, and $k_B$ has units $\mathrm{J}{\mathrm{K}}^{-1}$; the particle count is dimensionless. The pole at $v=b$ excludes smaller volume values from this model.

The $b$ correction represents excluded volume; the attraction term lowers pressure at fixed temperature and particle number density $1/v$. This is a mean-field theory, replacing detailed intermolecular structure by two parameters. At $v\gg b$ with $a/v^2\ll k_{B}T/v$, it approaches the ideal gas law $pv=k_{B}T$. It supplies a useful qualitative route from a dilute gas to phase coexistence, without a claim of quantitative accuracy for every fluid.^[ref]

Do not mix this particle convention with the molar volume convention elsewhere in this article. With amount of substance $n_{\mathrm{mol}}=N/N_A$, Avogadro constant $N_A$, and molar gas constant $R=N_A{k}_B$, one has $v_m=N_{A}v$, $b_m=N_{A}b$, and $a_m=N_A^{2}a$, so $p=RT/(v_m-b_m)-a_m/v_m^2$. The molar parameters have units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$, respectively.

Johannes Diderik van der Waals introduced this equation of state in 1873, connecting molecular attraction with departures from the dilute ideal gas law. Its historical significance is the unified treatment of gases and liquids, rather than a universally exact fit.^[ref]

Volume magnetic susceptibility is the dimensionless ratio of longitudinal magnetization to internal magnetic field intensity in a linear regime through the origin. Positive response gives paramagnetism. Always distinguish magnetization $M=\mathcal{M}/V$ from total magnetic moment $\mathcal{M}$: dividing by volume changes the normalization. Mass-normalized and mole-normalized values likewise require the mass density or the volume per amount of substance before comparison with ${\chi }_v$.

In the existing Pauli paramagnetism formula, $M$ denotes total magnetic moment, and $g(E_F)$ is the total spin-summed density of states. Relabel that response asHere ${\mu }_B$ is the Bohr magneton. If its $B$ is a prescribed vacuum/applied magnetic flux density $B_0={\mu }_0{H}_{\mathrm{app}}$, the dimensionless apparent slope is ${\mu }_0{\chi }_B^{\mathrm{vol}}$. It agrees approximately with intrinsic ${\chi }_v$ only when the demagnetizing field correction and magnetic feedback are negligible. The existing Curie's law formula also uses a total-magnetic moment, $B$-derivative convention; its $N$ counts moments rather than giving a number density.

The volume divided by the positive number of particles $N$. The Van der Waals equation uses this convention for $v$, whereas specific volume divides by mass and molar volume divides by the amount of substance. With the Avogadro constant $N_A$, the molar volume is $v_m=N_{A}v$.

A pH measurement becomes evidence for a weak acid chemical equilibrium when paired with a known analytical concentration and a calibrated glass electrode. For the pure-weak acid model, measured pH estimates $h=10^{-\mathrm{pH}}$ only in the ideal thermodynamic activity approximation. Then $o=K_w/h$, $x=h-o$, $y=c-x$, and $K_a=hx/y$. Require $0<x<c$; values outside that range signal inconsistent measurements or an inadequate model. Repeating this reconstruction across a dilution series tests whether one acid dissociation constant describes the data, rather than merely fitting one point.

The Analytical Sciences Digital Library's Potentiometric Titration of an Unknown Weak Monoprotic Acid provides a complementary acid-base titration experiment. Its apparatus includes a glass electrode and meter, a stirring magnet, a volumetric flask, pipettes, and a burette containing standardized sodium hydroxide. Prepare an aqueous acid stock of known analytical concentration, transfer a measured portion, and stir gently with the glass electrode clear of the magnet. Record added volume and stable pH after each addition; use a preliminary run to locate the steep region and smaller additions there on a fresh portion. The readout is the pH versus added volume curve, not a visual count of dissociated molecules.

Calibrate with fresh standard buffer solutions bracketing the pH range being interpreted. Use their assigned values at the actual temperature; a pair near pH $4$ and $7$ can cover the midpoint of many weak acid examples, but does not cover an entire alkaline endpoint region. Rinse between standards and samples to limit carryover, keep the sensing bulb and reference electrode junction immersed, and wait for a stable reading. These instrument practices are documented in HI5221 & HI5222 Instruction Manual: pH and mV and ISE and Temperature Bench Meters by Hanna Instruments (pp. 45–47, 49–50 and 73–74: calibration, measurement, rinsing and electrode conditioning).

Locate the acid-base titration equivalence region from the steepest change of pH with added volume. For one-to-one neutralization, standardized sodium hydroxide concentration times equivalence volume estimates the initial amount of substance of acid. Half that volume gives approximately equal $[\mathrm{HA}]$ and $[{\mathrm{A}}^{-}]$ only when subsequent chemical equilibrium redistribution is small. The Henderson-Hasselbalch equation then gives $\mathrm{pH}\approx \mathrm{p}{K}_a$. Unequal activity coefficients or appreciable hydronium or hydroxide inventories can shift this estimate. The full chemical equilibrium curve supplies a stronger consistency check than its midpoint alone.

Useful controls for this model test are a fresh independently prepared acid portion, a water blank, a check standard after measurement to detect instrument drift, and fixed recorded temperature. A blank diagnoses contamination but should not be subtracted directly from pH, which is a logarithm. Keep exposure to atmospheric carbon dioxide comparable, especially at high dilution, and record any background salt that changes ionic strength. These are experimental-design controls suggested by the model's assumptions. The observed curve and its uncertainty determine whether the model is adequate; the diagrams here contain no experimental readouts. An aqueous acid dissociation constant cannot be inferred unchanged from a sample requiring a different solvent mixture, and a nearly neutral dilute sample makes the difference $h-K_w/h$ particularly sensitive to pH error.