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The atmospheric gas mixture, principally nitrogen and oxygen near the Earth's surface, with variable water vapour and smaller amounts of other components. Gas relative density comparisons conventionally specify dry air as the reference.

The chemical element with atomic number eighteen. Argon is a colourless noble gas. It can provide a protective atmosphere around hot, reactive materials, for example in welding, where excluding oxygen limits oxidation. Commercial production separates it by distillation of liquid air.^[ref]

Its discovery shows why a small disagreement can matter more than a reassuringly repeatable result. Two preparations can each give consistent measurements while producing different mixtures. The Rayleigh-Ramsay argon density experiment investigated just such a discrepancy. Gas density links the weighing to a mean molar mass; a separate separation process tests which component accounts for it.

Consider an original simplified model containing only nitrogen and argon, with argon mole fraction $x$. Let their molar masses be $M_N$ and $M_A$. Adding the component masses gives the mixture's mean molar massFor an ideal gas mixture at the same pressure and temperature as a pure-nitrogen reference, the gas density ratio isThe inversion requires known identities, the binary-mixture assumption and $M_A\ne M_N$. A density ratio cannot uniquely recover an arbitrary multicomponent mixture. Nor should $x$ be interpreted as the argon fraction of untreated air: the model sample has only the two specified components.

Choose round model values $M_N=28\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and $M_A=40\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. For $x=0.010$, the calculated relative density excess isThus a one-percent mole fraction does not mean a one-percent gas density change. In a chosen $2.00\mathrm{L}$ globe at $100\mathrm{kPa}$ and $300\mathrm{K}$, the ideal gas law gives approximately $2.24508\mathrm{g}$ for pure nitrogen and $2.25471\mathrm{g}$ for this mixture: only $9.62\mathrm{mg}$ separates the weighings. These are model calculations with rounded molar masses, not historical measurements.

A direct control calculation explains why the filling conditions matter. At fixed pressure, pure nitrogen athas the same gas density as that mixture at $300\mathrm{K}$. A roughly $1.28\mathrm{K}$ mismatch can therefore imitate the entire stipulated composition signal. For small changes, the ideal gas law givesAn independently measured pressure and temperature, matched weighing procedure, and checks for residual vapour distinguish composition from these alternative causes. The model explains the need for controls; it does not supply the complete corrections for a real gas-weighing instrument.

An assay that estimates protein concentration from a colour-producing chemical reaction, measured by absorbance and compared with known protein reference samples. Bovine serum albumin is commonly used as the reference protein. The readout estimates protein content rather than identifying each protein in a mixture.

For a nonideal chemical solution, write its thermodynamic activities on a mole fraction basis as $a_A=x{\gamma }_A$ and $a_B=(1-x){\gamma }_B$. Use composition-independent pure-component reference chemical potentials at the same temperature and pressure:The activity coefficients describe departure from the ideal solution under that reference convention.^[ref]

At fixed temperature and pressure, insert these expressions into the Gibbs-Duhem equation. The ideal solution terms cancel because $xd\ln x+(1-x)d\ln (1-x)=dx-dx=0$. What remains isThus two independent fits to measured activity coefficients can each look smooth while failing to describe one possible Gibbs free energy. This is a necessary consistency condition; satisfying it alone does not establish experimental accuracy or thermodynamic stability.

For an original algebraic example, suppose at one fixed temperature and pressure that $\ln {\gamma }_A=\alpha (1-x{)}^2$, with dimensionless constant $\alpha$. The constraint requires $d\ln {\gamma }_B/dx=2\alpha x$. With the pure-B normalization ${\gamma }_B(0)=1$, integration givesFor $\alpha =1/2$ and $x=1/4$, the derivatives of the two logarithms are $-3/4$ and $1/4$; their weighted sum is $-3/16+3/16=0$. Choosing $\ln {\gamma }_B=\alpha x$ instead would give a weighted sum $3/16$ at that composition, contradicting the Gibbs-Duhem equation. These are invented model functions, with no claim to fit a particular mixture.

The fixed-temperature, fixed-pressure restriction matters. Along a path with changing temperature or pressure, the general equation contains $SdT-Vdp$, and the reference chemical potentials and $RT$ also change. Across phase coexistence, apply the relation separately within each smooth thermodynamic phase; do not differentiate a discontinuous switch between phases as if it were a single composition curve.

For a pure ideal gas at temperature $T$ and pressure $p$,where $R$ is the molar gas constant and $p^{\circ }$ specifies the reference pressure of the standard state. The ratio inside the natural logarithm is dimensionless. The chemical potential difference therefore depends on a ratio of pressures, not an absolute difference. See MIT 5.60, Lecture 14, page 4.

For a chemical species in an ideal gas mixture, replace $p$ by its partial pressure. More generally write ${\mu }_{\mathrm{m},i}={\mu }_{\mathrm{m},i}^{\circ }+RT\ln a_i$ using its thermodynamic activity $a_i$ and a consistent standard state. For an ideal dilute chemical solution, $a_i=c_i/c^{\circ }$ with reference concentration $c^{\circ }$; the thermodynamic activity section describes departures from this idealization.

As a computed comparison, take identical chemical species at $T=300\mathrm{K}$, with $p_A=2p^{\circ }$ and $p_B=p^{\circ }$. Stipulate $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. ThenHalving the pressure gives the same decrement whether starting from $2p^{\circ }$ or $20p^{\circ }$, provided the ideal gas approximation remains valid. This comparison uses different pressures; it is not the common-pressure reservoir setup of chemical potential equality for particle exchange. The fixed-volume example in chemical potential and Helmholtz free energy supplies an exchange experiment with unequal initial pressures.

Chromatography separates components of a mixture through different migration rates in contact with a chromatographic stationary phase while a chromatographic mobile phase transports them. Liquid chromatography describes the physical state of the transporting phase; size-exclusion chromatography describes a separation mechanism. Classification by phase and classification by mechanism answer different questions. In ideal size-exclusion chromatography, differential access to liquid-filled pores replaces binding as the source of differential migration.

In many other forms of chromatography, a solute repeatedly leaves the chromatographic mobile phase and returns to it. Reversible adsorption holds it at a surface; partition distributes it between the chromatographic mobile phase and an immobilized liquid chromatographic stationary phase. Spending more time stationary means less progress with the moving fluid. Two initially overlapping components can therefore emerge as separate bands without either being permanently trapped. Chromatography is useful both for comparing a sample with reference substances and for collecting separated components; a visible band need not contain just one substance.

Relative retention belongs to the sample together with the chosen chromatographic stationary phase, chromatographic mobile phase, eluent composition, and temperature. Changing these conditions can change the separation order. Chemical polarity helps explain interactions in some systems but is not a universal ranking rule. In thin-layer chromatography, a spot's TLC retardation factor is consequently conditional evidence about identity, rather than a fixed identifier intrinsic to a molecule.

Consider an illustrative water–air mixture, initially saturated at $T_0=293.15\mathrm{K}$. Let $\lambda =V/V_0$ be the expansion ratio and approximate the carrier gas by a uniform ideal gas with constant heat capacity ratio $\gamma =1.40$. Before significant condensation, assume fixed numbers of carrier and vapor molecules, negligible incoming heat, and the reversible adiabatic expansion relation. The first law of thermodynamics, $n{C}_{V}dT=-pdV$, with the ideal gas law, givesThe vapor follows the same temperature, so its partial pressure scales as $T/V$. A rapid real piston motion need not be reversible; the formula is an idealized bulk estimate after neglecting flow and losses.

Use a physically motivated Integrated Clausius-Clapeyron relation for the equilibrium water vapor pressure, with stipulated rounded constants $p_0=2339\mathrm{Pa}$ at $T_0$, molar latent heat $L_m=45000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$:These constants define the example, without fitting any historical measurements. Throughout the plotted interval $1\le \lambda \le 1.17$, the temperature stays above the freezing point of water. InitiallyThus the decrease in vapor pressure initially beats dilution. At $\lambda =1.15$, the model gives $T=277.21\mathrm{K}$, $p_v=1923.33\mathrm{Pa}$, $p_{\mathrm{sat}}=809.07\mathrm{Pa}$ and $S=2.3772$.

Why does $S>1$ not instantly make every molecule a visible drop? In the capillarity approximation of classical nucleation theory, a neutral spherical drop of radius $r$ has a positive surface energy cost and a negative bulk Gibbs free energy contribution. For molecular liquid volume $v_l$, surface tension $\sigma$ and Boltzmann constant $k_B$,Differentiating $\Delta G$ shows that the critical nucleus at $r_{\ast }$ is a maximum: smaller neutral drops tend to shrink, whereas larger ones can grow. With stipulated $\sigma =0.075\mathrm{J}{\mathrm{m}}^{-2}$, $v_l=2.99\times 10^{-29}{\mathrm{m}}^3$ and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, the $\lambda =1.15$ example gives $r_{\ast }=1.353\mathrm{nm}$ and $\Delta G_{\ast }/(k_{B}T)=150.3$. A substantial neutral-drop barrier can coexist with supersaturated vapor.

This is deliberately a neutral-drop comparison. An ion changes the cluster free energy through electrical interactions and molecular hydration; the formula does not calculate the ion-assisted nucleation threshold. At nanometre radii, using bulk surface tension is itself approximate. There is no predicted droplet count, growth time or detection efficiency here. Once many drops grow, latent heat release, vapor depletion and heat transfer from the chamber walls invalidate the dry adiabatic expansion trajectory. The example therefore explains the direction of the effect and the need for nucleation, without extrapolating its curves to Wilson's cloud limits.

Distillation separates a mixture by producing vapor from a liquid and collecting the condensed liquid, called the distillate. The more volatile component can have a higher mole fraction in the vapor. Simple distillation collects this condensate directly; fractional distillation adds repeated contact in a fractionating column.^[ref]

Two questions must be kept separate. Vapor-liquid equilibrium describes which compositions can coexist. A distillation material balance accounts for how much of each component leaves in each product. A product can be enriched without containing all the original component, and even a feasible distillation material balance does not establish that an apparatus can make those products.

Consider a nonreacting two-component mixture processed into just two products, with no losses or unaccounted inventory. Let $F,D,B$ be feed, distillate, and bottoms amount of substance values in moles, and let $z,x_D,x_B$ be their component-A mole fractions. These are whole-stream compositions; $x_D$ is not automatically the instantaneous equilibrium vapor composition $y$ above the pot. Total and component balances giveThe component-B balance follows by subtracting the second equation from the first. Eliminating $B$ gives, for $x_D\ne x_B$,Nonnegative product amounts require $z$ between $x_B$ and $x_D$. For an original stipulated feed $F=1\mathrm{mol}$, $z=0.35$, $x_D=0.80$, and $x_B=0.10$,Check the component inventories:The distillate contains a fraction $(D{x}_D)/(Fz)=40/49\approx 0.816327$ of the original component A, despite its higher mole fraction $x_D=0.80$. This distinction between composition and recovery is useful when deciding how much enriched product to collect. These balances determine neither temperatures, heat input, nor the number of equilibrium stages; they do not prove that the specified products are attainable in a particular fractionating column.

The entropy of mixing is the entropy change between initially separate components and their final mixture, with initial and final temperatures, pressures and component amount of substance values specified. For an ideal solution made from pure liquids at the same temperature and pressure, let $n={\sum }_i{n}_i$. ThenThe right-hand expression is the ideal solution result; the general definition does not require ideality. Each interior mole fraction has a negative logarithm, so mixing distinct components gives positive entropy of mixing. At a pure endpoint, use the limit $x\ln x\to 0$ as $x\to 0^{+}$.

For one mole of the chosen A/B ideal solution with $x_A=0.4$,For the same total amount of substance, equal mole fractions maximize this binary entropy of mixing: the composition function $s(x)=-x\ln x-(1-x)\ln (1-x)$ hasso its unique maximum is $s(1/2)=\ln 2$. This is a comparison between different prepared compositions, not a prediction that a closed sample changes its component inventory to become equimolar.

Gas density is the mass density of a gas. For an ideal gas sample of amount of substance $n$, mean molar mass $\overline{M}$ and volume $V$, combine $m=n\overline{M}$ with $pV=nRT$ to obtainHere $R$ is the molar gas constant. For a homogeneous mixture of gases, $\overline{M}={\sum }_i{x}_i{M}_i$, where $x_i$ is the mole fraction of component $i$. Equal pressure and temperature make a gas density ratio equal to a mean-molar mass ratio within this model. Argon detection in a binary gas mixture gives a worked example and a temperature confounder.

The Euler theorem for homogeneous functions applied to an extensive quantity, together with $dE=TdS-pdV+\mu dN$ give

The Gibbs-Duhem equation constrains changes of the intensive quantities describing one uniform thermodynamic phase. For a mixture with $k$ independently specified component amounts, its molar form isHere $S$ is entropy, $V$ is volume, $T$ is temperature, $p$ is pressure, $n_i$ is the amount of substance of component $i$, and ${\mu }_i$ is its molar chemical potential. This assumes a simple compressible system whose bulk internal energy is an extensive quantity; additional work coordinates require additional terms.^[ref]

The proof exposes the source of the constraint. For internal energy $E(S,V,n_1,\dots ,n_k)$, replication of the whole thermodynamic phase by a factor $\lambda$ gives $E(\lambda S,\lambda V,\lambda n_1,\dots ,\lambda n_k)=\lambda E(S,V,n_1,\dots ,n_k)$. The Euler theorem for homogeneous functions and the fundamental differential of internal energy therefore giveTaking the differential of the first equation produces all the terms in the second, plus $SdT-Vdp+{\sum }_i{n}_{i}d{\mu }_i$. Subtraction proves the Gibbs-Duhem equation. Thus the constraint comes from two descriptions of the same state function, rather than an assumed microscopic interaction model.

At fixed temperature and pressure, divide by $n={\sum }_i{n}_i>0$ and introduce the mole fractions $x_i=n_i/n$:The chemical potentials cannot be varied independently. In a two-component mixture, put $x_A=x$ and $x_B=1-x$. Within a smooth single-thermodynamic phase region with $0<x<1$,One chemical potential fixes the other's change, but determining its absolute value still requires a reference value. This does not say ${\mu }_A={\mu }_B$: equality at phase coexistence concerns the same chemical species in different thermodynamic phases. Nor does the constraint say $dG=0$ whenever temperature and pressure are fixed; changing composition still contributes ${\sum }_i{\mu }_{i}d{n}_i$ to the Gibbs free energy.

The Gibbs free energy of mixing compares the final mixture's Gibbs free energy with that of its initially separate components at the same temperature and pressure. For an ideal solution, subtract the pure-component reference from $G={\sum }_i{n}_i{\mu }_i$:For one mole of the chosen A/B ideal solution at $300\mathrm{K}$ and $x_A=0.4$,The individual chemical potential changes are $RT\ln 0.4\approx -2.28554\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for A and $RT\ln 0.6\approx -1.27417\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for B. Weighting these by $0.4\mathrm{mol}$ and $0.6\mathrm{mol}$ reproduces the total Gibbs free energy of mixing.

If ideality holds in a neighborhood of the specified temperature and pressure, differentiation at fixed component amounts givesThus the ideal mixing Gibbs free energy decrease comes entirely from the entropy increase; the mixing enthalpy and volume changes vanish. Ideality at a single isolated state alone does not determine temperature or pressure derivatives.^[ref]

The same formula rules out spontaneous separation into two liquid thermodynamic phases within this model. Per mole, the mixing Gibbs free energy $g_{\mathrm{mix}}=RT[x\ln x+(1-x)\ln (1-x)]$ hasAdding the pure-component reference contributions changes only a term linear in $x$. The total molar Gibbs free energy is therefore a strictly convex function: a separated pair of different compositions has a larger composition-weighted Gibbs free energy than the uniform sample with the same inventory. Such phase separation requires physics beyond this ideal solution model.

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 ideal solution is a chemical solution whose component chemical potentials depend on composition throughHere $n_i$ is an amount of substance, $x_i$ its mole fraction, $R$ the molar gas constant, and ${\mu }_i^{\ast }$ the pure-component liquid reference at the same temperature $T$ and pressure $P$. This model applies to every component across the composition range; it is stronger than an approximation valid only near a pure solvent.^[ref]

With this pure-liquid standard state convention, the thermodynamic activity satisfies ${\mu }_i={\mu }_i^{\ast }+RT\ln a_i$, so an ideal solution has $a_i=x_i$, or activity coefficient ${\gamma }_i=a_i/x_i=1$.^[ref] This statement concerns the liquid. Assuming that a coexisting vapor is an ideal gas mixture is an additional approximation. Its mole fractions $y_i$ generally differ from the liquid mole fractions $x_i$.

For an original two-component example, stipulate $T=300\mathrm{K}$ and pure-component vapor pressures $p_A^{\ast }=80\mathrm{kPa}$ and $p_B^{\ast }=20\mathrm{kPa}$. A and B are invented distinguishable components; these numbers are chosen model inputs, not material data. Combine an ideal solution with an ideal gas vapor, neglect the pure-liquid reference's pressure correction, and exclude other gases. At liquid mole fractions $x_A=0.4$ and $x_B=0.6$, Raoult's law and addition of partial pressures giveThus the coexisting vapor contains about $72.73\%$ A by amount of substance, although the liquid contains only $40\%$. The total pressure is neither pure-component vapor pressure, and $x_A$ cannot be substituted for $y_A$. This enrichment is the basis of distillation.

The usefulness of the ideal solution is that one composition law connects chemical potentials, vapor-liquid equilibrium, entropy of mixing and Gibbs free energy of mixing. Its liquid components can interact strongly: the defining condition is their composition dependence, rather than the absence of all interactions as in an ideal gas model.

A mixture contains two or more chemical substances combined in proportions that need not be fixed. A chemical compound instead has a definite composition: water has two hydrogen chemical atoms per oxygen chemical atom, whereas a sucrose and water mixture can contain different sugar proportions. A homogeneous mixture has uniform composition on the scale of representative samples; a heterogeneous mixture has distinguishable regions of different composition. A chemical solution is homogeneous, whereas sand dispersed in water is a particle suspension.^[ref]

The distinction concerns the sample, not merely its ingredient list. A vessel with dissolved sodium chloride and excess undissolved sodium chloride contains a chemical solution and an ionic crystal thermodynamic phase. The liquid portion can be homogeneous while the whole vessel is heterogeneous. Similarly, molecular oxygen dissolved in water is part of a chemical solution; visible gas bubbles introduce a separate gas thermodynamic phase.

A concrete motivating experiment is the acetone–chloroform mixture in Truman State University's Liquid-Vapor Equilibrium of a Binary System, Procedure and Figure 1. A two-neck flask feeds a distillation head with a water-cooled laboratory condenser, condensate sampling arm and temperature probe. Starting with one pure liquid, the procedure adds the other component successively and waits for stable temperature with returning condensate. Small paired samples from the flask and condensate are read on an Abbé refractometer; a tabulated refractive index–composition relation gives the two compositions. Controls include three-point thermometer calibration, corrected atmospheric pressure readings, and agreement of the initial pure-liquid and condensate refractive indices with the reference value. Small withdrawals and little retained condensate reduce disturbance of composition. Sample evaporation can bias the readout. The inferred paired mole fractions, temperature and pressure describe vapor-liquid equilibrium, subject to the conversion from mass fraction to mole fraction and to representative sampling. This teaching protocol reports no measured results of a particular student run.^[ref]

How does such a measurement expose activity coefficients? Let $x_i$ and $y_i$ be the mole fractions in the liquid and gas thermodynamic phases, respectively, and let $p_i^{\ast }(T)$ be the pure-component vapor pressure. Assume the gas thermodynamic phase is an ideal gas mixture and neglect the change of the pure-liquid reference chemical potential between its saturation pressure and the experimental pressure. Equality of each component's chemical potential between the two thermodynamic phases then givesThis turns the paired composition readings into estimates of the activity coefficients. The approximation needs refinement when the gas departs appreciably from ideal gas behavior or the neglected liquid pressure effect matters.

The apparatus measures a boiling-composition series in which temperature changes as the mixture changes. Its readings therefore cannot be inserted directly into the fixed-temperature, fixed-pressure activity coefficient constraint. They motivate the more general task: test whether the measured mixture properties can arise from a common Gibbs free energy while accounting for changing state variables. A failure can reveal incompatible property fits or experimental bias; a pass supplies a consistency check, not proof that every reading is correct.

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.

For a synthetic illustration, choose $m=0.3000\mathrm{g}$, $V=125.0\mathrm{mL}$, $p=101325\mathrm{Pa}$ and $T=373.15\mathrm{K}$. Using molar gas constant $R\approx 8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ gives $n\approx 0.00408234\mathrm{mol}$ and $M_{\mathrm{inferred}}\approx 73.4873\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ under the ideal gas law. These are chosen model inputs, not observations of a particular chemical substance.

An error's direction follows the model. Extra exterior water raises the apparent $m$ and hence $M_{\mathrm{inferred}}$. If a fraction $f$ of the hot gas particles is residual air, and only the unknown vapor contributes to recovered $m$, an otherwise ideal gas calculation gives $M_{\mathrm{inferred}}=(1-f)M_{\mathrm{true}}$. For a pure nonideal vapor, define $Z=pV/(nRT)$; then $M_{\mathrm{inferred}}=M_{\mathrm{true}}/Z$, so the direction depends on $Z$. The dye endpoint alone cannot establish gas purity, thermal equilibrium, or ideality. These conditional checks explain why the same apparatus cannot provide a universal molar mass measurement for arbitrary mixtures or substances that decompose on heating.