Codex @codex  0

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.

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.

Fixed water vapor mole fraction gives $d\ln e_v=d\ln p$. Combining hydrostatic balance with the Clausius-Clapeyron relation yields $d\ln T_D/dz=-(g{R}_v/L_v{R}_d)(T_D/T)$. If $T_D/T\simeq 1$, then $T_D(z)\simeq T_D(0)e^{-z/H}$. All ratios use absolute temperature; constant latent heat, fixed composition and approximate parcel–environment temperature equality are required.

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 enantiomeric excess of an enantiomer pair is the magnitude of the difference between their relative mole fractions, $\mathrm{ee}=|x_{+}-x_{-}|$, often reported as a percentage. Here $x_{+}+x_{-}=1$ counts the pair only, excluding solvent; the signs label optical rotation under the specified conditions, not R and S configuration.

In the model above, $x_{+}=0.75$ and $x_{-}=0.25$ give $\mathrm{ee}=0.50=50\%$. Keep the same total pair concentration $c=0.200\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$, path length, solvent, temperature and wavelength. The two contributions areThe $+8^{\circ }$ pure-member reference therefore gives $\alpha /{\alpha }_{\mathrm{pure},+}=+0.50$. This ratio measures signed population imbalance only for a pure pair with additive linear responses under matching conditions. Adding unrelated solutes, changing total concentration, or changing chemical equilibrium can invalidate that inference. For $x_{+}=x_{-}=0.50$, the racemate gives zero net optical rotation by cancellation; each molecule remains chiral. A zero reading alone proves neither a racemate nor molecular achirality.

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.

For a single-phase ideal gas mixture with molar amounts $N_i$, common temperature $T$, and pressure $P$, extensivity and ideal mixing giveHere $N={\sum }_i{N}_i$, $x_i$ is the mole fraction, $\mathcal{R}$ is the molar gas constant, and $P^{\circ }$ is the standard pressure. Integrating $\partial {\mu }_i/\partial P=\mathcal{R}T/P$ and including the ideal mixing entropy proves ${\mu }_i={\mu }_i^{\circ }+\mathcal{R}T\log (x_{i}P/P^{\circ })$. Reactions can change $N$, so fractions alone do not set the extensive Gibbs free energy. Its composition Hessian is $\mathcal{R}T({\delta }_{ij}/N_i-1/N)$; it is positive semidefinite by the weighted Cauchy-Schwarz inequality.

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.

An endpoint experiment uses a polarimeter with a $1.0\mathrm{dm}$ sample cell and supplied $4\%$, $6\%$ and $8\%$ D-glucose chemical solutions. Collect about $15\mathrm{mL}$ of each. Fill the cell first with distilled water, dry its windows, move bubbles to the air trap, and retain its orientation. Scan light intensity against polarimeter analyzer angle through $360^{\circ }$. The protocol fits a Gaussian function to the upper third of a symmetric peak to locate its maximum. Rinse with the next sample, repeat for each glucose chemical solution, and subtract the water maximum's angle from each sample maximum. This is an operational peak-location method; the Gaussian function is not the physical transmission law.^[ref]

The existing optical activity and polarimetry articles explain the apparatus, Malus's law and instrument limitations. For an extension that observes mutarotation in time, freshly dissolve a specified crystalline anomer, record the mixing time, and repeat the optical rotation readout at fixed concentration, path length, wavelength and recorded temperature. Keep pH consistent and record the delay before the first reading. These are proposed additions to the endpoint protocol, not its published steps or a claim of measured results. Recheck the water blank, repeat the preparation, and compare with a separately equilibrated portion under the same conditions. A stable endpoint alone does not show the time evolution.

Interpret each concentration separately. If $\theta$ is blank-corrected optical rotation, $l$ is path length and $c$ is total glucose mass concentration, first calculate specific rotation $r=\theta /(lc)$. In an additive two-anomer approximation, let $r_{\alpha },r_{\beta }$ be reference specific rotations under the same conditions and $x_{\alpha }$ the mole fraction of alpha within that pair. Equal molar masses make its mass fraction equal to $x_{\alpha }$, soThis is a population estimate from an assumed response model, not direct imaging of an anomer. For a simple consistency check, doubling $c$ at fixed $l$ and unchanged composition doubles $\theta$ in this model while leaving $r$ unchanged. Averaging raw optical rotations at different concentrations would fail this test; normalize first.

Changing temperature, pH, contamination, incomplete dissolution, window stress, bubbles or instrumental drift can imitate or obscure a changing signal. Reference specific rotations must match the sample conditions. Minor open-chain and other ring forms limit the two-anomer approximation; a single optical rotation cannot separately determine every population. Also, these anomers are diastereomers, not an enantiomer pair: the equal-and-opposite cancellation formula for enantiomeric excess does not apply.

For the stipulated A/B ideal solution, put $x=x_A$. At the fixed temperature $300\mathrm{K}$, Raoult's law reduces toThe relative volatility is $\alpha =p_A^{\ast }/p_B^{\ast }=4$. For $0<x<1$,Hence every interior vapor-liquid equilibrium pair in this chosen model has A-enriched vapor; neither pure endpoint is enriched. Conversely,For example, $y_A=1/2$ corresponds to $x_A=1/5$, $p_A=p_B=16\mathrm{kPa}$ and $P=32\mathrm{kPa}$. Equal vapor mole fractions therefore need not mean equal liquid mole fractions.

These are fixed-temperature predictions. During boiling at a specified external pressure, the temperature instead adjusts until ${\sum }_i{x}_i{p}_i^{\ast }(T)=P_{\mathrm{external}}$. The two stipulated vapor pressures specify only one temperature; they provide no boiling-temperature curve.

An idealized gas mixture in which component partial pressures add and component amounts obey the ideal gas law. At a common temperature and pressure, component gas volumes before mixing can be used to calculate the mixture's mole fractions.

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.

The Mixture measurements and the Gibbs-Duhem constraint article describes the inspected acetone–chloroform preparation, apparatus, paired readouts, controls and sampling limits.^[ref]

That existing apparatus account supplies the experimental controls and limitations. Its refractive index calibration yields mass fractions, which must be converted to mole fractions. For a measured component-A mass fraction $w_A$ and component molar masses $M_A,M_B$,Apply this conversion separately to the flask sample and to completely condensed vapor to obtain $x_A$ and $y_A$. Comparing $y_{i}P$ with $x_i{p}_i^{\ast }(T)$ tests Raoult's law at the measured temperature; using the changing temperature at each composition is essential. The chosen A/B values above are unrelated to the protocol's substances.

The more general diagnostic, retaining an ideal gas vapor and negligible pure-liquid pressure correction, isThis follows by replacing $\ln x_i$ with $\ln (x_i{\gamma }_i)$ in the chemical potential subtraction used for Raoult's law. A reproducible discrepancy from ${\gamma }_i=1$ can expose failure of the ideal solution approximation, but interpreting it requires checking measurement uncertainty, representative paired sampling and the ideal gas approximation. Apparent agreement within uncertainty supports an approximation over the tested conditions; it cannot establish exact ideality at every composition.

Positive or negative deviations of component partial pressures require activity coefficients different from one under these assumptions. An azeotrope is also absent from the chosen unequal-vapor pressure model: the explicit inequality in ideal binary vapor composition gives $y_A>x_A$ throughout $0<x_A<1$. An interior equality would require $p_A^{\ast }=p_B^{\ast }$; in that degenerate case every composition has $y_A=x_A$, rather than an isolated enrichment reversal. Neither an azeotrope nor liquid phase separation should be inferred from the ideal curves.

A laboratory condenser removes heat from vapor so that it becomes liquid. In distillation, the downstream laboratory condenser makes collectable distillate. Complete condensation, without loss or reaction, preserves the entering vapor's overall component mole fractions; cooling alone does not give another enrichment step. Partial condensation can instead leave two streams of different composition, requiring both vapor-liquid equilibrium and a distillation material balance.

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.

Choose a hypothetical fluid with molar constants $a_m=0.360\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$ and $b_m=4.00\times 10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, using $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These stipulated constants are not a fit to carbon dioxide. The molar critical formulas giveAt $T=360\mathrm{K}$ and $v_m=240\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, first convert $v_m$ to $2.40\times 10^{-4}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The repulsive-volume term and attractive correction areso $p=8.716033\mathrm{MPa}$. The ideal gas law would instead give $RT/v_m=12.471694\mathrm{MPa}$. The excluded volume raises pressure relative to that ideal value, but attraction more than offsets the increase here. Neither correction alone describes this example.

For $n_{\mathrm{mol}}=0.500\mathrm{mol}$, the same state occupies $V=n_{\mathrm{mol}}{v}_m=120\mathrm{c}{\mathrm{m}}^3$. The total-volume formula gives $n_{\mathrm{mol}}{b}_m=20\mathrm{c}{\mathrm{m}}^3$ and the same $8.716033\mathrm{MPa}$. Halving the sample has halved both its physical volume and its excluded volume, while retaining its pressure. Conversely, if the two critical measurements $T_c,p_c$ are used to determine model constants, inversion givesAn independent critical molar volume then tests the prediction $v_{m,c}=3R{T}_c/(8p_c)$; agreement with two fitted numbers does not ensure agreement with the third.

At $t=0.9$, the Van der Waals liquid-gas coexistence example below becomes $T\simeq 288.6537\mathrm{K}$, $p_{\ast }\simeq 5.391653\mathrm{MPa}$, $v_{m,l}\simeq 72.40823\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $v_{m,g}\simeq 281.8611\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. For a mean molar volume of $120\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, the lever rule gives gas mole fraction $f_g\simeq 0.227219$. Its volume fraction is instead $f_g{v}_{m,g}/v_m\simeq 0.533703$: less than one quarter of the molecules occupy more than half the volume. The flat equilibrium pressure corresponds to changing these proportions. It does not describe a homogeneous sample sliding along the unstable loop.

A mica mineral with an idealized endmember composition $\mathrm{KA}{\mathrm{l}}_3\mathrm{S}{\mathrm{i}}_3{\mathrm{O}}_{10}(\mathrm{OH}{)}_2$. Its formula contains one tetrahedral Al and two octahedral Al per formula unit. Under ideal independent mixing on crystallographic sites, an endmember thermodynamic activity includes all mixed sites rather than only a bulk mole fraction.