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McMaster University's 1997/98 laboratory manual compares simple distillation and fractional distillation using charges of $25\mathrm{mL}$ each of cyclohexane and p-xylene. Part B uses a $100\mathrm{mL}$ round-bottom flask and a fractionating column. The protocol specifies a boiling chip, a heating mantle on a ring support, and the top of the thermometer bulb midway in the opening toward the laboratory condenser. Heating is adjusted for roughly one drop per second. Students record head temperature every $2\mathrm{mL}$ collected, up to $45\mathrm{mL}$, and plot both runs against collected volume on the same axes.^[ref]

Charge, atmospheric operation, and collection rate provide comparison conditions; the fractionating column is changed. A sharper separation of temperature regions would support improved fractionation. It would not directly measure product mole fractions or prove purity. The manual distinguishes collected volume from composition and states that this pair forms no azeotrope. No student dataset is supplied here, and the constant-relative volatility example is not assigned to these liquids.^[ref]

Prepare carbon dioxide, make measured mixtures with air, and test a standardized lighted taper in fresh mixtures to find an extinction boundary. In this taper extinction experiment, “carbonic acid gas” means gaseous carbon dioxide, not liquid carbonic acid. The proportion should be reported as a fraction of the final mixture, with the conditions of the test.

A documented sufficient mixture is available from a later first-hand experiment: one volume of carbon dioxide with two volumes of air extinguished the introduced taper in the apparatus used. ThusThis is a conditional literature result, not an invented observation from the Oxford examination. In The True Position of Oxygen as a Restorative in Carbonic Acid Poisoning by W. Ernest Thomson, Experiment XLVIII directly compares one-third carbon dioxide mixed with two-thirds oxygen and with two-thirds air; the latter extinguished the taper. The experiment was published in 1894, so it supplies a later physical comparison rather than evidence of what the 1866 candidate measured.^[ref]

The question asks for an approximately sufficient proportion; it does not explicitly require a universal minimum. The cited condition supplies one sufficient point. To find approximately the least sufficient proportion for the proposed apparatus, determine the lower boundary by the following procedure instead of treating one-third as that boundary.

Prepare the carbon dioxide in a small gas-generating flask by adding dilute hydrochloric acid to calcium carbonate:Fit a delivery tube, keep the outlet unobstructed, and discard the first gas while the original air is being displaced from the apparatus. A wash bottle and droplet trap can reduce acid carry-over before collection in a freely moving gas syringe or suitable gas holder. This acid-carbonate preparation is also used in a current Royal Society of Chemistry demonstration.^[ref] A separate sample may be checked with limewater:The turbidity is a qualitative check for carbon dioxide; it is not a measurement of the fraction in the flame-testing mixture. Avoid using a water-filled collecting vessel for the quantitative mixing stage without allowing for dissolved carbon dioxide and water vapour.

For the mixing stage, use two calibrated gas syringes and a closed mixing vessel with a means of transferring the mixture into a larger flame-testing jar. Measure the component gas volumes at the same temperature and pressure. Mix the gases thoroughly before the flame test. If the gases behave approximately as an ideal gas mixture, their measured volume fractions then approximate their mole fractions. For example, an intended one-third mixture can be prepared from one measured portion of carbon dioxide and twice that volume of air. These are preparation instructions, not recorded trial outcomes.

Keep the taper's material, exposed wick, initial candle flame, insertion depth, gas volume and test duration fixed. Give the flame a reproducible exposure, using a gas-filled jar and a narrow insertion opening with a loose collar to limit entry of fresh air. The observation period must be specified before testing; a fixed five-second exposure is one possible chosen protocol, not a reported condition of the historical reference experiment. Use a large mixture volume relative to the small taper, introduce it briefly, and use a fresh mixture for every trial; prolonged burning changes the mixture by consuming oxygen and producing carbon dioxide and water vapour. Do not seal a burning vessel into an improvised pressure-tight apparatus.

Start with an air control that sustains the taper. Test successively richer, separately prepared carbon dioxide mixtures until extinction occurs within the specified observation period. Retest with a less rich mixture and subdivide the interval. Repeat mixtures near the boundary to detect inconsistent results. A completed experiment would report, for example, the greatest tested fraction $x_b$ at which burning consistently persists and the smallest tested fraction $x_e$ that consistently extinguishes the taper. If the responses are ordered and reproducible under the fixed protocol, reportA midpoint and half-interval may describe the resolution of that bracket; they are not a statistical confidence interval. No numerical values of $x_b$ or $x_e$ are supplied by the source, and none are asserted here.

For a useful model check, let $y_{{\mathrm{O}}_2,0}$ denote the initial oxygen fraction in air, and let $x$ denote the fraction of added pure carbon dioxide in the final mixture. Before the taper changes the gas, an ideal, uniformly mixed replacement model givesTaking dry air to have $y_{{\mathrm{O}}_2,0}=0.2095$ is an explicit model input. For the documented one-third mixture, this gives $y_{{\mathrm{O}}_2}\approx 0.140$, or about $14\%$ oxygen. It is a calculation from an assumed starting composition, not a measured oxygen analysis or a general extinction rule. Also distinguish a fraction of the final mixture from an addition relative to the original air: the volume ratio $r=V_{\mathrm{C}{\mathrm{O}}_2}/V_{\mathrm{air}}$ satisfies $x=r/(1+r)$, so a 1:2 addition ratio means $x=1/3$, not $x=1/2$.

Carbon dioxide affects the thermal balance of a diffusion flame as well as the oxidizer fraction. Extinction therefore depends on the taper, mixing, gas flow, temperature, pressure and heat loss; an oxygen-dilution calculation alone cannot predict its value. The official account of carbon dioxide suppression describes its predominantly thermophysical action.^[ref]

Finally, distinguish this experiment from covering a burning candle and waiting for it to go out. That procedure simultaneously changes gas composition, temperature and spatial distribution. A modern experimental study using gas and temperature sensors found different conditions at different heights and extinction before all oxygen was consumed.^[ref] A pouring demonstration with unmixed carbon dioxide likewise does not determine a uniform bulk fraction. A sufficient, condition-labelled literature mixture and a reproducible bracketing procedure answer the available task; an actual examination threshold or a universal minimum is not claimed.

Notation. The source uses $p_s$ for both the hypothetical dry surface pressure and the actual surface pressure after evaporation. Distinguish them as $P_d$ and $P$, respectively, and interpret the specified pressure thickness $\Delta p$ as fixed. This conserves the original hydrogen inventory and permits $P$ to vary. Reading both occurrences as one fixed number would be inconsistent with adding water vapor to a closed atmospheric column mass budget.

Let $E(T)$ denote the saturation pressure of water, let $x$ be the mole fraction of water vapor in the mixed layer, and let $M_d,M_v$ be the molar masses of hydrogen and water. Surface phase equilibrium and the ideal gas law giveBecause the layer has constant temperature, its relative humidity isThus every level strictly above the surface is subsaturated: hydrostatic balance makes $p<P$, whereas the saturation pressure stays constant. The result does not require a dilute water vapor approximation.

The Clausius-Clapeyron relation, neglecting the specific volume of liquid water and treating the water vapor as an ideal gas, givesHere $T_t,E_t$ are the triple point values, $L_v$ is the specific latent heat, and $R$ is the molar gas constant. All temperatures in these expressions are absolute. As throughout the question, this is a constant-latent heat, ideal gas model, not an extrapolation valid through the real critical point of water.

For a well-mixed ideal gas mixture, its mean molar mass and water vapor mass fraction areBy hydrostatic balance, the mixed layer has total mass per unit area $\Delta p/g$, of which $w_v\Delta p/g$ is water vapor. The remaining atmosphere contains no water vapor. Consequently conservation of hydrogen mass givesEquivalently the requested quadratic equation isFor each $T_s$, calculate $E$, take the positive root of this quadratic equation, and retain it only if $0\le x<1$ and $P>\Delta p$. The last condition leaves a nonempty dry layer. In particular the dry constituent's surface partial pressure is not generally $P_d$: $P_d$ measures its column weight, and the hydrogen and water vapor have different molar masses.

A convenient parameter is the water vapor mole fraction itself. The mass conservation calculation yieldsBoth functions increase for $0\le x\le 1$. At the endpoint,Therefore the physically admissible hydrogen–water vapor mixture requiresBeyond this bound, the formal positive quadratic equation root requires $x>1$, equivalently a negative hydrogen partial pressure. It is the physical constraints, rather than loss of real roots, that end the solution branch. For a concrete example, put $P_d=\Delta p=1$ in consistent pressure units, $ϵ=9$, and $E=3$. The positive root of $P^2+23P-51=0$ is about $2.037$, giving $x\simeq 1.473$, which is impossible.

The limiting temperature is obtained from $E(T_{\ast })=P_d+\Delta p$:This finite limit exists above $T_t$ if $E_t<P_d+\Delta p<E_t\exp [L_v/(R_v{T}_t)]$. The upper qualification matters mathematically: the stipulated constant-latent heat formula has a finite limiting saturation pressure as $T\to \infty$, so for an arbitrarily large unspecified $P_d+\Delta p$ the printed assertion of a finite breakdown temperature is not unconditional. For a concrete formal counterexample, choose $P_d+\Delta p=10^{15}\mathrm{Pa}$: with the usual triple-point constants the model has $E(\infty )\simeq 2.5\times 10^{11}\mathrm{Pa}$, so it never reaches the proposed inventory limit. This is a counterexample inside the stipulated equations, not a physically credible high-pressure ideal gas ocean. Actual liquid water also ceases to exist above its critical point, before such an extrapolation would make sense.

Physically, the imposed finite-pressure-thickness layer can hold no more than $\Delta p/g$ of water vapor per unit area. As $x\to 1$, it expels all its hydrogen into the dry upper layer. A still larger surface saturation pressure cannot coexist with this geometry. The layer must deepen, the temperature profile must change, or the assumed phase equilibrium must fail. If $P_d<\Delta p$, the additional restriction $P>\Delta p$ can also exclude the low-temperature end of the branch.

At the interface the pressure is continuous. Immediately below it the temperature is $T_s$ and mean molar mass is $M_d(1+ax)$; immediately above it the dry hydrogen has temperature $T_1$. The ideal gas law therefore givesAn infinitesimal interchange is unstable to convection when the lower fluid is less dense than the upper fluid. HenceEquality gives neutral interfacial buoyancy; reversing the inequality gives a stable density jump. This is an interfacial criterion, not a claim that the whole dry upper atmosphere is stable. Its vertical temperature profile has not been specified. Also, $\mathrm{RH}(p_1)=p_1/P<1$, so infinitesimal displacement at the interface initially involves no condensation or release of latent heat.

There are two competing effects. Increasing $T_s$ directly reduces the lower density, favouring convection. Increasing the mole fraction of heavy water vapor increases its mean molar mass, opposing convection. With $x\ll 1$, the criterion is approximately $T_s>T_1$. With $x\to 1$, it tends to $T_s>ϵT_1\simeq 9T_1$: substantial surface heating need not overcome the compositional stabilization in hydrogen.

For a specified $P_d,\Delta p,T_1$, solve $T_s/T_1-1-ax(T_s)=0$ on the admissible branch to locate the transitions. Its derivative is $1/T_1-a{x}^{\prime }(T_s)$, so there is no universal monotone trend: rapid growth of the saturation pressure can suppress an instability that was present at lower $T_s$. Without numerical parameter values, neither the number of crossings nor a unique interval of unstable temperatures can be asserted.

Let $e_v$ be the parcel's water vapor partial pressure. Fixed mole fraction before condensation means $e_v/p$ is constant. Its dew point $T_D$ is defined by $E(T_D)=e_v$, where $E$ is the saturation pressure. The Clausius-Clapeyron relation givesThe last approximation uses hydrostatic balance, the ideal gas law, and an environmental temperature close to the parcel's. Rearrangement yieldsFor a modest initial dew-point depression, $T_D/T\simeq 1$ in absolute temperature, soWith $L_v\simeq 2.5\times 10^6\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, $R_d\simeq 287\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and $R_v\simeq 461.5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, $H\simeq 1.58\times 10^5\mathrm{m}$. This dew point scale is much longer than the atmospheric pressure scale height.

The exponential function is an approximation, not the exact consequence of a dry adiabatic process. Retaining $T=T_s-{\Gamma }_{d}z$ gives $p/p_s=(T/T_s{)}^{c_p/R_d}$ and the useful comparisonIt also shows explicitly that expansion lowers the dew point even before any water vapor condenses.

In a regular solution, ideal combinatorial entropy competes with a nonzero unlike-contact energy. For equal molecular volumes and mole fraction $\varphi$ of species A, write mixing free energy per site aswhereHere $z$ is the coordination number and ${ϵ}_{ij}$ a pair-contact energy counted consistently in the lattice model. A positive $\chi$ penalizes unlike contacts; a negative value favors them. The logarithmic terms lower free energy on mixing. The sign of the enthalpy of mixing alone therefore does not determine miscibility.

At fixed temperature and composition, a homogeneous phase is globally stable only if no division into two compositions lowers its free energy. The common-tangent construction selects coexisting compositions ${\varphi }_a,{\varphi }_b$, equivalent to equality of both species' chemical potentials. Local stability to infinitesimal fluctuations instead requiresZero curvature gives the spinodal, while the coexistence boundary is the binodal. Between them a homogeneous state can be metastable and require nucleation; inside the spinodal its small composition fluctuations grow. The symmetric regular solution hasBelow this contact penalty, the free energy is convex at every composition. Above it a miscibility gap is possible around the central composition. If $\chi =A/T$ with $A>0$, heating promotes mixing, but specific interactions or temperature-dependent noncombinatorial terms can produce other trends.

For two polymers, let $N_A,N_B$ count equal-volume segments per chain, and now let $\varphi$ be the segment volume fraction of A. Connecting segments into chains reduces their independent translational entropy. The binary Flory-Huggins polymer blend hasThe energetic contact term remains extensive in the number of segments, while the favorable entropy per segment is reduced by $1/N_A$ and $1/N_B$. This is the key long-chain limit of polymer-blend miscibility: a small positive incompatibility can overcome the mixing entropy. The lattice treatment is a controlled model for this argument, rather than a claim that every real blend has composition-independent $\chi$.^[ref]

The local stability condition and critical interaction parameter becomeFor equal chain lengths, ${\chi }_c=2/N$. Similar segment chemistry and matched cohesive interactions can reduce the unlike-contact penalty; favorable specific interactions, such as complementary hydrogen bonding, can make mixing favorable even for long chains. Lower molar mass increases mixing entropy. temperature can help when it lowers the effective $\chi$, but heating is not universally sufficient. A suitable common solvent can facilitate processing or ternary mixing, and a block-copolymer compatibilizer can stabilize fine domains; neither guarantees that the final solvent-free binary material is one equilibrium molecular phase.

An oil–water emulsion illustrates emulsion stability versus molecular miscibility. A surfactant adsorbs with its water-affine head toward water and its oil-affine tail toward oil, lowering interfacial tension and the work $\gamma dA$ needed to make additional interface. Breaking a fixed oil volume $V_o$ into spherical droplets of radius $R$ gives total area $3V_o/R$, so creating small droplets still requires substantial area and usually mechanical work.

Adsorbed surfactants can oppose droplet coalescence through charged double-layer repulsion, hydrated-head interactions or steric repulsion from sufficiently extended groups. Adequate coverage, adsorption time, ionic conditions and film drainage affect the lifetime. A lower interfacial tension does not itself remove the bulk miscibility gap, and a protected emulsion is commonly kinetically stabilized. A thermodynamically stable microemulsion is a distinct situation requiring an appropriate composition and free-energy balance. The two liquids must not be called molecularly miscible simply because droplets disperse visibly.

For the capillary, balance upward contact-line force against hydrostatic column weight, neglecting meniscus volume and gas density. For inner radius $r$,The contact angle is measured through the liquid. complete wetting means ${\theta }_2=0$, so the contact-angle change of capillary rise givesThis follows from the stipulated wetting change, holding radius, bulk liquid surface tension and density fixed. One should not replace that premise with a general claim that grease always improves glass wetting; if contamination also changes the liquid surface tension, the original height and new angle alone would not determine the new height.

For two wide vertical plates separated by gap $d$, use unit width. There are two upward contact-line contributions and liquid cross-sectional area $d$, giving the capillary rise between parallel plates:Only one principal meniscus curvature contributes between wide plates, whereas a circular tube has two. For the same fluid and angle,If the plate gap equals the tube diameter, $d=2r$, the rise is half as large: the initial and completely wetting heights would be $2.50$ cm and $3.05$ cm. If instead $d=r$, the heights are equal. In either geometry, changing only the angle from $35^{\circ }$ to zero multiplies the height by $1/\cos 35^{\circ }$. Since the question specifies no plate spacing, the general gap-dependent relation is the unambiguous result; the half-height answer requires the equal-gap-to-diameter convention.

Under ideal independent mixing on crystallographic sites, a mineral endmember's thermodynamic activity is the product of its required site fractions, one factor per mixed site. Muscovite has one K and three Al per formula unit, sanidine has one K and one Al, and corundum has two Al. Interpreting the given Al fractions as the fractions on every relevant Al/Fe site givesThus $A_s=a_{\mathrm{san}}{a}_{\mathrm{cor}}/a_{\mathrm{mus}}=0.6237816764$. The cubic power for muscovite includes its tetrahedral Al site and two octahedral Al sites; if site-specific fractions differed, bulk Al/(Al+Fe) alone would not determine this activity. The calculation uses the uniform-site ideal-mixing interpretation intended by the given data.

At $P=5000\mathrm{bar}$ with ideal water, solvegiving $T_{\mathrm{eq}}=979.40\mathrm{K}=706.25{}^{\circ }\mathrm{C}$. Using the bulk mole fraction only once for a multisite mineral endmember would omit part of its ideal mixing chemical potential.

Choose two hypothetical environments A and B of one chemical element, with line centers ${\mu }_A=284\mathrm{eV}$ and ${\mu }_B=286\mathrm{eV}$. Model each line by a normalized Gaussian function with common standard deviation $\sigma$, and stipulate integrated count weights $w_A=3/4$, $w_B=1/4$. On the binding energy axis $E$, the normalized count density isAll energies in this expression use electronvolts. Each component's integrated area equals its count weight, and $\int I(E)dE=1$. There is no background, statistical counting fluctuation, or additional emission feature in this stipulated model. These choices deliberately isolate photoelectron peak overlap and resolution.

Direct integration, or the law of total expectation applied to the component label, gives the centroidThe centroid lies between the chemical environments and is not either component's binding energy. Expanding $(E-\bar{E}{)}^2$ within each component, the odd cross term integrates to zero. The resulting variance isFor $\sigma =0.25\mathrm{eV}$ the variance is $0.8125{\mathrm{eV}}^2$ and the two maxima are separated. For $\sigma =1.00\mathrm{eV}$ it is $1.7500{\mathrm{eV}}^2$ and there is one maximum. Each Gaussian's full width at half maximum is $2\sqrt{2\ln 2}\sigma$, approximately $0.589$ and $2.355\mathrm{eV}$ respectively. The chemical separation remains $2\mathrm{eV}$ in both cases. The number of visible maxima therefore does not equal the number of chemical components.

One interpretation of this change is additional independent Gaussian instrument blur. The convolution of probability densities adds the variances, so broadening $\sigma =0.25$ to $1$ requires response standard deviation $\sqrt{1-0.25^2}=\sqrt{15}/4\simeq 0.968246\mathrm{eV}$. A normalized response with zero mean preserves the centroid and integrated counts. Real line shapes and backgrounds can depart from this model, and separating hidden components is an inverse problem; a plausible fit does not alone establish uniqueness.

Finally, count weights describe detection. In a simple response model, let $n_i$ denote the number of emitting sites sampled in environment $i$, $s_i$ its effective photoionization cross section, and $T_i$ the probability that its emitted electron contributes to the selected measured line. ThenThe common factor $C$ includes exposure. Here $T_i$ includes escape from the sample, analyzer transmission and detection. Count weights equal site fractions only if the relevant response factors are equal or corrected. They are not automatically molecular mole fractions; different molecules can supply different numbers of the selected emitting sites. In an original example, equal site numbers with $s_A{T}_A=3s_B{T}_B$ give the plotted $3:1$ count ratio despite a $1:1$ site ratio.

Polarimetry measures optical rotation with a polarimeter. The vertical Vernier Go Direct instrument uses a $589\mathrm{nm}$ light-emitting diode source, fixed polarizer, sample cell, motor-driven polarimeter analyzer, photodetector and optical rotary encoder. Calibrate using the same solvent at $10\mathrm{cm}$ fill height; rinse with sample, fill to $10.0\mathrm{cm}$, and read the meniscus to $0.1\mathrm{cm}$. The readout plots illumination against analyzer angle; a maximum near zero or a cosine-squared fit locates $\alpha$.^[ref]

An actual teaching experiment in the Truman State University-hosted Optical Activity Laboratory handout, revised July 2013, uses sucrose chemical solutions. Weigh the sucrose, dissolve completely without heating, and measure final solution volume for concentration. Use one polarimeter throughout; measure the water blank's extinction reading, then each sample's extinction reading and actual liquid path length near $10\mathrm{cm}$. Subtract the blank and plot corrected optical rotation against concentration. The attached apparatus description specifies a vertical light path, an incandescent or tungsten source with yellow filter, sheet polarizers and a rotating eyepiece polarimeter analyzer. Its coarse demonstration permits temperature control to the nearest $5^{\circ }\mathrm{C}$, path length to the nearest millimeter and approximately $1^{\circ }$ angular precision. These are design tolerances, not measured results.^[ref]

For a quantitative comparison, equilibrate blank, reference and sample at the same recorded temperature; use a thermostatted cell when needed. This is a recommended control, not a claim that the teaching instruments provide a thermostat. A yellow filter also passes a band of wavelengths, so its result need not match a narrow-source reference exactly. The laboratory's concentration series tests the linear model rather than establishing it in advance; a second series varying only path length tests the other factor.

For our original ideal model, Malus's law giveswhere $I_0$ is light intensity immediately before the ideal polarimeter analyzer, after sample transmission. It is not the intensity of the original unpolarized source. The angle difference is evaluated in consistent units; the plotted degree values are converted to radians before computing the cosine. Assume transmitted light retains linear polarization, with negligible scattering and no polarization-dependent absorption. The pure positive model, $75:25$ mixture and racemate shift the maxima to $8^{\circ }$, $4^{\circ }$ and $0^{\circ }$, respectively. Minima shift by the same amount, to $98^{\circ }$, $94^{\circ }$ and $90^{\circ }$. Thus an extinction measurement and a maximum measurement recover the same shift after their respective blanks are subtracted.

Repeat blanks and use a reference to check drift and scale. Bubbles, particles, stressed windows and sample absorption can spoil the ideal curve. Vernier specifies $\pm 1^{\circ }$ accuracy; the manual's $\pm 0.1^{\circ }$ fit reproducibility does not establish that accuracy. It warns against interpreting its illumination readout quantitatively and requires a transparent homogeneous sample.^[ref] Neither a zero reading nor a matching specific rotation uniquely identifies a molecule.

In the ideal solution and ideal gas vapor approximation, each component's partial pressure equals its liquid mole fraction times its pure-component vapor pressure at the same temperature:These pressure formulas neglect the change in the pure-liquid reference between its saturation pressure and $P$; appreciable real gas behavior also requires a more general treatment. The ideal solution definition in terms of chemical potentials does not require either approximation.^[ref]

A short derivation explains the reference choice. At vapor-liquid equilibrium, each component has the same chemical potential in the liquid and vapor. For an ideal gas mixture, chemical potential of an ideal gas giveswhere $p^{\circ }$ is a reference pressure. Approximate ${\mu }_i^{\ast }(T,P)$ by ${\mu }_i^{\ast }(T,p_i^{\ast })$ and subtract the pure-component equation from the mixture equation:Taking the exponential function gives Raoult's law. Summing partial pressures and normalizing gives $P$ and $y_i$; this last step uses the ideal gas vapor assumption separately from the liquid model.

The law is named after François-Marie Raoult, who measured vapor pressures and freezing behavior of chemical solutions.^[ref]

For a two-component mixture, let $x$ be component A's mole fraction in the liquid and $y$ its mole fraction in the coexisting vapor. At interior compositions, define the relative volatility of A with respect to B by the ratio of composition odds:Thus $\alpha >1$ means A is enriched in the vapor, while $\alpha =1$ means $y=x$. Solving this definition givesHere a constant relative volatility is a stipulated model of vapor-liquid equilibrium, not a measured property of a named chemical solution. Real relative volatility can vary with composition, temperature, and pressure. The relation specifies no absolute temperature scale: even fixing $\alpha =2$ does not determine a boiling-temperature curve or a heat requirement.

Stipulate a constant relative volatility $\alpha =2$. The vapor-liquid equilibrium model then becomesAn initial liquid mole fraction $x_0=3/10$ gives an equilibrium vapor mole fraction $y_0=6/13\approx 0.461538$. Now imagine completely condensing a small sample of that vapor and using this condensate as a fresh liquid for another equilibration. Repeating this formal operation givesThese are original model calculations of ideal-stage enrichment. They are not measured compositions, a reflux design, or successive compositions of one pot undergoing simple distillation. That pot loses the more volatile component and becomes depleted; each step here instead starts from the previous enriched condensate. The displayed sequence specifies compositions, not the recoverable amount at each step.

The algebra makes the enrichment especially transparent. With composition odds $r_n=x_n/(1-x_n)$, every formal step doubles the odds, soFor every finite $n$, $x_n<1$, although $x_n\to 1$ as $n\to \infty$. This explains why repeated ideal equilibration can improve composition without making finite-step perfect separation a consequence of the model.

In simple distillation, pot vapor passes directly to the laboratory condenser. An initial small condensate sample can approximate vapor-liquid equilibrium with the current pot liquid. Removing enriched distillate depletes the pot's more volatile component. A collected fraction averages the changing condensate composition, rather than preserving the first-drop mole fraction.^[ref]

For a chemical solution in a validated linear regime,where $l$ is path length in $\mathrm{dm}$ and $c$ is solute mass per final solution volume in $\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$. These conventions give specific rotation units $\mathrm{deg}\mathrm{mL}{\mathrm{g}}^{-1}\mathrm{d}{\mathrm{m}}^{-1}$. Specify solvent, temperature $T$ and wavelength $\lambda$: specific rotation is conditional on the sample environment.^[ref]

For an original model, stipulate an unnamed enantiomer pair in the same achiral solvent at $T=20^{\circ }\mathrm{C}$ and wavelength $\lambda =589\mathrm{nm}$. Assign the positive member specific rotation $+40\mathrm{deg}\mathrm{mL}{\mathrm{g}}^{-1}\mathrm{d}{\mathrm{m}}^{-1}$, assumed constant over the modeled concentrations. With $l=1\mathrm{dm}$ and $c=0.200\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$,These are chosen model parameters, not measured properties of a named material. Doubling only $l$ or only $c$ doubles $\alpha$ within this stipulated linear model; doubling both multiplies it by four. A real chemical solution must be checked for concentration dependence rather than assuming that $0.200\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$ is dilute enough.

For an additive mixture containing only this enantiomer pair as solutes, put $k=[\alpha {]}_{+}$ and write $c_{+},c_{-}$ for their mass concentrations. ThenBecause enantiomers have the same molar mass, their relative mole fractions within the pair also equal their relative mass fractions: $c_{+}=x_{+}c$, $c_{-}=x_{-}c$. Thus $\alpha =klc(x_{+}-x_{-})$, which connects optical rotation to enantiomeric excess.

For an ideal gas mixture, the volume mixing ratio $x_i$ equals the mole fraction, number fraction, and ratio of partial to total pressure: $x_i=N_i/N=n_i/n=P_i/P$, with ${\sum }_i{x}_i=1$. This equivalence fails for general nonideal mixtures.