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An adsorption isotherm relates equilibrium uptake to surrounding pressure or concentration at fixed temperature. For a gas, specify the partial pressure of the adsorbing chemical species. An uptake trace during incomplete equilibration need not be an adsorption isotherm.

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.

A chemoreceptor is a sensory cell or receptor system responding to chemical composition. In respiratory regulation, carbon-dioxide-related chemical signals contribute to feedback controlling ventilation. Blood transport and neural response introduce delays, so a lumped ventilation model may respond to past rather than present carbon-dioxide partial pressure.

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.

Competitive adsorption occurs when different chemical species seek the same adsorption sites. For intact molecules $A$ and $B$, retain identical adsorption sites and no interactions, but replace the vacant fraction by $1-{\theta }_A-{\theta }_B$. The same equilibrium balance givesHere $P_A$ and $P_B$ are partial pressures, not total pressure. Increasing $P_B$ can reduce ${\theta }_A$ at unchanged $P_A$: replacing $P_A$ by total pressure in a single-species Langmuir isotherm loses this competition.

The temperature at which an air parcel with a specified water vapor partial pressure first becomes saturated on cooling at fixed pressure. It solves $E(T_D)=e_v$ with saturation pressure $E$. A falling dew point during ascent does not by itself imply loss of water vapor.

The central mechanism is a competition between two falling pressures: expansion reduces the actual water partial pressure, but cooling can reduce the equilibrium vapor pressure still faster. The ratio can therefore rise above one even though the gas becomes more dilute. The cloud-chamber expansion model below makes that comparison explicit.

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.

Consider an ideal gas in chemical equilibrium with homogeneous, identical adsorption sites at fixed temperature. Each intact molecule occupies one site; only one monolayer is allowed. Occupation of distinct sites is independent: neighboring bound molecules do not change binding or release rates. The simplest Langmuir adsorption model has one adsorbing chemical species, no splitting of molecules, and no competitors. Irving Langmuir's 1918 paper treats simple adsorption alongside more complicated cases; its scope exceeds this particular model.^[ref]

The derivation is a balance between arrivals finding vacancies and departures from occupied adsorption sites. Let $k_{\mathrm{a}}P$ be the arrival-and-binding rate per vacant site and $k_{\mathrm{d}}$ the departure rate per occupied site. Thus $k_{\mathrm{a}}$ has units of inverse pressure per unit time, whereas $k_{\mathrm{d}}$ has units of inverse time. The mean surface coverage obeysAt chemical equilibrium, setting the two terms equal givesThe product $bP$ is dimensionless. Here $P$ is the adsorbing gas's partial pressure, expressed in modern bar, with $1\mathrm{bar}=10^5\mathrm{Pa}$, and $b$ is expressed in ${\mathrm{bar}}^{-1}$. The pascal is the SI unit of pressure. This dimensional $b$ is not itself a dimensionless thermodynamic equilibrium constant. A larger $b$ gives greater surface coverage at the same $P$ and temperature, but does not specify how quickly chemical equilibrium is reached: multiplying both rate coefficients by the same factor leaves $b$ unchanged.

Choose $b=0.5$, $2$, and $8{\mathrm{bar}}^{-1}$ as illustrative parameters at one fixed temperature, without assigning a particular material. At $P=0.5\mathrm{bar}$, substitution gives, respectively,Solving $\theta =1/2$ gives $P_{1/2}=1/b$: the three half-occupation pressures are $2$, $0.5$, and $0.125\mathrm{bar}$. For $b=2{\mathrm{bar}}^{-1}$, reaching $\theta =0.9$ instead requires $bP=9$, or $P=4.5\mathrm{bar}$. If the complete-monolayer capacity is stipulated to be $q_{\mathrm{m}}=10\mathrm{mmol}{\mathrm{g}}^{-1}$, then $q=q_{\mathrm{m}}\theta$ predicts $q=5\mathrm{mmol}{\mathrm{g}}^{-1}$ at $P=0.5\mathrm{bar}$. This last number is an original hypothetical example, not a measured uptake.

The shape follows directly from the finite number of adsorption sites. For $bP\ll 1$, $\theta \approx bP$: nearly every arriving molecule finds a vacancy. For $bP\gg 1$, $1-\theta =1/(1+bP)\approx 1/(bP)$: vacancies become scarce. More formally,Thus surface coverage rises with diminishing slope and approaches one only as $P\to \infty$. This saturation is a site-capacity limit, not a claim that the gas has reached its condensation pressure. A second layer, different site binding strengths, interactions between bound molecules, or a changing surface invalidate parts of the setup. A good fit alone does not establish identical adsorption sites: a restricted pressure range can hide such differences. The Langmuir adsorption model is useful because it makes the capacity and the pressure scale of filling separate, testable quantities.

For a dry background inventory of column weight $P_d$, a mixed layer of fixed pressure thickness $\Delta p$, and saturated surface vapor partial pressure $E$, put $x=E/P$ and let $ϵ$ be the vapor-to-background molar mass ratio. The vapor mass fraction is $ϵx/[1+(ϵ-1)x]$, so hydrostatic balance and mass conservation give the displayed quadratic equation. Physical solutions require $x\le 1$ and a nonnegative upper-layer pressure. At $x=1$, $P=E=P_d+\Delta p$. A real positive algebraic root beyond that value need not be physically admissible.

The linked equilibria among dissolved carbon dioxide, bicarbonate, carbonate and hydrogen ions in seawater. At a fixed alkalinity, added carbon changes speciation and acidity, limiting how much extra carbon can be held at a given atmospheric partial pressure. Exchange and chemical partitioning therefore do not behave like irreversible destruction by a single first-order sink.

The partial pressure of a component of a gas mixture is the pressure it would exert alone at the same temperature and total volume. For an ideal mixture, $p_i=n_{i}RT/V$ and $p={\sum }_i{p}_i$. The chemical potential of an ideal gas uses $p_i$ for that component rather than the total pressure $p$.

The two remaining partial pressures sum to $200-160=40$ bar. Their ratio therefore gives $p_{{\mathrm{N}}_2}=10$ bar and $p_{\mathrm{N}{\mathrm{H}}_3}=30$ bar. With the ideal-gas activity approximation $a_i=p_i/p^{\circ }$ and $p^{\circ }=1$ bar,The equilibrium constant is dimensionless: pressures enter as ratios to the standard pressure. Blindly substituting pressures in pascals into an unnormalized expression would give a different, incorrect thermodynamic $K$.

The standard reaction enthalpy is $2(-45.90)=-91.80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ per mole of reaction as written, because the elements are in their reference states. ThusThis calculation uses the supplied constant-enthalpy simplification. At 200 bar, an accurate thermodynamic treatment would use chemical fugacities rather than ideal partial pressures; no chemical fugacity coefficients are supplied. The iron catalyst changes the approach to equilibrium, not its thermodynamic value.

In delayed carbon-dioxide ventilation feedback, $P$ is the effective carbon-dioxide partial pressure of the lumped body compartment. Its stored carbon-dioxide amount is proportional to $P$, with effective capacitance $K$, so $KdP/dT$ is its accumulation rate. Metabolism produces carbon dioxide at rate $M$. Ventilation removes it at a rate proportional to partial pressure times ventilation, $P\dot{V}$, with gas-conversion constants absorbed into the model coefficients. The equation is therefore a one-compartment carbon-dioxide balance: production minus respiratory removal equals accumulation. It idealizes mixing and treats metabolic production as constant over the time scale of interest.

The historical names identify potassium chlorate, ammonium nitrate and calcium carbonate. Their expected net thermal decomposition chemical equations, written in modern formulae, areandThe first gives potassium chloride and oxygen. A potassium perchlorate intermediate can occur during uncatalysed heating; the first chemical equation gives the final overall balance. The second represents the familiar nitrous oxide pathway, with hot water initially as vapour. It does not mean that ammonium nitrate has only one decomposition pathway: conditions, impurities and confinement can produce different gases and violent decomposition.^[ref]

The third is calcination of limestone to quicklime. Its extent depends on temperature and the partial pressure of carbon dioxide; removing carbon dioxide favours further decomposition. Reversing the process gives carbonation of calcium oxide.

For dissolution in hydrochloric acid, the corresponding balances areand“Magnesia” here means magnesium oxide, and “alumina” means aluminium oxide. Zinc undergoes an oxidation-reduction reaction, reducing hydrogen ions to molecular hydrogen. The oxides instead undergo acid-base reactions: hydrogen ions convert oxide into water, without hydrogen evolution. The chlorides are dissolved species in water, rather than anhydrous molecular salts necessarily existing intact in the solution. Strongly ignited aluminium oxide may dissolve slowly; the balance does not determine the rate.

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.

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.