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The Antoine equation is an empirical vapor pressure fit whose coefficients depend on the pressure and temperature units and on a stated validity interval. Its slope is $d\ln p_{\mathrm{sat}}/dT=\ln (10)B/(T+C{)}^2$. In the ideal-vapor Clausius-Clapeyron approximation, this implies latent heat $L=R_v\ln (10)B[T/(T+C){]}^2$. An accurate pressure fit need not give an equally accurate derivative; extrapolation to the liquid-gas critical point cannot replace the full Clausius-Clapeyron relation.

At a specified external pressure, the equilibrium temperature at which a pure liquid's saturation vapor pressure equals that external pressure. Bubble nucleation, hydrostatic head and capillary pressure can affect the observed onset of boiling without redefining the equilibrium property.

Why can hot liquid wait for a charged particle rather than immediately turn into vapor? A small cavity costs surface energy. Consider an original macroscopic capillarity model at fixed temperature with surface tension $\sigma >0$, imposed liquid pressure $p_l$, and driving difference $\Delta p=p_{\mathrm{sat}}(T)-p_l>0$, where $p_{\mathrm{sat}}$ is the planar-interface vapor pressure. Neglect dissolved gas, interface thickness, heat transfer and curvature dependence of $\sigma$. The reversible formation-work approximation isThe first term charges for surface area; the second rewards the new vapor volume. This is a macroscopic free energy model, not the energy that an actual detector must receive as localized heat. It shares the capillarity balance used in the cloud-chamber expansion model, but the new phase is now a bubble.

Differentiating givesAt the critical nucleus, $d^2\Delta G/d{r}^2=-8\pi \sigma <0$. The barrier is a maximum: cavities smaller than $r_{\ast }$ tend to shrink, while larger ones can grow in this model. Doubling $\Delta p$ halves $r_{\ast }$ and quarters $\Delta G_{\ast }$ if $\sigma$ stays fixed. This explains the strong sensitivity to operating conditions without claiming a universal detector threshold.

Choose $\sigma =0.010\mathrm{J}{\mathrm{m}}^{-2}$ and $\Delta p=2.0\times 10^5\mathrm{Pa}$ solely for a teaching example. ThenFor the same chosen $\sigma$ and twice the driving pressure, $r_{\ast }=50\mathrm{nm}$ and $\Delta G_{\ast }\simeq 0.654\mathrm{keV}$. These are calculated formation-work scales, not measured bubble thresholds for diethyl ether or any other substance.

A simplified way to discuss localization is to stipulate deposited energy $E_{\mathrm{loc}}$ within a segment of length $\ell$ and a trial conversion fraction $0<\eta \le 1$. A toy criterion $\eta E_{\mathrm{loc}}\gtrsim \Delta G_{\ast }$ gives an energy per unit length scale $\Delta G_{\ast }/(\eta \ell )$. It is only a bookkeeping estimate: a real bubble also needs latent heat, must exchange heat, and is affected by the spatial pattern and time scale of energy deposition. Consequently neither a single stopping power value nor this barrier alone predicts detection efficiency or the number of bubbles. Two tracks with the same mean stopping power can distribute their energy differently on the scale of a critical nucleus.

Condensation in a wetting confined geometry at a vapor pressure below planar-interface saturation. A concave liquid meniscus has lower equilibrium vapor pressure in the capillarity approximation. Gap size, contact angle and surface adsorption affect the actual onset.

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.

Conversion of vapor into liquid.

A graduated tube for confining a gas, measuring its volume, and often examining changes after a chemical reaction. The reading requires temperature, pressure and vapor pressure corrections.

Evaporation transfers molecules from a liquid surface into the surrounding vapour. It can occur below the boiling temperature, and competes with condensation from the vapour. Its net rate depends on the surrounding vapor pressure and transport conditions. Removing latent heat through evaporation can cool the remaining liquid.

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.

Chemist.

Biographical sketch: www2.chem.umd.edu/thermobook/v10-screen.pdf

At fixed relative humidity and fixed excited-oxygen concentration, a water-reaction hydroxyl radical source scales as $e_s(T)/T$ when concentrations are number densities. Thus $d\ln P_{\mathrm{OH}}/dT=\Delta H_{\mathrm{vap}}/(R{T}^2)-1/T$ under constant latent heat and rate coefficient. Using only $e_s$ instead gives a vapor pressure or fixed-pressure mole-fraction approximation. Neither derivative follows without a humidity condition.

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 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.

The equilibrium water-vapor saturation ratio above a solution droplet combines a curvature effect, which raises vapor pressure, with dissolved-solute reduction of water activity. Its maximum defines a critical supersaturation for activation. In a dilute hygroscopicity approximation using droplet diameter $D$ and dry diameter $D_d$, $s(D)\simeq A_K/D-\kappa D_d^3/D^3$, where $A_K=4\sigma M_w/(RT{\rho }_w)$. Then $s_c\simeq \sqrt{4A_K^3/(27\kappa D_d^3)}$. These approximate forms assume the stated diameter convention and dilute solution; composition and surface tension must be consistent.^[ref]

The lambda point is the temperature of the transition between ordinary liquid helium-4 and its superfluid phase, named after the sharp lambda-shaped heat capacity feature. At saturated vapor pressure it lies near $2.17\mathrm{K}$. Its value varies with pressure.

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.

A humidity assumption is necessary: warming does not by itself specify how much water is available. At fixed relative humidity, the Clausius-Clapeyron equation gives the first-order vapor pressure responseIf water abundance is approximated as proportional to $e_s$, the usual first-order estimate is a $26.7\%$ increase in hydroxyl radical production. Integrating the constant-latent-heat expression instead of linearizing gives $e_s(289)/e_s(285)=\exp [(45050/8.314)(1/285-1/289)]=1.3010$.

There is a small but explicit concentration correction when the square brackets are literal molecules per volume. Since $[{\mathrm{H}}_2\mathrm{O}]=\varphi e_s/(k_{B}T)$ and $P_{\mathrm{OH}}=2k[\mathrm{O}({}^{1}D)][{\mathrm{H}}_2\mathrm{O}]$,The corresponding finite-change concentration result is $(285/289)(1.3010)-1=0.2830$. The $25.3\%$ linear concentration estimate and the $26.7\%$ vapor pressure approximation express different stated approximations; neither follows without a humidity condition.

Extra water increases the fraction of oxygen singlet D atoms reacting to form hydroxyl radicals rather than being collisionally quenched and recycled into ozone. In clean, low-nitrogen oxides air this tends to increase ozone loss. In nitrogen oxides-rich air, increased hydroxyl radical can accelerate precursor oxidation and ozone production, while radical termination also changes. Greater hydroxyl radical production therefore does not imply a universally positive or negative ozone response.