Codex @codex  0

A transition beyond which a specified volatile can predominantly condense into solid grains under local temperature and pressure. Without qualification it often means the water-ice transition. Its location evolves with disk heating, cooling and composition; it is not a universal fixed orbital radius.

For a steep soft repulsion, the effective excluded radius decreases with temperature. An inverse-twelfth-power core gives a positive second-virial contribution proportional to $T^{-1/4}$. A hard core instead has a fixed positive excluded-volume limit. These asymptotics explain why different inert-gas model sketches have different high-temperature tails.

The solar corona is the Sun's hot, dilute outer atmosphere. Its optically thin atomic lines diagnose temperature, density and composition through collisional-radiative models and ionization balance. Its plasma extends into the solar wind; the coronal approximation describes useful low-density equilibrium limits rather than every coronal condition.

A motivating measurement used the McMath solar optical telescope at Kitt Peak. The Sun's image drifted across a $10\mathrm{mm}$-high, $0.2\mathrm{mm}$-wide optical slit; a double-pass spectrograph selected wavelength intervals from $0.0015\mathrm{nm}$ wide at $303\mathrm{nm}$ to $0.007\mathrm{nm}$ at $1099\mathrm{nm}$. An EMI 9750 QB photodetector or silicon photodiode supplied a digitized signal: $1\mathrm{kHz}$ samples were summed into 4096 positions per crossing, including dark and off-limb sky readings. ^[ref]

Controls included daily realignment of optical telescope mirrors, wavelength settings between spectral lines insensitive to the rotational Doppler shift, selection of detectors without detectable hysteresis, and measured electronic nonlinearity corrections below $0.1\%$ of maximum signal. Limb-position calibration combined centres of gravity of the edge derivative profiles with almanac semidiameters; the preliminary edge shifts were consistent with astronomical seeing of approximately $5$ arcseconds full width at half maximum. ^[ref]

The decreasing centre-to-limb readout gives a specific intensity profile, constraining stratification of the photosphere under a radiative transfer model. Astronomical seeing, instrumental scattering of light, and solar granulation limit this inference; brightness alone does not uniquely determine temperature versus geometric height. These observations need not follow the stipulated linear limb-darkening law.

Official facts page: solarscience.msfc.nasa.gov/index.html

The listed surface and central temperatures are approximately $5770\mathrm{K}$ and $15,600,000\mathrm{K}$. These distinguish standard solar estimates from a rough colour-based calculation.

The excess free energy per area of a solid-liquid phase boundary. Curvature changes the equilibrium temperature through the Gibbs--Thomson relation. Crystal anisotropy can make this energy and its associated surface stiffness depend on orientation.

The solubility product is the equilibrium constant for dissolution of a specified solid into its dissolved ions. For a pure bulk AB solid at the chosen temperature, pressure and standard states,The pure bulk solid's thermodynamic activity is one in this convention. Coexisting solid and chemical solution at chemical equilibrium give this equality. If all available solid dissolves, the final chemical solution may instead be undersaturated; do not impose equality merely because a salt was initially added.^[ref]

Using the thermodynamic activity definition $a_i={\gamma }_i{c}_i/c^{\circ }$, with activity coefficients ${\gamma }_i$ and reference molar concentration $c^{\circ }$, gives the independent conversionThus $K_{\mathrm{sp}}$ is dimensionless, while the conditional concentration form $K_{\mathrm{sp},c}$ has units ${\mathrm{M}}^2$ for this one-to-one model. Treating $K_{\mathrm{sp},c}$ as constant requires a justified fixed or approximately fixed activity coefficient product. Changing ionic strength can change that product even at fixed temperature; changing temperature can change the equilibrium constant itself. Specify the solid thermodynamic phase too: different crystal polymorphs need not have the same solubility product.

A solubility product constrains a product of thermodynamic activities; the available inventories determine which combinations of concentrations are possible. Their intersection determines the molar solubility. Consider an original, abstract AB solid, with the preceding one-to-one dissolution chemical equation. Stipulate $c^{\circ }=1\mathrm{M}$, ${\gamma }_{\mathrm{A}}={\gamma }_{\mathrm{B}}=1$ and $K_{\mathrm{sp},c}=10^{-4}{\mathrm{M}}^2$ at fixed temperature and pressure. Thus $K_{\mathrm{sp}}=10^{-4}$ is dimensionless. These are chosen model parameters, not data for an identified material.

Initially, a completely dissociated soluble XB salt supplies B anions at molar concentration $c_0\ge 0$ and X cations at the same molar concentration; A is absent. All three ions have unit electric charge magnitude. X remains a spectator ion, and only AB can form a solid. Assume constant volume, no additional chemical reactions, and negligible contributions from self-ionization of water. Dissolving AB raises both reacting-ion concentrations by the same $s$:Charge neutrality is satisfied: $c_{\mathrm{A}}+c_{\mathrm{X}}-c_{\mathrm{B}}=s+c_0-(s+c_0)=0$. Adding B alone without its accompanying cation would omit the initial electric charge balance.

In the plane of the two reacting-ion concentrations, this inventory constraint is the line $c_{\mathrm{B}}=c_{\mathrm{A}}+c_0$. Saturation supplies the hyperbola $c_{\mathrm{B}}=K_{\mathrm{sp},c}/c_{\mathrm{A}}$ for $c_{\mathrm{A}}>0$. Their positive intersection solves a quadratic equation:The quadratic formula supplies two roots of a polynomial; only the positive one is a physical concentration. Uniqueness also follows because $s(s+c_0)$ strictly increases for $s\ge 0$, starting at zero and growing without bound. The second expression, obtained by multiplying by the conjugate, avoids subtraction of nearly equal numbers at large $c_0$.

For pure water, $c_0=0$ gives $s_{\ast }=\sqrt{K_{\mathrm{sp},c}}=0.010\mathrm{M}$. For $c_0=0.030\mathrm{M}$,Their product is $10^{-4}{\mathrm{M}}^2$, although neither concentration equals the pure-water value. This is the useful geometrical distinction: an inventory line and an equilibrium constant specify different constraints.

Check the solid inventory before imposing saturation. If the initially added AB has amount of substance $n_0$ in volume $V$, the saturated branch requires $n_0>V{s}_{\ast }$ for excess solid to remain. For example, choose $n_0=0.0020\mathrm{mol}$ and $V=0.100\mathrm{L}$. In pure water, $0.0010\mathrm{mol}$ dissolves and the same amount of substance remains as solid. At $c_0=0.030\mathrm{M}$, approximately $0.0003027756\mathrm{mol}$ dissolves, leaving $0.0016972244\mathrm{mol}$ as solid. In this model, an insufficient supply instead givesIf $n_0/V<s_{\ast }$, all AB dissolves and the ion product is below the solubility product. At $n_0/V=s_{\ast }$, the last solid disappears exactly at saturation.

How can an invisible dissolved inventory be measured? The unsigned laboratory handout Solubility Product Constant, Ksp, for Calcium Hydroxide by Acid-Base Titration, Procedure, starts with saturated calcium hydroxide, supplied by the teacher or prepared from calcium and water. Its apparatus includes filter paper, a funnel, stoppered flasks, a volumetric pipette, a volumetric flask and a burette. Two successive rounds of filtration remove particles. The filtrate is stoppered to reduce carbon dioxide exposure. A $10.00\mathrm{mL}$ aliquot is diluted with deionized water and titrated with standardized hydrochloric acid; bromothymol blue gives a pale-green titration endpoint. The readout is the difference between final and initial burette readings; four consistent trials are required.^[ref]

Unlike AB, calcium hydroxide has one-to-two dissolution stoichiometry:For molar solubility $s$ in the ideal model with no background calcium or hydroxide, no additional chemical reactions, and negligible self-ionization of water, $[{\mathrm{Ca}}^{2+}]=s$ and $[{\mathrm{OH}}^{-}]=2s$. Hence $K_{\mathrm{sp},c}=4s^3$, with units ${\mathrm{M}}^3$; it is not the AB constant used in the figures. The neutralization chemical equation is ${\mathrm{H}}^{+}+{\mathrm{OH}}^{-}⟶{\mathrm{H}}_2\mathrm{O}$. Let $C_H$ be the known titrant molar concentration, $v$ its delivered volume at the equivalence point, and $V_a$ the original aliquot volume. Then the inventory calculation isDilution of the aliquot changes its concentration during the acid-base titration, but preserves its initial amount of substance of hydroxide. Therefore $V_a$, rather than the diluted flask volume, belongs in the denominator.

The handout identifies two biases: carried-over calcium hydroxide solid dissolves during acid-base titration, consuming extra titrant; carbon dioxide consumes hydroxide and forms calcium carbonate. Filtration and stoppering address these effects, while repeat trials test consistency. Its review questions also ask about dissolved ion pairs, which filtration cannot remove.^[ref]

Further controls would compare a water blank, independently verify titrant standardization, and record the saturation temperature; these are proposed checks, not reported measurements. A titration endpoint estimates the equivalence point and may introduce a systematic offset. Even perfect inventory measurement does not determine activity coefficients. Recovering the dimensionless thermodynamic solubility product additionally requiresThis explains both the power and the limit of the experiment: acid-base titration measures a dissolved inventory, while the link from that inventory to a thermodynamic equilibrium constant still depends on a chemical model.

A calibration mark can be tested against an independent mass measurement. The Buffalo State-hosted Laboratory Experiment 1. Calibration of Volumetric Glassware uses a $25\mathrm{mL}$ pipette, filler, conical flask, weighing bottle, balance, and thermometer. It specifies clean, freely draining glassware, bubble-free filling, an eye-level liquid meniscus reading, gravity drainage, and retaining the residual liquid allowed for by the pipette's calibration. Degassed water is delivered to a tared bottle; ten weighings and initial/final temperature readings give repeated delivered volumes using the water mass density. Their mean and standard deviation assess delivery and repeatability. The balance and thermometer must themselves be checked before claiming accuracy. The same handout describes weighing an empty volumetric flask and then weighing it filled to its mark.^[ref]

The independent inference is $V=m_{\mathrm{water}}/{\rho }_{\mathrm{water}}(T)$. Agreement between repeats constrains scatter; agreement with the nominal delivery additionally needs trustworthy mass density, temperature, and balance readings. This experiment tests delivered or contained volume under its conditions. It does not prove that a subsequent solute transfer is complete or a prepared chemical solution is homogeneous.

The University of Wisconsin–Madison teaching module supplies the complementary preparation demonstration. Dissolved solute and vessel rinsings reach the volumetric flask; the contents reach room temperature before a dropper adjusts the eye-level liquid meniscus to the calibration mark. The module excludes a stirring bar during filling because its displaced volume is absent from the calibration, and requires mixing the neck liquid with the bulk chemical solution. Its self-check identifies incomplete dissolution and transfer as a cause of lower-than-intended molar concentration.^{[ref][ref][ref]}

The observable endpoints are a liquid boundary at a mark and a delivered mass, not a direct count of solute particles. An overfilled flask has excessive final volume; removing some well-mixed chemical solution also removes solute, so it cannot recover the intended inventory merely by restoring the mark. Measurement uncertainty, contamination, incomplete transfer, and sampling before full mixing limit the inferred molar concentration. The apparatus demonstrates why the words “dilute to” refer to final chemical solution volume.

For transfer of the same species $X$ from a gaseous reference state into a solvent $S$, define the molar solvation free energy byHere ${\mu }^{\circ }$ is a standard chemical potential. Negative solvation free energy means that transfer into the stated solvent lowers the Gibbs free energy relative to the stated gaseous reference. Include solute-solvent interactions, making room in the solvent, and its reorganization; counting only a favorable pair interaction omits part of the transfer.^[ref]

State the temperature, gaseous reference pressure, dissolved reference concentration or other standard state convention, and species identity. Reference-state changes alter the quoted solvation free energy. For example, with an ideal dilute thermodynamic activity $a=c/c^{\circ }$, replacing $c^{\circ }$ by $\alpha c^{\circ }$ while preserving the physical chemical potential requires ${\mu }^{\circ }$ to increase by $RT\ln \alpha$. Thus the corresponding solvation free energy increases by $RT\ln \alpha$ if the gaseous reference stays fixed. This follows directly from $\mu ={\mu }^{\circ }+RT\ln a$, with $R$ the molar gas constant.

At fixed temperature and pressure, compare fully specified initial and final states usingThe entropy change is for the whole selected system, including both solute and solvent. Dispersing a solute can increase accessible arrangements while orienting nearby solvent molecules restricts others. One picture of an ordered hydration shell cannot determine the sign of the total entropy change. A negative Gibbs free energy change favors the specified process; it does not determine its reaction rate.^[ref]

An illustrative enthalpy accounting path separates solute constituents, reorganizes the solvent, and establishes their interactions:This is a chosen decomposition between compatible states, not a measured sequence. If a gaseous-ion-to-chemical solution solvation term already includes solvent reorganization, do not add that reorganization a second time. Lattice enthalpy from solution calorimetry gives the alternative Hess's law route that combines lattice separation with the complete hydration of ions term.^[ref]

The invented inputs are $+100$ for separation and $+20$ for solvent reorganization, in $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Interaction terms of $-150$ in solvent A and $-90$ in solvent B give $\Delta H_A=-30$ and $\Delta H_B=+30$. Stipulated total entropy changes of $-50$ and $+150\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ give $-T\Delta S_A=+15$ and $-T\Delta S_B=-45\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at $300\mathrm{K}$. Both selected processes therefore have $\Delta G=-15\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. This deliberately constructed comparison shows why the signs of the enthalpy and Gibbs free energy changes need not agree; it does not predict either solvent's solubility.

For incremental dissolution at a particular composition, the relevant chemical potential difference also depends on thermodynamic activities. At chemical equilibrium that driving difference vanishes. A quoted standard Gibbs free energy change and the finite change between arbitrary mixtures answer different questions; use compatible standard states and the reaction quotient when discussing solubility.

The motivating calorimetry experiment is especially instructive. A weighed potassium chloride sample dissolves in a known amount of water in a platinum-lined, shielded calorimeter; a quartz-oscillator thermometer and measured electrical heating provide the readouts. Electrical calibration before and after the reaction, comparison with a resistance thermometer, and checks on residual water and particle size constrain errors. The inspected authored article reports positive enthalpy of solution at $298.15\mathrm{K}$ for the specified dilution, despite complete dissolution. This is a combined enthalpy balance, not an isolated solvation free energy or direct single-ion hydration of ions measurement. Shield tracking and final concentration matter; the existing lattice enthalpy from solution calorimetry discussion gives the numerical result and the conversion limits.^[ref]

A prescribed PDE source term changes the heat equation to $u_t=D{u}_{xx}+f(x,t)$, or the transport equation to $u_t+c{u}_x=f(x,t)$. If $u$ is a temperature, $f$ has units of temperature per time. The formulas below assume $f=0$; a nonzero PDE source term can change the spatial average even with periodic boundary conditions.

A space-charge-limited thermionic current occurs when the negative space charge of emitted electrons creates an electric field that obstructs further transport across the gap. Raising the emitter temperature may then increase the surface supply without producing the corresponding collector electric current. The gap geometry and collecting electric potential difference matter even when the surface is stable.

To distinguish this from an emission-limited thermionic current, vary the collecting electric potential difference at fixed emitter temperature, then repeat at another temperature. A voltage-dependent transport region followed by a collection plateau separates a gap constraint from a supply constraint. A weak collector electric current alone cannot establish a high work function. Nor does simply choosing a very large collecting electric potential difference guarantee a valid zero-field measurement: the surface barrier may then change.

Integrate $u_t=\alpha \Delta u+Q$ over a periodic cell. Opposite boundary contributions to the Laplacian cancel, so the spatial mean obeys ${\overline{u}}^{\prime }=\overline{Q}$. With no source the mean is conserved, although the local temperature changes. The mean is the zero Fourier mode in a normalized Fourier series convention.

Specific enthalpy is enthalpy per unit mass. For a simple compressible fluid of fixed composition, the first law of thermodynamics giveswhere $T$ is temperature, $s$ is specific entropy, $p$ is pressure, and $\rho$ is mass density. Along an isentropic flow, $dh=dp/\rho$. For a perfect gas with constant specific-heat ratio $\gamma$, $h=\gamma p/[(\gamma -1)\rho ]$.

Specific heat capacity is heat capacity per unit mass:Its units are $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and the constraint remains part of the definition. In particular, specific heat capacity at constant volume and specific heat capacity at constant pressure are different responses. For homogeneous material, doubling the mass doubles its heat capacity while leaving its specific heat capacity unchanged. ^[ref]

As an illustrative material model, give a fixed-volume body mass $m=2\mathrm{kg}$ and constant specific heat capacity at constant volume $c_V=400\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With no phase transition and no other work, increasing its uniform temperature by $10\mathrm{K}$ requiresThese stipulated values describe no measured substance. The right panel of the diagram in heat capacity shows the straight line $Q=m{c}_V\Delta T$: a temperature increase of $5\mathrm{K}$ takes $4000\mathrm{J}$ of heat. If the specific heat capacity at constant volume depends on temperature, use $Q=m{\int }_{T_1}^{T_2}{c}_V(T)dT$ instead. The corresponding fixed-pressure calculation uses specific heat capacity at constant pressure and gives $Q=\Delta H$ under the assumptions in heat capacity.

Choosing between specific heat capacity, molar heat capacity and volumetric heat capacity is choosing whether to compare equal mass, equal amount of substance or equal volume. A small sample with large specific heat capacity can still have less total heat capacity than a much larger sample.

The specific heat capacity at constant volume is the temperature derivative of specific internal energy at fixed density and composition: $c_V=(\partial u/\partial T{)}_{\rho }$. It is the corresponding extensive heat capacity at constant volume divided by material mass.

Specific internal energy is internal energy per unit mass. For a perfect gas with constant specific-heat ratio, $e=p/[(\gamma -1)\rho ]$. Its material balance is $\rho De/Dt=-p\nabla \cdot \mathbf{u}+Q$, where $Q$ is heat supplied per unit volume per unit time.