Advanced Chemistry - Mono Mole

June 28, 2025

Temperature-composition diagrams for ideal and non-ideal solutions

Temperature-composition diagrams are graphical representations that show the relationship between the boiling point (temperature) of a liquid mixture and the composition of the mixture at a constant pressure.

Ideal solution

Consider two liquids A and B (e.g. benzene and toluene) forming an ideal solution in a closed container at a constant pressure $p_T$ . The partial pressures of the components follow Raoult’s law:

$p_A=x_{A,l}p_A^0(T)\;\;\;\;\;\;\;\;p_B=x_{B,l}p_B^0(T)$

where

$p_i$ is the partial vapour pressure of component $i$
$p_i^0(T)$ is the vapour pressure of pure $i$ as a function of $T$
$x_{i,l}$ is the mole fraction of $i$ in the solution

The total vapour pressure of the ideal solution is equal to the fixed pressure: $p_T=x_{A,l}p_A^0(T)+(1-x_{A,l})p_B^0(T)$ , which rearranges to

$x_{A,l}=\frac{p_T-p_B^0(T)}{p_A^0(T)-p_B^0(T)}\;\;\;\;\;\;\;\;201$

Using known values of $p_i^0(T)$ at different $T$ , eq201 is used to plot the liquid composition curve for a graph of $T$ against $x_{A,l}$ (see diagram below). Substituting eq201 into $x_{A,g}=\frac{p_A}{p_T}=\frac{X_{A,l}\;p_A^0(T)}{p_T}$ yields

$x_{A,g}=\frac{p_T-p_B^0(T)}{p_A^0(T)-p_B^0(T)}\frac{p_A^0(T)}{p_T}\;\;\;\;\;\;\;\;202$

Eq202 describes the vapour composition curve. Unlike the pressure-composition diagram of an ideal solution, the vapour composition curve lies above the liquid composition curve (also known as the boiling point curve). The vapour phase corresponds to the region above the vapour composition curve, while the liquid phase lies below the liquid composition curve. Furthermore, neither curve is linear.

Question

Why is the liquid composition curve also called the boiling point curve?

Answer

At a fixed pressure $p$ , boiling in a temperature–composition diagram refers to the phase change from liquid to vapour that occurs when the mixture’s vapour pressure equals or exceeds $p$ . The liquid composition curve defines the boundary at which this phase change begins.

Similar to pressure-composition diagrams, the horizontal axis, after plotting the two curves, is relabelled as $x_A$ , representing the overall mole fraction of A in the system, including both the liquid and vapour phases. Consider the process of isobarically increasing the temperature from point a to i (see diagram above). At point a, the system is entirely in the liquid phase. As the temperature is increased to point b, the system reaches the liquid composition curve. Here, the first infinitesimal amount of vapour forms, and the system enters a two-phase equilibrium. At this point, the liquid composition is still equal to the overall composition ( $x_{A,l}=x_{A3}$ ), but the vapour phase has a different composition corresponding to point f ( $x_{A,g}=x_{A5}$ ). A tie-line bf on the diagram connects these two phase compositions, indicating that both phases coexist in equilibrium. The lever rule, as discussed in the article on pressure-composition diagrams, can likewise be used to determine the relative equilibrium amounts of the two phases present at any stage of heating, provided the temperature and overall composition fall between the liquid and vapour lines.

Continuing to increase the temperature brings the system to point e, which lies between the two curves. This is the two-phase region, where both liquid and vapour coexist in significant amounts. The overall composition remains fixed at $x_{A3}$ , but the compositions of the individual phases are $x_{A,l}=x_{A2}$ (point c) and $x_{A,g}=x_{A4}$ (point g) for the liquid phase and vapour phase respectively.

Question

Why is the overall composition of the system fixed at $x_{A3}$ when temperature is increased?

Answer

$x_{A3}$ is the overall mole fraction of A in the system, including both the liquid and vapour phases. In the closed container, the total number of moles of component A and B remains constant — just distributed differently between liquid and vapour as temperature increases.

When the system reaches point h on the vapour composition curve, the last drop of liquid evaporates. At this point, the liquid phase composition is ( $x_{A,l}=x_{A1}$ ), and the vapour phase composition matches the overall composition ( $x_{A,g}=x_{A3}$ ). Finally, as the temperature is increased even further to point i, the system exists as a single-phase vapour. No liquid remains and the composition remains constant at $x_{A,g}=x_{A3}$ . The line abehi is called an isopleth.

Question

Why does the mixture boil at a single temperature, instead of the more volatile component boiling off completely before the other begins to boil?

Answer

In a mixture, boiling occurs when the total vapour pressure — the sum of the partial vapour pressures of all volatile components — equals the external pressure. This means the mixture doesn’t boil when each component individually reaches its own boiling point, but rather when their combined vapour pressures are sufficient to match the external pressure. Both components vaporise simultaneously, but in proportions that depend on their relative volatilities. For example, although pure benzene boils at a lower temperature (80 °C at 1 atm) than pure toluene (111 °C at 1 atm), benzene’s vapour pressure in a mixture is reduced due to dilution by toluene, as described by Raoult’s Law. Benzene alone cannot reach the external pressure unless it is pure — and the same applies to toluene. Only together can their partial vapour pressures sum to equal the external pressure, allowing the mixture to boil.

Non-ideal solution

Temperature-composition diagrams for non-ideal binary solutions typically exhibit two characteristic shapes, depending on the degree and nature of deviation from Raoult’s law. When the deviation is small, the diagram features two smooth monotonic curves intersecting only at $x_{A}=0$ and $x_{A}=1$ (see diagram below). A typical example is the diethyl ether–ethanol system. Such a diagram has the same structure as that of an ideal solution, but differs in the curvature of the curves.

In contrast, when the deviation from Raoult’s law is significant, the curves may develop a stationary point. The nature of this point is governed by the intermolecular interactions between components A and B. When A-B interactions are weaker than A-A and B-B interactions, a minimum point may occur, as the mixture exhibits higher vapour pressures than predicted, resulting in lower boiling points. Conversely, if A-B interactions are stronger, a maximum point may form due to lower vapour pressures, which lead to higher boiling points than expected. Mixtures with such characteristics are called azeotropes. The ethanol–water system is an example of an azeotrope with a minimum point, while the HCl–water system has a maximum point (see diagram below). These diagrams are typically constructed using empirical data.

Question

Why must the two curves of an azeotrope intersect at the stationary point?

Answer

See this article for explanation.

Distillation

Temperature-composition diagrams are essential tools in understanding and designing simple and fractional distillation processes.

Consider the ethanol–water system (see above diagram). In simple distillation, the mixture $(x_{A1})$ is heated until it reaches its boiling point $(T_{1})$ and begins to vaporise. At this stage, the vapour composition of ethanol is $x_{A2}$ (point b), which is richer in ethanol. This vapour is then condensed and collected as the distillate, effectively separating it from the original mixture. Since this process involves just a single equilibrium between the liquid and vapour phases, it is typically considered a one-step separation. By using the diagram, one can estimate how effective the process will be for a single distillation step. However, simple distillation has its limitations. It provides only a partial enrichment of the more volatile component. For mixtures requiring higher purity, more advanced techniques like fractional distillation are necessary.

The fractional distillation apparatus used in the laboratory includes a fractionating column filled with glass beads, which provide a large surface area for repeated condensation and vaporisation cycles (see diagram above). Suppose the initial overall mole fraction of ethanol in the mixture is $x_{A1}$ (with reference to the minimum azeotrope diagram above). When the temperature reaches $T_{1}$ , the vapour in equilibrium with the liquid has an ethanol composition of $x_{A2}$ , which is richer in ethanol due to its greater volatility. As this vapour rises through the fractionating column, it encounters a cooler region at temperature $T_{2}$ , where part of it condenses into a liquid with the same composition $x_{A2}$ . The remaining uncondensed vapour becomes even more enriched in ethanol, now with composition $x_{A3}$ . This ethanol-riched vapour continues to ascend, repeating the cycle: it partially condenses into liquid of composition $x_{A3}$ , and leaves behind a vapour of even higher ethanol content, $x_{A4}$ .

The condensation and vaporisation cycle continues as long as the column is tall enough to provide sufficient equilibrium stages. Eventually, the vapour composition reaches $x_{A5}$ , which corresponds to the azeotropic point — the minimum boiling point of the ethanol–water system. At this composition, the mixture behaves like a pure substance and boils at a constant temperature, $T_{4}$ . No further separation by distillation is possible beyond this point. Each individual equilibrium step in the fractional distillation process is referred to as a theoretical plate. For an ethanol-water mixture to reach azeotropic purity, typically 8 to 10 theoretical plates are needed. This can be achieved in the lab using a well-packed fractionating column of 50 to 70 cm in height.

At the top of the column, the azeotropic vapour is directed into the condenser, while the condensed liquid within the column flows back down towards the boiling flask, diluting the boiling mixture. The final composition of the distilled ethanol depends on this azeotropic limit, which is approximately 95.6% ethanol by mass at atmospheric pressure ( $x_{A}=0.895$ ).

Question

What happens if the starting mixture, containing a 0.92 mole fraction of ethanol, is heating to a temperature between $T_{A}^0$ and $T_{4}$ ?

Answer

The final composition of the distillate contains 0.895 mole fraction of ethanol, while the residue will be richer in ethanol ( $x_{A} > 0.92$ ).

Temperature-composition diagrams are essential tools in understanding and optimising fractional distillation, particularly in complex mixtures like petroleum. In an industrial fractional distillation column (see above diagram), crude oil is first heated in a furnace until it partially vaporises. The resulting vapour enters the base of the column (reboiler section) and rises through a series of trays or packing. Components with lower boiling points condense on trays higher up in the column (where temperatures are cooler), while components with higher boiling points condense on trays lower down (where it is hotter).

Although temperature–composition diagrams are typically illustrated for binary mixtures, the same principles can be conceptually applied to the multi-component mixture of crude oil. At different heights in the column, the mixture can be approximated as a “pseudo-binary” system, where one component (A) represents a heavier fraction and the other component (B) represents a lighter fraction. At a given temperature (i.e. at a particular tray), the vapour phase will be richer in the more volatile component B (lower boiling point), while the liquid phase will be richer in component A (higher boiling point). This vapour–liquid equilibrium process repeats over many stages, progressively enriching the vapour in more volatile components and enabling effective separation based on differences in volatility.

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June 28, 2025

Liquid-liquid phase diagrams

Liquid-liquid phase diagrams are graphical representations that show the conditions under which two (or more) partially or fully immiscible liquids coexist in equilibrium. The most common type of liquid-liquid phase diagram is the temperature-composition diagram.

A typical temperature-composition diagram of two partially miscible liquids features a binodal curve (with two nodes at the ends of a tie-line) that divides the diagram into two regions (see diagram below). The region outside the curve corresponds to a single liquid phase, while the area enclosed by the curve represents a two-phase region, where the liquids separate into two immiscible layers

To interpret this diagram, consider starting at point e, which represents pure liquid A (hexane) at a fixed temperature $T_{1}$ and 1 atm. As small amounts of liquid B (nitrobenzene) are gradually added, it dissolves completely in A to form a single-phase solution between e and f. Upon further addition of B, the system reaches the limit of mutual solubility, and phase separation occurs (between points f and g, the system exists as two liquid phases). Along the tie line connecting f and g, each end corresponds to the composition of one of the two coexisting phases. A mixture composition closer to f indicates that the phase richer in A is more abundant, while the phase richer in B (composition at g) is less abundant. The opposite holds for mixtures closer to g. The lever rule can be applied to determine the relative amounts of each phase. At point g, enough B has been added to dissolve all of A, resulting in a saturated single-phase solution of A in B. Further addition of B beyond this point leads to continuous dilution of A, and the system remains in the single-phase region until pure B is reached at point h, where $x_{B}=1$ .

A liquid-liquid phase diagram like the one above is constructed with empirical data. It is interesting to note that the compositions of the two phases at equilibrium for the hexane-nitrobenzene system vary with temperature. The narrowing of the curve at higher temperatures indicates an increasing miscibility of the two components as temperature rises. The upper critical solution temperature $T_{uc}$ is the temperature above which the system exists as a single phase, with the two liquids miscible in all proportions.

Some binary liquid systems exhibit greater miscibility as temperature decreases (see diagram below). An example is the water–triethylamine system, in which the two components form a weak complex at lower temperatures. In such cases, a lower critical solution temperature $T_{lc}$ marks the temperature below which the system becomes fully miscible and exists as a single phase.

Occasionally, a system displays a combination of both behaviours, exhibiting both an upper and a lower critical solution temperature (see diagram above). In such systems, the components are completely miscible at both high and low temperatures, but become partially miscible over an intermediate temperature range. This results in a closed-loop miscibility gap on the temperature–composition diagram, bounded above by the upper critical solution temperature and below by the lower critical solution temperature. The unusual phase behaviour is typically due to competing molecular interactions, such as complex formation at low temperatures and thermal disruption at higher temperatures. Examples include nicotine–water and m-toluidine–glycerol systems.

Liquid–liquid diagrams are commonly used in extractive distillation processes, where the goal is to exploit both vapour–liquid equilibrium and liquid–liquid immiscibility to achieve better separation. The diagram above shows a mixture of two partially miscible liquids that form a low-boiling azeotrope. An example is the water–isobutanol system. If a water–isobutanol mixture of composition at point a is heated from $T_{2}$ to $T_{3}$ , the resulting vapour, upon cooling to $T_{1}$ will condense into a distillate consisting of two phases: one with composition $x_{B1}$ and the other $x_{B2}$ .

Beyond separation, liquid–liquid diagrams also play a crucial role in other scientific fields. In polymer blends and materials engineering, they help predict phase compatibility and guide the design of homogeneous or phase-separated materials with tailored properties. In food science, liquid–liquid behaviour underpins the formation and stabilisation of oil–water emulsions, which are essential in products like dressings, creams and sauces. Similarly, in environmental science, such diagrams aid in understanding the partitioning of pollutants between aqueous and organic phases, which is key to modelling contaminant transport and designing remediation strategies.

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Solid-liquid phase diagrams

Solid-liquid phase diagrams illustrate the relationship between temperature and composition in mixtures as they transition between solid and liquid states. These diagrams are essential for understanding melting behavior, phase equilibrium and solubility in binary or multi-component systems. They typically feature regions representing pure solid, pure liquid, and a mixture of both, separated by boundary lines such as the liquidus (liquid composition curve) and solidus (solid composition curve).

Ideal liquid solution and ideal solid solution

For instance, the silicon-germanium solid-liquid temperature-composition diagram resembles the liquid–vapour diagram of an ideal liquid solution (see above diagram). This similarity arises because the two elements form both an ideal solid solution and an ideal liquid solution. The region below the solidus corresponds to a single-phase solid solution, while the area above the liquidus represents a single-phase liquid solution. Between these two curves, the solid and liquid phases coexist in equilibrium, and the lever rule can be applied to determine the relative amounts of each phase at a given temperature.

Question

What is a solid solution?

Answer

A solid solution is a homogeneous mixture of two or more substances in the solid state, where one substance (the solute) is dissolved in another (the solvent) to form a single, uniform phase with a consistent crystal structure. Solute atoms either replace solvent atoms within the crystal lattice (substitutional solid solution) or occupy the spaces between them (interstitial solid solution). Essentially, it’s a “solid solution” of metals, often referred to as an alloy. For example, brass is a solid solution of copper (solvent) and zinc (solute).

Eutectics

The temperature-composition diagram of the gold-copper system is shown in the diagram below. In this system, gold and copper are completely miscible in both the liquid and solid phases, but they do not form an ideal solid solution due to differences in atomic interactions. The diagram features a eutectic point and resembles a minimum azeotrope in shape. The eutectic point corresponds to the lowest melting temperature of the alloy, occurring at a specific composition known as the eutectic composition. At this point, the homogeneous liquid solidifies into a single solid phase with a well-defined composition. Because the eutectic temperature is lower than the melting points of either pure component, this point is particularly useful in applications such as casting and welding. The term “eutectic” was coined in 1884 by British physicist and chemist Frederick Guthrie, derived from the Greek words eu (“well” or “good”) and têxis (“melting”), meaning “easily melted.”

Unlike the gold-copper system, the silver-copper system features complete miscibility of its components in the liquid phase but only partial miscibility in the solid phase. Based on the gold-copper phase diagram, we would expect the solid solution region to divide into multiple regions due to the limited solid-state solubility of silver and copper at lower temperatures (see diagram below).

Indeed, the temperature-composition diagram of the silver-copper system includes a single-phase liquid solution region ( $ls$ ), two single-phase solid solution regions ( $ss_{\alpha}$ and $ss_{\beta}$ ), a two-phase solid region ( $\alpha+\beta$ ), and two two-phase regions where a liquid solution is in equilibrium with a solid solution ( $ls+ss_{\alpha}$ and $ls+ss_{\beta}$ ). The eutectic point k, at 779^oC, represents the equilibrium state where the solid solutions $ss_{\alpha}$ and $ss_{\beta}$ coexist with the liquid phase .

When a liquid solution with composition at point d is cooled, the solid solution phase $ss_{\alpha}$ begins to precipitate at point e, with its composition corresponding to point h on the tie-line. As the two-phase mixtrure of $ls+ss_{\alpha}$ cools to point f, it reaches the eutectic composition. At this stage, the solid solution $ss_{\beta}$ also begins to form.

The Ag–Cu phase diagram is a powerful tool for understanding, designing and processing silver–copper alloys. It helps engineers to select the right alloy composition for specific properties (e.g. ductility and conductivity), and guide thermal treatments for practical applications in electronics, jewelry and joining technologies.

Compound formation (congruent melting)

Consider two substances, A and B, that react to produce a compound C in a 1:1 ratio. If C separately forms eutectic mixtures with both A and B, the solid-liquid phase diagram of A and B (e.g. aniline and phenol) will appear as follows:

Aniline $C_6H_5NH_2$ and phenol $C_6H_5OH$ combine to form an adduct $C_6H_5OH\cdot C_6H_5NH_2$ , which is a stable complex with its own characteristic melting point and crystal structure distinct from the individual components. Aniline and phenol are miscible in the liquid phase at higher temperatures, while the adduct does not exist in the liquid state. Furthermore, the solid forms of all three species are completely insoluble in one another. As a result, solid phenol coexists with the solid adduct in a two-phase region between $0\leq x_{aniline}\leq 0.5$ at lower temperatures. When $x_{aniline}\geq 0.5$ , all phenol molecules will react with aniline, leaving a two-phase region consisting of solid adduct and solid aniline. In other words, the maximum amount of adduct is formed when $x_{aniline}=0.5$ . Since the adduct forms eutectic mixtures with both aniline and phenol, the phase diagram can be viewed as two eutectic phase diagrams positioned side by side. Points a and b correspond to the eutectic points of the phenol-adduct and the adduct-aniline mixtures respectively.

Consider the isopleth defg. As the mixture cools from point e to point g, negligible amounts of phenol will be present, meaning that the tie-lines extend only from $x_{aniline}=0.5$ onwards. On the other hand, if a solution with $x_{aniline}=0.5$ is cooled, only pure solid adduct will separate out, without any change in composition. If the process is reversed, the liquid formed will have the same composition as the solid. This phenomenon is known as congruent melting.

In pharmaceutical and chemical industries, the diagram can help in purification processes or the synthesis of the adduct of aniline and phenol, which is a valuable intermediate in the production of various compounds, such as dyes, drugs, and plastics

Compound formation (incongruent melting)

Some stable solids of the form A₂B, each melts to give a liquid with a different composition. We call such a process incongruent melting. The phase diagram of an A₂B alloy resembles that of the aniline-phenol system, except that the melting point of one component is much higher the other (see diagram below).

The vertical line at $x_{A}=0.67$ in the phase diagram marks the composition of the pure solid compound A₂B. When the alloy at $x_{A}=0.67$ is heated, it begins to melt at temperature $T_{1}$ , producing pure solid A and a single-phase liquid containing both A and B. The resulting liquid has a mole fraction of A ( $x_{A'}$ ) that differs from that of A₂B.

An example of such a system is the Na₂K alloy, which is used in liquid metal coolants and metallic lubricants. Understanding the melting behavior of Na₂K is important for selecting suitable compositions in these applications. Although the phase diagram shows the presence of Na₂K, it is still a binary system diagram for the Na-K system. The compound Na₂K is simply an intermetallic phase that appears as part of the equilibrium between Na and K. In contrast, ternary phase diagrams involve three independent components.

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Ternary systems

A ternary system refers to a mixture composed of three components. Its phase diagram requires at least three dimensions, typically representing either temperature $T$ or pressure $p$ , along with two independent mole fractions.

Question

Show that the height $h$ of an equilateral triangle is $h=\frac{\sqrt{3}}{2}s$ , where $s$ is the side length, and hence, $DE+DF+DG=h$ .

Answer

Using Pythagorean theorem, $\biggr$\frac{s}{2}\biggr$^2+h^2=s^2$ , which rearranges to $h=\frac{\sqrt{3}}{2}s$ . The area of the triangle is $\frac{1}{2}sh=\frac{\sqrt{3}}{4}s^2$ . Since $\Delta_{BCD}+\Delta_{BAD}+\Delta_{ACD}=\frac{\sqrt{3}}{4}s^2$ , we have $\frac{s}{2}(DE+DF+DG)=\frac{\sqrt{3}}{4}s^2$ , which rearranges to $DE+DF+DG=h$ .

Since $DE+DF+DG=h$ , the perpendicular distances DE, DF and DG becomes $x_C$ , $x_B$ and $x_A$ respectively if we set $h=1$ . This normalisation allows us to represent the ternary system in an equilateral triangle, assuming both $T$ and $p$ are held constant (see diagram above). For instance, the red dot in the diagram corresponds to mole fractions $x_A=0.4$ , $x_B=0.2$ and $x_C=0.4$ . In fact, once two independent mole fractions are specified, the third is automatically determined due to the constraint $x_A+x_B+x_C=1$ . This effectively reduces the degrees of freedom to two, allowing the system to be represented on a two-dimensional diagram.

The ternary phase diagram below represents the acetone-water-diethyl ether system at 30^oC and 1 atm. Under these conditions, water and diethyl ether are only partially miscible, while the other two binary pairs are fully miscible. The region under the curve consists of two liquid phases in equilibrium, whereas the region above the curve represents a single-phase liquid.

Since temperature and pressure are constant, tie lines must remain in the plane of the diagram, but they are not required to be parallel. Their orientations are determined experimentally. The compositions of the two coexisting phases at a given point under the curve are found at the ends of the tie line that intersects the point. For example, point f lies within the two-phase region and corresponds to a water-rich, ether-poor phase ( $\alpha$ ) of composition e, and a water-poor, ether-rich phase ( $\beta$ ) of composition g. Moreover, point e is richer in acetone than point g. Finally, the lever rule also applies here: the relative amounts of the two phases are inversely proportional to the lengths of the tie line segments, i.e. $l_{ef}n_{\alpha}=l_{fg}n_{\beta}$ .

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Phase rule

The phase rule, formulated by Josiah Willard Gibbs, describes the number of degrees of freedom (independent variables) needed to define a multiphase, multicomponent system at equilibrium. It is given by

$F=C-P+2\;\;\;\;\;\;\;\;203$

where

$F$ is the number of degrees of freedom (i.e. independent intensive variables such as temperature, pressure or mole fraction). This value is a non-negative integer.
$C$ is the number of independent components that can describe the composition of a point in a phase diagram.
$P$ is the number of phases present in equilibrium.

For example, we need two independent mole fractions to describe a liquid mixture of three fully miscible components at constant temperature and pressure (single-phase ternary system) because $C=3$ and $P=1$ .

To derive eq203, assume that every chemical species is present in every phase of the system. Each phase has a composition defined by the mole fractions of $C$ components. Since mole fractions must sum to 1, each phase has $C-1$ independent composition variables. For $P$ phases, the total number of composition variables is $P(C-1)$ . Including temperature and pressure, the total number of variables is $P(C-1)+2$ .

However, not all of these variables are independent because the system is at equilibrium. At equilibrium, the chemical potential of each component must be equal in all phases. For example, for component 1,

$\mu_1^{\alpha}=\mu_1^{\beta}=\cdots=\mu_1^{p}$

This gives $P-1$ equations per component, resulting in a total of $(P-1)C$ independent equations. These equations represent constaints that reduce the number of independent variables because each chemical potential is a function of temperature, pressure and the mole fractions of all components in each phase, e.g. $\mu_1^{\alpha}=\mu_1^{\alpha}(T,p,x_1^{\alpha},x_2^{\alpha},\cdots,x_C^{\alpha})$ . If say, $\mu_1^{\alpha}=AT+Bp+Cx_1^{\alpha}+\cdots$ and $\mu_1^{\beta}=ET+Fp+Gx_1^{\beta}+\cdots$ , then one variable becomes dependent on the others when we equate $\mu_1^{\alpha}=\mu_1^{\beta}$ . Even if the functions are more complex, the value of any one variable remains dependent on the others.

Therefore, the number of independent variables needed to define a multiphase, multicomponent system at equilibrium is $F=P(C-1)+2-(P-1)C$ , which rearranges to eq203. In this derivation, we assumed that every chemical species is present in every phase of the system. What if one or more species is absent from one or more phases? Suppose species $i$ is absent from phase $\delta$ . In this case, the number of variables is reduced by one because $x_i^{\delta}$ is no longer a variable. At the same time, the number of independent chemical potential equations is also reduced by 1, e.g. $\mu_i^{\alpha}=\mu_i^{\beta}=\mu_i^{\delta}$ becomes $\mu_i^{\alpha}=\mu_i^{\beta}$ . Therefore, the phase rule still holds.

Question

Apply the phase rule to 1) a system where calcium carbonate, calcium oxide and carbon dioxide are at equilibrium, 2) points A to F indicated in the one-component phase diagram above, and explain why four phases cannot mutually coexist in equilibrium for this case.

Answer

There are clearly three phases ( $P=3$ ): $CaCO_3(s)$ , $CaO(s)$ and $CO_2(g)$ . Although there are three chemical species in the system, the reaction $CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$ introduces a constraint, reducing the number of independent components to two ( $C=2$ ). Therefore, $F=1$ . This implies that if one variable is fixed, the other is automatically determined. For instance, at a given temperature, there’s a unique equilibrium pressure of $CO_2$ where all three phases can coexist.

Point	C	P	F
A	1	1	2
B	1	2	1
C	1	3	0
D	1	1	2
E	1	1	2
F	1	2	1

When only one phase is present in a one-component system (points A, D and E), pressure and temperature can be varied independently within that single-phase region, giving two degrees of freedom. Points on phase boundaries (such as B and F) represent the coexistence of two phases in equilibrium. These points have only one degree of freedom, meaning that if pressure changes, temperature must adjust accordingly to stay on the phase boundary (and vice versa). Point C, the triple point, represents the unique conditions where three phases coexist in equilibrium. For a one-component system, this corresponds to zero degrees of freedom as temperature and pressure are fixed. If we consider four phases ( $P=4$ ) in a one-component system ( $C=1$ ), we find $F=-1$ , which is not physically meaningful. This confirms that no more than three phases can coexist in equilibrium in a one-component system.

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December 29, 2024February 8, 2025

Why does a current flowing through a wire generate a magnetic field?

Why does a current flowing through a wire generate a magnetic field? The answer lies in special relativity.

Consider a current flowing through a straight length of wire. In the rest frame of the wire, the positive ions (nuclei) are stationary, while the electrons are moving. Suppose a negative test charge is located at a distance from the wire, and moves parallel to the wire with the same velocity as the electrons. In this frame, the wire is electrically neutral overall because the number of positive charges equals the number of negative charges. Consequently, there is no net static electric field acting on the test charge in this frame. However, we observe that the test charge moves in response to a force towards the wire.

In the frame of the moving test charge, the test charge and the electrons are stationary, while the positive ions move in the opposite direction. Due to length contraction, the distance between the moving positive ions appears shorter to the test charge, leading to an increased positive charge density in the test charge’s frame. As a result, the wire has an apparent higher positive charge density compared to the negative charge density of the stationary electrons. This creates a net electric field, which exerts a force on the test charge towards the wire. In the rest frame of the wire, this force is referred to as a magnetic force. Therefore, both frames yield the same total force on the test charge, but it is described as a magnetic force in the wire’s rest frame and an electric force in the test charge’s frame. Since we can mathematically describe the magnetic force in the wire’s rest frame by defining a magnetic field, we say that a current flowing in a wire generates a magnetic field.

In summary, the electric field and the magnetic field are simply different observations of the same underlying phenomenon. Just as a coin has two faces that are inseparable yet different, electric and magnetic fields are inseparable parts of the same phenomenon, with their roles depending on the relative motion of the observer.

Question

Doesn’t length contraction also apply to the moving electrons in the wire’s rest frame? If so, why is there no net electric field acting on the test charge in that frame?

Answer

The rest frame of the wire is the initial premise. In this frame, the wire is observed to be electrically neutral, with length contraction already accounted for in this electrical neutrality. The question then shifts to how the situation differs in another reference frame, based on the initial premise.

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September 23, 2024

Laguerre polynomials

Laguerre polynomials $v(x)$ are a sequence of polynomials that are solutions to the Laguerre differential equation:

$x\frac{d^2v}{dx^2}+(1-x)\frac{dv}{dx}+nv=0\;\;\;\;\;\;\;\;420$

where $n$ is a constant.

When $x=0$ , eq420 simplifies to $\frac{dv}{dx}+nv=0$ . The solution to this first-order differential equation is $v(x)=Ae^{-nx}$ , which can be expressed as the Taylor series $v(x)=A\sum_{n=0}^{\infty}\frac{(-nx)^n}{n!}$ . This implies that eq420 has a power series solution around $x=0$ . To determine the exact form of the power series solution to eq420, let $v(x)=\sum_{j=0}^{\infty}a_jx^j$ .

Substituting $v$ , $v'$ and $v''$ in eq420 yields

$x\sum_{j=0}^{\infty}j(j-1)a_jx^{j-2}+(1-x)\sum_{j=0}^{\infty}ja_jx^{j-1}+n\sum_{j=0}^{\infty}a_jx^{j}=0\;\;\;\;\;\;\;\;421$

$\sum_{j=0}^{\infty}j^2a_jx^{j-1}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=0$

Setting $j=j+1$ in the first sum,

$\sum_{j=-1}^{\infty}(j+1)^2a_{j+1}x^{j}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=\sum_{j=0}^{\infty}(j+1)^2a_{j+1}x^{j}+\sum_{j=0}^{\infty}(n-j)a_jx^{j}=0$

$\sum_{j=0}^{\infty}\biggr\[(j+1)^2a_{j+1}+(n-j)a_j\biggr\]x^{j}=0\;\;\;\;\;\;\;\;422$

Eq422 is only true if all coefficients of $x$ in is 0 (see this article for explanation). So, $(j+1)^2a_{j+1}+(n-j)a_j=0$ , or equivalently,

$a_{j+1}=\frac{j-n}{(j+1)^2}a_j\;\;\;\;\;\;\;\;423$

Eq423 is a recurrence relation. If we know the value of $a_0$ , we can use the relation to find $a_1,a_2,...$ .

$\boldsymbol{\mathit{j}}$	Recurrence relation
$0$	$a_1=-na_0$
$1$	$a_2=\frac{1-n}{(1+1)^2}a_1=\frac{n(n-1)}{1^22^2}a_0$
$2$	$a_3=\frac{2-n}{(2+1)^2}a_2=-\frac{n(n-1)(n-2)}{1^22^23^2}a_0$
$\vdots$	$\vdots$

Comparing the recurrence relations, we have

$a_k=(-1)^k\frac{n(n-1)(n-2)\cdots (n-k+1)}{1^22^2\cdots k^2}a_0$

$a_k=(-1)^k\frac{n!}{(k!)^2(n-k)!}\;\;\;\;\;\;\;\;424$

where $a_0=1$ by convention (so that $L_n(0)=1$ ).

Letting $k=j$ in eq424 and substituting it in $v(x)=\sum_{j=0}^{n}a_jx^j$ yields the Laguerre polynomials:

$L_n(x)=\sum_{j=0}^{n}(-1)^j\frac{n!}{(j!)^2(n-j)!}x^j\;\;\;\;\;\;\;\;425$

where we have replaced $v(x)$ with $L_n(x)$ .

The first few Laguerre polynomials are:

$L_0(x)=1$

$L_1(x)=-x+1$

$L_2(x)=\frac{1}{2}(x^2-4x+2)$

$L_3(x)=\frac{1}{6}(-x^3+9x^2-18x+6)$

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September 23, 2024

Rodrigues’ formula for the Laguerre polynomials

The Rodrigues’ formula for the Laguerre polynomials is a mathematical expression that provides a method to calculate any Laguerre polynomial using differentiation.

It is given by

$L_n(x)=\frac{e^x}{n!}\frac{d^n}{dx^n}(x^ne^{-x})\;\;\;\;\;\;\;\;428$

To prove eq428, we apply Leibniz’ theorem as follows:

$L_n(x)=\frac{e^x}{n!}\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}\frac{d^kx^n}{dx^k}\frac{d^{n-k}e^{-x}}{dx^{n-k}}$

Substituting $\frac{d^kx^n}{dx^k}=\frac{n!}{(n-k)!}x^{n-k}$ and $\frac{d^{n-k}e^{-x}}{dx^{n-k}}=(-1)^{n-k}e^{-x}$ in the above equation and rearranging yields

$L_n(x)=\sum_{k=0}^n(-1)^{n-k}\frac{n!}{k!(n-k)!(n-k)!}x^{n-k}$

Letting $j=n-k$ , we have $L_n(x)=\sum_{j=0}^{n}(-1)^j\frac{n!}{(j!)^2(n-j)!}x^j$ , which is the expression for the Laguerre polynomials

Question

How do we change the variable in the summation $\sum_{k=0}^{n}$ by letting $j=n-k$ ?

Answer

When $k=0$ , $j=n$ , and when $k=n$ , $j=0$ . So, $\sum_{k=0}^{n}=\sum_{j=n}^{0}$ . Reversing the summation order, $\sum_{k=0}^{n}=\sum_{j=0}^{n}$ .

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September 23, 2024May 8, 2025

Generating function for the Laguerre polynomials

The generating function $G(x,t)$ for the Laguerre polynomials is a mathematical tool that, when expanded as a power series, produces Laguerre polynomials as its coefficients in terms of a variable.

$G(x,t)=\frac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^{\infty}L_n(x)t^n\;\;\;\;\;\;\;\;430$

Eq430 means that the cofficient of $t^n$ in the expansion of $\frac{e^{-\frac{xt}{1-t}}}{1-t}$ is $L_n(x)$ . To prove this, we expand the exponential term as a Taylor series:

$\frac{e^{-\frac{xt}{1-t}}}{1-t}=\frac{1}{1-t}\sum_{n=0}^{\infty}\frac{$-\frac{xt}{1-t}$^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^nt^n(1-t)^{-(n+1)}$

Expanding $(1-t)^{-(n+1)}$ as a binomial series gives

$\frac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^nt^n\sum_{k=0}^{\infty}\begin{pmatrix}-n-1\\k\end{pmatrix}(-t)^{k}$

Since

$\begin{align}\begin{pmatrix}-n-1\\k\end{pmatrix}&=\frac{(-n-1)!}{k!(-n-k-1)!}=\frac{(-n-1)(-n-2)\cdots (-n-k)}{k!}\\&=\frac{(-1)^k(n+1)(n+2)\cdots(n+k)}{k!}=\frac{(-1)^k(n+k)!}{n!k!}\end{align}$

we have

$\frac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^nt^n\sum_{k=0}^{\infty}\frac{(n+k)!}{n!k!}t^{k}=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}(-1)^n\frac{(n+k)!}{k!(n!)^2}x^nt^{n+k}$

Letting $n+k=m$

$\frac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^{\infty}\sum_{m=n}^{\infty}(-1)^n\frac{m!}{(m-n)!(n!)^2}x^nt^{m}\;\;\;\;\;\;\;\;431$

We now have a sum over $m$ and then over $n$ . Since $m=n+k$ and both $n$ and $k$ range from $0$ to $\infty$ , the sum over $m$ ranges from $0$ to $\infty$ . The new range of $n$ in the outer sum is determined by the conditions that $n\geq 0$ and $n=m-k$ , where $m$ and $k$ range from $0$ to $\infty$ . Consequently, $n$ has a lower limit of 0 and an upper limit of $m$ . Eq431, after swapping the order of summation, then becomes