Parallel axis theorem

The parallel axis theorem relates a body’s moment of inertia about any axis to its moment about a parallel axis through the centre of mass.

Consider the diagram above, where the green area represents a system of particles, each with mass , rotating about an axis perpendicular to point . Point serves as both the origin of the coordinate system and the centre of mass of the system. As a result, point has the coordinates , while particle at point has the coordinates .

By definition, the centre of mass with respect to the -coordinate satisfies . Similarly, for the -coordinate, we have . Expanding these two equations and rearranging them yields

where .

Since , we have

 

Question

Show that the moment of inertia of a system of particles, each with mass , and located at perpendicular distances from the axis of rotation, is given by .

Answer

From eq4, the total rotational kinetic energy of the particles is

where .

 

The moment of inertia about the axis perpendicular to point is

where is measured from to .

Substituting in eq7 and expanding gives

Substituting eq6 in eq8 yields

Since , where is measured from to , and , eq9 becomes

Eq10 is the mathematical expression of the parallel axis theorem.

 

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The rigid rotor

The rigid rotor is a fundamental model in quantum mechanics used to describe the rotational motion of a molecule, particularly diatomic molecules, where the two atoms are treated as point masses connected by a fixed-length bond. This simplification assumes that the molecule does not vibrate or stretch during rotation, allowing the system to be treated as a rigid body rotating in space. The model works reasonably well in the microwave frequency range of 1011 to 1012 Hz and provides critical insights into energy levels associated with molecular rotation.

For the more general case of a diatomic molecule’s internal motion (encompassing both rotational and vibrational motions), the Schrödinger equation for the two particles, can be derived by reducing the two-particle problem to a one-particle problem. The result is given by eq311:

where

is the kinetic energy operator of the internal motion of the system,
is the reduced mass of the molecule,
is the internuclear potential energy ,
is the molecular wavefunction in spherical coordinates,
is the eigenvalue corresponding to the wavefunction.

Within the rigid rotor approximation, the bond length is fixed ( is a constant) and . The above equation simplifies to a rotational Schrödinger equation:

Eq1 is equivalent to a Schrödinger equation for a particle of mass constrained to a spherical surface of zero relative potential and radius . The eigenfunctions of this system are the spherical harmonics , and the Hamiltonian can be explicitly written as (see eq50 of an eariler article):

Here, is the moment of inertia and is the square of the angular momentum operator, with eigenvalues (see eq133):

Replacing with , the rotational quantum number, the energy of the rigid rotor is:

 

Question

The eigenvalue equation is associated with the angular momentum eigenvalue of an electron rotating about an origin. Is it also applicable to a nucleus rotating around a point?

Answer

Yes. This is because the spherical harmonics are general solutions to the angular part of the Schrödinger equation and describe the angular momentum states of a rotating particle. These states can represent either an electron or a nucleus rotating about a point, depending on the system being modelled.

 

Eq3 can also be derived from classical mechanics, where the rotational kinetic energy of a particle with reduced mass is

with and , where is the magnitude of the angular velocity vector .

Substituting the classical angular momentum into eq4 yields

Using the quantum mechanics postulate that “to every observable quantity in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics”, and replacing the classical angular momentum with the quantum mechanical operator gives


which is equivalent to eq2, and hence allows us to derive eq3.

The same postulate also enables us to derive similar energy expressions for polyatomic molecules. However, we must first examine the different moments of inertia for some common classes of polyatomic molecules, and to do that, we need to understand the parallel axis theorem.

 

Question

How many axes of rotation does a diatomic molecule have? If it has more than one, why is there only one moment of inertia mathematical expression ?

Answer

A diatomic molecule has three principal axes of rotation that pass through its centre of mass and are mutually perpendicular, just like any rigid body. However, since the molecule is linear, the axis aligned with the bond (commonly taken as the z-axis) passes through both atoms, meaning the mass has no perpendicular displacement from this axis. As a result, the moment of inertia about this axis is zero: .

The other two axes, both perpendicular to the bond, are physically equivalent due to the molecule’s cylindrical symmetry. Rotation about either of these axes produces the same resistance to rotation, leading to equal moments of inertia: . Therefore, although the molecule has three rotational axes, only one unique, nonzero moment of inertia expression is needed.

 

In other words, different axes of rotation will have different moments of inertia. In general, for a rigid body rotating in 3D space, the moment of inertia can be conveniently expressed as the following matrix, called the inertia tensor:

with

where , and .

Eq4b states that , which is consistent with classical mechanics (see this article). It follows that:

 

Question

Why is , and not or ?

Answer

is a scalar, and only evaluates to a scalar. The other two expressions result in matrices and are not valid for representing energy.

 

Substituting , and  back into eq4c gives

and hence

 

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Permutation and combination

A permutation of a set is an ordered arrangement of its elements.

 

Distinct elements

Consider the task of filling  buckets, each with a ball chosen from a pool of distinct numbered balls. There are possible choices for the first bucket. Once the first ball is used, only choices remain for the second bucket, then choices for the third bucket, and so on. This continues until the last bucket, for which only 1 ball remains.

If you have ways to do a first task, and for each of those, ways to do a second task, then the total number of ways to perform both tasks is . This principle extends to any number of sequential tasks. Therefore, the total number of distinct arrangements (permutations) of the balls in buckets, denoted by or , is the product of the number of choices for each task:

Now, suppose there are distinct balls but only buckets, where . Again, there are choices for the first bucket, for the second, and so on until the -th bucket. Since we can also express the number of choices for the second bucket as , we have ways of filling the -th bucket. Hence,

Multiplying RHS of eq306 by gives

Eq307 gives the general formula for calculating permutations — the number of ways to arrange objects selected from a set of distinct objects.

 

Question

In a lottery draw for a four-digit winning number, each digit can be any number from zero to nine. How many permutations are possible?

Answer

Total permutations = 10 x 10 x 10 x 10 = 104. In general, the number of permutations for forming a sequence of positions, where each position can be filled with possible choices is .

 

 

Repeating elements (multinomial permutation)

How many distinct permutations are there of the letters in the word “MISSISSIPPI”? If all letters were distinct, the number of permutations would be 11! = 39,916,800. However, some letters are repeated in the word: 4 I’s, 4 S’s, 2 P’s and 1 M. To understand why we need to adjust for repeated letters, let’s consider the I’s. Suppose we label the four I’s as I1, I2, I3 and I4, treating them as distinct. In any arrangement, these four labelled I’s can be rearranged among themselves in 4! = 24 ways. For example, we could have:

M I1​ S S I2​ S S I3​ P P I4

M I1​ S S I2​ S S I4​ P P I3

M I2​ S S I1​ S S I3​ P P I4

…and so on.

But since these I’s are actually indistinguishable, all 24 of those arrangements represent the same permutation of the word. So, we’ve overcounted each unique arrangement by a factor of 4!. Similarly, the 4 S’s and 2 P’s can be rearranged in 4! ways and 2! ways respectively. Each of these reorderings also causes overcounting, resulting in a total overcounting factor of 4! x 4! x 2!. Therefore, to find the number of distinct permutations of the letters in “MISSISSIPPI”, we divide the total number of arrangements by the total repeated counts: .

In general, if you have a set of objects, where are identical to each other, are identical to each other but different from the first group, and so on until , each representing a distinct group of identical items,  then the number of distinct permutations is:

where and is also known as the multinomial coefficient.

Eq308 is used to derive the Boltzmann distribution and Raoult’s law.

 

A combination is a selection of items where the order does not matter. For example, AB and BA are two different permutations of the set {A,B}, but they represent the same combination. To calculate the number of combinations, we start by counting the permutations (where order does matter), and then correct for overcounting by dividing out the number of ways the selected items can be rearranged.

Suppose we are selecting objects from a set of distinct objects. The number of permutations is given by eq307: . However, items can be arranged in ways, all of which count as the same combination when order doesn’t matter. There, the number of combinations, denoted by or or , is

Interesting, the number of combinations is also the binomial coefficient.

 

Question

Show that a mutlinomial permutation can be expressed as a product of combinations.

Answer

Suppose we want to partition a set of distinct items into groups of specified sizes ​, ​, …, ​, where . The number of ways to select items for the first group is

The number of ways to select the remaining ​ items for the second group is

The number of ways to select the remaining ​ items for the third group is

and so on until the remaining the number of ways to select the remaining ​ items for the last group is

Therefore, the total number of ways to perform all the sequential tasks is

which simplifies to the RHS of eq308 after all the cancellations.

 

 

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Saturated Vapour Pressure

Saturated vapour pressure is the pressure exerted by a vapour in equilibrium with its liquid or solid phase at a given temperature in a closed system. It represents the point at which the rate of particles leaving the condensed phase to enter the vapour phase equals the rate at which vapour particles condense back. This balance depends critically on temperature, as the energy available to the particles governs their ability to escape into the gas phase.

In a liquid (or solid), the particles have a range of kinetic energies, with their average energy determined by the temperature. As temperature increases, so does this average energy. However, even at a fixed temperature, some particles possess more energy than others. Among these, the more energetic particles near the surface can overcome the intermolecular forces holding them in the liquid (or solid) and escape into the vapour phase (see diagram above). As more particles accumulate in the vapour, some will lose energy through collisions and return to the liquid (or solid). The vapour particles move freely and, when they strike the walls of the container, they exert a pressure. Eventually, the rates of evaporation and condensation become equal, establishing a dynamic equilibrium. At this point, the pressure exerted by the vapour remains constant and is known as the saturated vapour pressure.

 

Measuring saturated vapour pressure

One of the most common techniques for measuring the saturated vapour pressure of a liquid or solid is the manometric method (see diagram below). In a typical setup, a known amount of the substance (liquid or solid) is placed in a sealed, evacuated chamber that is connected to a manometer. The system is kept at a constant and controlled temperature using a thermostatic bath, ensuring that any pressure changes are solely due to the substance’s vapour pressure at that temperature. As the substance begins to evaporate or sublimate, its vapour fills the chamber. The manometer, often a U-tube containing mercury or another fluid, registers the pressure exerted by the vapour as it reaches equilibrium with its condensed phase.

As more molecules escape from the surface of the liquid (or solid), the pressure inside the chamber increases. Eventually, the rate of molecules entering the vapour phase equals the rate returning to the liquid or solid—the point of dynamic equilibrium. At this stage, the manometer reading stabilises, indicating that the saturated vapour pressure has been reached.

 

Applications

The boiling point of a liquid is the temperature at which its saturated vapour pressure equals the external atmospheric pressure. At this temperature, bubbles of vapour can form within the bulk of the liquid, leading to boiling. This relationship explains why liquids boil at lower temperatures at higher altitudes — where atmospheric pressure is reduced — and why pressure cookers, which increase external pressure, raise the boiling point and cook food faster. Mathematically, the Clausius-Clapeyron equation  describes the relationship between vapour pressure and temperature.

 

Question

Why do food cook faster when the boiling point of water is raised in a pressure cooker?

Answer

Inside a pressure cooker operating at increased pressure, the temperature of the steam and liquid water rises above 100°C — typically around 120°C at about 2 atmosphere. This means that the food inside the pressure cooker is essentially cooking in an environment at 120°C instead of 100°C.

 

Sublimation is the process where a solid transitions directly to a vapour without passing through the liquid phase. This occurs when the saturated vapour pressure of the solid equals or exceeds the surrounding atmospheric pressure at a given temperature. Just like with liquids, solids possess a saturated vapour pressure that increases with temperature. Substances like dry ice (solid carbon dioxide) and iodine readily sublimate at atmospheric pressure due to their relatively high vapour pressures even in the solid state. The point at which the solid and vapour are in equilibrium is also described by the Clausius-Clapeyron relation .

 

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Phase equilibria (overview)

Phase equilibria is a fundamental concept in thermodynamics that describes the balance between different phases of a substance or mixture at equilibrium. When a system reaches phase equilibrium, the rates of transition between phases are equal, resulting in no net change in the amount of each phase over time. This balance is governed by temperature, pressure and composition, and is represented using phase diagrams and equilibrium laws such as Raoult’s Law and the Clausius-Clapeyron equation.

Understanding phase equilibria is crucial in both theoretical and applied sciences, as it explains phenomena like boiling, melting, sublimation and condensation. It also plays a vital role in fields such as material science, chemical engineering, and environmental science, where control over phase transitions is essential for the design and optimisation of various processes and products.

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Phase Diagrams

A phase diagram illustrates the equilibrium conditions — typically pressure versus temperature — under which a substance exists in different phases.

Defined as a distinct, homogeneous part of a system, a phase has uniform physical and chemical properties. The most common phases are solid, liquid, gas and plasma. Among these, solid and liquid are referred to as condensed phases, while the gas phase is often called the vapour phase. In the figure below, the phase diagram displays temperature on the horizontal axis and pressure on the vertical axis. The diagram is divided into regions, each representing the stable phase of a substance under specific temperature and pressure conditions.

A phase transition takes place when a substance changes from one phase to another, typically due to variations in temperature or pressure. These transitions involve transformations in energy. For example, when a solid becomes a liquid, the process is known as melting. The temperature at which this change occurs under constant pressure is called the transition temperature.

Lines separating the regions in the diagram are known as phase boundaries. These boundaries represent the equilibrium conditions under which two phases coexist. For instance, the solid–liquid boundary marks the conditions at which solid and liquid phases are in equilibrium. At any point on a boundary, the substance can exist as a stable mixture of the two phases, provided that and change infinitesimally. Since there is no net change in the amount of substance transforming between phases at equilibrium, the tendency for each phase to change is the same. In other words, the chemical potentials of the two phases are equal at equilibrium

 

Question

Why must and change infinitesimally when a substance is on a phase boundary?

Answer

An infinitesimal change in or  allows the system to adjust reversibly. For instance, if the temperature is raised slightly at the melting point, a small amount of solid will melt into liquid — maintaining equilibrium and remaining on the phase boundary as the system shifts slightly towards the liquid phase. However, if or  changes by a noticeable amount, the substance will leave the phase boundary and enter a region where only one phase is stable. The system will then undergo a complete phase transition rather than coexistence.

 

A phase diagram contains valuable information about a substance. The triple point (TP) is the unique condition (specific temperature and pressure) at which all three phases — solid, liquid and gas — coexist in equilibrium. Every pure substance has a distinct triple point. This singular equilibrium state serves as a fixed reference for thermodynamic calculations (e.g. the triple point of water defines the Kelvin temperature scale). Located at the end of the liquid-gas boundary is the critical point (CT), beyond which the liquid and gas phases become indistinguishable (having equal density) and form a supercritical fluid.

The boiling point of the substance is the temperature at which the vapour pressure of a liquid equals the external pressure, resulting in vaporisation. It varies with pressure, with the normal boiling point (NBP) defined at 1 atm. Similarly, the melting point is the temperature at which a solid becomes a liquid under a given pressure, and the normal melting point (NMP) is again defined at 1 atm.

Sublimation can occur over a range of pressures and temperatures below the triple point. A solid sublimes when its vapour pressure equals or exceeds the surrounding partial pressure of its gaseous phase. Unlike melting and boiling points, sublimation does not have a single, universally agreed-upon “normal” condition. Nevertheless, the sublimation point of a substance is commonly referenced at 1 atm — for example, dry ice (solid CO₂) sublimes at -78.5 °C at 1 atm.

 

Liquid-gas and solid-gas boundaries

How is the phase diagram constructed? Each phase boundary is a mathematical function of temperature, with the Clasius-Clapeyron equation describing the liquid-gas boundary. Given that the enthalpy of vaporisation is positive and assuming it is independent of temperature, the equation can be integrated as follows:

which rearranges to

Eq170 is an exponential curve. If , it represents a smooth, upward-sloping curve, which is typical of the liquid-gas boundary. The curve truncates at the critical temperature, as the two phases become a uniform phase above this temperature.

Similarly, the solid-gas boundary is described by the Clasius-Clapeyron equation in the form:

Since the enthalpy of sublimation is also positive, and , the solid-gas boundary has a steeper gradient than the liquid-gas boundary.

 

Solid-liquid boundary

From eq148,

where the subscript denotes a molar quantity.

Substituting eq149a in eq172 gives

Since the chemical potentials of the solid phase and the liquid phase are equal along the solid-liquid boundary,

which rearranges into the Clapeyron equation:

Substituting eq147, where at equilibrium, in eq174 yields

Integrating eq175, and assuming and are independent of temperature, gives

Eq176 describes the solid-liquid boundary. For most substances, the change in molar volume upon melting is very small because solids and liquids are relatively incompressible. This makes the slope very steep, which can often be approximated as a straight line in some substances.

 

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Henry’s Law

Henry’s Law states that, at a constant temperature, the partial pressure of a gaseous species in equilibrium with its liquid form is directly proportional to its mole fraction in dilute solutions. Mathematically, Henry’s Law can be expressed as:

where:

is the partial pressure of the gas,
is the Henry’s Law constant (which varies with temperature),
is the mole fraction of the gas in the liquid.

The law was empirically derived in 1803 by William Henry, an English chemist. While Raoult’s law describes how the vapour pressure of a solvent is lowered when a non-volatile solute is added to an ideal solution, Henry’s law is primarily used to predict the solubility of a gas in real solutions, with important applications in environmental science, chemical engineering and biology. In other words, both the solute and solvent obey Raoult’s law in an ideal solution, while the solute follows Henry’s law in a real dilute solution (see diagram below). This is because, in a real dilute solution, each solute molecule is almost exclusively surrounded by solvent molecules, and its behaviour is determined by interactions with the solvent, not with other solute molecules. The proportionality constant  reflects these solute-solvent interactions.

Despite being an empirical law, Henry’s law can be derived using basic thermodynamics principles. At equilibrium, the chemical potential of the gas in the gas phase must be equal to its chemical potential in the solution. Rearranging eq182, where is defined as 1 atm (the convention for the standard state of gases), gives

Since the exponential factor is a constant for a given temperature, eq188 is equivalent to eq187, with .

A solute-solvent system in which the mole fraction of the solvent is close to 1, and the two species have different intermolecular interactions, is called an ideal-dilute solution (see diagram above). In such a solution, the solvent obeys Raoult’s law (ideal) and the solute follows Henry’s law (dilute). The distinct behaviours exhibited by the solute and solvent in such solutions can be attributed to fundamental differences in their molecular environments.

In a dilute solution, solvent molecules are predominantly surrounded by other solvent molecules. As a result, their immediate molecular environment closely approximates that of the pure liquid solvent. This preservation of their native environment explains why the solvent’s vapour pressure continues to align with Raoult’s law, which describes the vapour pressure of an ideal pure component.

Conversely, solute molecules in a dilute solution are almost exclusively surrounded by solvent molecules. This marks a significant departure from their pure state, where they would typically be surrounded solely by other solute molecules. Due to this altered molecular environment, the solute’s behaviour, particularly its partial vapour pressure, deviates considerably from what might be expected from its pure form. This distinct behaviour is precisely what Henry’s law describes.

Therefore, the solvent behaves effectively as a slightly perturbed pure liquid, while the solute exhibits fundamentally different behaviour. Many naturally occuring systems are ideal-dilute solutions. For example, the small amount of dissolved oxygen in natural waters or blood follows Henry’s Law, while the water (solvent) makes up the majority and its vapour pressure is essentially determined by Raoult’s Law. In carbonated drinks, the dissolution of CO2 is well-described by Henry’s Law, while the water acts as the Raoultian solvent.

 

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Pressure-composition diagrams for ideal and non-ideal solutions

Pressure-composition diagrams are graphical representations that illustrate the relationship between the vapour pressures of components in a liquid mixture and their compositions at a constant temperature. These diagrams are essential tools in physical chemistry and chemical engineering, particularly for understanding phase equilibria in binary mixtures.

 

Ideal solution

Consider two liquids A and B (e.g. benzene and toluene) forming an ideal solution in a closed container (e.g. a container with a movable piston) at a constant temperature above the freezing points of both species. The partial pressures of the components follow Raoult’s law:

where

is the partial vapour pressure of component
is the vapour pressure of pure
is the mole fraction of in the solution

The total vapour pressure of the ideal solution is:

According to Dalton’s law, the mole fractions and of the components in the gas are

Substituting eq189 and eq190 in eq191 results in

Substituting in eq189 into in eq191 gives

Substituting eq193 in of eq192 and rearranging yields

Eq190 and eq194 are plotted in a graph of against to produce the pressure-composition diagram of the ideal solution (see diagram above). Eq190, also known as the liquid composition curve, describes a straight line (blue) along which the liquid begins to vaporise. In other words, the high pressure region above this line corresponds to the liquid phase of the solution. In contrast, eq194 defines the curve (red) along which the last drop of the liquid vaporises. The region below this curve, also called the vapour composition curve, represents the vapour phase. The area between eq190 and eq194 marks the two-phase region where liquid and vapour coexist in equilibrium.

 

Question

What is ? Isn’t the graph plotted against and ?

Answer

is the overall mole fraction of A in the system, including both the liquid and vapour phases. It ranges from 0 to 1 and is given by:

When we plot eq190 and eq194, we are plotting them against and  respectively. These values also lie between 0 and 1. However, neither or  can define all the points in the liquid, vapour and two-phase regions. Therefore, after plotting the two equations, we replace the horizontal axis with , which allows a single, continuous representation of the system’s behaviour across all regions. When interpreting the horizontal axis at a point along eq190, we consider the case where , which gives


Similarly, at a point along eq194, we assume , resulting in .

 

Consider the process of isothermally lowering the pressure from point a to i (see diagram above), which can be achieved by drawing out the piston. At point a, the pressure is above eq190 at . Under these conditions, the system is entirely in the liquid phase. As the pressure is reduced 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 (), but the vapour phase has a different composition corresponding to point f (). A tie-line bf on the diagram connects these two phase compositions, indicating that both phases coexist in equilibrium.

Continuing to lower the pressure 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 , but the compositions of the individual phases are (point c) and (point g) for the liquid phase and vapour phase respectively.

​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 (), and the vapour phase composition matches the overall composition (). Finally, as the pressure is reduced even further to point i, the system moves below the vapour composition curve, where it exists as a single-phase vapour. No liquid remains and the composition remains constant at . The line abehi is called an isopleth.

 

Question

Why is the overall composition of the system fixed at when the pressure is reduced?

Answer

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 when pressure is reduced.

 

Interestingly, even though the compositions of the phases are fixed at a given pressure (e.g. and at ), the overall mole fraction of A can vary continuously between them, depending on the relative amounts of each phase. To understand this, let

be the number of moles of A in the liquid phase
be the number of moles of A in the vapour phase
be the total number of moles of A and B in the liquid phase
be the total number of moles of A and B in the vapour phase
be the total number of moles of A and B in the closed container

Clearly,

Substituting eq195 and eq196 into eq197 and rearranging gives

Since , the range of is , which corresponds to the domain of a tie-line at a particular pressure and temperature. Multiplying both sides of eq198 by  and rearranging yields

Substituting into eq199 and rearranging results in the lever rule:

Here, represents the length from the left end of a tie-line to the intersection of the tie-line and with the isopleth, while represents the length from that intersection to the right end of the tie-line. Eq200 allows us to determine the relative amounts of the two phases in equilibrium by measuring these two lengths. This principle has important applications in material science and chemical engineering — for instance, in predicting phase amounts, which is critical for the design of distillation columns.

 

Non-ideal solution

Pressure-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 liquid composition curve is monotonic, with no stationary points. In such cases, the diagram features two smooth curves that intersect only at and (see diagram below). A typical example is the carbon tetrachloride-toluene system, where carbon tetrachloride is the more volatile component.

In contrast, when the deviation from Raoult’s law is significant, the liquid composition curve may develop a stationary point. The nature of this point is governed by the intermolecular interactions between components A and B. A maximum point occurs when A-B interactions are weaker than A-A and B-B interactions, while a minimum point arises when A-B interactions are stronger than A-A and B-B interactions.

Now consider a liquid composition curve with a maximum point. Suppose the corresponding vapour composition curve intersects the liquid composition curve only at and . In this case, a tie-line drawn just below the maximum point would connect two points on the liquid composition curve (see diagram above), implying two coexisting liquid phases at different compositions — a physical impossibility for a binary system in vapour–liquid equilibrium. This contradiction indicates that the vapour composition curve must also touch the liquid composition curve at the maximum point, as shown in the diagram below. A solution that behaves this way is called an azeotrope. The ethanol–water system is an example of an azeotrope with a maximum point, while the HCl–water system forms an azeotrope with a minimum point. These diagrams are typically constructed using empirical data.

 

Distillation

A rotary evaporator (see diagram below) separates the components of a binary mixture by reducing the pressure above the mixture with a vacuum pump. Pressure-composition diagrams can help predict the operating pressure required for distilling the more volatile component when using a rotary evaporator.

Pressure-composition diagrams are also used in flash distillation, which involves the partial vaporisation of a liquid mixture to separate its components based on differences in volatility. In a flash distillation process, a pre-heated liquid mixture is introduced into a flash drum or separator operating at a reduced pressure (see diagram below). Upon entry, a portion of the liquid “flashes” into vapour due to the sudden pressure drop. This generates two phases in equilibrium: a vapour phase enriched in the more volatile component, and a liquid phase enriched in the less volatile one. These two phases are then physically separated, with the vapour removed at the top and the liquid removed at the bottom. The phases can undergo further distillation, depending on the desired separation efficiency and purity of the components.

Pressure-composition diagrams are crucial in this context because they show how the equilibrium compositions of the vapour and liquid phases vary with pressure at a fixed temperature. These diagrams help determine the operating conditions (such as temperature and pressure) at which the desired separation can occur, as well as predict the relative amounts of vapour and liquid formed.

Flash distillation is widely used in the chemical and petroleum industries — for example, in crude oil refining — where it serves as a rapid, energy-efficient method for achieving partial separation of multi-component mixtures without the complexity of a full distillation column.

 

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Colligative properties

Colligative properties are physical properties of solutions that depend solely on the number of solute particles present, not on their chemical identity. The term “colligative” comes from the Latin word colligatus, meaning “bound together”, reflecting the idea that these properties arise from the collective presence of solute particles, regardless of their type, size or chemical nature.

The classical treatment of colligative properties rests on two key assumptions:

    • The solute is non-volatile — only the solvent contributes to the vapour phase, so any change in vapour pressure (and related properties) comes solely from the dilution of the solvent.
    • The solution behaves ideally — solute and solvent molecules interact only through simple physical mixing, with no strong intermolecular forces altering the behaviour of the components. Dilute non-ideal solutions behave nearly ideally at low solute concentrations and also exhibit colligative properties.

The four principal colligative properties are vapour pressure lowering, boiling point elevation, freezing point depression and osmotic pressure. As vapour pressure lowering has been discussed in the article on Raoult’s law, we will elaborate on the remaining properties.

 

Boiling point elevation

The normal boiling point of a pure liquid or solution is the temperature at which its vapour pressure equals 1 atm. According to Raoult’s law, the addition of a non-volatile solute to a pure liquid lowers the vapour pressure of the solvent. As a result, a higher temperature is required for the solution’s vapour pressure to reach 1 atm. Therefore, the normal boiling point of the solution is elevated relative to that of the pure solvent.

To explain this mathematically, consider a volatile solvent and a nonvolatile solute in a solution at equilibrium with the gaseous solvent at a constant pressure. Raoult’s Law states that at any given temperature, the vapour pressure of the solvent in the solution is related to the vapour pressure of the pure solvent at that same temperature. Using eq178 at the solution’s boiling point , we have

where is the external pressure and denotes the vapour pressure of the solution at .

Since the pure solvent boils at , we have , which when combined with eq189 gives

As the Clausius-Clapeyron equation describes how the vapour pressure of a substance changes with temperature, we can express eq169 as:

Substituting eq190 into eq191 yields

For dilute solutions, . So, the Taylor expansion of becomes , and

where .

Assuming that for a dilute solution and is independent of temperature, eq192 becomes

where .

 

Question

If , why isn’t ?

Answer

Even if , there is still some small elevation in boiling point when increases. Another way to proceed from eq192 is to rearrange it to:

Let and

Using the Taylor expansion of , we have for small , and eq193a rearranges to eq193.

 

 

Freezing point depression

Freezing is the process by which a liquid transforms into a solid with an ordered, crystalline structure. When a non-volatile solute is added to the solvent, it effectively dilutes the solvent and increases the entropy of the solution. The reduction in the concentration of solvent molecules, along with the enhanced molecular randomness, opposes the tendency of the solution to freeze (freezing lowers the entropy of the system). As a result, a lower temperature is required for freezing to occur.

Consider a pure solid (A) at equilibrium with a solution containing the dilute solute (B). Since the chemical potentials of A in the two phases are equal at equilibrium, eq181 becomes

Dividing eq194 by and differentiating it with respect to at constant gives

 

Question

Show that .

Answer

Using the quotient rule,

From eq173, . So,

Combining eq143 and eq149a, we have , and

 

Therefore, eq195 becomes , or equivalently,

where .

Assuming is independent of temperature and integrating both sides of eq196 with respect to yields,

The limits of integration represent a transition from the normal freezing temperature of pure solvent A (where ) to the depressed freezing point of the solution (where ). Therefore,

For dilute solutions, . So, the Taylor expansion of and

Assuming that for a dilute solution,

where is the freezing point depression and .

 

Osmotic pressure

Osmosis is the tendency of pure solvent molecules A to diffuse across a semipermeable membrane into a solution of A and B. Consider two equal chambers separated by a rigid, thermally conducting, semipermeable membrane that allows only molecules of solvent A to pass through, but not solute B (see diagram below). In the left chamber is pure solvent A. In the right chamber, we have a solution of B in A. Initially, the heights of the liquids in the two capillary tubes connected to the chambers are equal, so the pressures in both chambers are the same (). Furthermore, the membrane conducts heat and thermal equilibrium is maintained.

The chemical potential of pure solvent A in the left chamber is denoted by . In the right chamber, the presence of solute B lowers the chemical potential of A, so . The resulting chemical potential difference indicates that the system, being out of equilibrium, will spontaneously evolve to minimise its Gibbs energy. This is achieved by solvent A flowing from the pure solvent side through the semipermeable membrane into the solution, thereby diluting the solute and raising the chemical potential of A in the right chamber.

As solvent flows into the right chamber, the liquid level in its capillary tube rises, increasing the hydrostatic pressure in that chamber. This pressure buildup continues until the chemical potentials of the solvent in both chambers are equal. At that point, equilibrium is restored and the additional pressure  generated in the right chamber is called the osmotic pressure. The magnitude of this pressure reflects the extent of osmosis that has occurred in the system and enables quantification of that extent.

When equilibrium is restored, . Using eq181, we have

Integrating eq173 at constant , we have  or

Substituting eq198 into eq199 yields

For dilute solutions, . So, the Taylor expansion of and


Substituting and for dilute solutions, and in eq200 gives

Eq201 is called the van’t Hoff law, which describes the osmotic pressure of an ideal-dilute solution. For non-ideal solutions, the osmotic pressure can be approximated using a power series expansion analogous to the virial expansion for real gases:

where , are osmotic virial coefficients that capture non-ideal interactions.

 

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Raoult’s Law

Raoult’s Law states that the partial vapour pressure of each volatile component in an ideal solution at equilibrium is equal to the product of its mole fraction in the liquid phase and the vapour pressure of the pure component at the same temperature. Mathematically:

where:

is the partial vapour pressure of component
is the mole fraction of component  in the liquid
is the vapour pressure of pure component

It is named after François-Marie Raoult, a French chemist who formulated it in the 1880s. Through his experiments on the vapour pressures of solutions, Raoult discovered that the presence of a non-volatile solute lowers the vapour pressure of the solvent in proportion to its mole fraction. His work laid the groundwork for colligative properties and was pivotal in the development of solution thermodynamics.

Let’s explain Raoult’s law in an intuitive way before we derive it from thermodynamic principles. Consider a sealed container with a pure volatile liquid A (see diagram below). At a given temperature, some molecules of A evaporate from the liquid surface into the gas phase, while other molecules of A from the gas phase condense back into the liquid. Eventually, a dynamic equilibrium is established, where the rate of evaporation equals the rate of condensation. The pressure exerted by the gaseous A at this equilibrium is the vapour pressure of pure A, .

If we add another volatile liquid component B to the container, forming an ideal solution (where A and B have similar intermolecular interactions), the surface of the liquid is now shared by molecules of both A and B. This means that, for any given area of the liquid surface, fewer molecules of A can escape into the vapour phase than when A was pure. When a new equilibrium is attained, the reduced partial pressure of A is expected to be equal to the product of its mole fraction in the liquid and . The same is true for B. This is the essence of Raoult’s law. The total vapour pressure in the container is then the sum of the partial pressures of A and B:

Similarly, if we add a non-volatile solute to a container of pure A to form an ideal solution, . However, the total vapour pressure is now equal to just , because the solute is non-volatile. Therefore, Raoult’s law can also be defined as follows: the vapour pressure of a solvent in a solution is directly proportional to its mole fraction in the liquid phase. In other words, when a non-volatile solute is added to a volatile solvent, the solvent’s vapour pressure decreases in proportion to the amount of solute present: .

Raoult’s law can be derived from thermodynamic principles. Consider a multi-component solution with a volatile liquid A at equilibrium with its vapour. At constant temperature eq148 becomes or its molar form . Substituting the ideal gas law in the molar form and integrating gives:

which leads to:

Substituting eq149a in eq179 yields

The equivalent expression for the chemical potential of the liquid-phase of A is

where is the mole fraction of A in the solution (see below for a complete derivation of eq181).

When a liquid and its vapour are in equilibrium, their chemical potentials are equal. Hence, , or

which rearranges to give the Raoult’s law because .

 

Question

Show that .

Answer

For a pure liquid A in equilibrium with its vapour, and in eq182, which simplifies to .

 

Raoult’s Law is most accurate when applied to ideal solutions, where the intermolecular forces between unlike molecules closely resemble those between like molecules. A classic example is the binary mixture of benzene and toluene, which closely follows Raoult’s Law. When the vapour pressure of toluene is plotted against its mole fraction in the solution, the resulting graph is a straight line, consistent with eq178. A similar linear plot can be made for benzene, using the inverse relationship between its mole fraction and that of toluene. The total vapor pressure of the ideal solution is obtained by summing the partial vapour pressures of both components, as illustrated in the diagram below.

However, the law fails or becomes less accurate for non-ideal solutions, in which the components interact more strongly or more weakly with each other than with themselves. An example is the acetone-chloroform system, where hydrogen bonding causes deviations from ideal behaviour (see diagram below).

Vapour pressure diagrams allow for a straightforward determination of the partial vapour pressure of each component and the total vapour pressure of the solution at a specific temperature. They also provide a basis for plotting boiling point diagrams and ultimately phase diagrams of mixtures.

 

Question

Derive eq181.

Answer

Consider a system with moles of solvent A and moles of solute B forming an ideal solution. is the Avogadro constant and the total number of moles is . From eq147, the change in Gibbs free energy upon mixing the two components is given by: . Since A and B have similar intermolecular interactions, there is no net release or absorption of heat upon mixing. So, and .

The number of ways, , to arrange molecules of A and molecules of B in a container with total sites is given by the multinomial permutation:

Substituting into the statistical entropy formula of eq30 yields , since . Applying Stirling’s approximation () for large and rearranging gives

where we have used (see this article for derivation ), and for the last equality.

Substituting eq183 into results in

The change in Gibbs energy upon mixing is

Substituting eq153 into the above equation yields

Equating eq184 and eq185:

For eq186 to hold, the coefficients of and on both sides must be equal. Therefore, and , with the general formula being eq181.

 

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