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The moments of inertia of a tetrahedral prolate rotor with symmetry (e.g. CHCl₃) are characterised by a unique moment of inertia around the principal axis and two equal moments of inertia perpendicular to the principal axis, where . They are derived using simple geometric considerations.

Since the rotation about the axis (-axis) of the molecule has the same symmetry as a trigonal pyramidal molecule, the moment of inertia for symmetric rotors like CHCl₃ along the -axis is given by eq22:

To derive the expression for , place the origin at the centre of mass of the molecule. The -coordinates of atom C, atom B and each of the A atoms are , and respectively. Then, the centre of mass of the molecule satisfies or

with

Using the and  coordinates of the three A atoms defined in the previous article, the positions of the atoms are:

Atom	Coordinates
C
B
A (point F)
A (point I)
A (point E)

Since , the moment of inertia of the molecule about the -axis is . Substituting the data from the above table and eq23 into this equation gives:

You’ll find that substituting the data from the above table into the expression for the moment of inertia of the molecule about the -axis , and then substituting eq23 into the resultant equation, yields the same expression as eq33. Therefore, eq22 and eq33 represent two distinct moments of inertia of a tetrahedral prolate rotor with symmetry.

 

 

 

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The moments of inertia of a trigonal pyramidal prolate rotor with symmetry (e.g. NH₃) are characterised by a unique moment of inertia around the rotational axis (-axis) and two equal moments of inertia perpendicular to the axis, where . They are derived using simple geometric considerations.

The structure on the right of the above diagram illustrates the geometry of the central atom B (with mass ) and two of the three A atoms, each with mass . Let be the angle between an A-B-A bond (between FG and GI), be the angle between a B-A bond and the -axis (GH) and be the length of each B-A bond. According to the VSEPR theory, .

Applying the cosine rule on gives . Since , we have

Using the cosine rule again on and noting that yields . Substituting eq20 into this equation results in

Hence, the moment of inertia for symmetric rotors like NH₃ along the -axis is:

To derive , we begin by noting that the centre of mass of the molecule (green sphere) lies along the -axis (one of three principal axes of rotation), between atom B and the plane formed by the three A atoms (see diagram below). The -axis and -axis are orthogonal to the -axis and to each other. They intersect at the centre of mass but do not intersect any of the B-A bonds.

The relationship between the angles and is established by letting atom B be the origin and the three A atoms at angles 0°, 120°, 240° around the -axis. Furthermore, let be the unit vector pointing from atom B to atom A at F, and be the unit vector pointing from atom B to atom A at I (see diagram below).

The dot product of the two vectors is given by or , which is equivalent to

Importantly, eq23 is independent of the choice of origin or reference frame even though it was derived with atom B as the origin.

Now, let’s place the origin at the centre of mass of the molecule. The -coordinates of atom B and each of the A atoms are and respectively. Then, the centre of mass of the molecule satisfies or

with  or

Since the position of the centre of mass relative to that of atom B only involves a shift in the -direction, the and coordinates of the three A atoms are those defined by the unit vectors above multiplied by the factor . Therefore, the positions of the atoms are:

Atom	Coordinates
B
A (point F)
A (point I)
A (point E)

 

 

Since , the moment of inertia of the molecule about the -axis is:

Substituting the data from the above table into eq26 and simplifying gives:

Substituting eq23 into eq27 and simplifying yields:

You’ll find that substituting the data in the above table into the moment of inertia of the molecule about the -axis results in the same expression as eq28. Therefore, eq22 and eq28 are the two distinct moments of inertia of a trigonal pyramidal prolate rotor with symmetry.

 

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The moments of inertia of symmetric rotors play a fundamental role in understanding their rotational dynamics and spectroscopic behaviour.

In such molecules, two of the three moments of inertia are equal, while the third is different. The unique moment of inertia, denoted by , describes rotation around an axis known as the principal axis. The other two, which are equal, are denoted by . If , the molecule is known as a prolate symmetric rotor (shaped like a cigar) and rotates more easily around the principal axis. Examples include NH₃ and CHCl₃. If , the molecule is called an oblate symmetric rotor (flattened like a disc), and it rotates more easily around an axis perpendicular to the disc. Examples include C₆H₆ and BF₃.

 

 

In the next few articles, we will derive the moments of inertia of a few common prolate and oblate symmetric rotors.

 

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The moments of inertia of triatomic linear molecules play a fundamental role in understanding their rotational dynamics and spectroscopic behaviour.

In such molecules — composed of three atoms aligned along a straight line — the distribution of mass relative to the molecular axis determines how the molecule responds to rotational motion. Since all atoms lie along the same axis, the moment of inertia about this axis is effectively zero due to the absence of perpendicular mass displacement. However, the moment of inertia about axes perpendicular to the molecular axis becomes significant and is central to characterising the molecule’s rotational energy levels.

Consider a linear molecule with three atoms A, B and C with masses , and respectively, and bond lengths and (see diagram above). The molecule rotates the centre of mass axis , which passes through the centre of mass of the molecule. If , and are the distances between the centre of mass and the respective atoms, then the parallel axis theorem states that:

where is the moment of inertia about the axis , is the moment of inertia about the axis ,  and .

Substituting in eq11 gives

Let’s assume that the molecular axis is the -axis with the origin at B. By definition, the center of mass with respect to the -coordinate satisfies , which rearranges to:

Substituting eq13 in eq12 yields

Eq14 is the moment of inertia of a triatomic linear molecule where . If , then and eq14 becomes

 

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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

 

 

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 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:

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.

 

 

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:

 

 

Substituting , and  back into eq4c gives

and hence

 

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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.

 

 

 

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.

 

 

 

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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.

 

 

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 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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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

 

 

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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