Advanced Chemistry

September 18, 2025

Mean molecular energies of non-interacting molecules

The mean molecular energy of a system of non-interacting molecules is defined as the average internal energy per molecule, calculated over all possible quantum states accessible to the molecule at a given temperature.

This average is weighted by the Boltzmann probability of each state and reflects the contributions from translational, rotational, vibrational and electronic motions, depending on the system’s complexity.

Mathematically, the mean molecular energy $\langle \varepsilon\rangle$ of a molecule in thermal equilibrium is given by:

$\langle \varepsilon\rangle=\sum_i\varepsilon_ip_i\;\;\;\;\;\;\;\;300$

where $\varepsilon_i$ is energy the $i$ -th state measured relative to the ground state energy of the molecule, and $p_i$ is the probability of the molecule being in that state.

Substituting eq251 into eq300 gives:

$\langle \varepsilon\rangle=\frac{1}{q}\sum_i\varepsilon_ie^{-\beta\varepsilon_i}\;\;\;\;\;\;\;\;301$

where

$q=\sum_ie^{-\beta\varepsilon_i}$ is the molecular partition function,
$k$ is the Boltzmann constant,
$T$ is the absolute temperature,
$\beta=\frac{1}{kT}$ .

Since $\frac{\partial q}{\partial\beta}=\sum_i\frac{\partial}{\partial\beta}e^{-\beta\varepsilon_i}=-\sum_i\varepsilon_ie^{-\beta\varepsilon_i}$ and $\frac{\partial lnq}{\partial q}=\frac{1}{q}$ , eq301 can be expressed as:

$\langle \varepsilon\rangle=-\frac{1}{q}\frac{\partial q}{\partial\beta}\;\;\;\;\;\;\;\;302$

$\langle \varepsilon\rangle=-\frac{\partial lnq}{\partial\beta}\;\;\;\;\;\;\;\;303$

Question

Why does eq303 (or eq302) involve a partial derivative?

Answer

A partial derivative is used because $q$ may be dependent on a few variables, such as $T$ and $V$ (see eq266).

Substituting eq257 into eq303 yields:

$\langle \varepsilon\rangle=\langle \varepsilon^T\rangle+\langle \varepsilon^R\rangle +\langle \varepsilon^V\rangle+\langle \varepsilon^E\rangle\;\;\;\;\;\;\;\;304$

where $\langle \varepsilon^T\rangle=-\frac{\partial lnq^T}{\partial\beta}$ , $\langle \varepsilon^R\rangle=-\frac{\partial lnq^R}{\partial\beta}$ , $\langle \varepsilon^V\rangle=-\frac{\partial lnq^V}{\partial\beta}$ and $\langle \varepsilon^E\rangle=-\frac{\partial lnq^E}{\partial\beta}$ .

Mean translational energy

The mean translational energy $\langle \varepsilon^T\rangle$ of a non-interacting molecule is derived by substituting eq266 into $\langle \varepsilon^T\rangle=-\frac{\partial lnq^T}{\partial\beta}$ to give:

$\langle \varepsilon^T\rangle=-\frac{\partial ln\biggr\[V\biggr$\frac{2\pi m}{h^2\beta}\biggr$^{\frac{3}{2}}\biggr\]}{\partial\beta}=-\frac{\partial ln\biggr$\frac{1}{\beta}\biggr$}{\partial \beta}^{\frac{3}{2}}=\frac{3}{2}kT\;\;\;\;\;\;\;\;305$

Question

Show that each of the three components of $\langle \varepsilon^T\rangle$ is equal to $\frac{1}{2}kT$ .

Answer

Substituting eq261 into $\langle \varepsilon^T\rangle=-\frac{\partial lnq^T}{\partial\beta}$ results in:

$\langle \varepsilon^T\rangle=-\frac{\partial ln q^T_x}{\partial \beta}-\frac{\partial ln q^T_y}{\partial \beta}-\frac{\partial ln q^T_z}{\partial \beta}\;\;\;\;\;\;\;\;306$

Let’s focus on the first derivative in eq306. Substituting eq265 into it yields:

$-\frac{\partial ln q^T_x}{\partial \beta}=-\frac{\partial ln\biggr\[a\biggr$\frac{2\pi m}{h^2\beta}\biggr$^{\frac{1}{2}}\biggr\]}{\partial\beta}=-\frac{\partial ln\sqrt{\frac{1}{\beta}}}{\partial\beta}=\frac{1}{2}kT$

Similarly, each of the other two components is equal to $\frac{1}{2}kT$ because $q^T_y=\sqrt{\frac{2\pi m}{h^2\beta}}b$ and $q^T_z=\sqrt{\frac{2\pi m}{h^2\beta}}c$ .

Thus, each translational degree of freedom contributes $\frac{1}{2}kT$ , consistent with the equipartition theorem.

Mean rotational energy

The mean rotational energy $\langle \varepsilon^R\rangle$ of a non-interacting heteronuclear linear molecule at low temperatures is derived by substituting eq270 into $\langle \varepsilon^R\rangle=-\frac{1}{q^R}\frac{\partial q^R}{\partial\beta}$ to give:

$\langle \varepsilon^R\rangle=-\frac{1}{\sum_J(2J+1)e^{-hcBJ(J+1)\beta}}\frac{\partial\sum_J(2J+1)e^{-hcBJ(J+1)\beta}}{\partial\beta}\;\;\;\;\;\;\;\;307$

At very low temperatures, almost all molecules occupy the ground state with $J=0$ , so eq307 reduces to $\langle \varepsilon^R\rangle\approx 0$ . As the temperature increases, $\langle \varepsilon^R\rangle$ is approximately given by expanding the summations in eq307 and differentiating to yield:

$\langle \varepsilon^R\rangle=-\frac{hcB(6e^{-2hcB\beta}+30e^{-6hcB\beta}+\cdots)}{1+3e^{-2hcB\beta}+5e^{-6hcB\beta}+\cdots}\;\;\;\;\;\;\;\;308$

At higher temperatures, we substitute eq275, where $\theta_R=\frac{hcB}{k}$ and $\beta=\frac{1}{kT}$ , into $\langle \varepsilon^R\rangle=-\frac{1}{q^R}\frac{\partial q^R}{\partial\beta}$ to give:

$\langle \varepsilon^R\rangle=-\sigma\beta hcB\frac{\partial}{\partial\beta}\frac{1}{\sigma hcB\beta}=kT\;\;\;\;\;\;\;\;309$

which is consistent with the equipartition theorem.

Since the symmetry number $\sigma$ cancels out, the mean rotational energy of a non-interacting symmetrical linear molecule at high temperatures is also $\langle \varepsilon^R\rangle=kT$ . Unlike linear rotors, which have two independent rotational degrees of freedom, spherical rotors, symmetric rotors and asymmetric rotors have three. Substituting eq280, eq286 and eq287 separately into $\langle \varepsilon^R\rangle=-\frac{1}{q^R}\frac{\partial q^R}{\partial\beta}$ and differentiating results in $\langle \varepsilon^R\rangle=\frac{3}{2}kT$ for each of the three types of rotors, which is again consistent with the equipartition theorem.

Mean vibrational energy

The mean vibrational energy $\langle \varepsilon^V\rangle$ of a non-interacting diatomic molecule oscillating harmonically at low temperatures is derived by substituting eq291 into $\langle \varepsilon^V\rangle=-\frac{1}{q^V}\frac{\partial q^V}{\partial\beta}$ , where $\theta_V=\frac{hc\tilde\nu}{k}$ , to give:

$\langle \varepsilon^V\rangle=-\frac{1-e^{-hc\tilde\nu\beta}}{e^{-\frac{1}{2}hc\tilde\nu\beta}}\frac{\partial}{\partial\beta}\frac{e^{-\frac{1}{2}hc\tilde\nu\beta}}{1-e^{-hc\tilde\nu\beta}}\;\;\;\;\;\;\;\;310$

If we compute the derivative in eq310 and multiply the result by $\frac{e^{hc\tilde\nu\beta}}{e^{hc\tilde\nu\beta}}$ , we get:

$\langle \varepsilon^V\rangle=-\frac{1}{2}hc\tilde\nu\frac{e^{hc\tilde\nu\beta}+1}{e^{hc\tilde\nu\beta}-1}\;\;\;\;\;\;\;\;311$

At high temperatures, $hc\tilde\nu\beta\ll 1$ , which allows us to expand $e^{hc\tilde\nu\beta}$ as a Taylor series ( $e^x=1+x+\cdots$ ) to give:

$\langle \varepsilon^V\rangle=\frac{1}{2}hc\tilde\nu\frac{(1+hc\tilde\nu\beta+\cdots)+1}{(1+hc\tilde\nu\beta+\cdots)-1}=kT+\frac{1}{2}hc\tilde\nu\;\;\;\;\;\;\;\;312$

As mentioned in an earlier article, eq291 evaluates absolute energy levels, including the zero-point energy. If we derive $\langle \varepsilon^V\rangle$ using eq293 instead of eq291, we obtain:

$\langle \varepsilon^V\rangle=kT\;\;\;\;\;\;\;\;313$

Although the equipartition theorem states that the mean vibrational energy of a classical oscillator at high temperatures is equal to $kT$ , both eq312 and eq313 are consistent with this result. This is because the theorem accounts only for the thermal contribution to vibrational energy; the zero-point energy term $\frac{1}{2}hc\tilde\nu$ is a quantum mechanical artifact that is independent of temperature.

For polyatomic molecules, each normal mode of vibration behaves approximately like a separate harmonic oscillator, with the total vibrational partition function given by eq294:

$q^V_{total}=\prod^f_{i=1}q^V_i$

where $f=3N-6$ for non-linear molecules, $f=3N-5$ for linear molecules, and $N$ is the number of atoms.

So,

$lnq^V_{total}=ln\prod^f_{i=1}q^V_i=\sum_{i=1}^flnq^V_i$

Since $\langle \varepsilon^V_i\rangle=-\frac{1}{q^V_i}\frac{\partial q^V_i}{\partial\beta}=-\frac{\partial lnq^V_i}{\partial\beta}$ ,

$\langle\varepsilon^V_{total}\rangle=-\frac{\partial lnq^V_{total}}{\partial\beta}=-\frac{\partial \sum_{i=1}^f lnq^V_i}{\partial\beta}=\sum_{i=1}^f\langle\varepsilon^V_i\rangle\;\;\;\;\;\;\;\;314$

In other words, the total mean vibrational energy of a non-interacting polyatomic molecule oscillating harmonically is the sum of the mean energy of each normal mode. This implies that

$\langle\varepsilon^V_{total}\rangle=(3N-6)kT\;\;\;\;\;non-linear\;molecule$

$\langle\varepsilon^V_{total}\rangle=(3N-5)kT\;\;\;\;\;linear\;molecule$

which are consistent with the equipartition theorem at high temperatures.

Mean electronic energy

The mean electronic energy $\langle \varepsilon^E\rangle$ of a non-interacting molecule is derived by substituting eq296, where $\beta=\frac{1}{kT}$ , into $\langle \varepsilon^E\rangle=-\frac{1}{q^E}\frac{\partial q^E_i}{\partial\beta}$ to give:

$\langle \varepsilon^E\rangle=\frac{g_1\varepsilon_1e^{-\varepsilon_1\beta}+g_2\varepsilon_2e^{-\varepsilon_2\beta}+\cdots}{g_0+g_1e^{-\varepsilon_1\beta}+g_2e^{-\varepsilon_2\beta}+\cdots}\;\;\;\;\;\;\;\;315$

At low temperatures, $\beta\rightarrow\infty$ . Therefore, all the numerator terms in eq315 approach zero, giving:

$\langle \varepsilon^E\rangle\approx 0\;\;\;\;\;\;\;\;316$

This can be explained qualitatively: the energy gap between the ground electronic state and the first excited electronic state of a typical molecule is very large, which results in the molecule occupying only the electronic ground state, which has zero energy.

At higher temperatures, the mean electronic energy increases as excited electronic states become thermally accessible. However, this regime is often not reached before the molecule dissociates

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September 18, 2025

The canonical ensemble

A canonical ensemble is a collection of identical, non-interacting copies of a physical system, each with a fixed number of particles $N$ , fixed volume $V$ , and constant temperature $T$ (maintained by thermal equilibrium with a heat bath), in which particles within each system may interact.

The word “canonical” means “according to a rule”, which, in this case, refers to fixed $N$ , $V$ and $T$ . As mentioned in the above definition, a canonical ensemble clearly states that even though the individual systems in the ensemble are identical from a thermodynamic perspective, they may not be identical at the molecular level. Such a theoretical construct serves to connect the microscopic behaviour of systems (governed by quantum mechanics or classical mechanics) to the macroscopic thermodynamic properties.

Question

Why do we only consider three parameters for each system?

Answer

While a physical system, such as a salt solution, may involve many microscopic variables (like individual particle positions and velocities), its macroscopic behaviour can often be fully characterised by a small set of thermodynamic parameters. In the canonical ensemble, fixing $N$ , $V$ and $T$ is sufficient to statistically describe the equilibrium properties of the system.

All thermodynamic properties are, in essence, measured as averages. Consider each system in the ensemble as a rigid container holding a gas. The pressure of the gas in each system is the time average of countless rapid collisions of gas particles with the walls of the container. The force on the wall fluctuates constantly, but a pressure gauge cannot register each individual fluctuation. Instead, the gauge averages the force over a short time to provide a stable reading. Similarly, while the kinetic energy of individual molecules varies moment to moment, the temperature reflects the average kinetic energy of all molecules at thermal equilibrium.

Due to fluctuations in properties such as pressure, kinetic energy or even quantum-mechanical variables, each system can occupy many possible energy states (microstates) at any given moment. In quantum mechanical systems, the energy corresponding to each microstate is determined by solving the Schrodinger equation $\hat{H}\psi_i=E_i\psi_i$ , where $\psi_i$ is the wavefunction that describes the quantum state resulting from the positions and momenta of all $N$ particles in the system, and $E_i$ is the total energy of system for the $i$ -th microstate.

If the number of systems $\tilde N$ in the ensemble is taken to be infinitely large, we postulate that:

Postulate 1

The macroscopic properties of a system are given by statistical averages over all possible microstates in the ensemble at any given moment (ensemble average).

For example, the thermodynamic internal energy $U$ is equal to the statistical average energy of the systems $\langle E\rangle$ in the ensemble:

$U=\langle E\rangle=\sum_ip_iE_i\;\;\;\;\;\;\;\;120$

where $p_i$ is the probability that a system in the ensemble is in the $i$ -th microstate with energy $E_i$ .

To determine $p_i$ , we further postulate that:

Postulate 2

All microstates with equal energy have equal probability of occurring for any system of fixed volume, composition and temperature.

Therefore, if we consider any such system in thermal equilibrium with a constant temperature heat bath, and identify two of its microstates, 1 and 2, with energies $E_1$ and $E_2$ respectively, the corresponding probabilities are given by $p_1=\frac{\tilde N_1}{\tilde N}=f(E_1)$ and $p_2=\frac{\tilde N_2}{\tilde N}=f(E_2)$ , where $\tilde N_i$ is the number of systems with energy $E_i$ . It follows that the relative probability is:

$\frac{\tilde N_2}{\tilde N_1}=f(E_1,E_2)\;\;\;\;\;\;\;\;121$

In thermodynamics and quantum mechanics, energy is always referenced to a zero of energy $E_0$ , which can be defined in different ways. Since the function $f(E_1,E_2)\equiv f(E_1-E_0,E_2-E_0)$ represents a physical quantity, it must yield the same result regardless of how $E_0$ is defined. This is only possible if the function $f(E_1,E_2)$ depends on the difference between $E_1$ and $E_2$ , i.e. $f(E_1,E_2)= f(E_1-E_2)$ .

Question

Show that the function $f(E_1-E_2)$ is invariant to $E_0$ .

Answer

Let $E_3=E_1-E_0$ and $E_4=E_2-E_0$ . We have $E_3-E_4=E_1-E_2$ , which does not depend on $E_0$ .

Multiplying $\frac{\tilde N_2}{\tilde N_1}=f(E_1-E_2)$ by $\frac{\tilde N_3}{\tilde N_2}=f(E_2-E_3)$ and using $\frac{\tilde N_3}{\tilde N_1}=f(E_1-E_3)$ gives:

$f(E_1-E_3)=f(E_1-E_2)f(E_2-E_3)\;\;\;\;\;\;\;\;122$

Letting $E_5=E_1-E_2$ and $E_6=E_2-E_3$ , we can rewrite eq122 as:

$f(E_5+E_6)=f(E_5)f(E_6)\;\;\;\;\;\;\;\;123$

Taking the natural logarithm on both sides of eq123 and then differentiating with respect to $E_5$ yields:

$\biggr\[\frac{\partial lnf(E_5+E_6)}{\partial E_5}\biggr\]_{E_6}=\frac{d lnf(E_5)}{d E_5}$

Using the chain rule,

$\frac{dlnf(E_5+E_6)}{d(E_5+E_6)}\biggr\[\frac{\partial (E_5+E_6)}{\partial E_5}\biggr\]_{E_6}=\frac{d lnf(E_5)}{d E_5}$

$\frac{dlnf(E_5+E_6)}{d(E_5+E_6)}=\frac{d lnf(E_5)}{d E_5}\;\;\;\;\;\;\;\;124$

Similarly, taking the natural logarithm on both sides of eq123 and then differentiating with respect to $E_6$ yields:

$\frac{dlnf(E_5+E_6)}{d(E_5+E_6)}=\frac{d lnf(E_6)}{d E_6}\;\;\;\;\;\;\;\;125$

Equating eq124 and eq125 results in:

$\frac{d lnf(E_5)}{d E_5}=\frac{d lnf(E_6)}{d E_6}\;\;\;\;\;\;\;\;126$

Each side of eq126 is a function of a different independent variable. The only way eq126 can be valid for all values of $E_5$ and $E_6$ is if both functions are equal to the same constant $\beta$ :

$\frac{d lnf(E_5)}{d E_5}=\frac{d lnf(E_6)}{d E_6}=\beta\;\;\;\;\;\;\;\;127$

Integrating $\frac{d lnf(E_5)}{d E_5}=\beta$ , i.e. $\int d lnf(E_5)=\int\beta dE_5$ , gives $lnf(E_5)=\beta E_5+c$ or equivalently,

$f(E_5)=ae^{\beta E_5}\;\;\;\;\;\;\;\;128$

where $a=e^c$ .

Accordingly, we have:

$f(E_h-E_i)=ae^{\beta (E_h-E_i)}\;\;\;\;\;\;\;\;129$

Substituting eq129 into $\frac{\tilde N_i}{\tilde N_h}=f(E_h-E_i)$ yields $\frac{\tilde N_i}{\tilde N_h}=ae^{\beta (E_h-E_i)}$ or:

$\tilde N_i=Ce^{-\beta E_i}\;\;\;\;\;\;\;\;130$

where $C=\tilde N_hae^{\beta E_h}$ .

$C$ can be evaluated by summing both sides of eq130 over $i$ :

$\sum_i\tilde N_i=C\sum_ie^{-\beta E_i}$

Since $\sum_i\tilde N_i=\tilde N$ ,

$C=\frac{\tilde N}{\sum_ie^{-\beta E_i}}\;\;\;\;\;\;\;\;131$

Substituting eq131 back into eq130 results in:

$p_i=\frac{e^{-\beta E_i}}{\sum_ie^{-\beta E_i}}\;\;\;\;\;\;\;\;133$

where $p_i=\frac{\tilde N_i}{\tilde N}$ .

Eq133 represents the general probability of finding a system, which is in thermal equilibrium with a constant temperature heat bath, in a specific microstate $i$ with energy $E_i$ . Its derivation relies on fundamental principles of probability and the nature of equilibrium, not on the specific nature of the particles themselves (e.g., whether they’re distinguishable or indistinguishable). In other words, eq133 is applicable to both classical and quantum mechanical systems. Therefore, eq120 becomes:

$U=\langle E\rangle=\frac{\sum_iE_ie^{-\beta E_i}}{Q}\;\;\;\;\;\;\;\;134$

where

$Q=\sum_je^{-\beta E_j}\;\;\;\;\;\;\;\;135$

is known as the canonical partition function.

The fact that eq133 yields the same expression as the Boltzmann distribution is not a coincidence. It demonstrates that the Boltzmann distribution naturally emerges from the statistical treatment of systems in thermal equilibrium, and illustrates the deep consistency between thermodynamic principles and statistical reasoning.

Question

Can two different microstates have the same energy?

Answer

Yes. Consider a system with two particles. In one microstate, particle A has an energy of 1 unit, and Particle B has an energy of 2 units. In another microstate, particle A has an energy of 2 units, while Particle B has an energy of 1 unit. These two distinct microstates (or configurations) clearly have the same total energy of 3 units. Microstates that share the same energy are called degenerate states.

Since different microstates can be degenerate, the probability that a system has energy $E_i$ is:

$p(E_i)=\frac{g_ie^{-\beta E_i}}{\sum_je^{-\beta E_j}}\;\;\;\;\;\;\;\;136$

where $g_i$ is the degeneracy of $E_i$ .

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September 18, 2025

Canonical partition function for a system of non-interacting particles

The canonical partition function is a central quantity in statistical mechanics, encapsulating the thermodynamic behaviour of a system in equilibrium with a heat reservoir at fixed temperature. It allows us to derive key thermodynamic quantities of a system, such as pressure (see eq146) and internal energy (see eq148). These relationships, e.g. $U=-\biggr$\frac{\partial Q}{\partial \beta}\biggr$_{V,N}$ , applies to both systems with interacting and non-interacting particles.

For a system with interacting particles, the Schrödinger equation is incredibly complex due to the interaction terms between the $N$ particles, making it difficult to calculate the exact energy $E_j$ of a given microstate . On the other hand, in a system of non-interacting particles, intermolecular forces are absent. This simplifies the evaluation of $Q$ significantly, since the Hamiltonian operator for a system of $N$ non-interacting particles can be expressed as $\hat{H}=\hat{H}_1+\hat{H}_2+\cdots+\hat{H}_N$ . The total energy of the system then becomes the sum of the individual particle energies,

$E_j=\varepsilon^j_{1,m}+\varepsilon^j_{2,n}+\cdots+\varepsilon^j_{N,y}\;\;\;\;\;\;\;\;160$

Here $=\varepsilon^j_{1,m}$ refers to the energy of particle 1 in quantum state $=m$ when the system is in microstate $j$ . The energy of each particle is determined by solving the corresponding one-particle Schrödinger equation, e.g. $\hat{H}_1\psi^j_{1,m}=\varepsilon_{1,m}\psi^j_{1,m}$ .

If the non-interacting particles are distinguishable (e.g. atoms or molecules with fixed positions in a solid), then the microstate of the system is defined by the quantum state of each individual particle. For example, microstate 1 might have the configuration $\langle m,n,\cdots,y\rangle$ , while microstate 2 is $\langle n,m,\cdots,y\rangle$ . The sum over all microstates $j$ of the system, by substituting eq160 into eq155, is given by:

$Q=e^{-\frac{E_1}{kT}}+e^{-\frac{E_2}{kT}}+\cdots=e^{-\frac{\varepsilon^1_{1,m}+\varepsilon^1_{2,n}+\cdots}{kT}}+e^{-\frac{\varepsilon^2_{1,n}+\varepsilon^2_{2,m}+\cdots}{kT}}+\cdots$

which is equivalent to:

$Q=e^{-\frac{\varepsilon^1_{1,1}}{kT}}e^{-\frac{\varepsilon^1_{2,2}}{kT}}\cdots+e^{-\frac{\varepsilon^2_{1,2}}{kT}}e^{-\frac{\varepsilon^2_{2,1}}{kT}}\cdots+\cdots\;\;\;\;\;\;\;\;161$

where we have set $m=1$ and $n=2$ for illustrative purposes.

Each term on the RHS of eq161 corresponds to a distinct microstate, represented by a specific permutation of $N$ exponential factors. Eq161 can also be written as:

$Q=\sum_me^{-\frac{\varepsilon^m_{1,m}}{kT}}\sum_ne^{-\frac{\varepsilon^n_{2,n}}{kT}}\cdots\sum_ye^{-\frac{\varepsilon^y_{N,y}}{kT}}\;\;\;\;\;\;\;\;162$

We can show that eq162 is equivalent to eq161 by expanding it:

$Q=\biggr$e^{-\frac{\varepsilon^1_{1,1}}{kT}}+e^{-\frac{\varepsilon^2_{1,2}}{kT}}+\cdots\biggr$\biggr$e^{-\frac{\varepsilon^1_{2,1}}{kT}}+e^{-\frac{\varepsilon^2_{2,2}}{kT}}+\cdots\biggr$\cdots\;\;\;\;\;\;\;\;163$

If we multiply out all the terms in eq163, each resulting term corresponds to a unique microstate, as expressed in eq161. Therefore,

$Q=\sum_je^{-\frac{E_j}{kT}}=q_1q_2\cdots q_N\;\;\;\;\;\;\;\;164$

where $q_1=\sum_me^{-\frac{\varepsilon^m_{1,m}}{kT}}$ , $q_2=\sum_ne^{-\frac{\varepsilon^n_{2,n}}{kT}}$ ,…, $q_N=\sum_ye^{-\frac{\varepsilon^y_{N,y}}{kT}}$ are known as the molecular partition functions.

Since each $q_N$ represents the total statistical weight of a single particle, the superscript $y$ in $\varepsilon^y_{N,y}$ becomes redundant. Thus, we can express a molecular partition function more simply as $q_N=\sum_ye^{-\frac{\varepsilon_{N,y}}{kT}}$ . If $N_1$ molecules are of the same species, where $N_1 < N$ , then

$Q=(q_1)^{N_1}q_2\cdots q_{N-N_1}\;\;\;\;\;\;\;\;165$

Question

Using the example of a system of 3 distinguishable, identical, non-interacting particles, show that the sum over all microstates of the system then becomes the product of the separate sums of the quantum states of each particle.

Answer

Suppose each particle can be in one of two quantum states with energy $\varepsilon_{0}=0$ and $\varepsilon_{1}=\varepsilon$ . The total possible microstates is $2^3=8$ , comprising of:

Microstate	Total energy
$\langle 0,0,0\rangle$	$0$
$\langle \varepsilon,0,0\rangle$	$\varepsilon$
$\langle 0,\varepsilon,0\rangle$	$\varepsilon$
$\langle 0,0,\varepsilon\rangle$	$\varepsilon$
$\langle \varepsilon,\varepsilon,0\rangle$	$2\varepsilon$
$\langle \varepsilon,0,\varepsilon\rangle$	$2\varepsilon$
$\langle 0,\varepsilon,\varepsilon\rangle$	$2\varepsilon$
$\langle \varepsilon,\varepsilon,\varepsilon\rangle$	$3\varepsilon$

So,

$Q=1+3e^{-\frac{\varepsilon}{kT}}+3e^{-\frac{2\varepsilon}{kT}}+e^{-\frac{3\varepsilon}{kT}}$

Similarly,

$Q=q^3=\biggr$1+e^{-\frac{\varepsilon}{kT}}\biggr$^3=1+3e^{-\frac{\varepsilon}{kT}}+3e^{-\frac{2\varepsilon}{kT}}+e^{-\frac{3\varepsilon}{kT}}$

It follows that for $N$ distinguishable, identical, non-interacting particles,

$Q=q^N\;\;\;\;\;\;\;\;166$

If the $N$ identical, non-interacting particles are indistinguishable, such as those of an ideal gas, and if each particle occupies a distinct quantum state, then the total number of microstates is no longer given by the number of permutations of the quantum states, but by the number of combinations. One way to approximate $Q$ in this case is to divide eq166 by a factor $N!$ :

$Q=\frac{q^N}{N!}\;\;\;\;\;\;\;\;167$

This gives $Q_{indist}=\frac{Q_{dist}}{N!}$ at high temperatures, where each particle is assumed to be in a distinct quantum state. It is important to note that the relation $Q_{indist}=\frac{Q_{dist}}{N!}$ is not a strict mathematical identity, because $Q_{dist}$ is not a literal count of microstates but a weighted sum of exponential terms, each corresponding to a microstate. Nevertheless, this approximation works well in the classical limit and successfully resolves the Gibbs paradox.

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Thermodynamic functions involving the canonical partition function

Thermodynamic functions involving the canonical partition function describe thermodynamic properties of a canonical ensemble.

In statistical mechanics, the canonical ensemble provides a powerful framework for connecting microscopic states to macroscopic thermodynamic properties. Central to this approach is the canonical partition function, $Q$ , which encapsulates the statistical behaviour of a system in thermal equilibrium at fixed temperature, volume, and particle number. From $Q$ , one can derive key thermodynamic functions such as pressure, internal energy and entropy through straightforward mathematical relations.

In the previous article, we derived the expressions for pressure and internal energy:

$P=\frac{1}{\beta}\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}\;\;\;\;\;\;\;\;146$

$U=-\biggr$\frac{\partial lnQ}{\partial \beta}\biggr$_{V,N}\;\;\;\;\;\;\;\;148$

where $\beta=\frac{1}{kT}$ and $Q=\sum_je^{-\frac{E_j}{kT}}$ .

Question

How does $P=\frac{1}{\beta}\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}$ connect quantum mechanics to thermodynamics?

Answer

$P$ , the pressure of a system, is a macroscopic thermodynamic property, while $E_i$ is the energy of a quantum mechanical microstate, which is an eigenvalue of the Hamiltonian. Therefore, the equation provides a bridge between the microscopic quantum description and macroscopic thermodynamic observables.

The corresponding expression for entropy $S$ can be derived by first dividing the fundamental equation of thermodynamics by $T$ to give:

$dS=d\biggr$\frac{U}{T}\biggr$+\frac{U}{T^2}dT+\frac{P}{T}dV\;\;\;\;\;\;\;\;156$

Applying the chain rule to eq148 gives:

$U=-\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}\biggr$\frac{d T}{d \beta}\biggr$=kT^2\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}\;\;\;\;\;\;\;\;157$

Substituting eq146 and eq157 into eq156 yields:

$\begin{align}dS&=d\biggr$\frac{U}{T}\biggr$+k\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}dT+k\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}dV\\&=d\biggr$\frac{U}{T}\biggr$+kdlnQ\;\;\;\;\;\;\;\;158\end{align}$

Integrating eq158 results in $S=\frac{U}{T}+klnQ+C$ , where $C$ is a constant. As $T\rightarrow 0$ , only the ground state $E_0$ is populated. Assuming the ground state is non-degenerate, the partition function becomes $Q=e^{-\frac{E_0}{kT}}$ , with $S=\frac{E_0}{T}-\frac{E_0}{T}+C=C$ . According to the third law of thermodynamics, entropy approaches zero as the temperature approaches absolute zero. Therefore, $C$ must be zero to satisfy this requirement, resulting in:

$dS=\frac{U}{T}+klnQ=kT\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}+klnQ\;\;\;\;\;\;\;\;159$

It follows that the Helmholtz energy expression is:

$A=U-TS=-kTlnQ\;\;\;\;\;\;\;\;159a$

Using the pressure and internal energy equations, the expression for enthalpy is:

$H=U+PV=KT^2\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}+VkT\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}\;\;\;\;\;\;\;\;159b$

Question

Does eq159b contradict the application of enthalpy, which describes a system at constant pressure?

Answer

The definition of enthalpy, $H=U+PV$ , holds for any thermodynamic state, regardless of whether the pressure is constant. The condition of constant pressure becomes relevant when evaluating the change in enthalpy, $\Delta H$ , which equals the heat exchanged in a constant-pressure process $q_p$ . Eq159b expresses the enthalpy of a system in terms of $N$ , $V$ and $T$ . It gives the enthalpy for a particular equilibrium state, not just under constant pressure. Therefore, we can calculate the enthalpy of a system at any given state using eq159b. By evaluating it at two different states (e.g. the initial and final states of a process), we obtain $\Delta H$ . If the process connecting those states occur at constant pressure, then $\Delta H=q_p$ . So, there is no contradiction.

The expression for constant-volume heat capacity is:

$\begin{align}C_V&=\biggr$\frac{\partial U}{\partial T}\biggr$_{V,N}\\&=k\biggr\[\frac{\partial T^2\biggr$\frac{\partial lnQ}{\partial T}\biggr$}{\partial T}\biggr\]_{V,N}\\&=KT^2\biggr$\frac{\partial^2 lnQ}{\partial^2 T}\biggr$_{V,N}+2kT\biggr$\frac{\partial lnQ}{\partial T}\biggr$_{V,N}\;\;\;\;\;\;\;\;159c\end{align}$

For Gibbs energy, we have $G=H-TS=U+PV-TS=A+PV$ . Hence,

$G=-kTlnQ+KTV\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}\;\;\;\;\;\;\;\;159d$

If we define chemical potential as $\mu=\biggr$\frac{\partial A}{\partial N}\biggr$_{T,V}$ , and $N$ is an integer-valued discrete variable, then the derivative becomes a finite difference, so:

$\mu\approx A_N-A_{N-1}=-kTln\biggr$\frac{Q_N}{Q_{N-1}}\biggr$\;\;\;\;\;\;\;\;159e$

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Vibrational molecular partition function

The vibrational molecular partition function $q^V$ is a measure of the number of thermally accessible vibrational energy states available to a single non-interacting molecule at a specific temperature.

Consider a diatomic harmonic oscillator with vibrational energies given by eq21: $\varepsilon_n=\biggr$n+\frac{1}{2}\biggr$\hbar\omega=\biggr$n+\frac{1}{2}\biggr$hc\tilde\nu$ , where $n=0,1,2,\cdots$ . Substituting this equation into the vibrational component of eq257 gives:

$q^V=e^{-\frac{hc\tilde\nu}{2kT}}\sum_{n=0}^{\infty}\biggr$e^{-\frac{hc\tilde\nu}{2kT}}\biggr$^n\;\;\;\;\;\;\;\;290$

Since $\frac{hc\tilde\nu}{kT}> 0$ for any temperature above absolute zero, we have $e^{-\frac{hc\tilde\nu}{kT}}< 1$ . Thus, the summation in eq290 can be evaluated as a binomial series, $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ , yielding:

$q^V=\frac{e^{-\frac{\theta_V}{2T}}}{1-e^{-\frac{\theta_V}{T}}}\;\;\;\;\;\;\;\;291$

where the characteristic vibrational temperature $\theta_V=\frac{hc\tilde\nu}{k}$ is the analogue to the characteristic rotational temperature.

The vibrational partition function of eq291 expresses the exponential terms in terms of the absolute energy levels of the quantum harmonic oscillator, where the ground state energy is the zero-point energy $\frac{1}{2}hc\tilde\nu$ . An alternative convention is to measure energies relative to the zero-point energy. In this case, the vibrational energy levels are given by:

$\varepsilon_n=\biggr$n+\frac{1}{2}\biggr$hc\tilde\nu-\frac{1}{2}hc\tilde\nu=nhc\tilde\nu$

This effectively sets the ground state energy to zero. The corresponding partition function becomes:

$q^V=\sum_{n=0}^{\infty}e^{-\frac{nhc\tilde\nu}{kT}}\;\;\;\;\;\;\;\;292$

Applying the same binomial series summation results in:

$q^V=\frac{1}{1-e^{-\frac{\theta_V}{T}}}\;\;\;\;\;\;\;\;293$

Both eq291 and eq293 are valid expressions of the vibrational partition function. Eq293 is usually preferred because it simplifies calculations by omitting the constant zero-point energy for the ground state. However, when computing absolute energy quantities, such as the internal energy, the zero-point energy must be added back in.

Question

Show that eq291 and eq293 converges to the same expression for low frequency modes at high temperatures.

Answer

For low frequency modes at high temperatures, $\frac{\theta_V}{T}\ll 1$ , and $e^{-\frac{\theta_V}{T}}$ can be approximated by the first-order Taylor series expansion: $e^{-\frac{\theta_V}{T}}\approx 1-\frac{\theta_V}{T}$ . Eq291 reduces to $q^V=\frac{1-\theta_V/2T}{\theta_V/T}=\frac{T}{\theta_V}-\frac{1}{2}$ . At high $T$ ,

$q^V\approx\frac{T}{\theta_V}$

Similarly, eq293 becomes $q^V=\frac{T}{\theta_V}$ .

The convergence of the two vibrational partition function conventions for low-frequency modes at high temperatures reflects the emergence of classical behaviour where quantum corrections like zero-point energy become negligible. This demonstrates how statistical mechanics unifies quantum mechanics and classical thermodynamics.

For polyatomic molecules, each normal mode of vibration behaves approximately like a separate harmonic oscillator. Assuming that these vibrational modes are independent and separable, with negligible anharmonicity effects, the total vibrational energy of the molecule is the sum of the energies of the normal modes. It follows from eq292 that the total vibrational partition function is the product of the individual partition functions for each mode:

$q^V_{total}=\prod_{1=1}^fq^V_i\;\;\;\;\;\;\;\;294$

where $f=3N-6$ for non-linear molecules, $f=3N-5$ for linear molecules, and $N$ is the number of atoms per molecule.

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Translational molecular partition function

The translational molecular partition function $q^T$ is a measure of the number of thermally accessible translational energy states available to a single non-interacting molecule in a given volume at a specific temperature.

Consider a particle of mass $m$ moving freely in a three-dimensional box of lengths $a$ , $b$ and $c$ . Its translational energy $\varepsilon^T$ is given by eq45h:

$\varepsilon^T=\varepsilon_{n_x}+\varepsilon_{n_y}+\varepsilon_{n_z}\;\;\;\;\;\;\;\;260$

where $\varepsilon_{n_x}=\frac{h^2n^2_x}{8ma^2}$ , $\varepsilon_{n_y}=\frac{h^2n^2_y}{8mb^2}$ , $\varepsilon_{n_z}=\frac{h^2n^2_z}{8mc^2}$ , with $n_x,n_y,n_z=1,2,\cdots$ .

Substituting eq260 into the translational component of eq257 gives:

$q^T=q^T_xq^T_yq^T_z\;\;\;\;\;\;\;\;261$

where $q^T_x=\sum_{n_x}e^{-\frac{\varepsilon_{n_x}}{kT}}$ , $q^T_y=\sum_{n_y}e^{-\frac{\varepsilon_{n_y}}{kT}}$ and $q^T_z=\sum_{n_z}e^{-\frac{\varepsilon_{n_z}}{kT}}$ .

To simplify eq261, we need to evaluate the three summations, each of which can be approximated as an integral. For example:

$q^T_x=\sum_{n_x}e^{-\frac{\varepsilon_{n_x}}{kT}}\approx \int^{\infty}_1e^{-\frac{h^2n^2_x}{8ma^2kT}}dn_x\;\;\;\;\;\;\;\;262$

This approximation is valid because the exponential term varies smoothly with $n_x$ for a particle in a laboratory-scale vessel, where $\frac{h^2n^2_x}{8ma^2kT}$ is typically on the order of 10^-20 or smaller. It follows that $e^{-\frac{h^2n^2_x}{8ma^2kT}}\approx 1$ for $0\leq n_x\leq 1$ . Therefore, $\int^{1}_0e^{-\frac{h^2n^2_x}{8ma^2kT}}dn_x\approx \int^1_0 dn_x=1$ . On the other hand, the full integral $\int^{\infty}_0e^{-\frac{h^2n^2_x}{8ma^2kT}}dn_x$ is a well-known Gaussian integral:

$\int^{\infty}_0e^{-Cn^2_x}dn_x=\frac{1}{2}\sqrt{\frac{\pi}{C}}\;\;\;\;\;\;\;\;263$

If $C$ is a very small number, then eq263 evaluates to a very large value (typically on the order of 10¹⁰). Therefore, we can change the lower limit of the integral in eq262 from 1 to 0 with negligible error:

$q^T_x\approx \int^{\infty}_0e^{-\frac{h^2n^2_x}{8ma^2kT}}dn_x\;\;\;\;\;\;\;\;264$

From the identity $\int^{\infty}_{-\infty}e^{-Cx^2}dx=\frac{\pi}{C}$ (see this article for derivation), we have $\int^{\infty}_{0}e^{-Cx^2}dx=\frac{1}{2}\frac{\pi}{C}$ since the integrand is an even function (symmetric about the vertical axis). Therefore, eq264 evaluates to:

$q^T_x= \sqrt{\frac{2\pi mkT}{h^2}}a\;\;\;\;\;\;\;\;265$

Similarly, we have $q^T_y= \sqrt{\frac{2\pi mkT}{h^2}}b$ and $q^T_z= \sqrt{\frac{2\pi mkT}{h^2}}c$ , with eq261 becoming:

$q^T=V\biggr$ \frac{2\pi mkT}{h^2}\biggr$^{\frac{3}{2}}\;\;\;\;\;\;\;\;266$

where $V=abc$ is the volume of the vessel.

By considering the energy levels that result from motion in three-dimensional space, the translational partition function enables the calculation of macroscopic properties such as pressure, entropy and internal energy for ideal gases. This function is especially crucial for connecting microscopic molecular behaviour with observable bulk thermodynamic quantities.

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Rotational molecular partition function

The rotational molecular partition function $q^R$ is a measure of the number of thermally accessible rotational energy states available to a single non-interacting molecule at a specific temperature.

Linear rotors

Consider a heteronuclear diatomic molecule with rotational energies given by eq45: $\varepsilon^R_J=hcBJ(J+1)$ . Substituting this equation into the rotational component of eq257, noting that each energy level is $(2J+1)$ -fold degenerate, gives:

$q^R=\sum_j(2J+1)e^{-\frac{hcBJ(J+1)}{kT}}\;\;\;\;\;\;\;\;270$

At low temperatures, $q^R$ is calculated by substituting experimental values of $\varepsilon^R_J$ into eq270. At higher temperatures, such as typical laboratory conditions at room temperature, many rotational energy levels become thermally accessible and populated because $kT\gg hcB$ . Since the exponential term varies smoothly with $J$ at room temperature when $\frac{hcBJ(J+1)}{kT}\ll 1$ , we can simplify eq270 by approximating the sum as an integral:

$q^R\approx \int_0^{\infty}(2J+1)e^{-\frac{hcBJ(J+1)}{kT}}dJ\;\;\;\;\;\;\;\;271$

Substituting $u=\frac{hcBJ(J+1)}{kT}$ and $du=(2J+1)\frac{hcB}{kT}dJ$ into eq271 yields $q^R=\frac{kT}{hcB} \int_0^{\infty}e^{-u}du$ , which evaluates to:

$q^R= \frac{kT}{hcB}\;\;\;\;\;\;\;\;272$

Since $\frac{k}{hcB}$ has units of Kelvin, eq272 can also be expressed as:

$q^R= \frac{T}{\theta_R}\;\;\;\;\;\;\;\;273$

where $\theta_R=\frac{hcB}{k}$ is known as the characteristic rotational temperature.

Eq273 is valid only when $T\gg \theta_R$ , which is generally true at room temperature. For this reason, it is often referred to as the high-temperature approximation to the rotational partition function.

As discussed in the article on the effect of nuclear statistics on rotational states, nuclear spins dictate which rotational states are populated for symmetrical molecules such as O₂ and CO₂, resulting in half of all possible rotational states being inaccessible. For such molecules, eq273 becomes:

$q^R= \frac{T}{2\theta_R}\;\;\;\;\;\;\;\;274$

In general, the rotational partition function for all linear rotors can be written as:

$q^R= \frac{T}{\sigma\theta_R}\;\;\;\;\;\;\;\;275$

where the symmetry number $\sigma$ is 1 for non-symmetrical linear molecules like HCl and 2 for for symmetrical linear molecules such as O₂ and CO₂.

Question

Is $\sigma$ unique to rotational partition functions?

Answer

Yes. The symmetry number appears specifically in rotational partition functions. Although we have explained it using quantum mechanics, $\sigma$ can also be understood within the framework of classical physics.

At high temperatures, the spacing between rotational energy levels becomes very small compared to $kT$ , making the discrete quantum levels effectively continuous. The rotational partition function is then approximated classically as an integral over phase space, i.e. over all possible orientations and angular momenta. Here, each energy state corresponds to a particular configuration of positions and momenta. However, for symmetric molecules like CO₂, certain rotational operations (such as a 180° rotation around the molecular axis) produce configurations that are physically indistinguishable. To avoid overcounting these indistinguishable orientations, the symmetry number is introduced as a correction factor in the rotational partition function.

Spherical rotors

The rotational partition function for spherical rotors is derived using the same logic as that for linear rotors. Since the rotational energy levels of a spherical rotor is given by the same expression as that for a linear rotor, the rotational partition function of a spherical rotor, noting that each energy level is $(2J+1)^2$ -fold degenerate, is given by:

$q^R=\sum_j(2J+1)^2e^{-\frac{hcBJ(J+1)}{kT}}\;\;\;\;\;\;\;\;276$

Similar to linear rotors, $q^R$ at low temperatures is calculated by substituting experimental values of $\varepsilon^R_J$ into eq276. At higher temperatures, we apply the high-temperature approximation, with eq276 becoming:

$q^R\approx \int_0^{\infty}(2J+1)^2e^{-\frac{\theta_RJ(J+1)}{T}}\;\;\;\;\;\;\;\;277$

The factor $(2J+1)^2$ in the function $f=(2J+1)^2e^{-\frac{\theta_RJ(J+1)}{T}}$ increases quadratically with $J$ , while $e^{-\frac{\theta_RJ(J+1)}{T}}$ decreases exponentially. It follows that the graph of $f$ versus $J$ forms a skewed bell curve with a maximum. If $T\gg \theta_R$ , the contribution of $f$ to the integral is negligible at low $J$ and becomes significant only at higher $J$ (see diagram below). Therefore, eq277 can be simplified to:

$q^R\approx 4\int_0^{\infty}J^2e^{-\frac{\theta_RJ^2}{T}}\;\;\;\;\;\;\;\;278$

Question

Prove that $\int_0^{\infty}x^2e^{-ax^2}dx=\frac{1}{4}\sqrt{\frac{\pi}{a^3}}$ for $a> 0$ .

Answer

We shall begin with the identity $\int_0^{\infty}e^{-ax^2}dx=\frac{1}{2}\int_{-\infty}^{\infty}e^{-ax^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}$ (see this article for derivation).

$\frac{d}{da}\int_0^{\infty}e^{-ax^2}dx=\int_{0}^{\infty}\frac{d}{da}e^{-ax^2}dx=-\int_{0}^{\infty}x^2e^{-ax^2}dx$

So,
$\int_{0}^{\infty}x^2e^{-ax^2}dx=\frac{1}{4}\sqrt{\frac{\pi}{a^3}}$

Eq278 evaluates to:

$q^R=\sqrt{\frac{\pi T^3}{\theta^3_R}}\;\;\;\;\;\;\;\;279$

Like linear rotors, the symmetry number is added to prevent overcounting of rotational states, giving:

$q^R=\frac{\sqrt{\pi}}{\sigma}\biggr$\frac{T}{\theta_R}\biggr$^{\frac{3}{2}}\;\;\;\;\;\;\;\;280$

Question

How do we determine the value of $\sigma$ for non-linear molecules?

Answer

As mentioned in the previous article, the symmetry number is a correction factor to prevent overcounting indistinguishable classical energy states. A straightforward way to determine $\sigma$ is to identify the rotational subgroup to which the non-linear molecule belongs. The order of this group, which is the number of unique proper rotational symmetry operations that map the molecule onto itself, is equal to $\sigma$ . For example, CH₄ belongs to the tetrahedral point group $T_d$ , whose rotational subgroup has an order of 12 (including 11 rotational symmetries and the identity), so $\sigma=12$ .

Symmetric rotors

The energy equation of a symmetric rotor depends on three quantum numbers: $\varepsilon(J,K,M_J)=hc\tilde BJ(J+1)+K^2hc(\tilde A-\tilde B)$ . Therefore, the rotational partition function for symmetric rotors involves three summations:

$q^R=\sum_{J=0}^{\infty}\sum_{K=-J}^J\sum_{M_J=-J}^Je^{-\frac{\varepsilon(J,K,M_J)}{kT}}\;\;\;\;\;\;\;\;281$

or equivalently,

$q^R=\sum_{J=0}^{\infty}\sum_{K=-J}^J(2J+1)e^{-\frac{\varepsilon(J,K,M_J)}{kT}}\;\;\;\;\;\;\;\;282$

Question

Is the total degeneracy of a symmetric rotor $2J+1$ or $2(2J+1)$ ?

Answer

The rotational partition function sums over all possible rotational states. The total degeneracy of a symmetric rotor is $2(2J+1)$ . Here, $2J+1$ comes from the number of $M_J$ values for a given $J$ , which is accounted for by the most inner sum. The additional factor of 2 comes from the quantum number $K$ , which ranges from $-J$ to $+J$ . For $K\neq 0$ , the energy levels associated with $\pm K$ are degenerate. However, these $K$ -dependent energy levels must be summed over explicitly, as not all values of $K$ correspond to the same energy.

The double summation in eq282 involves fixing a value of $J$ and summing the allowed values of $K$ . Since $J$ runs from $0$ to $\infty$ , with $-J\leq K\leq J$ for each $J$ , the minimum allowed value of $J$ for a given $K$ is $\vert K\vert$ . This allows us to change the order of summation in eq282 by fixing a value for $K$ and then summing over the allowed values of $J$ :

$q^R=\sum_{K=-\infty}^{\infty}\sum_{J=\vert K\vert}^{\infty}(2J+1)e^{-\frac{\varepsilon(J,K,M_J)}{kT}}\;\;\;\;\;\;\;\;283$

Substituting $\varepsilon(J,K,M_J)=hc\tilde BJ(J+1)+K^2hc(\tilde A-\tilde B)$ into eq283 and using the high-temperature approximation gives:

$q^R\approx \int_{-\infty}^{\infty}e^{-\frac{hc(\tilde A-\tilde B)K^2}{kT}}dK \int_{\vert K\vert}^{\infty}(2J+1)e^{-\frac{hc\tilde BJ(J+1)}{kT}}dJ\;\;\;\;\;\;\;\;284$

Using the earlier argument that if $T\gg \theta_R$ , the contribution of $(2J+1)e^{-\frac{hc\tilde BJ(J+1)}{kT}}$ to the 2^nd integral in eq284 is negligible at low $J$ and becomes significant only at higher $J$ , we have

$q^R\approx \int_{-\infty}^{\infty}e^{-\frac{hc(\tilde A-\tilde B)K^2}{kT}}dK \int_{\vert K\vert}^{\infty}2Je^{-\frac{hc\tilde BJ^2}{kT}}dJ\;\;\;\;\;\;\;\;285$

Since $\frac{d}{dJ}e^{-\frac{hc\tilde BJ^2}{kT}}=-\frac{2Jhc\tilde B}{kT}e^{-\frac{hc\tilde BJ^2}{kT}}$ , the 2^nd integral in eq285 evaluates to $\biggr\[-\frac{kT}{hc\tilde B}e^{-\frac{hc\tilde BJ^2}{kT}}\biggr\]^{\infty}_{\vert K\vert}$ and eq285 becomes:

$q^R=\frac{kT}{hc\tilde B}\int_{-\infty}^{\infty}e^{-\frac{hc(\tilde A-\tilde B)K^2}{kT}}e^{-\frac{hc\tilde BK^2}{kT}}dK=\frac{kT}{hc\tilde B}\int_{-\infty}^{\infty}e^{-\frac{hc\tilde AK^2}{kT}}dK$

Using the identity $\int_{-\infty}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$ , we have

$q^R=\frac{kT}{hc\tilde B}\sqrt{\frac{kT\pi}{hc\tilde A}}=\frac{\sqrt{\pi}}{\sigma}\frac{T}{\theta_{R,B}}\sqrt{\frac{T}{\theta_{R,A}}}\;\;\;\;\;\;\;\;286$

where we have added the symmetry number (e.g. $\sigma=3$ for NH₃), with $\theta_{R,A}=\frac{hc\tilde A}{k}$ and $\theta_{R,B}=\frac{hc\tilde B}{k}$ .

Question

What is the expression for the rotational partition function of an asymmetric rotor?

Answer

An asymmetric rotor has three distinct moments of inertia: $I_A\neq I_B\neq I_C$ . Its rotational energy is characterised by three rotational constants: $\tilde A$ , $\tilde B$ and $\tilde C$ . If two of the moments of inertia are equal, the molecule becomes a symmetric rotor. It follows that the rotational partition function of asymmetric rotor is:

$q^R=\frac{\sqrt{\pi}}{\sigma}\sqrt{\frac{T^3}{\theta_{R,A}\theta_{R,B}\theta_{R,C}}}\;\;\;\;\;\;\;\;287$

which reduces to eq286 when $\theta_{R,B}=\theta_{R,C}$ .

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Electronic molecular partition function

The electronic molecular partition function $q^E$ is a measure of the number of thermally accessible electronic energy states available to a single non-interacting molecule at a specific temperature.

It is defined by eq255:

$q^E=\sum_{level\;i}g_ie^{-\frac{\varepsilon_i}{kT}}\;\;\;\;\;\;\;\;295$

where

$g_i$ is the degeneracy of the $i$ -th electronic level,
$\varepsilon_i$ is the energy of the $i$ -th electronic state,
$k$ is the Boltzmann constant,
$T$ is the absolute temperature.

Expanding the above equation yields:

$q^E=g_0+g_1e^{-\frac{\varepsilon_1}{kT}}+g_2e^{-\frac{\varepsilon_2}{kT}}+\cdots\;\;\;\;\;\;\;\;296$

where the ground electronic energy level, $\varepsilon_0=0$ , is the lowest electronic energy level with rotational quantum number $J=0$ and vibrational quantum number $n=0$ .

For many diatomic molecules under standard laboratory conditions (e.g. room temperature), the electronic partition function simplifies significantly. This is because the energy gap between the ground state and the first excited electronic state is often much larger than $kT$ , leading to an exponentially small population in the excited states. In other words, $\varepsilon_i\gg kT$ for $i> 0$ at temperatures up to 10000 K, giving:

$q^E\approx g_0\;\;\;\;\;\;\;\;297$

Since the ground electronic level for most diatomic molecules, as well as monatomic species with filled subshells, is nondegenerate, $q^E=1$ .

Some exceptions to this generalisation include O₂and NO. The ground electronic level of O₂ consists of two unpaired electrons, each occupying one of two degenerate antibonding molecular orbitals, resulting in a triplet state. Therefore, $q^E=3$ . Furthermore, O₂ has an excited electronic state that becomes significantly populated above 1500 K, so $q^E> 3$ at higher temperatures.

In its ground electronic state, NO has a single unpaired electron occupying one of two degenerate $\pi^*$ antibonding molecular orbitals. These orbitals are formed by the overlap of the $p_x$ and $p_y$ atomic orbitals of N and O. Consequently, the unpaired electron can have an orbital angular momentum projection of either +1 or -1, giving rise to two degenerate orbital states. Additionally, the unpaired electron is characterised by a doublet state (spin degeneracy of 2). Due to spin-orbit coupling, each of the two orbital states is further split into two fine-structure levels, with a small energy separation of approximately 0.015 eV (or 121 cm^-1). Therefore, $q^E=2$ at $T=0$ , and it rapidly approaches 4 as the temperature increases.

Since the electronic partition function is tied directly to electronic energy levels, it reflects the electronic structure of the molecule. Spectroscopic data from techniques such as UV-Vis spectroscopy or photoelectron spectroscopy provide the necessary input for estimating the energy levels $\varepsilon_i$ , and hence computing $q^E$ . Self-consistency methods can also be used to numerically evaluate $q^E$ with accuracy, especially in systems where experimental $\varepsilon_i$ data are lacking.

In practice, $q^E$ becomes crucial in systems where excited electronic states are thermally accessible, such as:

- High-temperature systems (e.g. combustion, plasma),
- Photochemically excited molecules,
- Transition metals and radicals, which often have low-lying excited states,
- Astrophysical and atmospheric systems where thermal populations can include excited states.

In these cases, the electronic partition function influences calculations of internal energy, entropy, heat capacity and equilibrium constants.

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Gibbs paradox

The Gibbs paradox refers to the apparent contradiction between statistical and classical thermodynamics where the entropy change for mixing identical gases is nonzero unless quantum indistinguishability is considered.

In classical thermodynamics, the entropy of mixing two gases should be zero if the gases are identical (nothing changes), and positive if the gases are different (mixing different gases leads to a greater number of possible microstates). To analyse whether statistical thermodynamics aligns with this, let’s consider a system of single-species, non-interacting, distinguishable particles. The canonical partition function $Q$ is given by eq166:

$Q=q^N$

where

$q$ is the molecular partition function
$N$ is the number of particles

Substituting eq166 into eq159 gives:

$S=Nk\biggr\[T\biggr$\frac{\partial lnq}{\partial T}\biggr$_{V,N}+lnq\biggr\]\;\;\;\;\;\;\;\;340$

If the partition between two containers, each with $N$ identical gas particles occupying a volume $V$ , is removed, the entropy of mixing $\Delta S_{mix}$ is:

$\begin{align}\Delta S_{mix}&=S_{final}-S_{initial}\\&=S(2N,2V)-2S(N,V)\\&=2NK\biggr\[T\biggr$\frac{\partial lnq_{2V}}{\partial T}\biggr$_{V,N}+lnq_{2V}\biggr\]-2Nk\biggr\[T\biggr$\frac{\partial lnq_{V}}{\partial T}\biggr$_{V,N}+lnq_{V}\biggr\]\end{align}$

Question

Show that $\biggr$\frac{\partial lnq_{2V}}{\partial T}\biggr$_{V,N}=\biggr$\frac{\partial lnq_{V}}{\partial T}\biggr$_{V,N}$ .

Answer

From eq257, we have $q=q^Tq^Rq^Vq^E$ . The translational part of the molecular partition function $q^T$ of the gas is proportional to $V$ (see eq266), while the remaining components (rotational, vibrational and electronic) are independent of $V$ . So,

$\frac{\partial lnq}{\partial T}=\frac{\partial}{\partial T}ln\biggr\[V\biggr$\frac{2\pi mkT}{h^2}\biggr$^{\frac{3}{2}}q^Rq^Vq^E\biggr\]=\frac{3T^{\frac{1}{2}}}{2}\frac{\partial}{\partial T}ln(q^Rq^Vq^E)$

In other words, $\frac{\partial lnq}{\partial T}$ is independent of $V$ , and so, $\biggr$\frac{\partial lnq_{2V}}{\partial T}\biggr$_{V,N}=\biggr$\frac{\partial lnq_{V}}{\partial T}\biggr$_{V,N}$ .

Therefore,

$\Delta S_{mix}=2Nkln\frac{q_{2V}}{q_V}=2Nkln2\;\;\;\;\;\;\;\;341$

Eq341 suggests an increase in entropy simply from removing a partition between two identical gases, even though nothing physical has changed (the final state is macroscopically the same as the initial one). This is not possible and leads to the Gibbs paradox.

To resolve the paradox, the partition function must be divided by $N!$ . Substituting eq167 into eq159 gives:

$S=NkT\biggr$\frac{\partial lnq}{\partial T}\biggr$_{V,N}+k(Nlnq-lnN!)$

So,

$\begin{align}\Delta S_{mix}&=S(2N,2V)-2S(N,V)\\&=2NKT\biggr$\frac{\partial lnq_{2V}}{\partial T}\biggr$_{V,N}+2Nkln\frac{q_{2V}}{q_V}-kln\frac{(2N)!}{N!^2}-2NkT\biggr$\frac{\partial lnq_{V}}{\partial T}\biggr$_{V,N}\end{align}$

Using Stirling’s approximation, $\Delta S_{mix}=0$ . Therefore, the entropy of mixing of two identical gases is zero only if the partition function correctly accounts for indistinguishability by dividing by $N!$ .

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Canonical partition function

The canonical partition function $Q$ is a statistical sum over all microstates in the canonical ensemble, providing the normalisation needed to determine the probability that each state is occupied in a system in thermal equilibrium at a fixed temperature.

It mathematically given by eq135. To explain why it is called a partition function, we need to evaluate $\beta$ in eq133, which defines the functional form of the probability distribution for a microstate in any system in thermal equilibrium with a heat bath.

Consider a thermal bath in contact with two distinct systems, $a$ and $b$ , where system $a$ has volume $V_a$ and particle number $N_a$ , while system $b$ has a different volume $V_b$ and particle number $N_b$ . The microstates of system $a$ are governed by a constant $\beta_a$ , such that $p_{i,a}\propto e^{-\beta_aE_{i,a}}$ , and those of system $b$ are governed by a constant $\beta_b$ , with $p_{i,b}\propto e^{-\beta_bE_{i,b}}$ . Similarly, the combined system $a+b$ is governed by a single constant, $\beta_{a+b}$ , with the total energy of a microstate for the combined system equal to the sum of the energies of the individual microstates, $E_{a+b}=E_a+E_b$ . According to the derived probability distribution, the probability of a microstate of the combined system is:

$p_{a+b}\propto e^{-\beta_{a+b}(E_a+E_b)}\;\;\;\;\;\;\;\;140$

On the other hand, the probability of finding the combined system in a specific state is also the product of the probabilities of finding each individual system in its corresponding state:

$p_{a+b}\propto p_{a}\cdot p_b\propto e^{-(\beta_aE_a+\beta_bE_b)}\;\;\;\;\;\;\;\;141$

For eq140 and eq141 to be consistent for all possible values of $E_a$ and $E_b$ , we must have $\beta_{a+b}=\beta_a=\beta_b=\beta$ . As explained in the previous article, eq133 is derived under the postulate that all microstates with equal energy have equal probability of occurring in any system with fixed volume, composition and temperature. This implies that $p_i$ , and possibly $\beta$ , may depend on $V$ , $N$ and $T$ . However, the two systems have different $V$ and $N$ , yet share the value of $\beta$ . Therefore, $\beta$ cannot depend on $V$ or $N$ . Since the only parameter common to both systems is the temperature of the thermal bath, $\beta$ can be a function of temperature alone: $\beta=\beta(T)$ .

Question

Explain why $E_i$ is a function of $V$ and $N$ only.

Answer

$E_i$ , the energy corresponding to each microstate, is determined by solving the Schrodinger equation $\hat{H}\psi_i=E_i\psi_i$ , where $\psi_i$ is the wavefunction that describes the quantum state resulting from the positions and momenta of all $N$ particles in the system. Therefore, $E_i$ is a function of $V$ and $N$ only.

Building on the above general argument that $\beta$ is solely a function of temperature, consider the special case of two systems with the same $V$ , $N$ and $T$ , as in the canonical ensemble. Applying the same reasoning here, we again find that the systems share the same $\beta$ , reinforcing that $\beta$ depends only on temperature. Thus, regardless of whether the systems have identical or differing volumes and particle numbers, as long as they are in thermal equilibrium with the same heat bath, $\beta$ is a function solely of temperature, independent of the systems’ internal structure or composition.

To determine the exact expression for $\beta$ , we refer to eq120, which is based on the postulate that the macroscopic properties of a system are given by the ensemble average. In other words, the equation applies to the average of any mechanical property $\langle M\rangle$ :

$M=\langle M\rangle=\sum_ip_iM_i\;\;\;\;\;\;\;\;142$

Consider the system’s pressure $P$ . When we change the volume of the system, the energy $E_i$ associated with a microstate of the system also changes. The work done on the system is $dW=-P_idV$ , where $P_i$ is the pressure contributed by the $i$ -th microstate of the system. Since the work done on the system corresponds to a mechanical change in the system’s internal energy, we can also write $dE_i=-P_idV$ , or equivalently,

$P_i=-\biggr$\frac{\partial E_i}{\partial V}\biggr$_N\;\;\;\;\;\;\;\;143$

Substituting eq143 back into eq142 gives:

$P=-\sum_ip_i\biggr$\frac{\partial E_i}{\partial V}\biggr$_N\;\;\;\;\;\;\;\;144$

Using the chain rule, the partial differentiation of eq135 yields:

$\biggr$\frac{\partial Q}{\partial V}\biggr$_{T,N}=\sum_i\frac{\partial e^{-\beta E_i}}{\partial E_i}\biggr$\frac{\partial E_i}{\partial V}\biggr$_N=-\beta\sum_ie^{-\beta E_i}\biggr$\frac{\partial E_i}{\partial V}\biggr$_N\;\;\;\;\;\;\;\;145$

Substituting eq133 and eq145 into eq144 results in:

$P=\frac{1}{\beta Q}\biggr$\frac{\partial Q}{\partial V}\biggr$_{T,N}=\frac{1}{\beta}\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}\;\;\;\;\;\;\;\;146$

The partial derivative of $Q$ with respect to $\beta$ is:

$\biggr$\frac{\partial Q}{\partial \beta}\biggr$_{V,N}=-\sum_iE_ie^{-\beta E_i}\;\;\;\;\;\;\;\;147$

Substituting eq147 into eq134 gives:

$U=-\frac{1}{Q}\biggr$\frac{\partial Q}{\partial \beta}\biggr$_{V,N}=-\biggr$\frac{\partial lnQ}{\partial \beta}\biggr$_{V,N}\;\;\;\;\;\;\;\;148$

It follows that:

$\biggr$\frac{\partial U}{\partial V}\biggr$_{T,N}=-\biggr\[\frac{\partial}{\partial V}\biggr$\frac{\partial lnQ}{\partial \beta}\biggr$_{V,N}\biggr\]_{T,N}$

Assuming that the function $lnQ$ is continuous over the domain of interest, we can interchange the derivatives:

$\biggr$\frac{\partial U}{\partial V}\biggr$_{T,N}=-\biggr\[\frac{\partial}{\partial \beta}\biggr$\frac{\partial lnQ}{\partial V}\biggr$_{T,N}\biggr\]_{V,N}\;\;\;\;\;\;\;\;149$

Substituting eq146 into eq149 yields:

$\biggr$\frac{\partial U}{\partial V}\biggr$_{T,N}=-\biggr\[\frac{\partial}{\partial \beta}(\beta P)\biggr\]_{V,N}=-P-\beta\biggr$\frac{\partial P}{\partial \beta}\biggr$_{V,N}\;\;\;\;\;\;\;\;150$

Comparing eq150 with the thermodynamic identity given by eq127 results in:

$-\beta\biggr$\frac{\partial P}{\partial \beta}\biggr$_{V,N}=T\biggr$\frac{\partial P}{\partial T}\biggr$_{V,N}\;\;\;\;\;\;\;\;151$

Using the chain rule $\frac{\partial P}{\partial T}=\frac{\partial P}{\partial (1/T)}\frac{\partial (1/T)}{\partial T}=-\frac{1}{T^2}\frac{\partial P}{\partial (1/T)}$ and substituting it into eq151 gives:

$\beta\biggr$\frac{\partial P}{\partial \beta}\biggr$_{V,N}=\frac{1}{T}\biggr\[\frac{\partial P}{\partial (1/T)}\biggr\]_{V,N}\;\;\;\;\;\;\;\;152$

Substituting $X=\frac{1}{T}$ into eq152 and using the reciprocal identity yields:

$\frac{\beta}{X}=\biggr$\frac{\partial \beta}{\partial P}\biggr$_{V,N}\biggr$\frac{\partial P}{\partial X}\biggr$_{V,N}\;\;\;\;\;\;\;\;153$

Since $\beta$ and $X$ are functions of $T$ only, the chain rule form of eq153 can be simplified to $\frac{\beta}{X}=\frac{d\beta}{dX}$ , or equivalently, $\frac{dX}{X}=\frac{d\beta}{\beta}$ . When integrated, this results in $lnX=ln\beta+C$ , or $X=\beta e^C$ , where $C$ is a constant. Therefore,

$\beta=\frac{1}{kT}\;\;\;\;\;\;\;\;154$

where $k=e^C$ is a constant.

Substituting eq154 back into eq135 gives the explicit form of the canonical partition function:

$Q=\sum_je^{-\frac{E_j}{kT}}\;\;\;\;\;\;\;\;155$

As discussed, the probability of finding a system, which is in thermal equilibrium with a constant temperature heat bath, in a specific microstate $i$ with energy $E_i$ is given by $p_i=\frac{e^{-\beta E_i}}{Q}$ . Here, the numerator represents the unnormalised statistical weight of microstate $i$ . This weight becomes a meaningful probability only after dividing by $Q$ , which represents the total statistical weight of all accessible microstates. In this sense, $Q$ enables the partitioning of the total probability of 1 across all microstates, in proportion to their energy. This is why it is called a partition function.

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