$$\newcommand{\xrightleftharpoons}[2]{\overset{#1}{\underset{#2}{\rightleftharpoons}}}$$ $$\newcommand{\conc}[1]{\left[\mathrm{#1}\right]}$$ $$\newcommand{\chem}[1]{\mathrm{#1}}$$ $$\newcommand{\expect}[1]{\left< #1 \right>}$$

# Partitioning the Energy

Shaun Williams, PhD

## Separation of Degrees of Freedom

### Nuclear Motion

• For atoms in chemical bonds we can break down their nuclear motion into
• Translational – motion of the center of mass along $$x$$, $$y$$, or $$z$$: 3 coordinates
• Rotational – changes in overall orientation of molecule: 2 angles for linear; 3 angles for nonlinear
• Vibrational – relative motions of the atoms: all remaining of the $$3N$$ coordinates

### Review of Canonical Ensemble

• Remember for the canonical ensemble \begin{align} \mathcal{P}(\varepsilon) &= \frac{g(\varepsilon)e^{-\bfrac{\varepsilon}{k_\mathrm{B}T}}}{q(T)} \\ q(T) &= \sum_{\varepsilon=0}^{\infty} g(\varepsilon) e^{-\bfrac{\varepsilon}{k_\mathrm{B}T}} \end{align}
• We can break up the degeneracy $$g=g_\chem{vib}g_\chem{rot}g_\chem{trans}$$
• So, $$\varepsilon = \varepsilon_\chem{vib} + \varepsilon_\chem{rot} + \varepsilon_\chem{trans}$$

### Evaluation of the Partition Function

• Adding our seperability to the partition function \begin{align} q(T) &= \sum_{\varepsilon=0}^\infty g(\varepsilon)e^{-\frac{\varepsilon}{k_\mathrm{B}T}} = \sum_{\varepsilon_\chem{vib}}^\infty \sum_{\varepsilon_\chem{rot}}^\infty \sum_{\varepsilon_\chem{trans}}^\infty g_\chem{vib} g_\chem{rot} g_\chem{trans} e^{-\frac{\left(\varepsilon_\chem{vib}+\varepsilon_\chem{rot}+\varepsilon_\chem{trans}\right)}{k_\mathrm{B}T}} \\ &= \sum_{\varepsilon_\chem{vib}}^\infty \left[ g_\chem{vib} e^{-\frac{\varepsilon_\chem{vib}}{k_\mathrm{B}T}} \right] \sum_{\varepsilon_\chem{rot}}^\infty \left[ g_\chem{rot} e^{-\frac{\varepsilon_\chem{rot}}{k_\mathrm{B}T}} \right] \sum_{\varepsilon_\chem{trans}}^\infty \left[ g_\chem{trans} e^{-\frac{\varepsilon_\chem{trans}}{k_\mathrm{B}T}} \right] \\ &= q_\chem{vib} q_\chem{rot} q_\chem{trans} \\ \mathcal{P}(\varepsilon) &= \mathcal{P}_\chem{vib}\left( \varepsilon_\chem{vib}\right) \mathcal{P}_\chem{rot}\left( \varepsilon_\chem{rot}\right) \mathcal{P}_\chem{trans}\left( \varepsilon_\chem{trans}\right) \end{align}

### Classical Average Value Theorem

• Measure a parameter $$A\left(\vec{s}\right)$$ where $$\vec{s}$$ are coords
• Find $$k$$ different values
• Average value is $$\expect{A} = \frac{\sum_i N_i A_i}{\sum_i N_i} = \sum_i \frac{N_i}{N} A_i = \sum_i \mathcal{P}\left(A_i\right) A_i$$
• $$A$$ is a continuous parameter so this becomes $$\expect{A} = \int_\text{all space} \mathcal{P}_\vec{s} \left(\vec{s}\right) A\left(\vec{s}\right)\, d\vec{s}$$

### Moving Towards a Partition Function

• Let s be a single coordinates $$\mathcal{P}\left( s_1 \le s < s_2 \right) = \int_{s_1}^{s_2} \mathcal{P}(s)\, ds = \int_{s_1}^{s_2} \frac{e^{-\frac{\varepsilon_s}{k_\mathrm{B}T}}}{q_s^(T)}ds$$
• Note: $$q$$’s will have the same units as $$s$$ \begin{align} & \int_\text{all space} \mathcal{P}(s)\, ds = \int_\text{all space} \frac{e^{-\frac{\varepsilon_s}{k_\mathrm{B}T}}}{q_s^(T)}ds = 1 \\ & \therefore q_s^(T) = \int_\text{all space} e^{-\frac{\varepsilon_s}{k_\mathrm{B}T}}\, ds \end{align}

### Translational Energy

• The total translational energy is $$\frac{mv^2}{2} = \frac{mv_x^2}{2} + \frac{mv_y^2}{2} + \frac{mv_z^2}{2}$$
• The average translational energy \begin{align} \expect{\frac{mv^2}{2}} &= \expect{\frac{mv_x^2}{2}+\frac{mv_y^2}{2}+\frac{mv_z^2}{2}} \\ &= \frac{m\expect{v_x^2}}{2} + \frac{m\expect{v_y^2}}{2} + \frac{m\expect{v_z^2}}{2} \end{align}

### Probability of Translational Energies

• Probability of a particle having $$v_x$$ \begin{align} \expect{\varepsilon_x} &= \int_{-\infty}^{\infty} \mathcal{P}\left( v_x \right) \frac{mv_x^2}{2} \,dv_x \\ &= \frac{m}{2}\frac{\int_{-\infty}^{\infty}e^{-\frac{mv_x^2}{2k_\mathrm{B}T}} v_x^2 \,dv_x}{q_v^(T)} \\ q_v^`(T) &= \int_{-\infty}^{\infty} e^{-\frac{mv_x^2}{2k_\mathrm{B}T}}\, dv_x \end{align}

### Solving the Integral

• We can use integral tables to solve the integrals \begin{align} & \int_0^\infty e^{-ax^2} \, dx = \frac{1}{2} \left( \frac{\pi}{a}\right)^{\bfrac{1}{2}} \\ & \int_0^\infty x^2 e^{-ax^2} \, dx = \frac{1}{4} \left( \frac{\pi}{a^3} \right)^\bfrac{1}{2} \end{align}

### Manipulating These Expressions

\begin{align} \int_{-\infty}^{\infty} x^2 e^{-ax^2}\,dx &= \int_{-\infty}^0 x^2 e^{-ax^2} \, dx + \int_0^\infty x^2 e^{-ax^2} \, dx \\ &= \int_0^\infty \left( -x \right)^2 e^{-a\left( -x\right)^2} \, dx + \int_0^\infty x^2 e^{-ax^2} \, dx \\ &= 2\int_0^\infty x^2 e^{-ax^2}\, dx = \frac{1}{2}\left( \frac{\pi}{a^3} \right)^\bfrac{1}{2} \\ \int_{-\infty}^\infty \, dx &= 2\int_0^\infty e^{-ax^2}\, dx = \left( \frac{\pi}{a} \right)^\bfrac{1}{2} \end{align}

### Solving Out Energy Equations

\begin{align} \expect{\varepsilon_x} &= \left( \frac{m}{2} \right) \frac{\int_{-\infty}^\infty e^{-\frac{mv_x^2}{2k_\mathrm{B}T}} v_x^2 \, dv_x}{\int_{-\infty}^\infty e^{-\frac{mv_x^2}{2k_\mathrm{B}T}} \, dv_x} \\ &= \left( \frac{m}{2} \right) \frac{\frac{1}{2} \left( \frac{\pi}{\left[ \frac{m}{2k_\mathrm{B}T} \right]^3} \right)^\bfrac{1}{2}}{\left( \frac{\pi}{\frac{m}{2k_\mathrm{B}T}} \right)^\bfrac{1}{2}} = \frac{m}{2}\frac{1}{2}\frac{2k_\mathrm{B}T}{m} = \frac{k_\mathrm{B}T}{2} \end{align}

### Analysis

• An energy that varies as the square of its coordinate, $$\varepsilon_x=ax^2$$, works out the same way $$\expect{\varepsilon_x} = \frac{k_\mathrm{B}T}{a}$$
• The average value of $$x^2$$ is $$\expect{x^2} = \frac{1}{a} \expect{\varepsilon_x} = \frac{k_\mathrm{B}T}{2a}$$
• Now we know $$\expect{\frac{mv^2}{2}} = \frac{m\expect{v_x^2}}{2} + \frac{m\expect{v_y^2}}{2} + \frac{m\expect{v_z^2}}{2} = \frac{3k_\mathrm{B}T}{2}$$

### Rotations

• We can approximate the rotational energy $$E_\chem{rot} = BJ \left( J+1 \right) \approx BJ^2$$
• Every rotational coordinate, just like translations, has an approximate average energy of $$\frac{k_\mathrm{B}T}{2}$$
• Unfortunately, the vibrational considerations is different

### Vibrations

• Consider vibrations where the thermal energy, $$k_\mathrm{B}T$$, is large compared to the energy spacing in that coordinate
• Under this condition $$\varepsilon_\chem{vib} = K+U = \frac{\mu v_\chem{vib}^2}{2} + \frac{k}{2}\left( R-R_e \right)^2$$
• We solve this just like translations and rotations $$\text{for }k_\mathrm{B}T \gg \omega_e:\; \expect{\varepsilon_\chem{vib}} = \frac{k_\mathrm{B}T}{2} + \frac{k_\mathrm{B}T}{2} = k_\mathrm{B}T$$

## The Equipartition Principle

### The Equipartition Principle

• $$N_\chem{ep}$$ is the number of equipartition energy contributions $$E=\\frac{1}{2}N_\chem{ep} Nk_\mathrm{B}T = \frac{1}{2} N_\chem{ep} nRT$$
• Note: $$R=\mathcal{N}_\mathrm{A} k_\mathrm{B} = 8.3145\, \chem{J\,K^{-1}\,mol^{-1}}$$
• At room temperature, all translations and rotations contribute to $$N_\chem{ep}$$
• At room temperature, only vibrations with roughly $$\omega_e < 300\, \chem{cm^{-1}}$$ contribute to $$N_\chem{ep}$$

### Motions Contributing to $$N_\chem{ep}$$

• For those vibrations that contribute to $$N_\chem{ep}$$ \begin{align} N_\chem{ep} = & \left( \text{number of translational degrees of freedom} \right) \\ & + \left( \text{number of rotational degrees of freedom} \right) \\ & + 2 \times \left( \text{number of vibrational degrees of freedom} \right) \end{align}
• At typical lab temperatures, few vibrational modes contribute to the number of equipartition degrees of freedom
• Vibrational spacings often more than 1000 K

### Types of Gases

Gas Trans Rotation Vibration Total Energy
Monatomic $$E=\frac{k_\mathrm{B}T}{2}\times \left( \right.$$ $$3$$ $$+0$$ $$+0$$ $$\left.\right)=\frac{3}{2}k_\mathrm{B}T$$
Diatomic $$E=\frac{k_\mathrm{B}T}{2}\times \left( \right.$$ $$3$$ $$+2$$ $$+1 \times 2$$ $$\left.\right)=\frac{7}{2}k_\mathrm{B}T$$
Linear Triatomic $$E=\frac{k_\mathrm{B}T}{2}\times \left( \right.$$ $$3$$ $$+2$$ $$+4 \times 2$$ $$\left.\right)=\frac{13}{2}k_\mathrm{B}T$$
Nonlinear Triatomics $$E=\frac{k_\mathrm{B}T}{2}\times \left( \right.$$ $$3$$ $$+2$$ $$+3 \times 2$$ $$\left.\right)=\frac{12}{2}k_\mathrm{B}T$$

### Example 3.1

From the equipartition principle, estimate the energies in $$\mathrm{J}$$ that must be carried away to cool (a) $$0.500\,\chem{mol\, He(g)}$$ and (b) $$0.500\, \chem{mol\, CO(g)}$$ from $$1000\, \mathrm{K}$$ to $$900\, \mathrm{K}$$ at constant volume.

## Vibrational and Rotational Partition Functions

### Vibrational Partition Function

• For a single, non-degenerate, harmonic vibrational mode $$q_\chem{vib}(T)=\sum_{v=0}^\infty e^{-\frac{v\omega_e}{k_\mathrm{B}T}}$$
• At very high temperatures or very low vibrational constants we can use the power series $$q_\chem{vib}(T) = 1+e^{-\frac{\omega_e}{k_\mathrm{B}T}}+e^{-\frac{2\omega_e}{k_\mathrm{B}T}}+e^{-\frac{3\omega_e}{k_\mathrm{B}T}}+\dots$$
• Taylor series expansion $$f(x)=\sum_{n=0}^\infty \frac{\left( x-x_0 \right)^n}{n!} \left. \left( \frac{d^n f(x)}{dx^n} \right) \right|_{x_0}$$

### The Partition Function

• From this we get the partition function $$q_\chem{vib}(T) = f\left( e^{-\frac{\omega_e}{k_\mathrm{B}T}} \right) = \frac{1}{1-e^{-\frac{\omega_e}{k_\mathrm{B}T}}}$$
• Polyatomic molecules have more than one vibrational mode so $$q_\chem{vib}(T) = q_\chem{vib}^{(\nu_1)}(T) q_\chem{vib}^{(\nu_2)}(T) \dots$$
• Degenerate modes must be treated as distinct motions in the product

### Select Vibrational and Rotational Constants

Molecule $$\mu \\ (\chem{amu})$$ $$R_e \\ (\AA)$$ $$B \\ (\textrm{cm}^{-1})$$ $$\alpha_e \\ (\textrm{cm}^{-1})$$ $$D \\ (10^{-6}\, \textrm{cm}^{-1})$$ $$\omega_e \\ (\textrm{cm}^{-1})$$ $$\omega_ex_e \\ (\textrm{cm}^{-1})$$
$$\chem{{}^1H{}^1H}$$ 0.50 0.742 60.8536 3.062 46660 4401.21 121.34
$$\chem{{}^1H{}^2D}$$ 0.67 0.742 45.6378 1.9500 3811.92 90.71
$$\chem{{}^2D{}^2D}$$ 1.01 0.742 30.442 1.0632 3118.46 117.91
$$\chem{{}^1H{}^{19}F}$$ 0.96 0.917 20.9557 0.798 2150 4138.32 89.88
$$\chem{{}^1H{}^{35}Cl}$$ 0.98 1.275 10.5934 0.3702 532 2990.95 52.82
$$\chem{{}^1H{}^{79}Br}$$ 1.00 1.414 8.3511 0.226 372 2649.67 45.21
$$\chem{{}^1H{}^{127}I}$$ 1.00 1.609 3.2535 0.0608 526 2309.60 39.36
$$\chem{{}^2D{}^{19}F}$$ 1.82 0.917 11.0000 0.2907 585 2998.19 45.76
$$\chem{{}^{12}C{}^{16}O}$$ 6.86 1.128 1.9313 0.0175 6 2169.82 13.29
$$\chem{{}^{14}N{}^{14}N}$$ 7.00 1.098 1.9987 0.0171 6 2358.07 14.19
$$\chem{{}^{14}N{}^{16}O^+}$$ 7.47 1.063 1.9982 0.0190 2377.48 16.45
$$\chem{{}^{14}N{}^{16}O}$$ 7.47 1.151 1.7043 0.0173 -37 1904.41 14.19
$$\chem{{}^{14}N{}^{16}O^-}$$ 7.47 1.286 1.427 1372 8

### More Select Vibrational and Rotational Constants

Molecule $$\mu \\ (\chem{amu})$$ $$R_e \\ (\AA)$$ $$B \\ (\textrm{cm}^{-1})$$ $$\alpha_e \\ (\textrm{cm}^{-1})$$ $$D \\ (10^{-6}\, \textrm{cm}^{-1})$$ $$\omega_e \\ (\textrm{cm}^{-1})$$ $$\omega_ex_e \\ (\textrm{cm}^{-1})$$
$$\chem{{}^{16}O{}^{16}O}$$ 8.00 1.207 1.4457 0.0158 5 1580.36 12.07
$$\chem{{}^{19}F{}^{19}F}$$ 9.50 1.418 0.8828 891.2
$$\chem{{}^{35}Cl{}^{35}Cl}$$ 17.48 1.988 0.2441 0.0017 0.2 560.50 2.90
$$\chem{{}^{79}Br{}^{79}Br}$$ 39.46 2.67 0.0821 0.0003 0.02 325.29 1.07
$$\chem{{}^{127}I{}^{79}Br}$$ 48.66 2.470 0.0559 0.0002 0.008 268.71 0.83
$$\chem{{}^{127}I{}^{127}I}$$ 63.45 2.664 0.0374 0.0001 -0.005 214.52 0.61
$$\chem{{}^{23}Na{}^{23}Na}$$ 11.49 3.077 0.1548 0.0009 0.7 159.13 0.73
$$\chem{{}^{133}Cs{}^{133}Cs}$$ 66.45 4.47 0.0127 0.00003 0.005 42.02 0.08

### Example 3.2

Find the abundance of $$\chem{ICl}$$ molecules in the ground and lowest excited vibrational states at $$300\,\mathrm{K}$$, given that the vibrational constant $$\omega_e$$ is $$384\,\mathrm{cm}^{-1}$$.

### Rotational Motion

• In the simplest case, the rotational energy of a linear molecule is approximated as $$E_\chem{rot}=BJ(J+1)$$
• Now we turn to evaluating $$\mathcal{P}(J)=\frac{g_\chem{rot} e^{-\frac{E_\chem{rot}}{k_\mathrm{B}T}}}{q_\chem{rot}}$$
• Recall that $$g_\chem{rot}=2J+1$$
• When $$B \ll k_\mathrm{B}T$$ $$q_\chem{rot} = \sum_{J=0}^\infty g_\chem{rot} e^{-\frac{E_\chem{rot}}{k_\mathrm{B}T}} \approx \int_0^\infty g_\chem{rot} e^{-\frac{E_\chem{rot}}{k_\mathrm{B}T}} \, dJ$$

### Simplification

$$q_\chem{rot} = \sum_{J=0}^\infty g_\chem{rot} e^{-\frac{E_\chem{rot}}{k_\mathrm{B}T}} \approx \int_0^\infty g_\chem{rot} e^{-\frac{BJ(J+1)}{k_\mathrm{B}T}} \, dJ$$

• Further simplification yields \begin{align} q_\chem{rot} \approx & \int_0^\infty g_\chem{rot} e^{-\frac{BJ(J+1)}{k_\mathrm{B}T}} \, d\left[ J(J+1) \right] = -\frac{k_\mathrm{B}T}{B} \left[ e^{-\frac{BJ(J+1)}{k_\mathrm{B}T}} \right]_0^\infty \\ &= \frac{k_\mathrm{B}T}{B} = \frac{\theta_\chem{rot}}{T} \end{align}
• This is called the integral approximation

### Combining Equations

• We can combine the vibrational and rotational probabilities \begin{align} \mathcal{P}(v,J) &= \mathcal{P}_\chem{vib}(v) \mathcal{P}_\chem{rot}(J) = \left( \frac{g_\chem{vib} e^{-\frac{E_\chem{vib}}{k_\mathrm{B}T}}}{q_\chem{vib}(T)} \right) \left( \frac{g_\chem{rot} e^{-\frac{E_\chem{rot}}{k_\mathrm{B}T}}}{q_\chem{rot}(T)} \right) \\ &= \frac{g_\chem{vib} g_\chem{rot} e^{-\frac{E_\chem{vib}+E_\chem{rot}}{k_\mathrm{B}T}}}{q_\chem{vib}(T)q_\chem{rot}(T)} \end{align}

### Example 3.3

Compare the rotational partition functions for $$\chem{CO}$$ ($$\frac{B}{k_\mathrm{B}} = 2.70\,\mathrm{K}$$) obtained by using the integral and by using the direct sum, at $$273\,\mathrm{K}$$ and at $$10\,\mathrm{K}$$.

### Example 3.4

Find the number density of $$\chem{OCS}$$ molecules in a $$100.0\,\mathrm{Pa}$$ sample at $$298\,\mathrm{K}$$ that are in the vibrational state $$(v_1, v_2, v_3) = (1, 0, 1)$$ and rotational state $$J = 10$$. The constants of $$\chem{CS}$$-stretch $$\omega_1 = 859\,\mathrm{cm}^{-1}$$, bend $$\omega_2 = 527\,\mathrm{cm}^{-1}$$, and $$\chem{CO}$$-stretch $$\omega_3 = 2080\,\mathrm{cm}^{-1}$$, $$B = 0.2028\,\mathrm{cm}^{-1}$$. The bending mode is a doubly degenerate vibrational motion, so it counts for two modes. Give the final value in molecules per cubic centimeter.

### Another Rotational Partition Function

• The rotational partition function of non-linear molecules are evaluated in terms of three rotational constants, $$A$$, $$B$$, and $$C$$ $$q_\chem{rot(nonlinear)} \approx \left( \frac{\pi k_\mathrm{B}^3 T^3}{ABC} \right)^\bfrac{1}{2}$$
• Works best for symmetric tops $$A=B$$ (oblate) or $$B=C$$ (prolate)
• Still useful for asymmetric tops $$A \ne B \ne C$$

### Complications of Rotations

• Molecules with proper rotation axes, $$q_\chem{rot}$$ depends on the molecule’s nuclear spin, ns, statistics
• Many molecules have atoms that are exactly identical, called equivalent atoms
• If a rotation exchanges equivalent atoms, we need to worry about the symmetrization principle
• We can write the total wavefunction as $$\Psi_\chem{total} = \psi_\chem{elec} \psi_\chem{vib} \psi_\chem{rot} \psi_\chem{ns}$$

### Example: $$\chem{{}^1H_2}$$

• There are two spin possibilities $$+\frac{1}{2}$$ ($$\alpha$$) or $$-\frac{1}{2}$$ ($$\beta$$)
• Total nuclear spin: $$\left| I_T \right| = \left| \vec{I}_A + \vec{I}_B \right| = 0\text{ or }1$$
• $$\left| I_T \right|=1$$ is symmetric to permutation operator
• $$\left| I_T \right|=0$$ is antisymmetric to permutation operator
• $$\chem{{}^1H}$$ nuclei are fermion and therefore must be antisymmetric to exchange overall
• This requirement gives rise to two sets of states, three triplet states and one singlet state

### Nuclear Spin States

• The singlet state mathematically is $$\psi_\chem{ns} = \frac{1}{\sqrt{2}} \left[ \alpha(A)\beta(B) - \beta(A)\alpha(B) \right]\;\text{note: }I_T=0$$
• The triplet set mathematically are, $$I_T = 1$$ $$\psi_\chem{ns} = \begin{cases} \alpha(A)\alpha(B) \\ \frac{1}{\sqrt{2}} \left[ \alpha(A)\beta{B} + \beta{A}\alpha{B} \right] \\ \beta(A)\beta(B) \end{cases}$$

### Symmetric vs Antisymmetric

• The singlet state is antisymmetric to exchange
• The triplet states are symmetric to exchange
• The overall wavefunction must be antisymmetric
• Now we have to check the symmetries of the other degrees of freedom
• Consider rotational wavefunctions for $$J = 1$$ and $$2$$, $$M_J = 0$$ \begin{align} Y_1^0 (\Theta) &= \cos \Theta \\ Y_2^0(\Theta) &= \left( 3\cos^2 \Theta - 1 \right) \end{align}

### Exchange Symmetry of Rotations

• Exchanging the labels, reverses the direction $$\Theta' = \pi - \Theta$$
• This means
• $$\cos \Theta$$ changes sign
• $$\cos^2\Theta$$ does not change sign

### Overall Rotational Symmetry

• From this analysis we know that:
• Odd J states are antisymmetric with respect to exchange
• Even J states are symmetric with respect to exchange
• For the $${}^1\Sigma_g^+$$ ground state $$\chem{H_2}$$, the electronic wavefunction is totally symmetric
• For a diatomic molecule, the vibrational wavefunction is always totally symmetric

### Meaning

• What we now know is that for $$\chem{{}^1H_2}$$:
• The singlet nuclear spin state (antisymmetric) may occupy only even $$J$$ states (symmetric)
• The triplet nuclear spin state (symmetric) may occupy only the odd $$J$$ states (antisymmetric)
• The degeneracy of the rotational state now depends on the nuclear spins
• Note: $$g$$ electronic and vibrational wavefunctions are symmetric and $$u$$ are antisymmetric

### Spin Statistics Effects on the Partition Function

• We found that the triplet state has an added factor of three in its degeneracy of $$\chem{{}^1H_2}$$
• Different nuclear spin states of a molecule act almost independently of each other
• They are best described by separate partition functions
• Energies are referenced to the lowest energy of the particular spin state

### Rotational Partition Function for $$\chem{{}^1H_2}$$

\begin{align} q_\chem{rot}\left(\chem{H_2}\right) =& q_\chem{rot}\left(I_T=0\right) + q_\chem{rot}\left(I_T=1\right) \\ =& \left[ 1+5e^{-\frac{6B}{k_\mathrm{B}T}}+9e^{-\frac{20B}{k_\mathrm{B}T}}+\dots \right] \\ & +3 \left[ 3e^{-\frac{(2-2)B}{k_\mathrm{B}T}}+7e^{-\frac{(12-2)B}{k_\mathrm{B}T}}+11e^{-\frac{(30-2)B}{k_\mathrm{B}T}}+\dots \right] \end{align}

• This is a dramatic effect in hydrogen
• Para-hydrogen – $$I_T = 0$$
• Ortho-hydrogen – $$I_T = 1$$

### Other Systems

• For homonuclear boson diatomics, such as $$\chem{O_2}$$, or other symmetric linear systems, such as $$\chem{CO_2}$$, the overall wavefunction must be symmetric
• As a general (approximate) rule, rotational partition functions in molecules where rotation exchanges $$n$$ equivalent $$I = 0$$ nuclei will be reduced by a factor of $$\frac{1}{n}$$.

## The Translational Partition Function

### Translations

• The quantum energies of translational energies $$\varepsilon = n^2 \frac{h^2}{8mV^\bfrac{2}{3}} = n^2 \varepsilon_0$$
• Molecules are rarely confined to small enough volumes for the energies levels to become important
• The classical limit is an approximation of a degree of freedom that remains quantum mechanical

### Counting the Energy Levels

• How many values of $$n_x$$, $$n_y$$, and $$n_z$$ that give the same value of $$n^2$$ (proportional of $$\varepsilon$$)? $$n\equiv \sqrt{n^2} = \left( n_x^2 + n_y^2+n_z^2 \right)^\bfrac{1}{2}$$
• The degeneracy is roughly the number of distinct points in a shell of width $$dn$$ that lies between a radius of $$\sqrt{n^2}$$ and $$\sqrt{n^2+1}$$

### The Number of Points

• We can estimate the $$dn$$ by \begin{align} d\left( n^2 \right) &= 2n\,dn =1 \\ dn &= \frac{1}{2n} \end{align}
• The degeneracy is the same as the ensemble size $$\Omega$$ for the system at this energy $$\Omega = \frac{4\pi n^2}{8}dn = \frac{\pi}{2}n^2\,dn = \frac{\pi}{2}n^2\frac{1}{2n} = \frac{\pi}{4}n = \frac{\pi}{4}\left(\frac{\varepsilon}{\varepsilon_0}\right)^\bfrac{1}{2}$$

### Density of States

• Now we can find the density of quantum states, $$W$$ $$W(\varepsilon) = \frac{\Omega}{\varepsilon_0} = \frac{\pi}{4\varepsilon_0^\bfrac{3}{2}} \varepsilon^\bfrac{1}{2}$$
• Now we can build the partition function
• Due to the involved derivation, here’s the results $$q_\chem{trans}(T,V) = q'_K(T) q'_U(T,V)=\underbrace{\left(\frac{2 \pi m k_\mathrm{B}T}{h^2}\right)^\bfrac{3}{2}}_{q'_K(T)} \underbrace{V}_{q'_U(T,V)}$$

## Temperature and the Maxwell-Boltzmann Distribution

### The Energies

• At low energy, energy can only be stored as translational degrees of freedom so that $$E=\sum_{j=1}^N \left( \frac{1}{2} mv_j^2 \right) = \frac{1}{2}mN\expect{v^2}$$
• Rewritten $$E=\frac{1}{2}mN\expect{v_x^2+v_y^2+v_z^2} = \frac{3}{2}mN\expect{v_x^2}$$
• Remember previous equations $$\expect{x^2} = \frac{k_\mathrm{B}T}{2a} \therefore E=\frac{3}{2}mN\left( \frac{k_\mathrm{B}T}{m} \right) = \frac{3}{2}Nk_\mathrm{B}T = \frac{3}{2}nRT$$

### Example 3.5

Calculate the root mean square momentum along the $$x$$ axis $$\expect{p_x^2}^\bfrac{1}{2}$$ in $$\mathrm{kg\, m\, s^{-1}}$$ of a helium atom in the gas-phase when the temperature is $$312\, \mathrm{K}$$. (We don’t want to calculate $$\expect{p_x}$$ because that should average to zero, motion in the $$+x$$ and $$–x$$ directions being equally likely.)

### Questions

1. Why does $$T$$ depend on only the easily excited degrees of freedom?
• Degree of freedom that isn’t excited doesn’t contribute to the degeneracy of quantum states.
2. Why does $$T$$ depend on only the kinetic energy?
• Once we have established the confines of the system and the forces, the potential energy becomes a fixed parameter and the kinetic energy can adjust to change the degeneracy.

### Boltzmann Constant and Temperature

• The Boltzmann constant is just a conversion factor between temperature and kinetic energy $$\frac{E}{N} = \expect{\varepsilon} = \frac{3k_\mathrm{B}T}{2}$$
• Temperature is an intensive parameter, like pressure

### Example 3.6

Find the average translational kinetic energy (in $$\mathrm{kJ\, mol^{-1}}$$) and the rms speed along the $$x$$ axis $$\expect{v_x^2}^\bfrac{1}{2}$$ for a nitrogen molecule at $$298\, \mathrm{K}$$.

### The Probability Distribution

• We can now work on the probability distribution \begin{align} \mathcal{P}_\vec{v}(\vec{v}) &= \frac{e^{-\frac{mv^2}{2k_\mathrm{B}T}}}{\int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-\frac{mv^2}{2k_\mathrm{B}T}} \, dv_x\,dv_y\,dv_z} \\ &= \frac{e^{-\frac{mv^2}{2k_\mathrm{B}T}}}{\frac{h^3 q'_K(T)}{m^3}} \\ &= \left( \frac{m}{2\pi k_\mathrm{B}T} \right)^\bfrac{3}{2} e^{-\frac{mv^2}{2k_\mathrm{B}T}} \end{align}

### Moving to Speed Instead of Velocity

• We can integrate away the directions to get to speed instead of velocity \begin{align} \mathcal{P}_v(v) &= \int_0^{2\pi} \int_0^\pi \mathcal{P}_\vec{v}(\vec{v}) v^2\, dv\, \sin \Theta\, d\Theta\, d\Phi \\ &= 4\pi \mathcal{P}_\vec{v}(\vec{v}) v^2\,dv \\ \mathcal{P}_v(v) &= 4\pi \left( \frac{m}{2\pi k_\mathrm{B}T} \right)^\bfrac{3}{2} v^2 e^{-\frac{mv^2}{2k_\mathrm{B}T}} \end{align}

### Exchange Symmetry of RotationsQuantizer Energy

1. Little population at the lowest speeds
2. Little population at the highest speeds
3. As temperature increases, the distribution shifts towards higher $$E$$

### Example 3.7

In a $$1.00\, \mathrm{mole}$$ sample of $$\chem{N_2}$$ at $$298\, \mathrm{K}$$, calculate the number of molecules that will have speeds $$v$$ within $$0.005\, \mathrm{m\, s^{-1}}$$ of $$10.0\, \mathrm{m\, s^{-1}}$$.

### Maxwell-Boltzmann Distribution: Average Velocity

\begin{align} \expect{v} &= \frac{\int_0^\infty \mathcal{P}_v(v) v\,dv}{\int_0^\infty \mathcal{P}_v(v)\,dv} \\ &= \frac{4\pi \int_0^\infty e^{-\frac{mv^2}{2k_\mathrm{B}T}} v^3 \,dv}{4\pi \int_0^\infty e^{-\frac{mv^2}{2k_\mathrm{B}T}} v^2 \,dv} \\ &= \frac{\frac{1}{2}\left( \frac{2k_\mathrm{B}T}{m} \right)^2}{\frac{1}{4}\left( \frac{2\pi k_\mathrm{B}^3T^3}{m^3} \right)^\bfrac{1}{2}} \\ &= \left( \frac{8k_\mathrm{B}T}{\pi m} \right)^\bfrac{1}{2} = \sqrt{\frac{8k_\mathrm{B}T}{\pi m}} \end{align}

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