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

Equipartition Energy and Degrees of Freedom

Excitations

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

More Select Vibrational and Rotational Constants

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

The Integral Approximation To \(q_\chem{rot}\)

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 Integral Approximation To \(q_\chem{rot}\)

Counting Points on a Surface

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.

Easily Excited Degrees of Freedom

Temperature and Kinetic Energy

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

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