Ideal Diatomic Gas
In addition to translational and electronic degrees of freedom, diatomic molecules possess vibrational and rotational degrees of freedom as well. A series of very good approximatinos can be used to reduce a complicated two-nuclei, n-electron problem to a set of simpler problems. The simplest is the rigid roto-harmonic oscillator approximation.
6-1 The Rigid Rotor-Harmonic Oscillator Approximation
The physical basis of the Born-Oppenheimer approximation is that the nuclei are much more massive than the electrons, and thus move slowly relative to the electrons. The electrons can be considered to move in a field produced by the nuclei fixed at some internuclear separation. The Schrodinger equation approximately separates into two separate equations. One equation describes the motion of the electrons in the field of the fixed nuclei. Denote the eigenvalues of this equation bu <tex>u_j (r)</tex>, where <tex>r</tex> is the internuclear separation. The other equation describes the motion of the nuclei in the electronic potential <tex>u_j(r)</tex>, which is the potential set up by the electrons in the electronic state <tex>j</tex>.
<p>
</p>
The motion of two masses in a spherically symmetric potential can be rigorously separated into two separate problems by the introduction of center of mass and relative coordinates. The Hamiltonian can be written as below
<center>
<br>
<tex>H = H_
+ H_
</tex>
<br>
<tex>\epsilon = \epsilon_
+ \epsilon_
</tex>
<br>
<tex>q = q_
q _
</tex>
<br>
<tex>q_
= \left [\frac
\right]^
</tex>
<br>
<tex>Q(N, V, T) = \frac{q_
^N q_
^N}
</tex>
<br>
</center>
The relative motion of the two nuclei in the potential <tex>u(r)</tex> consists of rotary motion about the center of mass and relative vibratory motion of the two nuclei. The amplitude of the vibratory motion is very small. Expand the internuclear potential about <tex>r_e</tex>.
<center>
<br>
<tex>u(r) = u(r_e) + \left ( r - r_e \right ) \left ( \frac
\right )_
+ \frac
\left (r-r_e \right )^2 \left ( \frac
\right )_
+ ...</tex>
<br>
<tex>u(r) = u(r_e) + \frac
k \left (r-r_e \right )^2</tex>
<br>
</center>
The parameter <tex>k</tex> is a measure of the force constant.
With the rigid rotor harmonic oscillator approximation, the Hamiltonian of the relative motion of the nuclei can be written as below.
<center>
<br>
<tex>H_
= H_
+ H_
</tex>
<br>
<tex>\epsilon_
= \epsilon_
+ \epsilon_
</tex>
<br>
<tex>q_
= q_
q_
</tex>
<br>
<tex>\mbox
</tex>
<br>
<tex>\epsilon_J = \frac
</tex>
<br>
<tex>\omega_J = 2J + 1</tex>
<br>
<tex>\mbox
</tex>
<br>
<tex>\epsilon_
= \hbar \nu \left ( n+ \frac
\right )</tex>
<br>
<tex>\omega_n = 1</tex>
<br>
<tex>\nu = \frac
\left ( \frac
\right )^
</tex>
<br>
</center>
Transitions from one rotational level to another can be induced by electromagnetic radiation. There must be a permanent dipole moment of the molecule, and the the change in <tex>J</tex> must be plus or minus one. Below is an expression of the frequency of radiation absorbed in the process of going from a level <tex>J</tex> to <tex>J + 1</tex>.
<center>
<br>
<tex>\nu = \frac{ \epsilon_
- \epsilon_
}
</tex>
<br>
<tex>\nu = \frac
(J + 1)</tex>
<br>
</center>
Expect a set of equally spaced spectral lines.
<center>
<br>
<tex>\mbox
</tex>
<br>
<tex>\bar \epsilon_J = \bar B J(J+1)</tex>
<br>
</center>
For a molecule to change it vibrational state by absorbing radiation, it must change its dipole moment when vibrating and obey the selection rule that the change in <tex>n</tex> is equal to plus or minus one.
<center>
<br>
<tex>\nu = \frac{\epsilon_
- \epsilon_n}}
</tex>
<br>
<tex>\nu = \frac
\left ( \frac
\right )^
</tex>
<br>
</center>
The vibrational spectrum of a diatomic molecule consists of just one line.
Hamiltonian
<center>
<br>
<tex>H = H_
+ H_
+ H_
+ H_
+ H_
</tex>
<br>
<tex>\epsilon = \epsilon_
+ \epsilon_
+ \epsilon_
+ \epsilon_
+ \epsilon_
</tex>
<br>
<tex>q = q_
q_
q_
q_
q_
</tex>
<br>
<tex>Q(N, V, T) = \frac{ \left (q_
q_
q_
q_
q_
\right )^N }
</tex>
<br>
</center>
The zero of rotational energy is usually taken to be the <tex>J=0</tex> state. Choose the zero of the vibrational energy to be the bottom of the internuclear potential well of the lowest electronic state. Take the zero of the electronic energy to be the separated, electronically unexcited atoms at rest. Denote the depth of the ground electronic state potential by <tex>D_e</tex>, and the energy of the ground electronic state is <tex>-D_e</tex>. The electronic partition function is below.
<center>
<br>
<tex>q_elec = \omega_
e^
+ \omega_
e^{- \epsilon_2 / k T}</tex>
<br>
</center>
6-2 The Vibrational Partition Function
<center>
<br>
<tex>\epsilon_n = \left (n + \frac
\right ) h \nu</tex>
<br>
<tex>\nu = \left ( \frac
\right )^
\frac
</tex>
<br>
<tex>q_
(T) = \sum_n e^{-\beta \epsilon_n}</tex>
<br>
<tex>q_
(T) = e^{-\beta h \nu / 2} \sum_
^
e^{-\beta h \nu n}</tex>
<br>
<tex>q_
(T) = \frac{ e^{\beta h \nu / 2} } { 1 - e^{\beta h \nu}}</tex>
<br>
<tex>kt >> h \mu </tex>
<br>
<tex>q_
(T) = e^{-\beta h \nu / 2} \int_0^
e^{-\beta h \nu n} \,dn</tex>
<br>
<tex>q_
(T) = \frac
</tex>
<br>
</center>
Vibrational Temperature
<center>
<br>
<tex>E_
= NkT^2 \frac{d \ln q_
}
</tex>
<br>
<tex>E_
= Nk \left ( \frac{\Theta_{\nu}}
+ \Theta_{\nu}}{e^{\Theta_
/ T} - 1}} \right )</tex>
<br>
</center>
The vibrational contribution to the heat capacity is below.
<center>
<br>
<tex> \left ( \frac{\partial E_{\nu}}
\right )_
= Nk \left ( \frac{\Theta_{\nu}}
\right )2 \frac{ e{\Theta_
/ T} }{(e^{\Theta_
/ T} - 1)^2} </tex>
<br>
</center>
Below is the fraction of molecules in excited vibrational states.
<center>
<br>
<tex>f_n = \frac{ e^{-\beta h \nu ( n+ 1 / 2) }}{q_{vib}}</tex>
<br>
<tex>f_
= \sum_
^
\frac{ e^{-\beta h \nu ( n+ 1 / 2) }}{q_{vib}}</tex>
<br>
<tex>f_
= 1 - f_0</tex>
<br>
<tex>f_
= e^{-\beta h \nu}</tex>
<br>
<tex>f_
= e^{-\Theta_
/ T} </tex>
<br>
</center>
6-3 The Rotational Partition Function of a Heteronuclear Diatomic Molecule
For heternuclear diatomic molecules, the calculation of the rotation partition function is straightforward. The rotational partition function is given by the expression below.
<center>
<br>
<tex>q_
(T) = \sum_
^
(2J + 1) e^{- \beta \bar B J (J + 1)}</tex>
<br>
<tex>\mbox
</tex>
<br>
<tex>\Theta_r = \frac
</tex>
<br>
</center>
Approximate the sum by an integral.
<center>
<br>
<tex>q_
(T) = \int_0^
(2J+ 1) e^{- \Theta_r J (J+1)/ T} /, dJ</tex>
<br>
<tex>q_
(T) = \int_0^
e^{- \Theta_r J (J+1)/ T} d
</tex>
<br>
<tex>\Theta_r << T</tex>
<br>
<tex>q_
(T) = \frac
</tex>
<br>
</center>
Use the sum directly for low temperatures or for molecules with large values of <tex>\Theta_r</tex>.
Euler-MacLaurin summation formula
<center>
<br>
<tex>\sum_
^b f
= \int_a^b f dn + \frac
+ \sum_
^
^j \frac
{(2j)! {f^(2j - 1) (a) - f^
(b) }</tex>
<br>
</center>
Apply this formula to <tex>q_
(T)</tex>.
<center>
<br>
<tex>q_
(T) = \frac
\left ( 1 + \frac
\left ( \frac
\right ) + \frac
\left ( \frac
\right )^2 + \frac
\left ( \frac
\right )^3 + ...</tex>
<br>
<tex>\mbox
</tex>
<br>
<tex>E_
= NkT^2 \left ( \frac{\partial \ln q_
}
\right )</tex>
<br>
<tex>E_
= NkT + ...</tex>
<br>
<tex>\mbox
</tex>
<br>
<tex>C_
= Nk + ...</tex>
<br>
<tex>\mbox
</tex>
<br>
<tex>\frac
= \frac{(2J + 1) e^{-\Theta_r J (J + 1) / T}}{q_
(T)}</tex>
<br>
</center>
Contrary to the vibrational case, most molecules are in excited rotational levels at ordinary temperatures.
<center>
<br>
<tex>J_
= \left ( \frac
\right )^
- \frac
</tex>
<br>
<tex>J_
= \left (\frac
\right )^
</tex>
<br>
<tex>\bar B \prop \frac
</tex>
<br>
</center>
Rotational partition function of homonuclear diatomic molecules
There is a profound effect by the symmetry requirement on the rotational energy levels of a homonuclear diatomic molecule, which can be understood only by understanding the general symmetry properties of a homonuclear diatomic molecule. Below is the approximation of the partition function of a homonuclear diatomic molecule.
<center>
<br>
<tex>q_
(T) = \frac
\left ( 1 + \frac
\left ( \frac
\right ) + \frac
\left ( \frac
\right )^2 + ...</tex>
<br>
</center>
The extra factor is due to the symmetry of the homonuclear diatomic molecule an, in particular, to the fact that there are two indistinguishable orientations of a homonuclear diatomic molecule. There is a symmetry number that represents the number of indistinguishable orientations of a molecule.
6-4 The Symmetry Requirement of the Total Wave Function of a Homonuclear Diatomic Molecule
The calculation of the rotational partition function is not quite so straightforward for homonuclear diatomic molecules.
<p>
</p>
Imagine the interchange of two identical nuclei. Imagine this as a result of (1) an inversion of all the particles, electrons and nuclei, through the origin, and then (2) an inversion of just the electrons back through the origin. The rotational portion of the wavefunction controls the symmetry of the wavefunction
<p>
</p>
There is a profound effect on the low-temperature thermodynamics of <tex>H_2</tex> due to the weighting of the rotational states.
<center>
<br>
<tex>\mbox
</tex>
<br>
<tex>I(2I + 1)</tex>
<br>
<tex>\mbox
</tex>
<br>
<tex>(I+1)(2I + 1)</tex>
<br>
<tex>\mbox
</tex>
<br>
<tex>\mbox
</tex>
<br>
<tex>I(2I + 1)</tex>
<br>
<tex>\mbox
</tex>
<br>
<tex>(I+1)(2I + 1)</tex>
<br>
<tex>\mbox
</tex>
<br>
</center>
6-5 The Rotation Partition Function of a Homonuclear Diatomic Molecule
Below is an equation of <tex>q_
(T)</tex> in the case of homonuclear diatomic molecules with integral spin.
<center>
<br>
<tex>q_
(T) = (I+1)(2I+1) \sum_
(2J+1)e^{-\Theta_r J(J+1)/T} + I(2I+1) \sum_
(2J+1)e^{-\Theta_r J(J+1)/T}</tex>
<br>
<tex>q_
(T) = I(2I+1) \sum_
(2J+1)e^{-\Theta_r J(J+1)/T} + (I+1)(2I+1) \sum_
(2J+1)e^{-\Theta_r J(J+1)/T}</tex>
<br>
</center>
For most molecules at ordinary temperatures, <tex>\Theta_r << T</tex>, which means the sum can be replaced with an integral. This is true when <tex>\Theta_r / T</tex> is less than about <tex>0.2</tex>. This is the requirement for the symmetry factor to appear.
<center>
<br>
<tex>\sum_
\approx \sum_
\approx \frac
\int_0^
(2J + 1) e^{-\Theta_r J (J + 1) / T} /,dJ = \frac
</tex>
<br>
<tex>q_
(T) = \frac
</tex>
<br>
<tex>q_
(T) = \frac
</tex>
<br>
<tex>q_
= (2I + 1)^2</tex>
<br>
</center>
The factor of two in the high-temperature limit takes into account that the molecule is homonuclear.
<center>
<br>
<tex>q_
(T) \approx \frac
\approx \frac
\sum_
^
(2J + 1) e^{-\Theta_r J (J + 1) / T}</tex>
<br>
</center>
Hydrogen is an important case where the value of <tex>\Theta_r / T</tex> is not small. There is a spin of <tex>\frac
</tex> associated with each nucleus of <tex>H_2</tex>. The hydrogen with only even rotational levels allowed is called para-hydrogena and that with only odd rotational levels allowed is called ortho-hydrogen. The ratio is below.
<center>
<br>
<tex>q_
= \sum_
(2J+1) e^{-\Theta_r J (J+1)/T} + 3 \sum_
(2J+1) e^{-\Theta_r J (J+1)/T}</tex>
<br>
<tex>\frac{N_{ortho}}{N_{para}} = \frac{ 3 \sum_
(2J+1) e^{-\Theta_r J (J+1)/T} }{ \sum_
(2J+1) e^{-\Theta_r J (J+1)/T} }</tex>
<br>
</center>
In principle, nuclear spin effects should be observable in other homonuclear molecules, but the characteristic rotational temperature for all other molecules is so small that the molecules reach the "high-temperature limit" while still in the solid state. Hydrogen is somewhat unusual in that its rotational constant is so much greater than its boiling point. For most cases, the Euler-MacLaurin expansion can be used. These formulas are valid when <tex>\Theta_r < 0.2 T</tex>.
<center>
<br>
<tex>q_
(T) = \frac
\left ( 1 + \frac
\left ( \frac
\right ) + \frac
\left ( \frac
\right )^2 + \frac
\left ( \frac
\right )^3 + ...</tex>
<br>
<tex>E_
= NkT \left ( 1 - \frac
- \frac
\left ( \frac
\right )^2 + ... \right )</tex>
<br>
<tex>E_
= Nk \left ( 1 + \frac
\left ( \frac
\right )^2 + ... \right )</tex>
<br>
<tex>S_
= Nk \left ( 1 - \ln \left ( \frac
\right ) - \frac
\left ( \frac
\right )^2 + ... \right )</tex>
</center>
6-6 Thermodynamic Functions
Write the harmonic oscillator-rigid rotor approximation.
<center>
<br>
<tex>q(V, T) = \left ( \frac
\right )^{\frac
{2}} V \frac
e^{\beta h \nu /2} \left (1 - e^{\beta h \nu} \right )^{-1} \omega_
e^
</tex>
<br>
</center>
This requires that <tex>\Theta_r << T</tex>, that only the ground state electronic state is important, and that the zero of energy is taken to be the separated states at rest in the ground electronic states.
<center>
<br>
<tex>\frac
= \frac
+ \frac
+ \frac
{e^
- 1} - \frac
</tex>
<br>
<tex>\frac
= \frac
+ \left ( \frac
\right )2 \frac{e{h\nu / kT}}{(e^
- 1)^2}</tex>
<br>
<tex>\frac
= \ln \left [\frac
\right]^
\frac{Ve^{5/2}}
+ \ln \frac
+ \frac
{e^
- 1} - \ln \left ( 1 - e^{- h \nu / kT } ) + \ln \omega_
</tex>
<br>
<tex>pV = NkT</tex>
<br>
<tex>\frac
= -\ln \left [\frac
\right]^
kT - \ln \frac
+ \frac
+ \ln \left ( 1 - e^{- h \nu / kT } ) - \frac
- \ln \omega_
</tex>
<br>
</center>
Consider the electronic partition function when <tex>T >> \Theta_r</tex>.
<center>
<br>
<tex>q_e (T) = \omega_
e^
+ \omega_
e^{-\epsilon_2 / kT}</tex>
<br>
</center>