Ideal Monatomic Gas
An ideal gas is a gas that is dilute enough that intermolecular interactions can be neglected. The results in this chapter are applicable to gases at pressures below 1 atmosphere and temperatures greater than room temperature. Write the partition function of the entire system in terms of the individual atomic partition functions. This applies when the number of available quantum states far exceeds the number of particles for an ideal gas.
<center>
<br>
<tex>Q(N, V, T) = \frac
</tex>
<br>
</center>
There are translational, electronic, and nuclear degrees of freedom.
<center>
<br>
<tex>q(V, T) = q_
q_
q _
</tex>
<br>
</center>
5-1 The Translational Partition Function
Substitute energy states into the translation partition function.
<center>
<br>
<tex>\epsilon_
= \frac
\left (n_x^2 + n_y^2 + n_z^2 \right )</tex>
<br>
<tex>q_
= \sum_
^
e^{- \beta n_x n_y n_z}</tex>
<br>
<tex>q_
= \left ( \sum_
^
\exp \left ( - \frac
\right ) \right )^3</tex>
<br>
</center>
Replace by an integral
<center>
<br>
<tex>q_
{ (V, T) = \left ( \int_0^
e^{-\beta h^2 n^2 / 8 m a^2} dn \right )^3</tex>
<br>
<tex>q_
{ (V, T) = \left ( \frac
\right )^
V</tex>
<br>
</center>
Write as a sum over levels.
<center>
<br>
<tex>\omega(\epsilon) d \epsilon = \frac
\left ( \frac
\right )^
\epsilon^
d \epsilon</tex>
<br>
<tex>q_
(V, T) = \frac
\left ( \frac
\right )^
\int_0^
\epsilon^
e^{-\beta \epsilon} d \epsilon</tex>
<br>
<tex>q_
(V, T) = \left ( \frac
\right )^
V </tex>
<br>
<tex>q_
= \frac
</tex>
<br>
<tex> \bar \epsilon_
= kT^2 \left (\frac{\partial \ln 1_
}
\right ) </tex>
<br>
<tex> \bar \epsilon_
= \frac
kT</tex>
<br>
<tex> \epsilon_
= \frac
</tex>
<br>
<tex> p_
= (mkT)^
</tex>
<br>
</center>
The term <tex>\wedge</tex> is the thermal De Broglie wavelength of the particle.
5-2 The Electronic and Nuclear Partition Functions
Write the electronic partition function as a sum of levels rather than a sum over states. The term <tex>\omega_
</tex> is the degeneracy, and \epsilon_i the energy of the _i_th electronic level. Measure all of our electronic energies relative to the ground state.
<center>
<br>
<tex>q_
= \sum \omega_
e^{-\beta \epsilon_i}</tex>
<br>
<tex>q_
= \omega_
+ \omega_
e^{-\beta \Delta \epsilon_
</tex>
<br>
</center>
At ordinary temperatures, only the first term in the summation for <tex>q_
</tex> is significantly different from zero. At ordinary temperatures the electronic partition function of the rare gases is essentially unity and that of the alkali metals is 2, while those for halogen atoms consist of two terms. Calculate the fraction of <tex>He</tex> atoms in the lowers triplet state.
<center>
<br>
<tex>f_2 = \frac{\omega_
e^{-\beta \Delta \epsilon_
}}{\omega_
+ \omega_
e^{-\beta \Delta \epsilon_
+ \omega_
e^{-\beta \Delta \epsilon_
</tex>
<br>
<tex>f_2 = \frac{3 e^{-\beta \Delta \epsilon_
}}{1 + 3 e^{-\beta \Delta \epsilon_
+ \omega_
e^{-\beta \Delta \epsilon_
</tex>
<br>
</center>
Write the electronic partition function as written below.
<center>
<br>
<tex>q_
(T) \approx \omega_
+ \omega_
e^{- \beta \epsilon_
</tex>
<br>
</center>
The nuclear partition function, q_n = \omega_
, contributes only a multiplicative constant to <tex>Q</tex>, and hence affects only the entropy and free energies by a constant additive factor. Since the nuclear state is rarely altered in any chemical process, it does not contribute to thermodynamic changes, and so shall usually not be included in <tex>q</tex>. This cannot be done in the case of the electronic contribution since there are many chemical processes in which the electronic states change.
<p>
</p>
Summary
<center>
<br>
<tex>Q = \frac{ (q_
q_
q_
)^N}
</tex>
<br>
<tex>q_
(V, T) = \left ( \frac
\right )^
V </tex>
<br>
<tex>q_
= \frac
</tex>
<br>
<tex>q_
(T) \approx \omega_
+ \omega_
e^{- \beta \epsilon_
+ ...</tex>
<br>
<tex>q_
= \omega_
+ ...</tex>
<br>
</center>
5-3 Thermodynamic Functions
The Hemholtz free energy is given by the expression below.
<center>
<br>
<tex>A(N, V, T) = -kT \ln Q</tex>
<br>
</center>
The argument of the first logarithm here is much larger than the argument of the second logarithm, and so the electronic contribution to <tex>A</tex> is quite small.
<p>
</p>
The thermodynamic energy is given by the expression below.
<center>
<br>
<tex>E = kT^2 \left ( \frac
\right )_
</tex>
<br>
<tex>p = kT \left ( \frac
\right )_
</tex>
<br>
<tex>p = \frac
</tex>
<br>
</center>
The only contribution to the pressure is from the translational energy of the atoms. Consider the entropy below.
<center>
<br>
<tex>S = kT \left ( \frac
\right )_
+ k \ln Q</tex>
<br>
<tex>S = N k \ln \left [\left ( \frac
\right )^
\frac{V e^{5/2}}
\right] + S_
</tex>
<br>
</center>
Consider the chemical potential.
<center>
<br>
<tex>\mu(T, p) = -kT \left ( \frac
\right )_
</tex>
<br>
<tex>\mu(T, p) = -kT \ln \frac
</tex>
<br>
<tex>\mu(T, p) = -kT \ln \left [\left ( \frac
\right )^
\frac
\right] - kT \ln q_e q_n</tex>
<br>
<tex>\mu(T, p) = -kT \ln \left [\left ( \frac
\right )^
\frac
\right] - kT \ln q_e q_n</tex>
<br>
<tex>\mu(T, p) = -kT \ln \left [\left ( \frac
\right )^
kT \right] - kT \ln q_e q_n + kT \ln p</tex>
<br>
<tex>\mu_0 (T) + kT \ln p</tex>
<br>
<tex>\mu_0 (T) = -kT \ln \left [\left ( \frac
\right )^
k T \right] - k T \ln q_e q_n</tex>
<br>
</center>
</center>