# 5.6: Translational Partition Functions of Monotonic Gases

Let consider the translational partition function of a monatomic gas. Consider a molecule confined to a cubic box. A molecule inside a cubic box of length $$L$$ has the translational energy levels given by

$E_{tr}= \dfrac{h^2 \left(n_x^2+ n_y^2+ n_z^2 \right)}{8 mL^2}$

where $$n_x$$, $$n_y$$ and $$n_z$$ are the quantum numbers in the three directions. The translational partition function is given by

$q_{tr} = \sum_i e^{−\epsilon_i/k_BT} \nonumber$

which is the product of translational partition functions in the three dimensions.

\begin{align} q_{tr} &= q_{x} q_{y} q_{z} \label{times1} \\[4pt] &= \sum_{n_x=1}^{\infty} e^{−\epsilon_x/k_BT} \sum_{n_y=1}^{\infty} e^{−\epsilon_y/k_BT} \sum_{n_z=1}^{\infty} e^{−\epsilon_z/k_BT} \label{sum1} \end{align}

Since the levels are very closely spaced, we can replace each sum in Equation $$\ref{sum1}$$ with an integral, for example:

\begin{align} q_{x} &= \sum_{n_x=1}^{\infty} e^{−\epsilon_x/k_BT} \\[4pt] &\approx \int_{n_x=1}^{\infty} e^{−\epsilon_x/k_BT} \label{int1} \end{align}

and after substituting the energy for the relevant dimension

$\epsilon_x = \dfrac{h^2 n_x^2}{8mL^2} \nonumber$

we can extend the lower limit of integration in the approximation of Equation $$\ref{int1}$$

$q_x= \int_{1}^{\infty} e^{− \frac{h^2 n_x^2}{8mL^2 k_BT}} \approx \int_{0}^{\infty} e^{− \frac{h^2 n_x^2}{8mL^2 k_BT}}$

we then use the following solved Gaussian integral

$\int_o^{\infty} e^{-an^2} dn = \sqrt{\dfrac{\pi}{4a}} \nonumber$

with the following substitution

$a = \dfrac{h^2}{8mL^2 k_BT} \nonumber$

we get,

$q_x= \dfrac{1}{2} \sqrt{\dfrac{π}{a}} = \dfrac{1}{2} \sqrt{\dfrac{π 8m k_BT }{ h^2 }} L \nonumber$

or more commonly presented as

$q_x = \dfrac{L}{\Lambda} \nonumber$

where $$\Lambda$$ is the de Broglie thermal wavelength and is given by

$\Lambda = \dfrac{h}{\sqrt{2 π 8m k_bT}}$

Multiplying the expressions for $$q_x$$, $$q_y$$ and $$q_z$$ (Equation $$\ref{times1}$$) and using $$V$$ as the volume of the box $$L^3$$, we arrive at

$q_{tr} = \left( \dfrac{\sqrt{2 π 8m k_bT}} {h} \right)^{3/2} V = \dfrac{ V}{\Lambda^3} \label{parttransation}$

This is usually a very large number (1020) for volumes of 1 cm3 for a typical small molecular masses. This means that such a large number of translational states are accessible available for occupation by the molecules of a gas. This result is very similar to the result of the classical kinetic gas theory that said that the observed energy of an ideal gas should read as

$U=\dfrac{3}{2} nRT$

We postulate therefore that the observed energy of a macroscopic system should equal the statistical average over the partition function as shown above. In other words: if you know the particles your system is composed of and their energy states you can use statistics to calculate what you should observe on the whole ensemble.

##### Example

Calculate the translational partition function of an $$I_2$$ molecule at 300K. Assume V to be 1 liter.

Solution

Mass of $$I_2$$ is $$2 \times 127 \times 1.6606 \times 10^{-27} kg$$

\begin{align*} 2πmk_BT &= 2 \times 3.1415 \times (2 \times 127 \times 1.6606 \times 10^{-27}\, kg) \times 1.3807 \times 10^{-23} \, J/K \times 300 K \\[4pt] &= 1.0969 \times 10^{-44}\; J\, kg \end{align*}

\begin{align*} Λ &= \dfrac{h}{\sqrt{2 π m k_BT}} \\[4pt] &= \dfrac{6.6262 \times 10^{-34}\;J\, s}{ \sqrt{1.0969 \times 10^{-44}\, J \, kg}} = 6.326 \times 10^{-12}\;m \end{align*}

The via Equation \ref{parttransation}

$q_{tr}= \dfrac{V}{Λ^3}= \dfrac{1000 \times 10^{-6} m^3}{(6.326 \times 10^{-12} \; m)^3}= 3.95 \times 10^{30}$

This means that $$3.95 \times 10^{30}$$ quantum states are thermally accessible to the molecular system