Let us consider the translational partition function of a monatomic gas particle confined to a cubic box of length \(L\). The particle inside the box has translational energy levels given by: \[E_\text{trans}= \dfrac{h^2 \left(n_x^2+ n_y^2+ n_z^2 \right)}{8 mL^2} \nonumber \] where \(n_x\), \(n_y\) and \(n_z\) are the quantum numbers in the three directions. The translational partition function is given by: \[q_\text{trans} = \sum_i e^{−\epsilon_i/kT} \nonumber \] which is the product of translational partition functions in the three dimensions. We can write the translation partition function as the product of the translation partition function for each direction: \[\begin{align} q_\text{trans} &= q_{x} q_{y} q_{z} \label{times1} \\[4pt] &= \sum_{n_x=1}^{\infty} e^{−\epsilon_x/kT} \sum_{n_y=1}^{\infty} e^{−\epsilon_y/kT} \sum_{n_z=1}^{\infty} e^{−\epsilon_z/kT} \label{sum1} \end{align} \] Since the levels are very closely spaced (continuous), we can replace each sum in Equation \(\ref{sum1}\) with an integral. For example: \[\begin{align} q_{x} &= \sum_{n_x=1}^{\infty} e^{−\epsilon_x/kT} \\[4pt] &\approx \int_{n_x=1}^{\infty} e^{−\epsilon_x/kT} \label{int1} \end{align} \] 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 \(\ref{int1}\): \[ q_x= \int_{1}^{\infty} e^{− \frac{h^2 n_x^2}{8mL^2 kT}} \approx \int_{0}^{\infty} e^{− \frac{h^2 n_x^2}{8mL^2 kT}} \nonumber \] We then use the following solved Gaussian integral: \[ \int_o^{\infty} e^{-an^2} dn = \sqrt{\dfrac{\pi}{4a}} \nonumber \] with the following substitution: \[a = \dfrac{h^2}{8mL^2 kT} \nonumber \] we get: \[q_x= \dfrac{1}{2} \sqrt{\dfrac{π}{a}} = \dfrac{1}{2} \sqrt{\dfrac{π 8m kT }{ h^2 }} L \nonumber \] or more commonly presented as: \[q_x = \dfrac{L}{\Lambda} \nonumber \] where \(\Lambda\) is the de Broglie wavelength and is given by \[ \Lambda = \dfrac{h}{\sqrt{2 π 8m kT}} \nonumber \] 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: \[q_\text{trans} = \left( \dfrac{\sqrt{2 π 8m kT}} {h} \right)^{3/2} V = \dfrac{ V}{\Lambda^3} \label{parttransation} \] This is usually a very large number (10 20 ) for volumes of 1 cm 3 for small molecular masses. This means that such a large number of translational states are accessible and 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 is: \[U=\dfrac{3}{2} nRT \nonumber \] 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. Calculate the translational partition function of an \(I_2\) molecule at 300K. Assume V to be 1 liter. Solution Mass of \(I_2\) is \(2 \times 127 \times 1.6606 \times 10^{-27} kg\) \[\begin{align*} 2πmkT &= 2 \times 3.1415 \times (2 \times 127 \times 1.6606 \times 10^{-27}\, kg) \times 1.3807 \times 10^{-23} \, J/K \times 300 K \\[4pt] &= 1.0969 \times 10^{-44}\; J\, kg \end{align*} \] \[\begin{align*} Λ &= \dfrac{h}{\sqrt{2 π m kT}} \\[4pt] &= \dfrac{6.6262 \times 10^{-34}\;J\, s}{ \sqrt{1.0969 \times 10^{-44}\, J \, kg}} = 6.326 \times 10^{-12}\;m \end{align*} \] The via Equation \ref{parttransation} \[q_\text{trans}= \dfrac{V}{Λ^3}= \dfrac{1000 \times 10^{-6} m^3}{(6.326 \times 10^{-12} \; m)^3}= 3.95 \times 10^{30} \nonumber \] This means that \(3.95 \times 10^{30}\) quantum states are thermally accessible to the molecular system