Statistical Thermodynamics and Rate Theories/Translational partition function

Derivation of Translational Partition Function

A Molecular Energy State or is the sum of available translational, vibrational, rotational and electronic states available. The Translational Partition Function gives a "sum over" the available microstates.

$q_{trans}={\left(\frac{2\pi m{k}_{B}T}{h^2}\right)}^{3/2}V$

The derivation begins with the fundamental partition function for a canonical ensemble that is classical and discrete which is defined as:

$q=\sum _j^{\mathrm{\infty }}{e}^{-\beta {ϵ}_j}$

where j is the index, $\beta =\frac{1}{k_{B}T}$ and ${ϵ}_j$is the total energy of the system in the microstate

For a particle in a 3D box with length $L$, mass $m$ and quantum numbers $n_x,n_y,n_z$ the energy levels are given by:

${ϵ}_{n_x,n_y,n_z}=\frac{h^2}{8m{L}^2}(n_x^2+n_y^2+n_z^2)$

Substituting the energy level equation ${ϵ}_{n_x,n_y,n_z}$ for ${ϵ}_j$ in the partition function

$q_{trans}=\sum _j^{\mathrm{\infty }}{e}^{-\beta [\frac{h^2}{8m{L}^2}(n_x^2+n_y^2+n_z^2)]}$

$q_{trans}=\sum _{n_x,n_y,n_z=1}^{\mathrm{\infty }}{e}^{-\beta [\frac{h^2}{8m{L}^2}(n_x^2)]}{e}^{-\beta [\frac{h^2}{8m{L}^2}(n_y^2)]}{e}^{-\beta [\frac{h^2}{8m{L}^2}(n_z^2)]}$

Using rules of summations we can split the above formula into a product of three summation formulas

$q_{trans}=\sum _{n_x=1}^{\mathrm{\infty }}{e}^{-\beta \frac{h^2}{8m{L}^2}(n_x^2)}\sum _{n_y=1}^{\mathrm{\infty }}{e}^{-\beta \frac{h^2}{8m{L}^2}(n_y^2)}\sum _{n_z=1}^{\mathrm{\infty }}{e}^{-\beta \frac{h^2}{8m{L}^2}(n_z^2)}$

Defining the dimensions of the box (Particle In A Box Model) in each direction to be equivalent $n_x=n_y=n_z$

$q_{trans}=[\sum _{n=1}^{\mathrm{\infty }}{e}^{-\beta \frac{h^2}{8m{L}^2}(n^2)}{]}^3$

Because the spacings between translational energy levels are very small they can be treated as continuous and therefore approximate the sum over energy levels as an integral over n

$\sum _{n=1}^{\mathrm{\infty }}{e}^{-\beta \frac{h^2}{8m{L}^2}(n^2)}\approx \underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\beta \frac{h^2}{8m{L}^2}(n^2)}dn$

Using the substitutions $\alpha =\beta \frac{h^2}{8m{L}^2}$ and $n=x$ The integral simplifies to

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\alpha x^2}dx$

From The list of definite integrals the simplified integral has a known solution:

$q_{trans}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\alpha x^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{\alpha }}$

Therefore,

$q_{trans}=(\frac{1}{2}\sqrt{\frac{\pi }{\alpha }}{)}^3$

Re-substituting $\alpha =\beta \frac{h^2}{8m{L}^2}$ and $\beta =\frac{1}{k_{B}T}$

$q_{trans}=(\frac{1}{2}\sqrt{\frac{\pi }{\beta \frac{h^2}{8m{L}^2}}}{)}^3$

$q_{trans}=(\sqrt{\frac{2\pi m{k}_{B}T{L}^2}{h^2}}{)}^3$

Since $L$ is length and $L^3=V$

$q_{trans}=(\frac{2\pi m{k}_{B}T}{h^2}{)}^{3/2}V$