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:

$$ q=\sum_{j}^\infty e^{-\beta \epsilon_j} $$

where j is the index, $$ \beta=\frac{1}{k_B T}$$ and $$ \epsilon_j $$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:

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

Substituting the energy level equation $$ \epsilon_{n_x,n_y,n_z} $$ for $$ \epsilon_j $$ in the partition function

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

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

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

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

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

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

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

$$ \sum_{n=1}^\infty e^{-\beta \frac{h^2}{8 m L^2} (n^2)} \approx \int_0^{\infty} e^{-\beta \frac{h^2}{8 m L^2} (n^2)}\, dn $$

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

$$ \int\limits_{0}^{\infty} e^{-\alpha x^2}\, dx $$

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

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

Therefore,

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

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

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

$$ q_{trans} = \Bigg( \sqrt{\frac{2 \pi m k_B T L^2}{h^2}} \Bigg)^3 $$

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

$$ q_{trans} = \Bigg( \frac{2 \pi m k_B T}{h^2} \Bigg)^{3/2} V $$