# 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.

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

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:

$\epsilon _{n_{x},n_{y},n_{z}}={\frac {h^{2}}{8mL^{2}}}(n_{x}^{2}+n_{y}^{2}+n_{z}^{2})$

Substituting the energy level equation $\epsilon _{n_{x},n_{y},n_{z}}$  for $\epsilon _{j}$  in the partition function

$q_{trans}=\sum _{j}^{\infty }e^{-\beta [{\frac {h^{2}}{8mL^{2}}}(n_{x}^{2}+n_{y}^{2}+n_{z}^{2})]}$

$q_{trans}=\sum _{n_{x},n_{y},n_{z}=1}^{\infty }e^{-\beta [{\frac {h^{2}}{8mL^{2}}}(n_{x}^{2})]}e^{-\beta [{\frac {h^{2}}{8mL^{2}}}(n_{y}^{2})]}e^{-\beta [{\frac {h^{2}}{8mL^{2}}}(n_{z}^{2})]}$

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

$q_{trans}=\sum _{n_{x}=1}^{\infty }e^{-\beta {\frac {h^{2}}{8mL^{2}}}(n_{x}^{2})}\sum _{n_{y}=1}^{\infty }e^{-\beta {\frac {h^{2}}{8mL^{2}}}(n_{y}^{2})}\sum _{n_{z}=1}^{\infty }e^{-\beta {\frac {h^{2}}{8mL^{2}}}(n_{z}^{2})}$

Defining the dimensions of the box (Particle In A Box Model) in each direction to be equivalent $n_{x}=n_{y}=n_{z}$

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

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

$\sum _{n=1}^{\infty }e^{-\beta {\frac {h^{2}}{8mL^{2}}}(n^{2})}\approx \int _{0}^{\infty }e^{-\beta {\frac {h^{2}}{8mL^{2}}}(n^{2})}\,dn$

Using the substitutions $\alpha =\beta {\frac {h^{2}}{8mL^{2}}}$  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 $\alpha =\beta {\frac {h^{2}}{8mL^{2}}}$  and $\beta ={\frac {1}{k_{B}T}}$

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

$q_{trans}={\Bigg (}{\sqrt {\frac {2\pi mk_{B}TL^{2}}{h^{2}}}}{\Bigg )}^{3}$

Since $L$  is length and $L^{3}=V$

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