# 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.

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

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

${\displaystyle q=\sum _{j}^{\infty }e^{-\beta \epsilon _{j}}}$

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

For a particle in a 3D box with length ${\displaystyle L}$ , mass ${\displaystyle m}$  and quantum numbers ${\displaystyle n_{x},n_{y},n_{z}}$  the energy levels are given by:

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle \int \limits _{0}^{\infty }e^{-\alpha x^{2}}\,dx}$

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

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

Therefore,

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

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

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

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

Since ${\displaystyle L}$  is length and ${\displaystyle L^{3}=V}$

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