91Ó°ÊÓ

The translational partition function is given by

$Z_{\mathrm{tr}}=\frac{1}{h^3}∫d^{3}r{d}^{3}p{e}^{-\raisebox{1ex}{$E_{\mathrm{tr}}$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}...........................\left(1\right)$

As the translational kinetic energy of the molecules does not depend on the position, the integrand is also independent of the position and the integral over the position variables only yields the total volume $V$of the box.

The expression for the translational energy of the gas molecules is given by

$E_{\mathrm{tr}}=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{p_z^2}{2m}..................\left(2\right)$

Here, x-,y- and z- denotes the three components of the momentum.

Substitute the expression of energy into equation (1) and solve to calculate the value of the translational partition function.

role="math" localid="1647147921072" $\begin{array}{rcl}{Z}_{\mathrm{tr}}& =& \frac{V}{h^3}{∫}_0^{∞}d{p}_x{e}^{-\raisebox{1ex}{$p_x^2$}\!\left/ \!\raisebox{-1ex}{$2mkT$}\right.}{∫}_0^{∞}d{p}_y{e}^{-\raisebox{1ex}{$p_y^2$}\!\left/ \!\raisebox{-1ex}{$2mkT$}\right.}{∫}_0^{∞}d{p}_z{e}^{-\raisebox{1ex}{$p_z^2$}\!\left/ \!\raisebox{-1ex}{$2mkT$}\right.}\\ & =& \frac{V}{h^3}\left(\sqrt{\mathrm{π}·2mkT}\right)\left(\sqrt{\mathrm{π}·2mkT}\right)\left(\sqrt{\mathrm{π}·2mkT}\right)\\ & =& \frac{V}{h^3}{\left(2\mathrm{π}mkT\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}........................\left(3\right)\end{array}$

The quantum volume of the gas molecules is given by

$v_Q={\left(\frac{h^2}{2\mathrm{Ï€}mkT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}........................\left(4\right)$

Substitute the value of the quantum volume from equation (4) into equation (3) to obtain the required translational partition function.

$\begin{array}{rcl}{Z}_{\mathrm{tr}}& =& \frac{V}{h^3}{\left(2\mathrm{Ï€}mkT\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& V{\left(\frac{1}{{\displaystyle \frac{h^2}{2\mathrm{Ï€}mkT}}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& \frac{V}{v_Q}\end{array}$