In this section we computed the single-particle translational partition function,$Z_{\mathrm{tr}}$, by summing over all definite-energy wave functions. An alternative approach, however, is to sum over all position and momentum vectors, as we did in Section 2.5. Because position and momentum are continuous variables, the sums are really integrals, and we need to slip a factor of $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$h^3$}\right.$to get a unitless number that actually counts the independent wavefunctions. Thus we might guess the formula

role="math" localid="1647147005946" $Z_{\mathrm{tr}}=\frac{1}{h^3}\int d^{3}r{d}^{3}p{e}^{-\raisebox{1ex}{$E_{\mathrm{tr}}$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}$

where the single integral sign actually represents six integrals, three over the position components and three over the momentum components. The region of integration includes all momentum vectors, but only those position vectors that lie within a box of volume $V$. By evaluating the integrals explicitly, show that this expression yields the same result for the translational partition function as that obtained in the text.

The translational partition function is given by

$Z_{\mathrm{tr}}=\frac{1}{h^3}\int d^{3}r{d}^{3}p{e}^{-\raisebox{1ex}{$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" $\begin{array}{rcl}{Z}_{\mathrm{tr}}& =& \frac{V}{h^3}{\int }_0^{\infty }d{p}_x{e}^{-\raisebox{1ex}{$p_x^2$}\!\left/ \!\raisebox{-1ex}{$2mkT$}\right.}{\int }_0^{\infty }d{p}_y{e}^{-\raisebox{1ex}{$p_y^2$}\!\left/ \!\raisebox{-1ex}{$2mkT$}\right.}{\int }_0^{\infty }d{p}_z{e}^{-\raisebox{1ex}{$p_z^2$}\!\left/ \!\raisebox{-1ex}{$2mkT$}\right.}\\ & =& \frac{V}{h^3}\left(\sqrt{\pi ·2mkT}\right)\left(\sqrt{\pi ·2mkT}\right)\left(\sqrt{\pi ·2mkT}\right)\\ & =& \frac{V}{h^3}{\left(2\pi 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\pi 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\pi mkT\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& V{\left(\frac{1}{{\displaystyle \frac{h^2}{2\pi mkT}}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& \frac{V}{v_Q}\end{array}$