Quantum-Mechanical Treatment of Ideal Gas

Let us calculate the partition function of an ideal gas from quantum mechanics, making use of Maxwell-Boltzmann statistics. Obviously, such a partition function is only applicable when the gas is non-degenerate. According to Equations (8.67) and (8.69), we can write the partition function in the form

$$Z = \frac{\zeta^{ N}}{N!},$$ (8.70)

where $N$ is the number of constituent molecules,

$$\zeta = \sum_r {\rm e}^{-\beta \epsilon_r}$$ (8.71)

is the partition function of an individual molecule, and the factor $N!$ is necessary to take into account the fact that the molecules are indistinguishable according to quantum theory.

Suppose that the gas is enclosed in a parallelepiped with sides of lengths $L_x$ , $L_y$ , and $L_z$ . It follows that the de Broglie wavenumbers of the constituent molecules are quantized such that

$$k_x = n_x \frac{\pi}{L_x},$$ (8.72)

$$k_y =n_y \frac{\pi}{L_y},$$ (8.73)

$$k_z =n_z \frac{\pi}{L_z},$$ (8.74)

where $n_x$ , $n_y$ , and $n_z$ are independent integers. (See Section C.10.) Thus, the allowed energy levels of a single molecule are

$$\epsilon_r = \frac{\hbar^{ 2} (k_x^{ 2} + k_y^{ 2}+k_z^{ 2})}{2 m},$$ (8.75)

where $m$ is the molecular mass. Hence, we deduce that

$$\zeta =\sum_{k_x,k_y,k_z}\exp\left[-\frac{\beta \hbar^{ 2}}{2 m} (k_x^{ 2}+k_y^{ 2}+k_z^{ 2})\right],$$ (8.76)

where the sum is over all possible values of $k_x$ , $k_y$ , $k_z$ . The previous expression can also be written

$$\zeta=\left(\sum_{k_x}{\rm e}^{-(\beta \hbar^{ 2}/2 m) k_x^{\... ...\right) \left(\sum_{k_z}{\rm e}^{-(\beta \hbar^{ 2}/2 m) k_z^{ 2}}\right).$$ (8.77)

Now, in the previous equation, the successive terms in the sum over $k_x$ (say) correspond to a very small increment, ${\mit\Delta}k_x=2\pi/L_x$ , in $k_x$ (in fact, the increment can be made arbitrarily small by increasing $L_x$ ), and, therefore, differ very little from each other. Hence, it is an excellent approximation to replace the sums in Equation (8.77) by integrals. Given that $1={\mit\Delta}n_x=(L_x/2\pi) dk_x$ , we can write

$$\sum_{k_x} {\rm e}^{-(\beta \hbar^{ 2}/2 m) k_x^{ 2}}\simeq ... ...\right)^{1/2}= \frac{L_x}{2\pi \hbar}\left(\frac{2\pi m}{\beta}\right)^{1/2}.$$ (8.78)

(See Exercise 2.) Hence, Equation (8.77) becomes

$$\zeta = \frac{V}{(2\pi \hbar)^{ 3}}\left(\frac{2\pi m}{\beta}\right)^{3/2} = V\left(\frac{2\pi m}{\beta h^{ 2}}\right)^{ 3/2},$$ (8.79)

where $V=L_x L_y L_z$ is the volume of the gas.

It follows from Equation (8.70) that

$$\ln Z=N (\ln\zeta-\ln N +1),$$ (8.80)

where use has been made of Stirling's approximation. Thus, we obtain

$$\ln Z =N\left[\ln \left(\frac{V}{N}\right) -\frac{3}{2} \ln\beta +\frac{3}{2} \ln\left(\frac{2\pi m}{h^{ 2}}\right)+1 \right],$$ (8.81)

and

$$\overline{E} =-\frac{\partial \ln Z}{\partial \beta} =\frac{3}{2} \frac{N}{\beta}=\frac{3}{2} N k T,$$ (8.82)

and, finally,

$$S = k\left(\ln Z+\beta \overline{E}\right) = N k \left[\ln \left(\frac{V}{N}\right) +\frac{3}{2} \ln T +\sigma_0 \right],$$ (8.83)

where

$$\sigma_0=\frac{3}{2} \ln\left(\frac{2\pi m k}{h^{ 2}}\right)+\frac{5}{2}.$$ (8.84)

These results are exactly the same as those obtained in Sections 7.7 and 7.8, using classical mechanics, except that the arbitrary parameter $h_0$ is replaced by Planck's constant, $h=6.61\times 10^{-34} {\rm J s}$ .