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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

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

where $ N$ is the number of constituent molecules,

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

$\displaystyle k_x$ $\displaystyle = n_x \frac{\pi}{L_x},$ (8.72)
$\displaystyle k_y$ $\displaystyle =n_y \frac{\pi}{L_y},$ (8.73)
$\displaystyle k_z$ $\displaystyle =n_z \frac{\pi}{L_z},$ (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

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

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

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

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

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

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

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

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

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

It follows from Equation (8.70) that

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

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

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


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

and, finally,

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


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

next up previous
Next: Derivation of van der Up: Quantum Statistics Previous: Quantum Statistics in Classical
Richard Fitzpatrick 2016-01-25