Ideal monatomic gases

Let us now practice calculating thermodynamic relations using the partition function by considering an example with which we are already quite familiar: i.e., an ideal monatomic gas. Consider a gas consisting of $N$ identical monatomic molecules of mass $m$ enclosed in a container of volume $V$. Let us denote the position and momentum vectors of the $i$th molecule by ${\bf r}_i$ and ${\bf p}_i$, respectively. Since the gas is ideal, there are no interatomic forces, and the total energy is simply the sum of the individual kinetic energies of the molecules:

$$E = \sum_{i=1}^N \frac{ p_i^{~2}}{2\,m},$$ (427)

where $p_i^{~2} = {\bf p}_i\!\cdot \!{\bf p}_i$.

Let us treat the problem classically. In this approach, we divide up phase-space into cells of equal volume $h_0^{~f}$. Here, $f$ is the number of degrees of freedom, and $h_0$ is a small constant with dimensions of angular momentum which parameterizes the precision to which the positions and momenta of molecules are determined (see Sect. 3.2). Each cell in phase-space corresponds to a different state. The partition function is the sum of the Boltzmann factor $\exp(-\beta \,E_r)$ over all possible states, where $E_r$ is the energy of state $r$. Classically, we can approximate the summation over cells in phase-space as an integration over all phase-space. Thus,

$$Z = \int\cdots \int \exp(-\beta \, E)\, \frac{d^3{\bf r}_1\cdots d^3{\bf r}_N\, d^3{\bf p}_1\cdots d^3{\bf p}_N}{h_0^{~3N}},$$ (428)

where $3N$ is the number of degrees of freedom of a monatomic gas containing $N$ molecules. Making use of Eq. (427), the above expression reduces to

$$Z = \frac{V^N}{h_0^{~3N}}\int\cdots \int \exp[-(\beta/2m)\,p... ...d^3{\bf p}_1 \cdots \exp[-(\beta/2m)\,p_N^{~2}]\,d^3{\bf p}_N.$$ (429)

Note that the integral over the coordinates of a given molecule simply yields the volume of the container, $V$, since the energy $E$ is independent of the locations of the molecules in an ideal gas. There are $N$ such integrals, so we obtain the factor $V^N$ in the above expression. Note, also, that each of the integrals over the molecular momenta in Eq. (429) are identical: they differ only by irrelevant dummy variables of integration. It follows that the partition function $Z$ of the gas is made up of the product of $N$ identical factors: i.e.,

$$Z = \zeta^N,$$ (430)

where

$$\zeta = \frac{V}{h_0^{~3}}\int \exp[-(\beta/2m)\,p^2]\,d^3{\bf p}$$ (431)

is the partition function for a single molecule. Of course, this result is obvious, since we have already shown that the partition function for a system made up of a number of weakly interacting subsystems is just the product of the partition functions of the subsystems (see Sect. 7.5).

The integral in Eq. (431) is easily evaluated:

$$\int \exp[-(\beta/2m)\,p^2]\,d^3{\bf p} = \int_{-\infty}^{\infty}\exp[-(\beta/2m)\,p_x^{~2}]\,dp_x\, \int_{-\infty}^{\infty}\exp[-(\beta/2m)\,p_y^{~2}]\,dp_y\,$$

$$\times \int_{-\infty}^{\infty}\exp[-(\beta/2m)\,p_z^{~2}]\,dp_z$$

$$= \left(\sqrt{\frac{2\,\pi\,m}{\beta}}\right)^3,$$ (432)

where use has been made of Eq. (79). Thus,

$$\zeta = V \left( \frac{2\,\pi\, m}{h_0^{~2}\,\beta}\right)^{3/2},$$ (433)

and

$$\ln Z = N\ln \zeta = N \left[\ln V - \frac{3}{2}\ln \beta +\frac{3}{2} \ln\!\left(\frac{2\,\pi\, m}{h_0^{~2}}\right)\right].$$ (434)

The expression for the mean pressure (414) yields

$$\overline{p} = \frac{1}{\beta}\frac{\partial\ln Z}{\partial V} = \frac{1}{\beta} \frac{N}{V},$$ (435)

which reduces to the ideal gas equation of state

$$\overline{p} \,V = N\, k\, T = \nu \,R \,T,$$ (436)

where use has been made of $N= \nu \,N_A$ and $R= N_A\, k$. According to Eq. (399), the mean energy of the gas is given by

$$\overline{E} = - \frac{\partial \ln Z}{\partial \beta} = \frac{3}{2} \frac{N}{\beta} = \nu\, \frac{3}{2} \,R\, T.$$ (437)

Note that the internal energy is a function of temperature alone, with no dependence on volume. The molar heat capacity at constant volume of the gas is given by

$$c_V = \frac{1}{\nu} \left(\frac{\partial \overline{E}}{\partial T} \right)_V = \frac{3}{2}\, R,$$ (438)

so the mean energy can be written

$$\overline{E} = \nu\, c_V \,T.$$ (439)

We have seen all of the above results before. Let us now use the partition function to calculate a new result. The entropy of the gas can be calculated quite simply from the expression

$$S = k\,(\ln Z + \beta\, \overline{E}).$$ (440)

Thus,

$$S = \nu R \left[ \ln V -\frac{3}{2} \ln \beta + \frac{3}{2}\... ...eft(\frac{2\,\pi\, m }{h_0^{~2}}\right) + \frac{3}{2} \right],$$ (441)

or

$$S = \nu R \left[ \ln V + \frac{3}{2}\ln T + \sigma\right],$$ (442)

where

$$\sigma = \frac{3}{2} \ln \!\left( \frac{2\,\pi\, m\,k}{h_0^{~2}}\right) + \frac{3}{2}.$$ (443)

The above expression for the entropy of an ideal gas is certainly new. Unfortunately, it is also quite obviously incorrect!