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Next: Gibb's paradox Up: Applications of statistical thermodynamics Previous: Partition functions

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},
\end{displaymath} (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}},
\end{displaymath} (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.
\end{displaymath} (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,
\end{displaymath} (430)

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

$\displaystyle \int \exp[-(\beta/2m)\,p^2]\,d^3{\bf p}$ $\textstyle =$ $\displaystyle \int_{-\infty}^{\infty}\exp[-(\beta/2m)\,p_x^{~2}]\,dp_x\,
    $\displaystyle \times \int_{-\infty}^{\infty}\exp[-(\beta/2m)\,p_z^{~2}]\,dp_z$  
  $\textstyle =$ $\displaystyle \left(\sqrt{\frac{2\,\pi\,m}{\beta}}\right)^3,$ (432)

where use has been made of Eq. (79). Thus,
\zeta = V \left( \frac{2\,\pi\, m}{h_0^{~2}\,\beta}\right)^{3/2},
\end{displaymath} (433)

\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].
\end{displaymath} (434)

The expression for the mean pressure (414) yields

\overline{p} = \frac{1}{\beta}\frac{\partial\ln Z}{\partial V} = \frac{1}{\beta}
\end{displaymath} (435)

which reduces to the ideal gas equation of state
\overline{p} \,V = N\, k\, T = \nu \,R \,T,
\end{displaymath} (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.
\end{displaymath} (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,
\end{displaymath} (438)

so the mean energy can be written
\overline{E} = \nu\, c_V \,T.
\end{displaymath} (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}).
\end{displaymath} (440)

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],
\end{displaymath} (441)

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

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

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

next up previous
Next: Gibb's paradox Up: Applications of statistical thermodynamics Previous: Partition functions
Richard Fitzpatrick 2006-02-02