4.9 The ideal gas

Take-home message: We can now derive the equation of state and other properties of the ideal gas.

We are now reaching the most important test of statistical physics: the ideal gas. For the moment we assume it is monatomic; the extra work for a diatomic gas is minimal.

Remember the one-particle translational partition function, at any attainable temperature, is

$$Z_1= V\left({mk_{\scriptscriptstyle B}T\over 2 \pi\hbar^2}\right)^{3/2}\equiv V n_Q.$$

From this we can obtain the average energy per particle, $\frac 3 2 k_{\scriptscriptstyle B}T$, and since the particles are non-interacting, the energy of $N$ particles in a box is just $\frac 3 2 Nk_{\scriptscriptstyle B}T$. This could be obtained from the expression we previously used for the $N$-particle partition function, $Z_N=Z_1^N$.

But if we follow this through and calculate the Helmholtz free energy and the entropy, we find that the results do not make sense: specifically, if one has double the number of particles, in double the volume, the entropy and the Helmholtz free energy, like the energy, should double. These are extensive variables. But if we go ahead and calculate based on $Z_N=Z_1^N$, we do not get extensive results, but terms like $V \ln N$.

However we shouldn't have expected $Z_N=Z_1^N$ to work, because the derivation was based on the idea that every one of the $N$ particles was distinguishable. But at a completely fundamental level, every molecule is exactly the same as every other molecule of the same substance.

The exact form of $Z_N$ in this case has no compact form. But there is an approximation which becomes exact in the limit of low number densities $n\equiv N/V$: specifically $n\ll n_Q$ where $n_Q\equiv(mk_{\scriptscriptstyle B}T/2 \pi\hbar^2)^{3/2}$ is the ``quantum concentration'' and is a measure of the number of energy levels available. It is also proportial to the inverse of cube of the thermal de Broglie wavelength (the wavelength of a particle of energy of order $k_B T$). The significance of this limit is that it is very unlikely that any two atoms are in the same energy level. This is called the classical limit.

$\mbox{\LARGE\colorbox{yellow}{\rule[-3mm]{0mm}{10mm} \ $\displaystyle Z_N={(Z_1)^N\over N!}$ }}$

More details of how this works here.

Now, using Stirling's approximation $\ln(N!)\approx N\ln N-N$, we find

$$\begin{eqnarray*} \left\langle F \right\rangle \!\!\!&=&\!\!\!-k_{\scriptscripts... ...iptscriptstyle B}T \left[\ln\left({ n\over n_Q}\right)-1\right]. \end{eqnarray*}$$

Also

$$\begin{eqnarray*} P\!\!\!&=&\!\!\!-\left(\partial F\over\partial V\right)_{\!\sc... ...tyle T,N}\\ \!\!\!&=&\!\!\!{ Nk_{\scriptscriptstyle B}T\over V} \end{eqnarray*}$$

and

$$\begin{eqnarray*} S\!\!\!&=&\!\!\!-\left(\partial F\over\partial T\right)_{\!\sc... ...ght)+\frac 5 2 \right] \qquad \hbox{The Sackur-Tetrode Equation} \end{eqnarray*}$$

Since $n_Q$ is composed only of constants and $T$, it is intensive; the number density $n\equiv V/N$ is the ratio of extensive quantities and so is also intensive. Hence $F$ and $S$ are clearly simply proportional to $N$, and so extensive as required, and $P$ is intensive. Since $n_Q\gg n$ if the result is to be valid, $S$ is also positive, as it should be!

The expression for $P$ is clearly experimentally verifiable: it is the ideal gas law. That's good, but we expected to get that. More interestingly the Sackur-Tetrode equation for $S$ can also be checked. First, if we unpick the dependence on $V$ and $T$, we get

$$S=Nk_{\scriptscriptstyle B}(\ln V+\frac 3 2 \ln T +\hbox{const.})$$

which is in accord with the form derived from classical thermodynamics (see here). But more importantly it predicts the absolute entropy of a gas at a certain temperature, and this can be checked experimentally too. If we start with the solid at some very low temperature $T_0$, at which the entropy can be assumed to be very small, and we know the experimental specific heat capacity as a function of temperature and the latent heats of melting and vaporisation, we can numerically calculate the integral

$$\int_{T_0}^T {{}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}Q\over T} =S(T)-S(T_0)\approx S(T)$$

Good agreement is found. An example with numerical details can be found on pages 5&6 here, from Edward J. Groth of Princeton University.

Finally, we include vibrations and rotations as well as translations: since the one-particle energies are independent and add, $\varepsilon=\varepsilon_{\rm tr}+\varepsilon_{\rm rot}+\varepsilon_{\rm vib}$, the partition functions multiply: $Z_1=Z_1^{\rm tr}Z_1^{\rm rot}Z_1^{\rm vib}$ (the argument is like that for the $N$-particle partition function for distinguishable particles and is given in more detail here) and so

$$\begin{eqnarray*} Z_N\!\!\!&=&\!\!\!{(Z_1^{\rm tr})^N(Z_1^{\rm rot})^N (Z_1^{\rm... ...rm vib})^N\\ F\!\!\!&=&\!\!\!F_{\rm tr}+F_{\rm rot}+F_{\rm vib} \end{eqnarray*}$$

and the energy and entropy also add.

References

• Mandl 7.1,7.4-6
• Bowley and Sánchez 6.5
• Kittel and Kroemer 3