The classical ideal gas

2.9 The classical ideal gas

Calculation of the micro canonical state sum (phase volume)
For the calculation of the thermodynamic potential of a classical ideal gas we need the partition function, i.e. the phase volume of the ”free particles”, which are in a box with volume \(V\):

\begin{equation*} \begin{split} \Phi(E, V_s, N) & = \frac{1}{N! h^{3N}} \int\limits_{H=\sum_i\frac{p_i^2}{2 m} + H_{wall}} d^3p_1...d^3p_N d^3r_1...d^3r_N\\ & = \frac{V^N}{N! h^{3N}} \frac{\pi^{\frac{3N}{2}}}{\left( \frac{3N}{2}\right)!} (2 m E)^{\frac{3N}{2}} \approx v^N \left( \frac{4 \pi m}{3 h^2} \epsilon \right)^{\frac{3N}{2}} e^{\frac{5N}{2}} \end{split} \end{equation*}

(2.58)

here we have used the Stirling formula \(N! \approx N^N e^{-N} = \left(\frac{N}{e} \right)^N\).
Finally we get with \(\epsilon := \frac{E}{N}\)

\begin{equation*} \begin{split} S(E, N, V) & = k \ln (\Phi(E, V_s, N)) = k \ln \left( v^N \left(\frac{4 \pi m}{3 h^2} \epsilon \right)^{\frac{3N}{2}} e^{\frac{5N}{2}} \right)\\ & = N k \ln(v) + \frac{3}{2} N k \ln(\epsilon) + k \ln \left( \left(\frac{4 \pi m}{3 h^2} \right)^{\frac{3N}{2}} e^{\frac{5N}{2}} \right) \end{split} \end{equation*}

(2.59)

By differentiating we find:

\begin{equation*} \frac{1}{T} = \frac{\partial S(E, N, V)}{\partial E} = \frac{3}{2} \frac{k N}{E} \qquad \mbox{leading to} \qquad E = \frac{3}{2} N k T \qquad , \end{equation*}

(2.60)

and

\begin{equation*} \frac{p}{T} = \frac{\partial S(E, N, V)}{\partial V} = \frac{k N}{V} \qquad \mbox{leading to} \qquad p V = k N T \qquad . \end{equation*}

(2.61)

Let

\begin{equation*} \lambda (T) := \frac{h}{\sqrt{2 \pi m k T}} \qquad , \end{equation*}

(2.62)

then we get

\begin{equation*} S(T, V, N) = k N \left[ \ln\left(\frac{v}{\lambda^3(T)} \right) + \frac{5}{2}\right] \qquad , \end{equation*}

(2.63)

\begin{equation*} F(T, V, N) = E - T S = \frac{3}{2} N k T - N k T \left[ \frac{5}{2} + \ln\left(\frac{v}{\lambda^3(T)} \right) \right] = - N k T \left[ 1 + \ln\left(\frac{v}{\lambda^3(T)} \right) \right] \qquad . \end{equation*}

(2.64)

The function \(\lambda(T)\) is except for the factor \(\sqrt{\frac{2 \pi}{3}}\) the de Broglie wavelength (see above) of a particle with the thermal energy \(\epsilon = \frac{3}{2} k T\). So our results only hold if \(\lambda^3(T) \ll \frac{V}{N}\).
Summing up all approximations:

Each particle occupies a volume which is much bigger than the uncertainty in space. That the particles are not distinguishable (the symmetry relations) is not important any more.
Not only the states with the lowest energy levels will be occupied in the phase space, i.e. we must not count the states exactly.
We need high temperatures and extremely diluted gases.

The grand canonical potential and variations of the number of particles
Starting with

\begin{equation*} \mu = \frac{\partial F}{\partial N} = \frac{F}{N} + k T = - k T \ln\left(\frac{v}{\lambda^3(T)}\right) \; \mbox{, i.e.} \quad N = \frac{V}{\lambda^3(T)} e^{\frac{\mu}{k T}} \quad , \end{equation*}

(2.65)

we get

\begin{equation*} \Omega(T, V, \mu) = F - \mu N = - N k T = - \frac{k T}{\lambda^3(T)} V e^{\frac{\mu}{k T}} \label{grand_c_pot} \quad . \end{equation*}

(2.66)

Generally holds:

\begin{equation*} \Omega(T, V, \mu) = - k T \sum_i \ln\left(\sum_{n_i}\exp\left(- n_i \frac{\epsilon_i - \mu}{kT} \right) \right) \; \mbox{and} \quad \left\langle N \right\rangle = - \frac{\partial \Omega}{\partial \mu} = \sum_i \frac{\sum_{n_i} n_i\exp\left(- n_i \frac{\epsilon_i - \mu}{kT}\right)}{\sum_{n_i}\exp\left(- n_i \frac{\epsilon_i - \mu}{kT}\right)} \quad. \end{equation*}

(2.67)

Consequently we find

\begin{equation*} \frac{\partial^2 \Omega}{\partial \mu^2} = - \frac{\partial \left\langle N \right\rangle }{\partial \mu} = \frac{1}{k T}\left[\left\langle N^2\right\rangle - \left\langle N \right\rangle ^2 \right] \qquad. \end{equation*}

(2.68)

Using Eq. (2.66) we get

\begin{equation*} \frac{\partial \Omega}{\partial \mu}(T, V, \mu) = - \frac{1}{\lambda^3(T)} V e^{\frac{\mu}{k T}}\quad \mbox{and} \quad \frac{\partial^2 \Omega}{\partial \mu^2}(T, V, \mu) = - \frac{1}{k T}\frac{1}{\lambda^3(T)} V e^{\frac{\mu}{k T}} \end{equation*}

(2.69)

The relative variance of the particle number is therefor

\begin{equation*} \frac{\left\langle N^2\right\rangle -\left\langle N \right\rangle^2}{\left\langle N \right\rangle^2} = \frac{k T}{\left\langle N \right\rangle^2} \frac{\partial^2 \Omega}{\partial \mu^2} = \frac{1}{\left\langle N \right\rangle} \qquad , \end{equation*}

(2.70)

which is the solution for the Poisson distribution. For macroscopic systems with \(N \approx 10^{23}\) particles the variance in \(N\) is negligible small.
Equivalent results hold for all other ”generalizes forces” in thermodynamics. Although we only fix the ”generalizes coordinates” of a system the forces are extremely well defined. The Legendre transformation just switches from coordinates to forces, which are in both contacts well defined and contain the same information. A random process leads to an extremely reliable result, if the involved numbers are large enough. Thermodynamic is just mathematics and its results are almost as exact as pure mathematics.