f ( ) = e(μ- /τ) = λe- /τ

Here we have used the symbol λ to mean eμ/τ.

### The Chemical Potential of an Ideal Gas

We will start using the term ideal gas to mean a gas of particles that do not interact with each other and are in the classical regime. Another way of expressing that a system is in the classical regime comes from the quantum concentration. We use n to mean N/V here. Then if a gas is less dense than the quantum concentration, nQ =   , we say it is in the classical regime.

Summing the particles over all orbitals of a system and setting this equal to N, the total number of particles, yields λ = . Expanding λ and solving for the chemical potential gives us:

μ = τ log   ### The Free Energy of an Ideal Gas

We spent a lot of time devising ways to relate the variables we need to the energies. We can utilize that now. Recall that μ =   . We can integrate to solve for F, and we obtain:

F = log   - 1 ### The Pressure of an Ideal Gas

We seek to get the pressure from the free energy. This is no problem though, since we can recall or rederive that p = -   . Looking at the expression for F above, we see that we can expand it to be the sum of many terms, most of which have no V dependence. The derivative becomes simple, and returns something familiar:

p = This is the ideal gas law. If it doesn't look familiar, recall that the chemistry version uses number of moles instead of number of particles, and replaces the temperature as we've defined it with the temperature in Kelvin. You might wish to work out the conversion to assure yourself.