I've been thinking again. The formula for the Maxwell Speed Distribution for a non-ideal gas is [itex]\displaystyle f(v) = 4\pi \left(\frac{M}{2\pi RT}\right)^{\frac{3}{2}} v^2 e^{\frac{-Mv^2}{2RT}}[/itex]. My derivation follows as such: [itex]\displaystyle f(v) = 4\pi \left(\frac{m}{2\pi nRT}\right)^{\frac{3}{2}} v^2 e^{\frac{-mv^2}{2nRT}}[/itex], where m is the mass of the gas and n is the number of moles. [itex]\displaystyle f(v) = 4\pi \left(\frac{m}{2\pi P_{ideal}V_{ideal}}\right)^{\frac{3}{2}} v^2 e^{\frac{-mv^2}{2P_{ideal}V_{ideal}}}[/itex], by the ideal gas law. [itex]\displaystyle f(v) = 4\pi \left(\frac{m}{2\pi (P + \frac{an^2}{V^2})(V-nb)}\right)^{\frac{3}{2}} v^2 e^{\frac{-mv^2}{2(P + \frac{an^2}{V^2})(V-nb)}}[/itex], through the Van der Waals equation. Factoring, we get [itex]\displaystyle f(v) = 4\pi \left(\frac{mV^2}{2\pi (PV^3-nbPV^2+an^2V-abn^3)}\right)^{\frac{3}{2}} v^2 e^{\frac{-mv^2}{2(PV^3-nbPV^2+an^2V-abn^3)}}[/itex]. As ridiculous as it looks, it probably isn't ridiculous enough. Would this work for modeling a non-ideal gas?
The maxwell boltzmann distribution is actually very general. It applies to any classical system that is at thermal equilibrium. It even applies to solid and liquid phases.
Maxwell Speed Distribution. Not the Maxwell-Boltzmann distribution. Apparently, they are two different distributions. The Maxwell-Boltzmann is given by [itex]\sqrt{\frac{2}{\pi}} \frac{x^2 e^{-x^2/(2a^2)}}{a^3}[/itex], where a is a scale parameter. Even then, Maxwell-Boltzmann applies to gases with free-moving particles that do not interact and experience completely elastic collisions (id est, ideal gases). It doesn't apply to solids and liquids.
I'm fairly certain as long as the canonical position and momenta are uncoupled, it applies to solids and liquids. Do you know any statistical mechanics?
A bit. I would not, however, consider myself an expert at it. Going back to its application to solids and liquids, a quick Google search yields multiple sources that are saying that it only applies to ideal gases. Then again, a couple of these sources are saying that the Maxwell Speed Distribution and the Maxwell-Boltzmann Distribution are the same, which may indicate that if you multiply the Maxwell-Boltzmann by the total amount of substance, there is some value of a such that they are equal.
From statistical mechanics one has, P(p,q) = exp(-BH)/Q. If momenta and position are uncoupled (ie H = g(p) + w(q) )then Q = QtransQconfigurational. The marginal probability distribution for momenta is P(p,q) where you integrate out position. p(p) = exp(-Bg(p))/Qtrans. So the maxwell boltzmann distribution applies regardless of phase - as long as the system can be treated classically and obeys a boltzmann distribution ie thermally equilibrated.