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MSD for nonideal gas?by Mandelbroth
Tags: nonideal 
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#1
Dec2412, 05:01 PM

P: 615

I've been thinking again. The formula for the Maxwell Speed Distribution for a nonideal 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})(Vnb)}\right)^{\frac{3}{2}} v^2 e^{\frac{mv^2}{2(P + \frac{an^2}{V^2})(Vnb)}}[/itex], through the Van der Waals equation. Factoring, we get [itex]\displaystyle f(v) = 4\pi \left(\frac{mV^2}{2\pi (PV^3nbPV^2+an^2Vabn^3)}\right)^{\frac{3}{2}} v^2 e^{\frac{mv^2}{2(PV^3nbPV^2+an^2Vabn^3)}}[/itex]. As ridiculous as it looks, it probably isn't ridiculous enough. Would this work for modeling a nonideal gas? 


#2
Dec2412, 06:31 PM

P: 1,042

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.



#3
Dec2412, 07:09 PM

P: 615

Even then, MaxwellBoltzmann applies to gases with freemoving particles that do not interact and experience completely elastic collisions (id est, ideal gases). It doesn't apply to solids and liquids. 


#4
Dec2412, 07:10 PM

P: 1,042

MSD for nonideal gas?



#5
Dec2412, 07:18 PM

P: 615

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 MaxwellBoltzmann Distribution are the same, which may indicate that if you multiply the MaxwellBoltzmann by the total amount of substance, there is some value of a such that they are equal. 


#6
Dec2412, 07:24 PM

P: 1,042

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. 


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