What is dielectric constant and relaxation time?

What is the relaxation time of an ideal gas?

  • Let us have a box separated into two equal parts. In one part, there is an ideal gas at a thermal equilibrium, the other part is empty. Let us remove the separator so quickly that it can be considered instantaneous. The gas expands to fill the whole box. What is the characteristic time after which we can assume that it reached a new equilibrium? I don't know the answer but I think it should be well-known in advanced thermodynamics. Please help if you know.

  • Answer:

    (I'll try to get something more concrete for you) Edit: This one is really tough to even define the problem well enough to have an unambiguous answer. I'm thinking along the lines of a long tube of length 2L, where the ideal gas is first contained in L, and then is allowed to expand into the rest of it. A pressure sensor would monitor the drop in pressure at the end of the tube. If there's indeed a nice "decay curve" to half of the pressure, then we can talk about time to reach a certain fraction of the way to final pressure. But I don't think it would be so simple because it probably does critically depend on the slenderness of the tube. In one extreme, the tube could virtually be a capilliary, in which the pressure drop would be quite steady, i.e. linear, and in other extreme, the container is just a short can, in which pressure wouldn't even be meaningful until the "explosion" is over. And then you have everything else inbetween, where Maxwellian distribution of velocities can play a role in how the gas fills the tube over time. Edit 2: I'm going to be out of town for a long weekend, maybe I'll come up with something. I keep thinking of a tie-in with the classic heat diffusion problem. Edit 3: (Pre-trip comment). In the simplest possible 1D heat diffusion case, the general solution is of the form T = ke^(-at), where k and a are constants of the system. Your "relaxation time" would then be the inverse of a. What I fear is that it could be difficult to even define a "relaxation time" for anyything more complicated than this. Let me think on a probabilisitic approach. Edit 4: Oh, a NUMERICAL estimate. Oh, totally different question. I was so bogged down on trying to even define the problem. This is a classic example of what I call, "jumping the manifold", i.e., where supposed well-behaved physical variables literally lose meaning and the problem has to be treated in a totally different way. But never mind that. I'll try to get an numerical approximation for you when I get back in a few days. I'll go by the drop in pressure (as a function of time) on the back wall of the box/tube, as far away as possible from the turbulence.

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Ideal gas is collision-less by definition of ideal gas. Hence you can simply take the Maxwell distribution of speeds and trace all the particles. Naturally the answer will depend on the precise geometry of the container. Maybe there is a funny integrable geometry. ***** No, I have to correct myself after giving it some thought. Molecules of ideal CAN collide, as long collisions are instantaneous point collisions. In this case we have free path λ and dimension of container A and size of the hole B. Then there are several cases. a) Vacuum : λ >> A and B In this case we must trace trajectories. The relaxation both spacial and in the space of momenta will occur simply due to cross-overs of geometric trajectories. b) Aerodynamic : λ << A and B. It is viscous gas, which good temperature T(x,y,z) and pressure P(x,y,z) everywhere, and the variation of T and P are not small. This the hell of full blown Navier-Stokes PDE. c) Mixed: λ ~ A or B. This is even worse that above mentioned hell. This in now not point PDEs anymore, but integral-and-differential equations like Boltzmann relaxation which depends both on time and space.

Alex Smirnoff

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