Hi all, This question is number 6.40 in the text book. Details in the question Y

Question

Hi all, This question is number 6.40 in the text book. Details in the question You might wonder why all the molecules in a gas in a thermal equilibrium don't have exactly the same speed. After all, when two molecules collide, doesn't the faster one always lose energy and the slower one gain energy? And if so, wouldn't repeated collisions eventually bring all the molecules to some common speed? Describe an example of a billiard-ball collision in which this is not the case: the faster ball gains energy and the slower ball loses energy. Include numbers, and be sure that your collision conserves both energy and momentum. Could you help me to find the example? Thanks

Explanation / Answer

I haven't completed this problem, and apparently no one else haseither. Here's my thoughts so far:    The idealized pool table example, without anydissipation (i.e. friction), is a deterministic model. Athermodynamic model is not deterministic. There is a randomelement in the form of heat, which spreads the equilibriumdistribution of speeds.    Imagine that you started out with a distribution of (unequal) speeds, and let the pool table equilibrate at the statewith all molecules having = speeds. Thermodynamically, thisrepresents a loss of entropy since you would have more informationabout the microscopic state of the system at equilibrium, where allspeeds are equal, than when the system had a distribution spread ofspeeds. So, in a thermodynamic system, equalization of speedswould never happen spontaneously. (Information theoreticinterpretation of entropy.) This assumes that thethermodynamic pool table is an isolated system, of course.    Ignoring rotational degrees of freedom, and othertypes of energy distributions internal to the particles, thermalenergy of the thermodynamic system is given by the equipartitionprinciple. = (d/2)*N*k*T in which                       is the average kinetic energy of the system,                      d is the number of degrees of freedom of the particles in thesystem,                      N = number of particles in the system,                      k = Boltzmann's constant,                      T= temperature in deg. K. At room temperature, each pool ball (particle) exchanges onlytranslational energy with the others. So d = the number oftranslational degrees of freedom. The particles are confinedto a 2-dimensional surface, so d=2 and = N*k*T. Also, we have = 1/2 * m *2, for = average particle speed, and m = particle mass. Thus 2 = 2*N*k*T/m. So the mean speed ofthe particles, = (2NkT/m)1/2 .    That is as far as I could get. I don't know howyou could create collisions in which the slower particle losesenergy to the faster one. Perhaps if you started my mixing 2populations, in which some particles were much heavier than therest, the heavier particles would have slower speeds than thelighter ones, but = mean kinetic energies. I can imagine thatthe larger, slower particles would lose throughcollisions with the lighter ones especially if the latter faroutnumbered the former. I can imagine the large particlessort of swimming through a viscous environment consisting of thelighter particles. But I haven't worked this outmathematically. If anyone has figured this part of theproblem out, please let me know.

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