|You didn't say that the gas was in thermal equilibrium but I'll assume
|that's what you had in mind.
Well, yes, I considered saying that all good and usual things are to be
assumed, Please do.
|On the other hand, since you are doing a quantum-mechanical calculation, I
|will *not* assume that the gas is classical.
This is an _ideal_ gas
|>Now Vp = (4/3)pi x p(max)^3 & p(max) = sqrt(2mE) (eq 3)
|
|This introduces another approximation. A thermal distribution doesn't have
|a "max" p. Perhaps "typical" p would be a better concept.
John, if I know the E of the system, how can the p of any particle be >
sqrt (2mE)? If it were, then E would be = p^2/2m ie > E. What has my
senile brain forgotten?