Microstates of an ideal gas

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etotheipi
Up to an undetermined constant ##a## the entropy of an ideal gas goes like$$S = k_B N\ln \left[ a^\frac{3}{2} \left(\frac{V}{N}\right) \left(\frac{E}{N}\right)^{\frac{3}{2}} \right]$$In some notes is written:
We can consider that E/N is the average energy of each molecule and V/N is the average volume occupied by each molecule (both are therefore intensive while S is extensive). How do we relate this equation to the Boltzmann’s law of Eq. (1) and how do we determine the constant a? Clearly, if we let$$\frac{(\Delta p)^2}{2m} = \frac{E}{N},\quad (\Delta x)^3 = \frac{V}{N}$$then, roughly speaking, ##\Delta p## is the magnitude of the change of momentum of the molecules and ##\Delta x## is that of the change of position

And then they identify ##\Omega = \left(\frac{\Delta x \Delta p}{w}\right)^{3N}## from Boltzmann's formula, with ##w=\sqrt{2m/a}##.

I am confused about what ##\Delta x## and ##\Delta p## represent here. At first glance it looks like the product ##\Delta x \Delta p## represents some area element in the ##x##/##p_x## phase space, but if that is true, I don't know how we obtain the relation ##\frac{(\Delta p)^2}{2m} = \frac{E}{N}##... if anything I would have thought it be something like$$\frac{1}{N} \sum_i^N \frac{p_i^2}{2m} = \frac{E}{N}$$I hoped someone could clarify. Thank you!
 
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  • #2
vanhees71
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##\Omega## is the phase-space measure. The problem in classical statistics is the lack of a physical determination of the phase-space measure. This is of course immediately cured by quantum theory. As to be expected it turns out that ##w=h=2 \pi \hbar##, counting the number of free-particle states per phase-space-volume element. Just put free particles in a finite-volume box like a cube and assume periodic boundary conditions (it should be periodic rather than rigid boundary conditions, because otherwise you don't have a cleanly defined self-adjoint momentum operator). Then you can calculate everything and finally take the "thermodynamic limit" ##V \rightarrow \infty## for intensive quantities like particle-number density, energy density etc. The limit is not always trivial, as already the example of Bose-Einstein condensation of a free gas shows.
 
  • #3
I think ##\Delta p## is considered as the variance of the momentum of the particle population. It should then be calculated as
$$(\Delta p)^2=\left( \frac{1}{N}\sum_{i=1}^{N}p_i^2\right)-\bar{p}^2$$
but ##\bar{p}=0## since it's an ideal gas.
 

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