Microstates of an ideal gas

• B
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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$$(\Delta p)^2=\left( \frac{1}{N}\sum_{i=1}^{N}p_i^2\right)-\bar{p}^2$$