MAXWELL-BOLTZMANN STATISTICS momentum space

eileen6a

here is the notes:
we have the MB statistics or distribution
ns = Agsexp(-es/kT) (1)
This is only for a set of discrete energy levels es. In this section, we shall see how (1) can be applied to a variety of situations. For instance, how can we use this distribution for an ideal gas which does not have truly discrete energy levels.

PHASE SPACE
Let us now consider an ideal gas which is monatomic and the only energy possessed by each particle is the ke by virtue of its momentum p
e = p2/2m = (px2 + py2 + pz2)/2m (2)
where m is the mass, p is the total momentum magnitude and px etc. are the momentum components in the Cartician frame. In Statistical Mechanics, the space containing statistical states is called the “phase space” J, where
J= dxdydz×dpxdpydpz (3)
Thus, the phase space has six dimensions, three for the real space and three for the momentum space. For an ideal gas of volume V, we know that the states are uniform throughout this volume V and thus we can integrate the first (spatial) part
int(dxdydz) = V (4)
Therefore the phase space element now only contains the momentum variable,
J = Vdpxdpydpz (5)
We now treat the gas as having a pseudo discrete energy levels system with energy and momentum given in (2). In this case, we can replace gs and es in (1) by g and e.
Further, let B be the size of the phase space occupied by one state. The relationship between the weight(degeneracy) g and the phase space element in (5) is
g = J/B = (V/B)dpxdpydpz (6)
The MB distribution of (1) now becomes
dn = ns = A(V/B)dpxdpydpzexp(-e/kT)
i know that ns = Agsexp(-es/kT)

My question is:
1.In momentum space why can we treat the gas as having a pseudo discrete energy levels system with energy and momentum and replace gs and es in by g and e?
2.why dn = ns?

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