# Maxwell's velocity/speed distribution

Maxwell's speed distribution for gas in thermal equilibrium follows: $$f(v)=\left(\frac{M_g}{2k_bT_g}\right)^{\frac{3}{2}}\exp\left(-\frac{M_gv^2}{2k_bT_g}\right)$$ Three queries:

$1.$Was this derived for gases with or without collision?

$2.$How can we be sure this breaksdown if collisions are considered?

$3.$What does it mean by particles are not interacting? Is collision not an interaction?

The Maxwell-Boltzmann distribution of velocities is correct for an atomic or molecular system with or without interactions, given that it is at equilibrium at temperature $T_g$ (adopting your notation). The only assumption is that classical (rather than quantum) statistical mechanics applies. So, it is true for ideal gases, non-ideal gases, liquids, solids (provided we are in the classical regime). The formula may be derived on the assumption of no collisions, but it may also be derived taking all the interatomic interactions into account.
The reason is that the total energy of a classical system may be written as $E=K+V$ where $K$, the kinetic energy, depends on atomic momenta or velocities, and $V$, the potential energy, depends on atomic positions. This immediately means that the probability distribution for positions and momenta factorizes into two independent terms $$\exp(-E/k_b T_g) = \exp(-K/k_b T_g) \times \exp(-V/k_b T_g)$$ and the first of these factorizes further into a separate contribution for each atom/molecule. This is the Maxwell-Boltzmann distribution. The translational kinetic energy may always be written in the form $\frac{1}{2}M_gv^2$, leading to the equation that you quote for the velocity distribution. For a system with interactions, the term in $\exp(-V/k_b T_g)$ is complicated, but this only affects the probability distribution of atomic coordinates, not their velocities.