You could read up on the gas law, e.g.
here
Your equations feature in there, but one appears molecular to me and the other macroscopic:
##{1\over 2} mv^2 = {3\over 2} kT## average kinetic energy for a molecule.
##P = \rho v_x^2 = {mass\over V} v_x^2 ## may be ok if in the right context.
I can wrangle a bit with formulas: $$ {1\over 2} mv^2 = {3\over 2} kT\ \Rightarrow\ {1\over 2} mv_x^2 = {1\over 2} kT$$ (the energy is equally distributed over the three degrees of freedom),
##\displaystyle \ \Rightarrow\ v_x^2 = {kT\over m} ##
mass = number of molecules * mass of a molecule = number of moles * ##N_A## * mass of a molecule (Avogadro number); write mass = ##n\; N_A\; m##
Leaves
##pV = n\;N_A\;m \ {kT\over m} = n\; N_A \;kT##;
then use
##R_G = N_A \; k## (gas constant) to get the ideal gas law
##pV = nR_GT##.
pressure times volume has the dimension of energy.
It's not the whole story, but quite a big part of it.