Pressure, Energy Per Unit Vol

From ideal gas,  This should not be ideal gas please refer to "Conceptual Mishap Clustering" dated 13 Dec 2017,

$T=\cfrac{1}{n}.\cfrac{PV}{R}$

$T=\cfrac{V}{n}.\cfrac{P}{R}=\cfrac{1}{\rho_n}.\cfrac{P}{R}$

so,

$P=\rho_nRT$

From the post "Going Back To Durian Constantly" dated 09 Nov 2017,

$T=\cfrac{(2c)^3}{3}*\cfrac{1}{2}m_av^2_{rms}$

so,

$P=\rho_nR.\cfrac{(2c)^3}{3}*\cfrac{1}{2}m_av^2_{rms}$

$P=\cfrac{1}{3}\rho_nR.{(2c)^3}*\cfrac{1}{2}m_av^2_{rms}$

If $R=2*\cfrac{4}{3}\pi$

$P=\cfrac{2}{3}\rho_n*\cfrac{4}{3}\pi{(2c)^3}*\cfrac{1}{2}m_av^2_{rms}$ --- (*)

where $\cfrac{1}{3}$ accounts for one of the three space dimensions.  The term,

$\rho_n*\cfrac{4}{3}\pi{(2c)^3}*\cfrac{1}{2}m_av^2_{rms}$

the total kinetic energy in a volume of radius $2c$.  It is likely that the factor $2$ accounts for both positive and negative directions.  The expression however is not dimensional-ly consistent.

Consider, $V=\cfrac{4}{3}\pi x^3$,

$\cfrac{dV}{dx}=3*\cfrac{4}{3}\pi x^2$

$\cfrac{d^2V}{dx^2}=2*3*\cfrac{4}{3}\pi x$

$\cfrac{d^3V}{dx^3}=1*2*3*\cfrac{4}{3}\pi=3R$

in which case $R$ is the change of a volume in all 3 space dimensions and has unit "$vol\,m^{-3}$" scaled by $\cfrac{1}{3}$ to consider only one direction of $x$.

This might account for the unit of the expression (*).  Then,

$P=\cfrac{2}{3}\rho_n*\cfrac{4}{3}\pi{(2c)^3}*\cfrac{1}{2}m_av^2_{rms}$

is energy per unit volume $r=2c$, or force per unit area as energy per meter is force.

$\cfrac{E}{m^3}=\cfrac{F}{m^2}$

$\because$

$\cfrac{E}{x}=F$

What is the factor $\cfrac{2}{3}$ for?