Otherwise An Idiot...

From the previous post "Match Making Kinetic Theory And Temperature Particles" dated 28 Dec 2017, volume $V$ is not explicit in the final expression for $T_k$,

$T_k=r*\cfrac{T}{T_g}$

where $r=\cfrac{q_g}{3\varepsilon_g k \Delta T}$

It went into hiding underneath "density",

$\rho_n=\cfrac{nN_A}{V}$

as per unit volume, where its surface area, $4\pi*1^2$ is three times its volume $\small{\cfrac{4}{3}\pi* 1^3}$ so much so that the surface charge density when all the temperature charge is pushed by an infinite pressure to the surface is three time smaller than the volume charge density when all the temperature charge is evenly distributed throughout its volume.

However, if we consider a volume of radius $r=3$ then,

$V=\cfrac{4}{3}*\pi 3^3=4\pi*3^2=surface\,area$

then both surface charge density and volume charge density will be the same.  There is then no need for infinite pressure, nor a very high pressure approximation.

This $V$olume with $r=3$ shall be called the Measurement Volume, $V_m$

Using this volume $V_m$,

$T_g=\cfrac{\rho_{\small{g\,P\rightarrow \infty}}}{\varepsilon_g}=\cfrac{\rho_n*q_g}{\varepsilon_g}$

where,

$\rho_n=\cfrac{nN_a}{V_m}=\cfrac{Q}{V_m}$

is the number of temperature charges per volume evenly distributed, and $q_g$ is the charge per gas particle, and $Q$ is the total number of temperature charge in the gas of volume $V_m$.

$T_k=r*\cfrac{T}{T_g}$

where $r=\cfrac{q_g}{\varepsilon_g k \Delta T}$

exactly,  irrespective of the pressure of the gas, and its approximation to infinite value.

Or is it...?