It should be,
\(kT_g=\cfrac{q_g}{\varepsilon_g *r_g}\)
where \(kT_g\) is in Joules per particle, and \(\cfrac{q_g}{\varepsilon_g *r_g}\) is potential (\(J\)) due to temperature charges on one particle.
Only if one started somewhere somehow, anywhere anyhow...
\(T_k=r*\cfrac{T}{T_g}\)
where \(r=\cfrac{q_g}{\varepsilon_g k\Delta T}\) or
\(T_k=r_g*\cfrac{T}{\Delta T}\)
where \(r_g=\cfrac{q_g}{\varepsilon_g kT_g}\)
(T_k\) is made up of \(\cfrac{T}{\Delta T}\) up to the length \(r_g\) of a field projected by the total temperature charge per particle, up to a potential of \(V=kT_g\). Where \(T_g\) is the temperature of the gas when the temperature particles it contains, fill the gas evenly at a volume of \(V_m\), radius \(r=3\). \(T\) is the measured temperature at the surface, and \(\Delta T\) is the difference between the measured temperature \(T\) and \(T_g\).
If the temperature charge particles are free from the gas particles, \(T=T_g\) where the measured temperature is the same as the temperature of the gas just below the inner surface of the containment. \(T\) is independent of \(T_k\).
If the temperature charge particles are bounded to the gas particles and moves to the interior of the gas as the gas particles moves, then \(T\ne T_g\) and \(T_k\) is related as above.
At high temperature and high pressure it is likely that \(T\) is independent of \(T_k\) as more temperature particles are free from the gas particle. \(T_g\) then measures the temperature charge content of the gas, as in the case of a conductor imbued with electric charges.
\(T_g\equiv\cfrac{\rho_{e\,area}}{\varepsilon}=D\)
Have a nice day.
