Match Making Kinetic Theory And Temperature Particles

How do Kinetic Theory of gasses and temperature particles coexist?

Consider a confinement of particles,

Given the surface density of temperature particle on the inner surface of the containment, collisions of the gas on the containment inner wall result in an exchange of temperature particles.  The gas carries a certain amount of temperature particles and on the wall there is a certain amount of temperature particle.

The pressure of the gas dictates the efficiency of this temperature particle exchange.

There are more temperature particles with the gas than on the wall of the containment.  Posted previously, temperature particles are caught in orbit around electron orbits (post "Not To Be taken Too Seriously, Please" dated 15 May 2016).

If the gas was a solid in thermal contact with the containment,

$\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_g}{\varepsilon_g}$ --- (**)

where $\rho$'s are the respective temperature charge density, and $\varepsilon$'s are the equivalents of dielectric constants.

But the gas is not in full contact with the containment wall.  The gas is in contact with the inner wall due to its pressure.  Given infinite pressure, the frequency of collision is so great that it is as if all the gas particles are in contact with the inner wall.

$\cfrac{nN_a}{A}*q_g=\rho_{\small{g,\,P\rightarrow \infty}}$

$n$ the number of moles of gas, $N_A$ Avogadro's number, $A$ area of the inner wall, $q_g$ the number of temperature particles caught in electron orbits per particle of gas.  And $\rho_{\small{g\,P\rightarrow \infty}}$, the surface charge density of the gas when its pressure is infinite, hypothetically.

$Q=nN_a*q_g$

where $Q$ is the total amount of temperature particles in the gas.  $q_{g}$ is the temperature of the gas particle.  $nN_A$ is the total number of gas particles in the containment.

When pressure is zero, hypothetically,

$\rho_{\small{g,\,P=0}}=0$

the charge density presented by the gas is also zero.  It is as if the gas is not in contact with the inner wall.

What charge density is between zero pressure and infinite pressure?

We shall try,

$\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\cfrac{P}{1+P}$ --- (+)

$\cfrac{\rho_c}{\varepsilon_c}({1+P})=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}{P}$

$\cfrac{\rho_c}{\varepsilon_c}+\cfrac{\rho_c}{\varepsilon_c}P=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}{P}$

we have,

$T=\Delta T*P$ --- (*)

where $\Delta T=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}-\cfrac{\rho_c}{\varepsilon_c}$  and

$T=\cfrac{\rho_c}{\varepsilon_c}$

this is temperature measured.

But $P=\cfrac{nRT_k}{V}=\rho_nkT_k$

where $\rho_n=\cfrac{nN_A}{V}$ is the number density of the gas in the volume $V$.

$T=\Delta T*\rho_n*kT_k$

where $T$ is the temperature of the gas as measured, and $T_k$ is the temperature of the gas calculated from Kinetic Theory of gases.

There is of course no reason for the expression 

$X=\cfrac{P}{1+P}$

except that,

$\left.X\right|_{P\rightarrow \infty}=1$  

$\left.X\right|_{P=0}=0$

and $P$ is linear after expansion, in the expression (*).

Does $T=\Delta T*\rho_n*kT_k$ make sense?

Let $T_g=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}$

with

$T=\cfrac{\rho_c}{\varepsilon_c}$

$T(1+\rho_n*kT_k)=T_g*\rho_n*kT_k$ 

$T=T_g*\cfrac{\rho_n*kT_k}{1+\rho_n*kT_k}$ 

so, $T_k\rightarrow \infty$

$T=T_g$

which is the same as (**), as if the gas is a solid at full thermal contact with the containment inner wall.

What is $T_g=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}$?

Consider,

$\cfrac{n}{\cfrac{4}{3}\pi(1)^3}$

and

$\cfrac{n}{4\pi(1)^2}$

for $\cfrac{n}{\cfrac{4}{3}\pi(1)^3}\rightarrow\cfrac{n}{4\pi(1)^2}$

we have to introduce a factor $\cfrac{1}{3}$.  ie,

$3\rho_A=\rho_n$

as $P\rightarrow \infty$

So,

$T=\Delta T*\rho_n*kT_k=\Delta T*3\rho_A*kT_k$

Since,

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

$T=3\Delta T*T_g*\cfrac{\varepsilon_gkT_k}{q_g}$

$\cfrac{T}{T_g}=\Delta T*3\varepsilon_gk*\cfrac{T_k}{q_g}$

where $\Delta T=T_g-T$ is the result of assuming,

$\cfrac{\rho_c}{\varepsilon_c}=\cfrac{\rho_{\small{g,\,P\rightarrow \infty}}}{\varepsilon_g}\cfrac{P}{1+P}$ 

It is the result of how temperature is measured.

$T_g$ is hypothetical but a constant given a volume of gas with a given amount of temperature particles.

$3\varepsilon_gk$ is a constant, if $\varepsilon_g$ exist.

If the gas is a solid, $T_g$ is just the particle temperature of the gas given $q_g$.  $q_g$ like electric charges on a conductor spread over the surface of the conductor.  But $T_k$ will not make sense, unless the temperature particles behave like a gas in the solid.

$T_k = \cfrac{T}{T_g}*\cfrac{q_g}{3\varepsilon_gk\Delta T}$

where $q_g$ exert a field in space $r$ and $\Delta T$ is the temperature potential across the gas-containment boundary.  Let,

$\Delta T=\cfrac{q_g}{3\varepsilon_gk*r}$

$r$ is the distance moved to provide for a potential change of $\Delta T$ in the presence of $q_g$.

${T_g}$

is temperature, if the gas is a solid imbue with $q_g$ amount of temperature particles per gas particle; a total amount of $Q$ temperature charges.

So,

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

Kinetic temperature is the moment of the measured temperature per temperature as a solid.  

What is $3\varepsilon_gk$, where $3$ has a unit of per meter?

Have nice day?