Gas Democracy

A plot of the two the two $\cfrac{pV}{T_n}$ variations, keeping in mind that the increase in $T_n$ reduces temperature, ie

$\cfrac{pV}{T_n}= -\cfrac{pV}{T}=ln(T^{ -A }_n)+C$

and

$\cfrac{pV}{T_n}= -\cfrac{pV}{T}=-\cfrac{T^{+}}{\tau_o}.\cfrac{V}{A_o}$

where the sign of $T^{-}$ is accounted for.

$\cfrac { V }{ A_o } =\cfrac { 1 }{ 2 } \cfrac { rh }{ r+h }$

$\cfrac { pV }{ T } =\cfrac { 1 }{ 2 } \cfrac { T^{ + } }{ \tau _{ o } } .\cfrac { rh }{ r+h }$

More importantly for the x variable axis,

$p=T_{E}=T_{E}=\cfrac{T_{n}.T^{-}}{A_o}\cfrac{1}{\tau_o }=T_n.T_{Ep}$

$p\propto T_{n}$  and $p\propto T^{-}$, $p\propto T^{+}$

We plot, 50*x/(x+50)*1/2 for r=50. h is a variable.  In the same plot graph of -log(x) and 50*x/(x+50)*1/2-log(x).

And we have the practical Gas Law!  What about adding $T^{+}$ particles?

If $T^{+}$ is added to a system, we start with,

$p_{ \small{T} }V_{ \small{T} }=pV+\Delta W$

that results in,

$\cfrac { pV }{ T_{p} }=\cfrac { pV }{ T}=+Aln(T_{n})+C=-ln(T^{ -A }_n)+C$

and

$\cfrac { pV }{ T_{p} }=\cfrac { pV }{ T}=\cfrac{T^{+}}{\tau_o}.\cfrac{V}{A_o}$

in both cases they are the same graphs as when temperature is increased by removing $T^{-}$.  By comparing the sum of the two plots for adding $T^{+}$ and subtracting $T^{-}$ with practical gas law plots

We have a visual match.  This means the two  variations are not equivalent at any level.  The expression,

$p=T_{E}=T_n.T_{Ep}=\left\{ln(T^{ -A }_n)+C\right\}\eta _{ T }$

is WRONG.

Both processes apply at the same time.  How is it possible that $T_{n}$ changes with $T$ held constant?  $T$ is not related to $T_{n}$, as yet.  It could be that, as the volume AND pressure changes the number of $T^{-}$ particles in the system changes to maintain $T$, because,

$\cfrac { pV }{ T_{n} } =\cfrac { p }{ \eta _{ T } }$

where $\eta _{ T }$ is the volume density of $T_{n}$ particles.  Changing volume changes the interior surface area and redistributes $T^{-}$.  As area changes pressure changes.  The total number of $T_n$ particles changes because pressure changes via,

$p=T_{E}=T_n.T_{Ep}$

The setup that maintains the temperature of the system adds or removes temperature particles.

It is possible that temperature $T$ is related to, the pressure component

$p_d=T_{E}=T_n.T_{Ep}$

due to the distribution of $T_n$ on the inner surface of the containment.  And not the free (gaseous) component,

$p_g=\left\{ln(T^{ -A }_n)+C\right\}\cfrac { T_{n} }{ V } =\left\{ln(T^{ -A }_n)+C\right\}\eta _{ T }$

from which negative work was derived.

Have a nice day.