\(\cfrac{pV}{T_n}= -\cfrac{pV}{T}=ln(T^{ -A }_n)+C \)
and
\( \cfrac{pV}{T_n}= -\cfrac{pV}{T}=-\cfrac{T^{+}}{\tau_o}.\cfrac{V}{A_o}\)
where the sign of \(T^{-}\) is accounted for.
\(\cfrac { pV }{ T } =\cfrac { 1 }{ 2 } \cfrac { T^{ + } }{ \tau _{ o } } .\cfrac { rh }{ r+h } \)
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.
