To examine this further, we look at the gaseous \(T_{n}\) expression for pressure,
\(p_{g-}=\left\{ln(T^{ -A }_n)+C\right\}\cfrac { T_{n} }{ V } =\left\{ln(T^{ -A }_n)+C\right\}\eta _{ T- }\)
and the corresponding expression of \(T^{-}\),
\(p_{g+}=\left\{ln(T^{ +A }_p)+C\right\}\cfrac { T_{p} }{ V } =\left\{ln(T^{ +A }_p)+C\right\}\eta _{ T+ }\)
Both expression applies. But the change in work done is first via the removal of \(T^{-}\) particles.
The change in work done is observed by adjusting pressure or temperature that remove the \(T^{-}\) particles.
The transition from \(p_{g-}\) to \(p_{g+}\) passes through \(T_{n}=1\) as pressure changes. After \(T_{n}=1\), only \(p_{g+}\) applies. At all temperature \(T\), the system starts with a number of \(T^{-}\) particles \(T_{n}\ne0\). This is possible when the experimental setup is first heated over the desired temperature \(T\) and then allowed to cool to \(T\) before the experiment starts. \(T^{-}\) enters into the system as it cools. If this is the case, there will be one common point,
\(p_{g-}=\left\{ln(T^{ -A }_n)+C\right\}\cfrac { T_{n} }{ V } =\cfrac{C}{V}\)
when \(T_{n}=1\).
And the pressure due to the distributed \(T^{-}\) particles on the interior surface of the containment is,
\(p_{d-}=T_{E}=T_n.T_{Ep}=T_{Ep}\)
which is the same across curves of all temperature \(T\). And the pressure \(p\) is,
\(p=p_{g-}+p_{d-}\)
Which is an assumption, because we can have,
\(p=p_{g-}=p_{d-}\)
where the pressure of the gas is the pressure on the containment wall. The pressure of the gas distributes \(T\) particles ON the surface of the containment which exert a force per unit temperature charge (like electron on the surface of a conductor) normal to the surface, outwards from the gas. This induced force acts against the containment (T^{-} particles IN the containment wall) and is equal to the pressure of the gas felt by the containment.
In both cases we have the same point \(p\) on the x-axis.
Then, in general,
\(\cfrac { pV }{ T_{n} } =\cfrac { p }{ \eta _{ T } }\)
where \( \eta _{ T }=\cfrac{T_{n}}{V} \).
\(\cfrac { pV }{ T_{n} }=\cfrac { pV }{ 1 }=C\)
This marks the same point on the y axis.
If \(p=p_{g-}+p_{d-}\) which is consistent with the way the practical plots are obtained theoretically,
\(pV=p_{g-}V+p_{d-}V=C+p_{d-}V\)
All the curves with different \(T\) coincide except for the an offset of \(p_{d-}V\). Since \(p_{d-}\) is due to one particle, the offset is small.
If \(p=p_{g-}=p_{d-}\) then, all the curves with different \(T\) coincide exactly.
This confusion arises because work done in the gas (\(\Delta W\)), and work done on the surface of the gas \(pV\) is assumed to be synonymous.
\(\Delta W=pV\)
whereas \(T^{-}\) removed from inside the gas and those removed from the surface of the gas just before the surface of the containment are two processes that effect \(pV\) through different dependence on \(T_{n}\) and \(T^{-}\) by equations
\(p=T_{E}=T_n.T_{Ep}\)
and
\(p=\left\{ln(T^{ -A }_n)+C\right\}\eta _{ T }\)
from the post "Add By Subtracting..." dated 23 May 2016.
It is likely that \(p=p_{g-}=p_{d-}\), where the pressure of the gas inside and that exerted by the gas on the containment at its surface is the same at an equilibrium, steady state. But the work done (\(pV\)) via the two processes (inside the gas and against the containment wall/movable piston) must be added for a total work done on the system.
And the rumble of a mad man goes on...
