Add By Subtracting...

From the post "Temperature Particle And The Dip" dated 13 May 2016, the work removed from the system due to a drop in the number of negative temperature particles (ie temperature increases) is,

$\Delta W=A\Delta T_{n}$

where $A$ is a constant.

The amount of energy remaining at a final temperature $T$,

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

$\cfrac { p_{ \small{T} }V_{ \small{T} } }{ T_{n} } =\cfrac { pV-A\Delta T_{n} }{ T_{n} } =\cfrac { pV }{ T_{n} } -\cfrac { A\Delta T_{n} }{ T_{n} }$

$\Delta \cfrac { pV }{ T_{n} } =-\cfrac { A\Delta T_{n} }{ T_{n} }$

$\cfrac { pV }{ T_{n} } =-Aln(T_{n})=ln(T^{ -A }_n)+C$ ---(1)

where $C$ is the integration constant.

$p=\left\{ln(T^{ -A }_n)+C\right\}\cfrac { T_{n} }{ V } =\left\{ln(T^{ -A }_n)+C\right\}\eta _{ T }$ --- (2)

where $\eta _{ T }=\cfrac { T_{n} }{ V }$ is the number of negative temperature per unit volume.   Substituting into (1),

$\cfrac { pV }{ T_{n} } =\cfrac { p }{ \eta _{ T } }$ --- (3)

In another view...

We postulate here that pressure is due to the $T$ field of the $T$ particles, behaving like charges on the surface of a conductor,

$E=\cfrac{\rho}{\varepsilon_o}$

in an analogous way,

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

acting normal to the surface of the containment, of area $A_o$, against other $T_{-}$ particles, outwards.  $T_n$ is the total number of negative temperature particles in the volume $V$.  The $T^{-}$ spread themselves over the surface of the containment like charges on the surface of a conductor.  This is the pressure on the containment.

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

If this is true, then pressure depends on the change in surface area of the containment and not on its volume.  Furthermore,

$pV=T_n.T_{Ep}V$

$\cfrac { pV }{ T_{n} } =T_{Ep}V=\cfrac{T^{-}}{\tau_o}.\cfrac{V}{A_o}$

and $\cfrac{V}{A_o}$ gives a linear dimension of the containment.  For example, in the case where a piston move to change the volume of a cylinder, $\cfrac{V}{A_o}$ is

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

More importantly, if these two view are the equivalent at some level,

This is wrong, both views apply and act simultaneously. 

But at an steady state, where the pressure in the gas is the same as the pressure on the containment wall, the rest still applies but for a different reason,

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

$T_{Ep}.V=\left\{ln(T^{ -A }_n)+C\right\}$

$T_{ n }.T_{ Ep }=-Aln(B.T_{ n }).\eta _{ T }=-Aln(B.T_{ n })\cfrac { T_{ n } }{ V }$

where $B=ln(C)$

$T_{ Ep }=-\cfrac { A }{ V } ln(B.T_{ n })$

$-\cfrac { T_{ Ep } }{ A_{ v } } =ln(B.T_{ n })$

where $A_{ v }=\cfrac { A }{ V }$ per unit volume

$T_{ n }=\cfrac { 1 }{ B } e^{ -\frac { T_{ Ep } }{ A_{ v } }  }$

$T_{ n }=Ke^{ -\frac { T_{ Ep } }{ A_{ v } }  }$

where $K=\cfrac{1}{B}$ is in units of $T_n$ and $A_v$ = energy per change in $T_n$ per unit volume.

$K=T_{ n }e^{ \frac { T_{ Ep } }{ A_{ v } }  }$

$K$ represents the effect of $T_n$ number of $T^{-}$ particles in the system.

This discussion is strictly in the presence of $T^{-}$, as such $T_n\gt0$.  When $T_n=1$,
$\require{cancel}$
$p_{n1}=\left\{\cancelto{0}{ln(T^{ -A }_n)}+C\right\}\eta _{ T }$

$p_{n1}=\cfrac{C}{V}$

using $\eta _{ T }=\cfrac{T_n}{V}=\cfrac{1}{V}$

$C$ is then the pressure due to one $T^{-}$ particle per given volume.  And

$p_{n1}V=C$

is a constant.  Energy capacity per unit volume, $A_v$

$A_v=-\cfrac{T_{ Ep }}{ln(B.T_{ n })}=\cfrac{T_{ Ep }}{ln(K.T^{-1}_{ n })}$

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

$p=T_{Ep}$,   for $T_n=1$

$A_v$ is the change in pressure per $ln(K.T^{-1}_{ n })$ per unit volume.

If there is no $T^{-}$ particle on the containment, there will be no positive pressure outward from the $T^{-}$ particle being contained.  If $T^{-}$ particles on the containment are driven away, negative pressure develops on the containment as the $T^{-}$ particles pulls the remaining $T^{+}$ particles on the containment in.

The big assumption $\Delta W=A\Delta T_{n}$ need to be examined closely.

And this is removing $T^{-}$ to raise the temperature in a volume of particles.