\(\cfrac{pV}{T_n}= -\cfrac{pV}{T}=ln(T^{ -A }_n)+C\)
\(C=(pV)_{\small{n=1}}\)
with \(T^{+}\),
\(\cfrac{pV}{T_p}= \cfrac{pV}{T}=ln(T^{ +A }_n)+C=-ln(T^{ -A }_n)+C\) --- (*)
\(\cfrac{pV}{T_n}= -\cfrac{pV}{T}=-\cfrac{T^{+}}{\tau_o}.\cfrac{V}{A_o}\)
which is a constant.
with \(T^{+}\),
\(\cfrac{pV}{T_p}= \cfrac{pV}{T}=\cfrac{T^{+}}{\tau_o}.\cfrac{V}{A_o}\)
Both positive and negative temperature have the same equations, but does \(T^{+}\) does negative work? No, a spinning \(T^{+}\) particles does not produce a \(g\) field that act against gravity, it does produce an \(E\) field. Unless in the presence of an existing \(E\) field, \(T^{+}\) does not perform positive work. \(T^{-}\) was removed from the system to decrease work done. Adding \(T^{-}\) to the system will increase work done, and reduces temperature. \(T^{-}\) itself does positive work against gravity. So, equation
\(\Delta pV=-A\Delta T_{p}\)
is not valid! And so equation (*) for \(T^{+}\) is not valid.
We plot \(\cfrac{pV}{T}\) vs \(T\), where \(T=T_{n}+T_{p}\),
Work done per temperature is a constant over temperature. The increase in \(\cfrac{pV}{T}\) after \(T=1+T_{p}\) is due to the removal of \(\cfrac{T^{-}}{\tau_o}.\cfrac{V}{A_o}\) which is expected to act against \(\cfrac{T^{+}}{\tau_o}.\cfrac{V}{A_o}\).
This graph suggests that on cooling a hot gas by adding \(T^{-}\) particles, \(\cfrac{pV}{T}\) increases, but cooling by removing \(T^{+}\) particles, \(\cfrac{pV}{T}\) remains constant.
There is no free gain. And removing negative temperature particles requires work.
