The negative gradient suggest \(n=Be^{-AT_{boom}}\) where \(A\) is the gradient of the line above and \(B=ln(y-intercept)\) a constant.
Then,
\(Temp=nT_{boom}=n.\cfrac{ln(\cfrac{B}{n})}{A}\) or \(Temp=-\cfrac{1}{A}n.ln(\cfrac{n}{B})\)
This is the way to relate \(n\) to Temp. But the density of the matter under study cannot be zero. For \(n\) to be valid \(T_{boom}\ne0\). The graph does not fit.
\(n\propto \cfrac{1}{T_{boom}}\)...
Given an average \(n\), \(Temp\propto T_{boom}\), that is why the previous posts on Methane suggests a series of \(ln(ln(ln(...))\).
\(n\) is the independent variable that determines \(Temp\) given \(T_{boom}\), something else set \(n\) and it is not \(Temp\). \(Temp\) is the result of \(n\) given \(T_{boom}\).
At the particle level, temperature jump at discrete steps as,
\(Temp=nT_{boom}\) or
\(\cfrac{E_a}{\cfrac{3}{2}kT_{boom}}=\cfrac{1}{n}\)
and \(n\) is an integer. But \(Temp\) given a volume of particles is a statistical average over all particles.
An we can all shower at the most comfortable temperature. Goodnight.
