Who Decides Temp

This is $ln(n)$  vs $T_{boom}$ for Methane,

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.