$T_{boom}$ To The Power Of $T_{boom}$

When we fit $n$ to a logarithmic function, because $n$ vs $Temp$ looks exponential, and we first convert to Kelvin,

Temp K	ln(n)
373.15	1.561
353.15	1.479
333.15	1.397
313.15	1.317
303.15	1.277
298.15	1.258
295.15	1.246
293.15	1.239
288.15	1.220
283.15	1.201
277.15	1.179
273.15	1.165
263.15	1.131
253.15	1.101
243.15	1.081

Mean x (x̄): 296.21666666667
Mean y (ȳ): 1.2568
Intercept (a): 0.1351233474904
Slope (b): 0.0037866763714947
Regression line equation: y=0.1351233474904+0.0037866763714947x

The regression function is,

$n=e^{0.1351}.e^{0.003787T}=1.1447e^{0.003787T}$

Which would then give,

$v^2_{rms}=v^2_{boom}=\cfrac{3RT_{boom}}{nM_m}$  --- (*)

where $n$ here is due to clustering, not the number of moles.

$v^2_{boom}=\cfrac{3R.T_{boom}}{1.1447e^{0.003787T_{boom}}M_m}$

$v^2_{boom}=\cfrac{2.2897R.T_{boom}e^{-0.003787T_{boom}}}{M_m}$

Furthermore,

$\cfrac{\cfrac{1}{2}{M_m}v^2_{boom}}{\cfrac{1}{2}*2.2897R.T_{boom}}={e^{-0.003787T_{boom}}}$

Let $M_m=m_a.N$,  $N$ is the number in one mole and $m_a$ the particle mass,

$\cfrac{\cfrac{1}{2}{m_a.N}v^2_{boom}}{\cfrac{1}{2}*3R.T_{boom}}={\cfrac{1}{1.1447}e^{-0.003787T_{boom}}}=\cfrac{1}{n}$

which is the same when we manipulate (*).  Let

$E_a=\cfrac{1}{2}{m_a}v^2_{boom}$

$\cfrac{E_a}{\cfrac{3}{2}\cfrac{R}{N}.T_{boom}}=\cfrac{1}{n}$

$\cfrac{E_a}{\cfrac{3}{2}k.T_{boom}}=\cfrac{1}{n}$

where $k=\cfrac{R}{N}$

This expression follows directly from (*) and has nothing to do with the regression line.

$\cfrac{3}{2}k.T_{boom}$ is thermal energy per particle.  $E_a$ is kinetic energy per particle.

$n$ is then the ratio of ratio of thermal energy over kinetic energy per atom.  Given a distribution, $n$ is the average number of times thermal energy is above kinetic energy.  Similarly, $\cfrac{1}{n}$ is the average fraction of times kinetic energy is below thermal energy.  When we use a poisson distribution, $\cfrac{1}{n}$ is the rate parameter, and the probability of a fraction $\cfrac{1}{m}$ is given by,

$P(\cfrac{1}{m})=e^{-\cfrac{1}{n}}.\cfrac{\cfrac{1}{n^m}}{m!}$ --- (**)

where $\lambda=\cfrac{1}{n}$, that is $n$ in $E_a$ and so, $m$ in $E_a$ is $\cfrac{1}{m}$.  Where the general interval of a Poisson distribution is replaced by $E_a$ and the average number of events in the internal is replaced by the average number of $n$ in $E_a$ and so the probability of $m$ events in the interval is the probability of $m$ number in $E_a$.  The probability mass function being,

$P(m\,in\,interval)=e^{-\lambda}.\cfrac{\lambda^m}{m!}$

From (**),

$\cfrac{E_a}{\cfrac{3}{2}k.T_{boom}}=\cfrac{1}{m}$

denotes a particular state of the particle where its thermal energy is $m$ times its kinetic energy,  (**) becomes,

$P(energy\,state\,m)=e^{-\cfrac{E_a}{k.\cfrac{3}{2}T_{boom}}}.\cfrac{\cfrac{1}{n^m}}{m!}$

$n$ here is the average number of times thermal kinetic energy of a particle is above its kinetic energy.  $P(energy\,state\,m)$ is the probability mass function of $m$ given $n$; $m$ is an integer.

Obviously,

$P(energy\,state\,m)\propto e^{-\cfrac{E_a}{k.\cfrac{3}{2}T_{boom}}}$

which would be the Boltzmann Distribution where

$F(state)\propto e^{-\cfrac{E}{kT}}$

if not for the factor $\cfrac{3}{2}$ on $T_{boom}$

Clustering does not propagate back to the equation,

$v_{boom}=3.4354*\cfrac{density}{Z}$

because this expression requires the molecule to behave like a quasi-nucleus, with a concentration $\psi$, subjected to collision.