| 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 (**),
\(\cfrac{E_a}{\cfrac{3}{2}k.T_{boom}}=\cfrac{1}{m}\)
\(n\) 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.
\(\cfrac{\cfrac{1}{2}{M_m}v^2_{boom}}{\cfrac{1}{2}*2.2897R.T_{boom}}={e^{-0.003787T_{boom}}}\)
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.
