Given,
\(v^2_{v\,p}=\cfrac{2RT_{boom}}{nM_m}\)
but
\(v^2_{boom}=\cfrac{3RT_{boom}}{nM_m}=v^2_{rms}\)
then in order for \(v_{rms}\rightarrow v_{v\,p}\) temperature has to change by,
\(T_{boom}\rightarrow \cfrac{3}{2}T_{boom}\) such that,
\(v^2_{v\,p}=\cfrac{2R}{nM_m}\cfrac{3}{2}T_{boom}=\cfrac{3RT_{boom}}{nM_m}=v^2_{boom}\)
So,
\(T_{v\,p}=\cfrac{3}{2}T_{boom}\); temperature is increased and the distribution shifted right until \(v_{p}\) is now \(v_{rms}\) previously, which was numerically, \(v_{boom}\).
It is wrong to use,
\(v^2_{boom}=\cfrac{3}{2}v^2_{p}\)
and so,
\(T_{boom}=v^2_{boom}*\cfrac{molar\,mass*10^{-3}}{3*8.3144}\)
\(T_{v\,p}=v^2_{p}*\cfrac{molar\,mass*10^{-3}}{2*8.3144}\)
\(T_{v\,p}=\cfrac{2}{3}v^2_{boom}*\cfrac{molar\,mass*10^{-3}}{2*8.3144}\)
\(T_{v\,p}=T_{boom}\)
In this case,\( v_{p}\) replaces \(v_{rms}\) on the same velocity probability distribution, the temperature of the profile remains the same. \(v_{boom}\) is still at \(v_{rms}\). What is needed is to increase temperature such that \(v_{p}=v_{boom}\), on a new distribution.
Have a nice day.
