Shifting Right With High Temperature

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.