# Will a gas sample have a root-mean-square speed of v_(RMS) = sqrt((3RT)/m)?

Nov 11, 2017

Yes, as long as the gas is reasonably isotropic (meaning it travels in any direction with no preference over another; this is provided the gas sample is homogeneous), and the temperature is reasonably high such that the gases are classical particles.

Those are the fundamental assumptions for a derivation of the Maxwell-Boltzmann distribution. That can be seen here:
https://socratic.org/questions/56cc484211ef6b489e5131f4?source=search

Note that your equation is incorrect. $R$ is paired with the MOLAR mass, $M$, in $\text{kg/mol}$, and not the mass in $\text{kg}$. ${k}_{B}$ is paired with the mass $m$.

Also, you cannot then invoke the ideal gas law to rewrite ${\upsilon}_{R M S}$ if it is a real gas. So this form as it is, works just fine.

Any homogeneous gas sample is reasonably isotropic, and so it would follow Maxwell-Boltzmann statistics as long as the temperature isn't too low.

When that is the case, the speed distribution is given by the Maxwell-Boltzmann distribution:

$f \left(\upsilon\right) \mathrm{du} \psi l o n = 4 \pi {\left(\frac{m}{2 \pi {k}_{B} T}\right)}^{3 / 2} {\upsilon}^{2} {e}^{- m {v}^{2} / 2 {k}_{B} T} \mathrm{du} \psi l o n$

To determine the RMS speed, which is what you have, one would evaluate the integral for the average squared speed and then take its square root:

$\textcolor{g r e e n}{{\upsilon}_{R M S}} = \sqrt{\left\langle{\upsilon}^{2}\right\rangle} = {\left({\int}_{0}^{\infty} {\upsilon}^{2} f \left(\upsilon\right) \mathrm{du} \psi l o n\right)}^{1 / 2}$

$= \textcolor{g r e e n}{\sqrt{\frac{3 {k}_{B} T}{m}} = \sqrt{\frac{3 R T}{M}}}$

That was done here:
https://socratic.org/questions/570f1ce97c014948061c530c?source=search

And so, you should see that ${\upsilon}_{R M S}$ comes from this distribution and must have obeyed the assumptions made in order to be valid.