Question

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

Answer 1

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 `"kg/mol"`, and not the mass in `"kg"`. `k_B` is paired with the mass `m`.

Also, you cannot then invoke the ideal gas law to rewrite `upsilon_(RMS)` if it is a real gas. So this form as it is, works just fine.

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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(upsilon)dupsilon = 4pi(m/(2pik_BT))^(3//2)upsilon^2e^(-mv^2//2k_BT)dupsilon`

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:

`color(green)(upsilon_(RMS)) = sqrt(<< upsilon^2 >>) = (int_(0)^(oo) upsilon^2f(upsilon)dupsilon)^(1//2)`

`= color(green)(sqrt((3k_BT)/m) = sqrt((3RT)/M))`

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

And so, you should see that `upsilon_(RMS)` comes from this distribution and must have obeyed the assumptions made in order to be valid.