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

##### 1 Answer

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.

Also, you cannot then invoke the ideal gas law to rewrite

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