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In kinetic molecular theory, the average velocity of gas particle is zero since the molecule move in different directions, and the overall effect is zero. Howeever, you can calculate different speeds.

Specifically, RMS (Root Mean Square) speed is the speed of a particle in a gas with the average kinetic energy. This is $$u_{rms}=\sqrt{\frac{3RT}{M}}$$

Conceptually, if $v_1, v_2,...,v_n$ are the speed of $n$ particles in a gas, the RMS is
$$u_{rms}=\sqrt{\sum\frac{v^2_i}{n}}$$

As the Wikipedia page on RMS points out, RMS is a special case of standard deviation when the mean is zero. Hence:

  1. Is it correct to say that $u_{rms}$ is the standard deviation of the magnitudes of the velocity (not speed).
  2. If it is correct, is there any significance/use of this interpretation?
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  • $\begingroup$ Speed IS the magnitude of velocity. // As there are no negative values, the mean cannot be zero.// The formal similarity with formula of s as sigma estimation with zero mean is coincidental, as any square root of the arithmetic mean of quadrates. $\endgroup$
    – Poutnik
    Jul 1, 2023 at 7:26
  • $\begingroup$ Don't confuse the vector of velocity (which averages to zero) with the magnitude of the velocity (which clearly isn't). And the speed distribution is not gaussian. $\endgroup$
    – matt_black
    Jul 1, 2023 at 9:13

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I'll answer in reverse.

(2) There is so-called Central limit theorem in math, which basically says that under certain precise conditions (which I won't discuss here) the sum of many independent random values behaves according to the Gaussian, or normal, distribution. It is enlightening to know that the gas molecule velocity, being a result of many independent random collisions, behaves somewhat similarly, in that each component ($v_x,\,v_y,\,v_z$) is distributed by Gauss. Of course the average values is 0 (since the whole body of gas is not moving anywhere), hence the RMS is the same as the standard deviation. Actually, the latter would be true for any other kind of distribution, so I don't even know why I brought up all that Gaussian stuff.

(1) One may say that the velocity has a 3D normal distribution with average being the zero vector and RMS being our $u_{rms}$. The magnitude of velocity, AKA speed, is another thing entirely. To begin with, it is always positive (except when 0), hence its average is positive and not 0. In fact, it is distributed by Maxwell-Boltzmann, and its variance (which you may see by following the link) is, naturally, the difference between the mean square speed and the square of the mean speed.

So it goes.

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  • $\begingroup$ Velocity components have the Bolzmann distribution $p(v_x) \propto \exp{(\frac{-mv_x^2}{2kT})}$, which is the Gauss one just in different colour. Integrated over spherical surface, one gets the Maxwell-Boltmann one. $\endgroup$
    – Poutnik
    Jul 1, 2023 at 8:59

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