This is a question at one level beyond the standard introductions to kinetic theory. I want to know which average we are taking when we talk about a mean free path in a gas. I recently read
Steve T. Paik, Is the mean free path the mean of a distribution?, Am. J. Phys. 82, 602 (2014); https://doi.org/10.1119/1.4869185 also arXiv:1309.1197v2.
To answer my question you don't need the whole paper; the essential point is given (I think by the same author) at this answer: RMS Free Path vs Mean Free Path answer by user Steve.
The mean free path for molecules in a gas is defined by an average over two quantities: free path lengths and speeds. One might say it is a "mean mean free path". Introductory treatments commonly gloss over this fact.
First let's write down something concrete in order to get some sense of where we are. I claim that the collision rate for a molecule at speed $v_1$ is $$ \Gamma(v_1) = n \sigma \int d^3{\bf v} | {\bf v} - {\bf v}_1 | f_3({\bf v}). $$ where $\sigma$ is the collision cross section, $n$ the number density and $f_3({\bf v})$ the velocity distribution (Maxwell-Boltzmann). The argument is that if one adopts a frame where the target molecule is not moving, then the other molecules have velocities ${\bf v} - {\bf v}_1$ and this expression gives the probability that the target will be found in one of the collision cylinders swept out by the other molecules. Let's define what rate I think this is. It is a rate such that if one starts a clock at any arbitrarily chosen time zero, and then waits a short time $\delta t$, then the number of molecules of speed $v_1$ which undergo at least one collision during that time interval is $N \Gamma(v_1) \delta t$ where $N$ is the total number of molecules. When we say "of speed $v_1$" here we are referring to the speed before the collision.
Notice that in the above result there is no need to start the clock at any particular time (such as after some tagged molecule just finished a collision for example), and the clock is started at the same time for all the molecules. The fact that one can consider a constant collision rate is owing in part to the assumed uniformity of the gas and it will only be correct at low density where one can neglect the tendency of nearby molecules to shield each other from collisions with other molecules.
There are two ways to extract a mean collision rate from $\Gamma(v)$, and this is what my question is about. One has \begin{eqnarray} \langle \Gamma(v) \rangle_v &\equiv& \int \Gamma(v) f(v) dv, \\ \langle \Gamma(v) \rangle_{\rm coll} &\equiv& \int \Gamma(v) f_{\rm coll}(v) dv, \end{eqnarray} where $f(v)$ is the Maxwell-Boltzmann speed distribution and $f_{\rm coll}$ is another distribution described by Paik in the above paper and answer: $$ f_{\rm coll} = A \Gamma(v) f(v) $$ where the normalization constant $A$ is easily shown to be $A = 1 / \langle{\Gamma(v)\rangle}_v = (\sqrt{2} n \sigma \bar{v})^{-1}$, with $\bar{v} = \int v f(v) dv$.
The idea behind $f_{\rm coll}$ is that one might pick a short time interval $\delta t_1$ and take an interest in those molecules (of given speed) which have just experienced a collision during $\delta t_1$. The number of such molecules will be weighted in proportion to their collision rate, hence the factor of $\Gamma(v)$ in the formula for $f_{\rm coll}$. Another way of saying it is that if you pick a collision at random from among those happening during $\delta t_1$, then you are more likely to find you have picked a collision involving molecules with a higher collision rate (and these are the faster molecules by the way).
We can make similar arguments for mean free path. The mean free path $\lambda(v)$ for molecules of speed $v$ is $$ \lambda(v) = \frac{v}{\Gamma(v)} $$ The mean mean free path is one of \begin{eqnarray} \langle \lambda(v) \rangle_v &\equiv& \int \lambda(v) f(v) dv, \\ \langle \lambda(v) \rangle_{\rm coll} &\equiv& \int \lambda(v) f_{\rm coll}(v) dv \end{eqnarray} but which one? That is my question. The values may also be of interest: \begin{eqnarray} \langle \lambda(v) \rangle_v &\simeq& 0.6775 \frac{1}{n \sigma}. \\ \langle \lambda(v) \rangle_{\rm coll} &=& \frac{1}{\sqrt{2} n \sigma} \; \simeq 0.7071 \frac{1}{n \sigma}. \end{eqnarray}
Of course one correct answer is "the mean free path is whichever one is the right one to use in whatever further calculation you propose to do". Another answer could be "well when we do more advanced calculations we don't need to use the concept of mean free path as such, except as an aid to physical intuition, so all we need to know is that it is of the order of $1/n\sigma$."
However the concept of weighting the distribution with a factor of $\Gamma(v)$, yielding $f_{\rm coll}$, has the pleasant effect of making the integrals easy to do (that's why you see an exact $1/\sqrt{2}$ in one of the results above, but just an approximate decimal value for the other). So it would be nice to use $f_{\rm coll}$ if in fact it is justified. But I think that the phrase "mean free path" is commonly understood to mean $\langle \lambda(v) \rangle_v$ not $\langle \lambda(v) \rangle_{\rm coll}$ and the only reason you see the $\sqrt{2}$ being widely used is that people did a sloppy calculation, such as $$ \langle \lambda(v) \rangle = \langle \frac{v}{\Gamma(v)} \rangle \sim \frac{\langle v \rangle}{\langle \Gamma(v) \rangle} \;=\; \frac{ \bar{v} }{\sqrt{2} n \sigma \bar{v}} $$ (or else they knew the number is of this order and just put a $\sqrt{2}$ as a rough figure).
I am hoping that someone out there who does calculations of this type will be able to guide me on whether this $f_{\rm coll}$ idea is a good idea and is widely appreciated and used, or whether it is more a curiosity which lets us know what exactly the formula $1/\sqrt{2} n \sigma$ is an average of, but is not of much use for anything else, and therefore the mean free path which we should teach students is approximately $0.6775 / n \sigma$.