Why is there a factor of $\pi$ in the average velocity of gas? Consider the most probable and RMS speeds for a kinetic gas:
$$ v_{mps} = \sqrt{\frac{2RT}{M}}$$
$$ v_{rms} = \sqrt{ \frac{3RT}{M}}$$
Now consider the average speed:
$$ v_{avg} = \sqrt{\frac{8RT}{\pi M}}$$
Why does a factor of $\pi$ pop up in the above equation? I know the mathematical reason is due to integrations of gaussian integrals but usually $\pi$ we associate with circles/spheres. So, is there a physical way to give meaning as to why we should expect a $\pi$ in the formula here?
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 A: Based on the $\pi$ term in the given $f_V(v)$, the question is not so much where the $\pi$ comes from in $v_{avg}$, but rather, “Where does the $\pi$ go in the others?”
First, note the $\pi$ term in our velocity pdf:
$$f_p(v)=4 \pi v^2 (\frac{1}{\pi} \frac{m}{2k_bT})^{1.5} \exp(\frac{-mv^2}{2k_bT}) $$ $$ f_p(v)=  j \frac{v^2}{\sqrt{\pi}}  \exp(-kv^2)$$
$$\text{with }~k \equiv \frac{m}{2k_bT} ~,~ j\equiv 4k^{1.5}$$
First, it’s disappearing from $v_{mp}$ is maximization:

1.$v_{mp}$: The most probable speed is determined by setting the derivative of the pdf equal to zero (We are finding the maximum density point):
$$ \frac{\partial f_p}{\partial v} = \frac{j}{\sqrt{\pi}} (2v e^{-kv^2} -2kv^3 e^{-kv^2})=0$$
$$\implies \frac{2 j v e^{-kv^2}}{\sqrt{\pi}} (1-kv^2)=0$$
The resulting $v_{mp}=\operatorname{argmax}[f_p(\cdot)]=\frac{1}{\sqrt{k}}= \sqrt{\frac{2k_bT}{m}}$ does not include a $\pi$.

2.Now for $v_{avg}$, where the $\pi$, perhaps unsurprisingly, ends up as a radical in the denominator, as it is in the distribution. $$v_{avg} = \int vf_p(v)dv$$
With a very minor notation abuse.
$$ v_{avg}= \frac{j}{\sqrt{\pi}}   \int  v^3 \exp(-kv^2)dv$$
Based in this link from Vincent Thacker in the comments https://math.stackexchange.com/q/28558 , we see a $\pi$ will not appear with an exponent of $3$, hence the original $f_V(v)$’s $\frac{1}{\sqrt{\pi}}$ will go through to the $v_{avg}$.
Specifically:
$$ v_{avg}= \frac{j}{\sqrt{\pi}}   \int  v^3  ~ e^{-kv^2}dv$$
$$v_{avg}= \frac{-j}{2k^2\sqrt{\pi}}(kv^2+1) e^{-kv^2}  \Big|_0^{\infty}$$
$$= 0 - \frac{-4k^{1.5}}{2k^2\sqrt{\pi}} ~ \cdot 1=\frac{2}{\sqrt{k \pi}} =\sqrt{\frac{8k_bT}{\pi m}}$$

3.Consider the $f_p(v)$ to $f_p(v^2)$ general transition with $x,y$: Note that $$F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(X\leq\sqrt{y})=F_X(\sqrt{y})=\int_0^\sqrt{y}3t^2\,dt=y^{3/2}$$ hence $$f_Y(y)=\frac{3}{2}\sqrt{y}\implies v^4$$
Based again in this link from Vincent Thacker in the comments https://math.stackexchange.com/q/28558 , we see a $\sqrt{\pi}$ will appear in the numerator with an exponent of $4$, hence the original $f_V(v)$’s $\frac{1}{\sqrt{\pi}}$ will not be in the expressions for $v_{rms}^2$ and then $v_{rms}$.
A: OP is asking for prose description of my other answer saying where the $\pi$ came from. So here that is - the prose part of the other answer expanded, plus a couple summary sentences at the end (jump to that if you wish):

Based on the $\pi$ term in the given $f_p(v)$, the question is not so much where the $\pi$ comes from in $v_{avg}$, but rather, “Where does the $\pi$ go in the others?”
First, note the $\pi$ term in our velocity pdf: $$f_p(v)=4 \pi v^2 (\frac{1}{\pi} \frac{m}{2k_bT})^{1.5} \exp(\frac{-mv^2}{2k_bT}) $$ If we plug in constants for the unchanging stuff and simplify, the $\pi$ becomes a radical in the denominator and the functional form is easier to see: $$ f_p(v)=  (\tfrac{j}{\sqrt{\pi}}) ~ v^2  e^{-kv^2}$$
Where did this $\frac{1}{\sqrt{\pi}}$ go in two of our answers? After all, $\frac{1}{\sqrt{\pi}}$ appears exactly in $v_{avg}$. Why not in $v_{mp}$ and $v_{rms}$?
First, it’s disappearing from $v_{mp}$ is due to maximization.

1.$v_{mp}$: The most probable speed is determined by setting the derivative of the pdf equal to zero (We are finding the maximum density point):
The resulting $v_{mp}=\operatorname{argmax}$ does not include a $\pi$.
Generally when you maximize, the constants out front dont matter. They don’t affect the maximizing value $x^*$ that you pick. Whatever maximizes some $f(x)$ also maximizes $k ~f(x)$. So it’s unsurprising to not see a $k$ in the result.
That’s true in our maximization. Meaning $k$ is $\frac{1}{\sqrt{\pi}}$ here. So that explains why the $\pi$ term disappeared from the given pdf when finding $v_{mp}$. We didn’t actually integrate any $f_p$ form like with the other two, just picked a $v$ to maximize it.
2.Now for $v_{avg}$, where the $\pi$, perhaps unsurprisingly, ends up as a radical in the denominator, as it is in the distribution function $f_p$, because: $$v_{avg} = \int vf_p(v)dv$$
Based on this link from Vincent Thacker in the comments https://math.stackexchange.com/q/28558 for how to integrate that, we see a $\pi$ will not appear just from integrating with an exponent of $3$ (the $\int v^3 e^{-kv} dv$ form), hence the original $f_p(v)$’s $\frac{1}{\sqrt{\pi}}$ will go through to the $v_{avg}$. Because integrating did not create a $\pi$ to cancel it out.
3.Based again on the link for how to integrate that, we see a $\sqrt{\pi}$ will appear in the numerator with an exponent of $4$, (the $\int v^4 e^{-kv} dv$ form), and that is available to cancel out, hence the original $f_p(v)$’s $\frac{1}{\sqrt{\pi}}$ will not be in the expressions for $v_{rms}^2$ and then $v_{rms}$.
In Summary, the given pdf for the velocity had a $\sqrt{\pi}$ in the denominator, so it was initially surprising to not see it in all three results. For $v_{mp}$, the $\frac{1}{\sqrt{\pi}}$ was a leading coefficient which did not affect the selection of the pdf’s maximizing $v$ (where the slope is zero). For $v_{rms}$, it disappeared because integration of a form $\int v^4 e^{-kv} dv$ creates a $\sqrt{\pi}$ as a result, canceling the $\frac{1}{\sqrt{\pi}}$ in $f_p$.
A: To evaluate $I=\int_{-\infty}^\infty dx e^{-x^2}$ the trick is often to evaluate
$$
I^2=\int_{-\infty}^\infty dx dy e^{-x^2}e^{-y^2}= 
2\pi \int_{0}^\infty \rho e^{-\rho^2} = \pi
$$
after the change to plane polar coordinates, so that $I=\sqrt{\pi}$.  The $\pi$ factor comes from integrating over the polar angles.
A lot of non-zero Gaussian integrals contain $\sqrt{\pi}$ factor since they can be obtained by parametric differentiation w/r to $\lambda$ of the basic result
\begin{align}
\int_{-\infty}^\infty e^{-\lambda x^2}=\sqrt{\frac{\pi}{\lambda}}\, ,
\end{align}
v.g.
\begin{align}
\frac{d}{d\lambda}\int_{-\infty}^\infty e^{-\lambda x^2}=
-\int_{-\infty}^\infty x^2 e^{-\lambda x^2}
=-\frac{\sqrt{\pi}}{2\lambda^{3/2}}\, ,
\end{align}
