Gas pressure in terms of particle number, volume, mass, and velocity fluctuation How do we derive
$$P = \frac{2N}{3V} \left( \frac{1}{2} m \sigma_v^2 \right)$$
where $\sigma_v^2 = (\langle v^2 \rangle - \langle v \rangle ^2)$ is the velocity fluctuation?
I would suppose that we would have to derive something of the following form:
$$P = \frac{2N}{3V} \left( \frac{1}{2} m \left( \langle v^2 \rangle - \langle v \rangle^2 \right) \right)$$
before concluding that $\sigma_v^2 = (\langle v^2 \rangle - \langle v \rangle ^2)$.
 A: *

*The equation from the kinetic theory (of ideal gases) 
$$p = \frac{Nm}{3V} \langle v^2 \rangle$$
assumed that the mean velocity of the container is zero, $\langle v_0 \rangle = 0$. 

*If the mean velocity of the container is non-zero, $\langle v_0 \rangle \ne 0$ this equation has to be modified. I would modify it by replacing $\langle v^2 \rangle \to \langle (v - v_0)^2 \rangle$. This yields 
\begin{align}
\langle (v - v_0)^2 \rangle &= \langle v^2 - 2 v\cdot v_0 + v_0^2 \rangle \\
&=\langle v^2 \rangle - 2 \langle v \rangle \cdot v_0 + v_0^2 \\
&= \langle v^2 \rangle - v_0^2
\end{align}
Finally, because the mean value of each particle, $\langle v\rangle$, will be equal to the velocity of the container, $\langle v_0 \rangle$, we get $\langle v\rangle^2 = v_0^2$. Hence your equation follows.

A: Just to expand on Semoi's answer,


*

*Let's go back to the one-dimensional case for the particle in a box model. Imagine that we're in a frame of reference where the box is not moving, while the particle is moving with the horizontal velocity ${v_x}'$:

Here the change in momentum is $\Delta p_x = 2m{v_x}'$. After extending this to the third dimension, and through Newton's law we would form the pressure equation (I shall assume you're already clear about this part).


$$P=\frac{2N}{3V}\Big(\frac{1}{2}m \langle {v'}^2 \rangle \Big)$$


*

*Now consider the same system but in an inertial frame of reference where the box is moving with a horizontal velocity ${u_x}$. In this frame, the particle is moving at $v_x = {v_x}' + u_x$:

What would be the change in momentum in this frame? Well, Galilean invariance means that the change in momentum would still be the same. Thus $\Delta p_x = 2m{v_x}' = 2m (v_x-u_x)$. Following the same steps as above, we will get:


$$P=\frac{2N}{3V}\Big(\frac{1}{2}m \, \langle \, (v-u)^2 \, \rangle \Big)$$


*

*What then is the value $u$? If the entire box is moving with the velocity $u$, this means that all the particles, on average, are moving with the velocity $u$. In other words, $u$ is merely the mean velocity of all the particles: $u = \langle v \rangle$.

*Lastly, all that is needed to be done is the expansion, as @Semoi has already done. I'd offer an alternative notation that allowed me to see the expansion more clearly:
$$
\begin{align}
\langle \, (v - u)^2 \, \rangle
&= \frac{\sum (v-u)^2}{N} \\[3mm]
&= \frac{\sum (v^2-2vu+u^2)}{N} \\[3mm]
&= \frac{\sum (v^2)}{N} - \frac{\sum (2vu)}{N} + \frac{\sum (u^2)}{N} \\[3mm]
&= \langle v^2 \rangle - 2u \frac{\sum v}{N} + \frac{N \cdot u^2}{N} \\[3mm]
&= \langle v^2 \rangle - 2 \langle v \rangle \langle v \rangle + {\langle v \rangle}^2 \\[4mm]
&= \langle v^2 \rangle - {\langle v \rangle}^2
\end{align}
$$
