# 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 $\sigma_v^2 = (\langle v^2 \rangle - \langle v \rangle ^2)$.

• Welcome to Physics Stack Exchange. Note that I've edited this post in a few ways: 1) Better title. Titles should deliver as much information as possible. See our FAQ on writing good question titles. 2) Replaced images of math with mathjax. Mathjax is better because it is i) editable, ii) searchable, and iii) easier to read. – DanielSank Aug 17 '17 at 16:23
• Could you please tell us more about the context of this equation. I know that in the kinetic theory of an ideal gas we get something like $p = \frac{N m}{3 V} \langle v^2_{\textrm{rms}}\rangle$. However, here $\langle v^2\rangle$ is the mean square expectation value and not the variance of the velocity distribution. – Semoi Aug 17 '17 at 20:15
• @Semoi this is the equation used when the container of gas is moving in a particular direction. Hence, the movement of individual particles are no longer totally random but instead have a tendency to move with the container. – Limzy Aug 17 '17 at 23:05

• 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.

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}