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Physical Audio Signal Processing
    Elementary Physics, Mechanics, and Acoustics
       Properties of Gases
          Pressure is Confined Kinetic Energy

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Pressure is Confined Kinetic Energy

According the kinetic theory of ideal gases [180], air pressure can be defined as the average momentum transfer per unit area per unit time due to molecular collisions between a confined gas and its boundary. Using Newton's second law, this pressure can be shown to be given by one third of the average kinetic energy of molecules in the gas.

$\displaystyle p = \frac{1}{3}\rho \left<u^2\right> $

Here, $ \langle u^2\rangle $ denotes the average squared particle velocity in the gas. (The constant $ 1/3$ comes from the fact that we are interested only in the kinetic energy directed along one dimension in 3D space.)



Proof: This is a classical result from the kinetic theory of gases [180]. Let $ M$ be the total mass of a gas confined to a rectangular volume $ V = Aw$, where $ A$ is the area of one side and $ w$ the distance to the opposite side. Let $ \overline{u}_x$ denote the average molecule velocity in the $ x$ direction. Then the total net molecular momentum in the $ x$ direction is given by $ M\vert\overline{u}_x\vert$. Suppose the momentum $ \overline{u}_x$ is directed against a face of area $ A$. A rigid-wall elastic collision by a mass $ M$ traveling into the wall at velocity $ \overline{u}_x$ imparts a momentum of magnitude $ 2M\overline{u}_x$ to the wall (because the momentum of the mass is changed from $ +M\overline{u}_x$ to $ -M\overline{u}_x$, and momentum is conserved). The average momentum-transfer per unit area is therefore $ 2M\overline{u}_x/A$ at any instant in time. To obtain the definition of pressure, we need only multiply by the average collision rate, which is given by $ \overline{u}_x/(2w)$. That is, the average $ x$-velocity divided by the round-trip distance along the $ x$ dimension gives the collision rate at either wall bounding the $ x$ dimension. Thus, we obtain

$\displaystyle p \isdefs \frac{2M\overline{u}_x}{A}\cdot \frac{\overline{u}_x}{2w} \eqsp \rho \overline{u}_x^2 $

where $ \rho=M/V$ is the density of the gas in mass per unit volume. The quantity $ \rho\overline{u}_x^2/2$ is the average kinetic energy density of molecules in the gas along the $ x$ dimension. The total kinetic energy density is $ \rho\overline{u}^2/2$, where $ \overline{u}=\sqrt{\overline{u}_x^2+\overline{u}_y^2+\overline{u}_z^2}$ is the average molecular velocity magnitude of the gas. Since the gas pressure must be the same in all directions, by symmetry, we must have $ \overline{u}_x^2=\overline{u}_y^2=\overline{u}_z^2 = \overline{u}^2/3$, so that

$\displaystyle p = \frac{1}{3}\rho \overline{u}^2. $

Previous: Volume Velocity of a Gas
Next: Bernoulli Equation

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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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