Derive Acoustic Pressure Relation from Ideal Gass Law

dimensionless

1. The problem statement, all variables and given/known data
Derive [tex]p(r) = \frac{A}{r}e^{j(\omega t - kr)}[/tex] from [tex]pV = nRT[/tex]

2. Relevant equations

3. The attempt at a solution
From ideal gas law I have

[tex]p(r) = \frac{nRT}{V}[/tex]

R and T are constant, so I can pull them out now and replace them with A. If V is the volume of a spherical shell of thickness dr, I get

[tex]p(r) = \frac{n A}{4 \pi r^{2} dr}[/tex]

This means that the only thing that changes is the net flow of mass flowing into and out of my spherical shell. Which lead to

[tex]p(r) = \frac{A cos(\omega t - k r)}{4 \pi r^{2} dr}[/tex]

Putting this in exponential form I get

[tex]p(r) = \frac{A}{4 \pi r^{2} dr} e^{j(\omega t - k r)}[/tex]

Because the atoms are only moving radially, I can ignore the dr part. This leads to

[tex]p(r) = \frac{A}{4 \pi r^{2}} e^{j(\omega t - k r)}[/tex]

...I'm still stuck with an [tex]r^{2}[/tex] in the denominator instead of just [tex]r[/tex]. I think I did something wrong somewhere. I'm not sure where. Any help is appreciated.

jdwood983

Not sure I see how this is done. The Euler relations are

[tex]
\cos[x]=\frac{1}{2}\left(\exp[ix]+\exp[-ix]\right)
[/tex]

Can you clarify why you ignore [itex]dr[/itex]?

dimensionless

I know that the Euler relation is

[tex]

\cos[x]=\frac{1}{2}\left(\exp[ix]+\exp[-ix]\right)

[/tex]

but quite I'm sure I've seen the expression

[tex]
\exp[ix] = \cos[x] + i \sin[x]
[/tex]

used in many places but ignoring the imaginary part so that it just becomes

[tex]
\exp[ix] = \cos[x]
[/tex]

Actually, it really bothers me, but I've seen it in so many places.

I'm ignoring dr, in part because I'm seeking a particular result. As for rational, the particles are moving only as part of a sound wave, so they move parallel to dr. The pressure is force per unit area, and the area is perpendicular to dr...in other words, I need the pressure on a spherical surface rather than in a volume ( p(r) is net air pressure above the equilibrium).

jdwood983

This is true, and I didn't quite think of this part, mostly because the relation actually is

[tex]
\Re\left[\exp[ix]\right]=\cos[x] [/tex]

That is, the real component of the exponential is the cosine term, while the imaginary, [itex]\Im[/itex], is sine.

The pressure does apply itself parallel to the radial component, not perpendicular--draw yourself a picture of a spherical wave, you'll see ;)

I think, rather than using volume, you should consider the density:

[tex]
P=\frac{nkT}{V}=\frac{nkT}{m}\rho
[/tex]

Then consider how the density [itex]\rho[/itex] would change with a wave. It is possible that you will actually want to integrate, rather than just ignore the infinitesimal radial projection.

dimensionless

I think it might be simpler if I defined density [tex]n/V[/tex], so that

[tex] P=\frac{n}{V} kT= \rho kT [/tex]

which leads to

[tex]\rho = \frac{n}{4 \pi r^{2} dr}[/tex]

Since the number of mols varies with time, I get

[tex]\rho \cos(\omega t)= \frac{n \cos(\omega t)}{4 \pi r^{2} dr}[/tex]

Conversely, I could hold the volume constant and let the number of mols (or the mass) per unit volume vary so that I get

[tex]\rho = \frac{n}{r^{2} V}[/tex]

At the moment though, I'm still stuck.

dimensionless

I also have

[tex]\frac{1}{2}m v_{average}^{2} = \frac{3}{2}kT = \frac{3}{2}\frac{R}{n}T[/tex]

this leads to

[tex]v = \sqrt{3 \frac{RT}{nm} }[/tex]

and

[tex]T = \frac{n v^{2}}{3mR}[/tex]

where m is the mass of a particle and v is the velocity. I also have

[tex]F = \frac{dp}{dt}[/tex]

Where p=mv is the momentum. I still can't derive this equation though.