Derive Acoustic Pressure Relation from Ideal Gass Law

[dimensionless]

Homework Statement

Derive p(r) = \frac{A}{r}e^{j(\omega t - kr)} from pV = nRT

Homework Equations

The Attempt at a Solution

From ideal gas law I have

p(r) = \frac{nRT}{V}

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

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

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

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

Putting this in exponential form I get

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

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

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

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

 

[jdwood983]

Which lead to

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

Putting this in exponential form I get

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

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

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

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

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

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

Can you clarify why you ignore dr?

 

[dimensionless]

I know that the Euler relation is

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

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

<br /> \exp[ix] = \cos[x] + i \sin[x]<br />

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

<br /> \exp[ix] = \cos[x] <br />

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]

I know that the Euler relation is

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

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

<br /> \exp[ix] = \cos[x] + i \sin[x]<br />

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

<br /> \exp[ix] = \cos[x] <br />

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

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

<br /> \Re\left[\exp[ix]\right]=\cos[x]

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

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).

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:

<br /> P=\frac{nkT}{V}=\frac{nkT}{m}\rho<br />

Then consider how the density \rho 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 n/V, so that

P=\frac{n}{V} kT= \rho kT

which leads to

\rho = \frac{n}{4 \pi r^{2} dr}

Since the number of mols varies with time, I get

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

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

\rho = \frac{n}{r^{2} V}

At the moment though, I'm still stuck.

 

[dimensionless]

I also have

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

this leads to

v = \sqrt{3 \frac{RT}{nm} }

and

T = \frac{n v^{2}}{3mR}

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

F = \frac{dp}{dt}

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