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Homework Help: Derive Acoustic Pressure Relation from Ideal Gass Law

  1. Feb 13, 2010 #1
    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.
  2. jcsd
  3. Feb 13, 2010 #2
    Not sure I see how this is done. The Euler relations are


    Can you clarify why you ignore [itex]dr[/itex]?
  4. Feb 13, 2010 #3
    I know that the Euler relation is




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

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

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

    \exp[ix] = \cos[x]

    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).
  5. Feb 13, 2010 #4
    This is true, and I didn't quite think of this part, mostly because the relation actually is

    \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:


    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.
  6. Feb 16, 2010 #5
    Re: Derive Acoustic Pressure Relation from Ideal Gas Law

    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.
    Last edited: Feb 16, 2010
  7. Feb 27, 2010 #6
    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]


    [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.
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