1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

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

    Can you clarify why you ignore [itex]dr[/itex]?
     
  4. Feb 13, 2010 #3
    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).
     
  5. Feb 13, 2010 #4
    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.
     
  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]

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook