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Acoustics/Thermodynamics Question

  1. Aug 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Given that
    [tex]P V^{\gamma} = P_{0} V_{0}^{\gamma} [/tex]
    Show that
    [tex]dP \doteq - \frac{\gamma P_{0}}{V_{0}} dV [/tex]

    2. Relevant equations
    [tex]c^{2} = \left( \frac{\partial P}{\partial \rho} \right)_{\rho_{0}}[/tex]
    [tex]P = P_{0} \frac{\rho}{\rho_{0}}[/tex]
    [tex]\gamma = \frac{c_{p}}{c_{v}}[/tex]
    [tex]P = \rho r T_{k}[/tex]

    3. The attempt at a solution
    I'm not sure what the [tex]\doteq[/tex] means, but I can pressume that it is the same thing as an equal sign. Starting with the first equation I can get
    [tex]P = \frac{P_{0} V_{0}^{\gamma}}{V^{\gamma} } [/tex]
    At this point I get stuck. I could try to relate the volume to the density, but I'm not sure that that would lead anywhere.
  2. jcsd
  3. Aug 10, 2007 #2
    from my last equation I can get
    [tex]dP = \frac{-\gamma P_{0} V_{0}^{\gamma}}{V^{\gamma-1} }dV [/tex]
    But this is not a huge help.
    Last edited: Aug 10, 2007
  4. Aug 10, 2007 #3


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    Staff: Mentor

    [tex]P V^{\gamma} = P_{0} V_{0}^{\gamma} [/tex]

    by the right hand side is a constant, so derivative or differential is zero.

    So one starts with

    [tex] d(P V^{\gamma})\,=\,0[/tex] and expand, and then put it in a form

    dP = ?, but that seems where one go to.

    Also I think its

    [tex]dP = \frac{-\gamma P_{0} V_{0}^{\gamma}}{V^{\gamma+1} }dV [/tex]

    and then what happens if V ~ Vo?
    Last edited: Aug 10, 2007
  5. Aug 10, 2007 #4


    User Avatar
    Homework Helper

    That's wrong. It should read
    dP = \frac{-\gamma P_0 V_0^{\gamma}}{V^{\gamma + 1}}dV
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