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Adiabatic compression at a constant rate

  1. Jun 2, 2010 #1
    I came up with this question a while ago, and my Physics teacher and I just can't get the math to work.

    Here it is:

    A cylindrical container of base area A and height h holds some number of moles, n, of an ideal gas. One end of the container is movable, and a force F is applied to that end. If the rate of compression of the gas is constant, i.e. the derivative of h with respect to time is constant, and the process is adiabatic, what is the equation of F as a function of time?

    If this isn't clear enough, let me know.
     
  2. jcsd
  3. Jun 2, 2010 #2
    Using

    P V^gamma = constant

    seems to solve the problem in a straightforward way...
     
  4. Jun 2, 2010 #3
    I don't see how. I was unfamiliar with that formula, so I looked it up and read the proof, and I don't really see how it relates to the question. Could you be more explicit, please?
     
  5. Jun 2, 2010 #4

    Q_Goest

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    gamma = ratio of specific heats for an adiabatic compression.

    So you have the volume as a function of time or did I misunderstand that? So A is a constant and volume is a function of h. If you have h, you have volume, and you have pressure also because pressure is dependent only on volume. Pressure is a function of h.

    It doesn't matter how fast h changes. Pressure (and thus the force) is a function only of volume. You should be able to come up with an equation that relates pressure (and thus the force) with volume, right? From there, it should be easy to relate volume to time. They are linearly dependent.
     
  6. Jun 2, 2010 #5

    jack action

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    Adiabatic process definition:

    [tex]P_{\left(t\right)}V_{\left(t\right)}^{k}=P_{\left(0\right)}V_{\left(0\right)}^{k}[/tex]

    Volume definition:

    [tex]V_{\left(t\right)}= A h_{\left(t\right)}[/tex]

    Gauge pressure definition:

    [tex]P_{\left(t\right)}=P_{\left(0\right)} + \Delta P_{\left(t\right)}[/tex]

    Force definition:

    [tex]F_{\left(t\right)}=\Delta P_{\left(t\right)}A[/tex]

    Height definition (given; C = constant):

    [tex]h_{\left(t\right)}=h-Ct[/tex]

    Then (if I didn't make any mistake):

    [tex]F_{\left(t\right)}=P_{\left(0\right)}A\left[ \left( \frac{h}{h-Ct} \right)^{k} -1 \right][/tex]
     
  7. Jun 7, 2010 #6
    This almost clears it up, the graph looks roughly like what we expected and everything. My one question is this: How did you get algebraically from the given equations you posted to this one? The unit-less factor, h/h-Ct etc, specifically.


    Thanks for your help and patience.
     
  8. Jun 7, 2010 #7

    jack action

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    Just start with this:

    [tex]F_{\left(t\right)}=\Delta P_{\left(t\right)}A[/tex]

    Then use [tex]P_{\left(t\right)}=P_{\left(0\right)} + \Delta P_{\left(t\right)}[/tex] to replace [tex]\Delta P_{\left(t\right)}[/tex] and so on. That's pretty straightforward manipulation.

    Just remember that [tex]h_{\left(0\right)}=h_{\left(t=0\right)}=h-C \left(0\right) = h[/tex]
     
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