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Can a volume of gas exert pressure on itself ?

  1. Dec 20, 2012 #1
    So, I was going through the solved problem sets and encountered this question which is to find the relation between two ends of a cylinder filled with gas rotating about a vertical axis fixed at one end (the red axis) .

    n2KBA.png

    I understood the solution but there is one thing still confusing me, the solution said that the force on the dx part would be one due to its rotation (let's say theres dm mass of gas there)
    [itex]m\omega^{2}x[/itex] that's alright but the solution also said that there'll be a force towards the inside which will be( A is the area of cross section of cylinder) [itex]Adp [/itex] . Where does this come from ? Does the gas exerts force on itself ?

    This is not a homework question .
     
  2. jcsd
  3. Dec 20, 2012 #2

    mfb

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    It does - just look at our atmosphere, where pressure is decreasing with increasing height. It is the same concept. The decreasing pressure is in equilibrium with the gravitational force on the gas - which gets replaced by centrifugal forces in your setup.
     
  4. Dec 20, 2012 #3

    CWatters

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    That sounds like the centripetal vs centrifugal argument eg The one that says centrifugal force doesn't exist and that there is only centripetal force and that acts inwards.

    The gas is rotating around the red axis so the direction component of it's velocity is changing. That implies an acceleration towards the centre. Essentially the base of the tube is pushing the gas sideways forcing it to move in a circle and preventing it flying off at a tangent.
     
  5. Dec 20, 2012 #4
    In this system, the pressure varies with x: P = P(x)

    Take as a free body the mass between cross sections x and x + dx:

    dm = ρAdx

    The force exerted by the gas located beyond (x + dx) on the mass dm is -P(x+dx)A. The force exerted by the gas located closer toward the axis on the mass dm = P(x)A. The net force on the mass dm must be: A (P(x) - P(x + dx)). This must be equal to the mass dm times the acceleration:

    -(ρAdx)ω2x = A (P(x) - P(x + dx))

    Expressing this as a differential equation, it becomes:

    ρω2x = dP/dx

    If we express the density using the ideal gas law, we get:

    PMω2x /(RT) = dP/dx
    or
    dlnP/dx2 = 2Mω2/(RT)

    where M is the molecular weight of the gas.

    so P(x) = P(0) exp (2Mω2x2/(RT) )
     
  6. Dec 21, 2012 #5
    Thanks everyone I got it finally !
     
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