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Air bubble in water at 50m

  1. May 16, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem is at
    http://dl.getdropbox.com/u/175564/volumePressureTemperature.JPG [Broken]


    2. Relevant equations
    I know the equation for ideal gases
    pV = nRT
    and
    the definition of pressure
    p = F / A


    3. The attempt at a solution

    The change in T is 10.
    V_i = 1 cm^3

    We need to first consider how the change in Temperature affects
    the depth at which the bubble is.
    It will goes up because the volume of the bubble increases.

    The increase in Volume causes the boyonce force to increase.
    Thus, the net force in f direction is zero at both depths:
    F_boyonce - F_gravity = 0
    g( m_Water - m_airBubble ) = 0

    There is a linear relation between p and T if we assume the
    situation for ideal gases.

    I have apparently ignored some critical part.
    It seems that the ideal Gas equation is not valid here.
    The right answer is 6.2cm^3 that is (f).
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 16, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Sure it is--it's all you need for this problem. How does the pressure change? The temperature? Use the ideal gas law to determine how the volume changes. (Set up ratios.)
     
  4. May 16, 2009 #3
    I have the following

    V1 / V2 = [tex] \frac { T1 / P1 } { T2 / P2} [/tex]


    V1 / V2 = [tex] \frac {290(p1 + 5atm)} {p1 * 300}[/tex]
    = 290/300 + 1450/(P1 * 300)

    The problem is now P1 which is the initial pressure.
    It seems that we need to approximate it, since we do not know it.
     
  5. May 16, 2009 #4

    Doc Al

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    Staff: Mentor

    You have the ratios a bit backwards. Start with this:
    P1V1/T1 = P2V2/T2

    You should know it. What's the pressure at the surface?
     
  6. May 16, 2009 #5
    It is one atm :)

    I get the following result
    V1 / V2 = 6 * 290/300
    = 5.8

    whrere 290 kelvins is 17 celcius and 300 kelvins is 27 celcius.

    Taking inverse

    V2 / V1 = 0.17
    where V2 is the volume at the surface.

    Both answers seem to be wrong: I cannot get 6.2.
    I have likely ignored some small volume.
     
  7. May 16, 2009 #6

    LowlyPion

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    Homework Helper

    Are you sure it's not 6*300/290 ?
     
  8. May 16, 2009 #7

    Doc Al

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    Staff: Mentor

    Good. :wink:

    Once again, you're flipping things around somehow.

    Start with this:
    (P1V1)/T1 = (P2V2)/T2

    Then solve for V1.
     
  9. May 16, 2009 #8
    I finally got the right answer.

    V1 = P2 V2 T1 / T2 P1

    = 6 * 30 / 29
    = 6.2

    Thank you for your answers!
     
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