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A diver and an air bubble

  1. Jun 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A diver 38.8 m deep in 14°C fresh water exhales a 2.19 cm diameter bubble. What is the bubble's diameter just as it reaches the surface of the lake, where the water temperature is 25°C?


    2. Relevant equations

    PV=nRT


    3. The attempt at a solution

    I dont know where to start on this problem at all I can find the intial pressure of the bubble but I dont see how that matters?
     
  2. jcsd
  3. Jun 29, 2009 #2

    cepheid

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    Helium balloons expand as they rise in the atmosphere, because the air gets thinner (and the air pressure reduces as a result). The same thing happens here. The pressure falls with rising height. So you can calculate the pressure at the two depths and see how it changes. However, the temperature also changes. As you know, from the ideal gas law, both of these things affect the volume.
     
  4. Jun 29, 2009 #3
    I didn't think about that way. Let me work through it and then post what I got, thanks
     
  5. Jun 29, 2009 #4
    What would be the initial pressure be? I know the final would be sea level of 101300 Pa.
     
  6. Jun 29, 2009 #5
    I found Pi should be 3.38 atm x 101300 Pa= 393044 Pa because since he is 38.8 meters below sea level the pressure is 3.38 atm
     
  7. Jun 29, 2009 #6

    cepheid

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    Yeah, I guess the pressure at the surface would be be atmospheric pressure. Do you know how the pressure varies with depth in a fluid?
     
  8. Jun 29, 2009 #7
    i approximated it that ever 10 meters below sea level= 1 atm is there a way to solve it for another way?
     
  9. Jun 29, 2009 #8

    cepheid

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    How do you figure that?
     
  10. Jun 29, 2009 #9

    cepheid

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    Interesting. Where did you find this approximation? It turns out to be more or less correct (for water). However, it might be helpful for you to know where it came from. Have you seen this before?

    http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html#fp

    Now you know the principle behind it, so you can figure it out precisely (it's not exactly 1 atm per 10 m) and you can also do this for any fluid.
     
  11. Jun 29, 2009 #10
    I am still wrong here is my work....
     
  12. Jun 29, 2009 #11

    cepheid

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    By the way, 38.8 m / 10 m = 3.88, not 3.38 as you have written.
     
  13. Jun 29, 2009 #12
    it is attached
     

    Attached Files:

  14. Jun 29, 2009 #13

    cepheid

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    Your attachment won't be approved right away (I can't see the image). If you want immediate help, use a free online image hosting service to post the image, or type up your work.
     
  15. Jun 29, 2009 #14
    PiVi/Ti=PfVf/Tf solved for Vf

    Vf=PiViTf/TiPf then plugged in 1/6d^3 for the volumes and then found out that.....

    df^3=PiTfdi^3/TiPf.......(380240*298*.0219^3/(285*101300))^1/3 and then got .125 m
     
  16. Jun 29, 2009 #15

    cepheid

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    The volume of a sphere is given by (4/3)πr3

    Even converting to diameter, that turns into (π/6)d3, which is not the same as what you have written there.

    I haven't checked your other work closely yet.
     
  17. Jun 29, 2009 #16
    Oh right, but it cancels either way
     
  18. Jun 29, 2009 #17
    2physics.jpg

    that is the work i did before i switched the pressure so the anwser i really got is not .0344 m( either way thats wrong too) but .1235 m
     
  19. Jun 29, 2009 #18

    cepheid

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    The initial temperature is 287 K, not 285 K.

    I should have an answer soon, and then we can compare.
     
  20. Jun 29, 2009 #19
    ok thank you
     
  21. Jun 29, 2009 #20

    cepheid

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    I'm not gonna reveal my answer yet ;)

    Here's what I have:

    [tex] V_f = \frac{P_i}{P_f}\frac{T_f}{T_i}V_i [/tex]

    [tex] \frac{\pi}{6}d_f^3 = 3.88\frac{298}{287} \frac{\pi}{6}d_i^3 [/tex]

    [tex] d_f = \sqrt[3]{3.88\frac{298}{287}d_i^3} [/tex] ​

    Is it any different from your work so far?

    EDIT: I see you have exactly the same thing as I do. So, do you get the right answer with the corrected numbers?
     
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