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Question Regarding Bubble in Kerosene Tank

  1. Nov 13, 2009 #1
    1. The problem.
    Okay, here's the question:
    'An air bubble situated at the bottom of an open kerosene tank rises to the top surface. It is observed that at the top, the volume of the bubble is thrice it's initial volume. If the atmospheric pressure is 72cm of Hg, and mercury is 17 times heavier than kerosene, then what is the depth of the tank?'



    2. Relevant equations.
    As far as I think, we'll be using Boyle's Law:
    P'V'=P''V''
    Pressure=Density(of Kerosene)*gravitational acceleration*depth(to where the bubble is)



    3. The attempt at a solution
    Well, it says that at the top, the volume of the bubble triples, so the pressure must become one-by-three of the initial(and I think we're supposed to consider the pressure exerted by the kerosene on the bubble). But, I don't understand how we're supposed to use the relation between the weights of kerosene an mercury to find the the pressure exerted by kerosene. A;so, what do they mean by '72 cm of Hg'....I've never really understood the concept of mercuric barometers.
    Any help will be appreciated. Thank you.
     
  2. jcsd
  3. Nov 13, 2009 #2

    Delphi51

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    Homework Helper

    Looks like you can use the first formula to find the pressure at the bottom of the tank. The second can then be used to find the height.
    There is a pressure unit conversion table here:
    http://en.wikipedia.org/wiki/Pressure#Units
     
  4. Nov 13, 2009 #3

    tiny-tim

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    Science Advisor
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    Hi modulus! :smile:

    As you say, pressure = ρkerosenegh.

    They give you ρkeroseneHg, and "72cm of Hg" means the pressure at a depth of 72cm below the surface in Hg. :wink:
     
  5. Nov 14, 2009 #4
    Yeess!! I got it!
    You guys have no clue how easy you've made this for me!
    Thanks a lot!
     
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