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Mercury barometer problem

  1. Nov 21, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    To construct a barometer, a tube of length 1m is filled completely with mercury and is inverted in a mercury cup. The barometer reading on a particular day is 76cm. Suppose a 1m tube is filled with mercury up to 76cm and then closed by a cork. It is inverted in a mercury cup and the cork is removed. The height of mercury column in the tube over the surface in the
    cup will be

    a)zero b)76cm c)>76cm d)<76cm

    2. Relevant equations

    3. The attempt at a solution
    Since the tube has been partially filled thus air is present in it. When it is inverted the air exerts a pressure on the mercury filled in it. Also the atm .pressure is exerted at the surface of mercury present in cup. But which one is greater?
     
  2. jcsd
  3. Nov 21, 2012 #2
    The volume of the 24 cm column of air will probably increase once the tube is inverted.
     
  4. Nov 21, 2012 #3
    The next question is, to precisely what height will the mercury drop to?
     
  5. Nov 22, 2012 #4

    utkarshakash

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    Why?
     
  6. Nov 22, 2012 #5
    The downwards pressure exected by the air and mercury column is balanced by the upwards atmospheric pressure.
     
    Last edited: Nov 22, 2012
  7. Nov 22, 2012 #6
    To answer the question in my previous post, you need to use the ideal gas law. You end up with the equation

    100 - x + (24)(76)/x = 76

    where x is the distance that the 24 cm increases to. The answer is x = 56.4 cm. That leaves a Hg column of 43.6 cm.
     
  8. Nov 22, 2012 #7

    utkarshakash

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    Can you please make it more clear by stating what those numbers mean?
     
  9. Nov 22, 2012 #8
    The (100 - x) is the hydrostatic pressure of the Hg column in cm of Hg.

    The (24)(76)/x is the pressure of the air bubble in cm of Hg. The 76 in this expression is the pressure of the air in the bubble before the bubble expands downward.

    The pressure of the Hg column plus the pressure of the air bubble must equal the pressure of the atmosphere outside the tube at the Hg surface, which is equal to 76 cm of Hg.
     
  10. Nov 23, 2012 #9

    utkarshakash

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    Are you using PV=const equation?
     
  11. Nov 23, 2012 #10
    Yes. Before the Hg level drops, the pressure in the air bubble is 76 cm Hg, and the length of the air bubble is 24 cm. If the area of the cylinder is A, the initial volume of the air bubble is 24A. After the Hg level drops, the length of the air bubble is x, and the volume of the air bubble is xA. So the volume ratio is (24A)/(xA). The new pressure in the expanded air bubble is

    (76)(24A)/(xA) = (76)(24)/x cm Hg.
     
  12. Nov 23, 2012 #11

    utkarshakash

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    OK I have understood.
     
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