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Pressure in Liquids

  1. Jul 20, 2012 #1
    I can't seem to get a clear picture of how the pressure changes in liquids.

    Here is a small self made file, made to make myself understand this topic .

    Click on the attachment.

    Obviously as the water rose up, air pressure in the small tube rose up. (Due to Boyle's Law)

    Now lets assume (for the sake of better understanding) that i pushed the small tube half a meter inside the large beaker.

    Now what is the pressure on the yellow mark ?
    Is it equal to the air pressure inside the small tube, because that only makes sense. If they were not equal then there would be no Equilibrium and the water would continue to rise up/down.

    And, what is the pressure at the pink mark ?
    Is it :

    Pressure = Desnsity x g x h + atm
    = 1000*9.8*0.5 +1atm

    Is the pressure on both marks the same ?

    Attached Files:

  2. jcsd
  3. Jul 20, 2012 #2


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    Why did the water rise up the tube? I can't see a reason for that unless the tube is sealed at the top and the air in it was cooled between 1 and 2?

    The only other way would be if the tube was open at the top and capillary action caused the water to rise a little way up the tube?
  4. Jul 21, 2012 #3
    So what u are suggesting is that the water level in the beaker remains constant, thus the awter in the small tube will not rise ?

    If that is the case, then the pressure at the yellow spot must be the same as the pressure of the air in the small tube, while Pressure at the pink mark will be ?

    1.Air pressure in the small tube(which is higher than 1atm) plus the pressure due to liquid water
    2.Air pressure (1 atm) plus the pressure due to liquid water

    I think it would be 1.

    Am i correct ?
  5. Jul 23, 2012 #4

    Is there no one who can help me ? :(
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