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Boyle's law problem.

  1. Feb 5, 2012 #1
    A narrow uniform glass tube contains air enclosed by a thread of mercury 15cm long.
    When the tube is held vertically, with the closed end at the bottom,the air column is 20.0cm long, but when it is held horizontally the air column is 24.0cm long. Calculate the atmospheric pressure.

    357f5sj.png



    I drew a diagram I'm not sure if it's correct, I know that I'm supposed to subtract the mercury height from the barometric height but I don't get the correct answer. Please help me.
     
  2. jcsd
  3. Feb 6, 2012 #2

    tiny-tim

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    hi lionely! :smile:

    call the atmospheric pressure "A" (in cm of mercury)

    since the mercury is stationary, what must be the pressure exerted on it by the enclosed air in each case? :wink:
     
  4. Feb 6, 2012 #3
    A + the mercury heights?
     
  5. Feb 7, 2012 #4

    tiny-tim

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    (just got up :zzz: …)

    what about the weight of the mercury? :confused:
     
  6. Feb 7, 2012 #5
    Can't we just work it out in cmHg?
     
  7. Feb 8, 2012 #6

    tiny-tim

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    (just got up :zzz: …)

    we are!

    but we need to know what forces (or pressures) we have to include, and whether to add or subtract …

    for that, you need to decide which forces are acting on the mercury

    soo … what about the weight of the mercury (in either case)?​
     
  8. Feb 8, 2012 #7
    well mg acts down, so in the second case I would assume the effect is not as great.
     
  9. Feb 9, 2012 #8

    tiny-tim

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    add the forces, luke! o:)
     
  10. Feb 9, 2012 #9
    P1V1 = P2V2

    (p+15)(20A) = P x 24A

    20P + 300 = 24P

    P = 300/4 = 75cmHg

    Is that it ?
     
  11. Feb 10, 2012 #10

    tiny-tim

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  12. Feb 10, 2012 #11
    Thanks for your help man XD
     
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