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Pressure+pascal's principle

  1. Jan 4, 2010 #1
    Help! What is the fractional decrease in pressure when a barometer is raised 35m to the top of a building? (assume that the density of air is constant over that distance)
     
  2. jcsd
  3. Jan 4, 2010 #2

    rock.freak667

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    You must post some attempt before anyone at PF can help you (forum rules).

    What I can do though is ask you, how would find the change in pressure at the height of 35m?
     
  4. Jan 4, 2010 #3
    I have tried to work this problem in different ways and I have gotten different answers. This was my last attempt but it still does not seem right.

    35 m of air at approx 29 gm/mole at 24.1 liter per mole at room temperature weighs 4.2 gram per cm squared.

    The air pressure at sea level is 1035 grams per cm squared.

    Raising the barometer 35 meters lowers the pressure 4.2/1035 or 0.4%.
     
  5. Jan 4, 2010 #4

    rock.freak667

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    0.4% looks fine to me using your method.
     
  6. Jan 4, 2010 #5
    here is another way with a different answer
    change in pressure , ΔP = ρgh

    where,
    ρ = density of medium = 1.184 kg/m3 (for air at 250 C)
    g = acc due to gravity = 9.8 m/s2
    h = height change = 35 m

    => ΔP = 1.184*9.8*35 = 406.112 Pa

    1 atm = 101325 Pa
    => 1 Pa = 1/101325 atm
    => 406.112 Pa = 406.112/101325 = 0.004 atm (approx)
     
  7. Jan 4, 2010 #6

    rock.freak667

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    Δp/p = 0.004 = 0.4%
     
  8. Jan 4, 2010 #7
    take a look at this; where did I go wrong this time

    ρgh = 1.15kg/m3*9.8*35 = 389.4*10-5 Pa
     
  9. Jan 4, 2010 #8

    rock.freak667

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    How did you get the 10-5? I got 394.8525 Pa
     
  10. Jan 4, 2010 #9
    I put it in wrong in my calculator! I just realized this. Thanks for your help!
     
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