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Pressure Problem

  1. Mar 23, 2008 #1

    bfr

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    [SOLVED] Pressure Problem

    SOLVED

    1. The problem statement, all variables and given/known data

    A vertical cylindrical tank is 42 cm tall, 2.00 cm in radius, and is open at the top. Atmospheric pressure is 1.013 x 10^5 N / m^2. A close-fitting cylindrical plug of mass 5 kg is inserted at the top, and is let fall inside. If the temperature of the trapped air does not change, how far from the top of the cylinder is the base of the plug when it comes to rest?

    (This is the exact problem, word for word)

    2. Relevant equations

    (p1*v1)/t1=(p2*v2)/t2

    3. The attempt at a solution

    Should I compare the pressure caused by the plug to the atmospheric pressure? ((5*9.8*x)/((.02)^2 * pi))((.02)^2 * pi * x)=1.013 * 10^5 * (.02)^2 * pi * .42 ...or, in terms of variables... m*g*h/(pi*r^2) * (pi*r^2*h)=atmospheric pressure * (pi*r^2*height_of_cylinder)

    (pi*r^2*height_of_cylinder) is the volume of the cylinder, (pi*r^2*h) is the volume under the plug.

    EDIT: I think this is right. Thanks!
     
    Last edited: Mar 23, 2008
  2. jcsd
  3. Mar 23, 2008 #2

    Andrew Mason

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    What is the downward force on the plug? What is the upward force on the plug? How are the two related? (ie. does the plug move?). Translate the downward force into pressure and apply the ideal gas law: P1V1 = nRT1 = nRT2 = P2V2

    AM
     
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