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Pressure in a manometer

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A mercury-filled manometer with two unequal-length arms of the same cross-sectional area is sealed off with the same pressure p in the two arms. With the temperature constant, an additional 10.0cm3 of mercury is admitted through the stopcock at the bottom. The level on the left increases 6.00cm and that on the right increases 4.00cm. Find the pressure p.

    Then there is a picture of a two-armed manometer (sorry don't have a scanner but it's simple enough). At the initial pressure p, the left arm has 50cm in length filled with gas and the right arm has 30cm in length filled with gas.

    2. Relevant equations

    Boyle's Law: P1V1 = P2V2

    3. The attempt at a solution

    P1 is what we are solving for. So first find the cross sectional area, A. 10.0cm3 of Hg was added and that went entirely into the two arms. So:

    6A + 4A = 10
    A = 1cm2

    Then apply Boyle’s law to each arm:

    Let Pl be the final pressure of the left arm and Pr be the final pressure of the right arm.

    50P1 = (50-6)Pl
    30P1 = (30-4)Pr

    Now here comes the equation I’m not sure is entirely justified. 10.0cm3 of Hg was added which is the same as adding 10/A cm Hg or 10 cm Hg or 100 torr of pressure to the system, right? So we can write:

    P1 + 10 = Pl + Pr

    Combining the 3 equations, I can easily solve for the final answer, but I wanted to make sure my logic is correct. Thanks.
     
    Last edited: Nov 4, 2009
  2. jcsd
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