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A U-shaped Tube problem

  1. Feb 15, 2014 #1
    A U-shaped Tube problem....

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

    A u-shaped tube with a 1 square centimeter cross-section is filled with mercury to a depth of 50 centimeters on each side.

    One side is capped and the air pressure is 750 Torr. The other side is connected to a vacuum pump.

    a) how far does the mercury in the dupe drop on the left side? b) what is the final pressure of the air there?

    (We are assuming a constant temperature)

    2. Relevant equations

    I'm using the old perfect gas law, PV = constant.


    3. The attempt at a solution

    OK, so I know that the mercury will move h centimeters on one side (down) and h on the other (up). So the total distance the mercury moves is 2h.

    If I assume that the pressure in the part of the tube where the air is starts at P0V0[itex] and goes to PfinalVfinal I can do this:

    [itex]P_0V_0 = P_{final}V_{final}[/itex] because both PVs = nRT.

    I also know that the pressure exerted by the mercury is [itex]50{\rho}_{Hg}[/itex] because there are 50 centimeters worth of it at the start.

    Since the pressure of the air increases the volume on the left side I know that

    [itex]P_{final}V_{final} = P_{final}(V_0+h)[/itex] and since the cross section is 1 cm2 we don't have to worry about area here, it's just 1.

    Meanwhile, the mercury will (once everything settles to equilibrium) exerts the same pressure, but it will be [itex]{\rho}_{Hg}(50+2h)[/itex].

    That should give me [itex]P_{final}V_{final} = P_{final}(V_0+h) = {\rho}_{Hg}(50+2h)(50+2h)[/itex] and since we know V0 (It's 50) then:

    [itex]P_{final}V_{final} = P_{final}(50+h) = (50{\rho}_{Hg}+2h{\rho}_{Hg})(50+2h)[/itex]

    And this is where I suspect I am going wrong. Or maybe I just did something algebraically that messed it up. I don't thnik I should be getting h2 factors.

    I know the answer it supposed to be 25 centimeters and 500 Torr for the pressure of the air. And sort of intuitively I know it. But I am trying to work this out correctly. I've seen similar problems on the forum here but none quite like this.

    EDIT: thanks in advance.
     
  2. jcsd
  3. Feb 16, 2014 #2

    ehild

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    The mercury is at depth 50 cm in both tubes initially. At the end, it stands 2h higher in the right tube with respect to the mercury level in the left tube. With respect to that level, the pressure must be the same at both sides: At left, it is the pressure of the air. At right, it is the pressure of the 2h high mercury column.


    ehild
     

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  4. Feb 16, 2014 #3
    OK, then would that mean the expression I want is

    [itex](\rho_{Hg}2h)(2h) = P_{final}(50+h)[/itex]?

    because when I use that we get an h2 term and I have a quadratic with no real solutions.
     
  5. Feb 16, 2014 #4

    ehild

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    No. Pfinal is Po(50/(50+h))=2hρHg.

    ehild
     
    Last edited: Feb 16, 2014
  6. Feb 16, 2014 #5
    OK, then that means for the mercury side of the equation it's

    [itex] (\rho_{Hg}2h)(2h)[/itex] -- but since [itex](\rho_{Hg}2h)=P_{final}[/itex]

    the left expression is

    [tex]\frac{(P_0)(50)(50+h)}{50+h}=\rho_{Hg}2h(2h)[/tex]

    which becomes
    [tex]P_{final}(50)=P_{final}(2h)[/tex]

    Tht seems right yes?
     
    Last edited: Feb 16, 2014
  7. Feb 16, 2014 #6

    ehild

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    That equation is wrong. Write the equation for the pressures at both sides.

    ehild
     
  8. Feb 16, 2014 #7
    I'm a bi confused here. You said the [itex]P_{final} = \frac{(P_0)(50)}{50+h}[/itex]

    That means to me that since that pressure has to be the same as the pressure on the other side, and the pressure on that side is [itex](\rho_{Hg}2h)[/itex]

    you should get something like [itex](\rho_{Hg}2h) = \frac{(P_0)(50)}{50+h}[/itex]

    But that by itself doesn't work. If you solve for H you end up with

    [itex](2h)(50+h) = \frac{(P_0)(50)}{\rho_{Hg}}[/itex]
    [itex](100h+2h^2)=\frac{(P_0)(50)}{\rho_{Hg}}[/itex]

    Which wen you plug the numbers in (pressure is in Torr, so he Hg density factor cancels out)

    [itex](100h+2h^2)=(750)(50) [/itex]
    [itex]100h+2h^2 = 37500[/itex]

    That means to me that you need to multiply both sides by the volume (the PV expression) which gets me the cancellation I had before

    [itex](\rho_{Hg}2h\bf2h) = \frac{(P_0)(50)}{50+h}\bf(50+h)[/itex]

    Didn't I put the expression for pressure on both sides here? thanks
     
  9. Feb 16, 2014 #8

    ehild

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    Right, but the initial pressure is 750 torr which is the pressure of a 750 mm =75 cm high mercury column. The hight of the mercury in the tube is measured in cm-s: 2h=100 cm. Change 750 mm to 75 cm.

    The correct equation for h is [itex]100h+2h^2 = 3750[/itex] Solve it for h.

    ehild
     
    Last edited: Feb 16, 2014
  10. Feb 16, 2014 #9
    Arrggh! Thanks so much. That was a stupid stupid error.
     
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