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    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #1

    Oct 4, 2010, 10:00 AM
    termodynamics problem
    In the middle of a tube made of glass set horizontally (length: l=1m) there is a column of quicksilver (length: h=20 cm). When the tube is put vertically, the column descends with Δl=10 cm. What is the pressure of the tube in horizontal position?

    I tried to use the formula p1v1=p2v2
    v1=(l-h)Δl
    p2=ρgh
    then I wrote the two pressures in vertical position: (l-h)/2-Δl and (l-h)/2+Δl but I don't know how to put them together for v2.
    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #2

    Oct 4, 2010, 10:02 AM
    Δl means Delta and ρ means Rho
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #3

    Oct 4, 2010, 10:27 AM

    Do you have any data? I don't know where to start for this one?

    I think we'll need the value of


    Type in (math)\rho(/math) but replace the () by []

    Maybe like this:



    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #4

    Oct 4, 2010, 11:06 AM
    The value for is
    The final answer is
    I hope this helps.Thank you.

    this is the correct form

    or not... sorry I am in a hurry in the first parenthesis I should have written "-" instead of "+"; [\rhogh] should have been and [\delta?] should have been . I'm sorry again and thank you (I finally did it :)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Oct 4, 2010, 12:48 PM
    EDIT: Look at ebaines' answer. I was hesitant about whether to consider the effect of both gases and he confirmed that we needed to take both into consideration.


    I don't know what you did exactly :confused:

    Am I missing something?



    The pressure where the silver is on the gas that I marked on the arrow is where the pressure of the gas and the pressure of the silver are equal.

    So, pressure of gas =

    So, using



    Or am I missing something?

    I never had such a problem with pressure like this before though :(
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Oct 4, 2010, 02:29 PM

    It's hard to follow the math that you're trying to write out, but I would approach it like this:

    In the vertical position the quicksilcver is supported by the difference between the air pressure below and the air pressure above. You know that the pressure below is 4/3 times the original pressure P (this comes from the concept that P1V1 = P2V2), and similarly the air pressure above is 4/5 times the original. Hence the difference in air pressure times the area it works on must equal the weight of the quicksilver:




    Solve for P.
    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #7

    Oct 5, 2010, 06:09 AM
    I succeeded in solving it with both methods (mine and ebaines') and I got the same result, so I think it must be correct. Thank you for help.

    My method:

    where:
    is the pressure and the volume when the tube is placed horizontally
    the pressure above is A and the volume
    the pressure below is and the volume

    but ebaines' method is by far much more simple.

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