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U-Tube (Pv=nRT, Pascals) Problem

  1. Feb 10, 2015 #1
    Challenge problem

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


    A U-tube of uniform cross section contains mercury. The initial temperature of the system is 301K and the barometric pressure is 749 torr. Suppose the left side of the tube is closed at the top and the temperature is raised to a temperature such that the air column on the left is 60cm high. Determine the temperature at which this occurs.

    Initially, the mercury level is 50 cm on both sides. The height of each tube, as measured from the bottom, is 100 cm.


    2. Relevant equations

    Ideal gas law PV=nRT

    Pascals principle P = P0+ρgh

    3. The attempt at a solution

    I know what my initial pressure and temperature is. In addition I can write my initial volume on the left tube as A(.50) .

    My goal is to find the temperature when the the column of mercury on the left tube drops by 10cm (air column now 60cm).

    So now I want to develop an expression to describe the final state of the left side of the tube.

    Im assuming the number of moles will not change & R is a constant of course.

    PliVli/(Ti) = PlfVlf/(Tf)

    So now I have one equation and two unknowns ( I dont know the final pressure or temperature, the volume can be written in terms of AH). So I need to develop another expression. Here is where I can take full advantage of pascals principle.

    When the mercury goes down on the left side & goes up on the right side, the pressure on the left side of the tube, at the "surface" of the mercury, must be equal to the pressure on the right tube at the same point.

    The pressure on the right side of the tube can be written as

    P = Pairhggh

    Where, h= 0.20m . I choose 0.20m because when the left side goes down by .10m the right side goes up by .10m. Thus if I want to look at the pressure on the right side, I must acknowledge that the point is 0.20m below the surface of the liquid.

    So now I can say Plf = P = Pairhggh

    ∴ I am now ready to solve for Tf .

    Issue is, when I went to solve for Tf , I got an answer that did not correspond to one of the solutions.

    here are the values I used

    ρhg = 13500 kg/m^3
    h = 0.20m
    Pair = 101325 pa
    Pli = 749 torr = 99858 pa
    Vli[/SUB = A(0.50m)
    Vlf[/SUB = A(0.40m)

    And of course the initial temperature is 301 K.
     
  2. jcsd
  3. Feb 10, 2015 #2

    Bystander

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    Could we see your work, please.
     
  4. Feb 10, 2015 #3
    PliVli/(Ti) =PlfVlf/(Tf)

    Tf =PlfVlfTi /(PliVli)

    Note:
    Plf = Pair + ρhggh = (101325)+(13500)(9.80)(.20) = 127785 pa

    Subbing in values:


    Tf = (127785) (A(0.40))(301) / ((99858)(A(0.50)) = 308 K
     
  5. Feb 10, 2015 #4

    ehild

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    What is the final volume of the gas on the left?
     
  6. Feb 10, 2015 #5
    I argue that the final volume on the left side will be, Vlf = A(0.40) .

    The 0.40 is the height of the fluid on the left.

    Is this a correct statement?
     
  7. Feb 10, 2015 #6

    ehild

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    Why did you use the height of the fluid when the gas law applies to gases?????
     
  8. Feb 10, 2015 #7

    Bystander

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  9. Feb 10, 2015 #8
    Goodness me, I have no idea.

    I now used (0.60) instead of (0.40) , I managed to get about 460K now.
     
  10. Feb 10, 2015 #9
    I dont follow, is the pressure given not an absolute pressure?
     
  11. Feb 10, 2015 #10

    Bystander

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    Yes you do.
     
  12. Feb 10, 2015 #11
    749 torr =/= barometric 749 torr ?
     
  13. Feb 11, 2015 #12

    ehild

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    You can measure the pressure in torr-s - the height in mm of a mercury column of equivalent pressure. So the 20 cm mercury column means additional 200 torr pressure. No need to convert to pascals.
    The density of mercury depends on temperature, and so does its length, but assuming that only the air column is heated you can ignore it.
     
  14. Feb 11, 2015 #13
    This is quite new to me, so excuse me if I sound a bit ignorant, but

    1 cm of mercury column constitutes 10 torr pressure?

    Therefore when I consider the 20cm mercury column, I should add 200 torr pressure. Consequently, I should account for the 40cm mercury column on the left tube?
     
  15. Feb 11, 2015 #14

    ehild

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    The 40 cm column in the left tube is below the air column, it does not add to the pressure. See attachment. The downward force at the red level has to be equal in both tubes, so the air pressure in the left tube is equal to the atmospheric pressure+ the pressure of the 20 cm mercury column.

    mercurymanometer.JPG
     
  16. Feb 11, 2015 #15
    Very clear explanation. Got it, thank you!
     
  17. Feb 11, 2015 #16

    ehild

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    You are welcome :).
     
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