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Air Trapped In Capillary Tube

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Hey folks, here is a question:
    A thread of mercury of lenght 15cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the lenght of the trapped air collumn is 20cm. Calculate the lenght of the air column when the tube is held.
    (i) Horizontally,
    (ii) Vertically with the open end underneath.
    (Atmospheric pressure= 76cm of mercury)
    In this question, am intrested in knowing what will happen when the tube is held horizontal and vertical as it will help in the calculation. What do they mean or it imply by refering the tube as having "uniform cross-sectional area"?
    Thank you as reply.

    2. Relevant equations

    I know that this relates to boyle's law equation.
    P1V1=P2V2

    3. The attempt at a solution
    V2= P1V1/P
    P1=15cm i.e lenght of mecury because lenght is directly propotional to pressure.
    I cannot go further because I don't know how to get my P1 and P2 respectively in case
    (i)
    (ii).
     
  2. jcsd
  3. Apr 17, 2012 #2
    Does it mean no body has an answer to my question?
     
  4. Apr 17, 2012 #3
    With the open end upwards the gas pressure is equal to atmospheric pressure(P1=H) plus the pressure due to the column of mercury of length h,(P2=hpg............p= density of mercury).Total pressure(P) is given by
    P1+P2

    P=H+hpg

    With the open end downwards the gas pressure is given by:

    P=H-hpg

    With the tube horizontal the pressure is given by

    P=H

    (You could express the pressure in terms of centimetres of mercury.The average value of atmospheric pressureat sea level is taken to be 76cm Hg.If you do this the three pressures (in cm of Hg) will be given by 76+15,76-15 and 76)
     
    Last edited: Apr 17, 2012
  5. Apr 17, 2012 #4
    Thank you so much. But I still have some questions concerning your explaination:
    If the gas pressure is equal to the atmospheric pressure. Why or what principle support that?
    You said that (P1=H) plus the pressure due to the column of mercury of length h,(P2=hpg............p= density of mercury), I understand it to mean that P1, the initial pressure is equal to the atmospheric pressure, H in addition to the pressure of the mecury. Right? You then said that total pressure(P) is given by
    P1+P2. I understand that statement to mean that, initial pressure, P1 and final pressure, P2 to be equal to P the pressure.
    You then wrote;
    P=H+hpg.
    What is hpg? Can I express the statement like this;
    P1+P2 = H+hpg?
     
  6. Apr 17, 2012 #5
    The mercury at the open end of the tube is in contact with the atmosphere and therefore at atmospheric pressure.When the open end is facing upwards the trapped air is supporting the pressure due to the atmosphere plus the pressure due to the weight of the mercury column.When facing downwards the atmosphere is supporting the pressure due to the trapped air plus the pressure due to the mercury column.When horizontal we can ignore the pressure due to the mercury column because its weight has negligible horizontal component.
    For this problem it is quicker to express the pressures in terms of cm of mercury rather than use hpg.

    (I would advise you to look up where hpg comes from.Try googling "hydrostatic pressure")
     
  7. Apr 17, 2012 #6
    Am satisfied with the explaination concerning when the open end of the tube is uppermost.
    What about when the open end of the tube is facing downward?
    I understand atmopheric pressure as acting downward just like force of gravity.
    If the atmospheric pressure is still acting downward on the tube even when it's open end is facing downward, how come do you say that the atmosphere is supporting the pressure due to the trapped air plus the pressure due to the mercury column?
     
  8. Apr 17, 2012 #7
    The pressure at any point in a liquid or gas acts equally in all directions and increases with depth.It's like being in a swimming pool.The pressure of the water pushes on all submerged parts of your body but the deeper parts experience a greater pressure this providing an upward force.Its the principle behind floating.
    Another way of looking at it is in terms of gas molecules.These move randomly continually collide and exert a force and pressure on any surface they are in contact with.
     
  9. Apr 17, 2012 #8
    So how does that relate to the topic on discussion?
     
  10. Apr 18, 2012 #9
    It relates to the topic under discussion because it answers your previous question.You stated that "atmospheric pressure is acting downwards" and I said that at a point it acts "equally in all directions".
    It can be described in terms of the moving molecules colliding with and exerting a force on any surface,such as the mercury surface at the end of the tube.
    Expanding on the above when the open end of the tube is upwards the force due to the atmosphere is acting downwards on the mercury surface and when the open end is downwards the force due to the atmosphere is acting upwards on the mercury surface.
     
  11. Apr 18, 2012 #10
    Thanks so much for that explaination. Am now cleared with that one.
    From what we have discussed so far, how do I get my P1 and P2 in each case (i) when the tube is horizontal with one end open and (ii) when the tube facing downward with the open end.
     
  12. Apr 18, 2012 #11
    As I said earlier for this problem it would be quicker to consider the pressure in terms of cm of mercury(cm Hg).Average pressure of the atmosphere at sea level =76cm of Hg.Look again at post three and you will see that the pressures,in cm of Hg are:

    1. 91 (open end upwards)
    2. 76 (open end horizontal)
    3. 61 (open end downwards)
     
  13. Apr 18, 2012 #12
    Please can you classify them based on P1 and P2 in each case respectively?
     
  14. Apr 18, 2012 #13
    Chikis,before you go any further may I suggest that you check out "fluid pressure" by googling.I recommend that you go to "hyperphysics" and then search the index for "barometers".When you click you will find a string of short pieces,including one on barometers, which are relevant to your question.I just had a look and I think hyperphysics covers the topics very well.Afterwards re read my posts two and three.
    Without some knowledge of the principle of barometers and the like it will be difficult to understand some of the points I made in my earlier posts.If you get stuck then come back here.
     
  15. Apr 18, 2012 #14
    Okay, thanks, I will do as you have advised. You have really lectured me a lot. Am very grateful. I will be back later whenever am through with my googling exercise. Thanks!
     
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