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Air Traped In A Capillary Tube

  1. Jan 25, 2012 #1
    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. jcsd
  3. Jan 25, 2012 #2

    tiny-tim

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    hey chikis! :smile:

    this is a fairly straightforward mechanics problem …

    just draw a free body diagram for the forces, in each of the three different situations :smile:
    they mean it's an ordinary tube, the same shape all the way along :wink:
     
  4. Jan 25, 2012 #3
    First of all, I must thank you for your reply. With the answer you gave concerning the "uniform sectional area" I now understand that term clearly.
    But come think of it, talking about the main question, I think the answer you gave me does not in anyway relate to it. What do "free body diagram" has to do with the question. This question relate to Boyle's Law. I told you that am more intrested in knowing what will happen when the tube is held horizonally and then vertically and you are telling me about "free body diagram". What does that has to do with the question?
    I hope you are not offended by this reply. Am only baring my mind.
     
  5. Jan 25, 2012 #4

    tiny-tim

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    is that the one about pressure times volume equals constant?

    how are you going to find the pressure without a force diagram? :wink:
     
  6. Jan 25, 2012 #5
    All you need to do is to draw the capillary tube inform of a testube. In the diagram, indicate the thread of mecury and it lenght that is used in traping the air in inside the tube. Then still indicate the lenght of air collumn that is being traped by the thread of mercury.
    If the question says that the tube is horizontal and open at one end, then show it in your drawing. If the question says the tube is the other way round (vertical), then still show it in your drawing. Then apply Boyle's Law and solve it. I don't no wether that is what you mean by the "free body diagram" you are talking about.
    The issue here is that, I still maintain my point. Am intrested in knowing what will happen when the tube is horizontal and when it will be kept vertical as it will aid in the calculation.
     
  7. Jan 26, 2012 #6

    tiny-tim

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    what's holding the mercury in balance?

    a force on one side, and a force on the other side (and its own weight) :wink:
     
  8. Jan 26, 2012 #7
    I guess you are thinking about the force that is keeping the tube vetical or horizontal. What if the tube is kept in position by a testtube rack or moreover human fingers holding it in grip? How about that?
    Justice has not been done to my question yet. Am mainly intrested in knowing what will happen if the tube is held vertical with the open end uppermost. Am also intrested in knowing what will happen if the tube is held horizontal with one end open. That's what am intrested in for now. If you don't know it, go and find out and report back. Thank You!
     
  9. Jan 26, 2012 #8
    Chikis:
    You are correct to realise that this question relates to Boyles law which gives you the link between Pressure and Volume of a gas when the temperature remains constant.
    When the tube is vertical the mercury column exerts a pressure on the gas but when the tube is horizontal the mercury column does not exert a pressure.
    Does this enable you to get the 2 pressures for the trapped gas?
     
  10. Jan 26, 2012 #9
    Thank you for the reply. Let me use this your explaination and see what I will get as I proceed towards the calculation. Whatever I get I will relay it to the forum. Thank You!
     
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