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What happens when tube is immersed in a liquid?

  1. Aug 8, 2013 #1
    1. The problem statement, all variables and given/known data
    A tube of length L=1 m closed at its upper end is immerged vertically, with its openend, in water by a distance l=27 cm. The atmospherical pressure is the normal one. Find the length of the water column that enters the tube. Consider the process isothermal.


    2. Relevant equations

    Since the process is isothermal the produc between the pressure of the gas and its volume is constat. I belvie we also need the formula for the hydrostatic pressure.


    3. The attempt at a solution

    Here's what bugs me. When we Immerge a tube in the conditions of the problem, what is the correct situation? A or B? What I mean is, does the level of the liquid in the tube stabilize below the level of the liquid in the vase, or above it?

    whatiscorrect_zps468486ad.png

    I say that the correct situation is B, because we "push" the liquid with a certain force due tot the air molecules inside the tube, but I'm not sure. I found situation A quite plasauble.

    Please, help me! :(
     
  2. jcsd
  3. Aug 8, 2013 #2
    Explain why normally still liquid would "shoot up" if covered by a closed tube.
     
  4. Aug 8, 2013 #3
    Well, I can't find a reason. If the situation were a little bit different, then I would be able to explain. For example, consider a cilinder divided into two containers. The containers are separated by a movable piston. The reason why the half below the piston would "shoot up" is that the pressure of the gas molecules below has overcome the pressure of the gas molecules above.

    I can't find a reason why a normally still liquid would "shoot up" if covered by a closed tube, because the liquid dosen't exert a certian pressure to the gas. The liquid would exert a hydrostatic pressure, if it were above the gas. But it is below.
     
  5. Aug 8, 2013 #4

    HallsofIvy

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    First, is the tube closed on the top, as it appears to be in the pictures? Then the air in the tube must be compressed by some pressure greater than that acting on the top of the liquid. And that is only possible if the air presses down into the liquid.

    If the tube is NOT closed on the to then the surface tension of the liquid will cause it to "creep" slightly up the tube.
     
  6. Aug 8, 2013 #5
    Yes, the tube is closed.
     
  7. Aug 8, 2013 #6
    The liquid at the surface does have a certain pressure. This is the pressure of the atmosphere that surrounds it. Will that cause the liquid to shoot up if you insert a tube, closed at the top, with air at the atmospheric pressure?
     
  8. Aug 8, 2013 #7
    If any liquid enters the tube, the pressure within the air in the tube will be higher than atmospheric. That would mean that figure B is correct. Use the ideal gas law to express the new pressure within the gas in terms of the tube length L, and the distance d. How does the pressure at the bottom of the tube have to be related to the pressure in the remainder of the reservoir at the depth h?

    Chet
     
  9. Aug 8, 2013 #8
    How can that be? The liquid is below, it dosen't presses on the air...
     
  10. Aug 8, 2013 #9
    This would be easy, but I still don't understand why the liquid dosen' "shoot up".
     
  11. Aug 8, 2013 #10
    In a similar problem you solved recently, you used that :)

    The air above the liquid exerts some pressure on it, The liquid has no choice but to be at the same pressure just below the air, and develop additional pressure due to its own weight as depth increases.
     
  12. Aug 8, 2013 #11
    Ok, I totally agree, and I strongly belive that it has something to do with Newton's third law, right? but I can't understand why the liquid dosen't behave like in situation A. It's such a simple thing, and I don't ubderstand it.

    Just to be clear, do you mean that at the green line there is a pressure on the air due to the liquid? That is due to action-reaction theorem principle, right?
    whatiscorrect_zpse62490ac.png
     
    Last edited: Aug 8, 2013
  13. Aug 8, 2013 #12
    Something must force the liquid into the tube for situation A to happen. What could that be?
     
  14. Aug 8, 2013 #13
    A difference of pressure between the air in the tube and the liquid, at the contact point between them?

    And please, I want to know if there is a pressure at the green line.
     
  15. Aug 8, 2013 #14
    What pressure would you have in the air in the tube in situation A?

    I think we have already covered this. But again: what is the pressure at the surface of a liquid surrounded by atmosphere?
     
  16. Aug 8, 2013 #15
    No problem. Just work out the formulation of the equation. Here's a hint. In the formulation, it doesn't matter whether you assume that figure A prevails or figure B prevails. The formulation will be the same for both figures. The solution to the equation will tell you which picture is correct and, more importantly, why. So, as I said in my previous post, use the ideal gas law to express the new pressure within the gas in terms of the tube length L, and the distance d and atmospheric pressure pa. What is your equation for the pressure of the gas above the liquid interface in the tube?

    Chet
     
  17. Aug 8, 2013 #16
    Oops. I just saw that you added a question in an edit. I advice against adding things in edits, they are too easy to overlook. Just add another post.

    Consider a very small volume of liquid of a negligible mass just at its surface. We know it does not move with time, so its velocity and acceleration are zero. By Newton's second law, the sum of all the forces must then also be zero. But we have a downward force due to the atmospheric pressure. Something then must cancel this force out. What is it?
     
  18. Aug 8, 2013 #17
    (@Voko)

    Well, let's consider the lower end of the tube. At that level, the pressure must be the same everywhere. So, we have (consider p_0 the atmospherical pressure):

    [itex] p_0+\rho g h=p_A+ \rho g d => p_A=p_0-\rho g (h-d) [/itex].

    Well, the atmospherical pressure.

    I belive is the force that the surface of the container exerts on the liquid, a.k.a the normal force?


    Then, let's take the A situation. Consider a intial situation, when the tube begins to make contact with the surface of the water, and the A situation as the final one. In the initial situation, a volume [itex]Sl[/itex] of air at atmospherical pressure, [itex] p_A[/itex], will be enclosed in the tube. So we have:

    [itex] p_1=p_a
    V_1=Sl
    [/itex]

    For the final state:
    [itex]
    p_2=\rho g(h-d)+ p_a

    V_2=(L-d)S
    [/itex]

    As this is an isothermical transformation we have that: [itex] p_1V_1=p_2V_2=>

    0= \rho ghd^2-d(\rho gL+\rho gh+ p_a)+\rho ghL [/itex] which is a quadratic equation.
     
  19. Aug 8, 2013 #18
    No, that should be ## p_A=p_0 + \rho g (h-d) ##. And because ## d > h ##, you have ## p_A < p_0 ##. Now, how could that happen in a tube with its top closed?
     
  20. Aug 8, 2013 #19
    How can the surface of the container exert a force on the surface of the liquid?
     
  21. Aug 8, 2013 #20
    In both situations A and B, by the ideal gas law, the pressure of the air within the tube is [itex]p_a\frac{L}{L-d}[/itex]. The pressure at the bottom of the tube is then [itex]p_a\frac{L}{L-d}+ρgd[/itex]. This must match the pressure within the reservoir at the depth of the bottom of the tube [itex]p_a+ρgh[/itex]. Equating these pressures gives (at static equilibrium):
    [tex]p_a\frac{L}{L-d}+ρgd=p_a+ρgh[/tex]
    All that needs to be done us to solve this equation for d, using the quadratic formula.
     
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