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A thermodynamics problem

  1. Aug 6, 2013 #1
    1. The problem statement, all variables and given/known data
    A test tube of length [itex] l=16 cm[/itex] is completely immersed with its uncovered end in a mercury vase. The initial pressure of the air is the atmospherical pressure, and its initial temperature is [itex] t=27 degrees Celsius[/itex] so that the column of air has a height [itex] h= 40 mm[/itex]. We extract the eprubete so that its uncovered remains in contact with the mercury. How much do we have too cool off the temperature so that the height of the column of air stays the same?

    To better ilustrate the problem I drew a picture.
    untitled_zps782cb277.png

    2. Relevant equations

    Well, if the height of the column of air remains the same we have an izocore transformation.

    3. The attempt at a solution

    I belive that the solution of this problem is found writing and solving the equation of the izocore transformation. But I don't know how to find the initial and the final pressure.
     
    Last edited: Aug 6, 2013
  2. jcsd
  3. Aug 6, 2013 #2
    The correct spelling is isochoric.

    The overall process is not isochoric, but if you consider its start and end, they do have equal volumes, and that may be useful.

    To find the pressures, consider what pressure is required to displace the given height of mercury in the first configuration, and what pressure is required to hold the same height of mercury in the second configuration.
     
  4. Aug 6, 2013 #3
    I get it. Sorry for my English, I'm not a native speaker.

    I get that too.

    I don't know if I can find out that. I don't know how to think the phenomenon. Can you help me?
     
  5. Aug 6, 2013 #4
    Let's consider the submerged situation first. Anywhere below the surface, mercury has some particular hydrostatic pressure. Do you know how to compute it?

    Now, because the mercury is prevented from occupying the whole tube, something must oppose its pressure - what is it?
     
  6. Aug 6, 2013 #5
    Yes, I know. But within the volume of liquid that the tube covers is there a hydrostatic pressure? As far as I know, the hydrostatic pressure is due to the column of liquid found above a certain level. The only hydrostatic pressure I see in that zone is:

    [itex] \rho g(l-h)[/itex]

    Well, that would pe the air pressure in the tube, correct?

    By the way, thank you very much for your answers!
     
    Last edited: Aug 6, 2013
  7. Aug 6, 2013 #6
    Where exactly do you think you have this pressure?
     
  8. Aug 6, 2013 #7
    Well, I've just drawn a black line where I think that pressure is. I know that I have some holes to fill in, concerning my reasoning for solving thermodynamic problems .That's why I have chosen to post on this forum.
    untitled_zps15fb3b13.png
     
  9. Aug 6, 2013 #8
    Let's see what's happening just at the edge of the tube. Inside the edge, the pressure, according to you, is ## \rho g (l - h) ##. What is it outside the edge (at the same depth)? Are the two equal? How is that possible? What about the pressure of air inside the tube - does it affect anything?
     
  10. Aug 6, 2013 #9
    Well, outside the edge there is a hydrostatic pressure: [itex]ρg(l−h)[/itex]

    Well no, there aren't equal. Inside the edge the pressure is:

    [itex] ρg(l−h) + p[/itex] where p is the pressure of the air in the tube.

    At the same level, outside the edge of the tube the presure is just [itex]ρg(l−h) [/itex].

    Correct?
     
    Last edited: Aug 6, 2013
  11. Aug 6, 2013 #10
    Why? The depth of the liquid here is ## l ##, not ## l - h ##. Plus, there is atmosphere over it.

    Correct.
     
  12. Aug 6, 2013 #11
    That's right, I was uncarfeull. So, the pressure outside the edge is:

    [itex] p_0 + ρ gl[/itex], where [itex] p_0[/itex] stands for atmospherical pressure.

    But I have a question. Shouldn't we take into account that inside the edge, besides the pressure due to the air in the tube there is atmospherical pressure?
     
  13. Aug 6, 2013 #12
    How? It was atmospheric before the tube was submerged. Then some volume was occupied by the mercury. Can the pressure remain the same, assuming the temperature did not change?
     
  14. Aug 6, 2013 #13
    What I meant is shouldn't the pressure inside the edge be something like:

    [itex] ρg(l−h)+p+p_0[/itex]. I mean, dosen't the atmospherical pressure above the tube has some sort of effect?
     
  15. Aug 6, 2013 #14
    The pressure on top of the tube does not have any effect, and here is why: the tube is held submerged by an additional force. The sum of the force due to the external atmospheric pressure and this additional force must be equal to the force exerted by the internal pressure (which is greater than atmospheric). Alternatively, we could say that the pressure just on top of the tube is exactly equal to the internal pressure inside the tube. Clearly, we cannot take this pressure into account two times, it would make no sense.
     
  16. Aug 6, 2013 #15
    I get it now! So, how do we continue the problem?
     
  17. Aug 6, 2013 #16
    So what is the pressure in the submerged case?

    How could you approach the other case?
     
  18. Aug 6, 2013 #17
    I belive I get it now. Regardin the line I drew; the pressure at the edge inside the tube, and the pressure at the edge outside it must be equal, otherwise the liquid wouldn't be in equilibrium, right? And I assume we're dealing with a liquid that is in equilibrium.

    For the final state, the things aren't different. At the surface of the liquid the pressure within the tube, and the one outside it must be also equal for the same reason stated before.

    Am I right?
     
  19. Aug 6, 2013 #18
    All you said is correct.
     
  20. Aug 6, 2013 #19
    Thank you for your guidance! I've managed to solve the problem. It was pretty easy, after you made me understand what's the deal with the pressures
     
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