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Boiling liquid in a sealed container

  1. Aug 12, 2014 #1
    Suppose I have a sealed container of fixed volume containing 3/4 air and 1/4 of some liquid by volume at room temperature (22C). I wish to boil the liquid by applying external heat.

    As I apply heat to the container, the pressure inside the container will rise, which will change the boiling point of the liquid. How do I calculate the temperature needed to boil the liquid under these conditions?

    If it matters, we can assume any boiling point in the range of Acetone (50C) to Ethyl Bromide (38C)

    FWIW, I'm playing with the idea of building a heat engine based on low temperature differentials and operating at near room temperature. Since this would be a hobby thing, the working fluid needs to be affordable, available, not encumbered by heavy regulation, and not TOO nasty to work with.

    Thanks In Advance.
  2. jcsd
  3. Aug 13, 2014 #2
    The equation is given in the Wikipedia article on boiling point. Heating a pressurised vessel of either acetone or ethyl bromide is not something you want to do as a "hobby thing".
  4. Aug 13, 2014 #3
    Mr. A:

    Thanks for your quick response. Check me on this:

    By Gay-Lussac's Law, and assuming the space above the liquid stays "air," the pressure (bar) goes up linearly with temperature (K), so 1.0 bar at 22 C or 295 K, and about 1.26 bar at 100 C or 373 K.

    Vapor pressure of the liquid is predicted by the Antoine formula, so for Acetone about .27 bar at 22 C/295 K, 1.0 bar at 56 C/329 K (boiling point at normal atmosphere), and about 3.7 bar at 100 C/373 K.

    The lines cross at about 60 C/333 K and 1.13 bar/16 PSI, which would be roughly the point I'm looking for, off by a bit because the "air" above the liquid isn't air any more.

    Right? Thanks.
  5. Aug 13, 2014 #4
    No. Not right. The space above the liquid does not "stay air." The partial pressure of the vapor in the head space will increase and be equal to the equilibrium vapor pressure at the liquid temperature. So the total pressure in the head space will be equal to the air partial pressure plus the equilibrium vapor pressure of the volatile component. The total pressure will always be greater than the equilibrium vapor pressure so that, unless the liquid is heated very non-uniformly, the liquid will not boil.

    Starting from your initial conditions, you can accurately calculate, for any given temperature of the vessel contents, (a) the amount of liquid (b) the amount of vapor in the head space (c) the partial pressure of the air (d) the total pressure. You can model this by treating the gases in the head space as an ideal gas, knowing the vapor pressure equilibrium relation, and assuming the density of the liquid is approximately constant.

  6. Aug 14, 2014 #5
    Chet, thank you.

    It's amazing what a little study will uncover. It appears that the physics of this thing are approximately those of a "two-phase thermosyphon" which is a wickless heat pipe. The big surprise for me is that such beasts do not need to operate at or near the boiling point of their working fluid if the tube is evacuated, and in fact can begin to function at just above the freezing/melting point of the fluid, which changes the whole dynamic.

    There is a parameter called the "merit number" of the working fluid, which depends on the operating temperature of the device. For near-room-temperature heat pipes both water and acetone have their supporters. (The merit number was developed for use with heat pipes that use wicks, and one of its factors has to do with capillary transport, which seems to be irrelevant to the two-phase thermosyphon, which has no wick, so the merit number may be misleading for this application.)

    Anyhow, I've put as much time into the subject as it warrants for now. My thanks to Mr. A and Chet for their responses.
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