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Eqilibrium between ice and water

  1. Jan 19, 2006 #1
    When ice is @ eqilibrium with water, the pressure is increased, what happens to the eqilibrium?
     
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  3. Jan 19, 2006 #2

    DaveC426913

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    I'm not an expert in this area, but I do know that when you apply pressure to ice, it will begin to liquify. This is the physics behind most winter sports - from skating to curling.
     
    Last edited: Jan 19, 2006
  4. Jan 19, 2006 #3

    vanesch

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    It shifts towards the water, and that is because the density of ice is smaller than the density of water (which is quite an exceptional feat : most materials are denser in their solid phase than in their liquid phase).

    It is a consequence of a general law of thermodynamical equilibrium (which you can derive more formally, but I cannot recall its derivation off the top of my head): "any action on a system in equilibrium will shift its point of equilibrium in such a way as to oppose the change you want to apply"

    So, when you increase pressure, the system can try to relieve pressure by shrinking in volume (hence shifting towards the most dense phase).
     
  5. Jan 21, 2006 #4
    I thought pressure was inversely propotional to volume. So if u increase pressure volume should also increase to decrease the pressure
     
  6. Jan 21, 2006 #5

    vanesch

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    Eh, your remark makes me see how badly I formulated my proposition.

    Imagine you have 2 phases F1 and F2, and imagine that the specific volume of phase F1, as a function of pressure p, is v1(p), and the specific volume of phase F2 is v2(p) ; now, assume that v1(p) is much smaller than v2(p).

    The total volume of the system is V = N1 v1(p) + N2 v2(p) for a given distribution of the substance over the two phases N1 in phase F1 and N2 in phase N2, which happens to be a certain equilibrium point.

    Now, when pressure is increased from p to p', one can show that in general, N1 will increase, and N2 will decrease, so that the phase with the smallest specific volume will gain in importance, hence lowering the overall volume MORE than if N1 and N2 remained constant. As such, the pressure needed to put the system in volume V' (with the changing N1 and N2), is LESS than the pressure that is needed to put the system in volume V' WITHOUT shifting the equilibrium. So it is loosely said that the increase in pressure is "resisted" by the system by shifting its own degree of freedom (the point of equilibrium N1/N2). So the shifting equilibrium will 'give in more' than you would expect if the ratios remained constant (if the equilibrium would not shift).
     
  7. Jan 23, 2006 #6
    If you assume water and ice are both incompressible, pressure will not decrease the volume of either. In general, extra pressure means extra heat (like putting air in a tire) so you might argue that the temperature of the mix will increase and that would cause some ice to melt. You could also argue that water is denser, so increased pressure will compress the ice more...that would also tend to increase the temperature of the ice, so either way it looks like ice loses!!
     
  8. Jan 23, 2006 #7

    vanesch

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    No, pressure does not always mean heat ; the "heat" comes from the work done by the pressure on the schrinking volume. If the volume doesn't change, then increasing the pressure does not do work. But in fact, it is the shifting in equilibrium from the ice to the water phase, which allows for some work to be done by the pressure (because of the decrease in volume, remember that water takes less volume than ice), and THIS work (the pressure times the decrease in volume) IS indeed converted into (latent) heat, to allow for a bit of the ice to melt (and to shift the equilibrium in the first place).

    But that was maybe what you wanted to say too:shy:
     
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