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Question about three states of matter

  1. Jul 11, 2015 #1
    as you have learned about the states of matter in high school, there are three states of matter (just assume i know nth about plasma or any other things else first), and when change of states occur energy is supplied to the matter to increase its potential energy while keep kinetic energy constant.
    my question is that, why the change of states happen discretely but not continuously as temperature increases?(energy can be supplied to increase both KE and PE) Why there are three states instead of two states?

    my guessed answer for the second question is:
    there are more than one kinds of intermolecular forces between each molecules , for example, water molecules have van der waals' force and hydrogen bond, so as temperature increases, van der waals' force is broken so ice is changed to liquid water, and when the temperature further increases hydrogen bond is also broken so there is no intermolecular forces anymore so liquid water is changed to water vapor. And substance like iodine has only van der waals' force, so as temperature increases, it sublimes to gas.

    Is my answer correct?
     
  2. jcsd
  3. Jul 11, 2015 #2
    It may be better to think of the three states resulting from different kinds of attachment.

    Attached and oriented molecules yield a solid.

    Attached and unoriented molecules yield a liquid.

    Unattached and unoriented molecules yield a gas.

    At some energy levels (relative to the bonds), materials fit into one of these three categories.
     
  4. Jul 11, 2015 #3
    The Van der Waals equation of state is useful for understanding the liquid-gas transition. It is more accurate than the ideal gas law, which doesn't have intermolecular forces which are needed for liquid formation.
    ##\left(p+\frac{n^2a}{V^2}\right)(V-nb)=nRT##
    a and b are constants which need to be tabulated for each molecule. Clearly, if a and b go to 0, then you get the ideal gas law.
    You get a PV curve which can look like https://en.wikipedia.org/wiki/Van_der_Waals_equation#/media/File:Van_der_Waals_Isotherms.PNG
    At temperature lower than the critical temperature, you can get a curve that has multiple values for V at a given P. The lower density solution is called a gas, and the higher density solution is called a liquid. The middle solution is unstable and isn't called anything. Now if you start with a liquid at high pressure (left side of the graph), you only have one solution. But if you gradually decrease the pressure, eventually you hit a value of pressure where a second solution for V suddenly appears. This is the boiling point for this value of temperature. As you try to decrease the pressure more, you can't, because the liquid converts to gas keep the pressure fixed. Even though the graph is fully continuous, the behavior is not. Finally, when all the liquid is converted, you can decrease the pressure more.
    There are many resources online which can describe the process more fully
    https://en.wikipedia.org/wiki/Van_der_Waals_equation
    Googling "van de waals phase transition" finds a bunch. Also searching for "3d phase diagram" can be useful.

    The van der Waals equation isn't complete enough to handle the solid-liquid-gas transitions. You need something more complicated. But the appearance of the non-continuous behavior is kind of similar.
     
    Last edited: Jul 11, 2015
  5. Jul 12, 2015 #4

    radium

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    Well actually there are much more than three states of matter. But back to your question, when talking about the conventional solid/liquid/gas phases, the solid/liquid and liquid/gas transition are actually very different kinds of phase transitions. The solid liquid one is a first order phase transitions and the liquid gas is second order phase transition. This means that in the solid-liquid case the free energy is discontinuous where in the second situation it is the derivative of the free energy that is discontinuous (the specific heat). This is because when a solid melts it (usually) goes from a crystal to a liquid breaking translation symmetry. In a solid you only have translation invariance going from cell to cell (the system is periodic) but in an isotropic liquid you have complete translation invariance.

    The liquid gas transition is much different. Liquids and gases differ only in their densities, in terms of symmetries they are the same phase. If you look at the water phase diagram you will see there is a critical point at a certain temperature and anything past that is a gas. At that point, the liquid and gas phases become the same. The Van der waals equation is a good way to look at this transition. So technically you can go around the critical point to get from one phase to the other

    Actually, there is also something called a universality class of a phase transition. It is determined by how observables like the specific heat diverge (there is a discontinuity) at the phase transition. Something that is incredibly interesting is that the liquid gas transition is in the same universality class as the Ising model of a ferromagnet! This is because you can map the models between each other.
     
  6. Jul 13, 2015 #5

    atyy

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    Last edited by a moderator: May 7, 2017
  7. Jul 13, 2015 #6
    I agree; a first order phase transition also (equivalently) means, that there is non-zero latent heat needed for the transition, so yes, a liquid-gas transition is first order (except at the critical point).

    For an imperfect analogy think of it this way: when you heat a pot of water to boiling temperature (100 Celsius), does "all" the water evaporate in unison upon an infinitesimal increase of temperature, or do you have to keep adding more and more heat (energy) to slowly boil the water away bit-by-bit? The answers is: you have to keep adding heat, thus forming bubbles of vapour/steam in the pot, little by little, because the latent (=needed) heat for this phase transition is non-zero, hence the transition is first order.

    Another way to tell that this is a first order phase transition: only first order phase transitions have the possibility to exhibit so-called overheated (superheated) and undercooled (supercooled) metastable states. This means, that as you add (or remove, respectively) heat from the system very gradually, the system could, in a narrow temperature range around the phase transition temperature, temporarily exist in the "wrong" state for a given temperature and pressure, and then suddenly undergo a phase transition upon any ("large enough"; typically even small perturbations will do) outside perturbation, without adding any more heat. (Or at least a part of the system will undergo the phase transition; the fraction of it that does depends on the amount of "extra" heat that was stored as increased/decreased temperature, that could then be used to "pay" for the latent heat needed for the phase transition [bringing the temperature back to the phase transition temperature in the process].) This is only possible for first-order phase transitions (Landau's phenomenological theory of phase transitions explains this very nicely).

    For a real world example: put a glass of water in a microwave and set the microwave to a long enough time, so that the water heats to the boiling temperature and slightly above (timing is everything). If the glass was clean enough (not many scratch marks or dirt) the water will still be liquid while its temperature will be above 100 Celsius! Now put something in the glass (like ground coffee, sugar, a spoon (not recommended, though; you might burn yourself), whatever). What happens is that the water starts to boil instantly upon (and at the point of) the perturbation, while it was completely calm and liquid before. (This has happened to me many times while heating water to make a cup of coffee; watch that you don't accidentally burn yourself!) This was overheated water, a metastable state (i.e. stable for short enough times and without outside interference, but not stable upon larger perturbations). Since overheated (and undercooled) states are only possible when the transition is first-order, the liquid-gas phase transition is, therefore, first order.

    Example:


    Another example (for the liquid-solid phase transition this time): put a clean glass (or a bottle) of distilled water in the freezer. Leave it undisturbed for a few hours (the exact time varies, of course). If you are lucky with the timing, the water will be cooled below the freezing point (0 Celsius), but will still be liquid and no ice crystals will have formed yet. Now take the glass out and: shake it, put a spoon in it, pour it from a height, whatever. The water, which was liquid before, will instantly freeze (at least a part, if not all, of it)! This was undercooled water, and the fact that it is even possible, proves that the liquid-solid phase transition is first order!

    Examples:



    A tutorial I found online:
    http://chemistry.about.com/od/chemistryhowtoguide/a/how-to-supercool-water.htm

    QED, I rest my case :)
     
    Last edited by a moderator: May 7, 2017
  8. Jul 13, 2015 #7

    radium

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    Yes I think that is actually correct when you do not go around the critical point. You do still need to make the distinction between the liquid gas and solid liquid transitions though. I referred to the liquid gas transition as a second order one since when you go around the critical point it's like an Ising model in a finite field. The two states up and down can be tuned into each other smoothly since there is always some magnetization. In that case the zero field line is the first order phase transition since there you start with no magnetization and then end up with one appearing at T=Tc. The two directions are degenerate without a field so you must choose one somehow.
     
    Last edited: Jul 13, 2015
  9. Jul 13, 2015 #8

    atyy

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    If one goes round the critical point, there is no discontinuity, so one usually says there is no phase transition. The second order transition is when one moves along the first order phase boundary, as one increases the temperature from below to above Tc, the first order boundary disappears. So the second order phase transition is at zero external field. Above Tc there is no spontaneous magnetization, but below Tc there is a spontaneous magnetization even at zero external field (technically, in some approaches there there is an external field, but one takes the thermodynamic limit then the zero field limit, eg. http://arxiv.org/abs/physics/0609177 Eq 12).
     
  10. Jul 13, 2015 #9

    radium

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    Yes, sorry for the confusion that makes sense since the magnetization is the derivative of the free energy. But something that is definitely true about the two states available to the Ising model below Tc is that you cannot tunnel from one to the other without breaking symmetry (at least not in an infinite system). So the different states are in different superselection sectors. They are topologically distinct.

    So just to clarify I assume you mean the second order transition in the liquid gas transition is when you cut through the critical point at constant pressure (not go around) and go from liquid to gas or vice versa?

    Another interesting thing is if the field is transverse. Then you only have a transition at T=0. Then The superselection applies when you tune the transverse field. You may ask if you do this can you in a sense rotate the up and down states from one to the other? The answer is no since spins flip in pairs (superselection).
     
  11. Jul 13, 2015 #10

    atyy

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    I don't think of it so much as constant pressure. Looking at http://www.nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_25/node1.html, say the P-T graph, it looks like if you are below Pc, then increasing T one will pass through the first order boundary. At Pc, as one goes from below to above Tc, I don't know if one will see anything "happening".

    I think the usual demonstrations are constant volume. So if one looks at the P-ρ graph, say if one is at the critical density, then below Tc, one should see liquid and gas coexisting. As one increases pressure and temperature, the phase boundary should disappear. Under some conditions, one should also see critical opalescence. This disappearance of the phase boundary is what I usually think of as second order.

    Here is a constant volume demonstration where there is no phase boundary initially, then as the material is cooled, the phase boundary appears:
    :)
     
    Last edited: Jul 13, 2015
  12. Jul 13, 2015 #11

    radium

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    Yes that's what I actually meant, the constant pressure I was referring to was the pressure at the critical point. Pressure is related to density so I guess moving along that represents the changing order parameter which is the difference between the liquid and gas, hence how you map the system onto the Ising model. So the density is like the difference between spin up and spin down but instead it is represented as an empty or occupied site. But then at the critical point the order parameter no longer makes sense as it no longer distinguishes the two phases. Just as there is no spontaneous magnetization above Tc in the Ising model. These phase transitions are most definitely path dependent which makes sense since the order of limits is always very important for these things.
     
  13. Jul 13, 2015 #12

    atyy

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    The discontinuous change relies on there being an infinite number of molecules.

    In real life, there aren't infinite number of molecules, and all "changes in state" are continuous as you imagine.

    However, one usually doesn't measure things with such precision, so the continuous change is well approximated by a discontinuous change (ie. a finite number of molecules is well approximated as an infinite number of molecules).
     
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