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Depression of freezing point

  1. May 20, 2017 #1
    When solute particles are added to a solvent boiling point increases, I am quite clear with its explanation (in terms of vapour pressure) but when it comes to freezing point, why does it decrease? What does it have to do with vapour pressure? My textbook says it's because liquid and solid state of the solvent exist in equilibrium at freezing point and there is a vapor pressure due to this, but how can a liquid or solid exert vapour pressure?
    Also, why doesn't increase in boiling point depend upon the nature of solute particles, I would imagine their size, volume and other properties would affect this but apparently they don't?
     
  2. jcsd
  3. May 20, 2017 #2

    Borek

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    Does the liquid dry without a boiling?

    Have you heard about sublimation?
     
  4. May 20, 2017 #3
    But sublimation happens only for a special category of solids: volatile solids. But I am not talking specifically about volatile solids, I want to know what is the cause of vapour pressure of any solution at its freezing point and why is a solution's freezing point lower than that of the original solvent's?
     
  5. May 20, 2017 #4

    Borek

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    No, it happens for every solid. Every phase diagram contains a solid/gas equilibrium line.

    If you have two phases at a given temperature, one with a higher saturated vapor pressure above, and the other with a lower saturated vapor pressure above, what will happen?
     
  6. May 21, 2017 #5
    I don't know what will happen then because I didn't know there were 2 vapor pressures involved in a phase change. But then when a liquid boils, I think what happens is some of the liquid evaporates and some of its vapour condenses and there's only one vapor pressure involved, that of the vaporized liquid. I have no clue regarding the freezing point.
     
  7. May 21, 2017 #6

    Borek

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    You are on the right track - there can't be two vapor pressures. That means if there are two phases with two different vapor pressures, one of the phases is unstable - it evaporates, and at the same time the vapors condensates producing the other phase.

    This is enough information to explain the depression in the freezing point.
     
  8. May 21, 2017 #7
    So, please
    Correct me when i'm wrong. When a pure solvent freezes, there exists an equilibrium between its liquid-vapour state change(and no equilibrium between its solid-vapour state change but we are talking about freezing here, what about a solid-liquid change?) and then when a solute is dissolved in it, what happens? You have not mentioned about this component, only about two phases (by which I understand you mean state).
     
  9. May 21, 2017 #8

    Borek

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    No idea what you mean by that. There can be an equilibrium between two different coexisiting phases, but "equilibrium between state change" doesn't mean anything.

    Phase is a concept more general than the state of matter - for example solid phosphorus is a state of matter (just like liquid and gas are), but this solid can be composed of several phases, that's what allotropy is about.

    Let's start with something simpler. Imagine you have a large, closed container. Inside there are two bakers, one containing 20% brine, the other containing 5% brine. What will happen and why? Hint: is the saturated vapor pressure over the brine the same in both bakers? Can a single container contain two different vapor pressures at the same time?
     
  10. May 21, 2017 #9
    According to me, No, the saturated vapor pressure will be different in both the cases: it will be higher in the 5% brine. Regarding the second hint, I think there will only be one vapor pressure but whether it will be the smaller, the larger or the sum of the 2 individual vapor pressures I don't know (though I think it should be the sum) ?
     
  11. May 21, 2017 #10

    Borek

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    You are right there can be only one pressure, and you are right when the concentrations are different there are two different pressures. Can you think of a process that will eliminate the problem?
     
  12. May 22, 2017 #11
    I can't think of a process to eliminate the problem other than mixing the 2 solutions.
     
  13. May 22, 2017 #12

    DrClaude

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  14. May 22, 2017 #13

    Borek

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    What about evaporation and condensation?
     
  15. May 22, 2017 #14
    So, are you saying that when we keep the two breakers then without any intervention needed, naturally the processes of evaporation and condensation will equalise the two vapor pressures, to maybe either of the individual vapor pressures or the sum of these pressures?
     
  16. May 22, 2017 #15

    Borek

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    Exactly.

    None of these. Vapor pressure depends on the concentration of the solution, doesn't it? When will the vapor pressure be identical over the content of each beaker?
     
  17. May 22, 2017 #16
    I think that will be when the concentrations are equal in both the beakers (but in this case they are both different so what will happen here?)
     
  18. May 22, 2017 #17

    Borek

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    Right.

    Think about evaporation and condensation.
     
  19. May 22, 2017 #18
    Okay, the vapor pressures will be equal over both of them but my question( this is out of curiosity ) is this: If we had put each of the beakers alone in the container before putting them together then they would have had different vapor pressures, right ( say, V1 & V2 )? Now, when we put both of them inside together, will this vapor pressure be the sum of V1 and V2,will be equal to either of them or will have some other relation to them depending on other factors?
     
  20. May 22, 2017 #19

    Borek

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    Initially it would equal the greater of the two. However, what happens when the vapor pressure is higher than the equilibrium pressure (in other words: what happens when the vapor is supersaturated)?
     
  21. May 22, 2017 #20
    Condensation would happen and thus we would obtain the saturated vapor pressure.
     
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