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I Compression of gases that liquefies

  1. Sep 22, 2018 #1
    For example we had a closed system with water at 298K, so it's pressure is 0.031atm. To it we add a gas, A such that we get 1atm of total pressure. That gas undergoes a phase transition at 1.5atm and 298K. So initially, the partial pressure of water is a little smaller than 0.031atm due to the dissolution of some of that gas, and the rest of the pressure is from that gas.

    As we compress this mixture of gases with 2atm of pressure what would happen? How would the A and water's partial pressure change?

    Since A is able to increase its partial pressure further before liquefying, I feel that it should increase to 1.5atm. But that results in a decrease in the water's partial pressure because it's mole fraction in the liquid continues to drop further. So the total pressure will be 1.5atm from A, and less than 0.031atm from water. But then since we're applying 2atm of pressure, how does that equal out?

    Something like the uploaded image but instead of air we replace it with a different gas.
     

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  2. jcsd
  3. Sep 23, 2018 #2

    sophiecentaur

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    Are you actually describing a stable final situation here? The final pressure in the cylinder has to be 2Atm .
     
  4. Sep 23, 2018 #3
    Are you saying that pure A would undergo a phase transition at 1.5 atm and 298K?
     
  5. Sep 23, 2018 #4
    Yes so in a phase diagram for A it would transition at 1.5atm. Does that matter if it's a mixture?
    I get that the final internal pressure has to be 2atm. But I don't see how that can be achieved with the parameters I've set because A would liquefy once it's pressure is at 1.5atm, and the water would liquefy at pressure less than 0.031atm ad well.
     
  6. Sep 23, 2018 #5
    Are the liquid phases of A and water miscible?
     
  7. Sep 23, 2018 #6
    Yes I would assume so for this scenario. So I guess I'd use Raoult's law for water and Henry's law for A to calculate their partial pressures at each point? So initially, when I first introduced A to the pure water system, how would I know how much it dissolves? We would just know that the total P is 1atm and initial P is 0.031atm for pure water vapour only.
     
  8. Sep 23, 2018 #7

    russ_watters

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    It really sounds to me like you are just arguing with your own constraints. If the final pressure is 2atm because you specified it to be 2atm, then the final pressure is 2atm.

    Perhaps what you really mean to specify (or find?) is a compression ratio...? I see your title says "compression", but then seemed to misuse the word in the description of the problem. Compression and pressurization are different things: compression is decreasing the volume by a specified amount whereas pressurization is increasing the pressure to a specified quantity.

    Also, you didn't specify what happens, if anything, to the energy of compression. Should we assume adiabatic or isothermal?
     
  9. Sep 23, 2018 #8
    You are assuming that the gas phase is an ideal gas mixture. That's fine.

    But, for the liquid phase, if you use Raoult's law, you are assuming an ideal liquid mixture (that applies to both species), and, if you assume A obeys Henry's law for the liquid, you are assuming a non-ideal liquid mixture model with a known activity coefficient. What liquid phase mixture model are you assuming (a) Raoult's law, (b) Henry's law, or (c) a different non-ideal fluid mixture model?
     
  10. Sep 23, 2018 #9
    Sorry for the wrong terminology used, I meant that the mixture is compressed with 2atm of pressure. But since gas A undergoes a phase transition at 1.5atm, and water undergoes a phase transition at 0.031atm in their pure states how can their gaseous pressures ever be equal to 2atm?

    Edit: it should be a isothermal process where temperature stays at 298K
     
    Last edited: Sep 23, 2018
  11. Sep 23, 2018 #10
    I was taught that an ideal dilute mixture would have the solvent follow Raoult's law, and the solute follow Henry's law. So I presumed that A being the added gas should follow Henry's law.

    But even if I were to use assume a Raoult's law for both, when I first introduced the gas water's mole fraction in the liquid phase should decrease so the pressure of 0.031atm should decrease. But I don't really know how to tell how much it decreases by and likewise for A, I'm unsure how much of it dissolves as well. So I know that P(H2O(g))<0.031atm and P(A(g))>0.969atm but is there a way to know for sure their actual pressures?

    And even if we do know it, when we compress this mixture with 2atm of pressure how would the system change with this pressure? I presumed that A could continue to increase in pressure up til it hit 1.5atm, because that's it's vapor pressure at 298K. But for water it's vapour pressure is 0.031atm at 298K, but since we increased the pressure of A to 1.5atm, by Raoult's law, it's pressure should decrease even more because more A would have dissolved right? I'm quite clueless as to how the system changes when such a scenario happens to be honest.
     
  12. Sep 23, 2018 #11
    Well, what does Raoult's law predict for the mole fractions of water and A in the liquid at 1 atm pressure?
     
    Last edited: Sep 23, 2018
  13. Sep 23, 2018 #12

    sophiecentaur

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    Because you have said it is in your initial description. Something may need to change state or dissolve . . . . . It's not an ideal situation.
     
  14. Sep 23, 2018 #13
    P(H2O(g))=P*(H2O(g))X(H2O) and P(A)=P*(A)X(A), but we only know the values of P*(H2O) and P*(A) so I don't get how we can predict their mole fractions though.
     
  15. Sep 23, 2018 #14
    Oh, wait we can tell it from the graph right? So we would know that the exact pressures of water and A at a total pressure of 1 atm?
    upload_2018-9-24_0-48-44.png
    So it will look something like this right? So we can get the mole fractions of either component?

    But continuing on, when we compress the gas with 2atm of pressure the mole fraction of water goes to 0? I don't really understand how the process continues on from here though.
     

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  16. Sep 23, 2018 #15
    We first have to specify how much water and how much A are present total. So, please specify. That way we can determine how much of each species is present in each phase.
     
  17. Sep 23, 2018 #16
    Let's say we have 20 moles of H2O in the liquid phase and 5 moles of A in the liquid phase initially. And from the graph, the partial pressure of water would be 0.0248atm and A would be 0.9752atm. Would this be alright?
     
  18. Sep 23, 2018 #17

    russ_watters

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    That's still the wrong terminology. Again, 2atm is a pressure, not a compression amount.
    They can't, but you could have all compressed liquid.
     
  19. Sep 23, 2018 #18
    No. The partial pressure of A would be (0.2)(1.5)=0.3 atm, and the total pressure would be 0.3248 atm. What number of moles of A would you need in the liquid for the total pressure to be 1 atm? Also, you would need to specify the total number of moles of vapor at 1 atm if we are going to compress the system and quantify what's happening next.

    Let me guess. You've been studying multicomponent phase equilibrium, but you haven't actually worked any practice problems yet.
     
  20. Sep 23, 2018 #19
    Haven't even gone there yet, we just got introduced to the topic of mixing but I've been thinking a lot to make sure I understood what's going on in the lectures. Sometimes I feel like I overthink too much but I'm just unsure if what I'm overthinking will be looked at further in the module or no, which leads me to these types of situations.

    But I'd need 37 moles of A to get 1 atm total pressure. 37/57.14x1.5=0.9752atm.
     
  21. Sep 24, 2018 #20
    Hmm, so everything would just liquefy?
     
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