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Thermodynamics conceptual problem

  1. Feb 27, 2016 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations



    3. The attempt at a solution

    1) I think the answer is c) . Since initially there is no pressure above water , it starts vaporizing . This causes loss of thermal energy from water . The heat lost results in water freezing .The process continues till all the water turns into either vapor or ice .

    2) The increasing vapor in the vessel causes the vapor pressure to increase till it is saturated .After that the vapor starts condensing turning into liquid .

    Are the answers as well as reasoning correct ?


    Many Thanks
     

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  3. Feb 28, 2016 #2

    ehild

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    What happens if the evacuated chamber is so big that the pressure is well below the triple point pressure even in case of all water turns into vapor ?
    That looks correct.
     
  4. Feb 28, 2016 #3
    Are you suggesting that option a) is a possibility ?

    Wouldn't the heat lost by water while turning into vapour simultaneously turn some fraction of water into ice ?

    Even though I have got it right , would you kindly explain why does vapour turns into liquid after saturation of vapour is reached .
     
  5. Feb 28, 2016 #4

    ehild

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    It was said that the water is kept at 0°
    http://www1.lsbu.ac.uk/water/water_phase_diagram.html
    I guess there will be only vapor if the pressure is very low, but the value of the initial pressure was not stated. So I am not sure if there will be some ice .

    The pressure can not go higher than the saturation pressure at the given temperature if the fluid and gas are in equilibrium. See the phase diagram again. If more vapor is injected that will not increase the pressure, but some gas turns into liquid which has much less specific volume .
     
  6. Feb 28, 2016 #5
    If I understand Q1 correctly, the chamber is adiabatic, and it is very large compared to the initial amount of liquid water initially present. It would seem that all the water could not evaporate because something would have to supply the required heat. So the part that doesn't evaporate must freeze and then get even colder. Why? Because the vapor pressure of the water in the gas phase would have to be consistent with the amount of vapor, the new system temperature, and the volume of the container. Therefore, the temperature would have to drop such that there could be some amount of condensed phase still present (since it has to supply the heat of vaporization) and also to make good on the vapor pressure, which would now be lower to be consistent with the chamber volume. So, the only answer that makes sense to me is c.
     
  7. Feb 28, 2016 #6

    ehild

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    The ice formed can sublimate even at freezing temperatures... Ice and vapor coexist only along the solid-vapor equilibrium line.
     
  8. Feb 28, 2016 #7
    This is correct. But, if the system is adiabatic, something must supply the heat for the vapor to form. So, the system keeps cooling until a low enough temperature is reached that (1) the pressure of the vapor is consistent with the tank volume and the mass of vapor, (2) the total mass of vapor and ice matches the initial mass of water in the tank, (3) the pressure of the vapor is equal to the ice-vapor equilibrium vapor pressure at the final temperature, and (4) the total internal energy of vapor and ice matches the initial internal energy of water in the tank. So, what I'm saying is that there has to be some ice left in the tank at the end.
     
  9. Feb 28, 2016 #8

    ehild

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    The energy of the water vapor is proportional to the absolute temperature - at very low pressure even water vapor can be considered ideal gas. Assume you have a container of volume 1000 m3, evacuated, and 1 g water, initially at 273 K. Do you think there will remain some ice?
     
  10. Feb 28, 2016 #9
    Yes. Definitely. Would you like to work this problem together to see how it plays out? I was thinking of using 1 m^3, but 1000 m^3 is also OK, if that's what you prefer.

    Chet
     
  11. Feb 28, 2016 #10
    Hi ehild,

    Well, I've thought about it some more, and I'm not so sure now. I'm going to do some modeling calculations to get a better handle on it.

    Chet
     
  12. Feb 28, 2016 #11
    OK. Some preliminary numbers.

    The change in internal energy of water at 0 C as a result of vaporization is about 2376 J/gm. The heat capacity at constant volume of water vapor in the temperature range of interest is about 1.39 J/(gm-K). So using as a reference a value of 0 for the specific internal energy of liquid water at 0 C, the specific internal energy (per gram) of water vapor at temperature T is ##u=2376+1.39(T-273)##. This would mean that, if all the liquid water went into water vapor, the temperature required for the internal energy of the 1 gm of water to remain zero would be -1436 K. So, it looks like all the original water can't change into water vapor.

    Chet
     
  13. Feb 28, 2016 #12

    ehild

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    Hi Chet,
    I have to think it over, but now I think you are right,
    Edit: It were my practical instincts which suggested me that the water could not cool down too much. I did not believe that it was thermally isolated completely. In case of an ideally adiabatic process, only a small amount turns to gas, and most of it becomes ice.
    In real life, there is some air left in the evacuated chamber and its walls have some temperature. As the water is at 0°C initially, the walls can be also at that temperature, and there can be some heat transition between the walls and the water. If that heat transition prevents the water from cooling down too much, it can evaporate completely.
    In very good vacuum and about complete isolation, like in the space, the water exist in form of ice, as it is in the comets.
     
    Last edited: Feb 28, 2016
  14. Feb 28, 2016 #13
    @Chestermiller , what do you think about my reasoning for Q2 in the OP ?
     
  15. Feb 28, 2016 #14
    Sounds correct to me.
     
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