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Can someone explain this? (water)

  1. Jun 5, 2008 #1
    Hi all.
    I've been thinking about this. When we take a warm shower we see water vapor coming out. First of all, why do we see this vapor? Isn't vapor supposed to come up at 100°C when at a pressure of 1atm? When I'm in the shower I'm at about 1atm of pressure I suppose and clearly not at 100°C!
    Also, when we go out of the tub we can see this condensated water (?) on our bathroom mirror. Why does this happen? My instinct tells me it's the water vapor being condensated by the cold mirror... but again, why was there water vapor in the first place?
    Any help in understanding this, is appreciated. Thanks in advance.
     
  2. jcsd
  3. Jun 5, 2008 #2
    Let's start here:

    "Water vapor is water in its gaseous state-instead of liquid or solid (ice). Water vapor is totally invisible. If you see a cloud, fog, or mist, these are all liquid water, not water vapor."
    Source: http://www.weatherquestions.com/What_is_water_vapor.htm

    That's a start anyway.
     
  4. Jun 5, 2008 #3
    a great start!
    Ok so what we see in a shower is liquid water... floating around in midair? How can that be?
     
  5. Jun 5, 2008 #4
    The humidity has to play some part in this. The smaller the room, the sooner you can see the vapor, which suggests that you reach a point where the air is so wet that it either creates liquid itself, or that it is so dense it can carry a few molecules of water, or it is perhaps the rogue water molecules bind among themselves once all the air is all ready occupied.

    I have NO idea what I am talking about, but I think one of those 3 should at least be close. And I'm almost certain humidity is a major factor.

    k
     
  6. Jun 5, 2008 #5

    russ_watters

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    Humidity is another way of describing the fact that water vapor always exists in the air. How much water vapor depends on the temperature, but the measurement of that is "vapor pressure". When something is boiling, that merely means that the vapor pressure is equal to atmospheric pressure and as a result, vapors can form inside the liquid (as opposed to just evaporating on the surface).
     
  7. Jun 5, 2008 #6

    Mapes

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    Your instincts are good. It's exactly the same, except the invisible water vapor (present at any temperature, but at a partial pressure of <1 atm at a temperature of <100°C) is hitting cold air instead of a cold mirror. At the lower temperature, less water vapor is stable. And so it condenses into a little cloud.

    Some water vapor exists above any water at any nonzero temperature because random fluctuations give some of the water molecules enough energy to leave the liquid. As Russ pointed out, the boiling point is defined as the point where the pressure of the water vapor reaches 1 atm.
     
  8. Jun 7, 2008 #7
    Thanks you all for your responses. I'm a little confused however. First, pallidiin said: "If you see a cloud, fog, or mist, these are all liquid water, not water vapor." But then some of you stated that what we see in a shower is in fact vapor. So what is it, water in its liquid or gas state?
     
  9. Jun 7, 2008 #8

    Vanadium 50

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    Who said that?
     
  10. Jun 7, 2008 #9

    Redbelly98

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    It's true that cloud, fog or mist are liquid water. However, they indicate that water vapor is present, and some of the vapor has condensed into a liquid state.
     
  11. Jun 7, 2008 #10

    uby

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    this can be generalized to all condensed species, not just liquid water:

    equilibrium is a concept which describes the balance of driving forces for a process. in the case of water evaporating, you might consider the equilibrium between the liquid water and gaseous water (called water vapor) via the following reaction:

    H2O(liquid) <---> H2O(gas)

    Note the double arrows. This means that, at equilibrium, the reaction proceeds at the same rate in both directions (this does NOT mean that no reaction occurs at all!).

    At 1atm total pressure, the amount of H2O(gas) in equilibrium with the H2O(liquid) at the liquid surface depends on the temperature of the surface. It is said that the liquid "boils" when the vapor pressure of H2O(gas) is the same as the total pressure of the environment. At 1atm total pressure, this temperature (called the normal [referring to 1atm total pressure] boiling point) is 100C.

    Note though that at temperatures lower than 100C, there is still H2O(gas) in equilibrium with H2O(liquid). In fact, there is ALWAYS some H2O(gas) in equilibrium with the condensed state H2O(liquid or solid!). To explain what the equilibrium vapor pressure means, consider a closed volume at a very good degree of vacuum. If liquid water were to suddenly appear inside this closed volume and remain at a certain temperature, it would give off water vapor until the volume has a total pressure equal to the equilibrium vapor pressure for that temperature. (you can, for a given volume, try approximating how much would evaporate using PV=nRT). Once the total pressure is equal to the vapor pressure, the NET amount of liquid would no longer change. This is easy to conceptualize once you get the hang of it, mostly because it is a closed system; however, your shower is more-or-less an open system.

    When you are in the shower, the temperature of the liquid water is enough to generate a substantial vapor pressure. However, the air temperature is not uniform with the temp. of the water - nor are most surfaces in your shower. Therefore, the water vapor, after traveling through the air as a gas will start to condense as its temperature drops. When it condenses, it is a liquid droplet that is still being carried by convection of the air in your shower.

    please let me know if any of this didn't make sense, i tried to be concise.
     
  12. Jun 8, 2008 #11
    Thanks a lot uby. So let me get some things straight:
    If at room temp., I have a bottle about 80% filled with water, 20% filled with air at a pressure of 1atm, water vapor shouldn't appear right? (The water and the air is at room temp.) For the vapor to be in equilibrium with the air, it would need a temp. of 100°C. However, wouldn't there be some very little amount of vapor as an effect of random fluctuations? Or not?

    Please tell me if that wasn't clear.

    Ahh I want to ask so many things, but I guess I'll wait until someone answers this... Thx in advance...
     
    Last edited: Jun 8, 2008
  13. Jun 8, 2008 #12

    Redbelly98

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    There will be water vapor in this example. At 20 C, the water vapor would have a partial pressure somewhere between 15 and 20 mm Hg. (and 1 atmosphere is 760 mm Hg).
     
  14. Jun 8, 2008 #13

    uby

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    to give you an idea of the numbers for the reaction:
    H2O(liquid) <--> H2O(gas)

    equilibrium partial pressure of H2O(gas) as function of temperature (deg C):
    T=25C ... p=0.0314atm
    T=77C ... p=0.4089atm
    T=97C ... p=0.8827atm
    T=100C ... p=1.0000atm

    if there were no air at all above the liquid water in your closed container, the net amount of liquid would change until the vapor accounted for (at 25C) a partial (and total) pressure of 0.0314atm. At that point, the liquid water would evaporate at the same rate as the water vapor condenses and the net amount of liquid water would remain constant. You can calculate the amount of liquid water lost to reach that equilibrium point using the ideal gas law PV=nRT if you assume the water vapor is an ideal gas and the entire system is isothermal.

    if there IS air above the water, the water will still evaporate until it reaches its equilibrium point, but the point will be shifted due to the resistance posed by the gas above the liquid. partial pressure is defined as the pressure the vapor would create if it were the only gas there -- which is why it is easy to speak of a gas evaporating into the vacuum. now, you must deal with a mixture of gases (water vapor mixing into air), which complicates things slightly but can be easily accounted for if you make assumptions as to how the gases mix. the simplest is an ideal mixture of ideal gases, where each component of the gas acts independent of the other and contributes partial pressures in a manner directly proportional to their relative mole fractions. this is probably putting the carriage ahead of the horse at this point, though! for now, just assume that the partial pressure of the water vapor in the air is essentially the same as it would be in any other gas, or, in other words, the total pressure it would exhibit in a vacuum sans any other gas.

    to be more concrete, at 25C, there would be water vapor in your sealed vessel and the total pressure would slightly rise due to the added gas molecules. assuming an ideal gas mixture, and air is essentially all N2 and O2, you could say:
    P (total pressure) = [sum over X] p(partial pressure of gas X) = p(N2) + p(O2) + p(H2O) = 1.0314atm. Calculating the amount of gas (mole fraction) can be a bit tricky, and I won't go into it.

    at 100C, the total pressure would be even higher. by my calculation, about 2.4atm. [1.000atm water vapor plus the original gas's increased pressure P2=P1(373K/273K)=1.4atm]

    above 100C, the total pressure rises RAPIDLY, and you risk an explosion if you don't vent the vapors. water boilers in the late 1800s/early 1900s were notorious for exploding and causing severe injury or death because of this. this is also why pressure vessels are not things to be taken lightly!
     
    Last edited: Jun 8, 2008
  15. Jun 8, 2008 #14

    Mapes

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    Nice explanation, uby.
     
  16. Jun 8, 2008 #15
    There is no distinct line in between evaporation and no evaporation. Read up on vapor pressure.
     
  17. Jun 10, 2008 #16
    Hiya, interesting thread...
    just to make sure I understand this:
    If you have a box with a wall in it
    and on one side you have water at 25 degrees
    and on the other side you have dry air at 1 atm
    What happens when you remove the wall?
    does the air pressure raise to about 1.0314atm (or maybe slightly less)?

    *pays homage to uby's explanatory skills*
     
  18. Jun 10, 2008 #17

    russ_watters

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    You didn't say what air pressure you had on the water side, but if you introduce water into a box with dry air at 1 atm, as the water evaporates, it will come into equilibrium at 1.0314 atm (assuming .0314 is the right vapor pressure....I didn't check).
     
  19. Jun 10, 2008 #18

    GT1

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    What about the size of the container ? if I put the same amount of water in a small container and in a big container, the pressure would be different for the same temperature.
     
  20. Jun 10, 2008 #19

    Mapes

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    No. At equilibrium, the vapor pressure over a liquid is constant at a given temperature, regardless of the amount of liquid. The only difference is that it would take longer to reach this pressure if the surface area of the liquid were small and/or the amount of space to fill were large.
     
  21. Jun 10, 2008 #20

    GT1

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    I still don't get it. according the ideal gas law P=(nRT)/V, if the volume is bigger the pressure is smaller for a given temperature. If I put a drop of water in a 1 cc empty container (vacuum) and if I put the same drop of water in a 100 m^3 empty container- how can the pressure be the same for both cases ?
     
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