Can someone explain this? (water)

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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.
 

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  • #2
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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.
 
  • #3
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a great start!
Ok so what we see in a shower is liquid water... floating around in midair? How can that be?
 
  • #4
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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
 
  • #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).
 
  • #6
Mapes
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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.
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.
 
  • #7
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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?
 
  • #8
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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?
Who said that?
 
  • #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.
 
  • #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.
 
  • #11
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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...
 
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  • #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).
 
  • #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!
 
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  • #14
Mapes
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Nice explanation, uby.
 
  • #15
There is no distinct line in between evaporation and no evaporation. Read up on vapor pressure.
 
  • #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*
 
  • #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).
 
  • #18
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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.
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.
 
  • #19
Mapes
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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.
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.
 
  • #20
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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.
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 ?
 
  • #21
Mapes
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It is somewhat counterintuitive. But if you use the ideal gas law you must acknowledge that n is changing as evaporation occurs. It's just like a chemical reaction; molecules evaporate at a certain rate and condense at a certain rate. Equilibrium occurs when the rates are equal. Size doesn't enter into it.

Note that this entire discussion assumes the continued presence of the liquid. If all of the liquid evaporates, then a liquid-vapor equilibrium cannot be reached. A drop of water will completely evaporate into a 100 m3 container before equilibrium can be reached.
 
  • #22
russ_watters
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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.
That's not what the ideal gas law says. The ideal gas law says that if you take a set of conditions and change one, the others have to change in proportion. But that really means changing one of the conditions, not just setting up a completely separate set of conditions, but actually taking a container and enlarging it (for example) while attempting to hold other conditions the same.
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 ?
In light of everything said above, the answer to that really depends on if a drop of water can fully evaporate into a 1L container. If it can, the resultant pressure will be lower. If it can't, the resultant pressure will be the same and the amount of liquid evaporated in each case will be different. (one will have 1 cc of gas at .03 ATM and the other will have 1L of gas at .03 ATM)

So for this case, you are choosing a larger container, but that isn't the only change: you are also changing the mass of gas evaporated and the volume of gas evaporated. But you'll have to do an equilibrium calculation to determine exactly how much evaporates and how much stays liquid in each case (not difficult).
 
  • #23
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Ok, I got it..
Thanks Russ and Mapes!
 
  • #24
uby
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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)?
assuming i understand your description correctly, then you are correct in your conclusion. once the wall is removed, you create an interface between the liquid water and the dry air. water vapor will leave the surface of the liquid water and a net loss of liquid water will occur until the equilibrium vapor pressure of water vapor is achieved (the amount [moles] of water needed to achieve this vapor pressure is size-dependent with respect to the size of the dry air volume.)

since the air pressure is not changing for a closed, fixed volume, isothermal system, the total pressure should increase by the added water vapor molecules. if you assume no interactions between the water vapor and the air molecules, the pressure should rise to about 1.0314atm. the expected pressure will be modified if you account for molecular interactions amongst the gas species. an analogy to help understand this comes in the form of mixing of liquids: when mixing two liquids that are miscible, the volumes are NOT additive and the new volume of liquid is greater or lesser than the sum of the individual liquid volumes depending on how the liquid molecules interact with each other relative to self-interactions. (i'm fairly certain that you'll tend to see a volume decrease in the solution upon mixing for any exothermic mixing process, although i'd have to review old notes to be sure!)

in the case of mixing two miscible liquids, the pressure of the liquid is maintained at a constant (temperature too of course!) and the volume is free to change based on molecular interactions. similarly, in the case of gas mixing, if you fix the volume (and temp) then the pressure will change based on molecular interactions. if the magnitude of these interactions is significant enough for a large change in the pressure, then you can (obviously) no longer make use of the ideal gas law and must use one of the many empirical equations of state that are out there.
 
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  • #25
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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!

Thanks uby. May I ask a fundamental question now... Just how is vapor created in the bottle with air on top. I mean, according to what I learned in school, it is created only with given characteristcis (pressure, temp.). I think I am understanding all the comments here about how water is evaporated to create an equilibrium. But still, how can that evaporation even exist? How does it get its necessary conditions for it to appear?
 

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