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Worked example: Vapor pressure and the ideal gas law

Vapor pressure example using the ideal gas law. Created by Sal Khan.

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  • leaf green style avatar for user allan1.davis
    It seems to me that you must be assuming that this room has no water molecules in it before this experiment begins. I live in Florida, and I know that air already has moisture of water molecules in it. Maybe, this experiment was conducted in New Mexico, where the humidity is much less.
    (35 votes)
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    • leaf red style avatar for user Paze
      Great point. The interesting conclusion to your good reasoning (that the room had to as dry as a desert), is that by knowing how much water actually evaporates from the bucket we can know the amount of moisture that was previously in the air – how many molecules of water were there.
      (9 votes)
  • blobby green style avatar for user Elsy Amaya
    where did the number 760 mm Hg come from?
    (17 votes)
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    • mr pants teal style avatar for user Matthew Belliveau
      1 mmHg = 1 torr, a unit named after Evangelista Torricelli, the Italian scientist who invented the first barometer, the instrument which measures pressure. He created it using a column of mercury (open at the bottom, closed at the top), sitting in a bath of mercury. When sitting in the bath, the atmospheric pressure pushes down on the bath of mercury and forces the mercury to rise up in the column. When atmospheric pressure is high, the column rises higher. When the pressure is low, the column does not rise as high. At standard atmospheric pressure, the column rises to a height of 760mm. You can convert between other common units of pressure using this conversion:

      760 torr = 760 mmHg = 1 atmosphere = 101.3 kilopascals
      (36 votes)
  • piceratops ultimate style avatar for user tfanaru
    At 25 degrees Celsius, shouldn't none of the water evaporate because the boiling point of water is 100 Celsius?
    (22 votes)
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    • leaf red style avatar for user Paze
      Yes, and this happens not only with water.
      For example: there are many reported cases of mercury intoxication due to breathing the metal’s vapor. (Interestingly, ingesting it doesn’t cause much harm).
      The point is the same as water: the mercury boiling point is around 357° C, but evaporation can happen all the time, slowly, on the surface, as the particles get enough energy to separate themselves from the forces of attraction that hold them together.
      (10 votes)
  • blobby green style avatar for user nizarahamedm33
    I can't still understand what vapour pressure is? please help me understanding it.
    (10 votes)
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    • duskpin ultimate style avatar for user Daniel Hollas
      Imagine you have a closed box partially filled with liquid. Initially, you can imagine, that there is a vaccuum above the liquid. Of course, the liquid starts to evaporate and there will be some gas above it (remember the box is closed so that the gas can't escape). At some point, the evaporation will stop and we say that the liquid is in an equillibrium with the gas phase......And finally, if you now measure the pressure of that gas, that;s your vapour pressure.

      In short, the vapour pressure is the pressure of the gas molecules of the given liquid in equillibrium with that liquid. It depends very much on temperature, when you increase the temperature, more liquid will evaporate and the vapour pressure will be higher.
      (37 votes)
  • blobby green style avatar for user ripstikboy28
    How do you measure mercury's pressure in millimeters?
    (6 votes)
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    • blobby green style avatar for user John Pfeifer
      If you take two separate cups of mercury and hold them side by side. Then connect a tube from the bottom of one to the bottom of the other. The two liquids will level themselves.
      Now put a cap with a pressure gage on top of one of the cups. Add one millimeter of extra mercury height to the open cup. The pressure in the capped cup should have increased by 1 mm Hg.
      (16 votes)
  • leaf green style avatar for user nittilina
    If the temperature of the room were to be 15 degrees Celsius instead of 25, then n = 56.3 mol, which means that more water would evaporize. How could this be when, intuitively, more water should evaporize at a higher temperature?
    (5 votes)
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    • blobby green style avatar for user kylebooh87
      You are completely correct that vapor pressure changes with the temperature. The problem is the logic that you used. At 15 degress celsius, you need more moles (56.3 moles instead of the 54.4 moles) to reach 23.8 mm Hg because the molecules have lower kinetic energy. That does not mean, that at 15 degrees, the equilibrium point (vapor pressure) is still 23.8 mm Hg. At 15 degress the vapor pressure of water is 12.8 mm Hg. So that does agree with your intuition. At 15 degrees, the vapor only creates a pressure of 12.8 mmg Hg while at 25 degrees, the vapor creates a pressure of 23.8 mm Hg. Meaning, at 25 degrees, more water vaporizes to create a greater pressure.
      (16 votes)
  • leaf green style avatar for user Craig Richer
    This scenario seems to be taking a non-ideal vessel (a dorm room) and applying ideal characteristics to it. Had this experiment been done in a carefully prepared (within the limits of a student's freedom to do so) dorm room, wouldn't all of the water evaporate? My reasoning is that the air temperature may be stable, but only because the heat losses 'to' balance the gains 'from' outside sources. Essentially, if the air was at some maximum humidity, I suspect that there would be condensation on the windows, leaving the water from the test vessel in another location. Leaving out the water in the plumbing waste traps, starting humidity, and hygroscopic substances (cereal) commonly found in the room, if the extra complication was put into the formula to make the vessel a "dorm room", then it can not simply be treated like an ideal vessel anymore.

    I imagine that this would not be what a teacher is looking for on a quiz, but my question remains. Wouldn't the water continue to leave the vessel that is at the average room air temperature to make up for condensing water on cooler areas?
    (4 votes)
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    • blobby green style avatar for user Bob Graff
      I had some of the same thoughts. I was imagining water condensing on the windows (leading to water running off and pooling on window sills), any mirrors or glass objects (more runoff), the paint, the floor (sometimes tile in a dorm room), and even slightly dampening the bed sheets and such. I was also thinking about how unlikely it would be for significant amounts of vapor to return to the original 2L container before condensing. In practice, I'd expect all of the water to evaporate, and for the student to return to a damp room, given enough time. It's still an interesting problem, though!
      (3 votes)
  • aqualine ultimate style avatar for user Stephanie A
    So, if I understand the principles behind this video, as temperature rises so does pressure, meaning that as temperature rises the amount of water able to be held in the air falls- that is, the equilibrium point mentioned in the video (where just as much water is condensing to liquid as is evaporating to gas) will happen at a smaller amount of water. Water is denser at higher temperatures, so any given volume of cold air will be able to hold more mols of water as a gas than an identical volume of warm air. Is this correct?

    How does this apply to winter-time, when it "feels" drier? Does the low temperature "overcome" the water vapor and freeze it out of the air to lie on the ground as frost?
    (3 votes)
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    • piceratops ultimate style avatar for user Just Keith
      The density of liquid water at any given temperature has little to no effect on how much water can exist in gaseous solution at that temperature. Thus, it is not correct to associate the density of liquid water with water vapor in air -- these are different phases of water and governed by different physical properties.
      (2 votes)
  • leafers ultimate style avatar for user DJash44
    Sal uses 0 degrees C = 273 Kelvins. I know it doesn't affect the final answer because of significant figures, but shouldn't we use 273.15? It's more accurate, so......
    (3 votes)
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    • piceratops ultimate style avatar for user Just Keith
      Sal does a lot of rounding to make the demonstrations easier to follow. You should not actually round that much when you do the computations. You should round off only for your reported values. For any intermediate computations, you should keep the correct number of significant digits plus two extra digits (what we call guard digits) to prevent round off error.
      (2 votes)
  • mr pink red style avatar for user Josh Talley
    At , Sal said three significant figures, but there was an edit that said he meant to say two significant figures. i thought 760 mm Hg had an infinite number of sig figs since it is a whole number. Am i wrong? and if so, could you explain why?
    (2 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      At one time, 1 atm as defined as exactly 760 mmHg. Thus the "760" had an infinite number of significant figures.
      Now, 1 atm is defined as 101 325.0 Pa or 101.3250 kPa. Also, 1 Torr = ¹/₇₆₀ atm.
      Historically, 1 Torr was meant to be exactly the same as 1 mm Hg. Subsequent redefinitions of the two units made them differ slightly (by less than 0.000 015 %).
      For all ordinary purposes, 1 atm = 760.0000 mm Hg (7 significant figures).
      (3 votes)

Video transcript

This exercises is from chapter 12 of the Kotz, Treichel in Townsend and Chemistry and Chemical Reactivity book, and I'm doing it with their permission. So they tell us you place 2 liters of water in an open container in your dormitory room. The room has a volume of 4.25 times 10 to the fourth liters. You seal the room and wait for the water to evaporate. Will all of the water evaporate at 25 degrees Celsius? And then they tell us at 25 degrees Celsius, the density of water is 0.997 grams per milliliter. And its vapor pressure is 23.8 millimeters of mercury. And this is actually the key clue to tell you how to solve this problem. And just as a bit of review, lets just think about what vapor pressure is. Let's say it's some temperature, and in this case we're dealing at 25 degrees Celsius. I have a bunch of water, and let me do that in a water color. I have a bunch of water molecules sitting here in a container. At 25 degrees Celsius, they're all bouncing around in every which way. And every now and then one of them is going to have enough kinetic energy to kind of escape the hydrogen bonds and all the things that keep liquid water in its liquid state, and it will escape. It'll go off in that direction, and then another one will. And this'll just keep happening. The water will naturally vaporize in a room. But at some point, enough of these molecules have vaporized over here that they're also bumping back into the water. And maybe some of them can be captured back into the liquid state. Now, the pressure at which this happens is the vapor pressure. As you can imagine, as more and more these water molecules vaporize and go into the gaseous state, more and more will also create pressure, downward pressure. More and more will also be colliding with the surface of the water. And the pressure at which the liquid and the vapor states are in equilibrium is the vapor pressure. And they're telling us right now. It is 23.8 millimeters of mercury. Now, what we need to do to figure out this problem is say, OK, if we could figure out how many molecules need to evaporate, how many molecules of water need to evaporate to give us this vapor pressure, we can then use the density of water to figure out how many liters of water that is. So how do we figure out how many molecules-- let me write this down-- how many molecules of water need to evaporate to give us the vapor pressure of 23.8 millimeters of mercury? So what, I guess, law or formula-- and I never like to just memorize formulas, but we've given this formal in the past and it's probably one of the top most useful formulas in chemistry, or really all of science-- what formula or law deals with pressure? They give us the volume of the room because that's where the pressure will be inside of. So we have pressure, the equilibrium vapor pressure. We have a volume of a room right over here. We know the temperature of the room right over there. And we're trying to figure out the number of molecules that need to evaporate for us to get that pressure in that volume at that temperature. So what deals with pressure, volume, number of molecules-- let's say in moles, so I'll write a lower case n-- number of molecules, and temperature? Well, we've seen this many, many times. It's the Ideal Gas Law. Pressure times volume is equal to the number of moles of our ideal gas-- in this case we're going to use water as our ideal gas, or vapor as our ideal gas-- times the universal gas constant times temperature. And this should never seem like some bizarre formula to you because it really, really makes sense. If your pressure goes up, then that means that either the number of molecules have gone up, and we're assuming the volume is constant. That means either the number of molecules have gone up, which makes sense-- more things bouncing onto the side of the container. Or your temperature has gone up-- the same number of things, but they're bumping with higher kinetic energy. Or if your pressure stays the same and your volume goes up, then that also means that your number of molecules went up, or your temperature went up. Because you now have a bigger container. In order to exert the same pressure you need either more molecules or more kinetic energy for the molecules you have. And you could keep playing around with this, but I just want to make it clear this isn't some mysterious formula. The first time I was exposed to this I kind of did view it as some type of mysterious formula. But it's just relating pressure, volume, number of molecules and temperature. And then this is just the universal gas constant. So let's just get everything into the right units here. And then what we're trying to solve for, we want to figure out the number of molecules of water. So we want to solve for n. And if we know the number of moles of water, we can figure out the number of grams of water. And then given the density of water we can figure out the number of milliliters of water we are dealing with. So let's just rewrite the Ideal Gas Law by dividing both sides by the universal gas constant and temperature. So that you get n is equal to pressure times volume, over the universal gas constant times temperature. Now, the hardest thing about this is just making sure you have your units right and you're using the right ideal gas constant for the right units, and we'll do that right here. So what I want to do, because the universal gas constant that I have is in terms of atmospheres, we need to figure out this vapor pressuree- this equilibrium pressure between vapor and liquid-- we need to write this down in terms of atmospheres. So let me write this down. So the vapor pressure is equal to 23.8 millimeters of mercury. And you can look it up at a table if you don't have this in your brain. One atmosphere is equivalent to 760 millimeters of mercury. So if we wanted to write the vapor pressure as atmospheres-- let me get my calculator out, get the calculator out, put it right over there-- so it's going to be 23.8 times 1 over 760, or just divided by 760. And we have three significant digits, so it looks like 0.0313. So this is equal to 0.0313 atmospheres. That is our vapor pressure. So let's just deal with this right here. So the number of molecules of water that are going to be in the air in the gaseous state, in the vapor state, is going to be equal to our vapor pressure. That's our equilibrium pressure. If more water molecules evaporate after that point, then we're going to have a higher pressure, which will actually make them favor more of them going into the liquid state, so we'll go kind of past the equilibrium, which is not likely. Or another way to think about it-- more water molecules are not going to of evaporate at a faster rate than they are going to condense beyond that pressure. Anyway, the pressure here is 0.0313 atmospheres. The volume here-- they told us right over here-- so that's the volume-- 4.25. 4.25 times 10 to the fourth liters. And then we want to divide that by-- and you want to make sure that your universal gas constant has the right units, I just looked mine up on Wikipedia-- 0.08-- see everything has three significant digits. So let me just allow that more significant digits and we'll just round at the end. 0.082057, and the units here are liters atmospheres per mole at kelvin. And this makes sense. This liter will cancel out with that liter. That atmospheres cancels out with that atmospheres. I'm about to multiply it by temperature right here in kelvin. We'll cancel out there. And then we'll have a 1 over moles in the denominator. A 1 over moles in the denominator will just be a moles because you're going to invert it again. So that gives us our answer in moles. And so finally our temperature-- and you've got to remember you've got to do it in kelvin. So 25 degrees Celsius-- let me right it over here-- 25 degrees Celsius is equal to, you just add 273 to it, so this is equal to 298 kelvin. So times 298 kelvin. And now we just have to calculate this. So let's do that. So let me clear this out. So we have-- let me use my keyboard-- so 0.0313 atmospheres times 4.25 times 10 to the fourth. That e just means times 10 to the fourth. That's just the way that it works on this calculator. And then divided by 0.082057 divided by-- actually, just to make it clear, let me show you that I'm dividing by this whole thing, so let me insert some parentheses right here. So in the denominator we also are multiplying by 298. And let me close the parentheses. And then we get 54.4. We only have three significant digits. So this is equal to 54.4 moles. And we could see this liters cancels out with that liters. Kelvin cancels out with kelvin. Atmospheres with atmospheres. You have a 1 over mole in the denominator. So then 1 over 1 over moles is just going to be moles. Now, this is going to be 54.4 moles of water vapor in the room to have our vapor pressure. If more evaporates, then more will condense-- we will be beyond our equilibrium. So we won't ever have more than this amount evaporate in that room. So let's figure out how much liquid water that actually is. Let me do it over here. So 54.4 moles-- let me write it down-- moles of H2O. That's going to be in its vapor form and its going to evaporate. But let's figure out how many grams that is. So what is the molar mass of water? Well, it's roughly 18. I actually figured it out exactly. It's actually 18.01 if you actually use the exact numbers on the periodic table, at least one that I used. So we could say that there's 18.01 grams of H2O for every 1 mole of H2O. And obviously, you can just look up the atomic weight of hydrogen, which is a little bit over 1, and the atomic weight of oxygen, which is a little bit below 16. So you have two of these. So 2 plus 16 gives you pretty close to 18. So this right here will tell you the grams of water that can evaporate to get us to that equilibrium pressure. So let's get the calculator out. So we have the 54.4 times 18.01 is equal to 970-- well, we only have three significant digits-- so 900, if your round this 0.7, it becomes 980. So this is 980 grams of H2O needs to evaporate for us to get to our equilibrium pressure, to our vapor pressure. So let's figure out how many milliliters of water this is. So they tell us the density of water right here. 0.997-- let me do this in a darker color-- 0.997 grams per millileter. Or another way you could view this is for everyone 1 milliliter you have 0.997 grams of water at 25 degrees Celsius. So for every milliliter-- this is grams per milliliter-- we want milliliters per gram because we want this and this to cancel out. So we're essentially just going to divide 980 by 0.997. So what is that? Get the calculator out. So we have 980-- not cover up our work-- divided by 0.997 is equal to 980-- we'll just round this-- 983. So this is equal to 983. This and this canceled out, or that and that canceled out. So 983 milliliters of H2O. So we've figured out, using the Ideal Gas Law, that at 25 degrees Celsius, which was 298 kelvin, that 983 milliliters of H2O will evaporate to get us to our equilibrium vapor pressure. Nothing more will evaporate, because beyond that if we have higher pressure than that, then you'll also have more vapor going to the liquid state. Because you'll have more stuff bouncing here. So if this much volume of water evaporates, we'll have the state where just as much is evaporating as just as much is condensing. So you will never get to a higher pressure than that at that temperature. So going back to the question, we figured out that 983 milliliters of water will evaporate. The question was is that we placed 2 liters of water in an open container. So we just figured out that only 983 milliliters of that-- so that's a little bit less than a liter. So this is a little bit less than 1,000 milliliters, and this is 1 liter. So a little bit less than half of this will evaporate for us to get to our vapor pressure. So to answer our question-- will all of the water evaporate at 25 degrees Celsius? No-- if we're assuming the room is sealed-- well, no, all of it will not. Only a little bit less than half of it will.