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ემზადებით გამოცდისთვის? მოემზადეთ ამ გაკვეთილის დახმარებით შემდეგ თემაზე: Gases and kinetic molecular theory
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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.