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## Chemistry library

### Course: Chemistry library > Unit 10

Lesson 1: Ideal gas equation- The ideal gas law (PV = nRT)
- Worked example: Using the ideal gas law to calculate number of moles
- Worked example: Using the ideal gas law to calculate a change in volume
- Gas mixtures and partial pressures
- Dalton's law of partial pressure
- Worked example: Calculating partial pressures
- Worked example: Vapor pressure and the ideal gas law
- Ideal gas law
- Calculations using the ideal gas equation

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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.

## Want to join the conversation?

- 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)
- 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)

- where did the number 760 mm Hg come from?(17 votes)
- 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)

- At 25 degrees Celsius, shouldn't none of the water evaporate because the boiling point of water is 100 Celsius?(22 votes)
- 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)

- I can't still understand what vapour pressure is? please help me understanding it.(10 votes)
- 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)

- How do you measure mercury's pressure in millimeters?(6 votes)
- 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)

- 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)
- 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)

- 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)- 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)

- 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)- 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)

- 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)
- 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)

- At7:25, 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)
- 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.