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.