Video Transcript
In this video, we’re going to learn
about Boyle’s law. This law shows us that when it
comes to certain kinds of gases, there’s a particular relationship between the
pressure of the gas on the one hand and the volume that the gas takes up on the
other. To get started talking about
Boyle’s law, imagine that we have this cloud of gas. As it is, this gas cloud is
completely uncontained. So it’s free to mix with other
gases in the atmosphere.
When it’s set up like this, we
can’t very well say what the volume of this gas is, how much space it takes up,
because there’s no boundary to it. Once the gas is within some sort of
container, it spreads out to fill that space. Once that’s done, we can now give a
clear answer for what the volume of this gas is. We know that it’s equal to the
width of our container multiplied by its height multiplied by whatever depth it
has.
We can write down a symbol for the
volume that this container of gas takes up. And since the gas perfectly fills
the container, we can say that the gas itself has a volume we’ll call 𝑉. Along with this volume, there’s
another property of this gas we like to know.
Say that we take a very up-close
view of a small section of this container wall. From this viewpoint, as we watch
over a span of time, we would see a number of molecules of this gas come in, bounce
off the wall, and then bounce back into the chamber. Each time a molecule bounces off
the wall, it exerts a very small force on the container. If we were to add up all the forces
experienced by the container wall over some interval of time and then divide that by
the area of the wall that’s experiencing these interactions, then we would’ve
calculated the pressure that this gas exerts on the container wall. We’ll symbolize that variable with
a capital 𝑃.
So we see that this gas has a
volume, an amount of space that it takes up. And it also exerts a pressure on
the walls of its container. And we saw that this pressure has
to do with the number of collisions between gas molecules and the container
wall.
Now that we’ve got those properties
figured out, let’s make some changes to our container. Let’s say that we keep the depth of
it the same but we move the other four walls in. When we do this, we’ll see that the
gas molecules become confined to a smaller space. When we make this change, one
question that comes up is, “How does this affect the volume and the pressure of the
gas?”
We can clearly see that the volume
of the gas has decreased. The depth of the container
containing it is unchanged. But the height and the width are
much smaller than before. So the gas is confined to a smaller
space. And as it fills it, it has a
smaller volume.
Okay, now what about the pressure
of the gas? If we again take a zoomed-in view
of a small section of the wall of our container, what we now see is that, over the
same interval of time as before, we have many more interactions between molecules of
the gas and the container wall. That’s because the gas molecules
are much more densely packed together. So overall, when we consider the
total or cumulative force of all these collisions on the area of this container
wall, we see that that force will be greater than before.
Since the force exerted over the
same area of wall has gone up, that means the pressure of the gas has risen. So we see that, by changing one
thing about this gas, the volume that it takes up, we’ve also changed its
pressure. And actually, this connection
between volume and pressure exists whether we shrink our container down, like we did
in this case, or if we expanded the container beyond its original volume.
By increasing the size of the
container like this, we make its volume go up. And now with the gas molecules all
spread out over this larger space, once more taking a zoomed-in view at a section of
the container wall, we can see that, over the same time period and over the same
wall area as before, there will just be fewer collisions between gas molecules and
this wall. And that’s because these molecules
are now so spread out within the large container volume. The overall effect is to decrease
the gas pressure from what it was before.
If we were to summarize the trend
that we’re seeing through these expansions and contractions of our gas container,
here is what we might say. We’ve seen that when our gas
container volume goes down, the pressure of the gas goes up and, on the other hand,
that when our volume increases, the pressure decreases. One way to express this type of
relationship between two variables is to say that they’re inversely proportional to
one another. We can write that like this. The gas volume is proportional to
one over the gas pressure.
And in fact, there’s another
mathematically equivalent way to write this out. When 𝑉 is proportional to one over
𝑃, we can say that 𝑉 is equal to some constant value — we’ll call it 𝐶 —
multiplied by one over 𝑃. We don’t know what this constant 𝐶
is quite yet. But because it makes 𝑉
proportional to one over 𝑃, sometimes it’s called the constant of
proportionality.
Anyway, taking this equation a step
further, look at what happens when we multiply both sides of the equation by the
pressure 𝑃. Doing that means that term cancels
out on the right-hand side. And we’re left with this expression
that says that the pressure of the gas multiplied by its volume is a constant.
Now at this point, it’s worth
mentioning a few assumptions we’ve made in this process. First off, we’ve assumed that the
mass of the gas we’re considering is constant. That means that as we expanded and
contracted our container, no gas molecules were able to escape or be added in. Overall, the mass was the same. We’ve also assumed that the
temperature of the gas involved is a constant value. We’re saying then that this gas
isn’t cooling down as it expands or heating up as it contracts.
In other words, all throughout
these changes of container size from smaller to larger, the average speed of the
molecules in our gas is staying the same. On average, they’re not moving any
more quickly or more slowly than they were before. So if these two assumptions hold,
if the mass of our gas and the temperature of the gas are both constant values, then
it’s the case that the gas volume is directly proportional to one over its
pressure. And like we said, that means that
the gas pressure at any one instant multiplied by its volume at that same moment is
equal to a constant value.
When we state the relationship this
way, we’re writing out the law known as Boyle’s law. Boyle’s law specifically has to do
with the gases that are considered ideal. For a gas to be considered an ideal
gas, that means the molecules of the gas occupy negligible space and also that they
don’t interact with one another, in other words attract or repel one another. We can see this idea of an ideal
gas is an approximation. After all, the molecules having
mass must interact with one another gravitationally. And also since they have mass, they
must take up some space.
But when these assumptions of
negligibly small molecule volume and negligibly small interactions between molecules
hold, then we can consider the gas an ideal gas. And once that’s the case, then if
these conditions we saw earlier are met, then we can say that the pressure of the
gas times its volume is equal to a constant value. As we’ll see, Boyle’s law is most
useful to us when something about our gas changes. For example, in this case, we’re
expanding the volume of the gas.
As we saw earlier, this change in
volume also creates a change in pressure, which means that the volume and pressure
of the gas initially — we can call them 𝑃 one and 𝑉 one — are different from the
volume and pressure of the gas after the change — we’ll call them 𝑃 two and 𝑉 two
to reflect this difference. Looking at Boyle’s law, this tells
us that if we take the pressure of this gas at any time and multiply it by the
volume of the gas at that same moment, the net product is equal to a constant
value. It doesn’t change.
Notice that this pressure and this
volume are general pressures and volumes of the gas. They occur at the same moment. That is, they correspond to one
another. But they could be at any time
either before or after the expansion of the gas container. So if this is true, and Boyle’s law
says it is, then that means we can relate 𝑃 one and 𝑉 one with 𝑃 two and 𝑉
two. After all, these are the pressure
and the volume of the same gas just at different instances.
Boyle’s law tells us that if we
have a gas that undergoes some change in its volume and pressure, then if that
change occurs while these assumptions of constant mass and constant temperature
hold. Then we can say that the pressure
of the gas initially before the change multiplied by its volume before that change
is equal to the pressure of the gas times the volume of the gas after the
change. And by the way, this law applies no
matter how many changes our gas goes through. Say after this initial change to a
larger volume, it changes to a smaller volume than before. So now we have a third pressure, 𝑃
three, and a third volume, 𝑉 three.
Well, Boyle’s law tells us that the
product of these two values, 𝑃 three and 𝑉 three, are equal to 𝑃 two times 𝑉
two. And that’s equal to 𝑃 one times 𝑉
one. That’s the meaning of this
expression. Pressure times volume is equal to a
constant.
Before we get a bit of practice
using Boyle’s law, let’s say a quick word about units. We see that this law involves
pressure and involves a volume. The standard units of volume are
cubic meters. And when we consider the units of
pressure, let’s recall our up-close view of the container wall with the gas
molecules running into it.
We saw that each time a gas
molecule runs into the wall, it exerts a force on it. When we consider the impact of many
gas molecules on a container wall, we know that that occurs over some area of the
wall. The pressure then that this gas
exerts on the wall is measured as a force, in units of newtons, over some area, in
units of square meters. And remember that this force in
newtons is the sum of the force of all the molecules spread over this particular
area as they run into the wall.
Now this unit, newton per meter
squared, is one way to express pressure. But there’s a shorthand unit for
this called the pascal. And it’s abbreviated Pa. Sometimes when we see a pressure,
it’s reported as 1000 pascals or 500 pascals. When we see that, we can recall
that one pascal is equivalent to one newton of force spread over a square meter of
area. Those units help remind us that
this truly is a pressure we’re talking about. Okay, with that said, let’s now
look at a couple of Boyle’s law examples.
A four-cubic-meter volume of gas is
at a pressure of 1000 pascals. The gas is allowed to expand at a
constant temperature until its pressure is half of the value before expansion
began. How many times greater is the
volume of the gas after its expansion?
Okay, in this scenario, we have a
gas which initially has a volume we’ll call 𝑉 one. And it has a pressure at first that
we’ll call 𝑃 one. Then we’re told that the gas
expands at a constant temperature so it fills some larger volume than before. After this expansion happens, we
know the volume of the gas will be different. We can call this new volume 𝑉
two. And the pressure of the gas will
change as well. We’ll call that 𝑃 two.
In our problem statement, we’re
told a relationship between 𝑃 one, the initial pressure of the gas, and 𝑃 two, its
final pressure. We’re told that the final pressure,
𝑃 two, is equal to one-half the initial pressure, 𝑃 one. Knowing that, we wanna figure out
how many times greater the volume of the gas is after its expansion compared to
before. In other words, we wanna solve for
the ratio 𝑉 two divided by 𝑉 one. That ratio will give us our
answer.
As we start working to figure that
ratio out, we notice in our statement that this gas expansion happens at a constant
temperature. This tells us that as the pressure
and the volume of this gas changes, that change is subject to a law known as Boyle’s
law. Boyle’s law tells us that, for an
ideal gas that expands or contracts at a constant temperature, the initial pressure
of the gas multiplied by its initial volume is equal to the final pressure times the
final volume. We can apply this relationship to
our scenario by writing 𝑃 one times 𝑉 one is equal to 𝑃 two times 𝑉 two.
Now let’s recall our goal. We wanna solve for the ratio 𝑉 two
divided by 𝑉 one. That will tell us how many times
greater the volume after the expansion of the gas is. We can set up that ratio using our
Boyle’s law relationship. If we divide both sides of the
equation by the initial volume of the gas, 𝑉 one, then that term cancels on the
left-hand side. And then if we divide both sides of
the equation by 𝑃 two, the pressure of the gas after the expansion happens, that
term cancels on the right-hand side.
Notice that, now on the right-hand
side of our equation, we have the ratio we want to solve for. And we can rewrite 𝑃 two, this
value on the left-hand side, in terms of 𝑃 one given by this relationship. When we substitute 𝑃 one divided
by two for 𝑃 two, then finally we can multiply the left-hand side of our equation
by two divided by two. The reason we do this is to cancel
out the factor of two in the denominator. This leaves us with the left-hand
side of two times 𝑃 one divided by 𝑃 one. So that factor 𝑃 one cancels as
well. And now we can see we’ve solved for
the ratio we wanted to solve for, 𝑉 two divided by 𝑉 one. That ratio is two, which means that
the volume of the gas is twice as big after the expansion as it was before.
Let’s summarize now what we’ve
learned about Boyle’s law in this lesson. Boyle’s law applies to ideal gases
that change while keeping their mass as well as their temperature constant. The law can be expressed by saying
the pressure of a gas multiplied by its volume at that same instant is equal to a
constant. For a gas that undergoes some type
of change in its pressure and volume, this can be rewritten to say that the initial
pressure times the initial volume is equal to the final pressure times the final
volume of the gas. And finally, we saw that the base
units of volume are cubic meters. And the pressure, which is equal to
a force in newtons divided by an area in square meters, is expressed in units called
pascals, abbreviated Pa.
