### Video Transcript

In this video, we’re learning about
Gay-Lussac’s law. This law helps us to understand
what happens to a particular type of gas when its temperature changes while it’s
held inside a fixed volume container.

To start getting a sense for this
law, let’s imagine that we have a container of fixed volume and inside the container
is some gas. We know that this gas because it is
a gas is made up of particles that are in motion in every sort of direction. We also know that as these gas
particles move around within this chamber. They move with some average or
overall speed. And it’s this speed which indicates
the temperature of this gas: the faster the average speed of the particles, the
higher the temperature of the gas. Because these particles are moving
around every which way, they collide with one another and especially they collide
with the walls of this container.

For example, if we were to take a
zoomed-in view of a section of the container wall and watch it for some amount of
time, then we would see particles of this gas approach the wall, run into it, and
bounce off. And as we kept watching, this will
continue to happen with other particles in the gas. Each one of those collisions
between a gas particle and the wall of the container exerts a force on that
container wall. In each case. there’s some
component of that force which points straight out perpendicular to the wall
surface. If we add up all the forces pushing
out on the wall over some amount of time and over some particular area of the wall
surface, then what we’re getting a measure of is the pressure that this gas
exerts.

We can recall that the units of
pressure where pressure is represented using a capital 𝑃 are equal to newtons, unit
of force, divided by square metres, the unit of area. And notice something about this
pressure on the container wall. When we consider the different gas
particles running into the wall, the faster they’re moving, the more force they
exert pushing outward. And as we said, the faster the
average speed of these particles, the higher the temperature of the gas. This tells us that there is a
relationship between the pressure a gas exerts and the temperature of that gas. These two quantities are
connected.

To see how, let’s imagine we do an
experiment with this gas. Let’s say that we start to heat
it. We start to put heat energy into
this gas in the chamber. When that happens, the average
speed of the particles picks up. Their velocity vectors are
longer. And that means when we look again
at an up-close view of our container wall, that the particles now running into the
wall are coming in with a higher speed and therefore they exert a stronger force on
the wall. In other words, the total outward
push on the container wall is increasing. Looking over here at our units for
pressure, we see that if the overall force is growing, but our area is staying the
same, then the pressure itself must be increasing.

When it comes to pressure and
temperature then of this gas in this fixed container, what we’re seeing is that when
temperature goes up, so does pressure. A higher temperature means faster
moving gas particles. And those faster moving particles
push harder on the container when they run into it, increasing the pressure. And if we continue this experiment,
say we increase the temperature of our gas even more by adding in more heat, then in
that case our particles will move even faster and therefore hit against the wall of
the container even harder. And so once again, as we increased
temperature, we would see an increase in pressure.

We can describe this relationship
between pressure and temperature this way. We can say that pressure is
proportional to temperature. That means that as temperature
changes, say we increase it by doubling it, then pressure will change the same
way. It will also double. Another way we can write this
expression, 𝑃 is directly proportional to 𝑇, is to say that pressure is equal to
some constant value — we’ll call it capital 𝐶 — multiplied by 𝑇. This constant capital 𝐶 is
sometimes called the constant of proportionality because it’s what makes 𝑃
proportional to 𝑇.

Now notice what happens if we do a
bit of rearrangement with this equation. Let’s say we divide both sides of
the equation by the temperature 𝑇. In that case, that term cancels on
the right-hand side. Then, what we find is that 𝑃
divided by 𝑇 is equal to some constant. An important thing to realise is
that for this relationship to be true, the volume of the gas we’re working with
needs to be held constant. It has to be fixed. In other words, the size of the
container the gas is in must stay the same. But if it does, then we can say
that as temperature goes up, pressure will go up by the same amount and vice versa,
as temperature goes down, pressure will also decrease by that amount.

When we write the relationship this
way using an equal sign, then this relationship goes by the name of the person who
discovered it. It’s called Gay-Lussac’s law, named
after Joseph Louis Gay-Lussac. We talked a bit earlier about the
units of one of the terms in this equation, pressure. In order to understand the law more
fully, it’s very helpful to also consider the other term temperature. Here’s a question: in the SI
system, what are the units of temperature? We might be most familiar with
temperature on the Fahrenheit scale, abbreviated degrees F. Or perhaps, we’ve seen how to
convert between degrees Fahrenheit and another temperature scale degrees
Celsius. In fact, neither one of these
scales represents the SI temperature scale.

The proper scale to use for SI
calculations involving temperature is the Kelvin scale, abbreviated capital K. And interestingly, while
temperatures in the Fahrenheit and Celsius scales are called degrees Fahrenheit and
degrees Celsius, in the Kelvin scale, we don’t use the word degrees. We just say kelvin. Because these three temperature
scales, Fahrenheit, Celsius, and Kelvin, are different, they report temperatures
differently. For example, on the Fahrenheit
scale, we know that water boils at 212 degrees. 212 degrees Fahrenheit is equal to
100 degrees Celsius and that’s equal to approximately 373 kelvin.

Now, let’s say we we’re doing a
calculation involving Gay-Lussac’s law. And we wanted to calculate the
ratio of a certain gas’s pressure to its temperature. If that gas was at 100 degrees
Celsius or 373 kelvin, we can see that when we go to actually substitute in the
numerical value of the temperature, it matters very much which temperature scale we
use. Depending on the scale we pick, we
get a very different answer. Whenever using Gay-Lussac’s law or
any equation involving temperature in the SI system, we always want to use the
Kelvin temperature scale rather than Celsius or Fahrenheit.

Before we look at an example to get
some practice with this idea, let’s look at a second way of writing out Gay-Lussac’s
law. And to do it, let’s go back to
considering this fixed-volume chamber of gas that we’ve been heating up. Let’s say that back towards the
beginning of the experiment when our heating flame with just this big and the
temperature of our gas wasn’t so high, let’s say the pressure of the gas at that
time was 𝑃 one and the temperature of the gas at that time was 𝑇 one. As we know though, we increased the
heating of this gas so that the average temperature went up and the pressure on the
wall of the container went up. We can say that at that latter
moment, our gas now has a pressure 𝑃 two and a temperature 𝑇 two.

Gay-lussac’s law in telling us that
pressure divided by temperature is a constant for a gas at fixed volume indicates
that if we take the ratio 𝑃 one divided by 𝑇 one, then that’s equal to the ratio
𝑃 two divided by 𝑇 two. In other words, if we have a gas
which goes through a change so that its pressure and therefore its temperature are
different, we can write that the ratio of pressure to temperature before the change
is equal to that same ratio after the change. This is what it means to say that
at any one instant, the pressure of the gas divided by its temperature is a
constant. Now, let’s look at an example of
Gay-Lussac’s law.

A gas is heated from 19 degrees
Celsius to 80 degrees Celsius, while the volume is kept constant. If the initial pressure was 2000
pascals, what is the pressure of the gas after it has been heated? Give your answer to three
significant figures.

Alright, so in this example, we
have a gas which starts out at 19 degrees Celsius, but then it’s heated and its
temperature goes up to 80 degrees Celsius. We’re told when the temperature of
the gas is 19 degrees Celsius, it has a pressure of 2000 pascals. Then as the temperature of the gas
changes, we want to see what impact — if any — this has on the pressure of the
gas. As we consider what kind of method
would help us figure out this final gas pressure, we find a clue in the problem
statement: this gas was heated and the volume was kept constant.

When the volume of a gas is held
constant over some process, that means that Gay-Lussac’s law applies to that
process. This law tells us that for a gas
held at constant volume, the pressure of that gas divided by its temperature is
equal to a constant value. In other words, even as the
pressure or the temperature of the gas will change, so long as the volume of the gas
is constant, that ratio 𝑃 divided by 𝑇 is always the same.

Another way of writing this is to
consider some change that happens in a gas and to say that the pressure divided by
the temperature before the change 𝑃 one divided by 𝑇 one is equal to the pressure
divided by temperature after the change. This second way of writing the law
is especially helpful for us because we indeed have a gas that has gone through a
change. It’s been heated. So we can say the initial pressure
of our gas divided by its initial temperature is equal to its final pressure —
that’s what we want to solve for — divided by its final temperature.

Since it’s 𝑃 two, the final gas
pressure that we want to solve for, let’s rearrange this equation to let us do
that. If we multiply both sides of the
equation by 𝑇 two, the temperature of the gas after it’s heated, then that term
cancels on the right-hand side and we have 𝑃 two all by itself. We can write the resulting equation
this way. We can say that 𝑃 two, the final
pressure of the gas, equals 𝑃 one times the temperature ratio 𝑇 two divided by 𝑇
one.

Now at this point in our solution,
it’s important to notice something. See the temperatures we’ve been
given in the problem statement, 19 degrees Celsius; 80 degrees Celsius. That is these temperatures are
given as values on the Celsius scale. But that’s not the SI unit of
temperature. The SI temperature unit is
kelvin. In other words, before we plug in
𝑇 two and 𝑇 one for our calculation, we want to convert these given temperatures
in to temperatures on the Kelvin scale. If we forgot to do that, our answer
would be off.

So we want to know how to convert a
temperature from degrees Celsius to kelvin. And roughly speaking, a temperature
in kelvin is equal to that same temperature in degrees Celsius plus 273. So for example, if we had a
temperature in Celsius of zero degrees, that is freezing, then the equivalent
temperature in kelvin would be zero plus 273 or 273 kelvin. Using this relationship, let’s
write out the initial and final temperatures of our gas in kelvin. 𝑇 one, the initial temperature of
the gas, will be equal to 19 plus 273 and all this will be in kelvin. All that together is 292
kelvin. This value in kelvin is the value
we’ll use to substitute in for 𝑇 one in our equation.

Next, what about 𝑇 two, the
temperature of our gas after it’s been heated? That’s equal to 80 plus 273 all in
kelvin. And this equals 353 kelvin. That’s the temperature we’ll use
for 𝑇 two in our equation. With these values plugged in,
notice that the units kelvin cancel out. But be careful. This doesn’t mean that if we had
used degrees Celsius instead of kelvin, the units would also have cancelled and
therefore would have made a difference. It’s true that if we had forgotten
to convert our initial temperatures in degrees Celsius into kelvin and simply plug
them into the equation, then those units would have also cancelled.

But notice what our fraction would
have been. Instead of 353 divided by 292, we
would have had a fraction of 80 divided by 19. These two fractions are not the
same thing. All that to say whenever we’re
working within the SI system as we usually are, then whenever an equation involves
temperature, we’ll want to make sure to use the Kelvin temperature scale in our
calculation. Okay, we’re almost ready to solve
for 𝑃 two, the final pressure of this gas.

In order to do so, let’s plug in
for 𝑃 one, the initial pressure. And that’s given us 2000
pascals. This unit, pascals, is the base
unit for pressure in the SI system. As the unit of pressure, a pascal
is equal to a newton of force divided by a square metre of area. So when we see 2000 pascals, that’s
another way of saying 2000 newtons of force per square metre of area. And that value can now go in for 𝑃
one in our equation. We’re now ready to calculate 𝑃
two, the pressure of the gas after it’s been heated. We see the units of this result
will be pascals. And when we calculate this
expression and round it to three significant figures, we find a result of 2420
pascals. This is the pressure of the gas
after it’s been heated.

Let’s summarise now what we’ve
learned in this lesson on Gay-Lussac’s law. Earlier on, we saw that for a gas
at constant volume, the gas pressure is proportional to the gas’s temperature. We can write that out this way: 𝑃
is proportional to 𝑇. And this means that for example if
the temperature doubled, then the pressure also would double.

Moreover, we saw that this
relationship implies another one that gas pressure is equal to a constant multiplied
by the gas temperature. And if we divide both sides of this
equation by the temperature 𝑇, we arrive at this formulation known as Gay-Lussac’s
law that the pressure of a gas divided by its temperature when that gas is held at
fixed volume is a constant. This implies that if we have a
system again held at constant volume with an initial and a final state, then we can
say that the pressure divided by the temperature of that initial system state is
equal to the pressure divided by the temperature at the final system state.

Along with that, we saw that
because we were working within the SI system, whenever we use temperatures in this
law, those temperatures must be in kelvin. If we’re given them in degrees
Fahrenheit or degrees Celsius, they’ll need to be converted to the Kelvin scale. And lastly, as we saw that the SI
unit of temperature is kelvin, we also learned that the unit of pressure is the
pascal, abbreviated Pa and that one pascal represents a newton of force spread over
one square metre of area.