### Video Transcript

A cubic container of volume 2.25
liters holds 0.750 moles of nitrogen gas at a temperature of 27.3 degrees
Celsius. What is the magnitude of the force
exerted on one wall of the container by the gas?

So in this question, we’ve got a
cubic container, which has a volume of 2.25 liters. In this container, we’ve got 0.750
moles of nitrogen gas. This gas is at a temperature of
27.3 degrees Celsius. We need to work out the magnitude
of force exerted on one wall of the container by the gas. So here’s a pictorial
representation of our cubic container, and let’s make it transparent so we can see
inside it as well.

Now we’ve been told that this
container is full of nitrogen gas, and we’ve been given certain information about
the container and the gas itself. For example, we’ve been told that
the volume of the container is 2.25 liters, the amount of gas that we have is 0.750
moles, and the temperature of the gas is 27.3 degrees Celsius. We’ve been asked to find the force
exerted on one wall, let’s say this top wall here, by the gas in the container. Now before we do anything with the
numbers that we’ve been given, let’s first convert them all to standard units. We’ve got volume in liters and a
temperature in degrees Celsius, and those are not standard units. We need to convert the volume into
meters cubed and the temperature into Kelvin. This way it saves us a lot of
bother later on where we don’t have to hassle with units if we’ve already converted
them.

So let’s start with the volume:
2.25 liters needs to be converted into meters cubed. The conversion is that one liter is
equal to 0.001 meters cubed. So 2.25 liters is 2.25 times 0.001
meters cubed. And this ends up being 2.25 times
10 to the power of negative three meters cubed. So we can say that the volume that
we’ve got is 2.25 times 10 to the power of negative three meters cubed. Now let’s look at the
temperature. We know that the temperature in
Kelvin 𝑇 sub K is equal to the temperature in degrees Celsius 𝑇 sub C plus
273. So we find that the temperature in
Kelvin is 27.3 plus 273, which ends up being 300.3 Kelvin. And so we can replace our
temperature as well with 300.3 Kelvin.

At this point, all of the
quantities we have are in standard units. We’ve got meters cubed, moles, and
Kelvin. So we can go on to using these
quantities to calculate what we’re looking for. Firstly, we can notice that we’ve
been given three of the quantities out of the ideal gas equation. The ideal gas equation says that
the pressure 𝑝 of a gas multiplied by the volume 𝑉 that it occupies is equal to
𝑛, the number of moles of gas that we have, multiplied by 𝑅, the molar gas
constant which is just a number, multiplied by 𝑇, the temperature of the gas. In this case, we already know what
𝑉, 𝑛, and 𝑇 are, which means that the only thing we don’t know is the pressure 𝑝
since of course the quantity 𝑅 is a constant. In fact, we can write it down here
on the left as well because we’ll be using it in calculations later.

So as we’ve already seen, we can
find out the pressure exerted by the gas. Why is that useful? Well if we know the pressure
exerted by the gas, then we can work out the force exerted by the gas. This is because pressure 𝑝 is
defined as the force exerted 𝐹 per unit area 𝐴. So if the gas exerts a certain
pressure, then we can work out the force that the gas exerts because that pressure
is the same as the force exerted per unit area. And since we’re looking for the
force exerted on one wall of the container, the area that we’re looking for in
question is the area of that wall 𝐴. So let’s go about doing this.

Firstly, we need to find the value
of 𝑝. To do this, we can rearrange this
equation here to give us 𝑝 is equal to 𝑛𝑅𝑇 over 𝑉. We know all the quantities on the
right-hand side. So all we need to do is to
substitute in the values. Plugging in our values of 𝑛, 𝑅,
𝑇, and 𝑉, we find that the pressure ends up being 831831 pascals, which means we
can write it down here. The pressure 𝑝 is 831831 pascals,
and we know that the pressure is in pascals because we’re working in standard
units. We converted everything into the
standard unit so we don’t have to faff about. We simply know that the calculation
will give us an answer in pascals. And at this point, we can start
looking at this equation.

We’re trying to find out that force
𝐹. We’ve just calculated the pressure
𝑝. In order to find out what that 𝐹
is therefore, we need to know the area 𝐴 of one wall of the container. To do this we can use the fact that
the container is a cube. In other words, its length in all
three dimensions is the same. Basically this length is the same
as this length and this length. So let’s just say that each length
is 𝑥 meters. Now we can recall that the volume
of a cube is found by multiplying the length by the height by the depth. In other words, the volume is given
by 𝑥 cubed, and what we’re trying to find is the area of one wall of the
container. This area is found by multiplying
this length which is also 𝑥 meters and this length, surprise surprise, 𝑥 meters
again. This is why having a cube makes
life a lot easier for us.

So the area of the wall that we’re
trying to find is the same as 𝑥 multiplied by 𝑥, which is 𝑥 squared. So we can find a relationship
between the volume of the cube 𝑉 and the area of the wall of the container 𝐴
because we’ve now got them both in terms of 𝑥. So let’s take our volume equation
and solve for 𝑥. To do this, we need to take the
cube root of both sides of the equation. This gives us the cube root of 𝑉
is equal to the cube root of 𝑥 cubed. But the cube root of 𝑥 cubed is
simply 𝑥. Now if we square both sides of the
equation, we get the cube root of 𝑉 squared is equal to 𝑥 squared. But we know from here that 𝑥
squared is equal to 𝐴. Therefore, the cube root of 𝑉
squared is equal to 𝐴. And so now we’ve got the area of
one wall of the container in terms of the volume of the container, which means that
we can go about finding out the force exerted by the gas on the wall of the
container.

First let’s rearrange this equation
to give us the force exerted is equal to the pressure multiplied by the area. We’ve done this by multiplying both
sides of the equation by 𝐴. And then we can substitute in the
fact that 𝐴 is equal to the cube root of 𝑉 all squared. At this point, all that’s left is
to plug in our numbers, which looks something like this and gives us an answer of
14283 point something something newtons. However, all of the quantities that
we’ve been given in the question have been given to three significant figures. So we need to give our answer to
three significant figures as well. So here’s the third significant
figure. The one after that is an eight. That’s larger than five, so this
third significant figure will round up to a three. Doing this gives us 𝐹 is equal to
14300 newtons to three significant figures. We can also give this answer in
kilonewtons if we wish because one kilonewton is the same as 1000 newtons. So we get 𝐹 is equal to 14.3
kilonewtons, and we can state either of these as the final answer to our
question.