Video: A Molecular Model of an Ideal Gas and Force due to Pressure

A cubic container of volume 2.25 L holds 0.750 mol of nitrogen gas at a temperature of 27.3°C. What is the magnitude of the force exerted on one wall of the container by the gas?

07:28

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 831831pascals, 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.

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