Question Video: Analyzing the Air in an Air Column | Nagwa Question Video: Analyzing the Air in an Air Column | Nagwa

# Question Video: Analyzing the Air in an Air Column Physics • Second Year of Secondary School

An air column is defined as all the air within a cuboid that has a cross-sectional area of one square meter that stretches from the ground to the top of Earth’s atmosphere, as shown in the diagram. The top of the atmosphere can be taken as 15 km above the surface of Earth. The pressure from the air at the base of an air column is 101 kPa. Find the weight of the air in an air column.

07:59

### Video Transcript

An air column is defined as all the air within a cuboid that has a cross-sectional area of one square meter that stretches from the ground to the top of Earth’s atmosphere, as shown in the diagram. The top of the atmosphere can be taken as 15 kilometers above the surface of Earth. The pressure from the air at the base of an air column is 101 kilopascals. Find the weight of the air in an air column.

In our diagram, we see what an air column looks like. From a different perspective, an air column could look like this. Both the top and the base of the air column have an area of one square meter and that column is 15 kilometers tall. An air column of course has air in it, and here we want to solve for the weight of that air. An important related fact is that the pressure at the base of the air column, the pressure on the one square meter here, is 101 kilopascals. Now, in general, a pressure is equal to a force spread out over some area. The pressure at the base of our air column is distributed over one square meter. And that pressure is due to the force of the weight of the air in the air column, what we’ll call 𝑊.

Note that the weight of the air is indeed a force. It’s the mass of the air in the column multiplied by the acceleration due to gravity 𝑔. Since we want to solve this equation for the weight of that air, let’s rearrange so that 𝑊 is the subject. We can begin doing this by reversing the sides of the equation and then multiplying both sides by one square meter, which cancels that factor on the left. We have then that the weight of the air in the air column equals 𝑃 𝑏, the pressure at the base of the column, times one square meter. And that pressure at the base is given as 101 kilopascals. Now, a kilopascal is equal to 1000 pascals, so we can rewrite the weight of the air in our air column as 101000 pascals times a meter squared.

And now let’s recall that a pascal is equal to a newton over a meter squared. Substituting a newton per meter squared in for pascals in our equation, notice that one over meter squared cancels with meter squared so that the final units in our result are newtons, units of force. And in fact, we get an answer of 101000 newtons, which is the weight of the air in the air column. And it’s also the force that’s exerted on the air column’s base.

Let’s store this result off to the side and then move on to part two of our question. Part two says this.

Find the mass of the air in an air column. Round your answer to the nearest kilogram.

So we’ve solved for the weight of the air in an air column. And now we want to solve for the mass of that air. We can begin to do this by recalling that the weight of an object equals that object’s mass times the acceleration due to gravity. In our case, since we’re fairly near Earth’s surface, we’ll take 𝑔 to be exactly 9.8 meters per second squared. So the weight of the air in our air column equals the mass of that air times 𝑔. Or dividing both sides of this equation by 𝑔 so that it cancels on the right, we find that 𝑚 is equal to 𝑊 divided by 𝑔. From before, we know that 𝑊 is 101000 newtons. And using our known value for 𝑔, we find that to the nearest kilogram, 𝑚 is 10306. This is the mass of the air in an air column, and we see it’s actually quite a lot.

Let’s also store this result off to the side, and now let’s move on to the next part of our question.

Part three says “Find the average density of the air in the air column. Round your answer to two decimal places.”

Our sketch of the distribution of air molecules in our column shows that the density of air in this column is not constant. Starting out at the top of the column, the closer we get to ground level, the more dense the air in the column is. Here, though, we’re looking to calculate the average density of the air in the column. Let’s recall that the density of an object equals that object’s mass divided by the volume that the object occupies. Therefore, the average density of air in the air column, we’ll call it 𝜌 sub avg, equals the mass of air in the column divided by the column’s volume.

We know that the base of the column, as well as its top, have one square meter of area. And if we multiply this by the height of the column, 15 kilometers, we’ll have the column’s volume. Before we find this volume though, let’s convert our height in kilometers to a height in meters. We recall that 1000 meters is a kilometer, which means if we multiply 15 by 1000, we’ll get the number of meters that’s equivalent to 15 kilometers. 15 times 1000 is 15000, and 15000 meters times one meter squared is equal to 15000 cubic meters. That’s the volume of our air column. If we then substitute in the value for the mass of air in an air column and round the resultant fraction to two decimal places, then we get a result of 0.69 kilograms per cubic meter. This is the average density of air in the air column, rounded to two decimal places.

We now copy this result off to the side and move on to the last part of our question.

Assume that the density of the air in the air column varies uniformly from the bottom to the top. What is the air’s density at the top of the air column if the density at the base of the column is 1.23 kilograms per cubic meter? Round your answer to two decimal places.

Earlier, we talked about how the density of air in the air column is not constant, that the air becomes more dense as we get closer to ground level. In this part of our question, we’re going to assume that as the density of air in the column varies, it varies uniformly from bottom to top. So far, we’ve calculated the average density of air in the column, and we’re now told the density of air at the base of the column. Based on all this, we want to solve for the density of air at the top of the column, 15 kilometers above ground level. Clearing some space to work, let’s label the density of air at the top of the air column 𝜌 sub 𝑡. Similarly, we’ll call the air density at the base of the column 𝜌 sub 𝑏.

Now, and this is a crucial point, because the density of air varies uniformly in this column, we know that if we go halfway up the column a height of 7.5 kilometers, then the air density at the midpoint of the column’s height will be the average density of air in the column, 𝜌 sub avg. We know this average density, and we’ve been given the density at the base of our air column. 𝜌 sub 𝑏 is 1.23 kilograms per cubic meter. So we know 𝜌 sub 𝑏, we know 𝜌 sub avg, and we want to solve for 𝜌 sub 𝑡. To sketch this out in a slightly different way, if we imagine starting at zero kilometers of elevation, we know the air density there is 𝜌 sub 𝑏. Then, 7.5 kilometers up from that, the air density is 𝜌 sub avg. And 7.5 kilometers up from their, 15 kilometers total, the air density is 𝜌 sub 𝑡.

The fact that air density varies uniformly throughout the height of this column means that if we take the difference between 𝜌 sub avg and 𝜌 sub 𝑏, that is, if we subtract the average air density from the air density at the base of the column, then whatever that difference is will be equal to the difference between 𝜌 sub avg and 𝜌 sub 𝑡. That is, this difference here is equal to this difference here. We can write this as an equation. We can say that the density of air at the base of the column minus the average density of air in the column equals the average air density in the column minus the density of air at the top of the column.

In this equation, we know 𝜌 sub 𝑏 and 𝜌 sub avg. And we want to solve for 𝜌 sub 𝑡. If we multiply both sides of this equation by negative one, then that changes all of the signs in front of all of the terms. Then, if we add 𝜌 sub avg to both sides of the equation, we get this result: the density of air at the top of the column equals two times the average air density minus the density at the base of the column.

We’re just about ready to substitute in and solve for 𝜌 sub 𝑡. As we do, an important note is that we won’t use this rounded answer that we got for 𝜌 sub avg. Instead, we’ll substitute in the exact fraction we used to calculate 𝜌 sub avg so that we don’t lose any accuracy. That fraction, we may recall, is equal to the mass of the air in the air column divided by the column’s volume. Here, we’ve also substituted in the density of air at the base of the air column. So when we enter this expression on our calculator, to two decimal places, we get a result of 0.14 kilograms per cubic meter. This is the density of the air at the top of the air column.

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