# Video: Comparing the Pressures Produced by Volumes of Air and of Water

Pressure is produced by fluids due to collisions between the particles in the fluid and objects that the fluid is in contact with. The smaller the average distance between particles is, the more frequently they collide. How does the average distance between particles in air compare to the average distance between particles in water? Figure 1 shows two graphs. One graph shows air pressure change with height above sea level. The other graph shows total pressure change with depth below sea level. How does the average distance between particles in a volume of water vary with the total weight pushing down from above the particles? How does the average distance between particles in a volume of air vary with the total weight pushing down from above the particles? Air particles exert a force of 2625 N on the surface of a high-altitude weather balloon. The balloon’s surface area is 750 cm². Use these values and the data shown in Figure 1 to show that the balloon is floating more than 5 km above sea level. Figure 2 shows a submarine’s floatation chamber, which can be filled with seawater through an outlet and with air through the cap of a compressed air tank. The area of the outlet is much larger than the area of the compressed air tank’s cap. The pressure due to the compressed air and the external seawater is the same. How does the force of the compressed air compare to the force of the external seawater?

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### Video Transcript

Pressure is produced by fluids due to collisions between the particles in the fluid and objects that the fluid is in contact with. The smaller the average distance between particles is, the more frequently they collide. How does the average distance between particles in air compared to the average distance between particles in water?

To consider this difference, let’s start out with a bit of a thought experiment. Imagine that we have a very tall cup which is currently empty; that is, it’s filled with air just like the cup is surrounded by air. Then, say that we fill the cup up with water. At this point, the walls of this cup have one kind of fluid on one side and another type on the other.

If we look at a zoomed-in view of the wall of the cup, we would see something like this, where the wall of the cup separates the water on one side and the air on the other side. We want to compare the average distance between particles in these two fluids. And the problem statement tells us that this average distance is related to the pressure exerted by the fluid.

If we consider the wall of this cup, we know that it’s under pressure from these two fluids. The water pushes outward on this wall and the air pushes inward. If we could figure out which of these pressures is greater, that would tell us about the relative average distance between particles in these two fluids.

So which fluid water or air exerts a greater pressure on the wall of the cup? One way to find this out is to continue our thought experiment by imagining that the cup suddenly disappears. If the cup were all of a sudden to vanish, there would no longer be a boundary between these two fluids. We can imagine that with the cup gone that this whole column of water would start to descend under the influence of gravity, but the bottom of this column would also start to push out backward against the air.

At this air-water interface then, we would see evidence of the water pushing on the air harder than the air is pushing on the water; that is, the water pressure is greater than the air pressure. And that tells us something about the relative average distance between particles in these fluids.

Here’s what we can write about that. We can write that the average distance between particles and air is much greater than the average distance between particles in water. That greater average distance between particles in air means there are far fewer collisions per unit time and therefore the pressure is less. Let’s continue thinking about these two different fluids. And specifically, we’ll consider whether this average distance between particles might vary in a column of air or in a column of water.

Figure one shows two graphs. One graph shows air pressure change with height above sea level. The other graph shows total pressure change with depth below sea level. How does the average distance between particles in a volume of water vary with the total weight pushing down from above the particles?

Looking at figure one, we see it consists of two graphs, each one showing us height above sea level on the vertical axis and pressure in kilopascals on the horizontal. That said as we look more closely, we see an important difference between the way these axes are labelled.

Notice in the graph on the left that the height is measured in kilometers and all the values on the axis are positive. But on the other graph, our heights above sea level are measured in meters and the values we see are negative, less than zero.

This tells us that our graph on the left describes the fluid which stays above sea level; that is, it describes the behavior of air. On the other hand, since our graph on the right shows negative height values relative to sea level, that means these are really depth values and that the chart refers to water.

Since our question has to do with water, that means we’ll be focusing on this graph on the right. Here’s what this question is getting at. What it means is as we start out at sea level and go further and further down, does the average distance between the water in this water column that’s getting longer and longer change as we go farther down?

For example, we might suspect that as we get farther and farther down the column, the increasing pressure crushes the water molecules closer together. That’s one possibility for what might happen. But to find out what really does happen, we’ll consult the curve shown on this graph.

Looking at this curve, perhaps the first thing that stands out is that it seems to be a straight line. It seems that the slope of this line is constant throughout. Here’s what that tells us: imagine that we had a column of water that was five meters deep. So on our curve, that would be halfway between our minimum and maximum depths. And imagine further that we were somehow able to measure the pressure at the very bottom of this column. And let’s say this pressure reading came back at a value we can call 𝑃 sub one.

Going back to our curve, the fact that the slope of this curve is constant means that if we added another five-meter-tall stack of water and we were able to measure the pressure at the bottom of this now 10-meter-tall stack, that pressure we’re told by our graph would be equal to twice 𝑃 sub one. This tells us that the weight of our bottom five-meter-tall stack of water is the same as the weight of our top stack.

In other words, the density of water stays constant throughout this 10-meter-tall column. If we were to compare the average distance between water particles in the top of this column to the average distance between particles at the bottom, we would find they’re effectively the same.

This happens because water is a nearly incompressible fluid. No matter how hard we press on it, the average distance between particles of water doesn’t change very much.

Here is what we can say then as an answer to our question. We can say that the average distance between particles in a volume of water remains constant as the total weight pushing down from above the particles increases. That’s how water behaves as the water column gets taller and taller.

Now, let’s see about air. The question is how does the average distance between particles in a volume of air vary with the total weight pushing down from above the particles. Since our focus is now on air in the graph on the left, we can clear away the graph on water to create some space for our discussion. Looking at this plot of air height versus pressure, something about it may seem confusing.

If we look at the heights of 10 and 20 kilometers above sea level, we notice that the pressure that’s associated with a 20-kilometer altitude is less than the pressure associated with a 10-kilometer altitude. Why might that be? We can sketch out both sea level as well as 10 kilometers elevation, 20 kilometers, and 30 kilometers above sea level.

On this sketch, let’s imagine that we’re at an altitude of 20 kilometers and that 30 kilometers is the top of an air column in this atmosphere. At this elevation of 20 kilometers, that means we have 10 kilometers of an air column above us pressing down on us at this altitude.

On the other hand, if we’re at an altitude of 10 kilometers above sea level, then that means there’s a 20-kilometer-tall column of air pressing down on us at this altitude. This explains why the pressure on our graph at lower altitudes is higher than the pressure at higher altitudes. It’s because the column of air pressing down on us is increasingly tall as we get closer and closer to sea level.

What we’re interested in understanding is does the average distance between particles of air in this air column vary as we move through the column up and down. To figure that out, we can refer to our graph and we’ll again use the attitudes of 10 and 20 kilometers above sea level, respectively.

Say that we had a 10-kilometer-tall stack of air and just like we did with our water column, we will somehow get to the bottom of it and measure the pressure of that stack. We’ll say that the pressure due to this 10-kilometer-tall column of air is 𝑃 sub two.

And then, just like we did for water, let’s add another stack of the same height on top of this one. And we’ll go to the bottom of the total combined height and again measure the pressure. What will we find to be the pressure reading at the bottom of this now 20-kilometer-tall column of air. Our answer based on the shape of our curve is that we’ll measure a pressure, which is greater than two times 𝑃 sub two.

In other words, the weight of the bottom 10 kilometers of air in our column is greater than the weight of the top 10 kilometers of air. That tells us that the density of air from top to bottom in this column is not constant like it was for water. Rather the farther down in the column we go, the more closely packed the air particles are. This tells us then that air, unlike water, is a compressible fluid.

And as regards the average distance between particles of air throughout this column, here’s what we can say. We can write that the average distance between particles in a volume of air decreases as the total weight pushing down from above the particles increases.

We know this is true based on the way that the slope of the curve in our graph changes. Because the slope flattens out as we move from left to right, that means that doubling the height of our air column will more than double the pressure at the bottom of that column.

Speaking of pressure, let’s consider an application of this concept next.

Air particles exert a force of 2625 newtons on the surface of a high-altitude weather balloon. The balloon surface area is 750 square centimeters. Use these values and the data shown in figure one to show that the balloon is floating more than five kilometers above sea level.

In figure one, we see the remaining plot which shows height above sea level in kilometers versus pressure in kilopascals. We want to apply the data we find in this plot to the scenario in order to show that the weather balloon is at least five kilometers above sea level.

To start out, let’s consider our balloon, which we know is surrounded by air particles. These air particles as they collide with the balloon create a force acting inward on it. And this force is distributed over the balloon’s surface area.

We can recall that a force spread over an area gives rise to pressure. In particular, pressure in units of pascals is equal to force in units of newtons divided by area in units of meter squared. As we look to calculate the pressure acting on this weather balloon, we’re given the force acting on it in units of newtons and we’re also given the area. But the area notice is given in units of square centimeters.

Before we use that value in our calculation, we want to convert it into units of meters squared. Here’s how we can do that. We recall that 100 centimeters is equal to one meter. If we then square both sides of this equation, we find that 10000 square centimeters is equal to one meter squared. That’s the conversion ratio between square centimeters and square meters.

So let’s use that ratio. We’ll take our given area of the weather balloon in square centimeters, 750, and we’ll multiply it by this conversion factor. Notice that as we do, the units of square centimeters cancel out and we’re left with units of square meters. We find a result of 0.075 square meters. That’s the area of the weather balloon in SI standard units.

Now that we know the area of the balloon and the force acting on it in the units we want, we’re ready to plug in and solve for the pressure acting on the balloon. 2625 newtons divided by 0.075 square meters is equal to 35000 pascals or — put in other way — 35 kilopascals.

Now that we have the pressure acting on our balloon in units according to our graph, we can use our graph to approximate the altitude of the weather balloon. Remember that we want to show that the balloon is at at least five kilometers above sea level.

If we find that elevation on our graph, it’s halfway between zero and 10. And if we find out the corresponding pressure to five kilometers of elevation, we see that it’s greater than 50 kilopascals. We don’t know exactly what that pressure is. But we can see it is greater than 50.

Now, remember our calculation which found that the pressure acting on this weather balloon is 35 kilopascals. If we locate that value on the horizontal axis of our plot, we don’t know exactly where will it lie, but we do know that it’s less than 50; it’s to the left of 50.

If we then look for the corresponding height above sea level for this pressure, we see that it’s in the neighborhood of 10 kilometers, again not an exact value. But here is what we do know for certain.

Because an elevation above sea level of five kilometers corresponds to a pressure of greater than 50 kilopascals and since the pressure we found acting on our weather balloon is less than 50 kilopascals, then based on the shape of our curve. That means the elevation of our weather balloon must be greater than five kilometers from sea level. With that then, we’ve demonstrated what we wanted to show.

Finally, let’s consider an example where we’re below sea level, under water.

Figure two shows a submarine’s flotation chamber, which can be filled with seawater through an outlet and with air through the cap of a compressed air tank. The area of the outlet is much larger than the area of the compressed air tank’s cap. The pressure due to the compressed air and the external seawater is the same. How does the force of the compressed air compared to the force of the external seawater?

Looking at figure two, we see a diagram of a floatation chamber, which is standard in submarines. We see that it’s possible for air from a compressed air tank to enter the floatation chamber and it’s also possible for seawater from the outside to enter the chamber through the outlet. The ratio of air and water in the flotation chamber is varied, depending on whether the submarine wants to ascend, descend, or keep at the same depth.

We’re told in the problem statement that the area of the outlet — the area shown here — is much larger than the area of the compressed air tank’s cap, shown here. However, we’re also told that the overall pressure exerted by the seawater to get into the floatation chamber and the overall pressure exerted by the compressed air is the same.

Knowing all that, we want to compare the force of the compressed air to the force of the external seawater. We can do this by making use of the relationship between pressure, force, and area. Recall that pressure, 𝑃, is equal to a force spread over some area, 𝐴.

In our situation, we have two pressures: the pressure of the seawater — we’ll call 𝑃 sub 𝑤 — and the pressure of the air — we’ll call 𝑃 sub 𝑎. And furthermore, we’re told that these two pressures are equal to one another. We can build off this fact by using our expression for pressure; that it’s a force over an area.

We can say that the force of the seawater on the outlet divided by the area that the water has to get through — that is the area of the outlet — is equal to the force of the compressed air acting on its cap divided by the area of that cap — we’ll call it 𝐴 sub 𝑎. We remember now that we’re also told something about 𝐴 sub 𝑤 and 𝐴 sub 𝑎 — the area that the water and the area that the air have to get into the floatation chamber, respectively.

In particular, we’re told that the area that the water has to get into the chamber is much bigger than the area that the air has. The area of the outlet is bigger than the area of the compressed air tank cap. We can use this fact to help us understand how the two forces 𝐹 sub 𝑤 and 𝐹 sub 𝑎 relate to one another.

To do that, let’s cross multiply so that our areas are on the right side of this equation and our forces are on the left. Our equation now tells us that the force of the water to the force of the air is equal to the area the water has to enter the chamber to the area the air has.

And since 𝐴 sub 𝑤 is greater than 𝐴 sub 𝑎, that means this fraction must be greater than one. Not only that, since the fraction is also equal to 𝐹 sub 𝑤 over 𝐹 sub 𝑎, that means that this ratio must also be greater than one.

If we multiplied both sides of this inequality by 𝐹 sub 𝑎, the force of the air on a cap, what we find is that 𝐹 sub 𝑤 is greater than 𝐹 sub 𝑎. This means that the force of the seawater pushing on the outlet is greater than the force of the compressed air pushing on the tank cap.