# GG 140: The Atmosphere, the Ocean, and Environmental Change

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# The Atmosphere, the Ocean, and Environmental Change

## GG 140 - Lecture 3 - The Perfect Gas Law

Chapter 1: SI System of Units [00:00:00]

Professor Ron Smith: Last time, we talked about how planets retain their atmospheres. And I wonder if there are any questions about that discussion we had last time? Just to review, we defined the escape velocity, we defined the molecular speed, and then we talked about the relationship between those two in regards to how an atmosphere can be retained by a planet. Anything on that?

OK, let’s get started then with a new subject. You know I promised I would spend a few minutes beginning today to talk about the system of units we’re going to be using. So if you don’t object, I’m going to spend about ten minutes or so on that. I’m sure it’s a review for most of you, but we’re going to talk about this thing called the SI system of units. We’ll be using that throughout the course primarily, although there are some traditional units that come up in meteorology and oceanography that don’t necessarily fit into this SI system.

SI is–well, it’s French. It’s Systeme International. We just say the International System of Units. It might also be called the metric system. And you’ll see why in just a minute. So the three basic–or the foundation blocks for the SI system of units are mass, for which we use the unit kilograms, length, for which we use the unit meter, and time, for which we use the unit seconds.

Now, there are some other fundamentals that involve electric field strength, magnetic field strength. We’re not going to be using those. So I’m going to just work with these three and then see what we can build up based on this foundation. So with these as a foundation, we’ve got a bunch of really simple things we can write down. For example, speed, the rate at which something is moving, is going to be meters per second, just taking the length of meters and the time of seconds.

By the way, in current scientific convention, it’s more appropriate to write this m, seconds to the minus one . Sometimes I’ll use this. Sometimes I’ll use that. Today, this is the more preferred. In the scientific literature, you’ll normally find this inverse operator used to indicate that it’s per second.

Acceleration, of course, is going to be meters per second per second, so we could write that as meters, second to the minus two . It’s how fast is the speed changing. A few other easy ones, area would be meters squared . Volume would be meters cubed .

And then we’ll move into the ones that are a little bit trickier. Let’s start with force. Now, the SI unit of force is the Newton, named after Sir Isaac, of course. But it’s not a fundamental unit. We can derive it from these. And the way I do that is just to remember Newton’s Law,. So the unit of force, which is a Newton, can also be written as the product of–well, I can write it this way. It’s going to be kilograms, meters, per second squared . So that’s another way to write a Newton.

We could do a pressure. Pressure is going to be the subject of most of today’s lecture. A pressure is a force per unit area, a force per unit area. The SI name for it is the Pascal, named after the French scientist who made a fundamental breakthrough in understanding pressure.

And of course, we can write it in terms of the three building blocks by realizing, if that’s a force per unit area, then a Pascal is going to be a kilogram, meter per second squared. And then per area, so it’s meters to the minus two . Let’s simplify that. So it’s kilograms, meters to the minus one, seconds to the minus two . That is the pressure unit called the Pascal.

Energy, well, the way I think of energy is that it’s the amount of work done as I push something along. It’s usually the product of the force times the distance. Force pushing over some distance is an amount of energy. Well, that makes it easy if you can remember that, because then I can take the unit of force–oh, by the way, the unit of energy in the SI system is going to be the Joule, J-O-U-L-E. And of course, that is going to be this times an extra distance, so it’s going to be kilograms, meters squared, seconds to the minus two . That will be a Joule.

Power, well, power is the rate of expending energy. How fast are you using or creating energy? The SI system unit for power is the familiar watt. You’ll find it stamped on your light bulbs. It’s a power unit. And we know immediately what that’s going to be, because we have energy here. But this is per unit time, so that’s going to be kilograms, meters squared, second to the minus three . I’ve changed that two to a three to get it into a per unit time system.

Any questions on this yet?

OK well, one we’ll be using today also is the mass density. It’s how much mass of a fluid or an object is there per unit volume. It’s a mass per unit volume. So this is a really trivial one. There’s no name for it, but it’s going to be kilograms per meter cubed . Or kilograms, meter to the minus three.

So that’s not bad once you get the hang of it. And what you’re going to be doing in this course is to be using various formulas, computing quantities. And to improve your odds of getting it right, I recommend that you check the units on every calculation that you do. If the units don’t work out right, then your numerical answer’s going to be wrong as well.

So let me give you an example of that. The one we’re going to be working on today is the perfect gas law. One way to write it is P equals rho (ρ) R T (P=ρRT). P is the pressure, rho (ρ) is the density, the mass density, R is the gas constant, and T is the temperature.

Now, let me write out the units for this. Pressure, we already know–where did I put pressure? There it is. Pressure is kilograms, meters to the minus one, seconds to the minus two. Now, that should be equal to the product of the units of all these other things. The units of mass density we’ve already said are kilograms per meter cubed.

The units of the gas constant I’ll give you. It’s Joules per kilogram per degree Kelvin. And the temperature will be in Kelvins as well. So is that going to cancel out? Well, it’s not completely clear yet, because we haven’t taken apart this Joule yet to see what’s inside that. But it looks like we’re going to get rid of the kilograms OK, and it looks like the Kelvins are going to cancel out.

But what is in that Joule? Well, that Joule is here. It is a kilogram, meter squared per second squared. And it looks like that’s going to work, because we’ve got a meters cubed downstairs. That’s going to take that meters squared and make it into a meters to the minus one, and then that’s going to exactly balance with the left-hand side. You see how that works?

So this is a calculation that should be going on in the background whenever you are working on a numerical problem to be sure you’ve got the units right. No questions on that?

Chapter 2: Pressure and the Ideal Gas Law [00:09:55]

Well, the focus today is talking about pressure and the perfect gas law. How many of you have seen the perfect gas law before in courses? Most of you. Well, let’s first imagine a box full of gas molecules, but they’re moving around. We know what the typical molecular speed is. They’re colliding off each other, but they’re also occasionally bouncing off the wall of that chamber.

And every time they do that, they impart a little bit of force to the wall that they bounce off of, and that’s called pressure. It’s the repeated bouncing of molecules off the side of a box that gives rise to this quantity we call pressure. It’s going to depend, of course, on the number of molecules, their speed, and their mass, in principle. At least, it might depend on these things.

Now, if you’ve taken a course in chemistry, you probably have seen the perfect gas law written this way: pressure times volume equals n R T (PV=nRT). That’s probably the most familiar way to write the perfect gas law in a chemistry course. Here, P is the pressure, V is the volume of the container that you have it in, n is the number of moles–let me write this out, number of moles. This is the gas constant. That’s the temperature, of course, in Kelvins. That’s the volume of the container. And that’s the pressure.

Do you remember what a mole is? A mole is a certain number of molecules. Avogadro’s number–which is, if I remember, it’s 6.02x1023–is a number of molecules of any gas in a mole. So this would be the number of moles you have in that box.

The interesting thing about this formula is that it seems to be independent of the mass of the molecule. While I speculated that this might depend on the mass, it turns out that it doesn’t depend on the mass. You would think that a molecule that has a heavier mass would impart more force as it bounces off the wall.

But you may remember from last time that at a given temperature, heavier molecules move more slowly. So in fact, those two factors cancel out in the perfect gas law. So the pressure you get depends only on the number of molecules that you have, not on the mass of those molecules. That’s a bit of a surprise, so be aware of that.

What gets a little bit confusing is that in atmospheric science, we don’t use the perfect gas law in this form. So I’m going to give you the form in which we will be using it in this class. Stop me if you have questions. We’re going to write the perfect gas law as P=ρRT, where this is the pressure again, this is the mass density, that is the gas constant for the gas in question–we call it the specific gas constant, not the universal gas constant–and that again is the temperature in degrees Kelvin.

What we’ve done here is to–I think you can see it if you compare the two–basically we’ve said that the air density, the mass density is going to be–it’s going to be the number of moles per unit volume times the molecular weight. So the more molecules you have and the heavier each molecule is in a given volume, that’s going to determine the mass density. So I’ve used this formula, if you like, to rewrite that so that it looks like this form that we use in atmospheric science.

Question?

Student: I have a question. For Avogadro’s number, is it 10 to the negative 3 or negative 23?

Professor Ron Smith: 10–what did I write there? Oh, thank you. 10 to the 23. 10 to the plus 23. That’s the number of molecules in a–thanks very much. Yeah, that’s your job out there, to keep me honest on this.

So what is this specific gas constant, then? If you follow the math through, you can see that we’ve defined the specific gas constant as being the universal gas constant divided by the molecular weight. So when you’re using air, that’ll be one number. When you’re using hydrogen, that’ll be a different number, and so on.

So what’s the advantage of this? We seem like we’ve made things more complicated, because we no longer have a universal gas constant. It’s because we want to get at this mass density. That’s important in atmospheric science. We want to know, how dense is the air? And that’s why we want it. We don’t want to work in terms of number of molecules. We want to work in terms of the mass of the air.

So let’s do an example. First of all, for air, then, let me put it subscript air there. The average molecular weight for air is 29. This is 8,314. And so that turns out to be roughly 287. And the units on that are Joules per kilogram per Kelvin. So that’s the specific gas constant for air, which is the gas we have most abundantly, of course, in our atmosphere.

Let’s work out a quick example of that. Let’s say–and I’ll try to make it somewhat similar to this room–let’s say the temperature is 15 degrees Celsius and the air density, I somehow know that, 1.2 kilograms per cubic meter. What will be the pressure?

Well, first of all, we have to convert this to Kelvins, so it’s going to be 15 plus 273.1. That’s going to be about 288.1. And then we’re ready to plug it into the formula. It’s going to be 1.2 times 287 times 288.1. And that comes out to be 99,221.7 Pascals. The unit on that is going to be Pascals. 1.2, which is the air density, the specific gas constant, 287, and the temperature expressed in Kelvins, 288.1. Questions on that?

Now, what good is this? This is a very useful formula, but it’s not as useful as one might think in every application. First of all, for air, we can take that as known. But in general, as you move around the atmosphere, the other three things will be changing. And so if I know one of these, like temperature, well, that formula’s pretty useless, because I don’t know either of the other two.

So this formula is best used when you know two of those quantities and need to get the third. For example, if you knew density and temperature, that would give you pressure. If you knew pressure and density, you could solve that for temperature, and so on. So it’s useful, but it’s not everything we would like.

Question?

Student: Just going back to the Pascals, do you want us to express it in Pascals or kilograms per meter cubed?

Professor Ron Smith: For pressure, you should express it in Pascals. Or what is sometimes a more frequent unit in meteorology is a millibar. Millibar, which is sometimes written as a hectoPascal, which is one hundred Pascals . So this would be 992.217 hPa. In the meteorological literature, you’ll often find hectoPascals used instead of Pascals. It’s easy to do the conversion. Just divide by 100 if you’re going that way, or multiply by 100 if you’re going that way.

Chapter 3: Buoyancy Force [00:20:39]

Now, there is an application, a direct application, for the perfect gas law that I’m going to show you now that is really of fundamental importance for how the atmosphere works. And so I’m going to go through this a little carefully, because it is something we’ll be needing over and over again in the course.

And there’s a very simple idea that I’m sure you are aware of, and that is warm air rises, and cold air sinks. I’d like to actually prove that to you. It seems like a trivial thing, but I’d like to prove that to you. And to do that, I’m going to have to define something called the buoyancy force. The buoyancy force is a pressure force on an object immersed in a liquid or a fluid in a gravity field.

Now, this is very easy to imagine, because if you’ve ever taken a basketball or a beach ball into a pool and tried to push it down in the water, you know there’s a rather large force resisting that trying to make that ball quickly lift back up to the top. That’s the buoyancy force.

And for example, here’s the top of the water. There’s your basketball or your beach ball. You’re trying to hold it down there with your hand. There’s something very strong pushing it up. What’s pushing that ball up? What’s the physics of that? Anybody? What’s pushing that ball up? Yeah.

Student: The displaced water?

Professor Ron Smith: Yes. But how does it work? In the back?

Student: Is it because the ball is less dense than the water is?

Professor Ron Smith: The ball is less dense than the water, but I’m looking for a more detailed mechanism. Yes.

Student: The same amount of force as the weight of the displaced water?

Professor Ron Smith: Yes. But actually how does it act? What’s the physics? How is acting on that ball? So you’re right. It depends on the water displaced. That’s going to be Archimedes’ Law. I’m going to put that on the board in just a moment. Yes.

Student:  Does it have to do with something like the force of the outside pressure pushing down on the water?

Professor Ron Smith: Well, it’ll have a bit to do with that, but that’s not going to be having to do with the pressure coming down here. It’s going to have to do with variations in pressure within that liquid. Anybody else? Yes.

Student: The pressure at the bottom is different?

Professor Ron Smith: Yes. So as you go down in this fluid, the pressure’s getting greater and greater. That means the pressure acting up on the bottom is greater than the pressure acting down on the top. So that’s why I said it’s got to be a liquid with some mass in a gravity field. Because only in a gravity field will there be that increase in pressure as you go down.

So when you push that beach ball down there, realize the pressure at the bottom of the ball is greater than the pressure at the top of the ball. And that is what’s causing this buoyancy force. So that’s step one.

By the way, let’s quantify that using the comment that was made earlier. What is Archimedes’ Law? Archimedes’ Law said that that buoyancy force is equal to the weight of the water displaced, or the weight of the– let’s call it water– the weight of the fluid displaced.

So in order to compute that force, we just have to know how much water would be there if the object were not there. In this case, it would be the volume of the object multiplied times the density of the fluid. But it’s the weight, not the mass, so this has to be multiplied by little g, the acceleration of gravity. If you have something of mass m, its weight is the product of mass and the acceleration of gravity. So that’s Archimedes’ Law. That’s the buoyancy force.

Now, it’s acting in the atmosphere all the time whenever you have a little parcel of air that’s at a different temperature than its surroundings. And that’s what I want to work out, and that’s where the perfect gas law is going to be a very nice thing to have.

So I’ve worked out an example here. I’ve imagined a little piece of air– maybe it’s about this big– that’s got a certain pressure. I’m going to use the subscript p, because I’m calling this a parcel, a little parcel of air. It’s got a density, and it’s got a temperature. And then surrounding it is the environment. That’ll be the pressure in the environment, the density of the environment, and the temperature of the environment. And my goal is to find out the buoyancy force acting on that parcel. I want to know, if it’s warm, is it going to rise, or is it going to sink? And so on and so forth.

Now, we’re going to have to make some assumptions, but they’re going to be very good assumptions. The first assumption is we’re going to assume that the pressure in the parcel is equal to the pressure of the environment. So the pressure here is equal to the pressure there. Why would that be? If you had air that was at different pressure than its environment, let’s say at greater pressure, it would immediately expand until the pressure matched.

If you don’t believe that, blow up a balloon so you got the pressure in that thing a little bit higher than the environment, and then pop it. Well, the instant you pop it, now the rubber is gone. You’ve got that high-pressure air, and what does it do? It immediately expands in order to equalize the pressure. So this idea of equalizing pressure happens very, very rapidly, and that’s why I can assume that these two pressures are equal.

Let me put in some numbers to this. Let’s say that these pressures are equal to 80,000 Pascals. The temperature of the environment let’s say is 275 Kelvin. The temperature of the parcel let’s say is 277 Kelvin, so just a two-degree difference between the two.

And I’m going to compute the density for both. ρ for the environment is going to be P for the environment over R and TE (ρ=PE/RTE). So it’ll be 80,000 divided by 287 divided by 275, and that’s going to be 1.0136. The units will be kilograms per cubic meter. That’s the density of air in the environment. The density of air in the parcel is going to be the same pressure, 80,000, the same gas constant, 287, but the temperature’s a little bit different. It’s 277. So that’s going to be 1.0063 kilograms per cubic meter.

So what I’ve shown you here is that the density of the environment is a little bit greater than the density of the parcel itself. Now, what does that mean in terms of buoyancy? I think I’ll move back over here. Here’s my parcel. The gravity force pulling down on that is going to be the mass of the parcel times gravity. We already talked about that, so it’s going to be the volume times the density of the parcel, ρp times g.

The buoyancy force acting up is going to be the volume– well, we’re using Archimedes’ Law now, so it’s going to be this quantity here. It’s going to be the volume again times the density of the environment– that’s the fluid that’s been displaced– times g.

Well, now you can see immediately what’s going to happen here. The V’s are the same for both, g is the same for both, but the densities appear differently. The down force is related to the density of the parcel. The up force is related to the density of the environment. In our case, the density of the environment is less, so– is that right? Greater. So this one is going to be a little bit less. That’s going to be a little bit greater. And the net buoyancy force is up.

So what I’ve proven here is that a parcel of air, if it’s equilibrated its pressure with the environment, is going to be less dense. Therefore, it’s going to have a buoyancy force that’s going to make it rise.

Well, this is probably the basic physics of what happens in the atmosphere to generate all the wind circulations, to generate clouds, sea breezes. Almost everything you can think of in the atmosphere, any air motion, probably can be tracked back to this simple little idea, that temperature differences, if the pressure is equilibrated, will generate buoyancy forces, either up or down.

Now, if I had chosen a cooler temperature for the parcel, let’s say 273, of course, then everything would be reversed. The parcel would be denser than air, this vector would be larger than that one, and the parcel of air would sink. So it works both ways. Now, this is a tricky argument, a number of steps. So I’d be pleased to stop for a minute or two and take questions on this.

Yes.

Student: So it equalizes pressure, but at the expense of equalizing temperature?

Professor Ron Smith: That’s right. So that’s a very good question. The question is, why does it equalize pressure? Why doesn’t it equalize temperature or density? Well, they’re different quantities. Pressure is a force per unit area, and that’s the thing that wants to equalize. There’s no quick process– temperature might equalize over an hour or two, because they’re in contact with each other, but not that instantaneous equilibration like you get with a balloon popping. That’s pressure equilibration, and it’s fast and it’s physical. And the other two either are slower, or just there’s no tendency for that at all.

But that’s right. That’s the key part of the argument, isn’t it? That of these three quantities we’re talking about, the pressure wants to equilibrate, but the other two do not. And that’s what leads rise to the whole concept of buoyancy, warm air rising, cold air sinking. It all comes from the way the pressure equilibrates. That’s key.

Other questions on this? Anything? Well, that went through pretty quickly. I wanted to– oh, wait. Let’s do another example. I want to do another example, because– let’s say that I’ve got my parcel, and everything’s defined as before. But now I’ve got– let’s say– what did I use? I’ve got helium in here, helium. And I’ve got air out here. The pressures are equal. In this case, I’m going to say the temperatures are equal as well. But are the densities equal? Do you think the densities are going to be equal in these two cases? No. And let’s see how that’s going to work out.

The density of the environment is going to be the pressure of the environment with the gas constant for air, 287, and then the temperature of the environment. The density of the parcel is going to be pressure of the parcel over– now, let’s see. What’s going to be the gas constant for helium? 8,314 divided by the molecular weight of helium, which you recall is four. The gas constant for helium’s about 2,079.

So look, even if the pressures are the same and the temperatures are the same, because they’re different gases, the densities are going to be very, very different. And so the helium balloon is going to be–have a smaller mass, smaller density than the air that it’s displaced. It’s going to rise. It’s going to have a buoyancy force that rises. And in this case, it comes in through the different gas constant, which in turn arises because of the different molecular weights.

So later on in the course, in the lab, we’ll be launching helium-filled balloons, and you can think back at that moment and realize, ah, that’s what’s going on. That’s why there’s a buoyancy force. That’s why that balloon wants to rise is because it has a different gas constant, because it has a different molecular weight. It’s a lighter gas. Each molecule has a smaller mass than the air molecules do. Any questions there?

I can’t leave this subject without mentioning mixtures of gases. So we imagine this same box, and it’s got some A molecules, and it’s got some B molecules. And they’re all bouncing around off the walls and so on. There’s a mixture of gas A and gas B. What is the deal there? When you mix two gasses together, what relationship do they have to each other?

I can tell you pretty clearly what’s going to happen. The temperatures are going to quickly equilibrate. Even if the masses are different, because they’re bouncing into each other frequently, thousands of times per second, the temperature of the A and B molecules will quickly come to the same value.

The pressure is additive. In other words, we can define the pressure that the A molecules are making, we can call that PA, and the pressure that the B molecules are making, that’s PB. And the total pressure, PTotal, is just the sum of the two. So we use this term partial pressure–partial pressure of A, partial pressure of B–and they add up to give the total pressure.

So for example, if the pressure in this room–let’s call it PTotal for the moment–is about 1,013 millibars–or that is to say 1,013 with two more decimal places [101,300] Pascals–part of that is due to the nitrogen molecules. That’s the partial pressure of the nitrogen. Part of it’s due to the oxygen molecules. Part of it’s due to the argon. There’s also some water vapor in this room. Water vapor is contributing something to that total pressure.

So when you’re measuring pressure in a gas, you’re measuring the sum of all the pressures of the components within that gas. Usually, we don’t need to know that, but occasionally, that’s the way we keep track of how much of these other gases you have. Someone might say, well, the partial pressure of water vapor is three millibars today or something like that. That’s the contribution water vapor is making to the total pressure on this particular day. So it’s a useful quantity.

Chapter 4: Composition of the Atmosphere [00:39:35]

Let me remind you what the atmospheric composition is for our atmosphere. For air on Earth, it’s primarily nitrogen, oxygen, and argon. I’m going to give you the number two ways: by volume, which is what the chemists say– I prefer to remember that that is by molecule, by the number of molecules– and I’m going to also give it to you by mass.

For nitrogen, it’s 78.1% by volume and 75.5% by mass. In other words, 78% of the molecules in this room are nitrogen, but 75.5% of the mass of the gas in this room is the nitrogen. Oxygen, 21.0% and 23.2%. Argon, 0.9% and 1.3%.

Just remember, there’s this difference because the molecules have different masses. Some are heavier. Some are lighter. So whether you’ve counted up the molecules and are representing the fraction that way, or whether you’re counting up the masses, you’re going to get slightly different numbers for the two.

Now I’ve chosen, and the convention is to define that part as being the air, because these proportions are constant everywhere you go in the atmosphere. If I go to the North Pole, the Equator, the South Pole, if I go high in the atmosphere, winter or summer, these proportions are unchanging. So we call that air.

But then there are other gases as well. And sometimes they’re called trace gases. Sometimes they’re just called variable gases. They’re found in varying proportions depending where you are. Let me give you an example. Probably the most important one is water, water vapor, H2O. And it’s found anywhere from– well, from let’s say one part per 100, 10-2, to really as small as you want to go, maybe 10-5, by volume.

CO2, another very important gas, is more thoroughly mixed, but not perfectly mixed. A typical value these days might be about 395 parts per million by volume, ppmv. So we’re using this method, we’re counting molecules. I could write that as 395x10-6 by volume.

That varies only up and down by about 5%, plus or minus 5%. So that’s nearly well mixed, but not quite as thoroughly mixed as these gases within the atmosphere– within the air.

Some other molecules I mentioned last time on the slide you should be aware of are methane, nitrous oxide–N2O– and ozone. And just be aware that those and a few other gases will pop up from time to time in this course, and we’ll be wanting to know what their partial pressure is, what their mass ratio is, what their ratio by molecules is. We can convert back and forth between these different measures using the formulas that I’ve given you today.

Any questions here? We’ve actually covered a lot of I think somewhat confusing material, so I want to be sure we take a few more minutes for questions. Yes.

Student: Wouldn’t the water temperature, the numbers there, 1019 varies to 10-5 ? [???]

Professor Ron Smith: Negative five. Thank you. For example, I don’t have an instrument with me to measure this– we’ll be doing it in lab– but yesterday and today have been rather humid days. So this means that this number is going to be a little larger than it would have been last week, when we had a drier atmosphere. So that’s an example of how that number fluctuates.

This fraction isn’t changed between last week and this week, but this one has. So these are variable ones, and these are constant proportions. Yes.

Student: When it’s 100% humidity, about where is that range?

Professor Ron Smith: Well, so that depends on the temperature. So the relative humidity– we’ll talk about this in great detail– the relative humidity is a measure of how much water vapor you have to the maximum that can be held in the vapor state. And because that second number is so strongly temperature-dependent, I can’t give you a fixed number for this. It’ll depend on the temperature. But we’ll talk about that later on, because that’s so important for how clouds form and so on.

Chapter 5: Density and Pressure Variations with Altitude [00:45:36]

I think I may have time to do one other thing before we quit today then, and that’s to talk about how density and pressure change with altitude. First of all, just some background information. The typical sea-level density, of course it varies from place to place. But if you want to work out a problem and you’re not given enough information, you should know, for example, in this room, the density is about 1.2 kilograms per cubic meter. And a typical sea-level pressure is about 1,013 millibars or 1,013 two more zeros (101,300) Pascals. So let’s take that as just basic climatological information.

But now I’m interested in how those numbers change with height. Typically, if I plot pressure on the x-axis and height using the letter z on the y-axis, it looks like this. It decreases rapidly at first, then less rapidly, and so on, asymptotically, but never actually reaching zero. And if I plot air density, it looks very much the same.

It’s such a simple relationship that we’d like to be able to have a formula for it. And there is a nice one, but we have to make an approximation now. We have to assume that the temperature is approximately constant with height, which is not a very good approximation, but we’re looking here just to get a rough– maybe an estimate at the 10% level or the 20% level, something in that range. But if this approximation is used, then I can write down a formula for the pressure as a function of height. It’s the pressure at sea level times e to the minus z over Hs.

And the density follows exactly the same formula. The density at sea level, sl– I’m using Greek letter rho for density–. Now, if you’re familiar with this exponential function, you would have already recognized it here. This is the behavior of the exponential function. It drops rapidly at first and then more slowly as you go on. This thing is called the density for the scale height, the scale height for the atmosphere. It is a measure of how fast the pressure and density decrease as you go up.

And there’s a very simple formula for it. It’s , the gas constant times the temperature divided by the surface gravity. Let’s work it out for Earth. Air has a gas constant of 287. Let’s take 288 for the temperature of the air and 9.81 for the acceleration of gravity. That turns out to be approximately 8,400 meters. Every time you go up 8,400 meters, you tick off another fractional decrease in atmospheric pressure and density.

We just have a minute left, so I can do a quick example of this. Let’s say we’ve got an aircraft flying at 37,000 feet. That’s typically what an airliner would fly at. And you’d like to know what is the pressure and density of the air just outside the cabin? Of course, the cabin itself is pressurized so you can breathe and maintain consciousness, but what is the air temperature and pressure just outside?

Well, first of all, we have to convert this to meters. That’s going to be 11,278 meters. And then I’m going to put it into this formula. So the pressure at that height z is going to be the pressure at sea level, 101,300, times  e to negative the altitude divided by 8,400. I hope you know how to do that in your scientific calculator with the minus sign in there. Practice that.

That comes out to be– let’s see– 26,524 Pascals. Well, that’s about a quarter of the pressure at sea level. And density would be something very similar. Here it’ll be 1.2 times e to negative the altitude over 8,400. And that’s going to come out to be 0.31. Units are kilograms per cubic meter. So that also is about a quarter of what you started with. So that’s not much. In other words, at typical airliner flight level, the density and pressure that you’re flying through is only about one quarter that you have here at sea level, and that’s why the cabin has to be pressurized.

We’re really out of time now, so we’ll move on to some new material on Wednesday.

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