RONALD 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. 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 10 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, F equals ma. 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? 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

trivia 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 RT. P is the pressure, rho 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? 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 mRT. 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, m 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

6.02 times 10 to the 23– 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 equals rho RT, 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– 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: 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’s 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 that’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: 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 one-hundredth of

a pascal [correction: one hundred Pascals]. So this would be 992.217

hectopascals. 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. 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

meeting 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: Yes. But how does it work? In the back? STUDENT: Is it because the

ball is less dense than the water is? PROFESSOR: The ball is less

dense than the water, but I’m looking for a more detailed

mechanism. Yes. STUDENT: The same amount of

force is the weight of the baseball bat? PROFESSOR: 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: [INAUDIBLE] outside pressure pushing

down on the water? PROFESSOR: 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: [INAUDIBLE] PROFESSOR: 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. 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. Rho for the environment is

going to be P for the environment over R and TE. 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,

rho sub p 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: 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 temperature 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 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,

P Total, 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 P Total

for the moment– is about 1,013 millibars– or that is to say 1,013

with two more decimal places 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. 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 to the minus two, to really as small as you want

to go, maybe 10 to the minus five, 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

395 times 10 to the minus six 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: For the water vapor,

the numbers there, it’s 10 to the negative 2 – there is

10 to the negative 5. PROFESSOR: 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: 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. I think I may have time to do

one other thing before we quit today, 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 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 H sub S. And the density follows exactly

the same formula. The density at sea level,

rho sub SL– I’m using Greek letter

rho for density– e to the minus z over H sub S.

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 RT over g, 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 the minus 11,278 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. It’ll be 1.2 times e to the

minus 11,278 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.