 # 03. The Perfect Gas Law

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
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
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
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
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
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
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–
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.