Monday, Feb. 1, 2016
Calexico
at Ancienne Belgique "Falling From the Sky" (0 - 4:20),
"Moon Never Rises" (27:00 - 31:30), "Beneath the City
of Dreams" (46:55 - 50:40)
We'll start with a topic
that we didn't have time to cover in class last Friday.
Changes in air density with
altitude
(see p. 34 in the ClassNotes)
We've spent a lot of time (too much?) looking at
air pressure and how it changes with altitude. Next
we'll consider air density.
How does air density change with increasing
altitude? You should know the answer to that
question. You get out of breath more easily at high
altitude than at sea level. Air gets thinner (less
dense) at higher altitude. A lungful of air at high
altitude just doesn't contain as many oxygen molecules as it
does at lower altitude or at sea level.
It would be nice to also understand why air density
decreases with increasing altitude.
The people pyramid reminds you that there is more
weight, more pressure, at the bottom of the atmosphere than there
is higher up.
Layers of air are not solid and rigid like in a stack of
bricks. Layers of air are more like mattresses stacked on
top of each other. Mattresses are compressible,
bricks (and people) aren't. Mattresses are also reasonably
heavy, the mattress at the bottom of the pile would be squished by
the weight of the three mattresses above. This is shown at
right. The mattresses higher up aren't compressed as much
because there is less weight remaining above. The same is
true with layers of air in the atmosphere.
The statement above is at the top of p. 34 in the photocopied
ClassNotes. I've redrawn the figure found at the bottom of
p. 34 below.
There's a surprising amount of information in this figure,
you need to spend a minute or two looking for it
1. You can first notice and remember that pressure
decreases with increasing altitude. 1000 mb at the
bottom decreases to 700 mb at the top of the picture. You
should be able to explain why this happens.
2. Each layer of air contains the same amount
(mass) of air. This is a fairly subtle point.
You can tell because the pressure drops by the same amount, 100
mb, as you move upward through each layer. Pressure depends
on weight. So if all the pressure changes are equal, the
weights of each of the layers must be the same. Each of the
layers must contain the same amount (mass) of air (each layer
contains 10% of the air in the atmosphere).
3. The densest air is found at the bottom of the picture.
The bottom layer is compressed the most because it is supporting
the weight of all of the rest of the atmosphere. It is the
thinnest layer in the picture and the layer with the smallest
volume. Since each layer has the same amount of air (same
mass) and the bottom layer has the smallest volume it must have
the highest density. The top layer has the same amount of
air but about twice the volume. It therefore has a lower
density (half the density of the air at sea level). Density
is decreasing with increasing altitude. That's the
main point in this figure.
4. A final point that you shouldn't worry too much about
yet. Pressure decreases 100 mb in a fairly short
vertical distance in the bottom layer of the picture - a rapid
rate of decrease with altitude. The same 100 mb drop takes
place in about twice the vertical distance in the top layer in the
picture - a smaller rate of decrease with altitude. Pressure
is decreasing most rapidly with increasing altitude in the
densest air in the bottom layer. We'll make use of
this concept again at the end of the semester when we try to
figure out why/how hurricanes intensify and get as strong as they
do.
Now on to the main topic of the day
and something I want to cover before the Experiment #1 reports are
due next Monday
Why does warm air rise and cold air sink?
Hot air balloons rise, so does the relatively warm air in a
thunderstorm updraft (it's warmer than the air around
it). Conversely cold air sinks. The surface
winds caused by a thunderstorm downdraft (as shown above) can
reach speeds of 100 MPH (stronger than most tornadoes) and are a
serious weather hazard that we'll come back to later in the
semester.
A full understanding of these rising and sinking motions is a
3-step process (the following is from the bottom part of p. 49
in the photocopied ClassNotes).
We will first learn about the
ideal gas law. It's is an equation that tells you
which properties of the air inside a balloon work to
determine the air's pressure. Then we will look at
Charles' Law, a special situation involving the ideal gas
law (air temperature volume, and density change together in
a way that keeps the pressure inside a balloon
constant). Then we'll learn about the 2 vertical
forces that act on air. I'm pretty sure you know what
the downward force is and am about equally sure you don't
remember what the upward force is (even though it is
something that has come up before in this class this
semester).
The ideal gas law - a microscopic scale explanation of air
pressure
We then went a bit further and tried to imagine the weight of
the atmosphere pushing down on a balloon sitting on the
ground. If you actually do push on a balloon you realize
that the air in the balloon pushes back with the same
force. Air pressure everywhere in the atmosphere pushes
upwards, downwards, and sideways.
These are large scale, atmosphere size, ways of thinking about
pressure. Next we are going to concentrate on just the
air in the balloon pictured above. This is more of a
microscopic view of pressure.
Imagine filling a balloon with air. If you could
look inside which picture below would be more realistic?
The view on the left is incorrect.
The air molecules actually do not fill the balloon and
take up all the available space.
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This is the correct
representation.
The air molecules are moving
around at 100s of MPH but actually take up little or no
space in the balloon.
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The air molecules are continually colliding
with the walls of the balloon and pushing outward (this force
divided by area is the pressure). Wikipedia
has a
nice animation. An individual molecule doesn't
exert a very strong force, but there are so many molecules
that the combined effect is significant.
We want to identify the properties or characteristics of the
air inside the balloon that determine the pressure and then put
them together into an equation called the ideal gas law
(actually there'll be two equations).
Step #1 The ideal gas law equation
You're not going to have to be able to figure out or remember
the ideal gas law equation. I'll give it to you. Here
is is:
You should know what the symbols in the equation
represent. Probably the most obvious variable is N the
number of air molecules. It's
the motions of the air molecules that produce pressure. No
air molecules (N = 0) means no pressure. The more air
molecules there are the higher the pressure.
Number of gas molecules or atoms
Pressure (P) is
directly proportional to Number of air molecules (N). If N
increases P increases and vice versa.
Here's an example. You're adding air
to a tire. As you add more and more air to something like a
bicycle tire, the pressure increases. Pressure is directly
proportional to N; an increase in N causes an increase in P.
If N doubles, P also doubles (as long as the other variables in
the equation don't change).
Temperature
Here's what I think is the next most obvious variable.
You shouldn't throw a
can of spray paint into a fire because the temperature
will cause the pressure of the gas (propellant) inside
the can to increase and the can could explode.
So T (temperature) belongs in the ideal gas law equation
Increasing the
temperature of the gas in a balloon will cause the gas
molecules to move more quickly (kind of like "Mexican
jumping beans"). They'll collide with the
walls of the balloon more frequently and rebound with
greater force - that will increase the pressure.
We've gotten a little bit ahead of the story. The
variable V (volume) has appeared in the equation and it's in the
denominator. A metal can is rigid. It's
volume can't change. When we start talking about volumes of
air in balloons or in the atmosphere volume can change. A
change in temperature or a change in number of air molecules might
be accompanied by a change in volume.
At this point we did a quick demonstration to show the effect
of temperature on the pressure of the gas in a rigid sealed
container (N and V in the ideal gas law equation stay constant,
just as in a can of spray paint). The description of the demonstration below
were added Tuesday morning, the day after class.
The container was a glass flask, sealed with a rubber
stopper. A piece of tubing with a valve was connected to the
flask. The valve was opened at the start of the
demonstration to be sure the pressures inside and outside the
flask were equal. The valve was then closed. The
manometer is a U-shaped tube filled with a liquid (transmission
oil) that can detect differences in pressure. Pressure from
the air inside the flask could enter one end of the manometer
tube. The other end was exposed to the pressure of the air
outside the flask.
Green in the figure indicates that the temperatures of the air
inside and outside the flask were equal. The manometer is
showing that the pressure of the air inside and outside the flask
were equal.
I wrapped my hands around the flask to warm the air inside very
slightly. The increase in air temperature caused a slight
increase in the pressure of the air inside the flask. The
air outside didn't change. Note the change in the levels of
the liquid in the manometer indicating the increase of the air
pressure inside the flask.
The valve was opened momentarily so that the pressures inside
and out would again be equal. The valve was then closed and
some isopropyl alcohol (rubbing alcohol) was dribbled on the
outside of the flask. As the alcohol evaporated it cooled
the flask and the air inside the flask. This caused the air
pressure inside the flask to drop. This change in air
pressure was again indicated by the liquid levels in the
manometer.
Volume
The effect of volume on pressure might be a little harder to
understand. Just barely fill a balloon with air, wrap your
hands around it, and squeeze it. It's hard and you don't
compress the balloon very much at all.
Or think of the bottom layer of the atmosphere
being squished by the weight of the air above. As the
bottom layer is compressed and its volume shrinks it pushes
back with enough force to eventually support the air above.
A decrease in volume
causes an increase in pressure, that's an inverse
proportionality.
It might take three or four
breaths of air to fill a balloon. Think about
that. You add some air (N increases) and the balloon
starts to inflate (V increases). Then you add another
breath of air. N increases some more and the balloon
gets a little bigger, V has increased again. As you
fill a balloon N and V are both increasing. What is
happening in this case is that the pressure of the air in
the balloon is staying constant. The
pressure inside the balloon pushing outward and trying to
expand the balloon is staying
equal to (in balance with) the pressure of the air outside
pushing inward and trying to compress the balloon.
Here's the same picture again except N and V
are decreasing together in a way that keeps pressure
constant. This is exactly what is happening in Experiment
#1.
Experiment #1
- P stays constant, N & V both decrease
Here's a little more detailed explanation of Expt. #1
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The object of Experiment #1
is to measure the percentage concentration
of oxygen in the air. An
air sample is trapped together with some steel wool
inside a graduated cylinder. The cylinder is
turned upside down and the open end is stuck into a
glass of water sealing off the air sample from the rest
of the atmosphere. This is shown at left
above. The pressure of air outside the cylinder
tries to push water into the cylinder, the pressure of
the air inside keeps the water out.
Oxygen in the cylinder reacts with
steel wool to form rust. Oxygen is removed from
the air sample which causes N (the total number of air
molecules) to decrease. Removal of oxygen would
ordinarily cause a drop in Pin
and
upset the balance between Pin
and Pout
. But, as oxygen
is removed, water rises up into the cylinder decreasing
the air sample volume. The decrease in V is what
keeps Pin
equal to Pout
.
N and V both decrease together in the same relative
amounts and the air sample pressure remains constant.
If you were to remove 20% of the air molecules, V would
decrease to 20% of its original value and pressure would stay
constant. It is the change in V that you can see, measure,
and use to determine the oxygen percentage concentration in
air. You should try to explain this in your experiment
report.
You might think that the mass of the gas molecules inside a
balloon might affect the pressure (big atoms or molecules might
hit the walls of the balloon harder and cause higher pressure
and vice versa).
The mass of the air molecules doesn't matter. The big ones
move relatively slowly, the smaller ones more quickly.
They both hit the walls of the balloon with the same
force. A variable for mass doesn't appear in
the ideal gas law equation.
The figure below shows two forms of the ideal
gas law. The top equation is the one we've been looking at
and the bottom is a second slightly different version. You
can ignore the constants k and R if you are just trying to
understand how a change in one of the variables would affect the
pressure. You only need the constants when you are doing a
calculation involving numbers and units (which we won't be
doing).
The ratio N/V is similar to density
(mass/volume). That's where the ρ (density)
term in the second equation comes from.
