Thursday Sep. 17, 2009
click here to download today's notes in a more printer friendly format.

Three more songs from Calexico today in honor of Mexican Independence Day yesterday and to celebrate the start of the new season of my favorite television show (Survivor).  The titles were "Cancion del Mariachi" (Ruben Moreno and Mariachi Luz de Luna), "Aires Del Mayab" (featuring Lulu Olivares), and "Corona." I haven't been able to find quality recordings of any of these songs online.  Here's another song "Crystal Frontier" that you might like and that we won't have time to play in class.

A preliminary version of the Quiz #1 Study Guide is now available. Quiz #1 is next Thursday (Sep. 24) and will cover material on both the Practice Quiz Study Guide and the Quiz #1 Study Guide.

The 1S1P Bonus Assignment was collected today.

The Expt. #1 reports are due next Tuesday.  Please try to return your materials this week (you'll need to bring them by my office in PAS 588).  The Experiment #2 materials will be handed out next Thursday.
There is also an Optional Assignment that is due next Tuesday.
The main goal of today's class is to understand why warm air rises and cold air sinks.


Hot air balloons rise (they also sink), so does the relatively warm air in a thunderstorm (its 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 and are a serious weather hazard.

A full understanding of this 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.  That is an equation that tells you which/how 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 and density change together in a way that keeps the pressure inside a balloon constant).  Then we'll learn about the vertical forces that act on air (an upward and a downward force)

The figure above makes an important point: the air molecules in a balloon "filled with air" really take up very little space.  A balloon filled with air is really mostly empty space.  It is the collisions of the air molecules with the inside walls of the balloon that keep it inflated.


Up to this point in the semester we have been thinking of pressure as being determined by the weight of the air overhead.  Air pressure pushes down against the ground at sea level with 14.7 pounds of force per square inch.  If you imagine the weight of the atmosphere pushing down on a balloon sitting on the ground you realize that the air in the balloon pushes back with the same force.  Air everywhere in the atmosphere pushes upwards, downwards, and sideways. 

The ideal gas law equation is another way of thinking about air pressure, sort of a microscope .  We ignore the atmosphere and concentrate on just the air inside the balloon.  We are going to "derive" an equation.  Pressure (P) will be on the left hand side.  Properties of the air inside the balloon will be found on the right side of the equation.
In A
the pressure produced by the air molecules inside a balloon will first depend on how many air molecules are there, N.  If there weren't any air molecules at all there wouldn't be any pressure.  As you add more and more add 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).

In B
air pressure inside a balloon also depends on the size of the balloon.  Pressure is inversely proportional to volume, V .  If V were to double, P would drop to 1/2 its original value.

Note
it is possible to keep pressure constant by changing N and V together in just the right kind of way.  This is what happens in Experiment #1 that some students are working on.  Oxygen in a graduated cylinder reacts with steel wool to form rust.  Oxygen is removed from the air sample which is a decrease in N.  As oxygen is removed, water rises up into the cylinder decreasing the air sample volume.  N and V both decrease 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.

Part C: Increasing the temperature of the gas in a balloon will cause the gas molecules to move more quickly.  They'll collide with the walls of the balloon more frequently and rebound with greater force.  Both will increase the pressure.  You shouldn't throw a can of spray paint into a fire because the temperature will cause the pressure inside the can to increase and the can could explode.  We'll demonstrate the effect of temperature on pressure in class on Friday. 

Surprisingly, as explained in Part D, the pressure does not depend on the mass of the molecules.  Pressure doesn't depend on the composition of the gas.  Gas molecules with a lot of mass will move slowly, the less massive molecules will move more quickly.  They both will collide with the walls of the container with the same force.

The figure below (which replaces the bottom of p. 51 in the photocopied ClassNotes) shows two forms of the ideal gas law.  The top equation is the one we just derived 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 (which we won't be doing).
Read through the explanation on p. 52 in the photocopied Classnotes.  In the atmosphere a parcel (balloon) of air will always try to keep its pressure the same as the pressure of the surrounding air.  If they aren't equal the parcel will either expand or shrink until they are again equal.

If you warm air it will expand and density will decrease until the pressure inside and outside the parcel are equal.
If you cool air the parcel will shrink and the density will increase until the pressures balance.

These two associations:
(i)   warm air = low density air
(ii)  cold air = high density air
are important and will come up a lot during the remainder of the semester.

Click here if you would like a little more detailed, more step-by-step, explanation of Charles Law.  Otherwise continue on and read a summary of the Charles' Law demonstration that we did in class.

Charles Law can be demonstrated by dipping a balloon in liquid nitrogen.  You'll find an explanation on the top of p. 54 in the photocopied ClassNotes.
The balloon had shrunk down to practically zero volume when pulled from the liquid nitrogen.  It was filled with cold high density air.  As the balloon warmed the balloon expanded and the density of the air inside the balloon decreased.  The volume and temperature kept changing in a way that kept pressure constant.  Eventually the balloon ends up back at room temperature (unless it pops).

Now finally on to step #3.  It's found on p. 53 in the photocopied ClassNotes.



Basically what it comes down to is this - there are two forces acting on a parcel (balloon) of air:
Gravity pulls downward.  The strength of the gravity force depends on the mass of the air inside the balloon.
There is an upward pointing pressure difference force.  This is caused by the air surrounding the balloon.

When the air inside a parcel is exactly the same as the air outside, the two forces have equal strengths and cancel out.  The parcel is neutrally bouyant and doesn't rise or sink.

If you replace the air inside the balloon with warm low density air, it won't weigh as much.  The gravity force is weaker.  The upward pressure difference force doesn't change and ends up stronger than the gravity force.  The balloon will rise.

Conversely if the air inside is cold high density air, it weighs more.  Gravity is stronger than the upward pressure difference force and the balloon sinks.

We used balloons filled with hydrogen instead of air (see bottom of p. 54 in the photocopied Class Notes).  Hydrogen is less dense than air even when the hydrogen has the same temperature as the surrounding air.  A hydrogen-filled balloon doesn't need to warmed up in order to rise.


We dunked the hydrogen-filled balloon in some liquid nitrogen to cool it and to cause the density of the hydrogen to increase.  When removed from the liquid nitrogen the balloon didn't rise, the gas inside was denser than the surrounding air (the purple and blue balloons in the figure above).  As the balloon warms and expands its density decreases.  The balloon at some point has the same density as the air around it (green above) and is neutrally bouyant.  Eventually the balloon becomes less dense that the surrounding air (yellow) and floats up to the ceiling.

Here's an example of where something like this happens in the atmosphere (this figure wasn't shown in class)

At (1) sunlight reaching the ground is absorbed and warms the ground.  This in turns warms air in contact with the ground (2).  Once this air becomes warm and its density is low enough, small "blobs" of air separate from the air layer at the ground and begin to rise (3).  Rising air expands and cools (we've haven't covered this yet).  If it cools enough (to the dew point) a cloud will become visible as shown at Point 4.  This whole process is called free convection; many of our summer thunderstorms start this way.

Here's some extra material that we didn't have time for in class, something that I promised to add to today's notes,  We'll come back to this briefly on Tuesday.  I have a really colorful demonstration that goes along with this topic.

We've just seen how the relative strengths of the downward graviational force and the upward pressure difference force determine whether a parcel of air will rise or sink.  Archimedes Law is another way of trying to understand the situation.

A gallon of water weighs about 8 pounds (lbs).

If you submerge a 1 gallon jug of water in a swimming pool, the jug becomes, for all intents and purposes, weightless.  Archimedes' Law (see figure below, from p. 53a in the photocopied ClassNotes) explains why this is true.

The upward bouyant force is really just another name for the pressure difference force covered earlier today (higher pressure pushing up on the bottle and low pressure at the top pushing down, resulting in a net upward force).  A 1 gallon bottle will displace 1 gallon of pool water.  One gallon of pool water weighs 8 pounds.  The upward bouyant force will be 8 pounds, the same as the downward force on the jug due to gravity.  The two forces are equal and opposite.

Now we imagine pouring out all the water and filling the 1 gallon jug with air.  Air is about 1000 times less dense than water;compared to water,  the jug will weigh practically nothing.

If you submerge the jug in a pool it will displace 1 gallon of water and experience an 8 pound upward bouyant force again.  Since there is no downward force the jug will float.

One gallon of sand (which is about 1.5 times denser than water) jug will weigh 12 pounds.

The jug of sand will sink because the downward force is greater than the upward force. 

You can sum all of this up by saying anything that is less dense than water will float in water, anything that is more dense than water will float in water.

The same reasoning applies to air in the atmosphere.

Air that is less dense (warmer) than the air around it will rise.  Air that is more dense (colder) than the air around it will sink.

 Here's a little more information about Archimedes.

So far we have looked at how pressure and air density change with increasing altitude.  Now we will look how air temperature changes with altitude. The figure drawn in class has been split into two parts and redrawn for improved clarity (actually this is a figure from the Fall 2008 semester)

The atmosphere can be split into layers depending on whether temperature is increasing or decreasing with increasing altitude.  The two lowest layers are shown in the figure above.  There are additional layers (the mesosphere and the thermosphere) above 50 km but we won't worry about them. 

1. We live in the troposphere.  The troposphere is found, on average, between 0 and about 10 km altitude, and is where temperature usually decreases with increasing altitude.  [the troposphere is usually a little higher in the tropics and lower at polar latitudes]

The troposphere contains most of the water vapor in the atmosphere (the water vapor comes from evaporation of ocean water) and is where most of the clouds and weather occurs.  The troposphere can be stable or unstable (tropo means to turn over and refers to the fact that air can move up and down in the troposphere).

2a. The thunderstorm shown in the figure indicates unstable conditions, meaning that strong up and down air motions are occurring.  When the thunderstorm reaches the top of the troposphere, it runs into the bottom edge of the stable stratosphere.  The air can't continue to rise into the stable stratosphere so the cloud flattens out and forms an anvil (anvil is the name given to the flat top of the thunderstorm).   The flat anvil top is something that you can go outside and see and often marks the top of the troposphere.

2b.  The summit of Mt. Everest is a little over 29,000 ft. tall and is close to the top of the troposphere.

2c.   Cruising altitude in a passenger jet is usually between 30,000 and 40,000, near or just above the top of the troposphere, and at the bottom of the stratosphere.

3.  Temperature remains constant between 10 and 20 km and then increases with increasing altitude between 20 and 50 km.  These two sections form the stratosphere.  The stratosphere is a very stable air layer.  Increasing temperature with increasing altitude is called an inversion.  This is what makes the stratosphere so stable.

4.   A kilometer is one thousand meters.  Since 1 meter is about 3 feet, 10 km is about 30,000 feet.  There are 5280 feet in a mile so this is about 6 miles (about is usually close enough in this class). 


5.   Sunlight is a mixture of ultraviolet (7%), visible (44%), and infrared light (49%).  We can see the visible light.

5a. Much of the sunlight arriving at the top of the atmosphere passes through the atmosphere and is absorbed at the ground.  This warms the ground.  The air in contact with the ground is warmer than air just above.  As you get further and further from the warm ground, the air is colder and colder.  This explains why air temperature decreases with increasing altitude.

5b. How do you explain increasing temperature with increasing altitude in the stratosphere.  

     The ozone layer is found in the stratosphere (peak concentrations are found near 25 km altitude).  Absorption of ultraviolet light by ozone warms the air in the stratosphere and explains why the air can warm.  The air in the stratosphere is much less dense (thinner) than in the troposphere.  So even though there is not very much UV light in sunlight, it doesn't take as much energy to warm this thin air as it would to warm denser air closer to the ground.

6. That's a manned balloon; Auguste Piccard and Paul Kipfer are inside.  They were to first men to travel into the stratosphere (see pps 31 & 32 in the photocopied Class Notes).  We'll see a short video showing part of their adventure at some point in the next week or so.  It really was quite a daring trip at the time at the time, and they very nearly didn't survive it.

We'll review this material on the troposphere and stratosphere in class on Tuesday.