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've spent a fair amount of time learning about pressure.  We first began 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.  That's a perfectly sound explanation.

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. 


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.





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





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.