Thursday Sept. 15, 2016

La Santa Cecilia: "Losing Game" (3:15), "La Negra" (3:16), "I Won't Cry for You (English)" (3:40), "No Te Lloro Mas (Espanol)" (3:21), "Tainted Love" (2:53), "Nunca Mas" (3:45)

An ambitious goal today: trying to understand why warm air rises & cold air sinks
Hot air balloons floating over the Rio Grande river during the Albuquerque Balloon Fiesta (source of the photo)
 Photograph of a microburst, a localized intense thunderstorm downdraft, that hit Wittmann Arizona in July 2015.  Surface winds of 55 MPH were measured. (source of the photo)

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, on Tuesday as a matter of fact).


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).


The ideal gas law equations
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 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.


Step #2 Charles Law

Charles Law means that P (pressure) in the ideal gas law stays constant.  Changing the temperature of a volume of air will cause a change in density and volume; pressure will stay constant.  This is an important situation because this is how volumes of air in the atmosphere behave. 




This is probably the most difficult part of today's class and is worked out in lots of detail.


We will use a balloon of air.  The air inside and outside the balloon (or parcel) are exactly the same. 

The pressure of the air surrounding the balloon pushing inward is balanced by the pressure of the air inside the balloon that is pushing outward.  If we change something inside the balloon that upsets this balance, the balloon would expand or shrink until the pressures are again in balance.

Volumes of air in the atmosphere  will behave the same way.

First let's imagine warming the air inside a balloon.  We'll won't change the temperature of the air outside the balloon.


Increasing the temperature will momentarily increase the pressure.  This creates an imbalance.  Now that P inside is greater than P outside the balloon will expand.




Increasing the volume causes the pressure to start to decrease.  The balloon will keep expanding until P inside is back in balance with P outside. 

We're left with a balloon that is larger, warmer, and filled with lower density air than it was originally. 






The pressures inside and outside are again the same.  The pressure inside is back to what it was before we warmed the air in the balloon.  You can increase the temperature and volume of a parcel together in a way that keeps pressure constant (which is what Charles' law requires).  This is equivalent to increasing the temperature and decreasing the density together and keeping the pressure constant.

In nature the change in temperature and volume occur simultaneously.  It's like jumping from the first to the last step above.



Warming a parcel of atmospheric air will cause the parcel volume to increase, the density of the air in the parcel to decrease, while pressure remains constant.

We can go through the same kind of reasoning and see what happens if we cool the air in a parcel.  Actually you should see if you can figure it yourself.  I've included all the steps below.  We'll just skip to the last step in class.

We'll start with a parcel of air that has the same temperature and density as the air around it.

We'll cool the air inside the parcel.  The air outside stays the same.

Reducing the air temperature causes the pressure of the air inside the balloon to momentarily decrease.  Because the outside air pressure is greater than the pressure inside the balloon the parcel is compressed.


The balloon will get smaller and smaller (and the pressure inside will get bigger and bigger) until the pressures inside and outside the balloon are again equal.  The pressure inside is back to the value it had before you cooled the air in the parcel.


The first and last steps, without all the intermediate and momentary details, are shown below.

Cooling some air will cause volume to decrease and density to increase while pressure stays constant.

If you want to skip all the details and just remember one thing, here's what I'd recommend






Demonstration of Charles Law in action
Parcels of atmospheric air and air in balloons behave the same way, they both obey Charles' Law.  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.


A balloon shrinks down to practically zero volume when dunked in the liquid nitrogen.  When pulled from the liquid nitrogen the balloon is filled with very cold, very high density air. 

Then the balloon starts to warm up.
The volume and temperature both increasing together in a way that kept pressure constant (pressure inside the balloon is staying equal to the air pressure outside the balloon).  Eventually the balloon ends up back at room temperature (unless it pops while warming up).

All of that was just Step #2, we've still got Step #3

Step #3 Two vertical forces acting on a parcel of air in the atmosphere
(see p. 53 in the ClassNotes)






Basically it comes down to this - there are two forces acting on a parcel of air in the atmosphere.  They are shown above.

The first force is gravity, it pulls downward.  Most everyone knows about this force.  The strength of the gravity force (the weight of the air in the parcel) depends on the mass of the air inside the parcel. 

Second there is an upward pointing pressure difference force.  Not too many people know about this one.  This force is caused by the air outside (surrounding) the parcel.  Pressure decreases with increasing altitude.  The pressure of the air at the bottom of a parcel pushing upward is slightly stronger than the pressure of the air at the top of the balloon that is pushing downward.  The overall effect is an upward pointing force.

When the air inside a parcel is exactly the same as the air outside (same densities), the two forces are equal in strength and cancel out.  The parcel is neutrally buoyant and it wouldn't rise or sink, it would just hover.

We'll replace the air inside the balloon with either warm low density air or cold high density air. 





In the first case, a balloon with warm low density air won't weigh as much.  The gravity force is weaker.  The upward pressure difference force doesn't change (because it is determined by the air outside the balloon which hasn't changed) 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.

It all comes down to a question of how the density of the air in a parcel compares to the density of the air surrounding the parcel.  If the parcel is filled with low density air it will rise.  A parcel full of high density air will sink.  That's true of things other than air.  Wood floats in water because it is less dense than water.

Here's a short demonstration of the role that density plays in determining whether a balloon will rise or sink (or hover)

Convection demonstration





We used balloons filled with helium (see bottom of p. 54 in the photocopied Class Notes).  Helium is less dense than air even when it has the same temperature as the surrounding air.  The downward gravity force (weight of the helium filled balloon) is weaker than the upward pressure difference force.  You don't need to warm a helium-filled balloon to make it rise.







We dunk the helium filled balloon in liquid nitrogen to cool it off.  When you pull the balloon out of the liquid nitrogen it has shrunk.  The helium is denser than the surrounding air.  I set it on the table (dark blue above) and it just sat there.

As the balloon of helium warms and expands its density decreases (light blue).  For a brief moment it has the same density as the surrounding air (green).  It's neutrally buoyant at this point.  Then it warms back to near room temperature where it is again less dense than the air and lifts off the table (yellow).

Free convection
Something like this happens in the atmosphere. 



Sunlight shines through the atmosphere.  Once it reaches the ground at (1) it is absorbed and warms the ground.  This in turns warms air in contact with the ground (2)  As this air warms, its density starts to decrease (pressure is staying constant).  When the density of the warm air is low enough, small "blobs" of air separate from the air layer at the ground and begin to rise, these are called "thermals."  (3) Rising air expands and cools (we've haven't covered this yet and it might sound a little contradictory).  If it cools enough (to the dew point) a cloud will become visible as shown at Point 4.  This whole process is called convection; many of our summer thunderstorms start this way.