Wednesday Feb. 27, 2008

The Experiment #2 reports are due next Monday.  Bring your materials in now and pick up the supplementary information sheet.

The revised Expt. #1 reports are also due next Monday.

1S1P Assignment #2 is now available.  Copies of the worksheets that accompany this assignment will be distributed in class on Friday.   The worksheets are an optional part of the assignment, they aren't required.


We are going to try to understand why warm air rises and cold air sinks (the figure above is on p. 49 in the photocopied Classnotes)..
It is always a good idea to have a picture in mind, a hot air balloon for example.  Cold air pours out of thundestorms in the summer.  When this descending air hits the ground it spreads our horizontally under the storm.  These surface winds can sometimes reach 100 MPH which makes them stronger than most tornadoes.

Hot air balloons do sometimes fall from the sky; most everyone would understand that gravity was the force responsible for bringing down a hot air balloon.
But what causes a hot air balloon to rise?  We will see that it is a pressure difference force.  Pressure decreases with increasing altitude.  This creates a force that points upward from high toward low pressure.

Understanding rising and sinking air is a 3-step process.  The first step is learning about the ideal gas law.


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 will give us an idea of what determines the strength of the pressure inside the balloon.
Step #1 - ideal gas law

The pressure produced by the air molecules inside a balloon will first depend on how many air molecules are there.  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).

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 of you 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.

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.


Surprisingly 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.

An attempt was made to demonstrate the effect of temperature on pressure.  An air-filled flask was connected to a manometer.

Initially the air in the flask was exactly the same as the air outside.  The levels of the red liquid in the manometer were the same indicating the Patmosphere and Pflask were the same.  Next we both warmed and cooled the air in the flask. 

Warming the air in the flask increased the pressure inside the flask.  Cooling the flask reduced the pressure of the air in the flask.  The changing liquid levels revealed these changes in pressure.


Two ideal gas law equations are shown: the one we just derived and a slightly different form of the same equation.  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.

Here is an ideal gas law animation that wasn't shown in class.  You can vary N, V, or T and see the effect on pressure.  Caution the animation sometimes takes quite a while to load.
Step #2 - Charles' law
We're still trying to understand why warm air rises and cold air sinks.  We've still got two more steps to go.
We now turn our attention to Charles' Law, a special situation involving the ideal gas law.  This is Step #2.

Read through the explanation on p. 52 in the photocopied Classnotes.  If you feel comfortable with that explanation then you can skip to the Charles Law demonstration further along in these online notes.  Otherwise you can read the following explanation.

Air in the atmosphere behaves like air in a balloon.  A balloon can grow or shrink in size depending on the pressure of the air inside. 

We start in the top figure with air inside a balloon that is exactly the same as the air outside.  The air inside and outside have been colored green.  The arrows show that the pressure of the air  inside pushing outward and the pressure of the air surrounding the balloon pushing inward are all the same. 

Next we warm the air in the balloon (Fig. 2).  The ideal gas law equation tells us that the pressure of the air in the balloon will increase.  The increase is momentary though. 

Because the pressure inside is now greater than the pressure outside, the balloon will expand.  As volume begins to increase, the pressure of the air inside the balloon will decrease.  Eventually the balloon will expand just enough that the pressures inside and outside are again in balance.  You end up with a balloon of warm low density air that has the same pressure as the air surrounding it (Fig. 3)

You can use the same reasoning to understand what happens when you cool the air in a balloon.

The air inside and outside are the same in Fig. 1.  Cooling the air inside the balloon in Fig. 2 causes a momentary drop in the inside pressure and creates a pressure imbalance.  The larger outside air pressure compresses the balloon.

As the balloon volume decreases, pressure inside the balloon increases.  It eventually is able to balance the outside air pressure.   You end up with a balloon filled with cold high density air.

These two associations:
warm air = low density air and
cold air = high density air
are important and will come up a lot during the remainder of the semester.
Demonstration of Charles' Law
Charles Law can be demonstrated by dipping a balloon in liquid nitrogen.

The balloon had shrunk down to practically no 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.
Step #3 - vertical forces acting on air parcels
Now the final step, we must learn about the vertical forces that operate on a parcel of air in the atmosphere.


Air has mass and weight   When an air parcel has the same temperature, pressure, and density as the air around it, the parcel will remain stationary (top picture below)

Gravity does pull down on the air in the balloon (middle figure).

With gravity pulling downward on the air, there must be another upward-pointing force of equal strength.  The upward force (bottom figure) is caused by pressure differences between the bottom (higher pressure pushing up) and top of the balloon (slightly lower pressure pushing down on the balloon). 


If the balloon is filled with warm, low density air the gravity force will weaken (there is less air in the balloon so it weighs less). The upward pressure difference force (which depends on the surrounding air) will not change.  The upward force will be stronger than the downward force and the balloon will rise. 

Conversely if a balloon is filled with cold high density air, the balloon gets heavier.  The upward pressure difference force doesn't change.  The net force is now downward and the balloon will sink.
Free convection demonstration

We modified the Charles Law demonstration  (performed in class last Friday).    We used a balloon filled with helium instead of air (see bottom of p. 54 in the photocopied Class Notes).  Helium is less dense than air even when the helium has the same temperature as the surrounding air.  A helium-filled balloon doesn't need to warmed up in order to rise.

We dunked the helium-filled balloon in some liquid nitrogen to cool it and to cause the density of the helium 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.