Fri., Feb. 6 notes

In A
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).

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 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. 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 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.

Read through the explanation on p. 52 in the photocopied Classnotes.

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 proceed on to 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 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.

Here's a summary (not shown in class)

Now finally on to step #3. 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 neturally 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 wins and 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.

To demonstrate free convection we modified the Charles Law demonstration. 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 hhydrogen-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.

Here's an example of where something like this happens in the atmosphere.

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). This is called free convection; many of our summer thunderstorms start this way.