Week 8 February 22

Week of February 22nd

Objectives:

1. Students will write a short constructed response on the relationship involved with Boyle's Law
2. Students will write a short constructed response on the relationship between the other law?

Agenda:

1. Pressure power point

2. How an internal combustion engine works

3. Boyles law, and see how different factors effect pressure

4. Scuba paperwork

5. Pressure Worksheet #1 & 2

6. Potato gun

Websites

http://phet.colorado.edu/simulations/sims.php?sim=States_of_Matter

Notes:

Intro to Boyle's Law
Pressure Volume Relationships

We are used to living at 1 ATM of pressure, so we rarely even take notice of it. We normally don't feel the pressure on us because the human body is primarily made up of liquid, and liquids are basically non compressible. At times, however, we do notice changes of pressure, primarily in our ears. You may have noticed your ears "popping" when flying, driving in the mountains, or even going up and down in elevators. This is because our ears have an air space in them, and air, like all other gases, is compressible. A gas will compress proportionately to the amount of pressure exerted on it. For example, if you have a 1 cubic foot balloon and double the pressure on it, it will be compressed to 1/2 cubic foot. Increase the pressure by 4, and the volume will drop to 1/4 the size etc. This theory was discovered by Sir Robert Boyle, a 17th century scientist. The theory known as Boyle's Law states: If the temperature remains constant, the volume of a given mass of gas is inversely proportional to the absolute pressure. Let's follow an example... Suppose you had a balloon measuring one cubic foot at the surface of the water. This balloon is under 1 ATM (14.7 psi) of pressure. If we push the balloon underwater, and take it to a depth of 33 feet, it is now under 2 ATM of pressure (29.4 lbs) - 1 ATM of pressure from the air, 1 ATM of pressure from the water. Boyle's Law then tells us that since we have twice the absolute pressure, the volume of the balloon will be decreased to one half. It follows then, that taking the balloon to 66 feet, the pressure would compress the balloon to one third its original size, 99 feet would make it 1/4 etc. If we bring the balloon in the previous example back up to the surface, it would increase in size due to the lessening pressure until it reached the surface and returned to its one cubic foot size. This is because the air in the balloon is compressed from the pressure when submerged, but returns to its normal size and pressure when it returns to the surface. We will achieve the same result with an open container, such as an inverted bottle, as we do with a balloon. By inverting a bottle at the surface and descending with it, the pressure from the surrounding water will compress the air and the bottle will start to fill with water. Even with no air escaping, the container will be half full of water at a depth of 33 feet due to the pressure compressing the air to half its original volume. Along with the volume of air in the balloon or container, the surrounding pressure will affect the density of the air as well. Density, simply stated, is how close the air molecules are packed together. The air in the balloon or container at the surface is at its standard density, but when we descend to the 33-foot level where its volume is reduced to one half, the density has doubled. At 66 feet, the density has tripled. This is because the pressure has pushed the air molecules closer together. Let's continue with this line of thinking and try an additional experiment. If we take our balloon and our open container down to 99 feet, we know that the density of air is four times what is was on the surface and the volume of air has been reduced to 1/4. Now at this depth, suppose we used a scuba tank and added air to the balloon until it returned to its original size. We will also blow air into the inverted container until it is completely full of air. We know the air at this depth is 4 times denser than at the surface. This means when we ascend with our balloon and container, the lessening pressure will make the air expand. This will have two different effects. The balloon will increase in size. It will attempt to grow to a size of 4 cubic feet by the time it hits the surface. If this is beyond the capability of the balloon, it will pop. The inverted container, however, will simply "bleed off" the expanding air that will harmlessly float away as bubbles. The main purpose of the proceeding material was to give you the theory behind the most important rule in scuba diving... "Never hold your breath!" Your lungs can act very much like a pair of balloons in your chest. As a breath hold diver (skin diver), if you fill your lungs with air at the surface, hold your breath, and dive to a depth of 33 feet, the surrounding pressure will compress your lungs to half of their original size. Upon ascending, they will return to normal size. If however, you descend to 33 feet and breath compressed air from a scuba tank, an ascent to the surface could cause you lungs to over expand and you could seriously injure yourself. This is easy to avoid, however, by simply not holding your breath which will let your lungs act like the open container in the preceding example, and you will simply "bleed off" the expanding air and maintain a normal lung capacity.



Charles' Law
Gas Volumes and Heat

There is another way to make a balloon smaller other than pushing it underwater. You can put it in the freezer. When you heat up a gas, the molecules that the gas is made up of move faster. In our balloon example, this increase in molecular motion causes the molecules to hit the sides of the balloon more often, and with more force, making the balloon expand. Cooling the gas would have the inverse effect, making the balloon smaller. While not too important when dealing with balloons, this concept has other applications. For example a full scuba tank, if left in the sun, will heat up. This causes the molecules in the air in the tank to move faster. Unlike the balloon which would expand, the tank is a rigid container that will not expand. This increase in motion then raises the pressure inside the tank. In fact, a full scuba tank will gain about 5-6 psi for every degree of temperature increase. This is one reason that full tanks should not be left in a hot trunk of a car. A tank filled to 3000 psi could easily reach 3500 psi if the temperature of it increased substantially. There have also been several cases where full scuba tanks, involved in boat fires, have exploded. This is due to the weakening the metal and the increased pressure from the heat. If go back to Boyle's Law and read a little more carefully we will see that Boyle's Law states: If the temperature remains constant, the volume of a given mass of gas is inversely proportional to the absolute pressure. Boyle did not concern himself much with changing temperature. This was, however, the main goal of a French scientist Jacques Charles. Charles showed that raising the temperature of a gas would tend to increase the volume of the gas, if its pressure remained constant. A few other laws, like Amonton’s Law, "The pressure of a fixed amount of a gas maintained at a constant volume is directly proportional to the gas pressure." or P µ T, P/T = constant or P1/P2 = T1/T2 all came together to help make what has come to be known as the General Gas Law: P1 x V1 P2 x V2 _______ = ________ T1 T2
T1 is the temperature at the first location, T2 is the temperature at the second. Remember when we were dealing with the pressure, we used the absolute pressure for our calculations. We must start at the zero mark. As stated in the Charles' law we must also use absolute temperature in the calculations as well. Absolute zero(** Raising the temperature of a gas causes the molecules in the gas to move faster. Lowering the temperature causes the molecules to slow down. Absolute zero is the theoretical temperature where all molecular motion stops.) is 460 degrees below Fahrenheit zero. This means we must add 460 to our temperatures before applying them to the formula. This is known as the Rankine scale. For example, 40 degrees Fahrenheit would be 460 + 40, or 500 degrees Rankine. Using the formula, we can explore the second half of Charles' law. This part of the law states that if the volume was kept constant, raising the temperature would increase the pressure of the gas. A good example of this can be found by an example with a scuba tank. Let us look at a scuba tank that shows it is filled to 3000 psi. We will assume the tank reads this pressure while sitting in an air-conditioned room which is at 70 degrees Fahrenheit. How much pressure would there be in the tank if it were left in a trunk of a car where the temperature climbed to 140 F? Let's put our numbers into the formula. The first thing we can do is take the V's out of the formula. Since the volume of the scuba tank will remain the same, we can cancel the V's and change our formula to: P1 / T1 = P2 / T2 Our starting pressure, P1, at first appears to be 3000 psi. We must remember to use absolute numbers though, so to obtain absolute pressure, we add in atmospheric pressure of 14.7. P1 = 3000 + 14.7. T1 would be our starting temperature in degrees Rankine. T1 = 460 + 70. And T2 would be 460 + 140. Let's now look at our formula with its numbers in place: 3000 + 14.7 P2 _____________ = _________ 460 + 70 460 + 140
adding up our numbers we get: 3014.7/530 = P2/600
We then multiply both sides of the equation by 600 to get the P2 on one side by itself. (600 x 3014.7) / 530 = P2 or P2 = 3412.8
It is important to note that 3412.8 is the absolute pressure at the second location. If we were asked "What is the gauge pressure at the second location?" we would subtract 14.7 from 3412.8 for an answer of 3398.1. This is a substantial increase in pressure. It turns out that a full 80 cubic foot scuba tank will have a pressure change of approximately 5-6 psi for every degree of temperature change. Since Charles' law also deals with changing the volume of a gas with a change in temperature, we can use the General Gas Law formula to determine the answer to the following question. If a balloon is inflated to one cubic foot at the surface, with air that is 85 degrees, how large would the balloon be if taken into 50 degree sea water to a depth of 40 feet? Solving for our Ps, Vs, and Ts, we get: P1 = 14.7 V1 = 1 T1 = 85 + 460 P2 = [40 x .445] + 14.7 T2 = 460 + 50
Using these numbers, we can solve for V2. (14.7 x 1) / 545 = ({[40 x .445] + 14.7} x V2) / 510
solving further we get: 14.7 / 545 = (32.5 x V2) / 510
Since we want to get the V2 by itself on one side of the equation, we will multiply both sides by 510 over 32.5. This will leave the V2 alone on the second side: (510 x 14.7) / (32.5 x 545) = V2
using a calculator we get: 7497 / 17712.5 = V2 or: V2 = 0.4232604093155
Thus the volume of our balloon at its second location would be about .42 cubic feet.