Nov. 5-9

This week in class, I learned how to measure atmospheric pressure using a barometer I made outside of class for a project, and how to mathematically find the pressure, temperature, volume, and number of particles.

Making my barometer for the project, I used the materials to make one: a jar, a balloon, scissors, an index card with mm markings, tape, and a straw. I used the scissors to cut off the round part of the balloon, blew into the balloon and then deflated it, wrapped the balloon on the rim of the jar, taped the straw onto the balloon, and taped the index card on the jar. From this project, I learned that a barometer measures air pressure.

On Wednesday, my class and I went outside to experiment with our barometers. As we went outside, the atmospheric pressure increased, so the barometer needle went up. The needle went up because outside, two factors played a role in atmospheric pressure: temperature and pressure. Since it was cold outside and there weren't many clouds in the sky, this meant that the cold air, denser than the warm air, pushed down on the warm air, or the warm air remained above the cold air. Thus, air pressure increased. Since it was cold outside, I figured the cold air pushed down, but not necessarily on the warm air. It may not have pushed down on the warm air because it is possible that the water evaporated because the air was warm enough to do so, and then the clouds condensed before the cold air pushed down.

I noticed that the needle on the barometer went up because the colder air exerted greater pressure on the balloon, thus accounting for the increasing downward force. Then, as the balloon is pushed downward, the needle goes up because of the elasticity in the balloon providing a counter upward force (torque) to lift up the balloon at the point where the balloon dips. I also learned that as we went inside, atmospheric pressure went down, since the atmosphere inside is mostly warm air. Thus, the decreasing pressure pushing onto the balloon causing the needle to go down.

This week, I learned how to do mathematical problems involving temperature, volume, number of particles, and pressure. I learned that the secret to do these problems is to keep in mind that when you are trying to find the change of one of the factors, you have to make that you keep them proportionately related to each other.

This problem involves volume and temperature. For example, a flask with water in it is on a hot plate (pressure and number of particles are constant). The temperature is 25ºC, and the volume is 50 mL. Then, in five minutes, the temperature is 50ºC. So, what is the volume? The fishy thing about this is that the temperatures are NOT in Kelvins. Thus, I converted 25ºC and 50ºC to Kelvins by adding them each to 273.15 since -273.15ºC is equal to 0 Kelvins.

Next, I set the temperature and volume as proportions since their rate of change is proportional. I then used the least precise number to make it easier to calculate the volume. So, I rounded 298.15 to 298 and 323.15 to 323.

Set them up in these steps: (Remember, x=temperature in Kelvins after the water has been heated. To set these proportionally, the initial temperature should be divided by the new temperature. So, the volumes have to be set up the same way since they both have direct relationships with each other.)

1. 298/323=50/x
2. 298x=50(323)
3. 298x=8075
4. x=52 mL.

The volume of the heated water after five minutes is 54mL, which supports the idea of expansion–the increasing temperature causes the liquid to rise, thus, the volume increases.

Next, I learned how to calculate pressure and the number of particles (temperature and volume are constant). For example, I have a syringe needle, and I have 5 puffs (particles) in the syringe needle. The pressure is 7 k/Pa. Then, I added 3 more puffs into the syringe needle in a closed system, thus increasing the number of particles. However, I had to convert to atm, or atmospheric pressure–air in the atmosphere exerts force per unit exerted onto the surface. 1 k/Pa=0.00986923266716 atm. So, 7 k/Pa=0.069 atm.

Next, divide the original number of particles by the new number of particles and set this equivalent to the original atm over x atm (x atm being the new pressure), since the number of particles and the pressure has a direct relationship.

Set them up in these steps:

1. 5/8=7/x
2. 56=5x
3. 11 atm=x

Thus, the new atmospheric pressure is 11 atm. Therefore, the greater the number of particles, the greater the pressure is.