I've inserted lots of notes. You're on the right track for the most part, just a few errors in details.
I expected the air column to shift slightly. When the pressure valve is opened it is allowing equal air in and out. When water is brought up through the vertical tube it removes some of the water and when the cap is not on the pressure indicating tube air takes up the space where the water was. When the cap is put back on and the water is released the pressure will build up because it cannot escape and the pressure that took the place of the water and the water are now in the bottle.
Honestly I did not see a significant change when I did this part. I understand what is taking place but I cannot physically see what is going on.
I would expect air to escape because there is more air in there now than there was before the vertical tube took water out of the system and then placed in back in the bottle. When I removed the cap I heard air escaping or a very faint pop as I removed it. I could not physically see the air escaping (of course :)) but I could tell that it was pushed out.
This introduced more air into the system. The air column meniscus went further away from the tube. When I quit making the bubbles the extra air forced water back up into the vertical tube. I would have anticipated for both to happen. I remember learning this lesson as I child when I would blow into my juice boxes and when I would quit the juice would shoot out the straw. When the bubbles are being blown they are releasing air into the bottle. With no where to escape the pressure builds up as indicated by the meniscus. When I removed my mouth the air could escape and it did so in such a manner that it forced water out of the bottle into the tube. This may have replaced the air that was being blown into the system.
.01 N/m^2
It would change by 1%, or in my case 0.5 cm
A 1% change in pressure can be translated directly into pressure change in N / m^2.
Atmospheric pressure is about 100 kPa, or 100,000 Pascals, i.e., 100,000 N/m^2. So a 1% change is 1 kPa, or 1,000 N/m^2.
How much altitude change in the tube is associated with a 1,000 N/m^2 change in pressure?
1 %
The temperature is 273 K. How many degrees would correspond to a 1% change in temperature?
100 kPa = 1 Pa 99 kPa = .99 Pa, 1 Pa = 1 N/m^2, so a change in .01 Pa is a change of .01 N/m^2. Everything would need to change by 1 %. My column was originally 50 cm. 1% of 50 cm is 0.5 cm.
100 kPa = 100,000 Pa or 100,000 N/m^2.
The water column height does not reflect the absolute pressure, only the difference between bottle pressure and atmospheric pressure. So a 1% change in absolute pressure would not result in a 1% change in the vertical column's height.
If the absolute pressure outside the bottle was zero (i.e., a vacuum) then the pressure exerted by the vertical column would equal the absolute pressure in the bottle, and a 1% change in pressure would result in a 1% change in the height of the vertical column.
3 degrees
.0033 N/m^2
.5 cm
If it 300 K then 1% is 3 degrees. 1 degree is a .033% change. .033% of .01 N/m^2 is .0033.
1/300 = .0033..., which is .33% rather than .033%.
It would be .0033 * 100,000 N/m^2, or about 330 N/m^2.
I am just estimating the third column, for I am not sure of how to determine this.
2 degrees
.2 degrees
It is a rough estimate but for every 2 degrees the system would rise about 1 cm. There are 10 mm in a cm so 10% of 2 is 0.2.
22, 0
22.1, 0
22.2, 0
22.2, 1
22.2, 1
22.2, 1
22.3, 2
22.3, 2
22.3, 2
22.3, 2
22.3, 2
22.2, 2
22.2, 1
22.2, 1
22.3, 2
22.3, 2
22.2, 2
22.2, 2
22.2, 1
22.2, 1
My eyes may have been playing tricks on me. I would record the temperature as close to 0.1 degrees as I could but I didn't want to get to close for fear that I would affect it. The water column was also hard to read.
This system is not very easy to decode because the themometer is senstive to me approaching it too closely. The temperature changed by 0.2 degrees throughout the 10 minutes. The water column also increased by about 2 cm.
The expectation would be that 2 cm corresponds to about .6 degrees.
The sensitivity of the thermometer to your presence probably does have a big effect on the accuracy of this result, but it's the right order of magnitude.
After 6 minutes the meniscus returned to its original position. It rose by 2 cm. It reached its peak at 2 minutes and 15 seconds. It took the rest of the time to return to normal.
Yes by about 5-6 cm. It occured much faster than the water bath the bottle just received. The heat from my hands increased the temperature of the water in the bottle and therefore increased the pressure inside the bottle making the altitude of water column rise.
22.1, 0
22.1, 0
22.2, 0
22.2, 1
22.2, 1
22.2, 1
22.1, 1
22.1, 1
22.1, 0
22.1, 0
This time the column increased by about 2 or 2.5 cm. It took about the same amount of time to observe a change but the change was not as significant.
2-3 degrees
70.65 cm^3
It would decrease by 70.65 cm^3
5 degrees
Everything would be multiplied by 2.
The water rose 5-6 cm which would coincide with 2-3 degrees. To find the volume I took the radius^2*pi*h which is 1.5^2*3.14*100 mm. This answer was 7065 mm which is 70.65 cm. The volume that shifted into the tube is equal to the volume that leaves the bottle. Each 1 degree changes the water by 2 cm. All the answers given were for 300 K, but 600 K would double all the answers.
This also had to do with the pressure in the bottle. I had water spilling into the bucket which leads me to believe this water change in not as important in determinig temperature and pressure.
6 N/m^2
3 Degrees
I am thinking that if it raises 6 cm that means the gas pushed the water that high up the tube. For every 2 cm it is 1 degree. I dont know exactly how to determine these answers. I read the Exp 8 and 9 but they arent really helping me answer these.
A 6 cm change in the height of the water column indicates a specific change in rho g h, therefore a specific change in pressure and therefore temperature.
How much would the pressure have to change?
What would be the new and the old pressure, the ratio of pressures, the ratio of temperatures and therefore the change in temperature? This is the change in temperature necessary to raise the water 6 cm in a vertical tube.
Now if volume was to increase by .7 cm^3, without changing the pressure, how much temperature change would be required? This is the temperature required to the meniscus to move 6 cm in a horizontal tube.
2 cm
Would it be approximately the same, but maybe a faster increase because the gas is not working against gravity to travel up?
That is correct.
If the tube was at a 45 degree angle it would make the system work half as hard as the vertical position.
Since, as we saw above before, the temperature change required to make a 6 cm change in the vertical position is to much greater than that required to make a 6 cm change in the horizontal position, we would expect that a the angle that equalizes the influence of altitude change and gas expansion would be much closer to horizontal than to vertical.
I know I have a lot of questions, I will print this out when you respond to refer to before the test.
I'll be glad to answer questions after you look over these responses.