Your 'bottle thermometer' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Your optional message or comment: **
** What happens when you pull water up into the vertical tube then remove the tube from your mouth? **
I think the air column will probably not move because the valve is open on the short tube. And when you hold the water in the tube and replace the valve, and then take the tube from your mouth the water will probably just fall right back down because it is like in the previous experiment when you let go of your squeeze, the water falls back down. But once the valve is in place may the air column gets shorter.
** What happens when you remove the pressure-release cap? **
Well, nothing happened because the water had already gone back down the tube. Maybe in the previous question the water was supposed to stay in the tube, but it didn't in my system.
** What happened when you blew a little air into the bottle? **
The length of the air column got smaller and stayed smaller, also the vertical tube filled with water after I took my mouth off of the tube.
The air column got smaller because the pressure inside the bottle was increased with the blowing. The air column did get slightly longer with time, but stayed relatively the same until I popped the pressure release valve off of the short tube.
** Your estimate of the pressure difference due to a 1% change in pressure, the corresponding change in air column height, and the required change in air temperature: **
1000 N/m^2
10.1 cm
1%
To find gas pressure I knew that 1 Pascal = 1 N/m^2. So I figured out that 100 kPa = 100000Pa or 100000 N/m^2, and 1% of 100000 N/m^2 = 1000 N/m^2. To find the change in height I just multiplied the 10 cm by 1% and figured out the change would be .1 cm, which would make the total height 10.1 cm. If the volume is constant the ratio will still be 1%, so temperature will still have to change by 1%.
If the temperature is around 300 K, then a 1% change would be about 3 Kelvin or Celsius degrees.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
.2685 degrees Celsius
3.7 kPa
10.35 cm
I made my estimate for the temperature by changing Kelvin to Celsius then multiplying by .01 to get the change in temperature. To find the pressure I used the formula P1/T1 = P2/T2 and solved for P2. To find the change in water column I figured out by what percentage the temperature changed and just used that as my percentage in change in water heigth, .035 * 10 cm.
when working with the gas laws you always work with Kelvin, not Celsius temperatures.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
.2685
.02685
I found the temperature changes by multiplying the Celsius form of 300 K by the decimal form of the percentages:
.01* 26.85 C
.001* 26.85 C
** water column position (cm) vs. thermometer temperature (Celsius) **
25.9, 10
25.8, 10
25.7, 9.5
25.5, 9.0
25.6, 9.0
25.6, 9.0
25.5, 9.0
25.5, 8.5
25.4, 8.5
25.4, 8.3
25.5, 8.3
25.5, 8.4
25.6, 9.0
25.7, 9.6
25.9, 10.1
25.95, 10.5
26.0, 11
26.1, 11.2
26.1, 11.5
26.2, 12
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The temperature decreased initially .5 degrees Celsius, then it increased .8 degrees Celsius. The maximum deviation was about .4 degrees Celsius on either side of 25.8 C. I just saw the highest temperature and the lowest temperature and used them as the extremes and found out how far away they were from the average temperature.
** Water column heights after pouring warm water over the bottle: **
25.4
50 cm
30 cm
17 cm
10 cm
7 cm
6 cm
3 cm
0 cm
-3 cm
** Response of the system to indirect thermal energy from your hands: **
My hands did warm the air in the bottle measurably.
5 cm
I was very careful not to touch the sides and as soon as my palms got near the sides the water started moving immediately and I watched it travel up the tube for about a minute before stopping.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
27.0, 11
26.9, 7
26.8, 6
26.8, 6
26.5, 5
26.2, 5
26.2, 5
26.2, 4
26.2, 3
26.0, 1
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
It took longer for the water to move up the tube for some reason. I am not really sure why. I would think the water would move faster since it didn't have to work against gravity.
Maybe my hands are colder not then they were earlier.
** Pressure change due to movement of water in horizonal tube, volume change due to 10 cm change in water position, percent change in air volume, change in temperature, difference if air started at 600 K: **
I think the pressure remained the same because it was still heating up and cooling by the same amounts.
I don't think the volume inside the system would change very much.
The percent is very small and insignificant.
I am not sure if you are asking me about the pressure of the air in the bottle, because I thought that the change in volume was insignificant and we didn't bother to measure it. I thought we just worried about the pressure and the temperature. I'm not sure what answers you are looking for here.
you can estimate the volume of gas originally in the bottle (maybe 1500 cm^3) and you have enough information about the tube to find the volume change when a 10 cm column of water is forced from the bottle into the tube. It's on the order of 1 cm^3, which is a change of about .06%.
** Why weren't we concerned with changes in gas volume with the vertical tube? **
The changes in volume were so small that it was not significant, so it was not important enough to factor in to the calculation. It would not have made a significant difference in our estimates of temperature change.
** Pressure change to raise water 6 cm, necessary temperature change in vicinity of 300 K, temperature change required to increase 3 L volume by .7 cm^3: **
488 kPa
51 K
I am not sure how to figure the Volume question.
I obtained the pressure change by using the formula P2 = P1 + (rho g h). I obtained the temperature change by using the formula P1/T1 = P2/T2
Right method, but P1 is atmospheric pressure, around 100,000 Pa, so P2 would be about 100,488 Pa (might as well say 100,500 Pa). T2 / T1 would be about 1.005, so the temp. change would be about 1.5 K.
** The effect of a 1 degree temperature increase on the water column in a vertical tube, in a horizontal tube, and the slope required to halve the preceding result: **
I am not sure how to go about answering these questions either.
In a vertical tube the volume change would be insignificant so P2 / P1 = T2 / T1 = 301 K / (300 K) = 1.0033, so pressure change would be .0033 atm or about 330 Pa; so the change in rho g h would be about 330 Pa, from which you can get the change in h. About 3 cm.
In a horizontal tube the volume of the gas would change by .0033 of the original volume; assuming 1500 cm^3 of gas in the bottle, this would be a change of about 5 cm^3. This would correspond to about 50 cm of tubing.
Good overall, but there were some errors you need to be aware of. See my notes and let me know if you have questions.