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? **
While I drew the water up the straw to 50cm, the water column in the pressure tube was pulled toward the bottle slightly. After I removed the tube and replaced the pressure-valve cap the water remained in the tube and the pressure tube remained at the position it was in. This is contrary to what I expected. I thought the pressure tube would return to its initial position.
** What happens when you remove the pressure-release cap? **
When I pulled water up the straw but left the pressure-valve open, the water column in the pressure tube did not move. The water ran back down the vertical tube and the water column in the pressure tube remained where it was. This is what I expected to happen because the pressure was allowed to equalize with the pressure-valve open.
** What happened when you blew a little air into the bottle? **
The air I blew into the system caused an increase in pressure. This pressure then pushed water up through the vertical tube to be relieved. The air column decreased when I blew air into the system due to the pressure pushing the water column away from the bottle. The air column returned to its initial position when I stopped blowing. This is because the pressure was being relieved by pushing water through the vertical tube. This is what I expected to happen.
** 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
1%, 0.3cm
1% (?)
A pressure of 100kPa is equal to 1.0*10^5 N/m^2 because 1Pa = 1N/m^2. To find 1% of this I did the following calculations:
100kPa = 1.0*10^5 Pa = 1.0*10^5 N/m^2
(1.0*10^5 N/m^2)(.01) = 1000 N/m^2
I am assuming that if the pressure of a system changed by 1%, the height of the supported air column in the pressure tube would change by the same percentage. Therefore, if the air column were 30cm long, it would change by 0.3cm (30cm*.01=0.3cm).
For the temperature, if V, n, and R are all held constant, then temperature is proportional to pressure (P/T). If pressure changes by 1%, temperature will change by 1%.
Honestly, I don't feel very comfortable with my answers here. I hate when I feel like I'm guessing. Could you tell me if I missed something here?
Your answers are very good.
** Your estimate of degrees of temperature change, amount of pressure change and change in vertical position of water column for 1% temperature change: **
3K
333 N/m^2
3.33cm
If the temperature is 300K, 1% change would be a change of 3K (300K * .01 = 3K).
I found the pressure two different ways. The first way I simply said that 1K change in temperature is a (1K/300K) .33% change. If the pressure is assumed to be 1.0*10^5 N/m^2, then .33% change would be 333 N/m^2 (1.0*10^5 N/m^2 * .0033 = 333 N/m^2. We could also set up a proportion for a 1K change, P2 = (P1*T2)/T1 = (1.0*10^5 N/m^2 * 299K)/300K = 99666 N/m^2. Therefore, the change in P is 1*10^5N/m^2 - 99666N/m^2 = 333 N/m^2.
My answer for the change in vertical position is a guess. I assumed that if the pressure change was 333N/m^2, then the change in height would be 3.33 cm. I divided the change in pressure by 100 to convert from meters to centimeters. I'm sure my method is not right, but I can't think of a way to calculate it.
** The temperature change corresponding to a 1 cm difference in water column height, and to a 1 mm change: **
0.3K
0.03K
If my above assumption is correct, that 1K change in temperature causes 3.33cm change in height, the change in temperature would be 0.3K.
(1K/3.33cm)(1cm) = 0.3K
One centimeter is equal to 10 millimeters. In order to find the change in temperature corresponding to a 1mm change in height, I just divided my previous answer by 10 to convert centimeters to millimeters.
(1K/33.3mm)(1mm)= 0.03K
** water column position (cm) vs. thermometer temperature (Celsius) **
Initial: 25.0, 30.0cm
24.9, 29.9
25.1, 29.9
25.3, 30.0
25.6, 30.2
25.4, 30.2
25.1, 29.9
25.0, 30.0
24.8, 29.9
24.9, 29.9
25.1, 30.1
25.3, 30.2
25.1, 30.0
25.0, 29.9
24.7, 29.8
24.7, 29.9
24.5, 29.8
24.6, 29.9
24.4, 29.8
24.5, 29.8
24.7, 29.9
** Trend of temperatures; estimates of maximum deviation of temperature based on both air column and alcohol thermometer. **
The temperature barely varied from the initial 25 degrees Celsius. I managed to find an area that the temperature was rather constant. When temperature increased, the water column rose. When the temperature decreased, the water column descended slightly. The biggest variations in temperature were at 25.6 degrees which caused a +.2cm change and at 24.4 degrees which caused a -.2cm change. These temperatures were both 0.6 degrees from the initial and average room temperature, 25.0 degrees Celsius.
** Water column heights after pouring warm water over the bottle: **
Initial thermometer reading, 26.1 degrees Celsius.
My height measurements at 15 second intervals are as follows:
67.0, 55.0, 45.0, 40.0, 36.0, 33.0, 31.0, 30.1, 29.5, 29.0, 28.6, 28.3, 28.0
This data shows that the pressure in the bottle greatly increased and pushed the water column up very quickly. As the bottle returned to room temperature the change in height began decreasing at a decreasing rate. The column stopped decreasing at 28 cm.
** Response of the system to indirect thermal energy from your hands: **
The altitude increased from 28cm to about 35cm. I got my hands very warm and held them near the sides of the bottle. After about 3 seconds, the water column began climbing very rapidly. After the peak at 35cm, I could feel my hands cooling as the column fell. I could also feel the heat from my hands bouncing off the sides of the bottle when I started. My hands were transmitting heat to the molecules in the bottle, causing the gas to expand, increase pressure and push the water up the tube.
** position of meniscus in horizontal tube vs. alcohol thermometer temperature at 30-second intervals **
Temperature for these trials remained at 26-26.1 degrees Celsius
26.0 C, 30.0cm
26.1, 29.8
25.9, 29.5
25.7, 29.2
25.6, 28.9
25.7, 28.7
25.8, 28.5
25.9, 28.4
26.0, 28.4
25.9, 28.4
The very slight changes in temperature did not seem to affect the water column as much in this position as it did when the tube was vertical. The column decreased to 28.4 cm and stopped.
** What happened to the position of the meniscus in the horizontal tube when you held your warm hands near the container? **
102cm
100
98.2
96
91
87
80.5
74
65
56
The water ran VERY quickly through the horizontal tube. It peaked at almost the end of the tube, 102cm. As the temperature began to return to room temperature, the water gradually picked up speed as the column returned to its initial position. This trial was slightly different from the same trial with the vertical tube.
** 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: **
2462 N/m^2 per 10cm
2.827cm^3
0.2827%
0.8482K
1.696K
If my calculations from the atmospheric pressure lab were correct, the pressure changed by about 2462 N/m^2 per 10cm change in water column position. Is there a better way to calculate this?
More accurate instruments could be used to measure the relevant quantities more accurately, but working within the limits of the instruments used here, there is no better way to calculate the results and yours are excellent.
The volume of the tube, given the above information, can be found by the equation V = pi*r^2*h = pi*(.3cm)^2(10cm) = 2.827cm^3.
The volume change in the container should be equal to the volume change in the tube because it is one system. Therefore, the volume of air in the container should increase 2.827cm^3. The container can hold 2 liters = 2000mL = 2000cm^3, and has 1000cm^3 of water in it. Therefore, there was initially 1000cm^3 of air in the container. The change in volume is then (2.827cm^3 / 1000cm^3)*100 = .2827%.
I set up a proportion to find the temperature change, assuming the initial temperature is 300K with the volume at 1000cm^3. If the volume increased 2.827cm^3, the temperature would increase by 0.8482K. (300K/1000cm^3)(2.827cm^3) = 0.8482K
If the temperature was 600K instead of 300K but the change in volume was still 2.827cm^3, the answer would change. My calculations would then be (600K/1000cm^3)(2.827cm^3) = 1.696K. Am I correct?
** Why weren't we concerned with changes in gas volume with the vertical tube? **
The changes in volume were too small to be of any significance and would have been difficult to accurately measure with the equipment I was using. As the calculations in the previous box show, a fairly significant change in volume does not affect temperature much, so an immeasurable change in volume would have virtually no effect.
** 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: **
I'm not sure I know what the situation is: Suppose that in the process of moving 10 cm along the tube, the meniscus 6 cm in the vertical direction.
1477 N/m^2
4.432K
0.07K
As mentioned earlier, I calculated in my last lab that for an altitude of 10cm, the pressure required was 2462N/m^2. Therefore, the pressure required for 6cm altitude would be 1477N/m^2. (2462N/m^2/10cm)(6cm) = 1477.2N/m^2.
To find the change in temperature required, I used the assumptions that at 300K the pressure is 1.0*10^5N/m^2. My calculations were: (300K/1*10^5N/m^2)(1477N/m^2) = 4.432K. This makes sense because temperature and pressure are directly proportional. In order to increase pressure, temperature must also increase by the same ratio when volume is constant.
The volume of the gas in the container is roughly 3000cm^3 at 300K. Therefore, the temperature increase necessary to increase the volume by .3cm^3 is 0.21K. (300K/3000cm^3)(0.7cm^3) = 0.07K.
** 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: **
3cm
30cm
45 degrees angle
It was found earlier that 1 degree change in temperature causes the altitude to change by about 3.33cm when the tube is vertical.
If the tube was perfectly horizontal, one degree change would cause a greater change in position. This is exhibited in my results of putting my warm hands next to the bottle when the tube was vertical and horizontal. I don't know the correct method to adjust for the angle the liquid travels, but from my data, the horizontal position causes the distance traveled to increase by about 10 times. This measurement is not exact however because my hands in these trials were probably not exactly the same temperature. (Vertical `ds = 7cm, Horizontal `ds = 74cm)
If a completely horizontal slope created a 30cm change in distance, then an angle halfway between horizontal and vertical would cause a change in distance that is 15cm.
** Optional additional comments and/or questions: **
For many of the pressure related questions, I referred to my Measuring Atmospheric Pressure labs. I hope that is alright.
As you can probably tell, I lost confidence in my answers as I got closer to the end. I was simply not sure how to calculate the exact numbers that were asked. However, I really do understand the relationships between temperature, pressure and volume.
You did superior work on this lab. I can't find a flaw in your data or your analysis.