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Phy 232
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.
** Bottle Thermometer_labelMessages **
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2hrs
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Instructions follow:
•Before you put your mouth on the tube, make sure it's clean and make sure there's nothing in the bottle you wouldn't want to drink. The bottle and the end of the tube can be cleaned, and you can run a cleaner through the tube (rubbing alcohol works well to sterilize the tube). If you're careful you aren't likely to ingest anything, but of course you want the end of the tube to be clean.
•Once the system is clean, just do this. Pull water up into the tube. While maintaining the water at a certain height, replace the cap on the pressure-valve tube and think for a minute about what's going to happen when you remove the tube from your mouth. Also think about what, if anything, is going to happen to the length of the air column at the end of the pressure-indicating tube. Then go ahead and remove the tube from your mouth and watch what happens.
Describe below what happens and what you expected to happen. Also indicate why you think this happens.
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I expected the water to go down, and as expected the water retreated back into the bottle when I removed my mouth. I think the shift was due to pressure trying to reach equilibrium between inside and out the bottle.
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Now think about what will happen if you remove the cap from the pressure-valve tube. Will air escape from the system? Why would you or would you not expect it to do so?
Go ahead and remove the cap, and report your expectations and your observations below.
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No apparent changed occurred as expected. One I removed my mouth earlier, pressure did its effects on the bottle so removing the cap was insignificant.
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Now replace the cap on the pressure-valve tube and, while keeping an eye on the air column in the pressure-indicating tube, blow just a little air through the vertical tube, making some bubbles in the water inside the tube. Blow enough that the air column in the pressure-indicating tube moves a little, but not more than half a centimeter or so. Then remove the tube from your mouth, keeping an eye on the pressure-indicating tube and also on the vertical tube.
•What happens?
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Water went up in the pressure indicating tube after blowing but did not retreat much when I stopped. When I removed my mouth water went up the vertical tube
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•Why did the length of the air column in the pressure-indicating tube change length when you blew air into the system? Did the air column move back to its original position when you removed the tube from your mouth? Did it move at all when you did so?
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In essence, I was blowing pressure into the bottle causing the shift. The air column did not move back to its original position when I removed the tube from my mouth. There was a faint movement but hardly seen.
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•What happened in the vertical tube?
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The water moved up the vertical tube
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•Why did all these things happen? Which would would you have anticipated, and which would you not have anticipated?
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Air is being forced out when the water shifs up the tube to return to equilibrium with atmospheric pressure. This was expected but the pressure indicating tube remaing the same was baffling.
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•What happened to the quantities P, V, n and T during various phases of this process?
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Pressured-increased, Volume-constant, n-increased, and temperature increased with pressure increase.
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Place the thermometer that came with your kit near the bottle, with the bulb not touching any surface so that it is sure to measure the air temperature in the vicinity of the bottle and leave it alone until you need to read it.
Now you will blow enough air into the bottle to raise water in the vertical tube to a position a little ways above the top of the bottle.
•Use the pressure-valve tube to equalize the pressure once more with atmospheric (i.e., take the cap off). Measure the length of the air column in the pressure-indicating tube, and as you did before place a measuring device in the vicinity of the meniscus in this tube.Replace the cap on the pressure-valve tube and again blow a little bit of air into the bottle through the vertical tube. Remove the tube from your mouth and see how far the water column rises. Blow in a little more air and remove the tube from your mouth. Repeat until water has reached a level about 10 cm above the top of the bottle.
•Place the bottle in a pan, a bowl or a basin to catch the water you will soon pour over it.
•Secure the vertical tube in a vertical or nearly-vertical position.
The water column is now supported by excess pressure in the bottle. This excess pressure is between a few hundredths and a tenth of an atmosphere.
The pressure in the bottle is probably in the range from 103 kPa to 110 kPa, depending on your altitude above sea level and on how high you chose to make the water column. You are going to make a few estimates, using 100 kPa as the approximate round-number pressure in the bottle, and 300 K as the approximate round-number air temperature. Using these ball-park figures:
•If gas pressure in the bottle changed by 1%, by how many N/m^2 would it change?
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1000N/m^2
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•What would be the corresponding change in the height of the supported air column?
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height of the column = (1000N/m^2)/(rho g)
assume g= 9.8m/s^2
= (1000N / m^2) / (1000 kg / m^3 * 9.8 m/s^2)
= (1000) / (9800m)
= .102m
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•By what percent would air temperature have to change to result in this change in pressure, assuming that the container volume remains constant?
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a pressure change of 1% would mean a temp change of 1% and vice versa in this case.
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Continuing the above assumptions:
•How many degrees of temperature change would correspond to a 1% change in temperature?
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.01*300K = 3K
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•How much pressure change would correspond to a 1 degree change in temperature?
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((1/3)*.01)*100kPa = .3kPa = .3 = 300Pa
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•By how much would the vertical position of the water column change with a 1 degree change in temperature?
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It would change by around 3cm
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•How much temperature change would correspond to a 1 cm difference in the height of the column?
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(1/10)*3K = .3K
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•How much temperature change would correspond to a 1 mm difference in the height of the column?
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(1/10)*.3K = .03K
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A change in temperature of 1 Kelvin or Celsius degree in the gas inside the container should correspond to a little more than a 3 cm change in the height of the water column. A change of 1 Fahrenheit degree should correspond to a little less than a 2 cm change in the height of the water column. Your results should be consistent with these figures; if not, keep the correct figures in mind as you make your observations.
The temperature in your room is not likely to be completely steady. You will first see whether this system reveals any temperature fluctuations:
•Make a mark, or fasten a small piece of clear tape, at the position of the water column.
•Observe, at 30-second intervals, the temperature on your alcohol thermometer and the height of the water column relative to the mark or tape (above the tape is positive, below the tape is negative).
•Try to estimate the temperatures on the alcohol thermometer to the nearest .1 degree, though you won't be completely accurate at this level of precision.
•Make these observations for 10 minutes.
Report in units of Celsius vs. cm your 20 water column position vs. temperature observations, in the form of a comma-delimited table below.
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26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.4 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.5 C, 40.6 cm
26.4 C, 40.6 cm
26.4 C, 40.6 cm
26.4 C, 40.6 cm
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Describe the trend of temperature fluctuations. Also include an estimate (or if you prefer two estimates) based on both the alcohol thermometer and the 'bottle thermometer' the maximum deviation in temperature over the 10-minute period. Explain the basis for your estimate(s):
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There was actually only a minor fluctuation in temperature. The max deviation for the bottle thermometer was .1 calculated from 26.5-26.4 = .1. The change was so minor and with the precision of the alcohol thermometer I was unable see any change
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Now you will change the temperature of the gas in the system by a few degrees and observe the response of the vertical water column:
•Read the alcohol thermometer once more and note the reading.
•Pour a single cup of warm tap water over the sides of the bottle and note the water-column altitude relative to your tape, noting altitudes at 15-second intervals.
•Continue until you are reasonably sure that the temperature of the system has returned to room temperature and any fluctuations in the column height are again just the result of fluctuations in room temperature. However don't take data on this part for more than 10 minutes.
Report your results below:
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50.2
48.1
45.9
43.7
40.9
39.7
39.0
38.6
38.3
38.1
37.8
37.5
37.2
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If your hands are cold, warm them for a minute in warm water. Then hold the palms of your hands very close to the walls of the container, being careful not to touch the walls. Keep your hands there for about a minute, and keep an eye on the air column.
Did your hands warm the air in the bottle measurably? If so, by how much? Give the basis for your answer:
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I'm sure my hands warmed the air in the bottle, however it was not by a measureable amount. Possibly energy transfer was so slight it wasn't picked up by the thermometer,
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Now reorient the vertical tube so that after rising out of the bottle the tube becomes horizontal. It's OK if some of the water in the tube leaks out during this process. What you want to achieve is an open horizontal tube,, about 30 cm above the level of water in the container, with the last few centimeters of the liquid in the horizontal portion of the tube and at least a foot of air between the meniscus and the end of the tube.
The system might look something like the picture below, but the tube running across the table would be more perfectly horizontal.
Place a piece of tape at the position of the vertical-tube meniscus (actually now the horizontal-tube meniscus). As you did earlier, observe the alcohol thermometer and the position of the meniscus at 30-second intervals, but this time for only 5 minutes. Report your results below in the same table format and using the same units you used previously:
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26.4 C, 12 cm
26.4 C, 12 cm
26.3 C, 11.9 cm
26.3 C, 11.9 cm
26.3 C, 11.7 cm
26.2 C, 11.4 cm
26.3 C, 11.4 cm
26.2 C, 10.9 cm
26.2 C, 10.6 cm
26.2 C, 10.3 cm
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Repeat the experiment with your warm hands near the bottle. Report below what you observe:
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I observed a significant change in the meniscus position.
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When in the first bottle experiment you squeezed water into a horizontal section of the tube, how much additional pressure was required to move water along the horizontal section?
•By how much do you think the pressure in the bottle changed as the water moved along the horizontal tube?
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No change because of the horizonatal setup there would be no change in the vertical water level
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•If the water moved 10 cm along the horizontal tube, whose inner diameter is about 3 millimeters, by how much would the volume of air inside the system change?
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pi r^2 = pi (.15 cm)^2 = .07 cm^2
(10 cm) * (.07 cm^2) = .7 cm^3
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•By what percent would the volume of the air inside the container therefore change?
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.7 cm^3 / (1500 cm^3) = .0005 = .05%
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•Assuming constant pressure, how much change in temperature would be required to achieve this change in volume?
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.05% change in volume means a .05%change in temperature.
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•If the air temperature inside the bottle was 600 K rather than about 300 K, how would your answer to the preceding question change?
.
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.05% of 600k = .0005*600K = .3K
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There were also changes in volume when the water was rising and falling in the vertical tube. Why didn't we worry about the volume change of the air in that case? Would that have made a significant difference in our estimates of temperature change?
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The minimal 1 cm change in vertical position would barerly change the volume. The insignificant change would have no impact on the results and can be ignored. No.
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If the tube was not completely horizontal, would that affect our estimate of the temperature difference?
For example consider the tube in the picture below.
Suppose that in the process of moving 10 cm along the tube, the meniscus moves 6 cm in the vertical direction.
•By how much would the pressure of the gas have to change to increase the altitude of the water by 6 cm?
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1000kg/m^3*9.8m/s^2*.06 m = 590N/m^2 = 590 Pa
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•Assuming a temperature in the neighborhood of 300 K, how much temperature change would be required, at constant volume, to achieve this pressure increase?
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.7/3000 = .000023
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•The volume of the gas would change by the additional volume occupied by the water in the tube, in this case about .7 cm^3. Assuming that there are 3 liters of gas in the container, how much temperature change would be necessary to increase the gas volume by .7 cm^3?
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.7/3000 * 300 K = .07 K
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Continue to assume a temperature near 300 K and a volume near 3 liters:
•If the tube was in the completely vertical position, by how much would the position of the meniscus change as a result of a 1 degree temperature increase?
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1/300 atmosphere = 330Pa which presumes a 3.3cm meniscus position as a result of a 1 degree temperature increase.
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•What would be the change if the tube at the position of the meniscus was perfectly horizontal? You may use the fact that the inside volume of a 10 cm length tube is .7 cm^3.
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Volume is .7 cm^3/10cm length of tubing, so 10 cm^3 would need about 140 cm
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At what slope do you think the change in the position of the meniscus would be half as much as your last result?
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I think a 30 degree slope would make it half.
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*#&!
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The change in position was much greater in the horizontal tube than in the vertical tube.
So we might expect that half the change in altitude, which is what would occur for a 30 degree slope, result in a drastic reduction of the change in meniscus position.
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&#Good work on this lab exercise. See my notes and let me know if you have questions.
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