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Phy 242
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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3hrs
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You can use the bottle, stopper and tubes as a very sensitive thermometer. This thermometer will have excellent precision, clearly registering temperature changes on the order of .01 degree. The system will also demonstrate a very basic thermal engine and its thermodynamic properties.
Set up your system with a vertical tube and a pressure-indicating tube, as in the experiment on measuring atmospheric pressure. There should be half a liter or so of water in the bottom of the container.
• Refer back to the experiment 'Measuring Atmospheric Pressure' for a detailed description of how the pressure-indicating tube is constructed for the 'stopper' version of the experiment.For the bottle-cap version, the pressure-indicating tube is the second-longest tube. The end inside the bottle should be open to the gas inside the bottle (a few cm of tube inside the bottle is sufficient) and the other end should be capped.
The figure below shows the basic shape of the tube; the left end extends down into the bottle and the capped end will be somewhere off to the right. The essential property of the tube is that when the pressure in the bottle increases, more force is exerted on the left-hand side of the 'plug' of liquid, which moves to the right until the compression of air in the 'plugged' end balances it. As long as the liquid 'plug' cannot 'leak' its liquid to the left or to the right, and as long as the air column in the plugged end is of significant length so it can be measured accurately, the tube is set up correctly.

If you pressurize the gas inside the tube, water will rise accordingly in the vertical tube. If the temperature changes but the system is not otherwise tampered with, the pressure and hence the level of water in the tube will change accordingly.
When the tube is sealed, pressure is atmospheric and the system is unable to sustain a water column in the vertical tube. So the pressure must be increased. Various means exist for increasing the pressure in the system.
• You could squeeze the bottle and maintain enough pressure to support, for example, a 50 cm column. However the strength of your squeeze would vary over time and the height of the water column would end up varying in response to many factors not directly related to small temperature changes.
• You could compress the bottle using mechanical means, such as a clamp. This could work well for a flexible bottle such as the one you are using, but would not generalize to a typical rigid container.
• You could use a source of compressed air to pressurize the bottle. For the purposes of this experiment, a low pressure, on the order of a few thousand Pascals (a few hudredths of an atmosphere) would suffice.
The means we will choose is the low-pressure source, which is readily available to every living land animal. We all need to regularly, several times a minute, increase and decrease the pressure in our lungs in order to breathe. We're going to take advantage of this capacity and simply blow a little air into the bottle.
• Caution: The pressure you will need to exert and the amount of air you will need to blow into the system will both be less than that required to blow up a typical toy balloon. However, if you have a physical condition that makes it inadvisable for you to do this, let the instructor know. There is an alternative way to pressurize the system.
You recall that it takes a pretty good squeeze to raise air 50 cm in the bottle. You will be surprised at how much easier it is to use your diaphragm to accomplish the same thing. If you open the 'pressure valve', which in this case consists of removing the terminating cap from the third tube, you can then use the vertical tube as a 'drinking straw' to draw water up into it. Most people can easily manage a 50 cm; however don't take this as a challenge. This isn't a test of how far you can raise the water.
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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When I moved my mouth from the system, the water returned to its original position, equalizing the pressure, as I expected it would.
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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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Nothing should happen because the pressures will continue to equalize. Like I predicted, nothing happened, the system remained the same.
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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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When I took my mouth off, water increased upward into the vertical tube. When I was blowing into the bottle, the pressure indicating tube showed the additional pressure in the movement of the water column. However, when the pressure was released, it did not return into the original position.
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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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By blowing air into the system I am increasing the pressure which pushes the water to compress the air in the air column. When I release the pressure the air does not return to its original volume
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• What happened in the vertical tube?
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Water was pushed by a greater pressure in the bottle which forced it higher into the 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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The water travels higher into the tube, which in turn allows less air in the tube. This is done to in reaction to the additional pressure. I anticipated this to happen and I also anticipated a small change in the pressure measurement due to the fact that there is means of escape in reference to the contents which pressure is being measured
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• What happened to the quantities P, V, n and T during various phases of this process?
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Pressure increased, volume remained constant, the amount of moles increased, and temperature is expected to have changed minimally in accordance to the gas law equations.
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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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(100,000)(.01) = 1000 Nm^-2
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• What would be the corresponding change in the height of the supported air column?
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P = rho*g*h
1000 = (1000)(9.8)*h
h = 0.1 m
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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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Because temperature and pressure are directly proportional to each other, the temperature would have to increase by 1% to compensate for pressure change.
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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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(300*.01 = 3)
So therefore the temperature change would be 3K.
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• How much pressure change would correspond to a 1 degree change in temperature?
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(1000/3= 333)
This means the pressure change is approximately 333Nm^-2
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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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P = rho*g*h
1000 = (1000)(9.8)*h
h = 0.1 m
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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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P = rho*g*h
P = (1000)(9.8)(.01 )
P = 98 Nm^-2
Due to the 98Nm^-2 increase, the temperature increase is .29 degrees Kelvin because 98Nm^-2 of pressure is 29% of the pressure change I calculated was require to change the temperature by 1 degree K.
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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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P = rho*g*h
P = (1000)(9.8)(.001 )
P = 9.8 N/m^2
With that values the temperature increase would be .029K.
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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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28.7 C, 35.2 cm
28.7 C, 35.2 cm
28.7 C, 35.2 cm
28.7 C, 35.2 cm
28.7 C, 35.2 cm
28.8 C, 35.2 cm
28.7 C, 35.2 cm
28.7 C, 35.2 cm
28.8 C, 35.2 cm
28.8 C, 35.2 cm
28.8 C, 35.2 cm
28.7 C, 35.2 cm
28.7 C, 35.2 cm
28.7 C, 35.2 cm
28.8 C, 35.2 cm
28.8 C, 35.2 cm
28.7 C, 35.2 cm
28.7 C, 35.2 cm
23.6 C, 35.2 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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The maximum deviation of the bottle thermometer was about 0.1. It was very hard to see much change at all, especially because there is most likely no change in room temperature.
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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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49.7
48.4
47.8
47.3
36.1
35.6
35.0
34.4
34.1
33.7
33.2
33.1
32.8
32.8
32.7
32.7
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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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At first the column didn’t move really, but then I warmed my hands by holding a hot coffee cup and when I repeated the exercise I was able to tell a still small, but a discernible amount of movement.
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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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23.7, 10.6
23.6, 10.4
23.6, 10.4
23.6, 10.4
23.6, 10.3
23.6, 10.1
22.8, 9.6
22.8, 9.5
22.8, 9.0
22.7, 8.9
22.7, 8.9
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Repeat the experiment with your warm hands near the bottle. Report below what you observe:
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The water changed much much more than it had when the system was upward and fighting against gravity harder.
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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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I assume that the pressure did not actually change very much. Maybe a very small amount.
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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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V = pi*r^2*h
V= pi*.0015 m*.1 m
V= 71*10^-8
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• By what percent would the volume of the air inside the container therefore change?
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At most I assume it would change approximately 1%
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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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The temperature would also be required to change by 1% due to the change of pressure. This would mean around 3K due to bottle being around room 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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The change would be more around 6K because 6K is 1% of 600K.
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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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I do not believe it would create a noteworthy difference.
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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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P = rho*g*h
P = 1000 *9.8 *06 m
P = 588 Nm^-2
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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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588/100,000 = 0.0058
300 *.0058 = 1.764K
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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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3 L = 3,000 cm^3
0.7 /3,000 = 0.023% dT
Approximately .07K
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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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Approximately 3 cm
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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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The volume ought to increase to 3.01L or 10cm^3. Since volume and temperature are directly related and the volume of the tube can be assumed to be .7cm^3 in 10cm, the meniscus position should change to 14.3cm.
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• A 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 at a slope of around 45 degrees because cos(45) and sin(45) are both one half so that would half the change in position.
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A change in pressure has a much greater effect on horizontal position than on vertical position. A much smaller slope would result in the desired change.
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*#&!*#&!
&#This lab submittion looks good. See my notes. Let me know if you have any questions.
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