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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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2 hours and 45 mins
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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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After doing the experiment, the water in th ecollumn dropped rapidly and leveled off at a distance above the bottle. Therefore, all of the water did not return back to the bottle. The water collumn stabilized in the vertical tube roughly 10cm above the bottle. I believe this happens because of the differences of pressure. At first, the force of the water and gravity in the vertical tube was enough to overcome the force of the water in the bottle. After dropping a reasonable amount, the two forces equaled out and the water in the vertical tube equalized. This is what I expected to happen.
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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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I expect the pressure of the water in the vertical tube to push the water out of the bottle through the pressure valve. After pulling out the cap, the water quickly fell into the bottle and the sound of air escaping the bottle was heard.
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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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The water in the pressure indicator increased by around a cm as the bubbles were being created, but went back to normal levels quickly after. The water level in the vertical tube increased dramatically and stayed at those levels even after the creation of bubbles ceased.
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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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The air column moved back to its normal position when removing it from my mouth. I believe the air column changed because air was added to a system that had already equalized its pressure. Therefore, the easiest way to equalize this added volume or pressure was to push the water in the pressure indicator.
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What happened in the vertical tube?
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The vertical tubes water level increased dramatically during this step of the experiment. However, unlike the pressure indicators difference, this water level difference in the vertical tube remained even after removing the tube from my mouth.
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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 air column moved back to its normal position when removing it from my mouth. I believe the air column changed because air was added to a system that had already equalized its pressure. I think the water in the vertical tube happened for siilar reasons. Even though I anticipated the water level to return to normal levels in the pressure indicator, I did not anticipate the water levels to remain the same in the vertical tube after removing the tube from my mouth.
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What happened to the quantities P, V, n and T during various phases of this process?
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When blowing into the tube, the pressure in the system increased. The volume of water decreased in the bottle. The temperature changed but this change would be extremely miniscule. Lastly, the moles in the system would have stayed constant.
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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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100kpa=100,000pa
100,000*.01=1000N/m^2 change
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What would be the corresponding change in the height of the supported air column?
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pressure diff=density*gravity*h
h=(1,000N/m^2)/((1000kg/m^2)(9.8m/s^2))
h=.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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If the volume remains constant, then the temp would change proportionally to the pressure. Since the pressure changed by 1% then the temperature would also change by 1%.
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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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Therefore, the temperature would raise by 300k*.01=3k.
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How much pressure change would correspond to a 1 degree change in temperature?
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1k/3k=.333 percent. Since the pressure and temperature change proportionally, then the pressure would change by roughly .333%.
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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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pressure diff=density*gravity*h
=.0033*100000pa=330pa
h=330pa/(1000kg/m^2*9.8m/s^2)
h=.034m
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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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pressure diff=density*gravity*h
press diff=1000kg/m^2*9.8m/s^2*.01m
98pa=press diff
98pa/100,000pa=.00098% change
.00098%*300k=.294k change
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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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pressure diff=density*gravity*h
press diff=1000kg/m^2*9.8m/s^2*.001m
9.8pa=press diff
9.8pa/100,000pa=.000098% change
.000098%*300k=.0294k change
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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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21.1,8.0
21.1,7.9
21.1,7.7
21.0,7.6
21.0,7.5
20.9,7.3
20.9,7.2
21.0,7.1
20.9,7.0
20.8,6.9
20.8,6.7
20.7,6.6
20.7,6.5
20.6,6.3
20.6,6.2
20.5,6.1
20.5,6.0
20.4,5.9
20.4,5.9
20.3,5.8
The first column of numbers was the temperature in celsius and the second column is the length of the water column in 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 max deviation for the temperatuve over these intervals is 0.1 degrees celsius. However, this is an estimate because the thermometer is not extremely accurate and we were told to only estimate to the tenths place so there could dramatic rounding error.
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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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23.1,11.0
23.0,10.9
22.8,10.7
22.7,10.5
22.5,10.4
22.3,10.2
22.1,10.1
22.0,9.9
21.8,9.7
21.6,9.6
21.4,9.4
21.2,9.3
21.1,9.1
20.9,8.8
20.6,8.5
20.4,8.2
The first number is the temperature in degrees celsius and the second column is the length of the water column in cm.
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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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My hands warmed the air in the bottle by a small amount. My basis for my answer is the water in the column increased by a small amount meaning the temperature must have increased.
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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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22.4,28.4
21.4,28.3
21.4,28.2
21.3,28.1
21.3,28.0
21.2,27.7
21.1,27.5
21.1,27.4
21.0,27.1
20.9,26.8
The first number is the temperature in degrees celsius and the second number is the length of the meniscus in cm.
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Repeat the experiment with your warm hands near the bottle. Report below what you observe:
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While observing the bottle as my hands warmed it, I noticed a slight increase in the distance of water in the tube. However, it seemed to move more in the horizontal tube than it did the first time I did this experiment in the vertical tube.
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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 am unsure of this but I believe that since the tube is horizontal the pressure will change by a small amount.
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Theoretically there would be no pressure difference, though the volume in the bottle would decrease slightly. Practically it does take a little pressure to overcome frictional resistance between the water and tube walls.
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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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.1m*.0015^2m*pi=7.07*10^-7m^3
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By what percent would the volume of the air inside the container therefore change?
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2 liters=.002m^3
7.07*10^-7m^3/.002m^3=.00035*100=.0353%
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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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Since we have the percent we can easily find this.
300k*.000353=.1059K
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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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600k*.000353=.2118K
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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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This would of made a difference, but it would not of been significant in the grand scheme of things. The small amount of this volume would not of made a significant 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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pressure diff=density*grav*h
pres diff=1000kg/m^2*9.8m/s^2*.06m
pres diff=590pa
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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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590pa/100,000pa=.0059%
.000059*300K=.0177K
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590 / 100 000 is .0059, or .59%
note that 1000 is 1% of 100 000.
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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 litters=3000cm^3
.7cm^3/3000cm^3=.023%
.00023*300k=.07K change
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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 degreeK/300degreeK=.33%
.0033*100,000pa=330pa incre
330pa=1000kg/m^2*9.8m/s^2*h
h=.0337m
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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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I understand that the change would be more drastic if the tube is horizontal, but I do not understand why. Do you say that it doesnt have the same effect caused by gravity and just divide the pressure change by the density?
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To raise the level of the water requires an increase in pressure. If you were doing this by heating, the need to increase the pressure would limit an increase in volume.
In a horizontal tube the expansion of the gas would occur freely, unlimited by the need to increase pressure.
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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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IT would make sense for the slope to be 45degrees for the meniscus to be half of the last result.
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This would only be the case if the two influences mention in the preceding note had equal effect on volume.
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