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course Phy 232
6-17-11 about 7:30pm
bottle thermometer#$&*
Phy 232
Your 'bottle thermometer' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Your optional message or comment: **
I'm having some trouble with the calculations, I left those blank. I'm not sure what formula to use to get the answers to the questions. Whether its a form of PV=nRT, or something else. I appreciate any help that you give me, and I will resubmit once I feel more confident about my answers.
#$&* What happens when you pull water up into the vertical tube then remove the tube from your mouth? **
#$&* What happens when you remove the pressure-release cap? **
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Data program is at http://www.vhcc.edu/dsmith/genInfo/labrynth_created_fall_05/levl1_15\levl2_51/dataProgram. exe
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.
After capping the pressure valve and releasing the tube from mouth, the water plug moved a little right then slowly moved left past its original location. The air column grew from what it was before. I expected the water plug to move left some, but I didn’t expect it to move right then slowly left. I also thought that the movement left would be fairly quick. Also, the bottle now is “sucked” in on the sides.
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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.
I expect the air to escape from the system and the bottle shape to return to normal. When I removed the plug the bottle’s shape returned to almost its original shape but was still slightly “sucked” in. The water plug remained at about the same place.
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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?
¥ 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?
¥ What happened in the vertical tube?
¥ Why did all these things happen? Which would would you have anticipated, and which would you not have anticipated?
Water shoots up the vert tube after removing the tube from my mouth, and the water plug returns to its previous position. By blowing air into the system the pressure in the bottle increases which pushes the water plug further right, shortening the air column by condensing the air in it. I didn’t expect the water to shoot back up the vert tube, I should have but didn’t. I expected the water plug to return, for the most part, to its original position.
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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?
¥ What would be the corresponding change in the height of the supported air column?
¥ By what percent would air temperature have to change to result in this change in pressure, assuming that the container volume remains constant?
Report your numbers in the first three lines below, one number to a line, then starting in the fourth line explain how you made your estimates:
1
.01
I’m not sure as to how to calculate these. I used dy= 101k Pa/(rho*g) for the 2nd #. The 3rd I’m not sure how to use PV=nRT to solve. V and R are constant, but n would change when I blow air into the tube, right?
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@& If atmospheric pressure is 100 kPa, then how many Pascals correspond to a 1% increase in pressure?
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1 kPa + 100 kPa= 101 kPa
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How high would a water column have to be to require this much pressure to support it? You will use your equation for dy to obtain this result.
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dy=101kPa/(1000kg/m^3*9.8m/s^2)
dy=10.31cm
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PV = n R T, which as you expected is useful here. If you raise the water by changing the temperature, which quantities change and which remain the same? You may regard the tube as having negligible volume, so the volume is among the quantities that don't change.
What therefore would be the temperature change required to increase the pressure by 1%, if the temperature is about 300 K?
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P and T are directly proportional if n, and V are constant, right?
101 kPa/100 kPa= 1.01
300 K * 1.01= 303 K
a change of 3 K is required
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Continuing the above assumptions:
¥ How many degrees of temperature change would correspond to a 1% change in temperature?
¥ How much pressure change would correspond to a 1 degree change in temperature?
¥ By how much would the vertical position of the water column change with a 1 degree change in temperature?
Report your three numerical estimates in the first three lines below, one number to a line, then starting in the fourth line explain how you made your estimates:
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3
333.34
3.44
1st # was made by the previous answer, 2nd was calculated by 1 deg change is 1/3 of the 1% pressure change, so 1/3 of 1%=.0034*100 kPa= 333.34 kPa. The 3rd was calculated by multiplying 10.31cm by 1/3
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How much temperature change would correspond to a 1 cm difference in the height of the column?
How much temperature change would correspond to a 1 mm difference in the height of the column?
Report your two numerical estimates in the first two lines below, one number to a line, then starting in the third line explain how you made your estimates:
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.294
.0294
By using P=rho*g*h and plugging the values of h into the equation. Since P and T are proportional I found the ratio that P increased and multiplied 300 K by the ratio.
100098 Pa/ 100 kPa= 1.00098
1.00098*300K= 300.294K
300.294K - 300K= .294K
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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 data, in the form of a comma-delimited table below.
21.8, 0
22, 0
22, 0
22.1, 0
22.1, 0
22, 0
21.4, .1
21.1, .1
20.9, .1
20.5, .15
20, .2
20, .2
19.8, .2
19.4, .2
19.1, .2
19, .2
19, .2
18.9, .25
18.7, .25
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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):
As the temp lowers, the water column moves further right shortening the air column.
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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.
.7
.5
.4
.25
.2
.15
.1
.1
.1
.05
0
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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:
Yes, by about .25cm. Before I put my hands closely around the bottle the water plug was at 0cm (35.9 on my paper ruler), after 1 min of my hands being closely around the bottle it moved .25cm (36.15 on my paper ruler)
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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:
21.1, 0
21.1, 0
21.3, 0
21.3, 0
21.4, 0
21.4, 0
21.5, .05
21.5, .05
21.5, .05
21.5, .05
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Repeat the experiment with your warm hands near the bottle. Report below what you observe:
21.5, .15
21.8, .1
22, .1
22, .1
22, .1
22, .1
21.9, .05
21.9, .05
21.9, .05
22, .1
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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?
¥ 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?
¥ By what percent would the volume of the air inside the container therefore change?
¥ Assuming constant pressure, how much change in temperature would be required to achieve this change in volume?
¥ If the air temperature inside the bottle was 600 K rather than about 300 K, how would your answer to the preceding question change?
Give your answers, one to a line, in the first 5 lines below. Starting in the sixth line, explain how you reasoned out these results:
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333 kPa
.00028cm^3
.00028%
.00084 K
it would double to .00168 K
the 1st is just a guess, 2nd using pi*r^2*h, 3rd by assuming the volume was 100, 4th multiplying 300 K by .00028%
@& A cylinder of diameter 3 mm and length 10 cm has volume approximately .7 cm^3. Your formula is right but your result isn't correct.*@
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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?
because the volume change was too small to matter, it would not have made a significant difference in the temp estimate.
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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 rises 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?
¥ Assuming a temperature in the neighborhood of 300 K, how much temperature change would be required, at constant volume, to achieve this pressure increase?
¥ 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?
Report your three numerical answers, one to a line. Then starting on the fourth line, explain how you obtained your results. Also make note of the relative magnitudes of the temperature changes required to increase the altitude of the water column, and to increase the volume of the gas.
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588 Pa
1.764 K
?
1st found by rho*g*h, 2nd found by taking the ratio of the 1st in regard to 100 kPa and multiplying this ratio by 300 K. The 3rd I do not understand, n= 3 liters and I need to find out what temp is necessary to get 3 liters +.7cm^3 of gas volume?
@& What is the .7 cm^3 volume increase as a percent of the 3 liter volume?
What would therefore be the percent change in the temperature?*@
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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?
¥ 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.
¥ A what slope do you think the change in the position of the meniscus would be half as much as your last result?
Report your three numerical answers, one to a line. Then starting on the fourth line, explain how you obtained your results. Also indicate what this illustrates about the importance in the last part of the experiment of having the tube in a truly horizontal position.
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I’m lost on how to calculate this, V and T are constant so P/n=constant? I’m thoroughly confused.
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@& T is not constant in any of these cases.
If the tube is completely vertical, then the water will rise a few centimeters. This does imply a slight change in volume, but for the moment assume that the volume change is negligible, so that you can for the moment assume that V is constant. n is constant since no air can enter or leave the system. The water rise is associated with a change in pressure, so it is P and T that vary. You are given the change in T, so you need to find the change in P and the resulting change in the height of the vertical water column.
If the tube is completely horizontal, then no pressure increase is necessary and the gas is free to expand. In this case V is not constant, but P is. n remains constant since the system is still sealed from the atmosphere. So V and T vary. What therefore is the change in V, and how far does water have to move along the tube to accomodate this change in V?
Comparing your answers, was the change in V really negligible in the first part?*@
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*#&!
@& You have good data.
Check my notes, and revise as indicated.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.
&#
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@& Check my notes. You shouldn't have too much trouble with an additional revision; this time use #### to denote insertions.*@