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phy 122
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.
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Overall I feel comfortable with this lab, having you show me how to set up the bottle helped get the lab going, and by that point my comfort level with the calculations involved increased making the lab run more smoothly. I added to some of my answers and denoted them with asterisks. I have a question for you, while trying to impress a lady friend of mine, I advised to her that when she opens a window it is better to leave it cracked then completely open (I had always heard this but proceeded to show off). Because, I said, at least in the case of a non compressible liquid, the continuity equation tells us that when the fluid goes from a large area to a small area, velocity increases. Is this the same with gases? Is it correct to say that because the air went from the large area of outside, to the smaller area of window, velocity will increase, and therefore a smaller crack in the window accounts for a greater decrease in area, and a greater increase in velocity, allowing more air in faster to circulate about the room? This is just something fun i happened upon and thought I would inquire. Thanks again for all your help with this lab.
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Pressure differences between inside and outside pretty much determine the volume of air that will enter. The smaller opening results in a greater air velocity, for pretty much the reasons you indicate. This increases air circulation in the room, which cools occupants more than still air and also better mixes the incoming air with the air present in the room.
The continuity equation has to be modified for gases, which expand and contract in response to pressure differences, some of which result from velocity changes. However for the flow you are describing, there won't be a lot of compression or expansion and the unadorned continuity equation gives a very good (if not quite perfect) result.
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about 2.5-3 hours total.
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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.
I covered the pressure release valve with my finger and put the long end of the vertical tube into my mouth and began to suck like a drinking straw. It took very little effort to bring the water up into the tube. Upon removing my finger from the pressure valve, or taking my mouth off the tube, some of the water flowed back into the bottle, the rest I blew back in with my mouth.
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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.
If I were to remove the cap and not hold the tube at such an angle as to prevent the water from escaping it would do so, air can escape the system when there is no water in the tube to prevent it from doing so.
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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?
When I blew on the tube with the pressure valve open, bubbles were made and it was clear air was being pushed into the bottle, but it was coming out the valve and not moving our pressure meter at all, if I held the pressure valve closed, the meter moved but only about a cm or so, any more would have been extremely difficult.
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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?
The air column moved slightly when the pressure valve was closed, upon releasing the valve or removing the tube from my mouth, the air column returned to its original position
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What happened in the vertical tube?
Any bubbles of water or bits of air in the column were forced back into the bottle, clearing out 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?
It was easy to see that blowing on the tube would increase the amount of air molecules and the overall pressure of the system. Its more difficult for me to describe the processes at work when one sucks water into the bottle, but here a form of pressure is also added. Perhaps since the tube is submerged, there is a decrease in volume of the water, but there may not be any forces acting on the bottle, just on the tube and the water.
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What happened to the quantities P, V, n and T during various phases of this process?
The number of moles of air could have likely increased upon blowing more air into the bottle, pressure would have also increased, temperature should have increased since a change in temp represents a change in pressure and vice versa. Volume would have decreased upon pulling more water into the tube, but more it less it stays constant. Less volume would have given more room for the air molecules to bounce around , there would be less collisions and less pressure. If air is blown into the system increasing the amount of air molecules and thus the number of collisions, this increase in collisions will increase the kinetic energy of the molecules in the bottle and will cause a temperature increase. **** If we take the system to be sealed and at constant volume we will see proportional changes between Pressure and temperature.
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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?
A 1 percent increase in pressure would increase the atmostpheric pressure of 100kPa by 1 to 101kPa.
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What would be the corresponding change in the height of the supported air column?
0.1 meters
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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?
I should be able to determine this by using the properties of PV = nRT however I am not comfortable enough with these properties yet to make an accurate prediction. The air temperature would change by the same amount since the quantities are proportional.
****The number of moles is constant and we take volume to be constant also, therefore we have a proportionality between Pressure and temperature. P1/P2 = T1/T2. Here we are asked for the temperature that would correspond to the pressure change, we arrange our proportionality too:
T2 = (P2/P1) T1
= (101kPa / 100kPa) (300K)
= 303 K
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Continuing the above assumptions:
How many degrees of temperature change would correspond to a 1% change in temperature?
3 degrees
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How much pressure change would correspond to a 1 degree change in temperature?
1 percent of 300 is 3, the temperature will change from 300 to 303 as our calculations predict.
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By how much would the vertical position of the water column change with a 1 degree change in temperature?
A temperature change of 1 degree would give us 301 degrees as our T2. The position will change by the same factor as the new pressure:
P2 = (T2/T1) P1
= (301/300) 100kPa
= 100.3 kPa
Or 0.03m
I previously recall a 111kPa pressure change accounting for 0.1m of change or better 0.11m, I assume our tactic is the same if the pressure changes from 100 to 100.3, accounting for only a 0.03m change. I cant really break this down right now to explicitly explain my reasoning, but this makes sense in my head
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How much temperature change would correspond to a 1 cm difference in the height of the column?
1cm = 0.01m
This factor would increase the pressure to 100.01kPa.
Again here I cant sort out the math between the pressure and height changes Im just going by what Ive seen so far. Perhaps you could break this down further for me.
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Based on your result that a 1% pressure change would correspond to a 10 cm rise in the water level, a 1% temperature change in a sealed bottle would raise pressure by 1% and therefore the level of the water would again increase by 10 cm.
And of course a 1% pressure increase is 1 kPa.
A 1 degree rise would be 1/3 of a 1% rise in temperature, and in a closed system would result in a pressure increase of 1/3 of 1%, or about 0.3%. And that would correspond to about a 3 cm rise (about 1/3 of 10 cm) in the water level, or .03 m, consistent with your response.
A 1 cm rise would correspond to 1/10 = 0.1 of a kPa, not .01 kPa.
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How much temperature change would correspond to a 1 mm difference in the height of the column?
1mm = 0.001m
This factor would increase the pressure to 100.001kPa
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I believe that would be .01 kPa, not .001 kPa.
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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. *****My numbers match this data very well
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.
****
19.2C
19C
19.1C
18.9C
18.9C
19.0C
19.0C
19.1C
19.1C
19.0C
18.9C
18.9C
19.0C
19.1C
19.2C
19.0C
19.1C
There were no significant changes in my plug of water, it stayed in the same position the whole time.
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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):
The temperature changes fluctuated between 18.9-19.2 at no real pattern of rate.
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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:
****I put the bottle into a pan of water that measured a temp of 46C. The plug of water moved some when I was positioning the bottle, but it didnt appear to skew results too much. At first I only put a Ό of the bottle in, the height of the plug raised by about 1 cm. Pushing the bottle ½ of the way in, the temperature climbed by 2-3 degrees and held there. Again the plug did move a little when the bottle was moved.
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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:
Putting my hands near the bottle did not produce any significant changes, a temperature change of 26C only moved the plug a few centimeters, I would not expect my hands to do much better. The water was 15 degrees warmer than body temperature, so it would have only accounted for an almost unobservable change.
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In the winter, with cold tap water, this effect is very noticeable. Your hands were probably at about the same temperature as the bottle, and would not have caused any significant change in the water level.
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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:
The thermometer remained at a steady 18.9-19.2C, the meniscus did not experience any noticeable changes
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Repeat the experiment with your warm hands near the bottle. Report below what you observe:
The meniscus made no significant changes.
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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?
The water moved a height of 100cm or 1m, this would result in a pressure increase from 100pKa to 110pKa
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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?
The 0.003m diameter times the 0.10m height gives a volume of 3E-4
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You want to multiply the cross-sectional area of the tube by the change in height.
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The main point is that the volume change is pretty insignificant compared to the volume of air in the container, which can therefore be considered nearly constant as long as the rising water stays confined to the thin tube.
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By what percent would the volume of the air inside the container therefore change?
Im not sure how to go about getting the original volume but if we divided two temperatures before and after the change T2/T1 and multiplied by the original volume this would give us our new volume.
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You probably have about 1500 cm^3 of air in the bottle. The volume of that section of tube is probably on the order of one cm^3.
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Assuming constant pressure, how much change in temperature would be required to achieve this change in volume?
Lets say original volume was 200mL, we increased the temp from 300 to 330 our new volume would be
V2 = (T2/T1) V1
= (330/300) (200mL
= 220mL
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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?
Im getting a little mixed up here, but a temp change of 300K would give a volume of 400mL, doubling the temperature would double the volume.
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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?
****
The volume change of the system would have been minute with our small temperature changes, it is easier to see with much larger temperatures how the properties change together.
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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?
There would be less work having to be done since the water doesnt have to rise to a greater height then from before. 6cm or 0.06m would give a pressure change of 100kPa to 100.06kPa.
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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?
T2 = (100.06/100) (300)
= 300.18K
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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?
0.7cm^3 is equal to 7E-7m^3
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3 liters is 3000 cm^3.
.7 cm^3 is about .0002 of this.
To increase volume by factor .0002, temperature would increase by factor .0002.
.0002 * 300 K = .06 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?
In this case we would have to account for the height to which the water is raised, this means that some work was done against gravity and this can be measured.
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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.
These numbers would be fairly minuscule but here we do not have to account for an excess amount of work being done to raise the water against gravity.
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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?
Maybe a 65-70 degree slope.
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&#Your work on this lab submission looks good. See my notes. Let me know if you have any questions.
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