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course phy122
OH my. I am not sure about this one professor. Those last few questions seemed to fool me.
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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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As I pull the water toward my mouth, I am able to maintain a certain level with minimal suction. I then replace the cap on the pressure valve, and when I remove my mouth from the tube, the water stays in the tube and doesnt completely retract into the bottle.
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Now think about what will happen if you remove the cap from the pressure-valve tube. Will air escape from the system? Why would you or would you not expect it to do so?
Go ahead and remove the cap, and report your expectations and your observations below.
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I thought that I would hear the air escape from the tube but I did not. I thought because the air chamber is directly involved with the pressure tube that there would be a big decrease in pressure, and I was incorrect. I am sure however, that some air did seep from the pressure valve.
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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 air column increases inside the pressure tube when I blow into the vertical tube. When I stop blowing the air column returns to its original length. I only have to blow a minimal amount of air into the tube to increase the water level significantly.
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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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When I blew into the tube the air column increased to proportion the new pressure. Yes the air column returned to its original position when I removed the tube from my mouth.
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What happened in the vertical tube?
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Initially the water went back into the bottle, but when I released the tube from my mouth, the water was forced back into the tube following air being blown into the bottle. In my opinion the system was trying to create some equilibrium. The pressure and volume had to proportion out.
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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 system needed to balance. The new pressure inside the system had to be in proportion with the volume in the system. I did not anticipate that when I released my breath from the vertical tube, that water would come shooting back into my mouth. Now that I have performed the experiment, I see why it did.
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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 with the forced expiration into the bottle and decreased with the release from my mouth. The pressure inside the tube increased with forced expiration as a result of increased pressure inside the bottle. The air column in the pressure tube increased in length inside the tube due to increased pressure inside the bottle.
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The air column would get shorted as pressure increased.
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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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If the original pressure is 103kPa, and increases by 1%, the new pressure will be 103.97088N/m^2
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Close, but 1% of 103 kPa is 1.03 kPa, which would increase the pressure to 103 kPa + 1.03 kPa = 104,03 kPa.
Accounting for significant figures this would be appropriately expressed as just 104 kPa.
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What would be the corresponding change in the height of the supported air column?
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The height of the supported column would either increase by 1% or decrease by 1%.
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Only the increase of 1 kPa would support a column. There would have been no column of water in the vertical tube prior to increasing the pressure. A 1% increase in zero height would be zero. That isn't the case, because we know an increase in pressure would result in some water being forced into the vertical tube.
How high would a water column have to be to require 1 kPa of pressure to support it?
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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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I am going to assume that the temperature would have to change equally.
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You don't need to assume that.
At constant volume, PV = nRT tells you that P / T remains constant. It follows that if P increases by a certain proportion, T would have to increase by the same proportion.
So a 1% increase in P would have to be accompanied by a 1% increase in T.
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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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The temperature would have to increase by 3 degrees.
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How much pressure change would correspond to a 1 degree change in temperature?
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0.333% or 34kPa
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0.333% is correct.
However 0.333% of 101 kPa is not 34 kPa. 34 kPa would be 33% of 101 kPa.
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By how much would the vertical position of the water column change with a 1 degree change in temperature?
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It would change by .333%.
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As in my previous note, this is not the case. The original height of the water column was 0.
You have to figure out the height of the column corresponding to the pressure change.
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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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First you have to find the pressure difference corresponding to a 1 cm column.
Having found that you can find it as a percent of the total pressure.
The temperature change will be the same percent of the temperature.
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I am not sure, I am going to guess and say 1 degree or perhaps 1/3 of a degree. Professor, do you care to elaborate? Im drawing a blank. Im not sure if its because its so late or because I still dont have a good understanding of this system.
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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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Please elaborate on this one also.
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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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19, 22
19, 21.7
19, 21.8
19, 21.4
19.1, 22
19.1, 21.8
19.1, 22
19.1, 21.9
19, 21.9
19.1, 22
18.9, 22
18.7, 21.9
18.8, 22.1
18.8, 22.2
19, 22.1
19, 21.8
19, 22
19, 21.9
19.1, 21.8
19, 21.7
My measurements are from the top of the bottlecap to the bottom of the first meniscus.
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The water column in the tube extends from the water surface in the bottle to the meniscus, not from the top of the cap to the meniscus.
You can correct this by making a good estimate of where the water was in the bottle, and adding the distance between the water and the bottlecap.
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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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Standard deviation: .1119; the fluctuations were minimal but there would be an increase up to 19.1 at the highest down to 18.7 at the lowest temperature Celsius.
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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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Temperature prior to starting timer: 19.8 C; At conclusion: 20.2 C
37.98
37.98
37.98
37.98
37.98
37.97
37.97
37.97
37.96
37.96
37.96
37.96
37.95
37.95
97.95
37.95
37.95
37.95
37.94
37.94
37.94
37.94
37.93
37.93
37.92
37.92
37.92
37.92
37.91
37.91
37.91
37.90
37.90
37.90
37.89
37.90
37.89
37.90
37.89
37.88
37.88
37.88
My measurements are from the bottlecap to the beginning of the meniscus.
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The water column appears to be between 15 and 20 cm higher than before.
However it shouldn't stay there. After pouring the water over the bottle, the water column will rise to a maximum, then fall back close to the original height.
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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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The meniscus moved about 1/10th of a centimeter with my hand placement near the bottle, but without touching the bottle. The warmth of my hands must have been conducted into the water to raise it fractionally.
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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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Measurements are in cm and are from the top of the bottlecap to the bottom of the meniscus.
36.9
36.9
36.9
36.9
36.9
36.92
36.98
37
36.99
36.99
36.9
36.9
36.8
36.88
36.9
36.9
36.9
36.92
36.9
36.9
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Repeat the experiment with your warm hands near the bottle. Report below what you observe:
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The meniscus moved minimally, if any at all. I could not notice even a fraction of movement between .9 and .8 on the centimeter ruler.
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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 think that the pressure in the bottle changed minimally as the water moved along the horizontal tube.
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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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It would drop by 2176.57kg/cm2
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Volume is measured in cm^3, or m^3, or liters. The kg unit would not appear in the volume units.
It's not clear how you got this. Can you elaborate?
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By what percent would the volume of the air inside the container therefore change?
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I would assume that it would change equally.
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The percent change in volume is equal to the change in volume divided by the original volume.
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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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By approximately 7 degrees K.
(2176.57/300K=7)
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Your 2176 quantity has units. Those units are not Kelvins. So this division by 300 K would not give you Kelvins.
You would need to know the two volumes. With constant pressure, PV = n R T implies that V / T is constant.
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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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It would change by 3.6 degrees K
(2176.57/600= 3.6K)
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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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Because the volume change of the air is negligible in this case, so it would be cancelled out. No it would have not made a significant difference in our estimates of temperature change.
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It wouldn't have cancelled out, but being negligible it wouldn't have 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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Im not really sure. I am assuming that it would not have to increase significantly, but I would not know without first performing this experiment. How can I figure this out?
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As on many of the questions you missed, you need to relate pressure change to depth change.
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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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I have no idea. If there are no constants or specific numbers, how can I solve this problem?
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The pressure increase in the preceding question can be easily calculated, if you understand the relationship between pressure and depth. Having calculated this, it becomes possible to answer the present question.
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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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Professor, where are you getting these numbers (.7 cm^3)
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That would be obtained by performing the previous calculations correctly.
For now you can accept that figure and use it to answer the present question.
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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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I am drawing a lot of blanks when it comes to solving these problems.
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The tempeature increase implies a pressure increase, which then implies a certain additional water depth in the tube.
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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 am going to assume that pressure is constant. So if I want to solve for this I can use the formula: .5rhov^2
So I would take 0.5(1000kg/m^3)(.7cm^3) ^2=245kg/m^3=245Pa
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The change in .5 rho v^2 is equal and opposite to the pressure change, provided altitude y is constant.
.5 rho v^2 is related to kinetic energy 1/2 m v^2.
In this context it should be clear that v stands for velocity, not volume.
Also, we use lower-case v for velocity, upper-case V for volume.
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There are no significant velocities involved in this system, so .5 rho v^2 is not particularly relevant to our analysis.
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I found the pressure based on the information but how do I find out the distance the meniscus moved. Should I know this?
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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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Professor, I really have no idea. Please help.
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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
4 hours"
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You are missing the relationship between pressure and water depth.
Look at the very first introductory problem set problem. What relationship exists between pressure and depth?
In Bernoulli's equation
.5 rho v^2 + rho g y + P
what are the meanings of rho, v, y and P? How are pressure P and vertical position y related when the fluid is stationary (or at low velocity, in which case .5 rho v^2 will be negligible)?
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The key to most of your errors is in these relationships. Submit and email your best answers to these questions. Once you understand these relationships, you should be able to correct most of your work without too much trouble.
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