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PHY 202
Your 'measuring atmospheric pressure' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Measuring Atmospheric Pressure_labelMessages **
** Measuring Atmospheric Pressure_labelMessages **
2 hours
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Preliminary Estimate of Atmospheric Pressure
In Brief Bottle Experiment 1b you estimated the percent change in the length of an confined air column for three different levels of squeeze.
In Brief Bottle Experiment 1d you found the heights to which water was raised in an open vertical tube, and estimated the squeeze for each.
As you should now know, the additional pressure caused by each squeeze in the 1d experiment can be easily calculated from the density of water and the height to which water was raised in the vertical tube (if necessary see Introductory Problem Set 5, Problem 1).
Based on your data from the 1d experiment:
Calculate the additional pressure for each of the observed heights.
Graph the additional pressure vs. the level of the corresponding squeeze.
Sketch a reasonable straight line to fit your data points as best as possible. (Note that there is no reason to expect that this graph should actually be linear; among other things your perception of how hard you squeeze is unlikely to be linearly related to how much pressure you create, since physical perceptions are not generally linear. For this reason alone you are unlikely to end up with a very good estimate of atmospheric pressure. However you will probably get a halfway reasonable ballpark figure.)
Report as follows below:
In the first three lines give your height vs. squeeze data from Experiment 1b, in tab-delimited format.
In the fourth line give the three pressures corresponding to your three observed heights, followed by the units of the pressure, in comma-delimited form.
In the next three lines give the resulting pressure vs. squeeze information.
In the next line report the slope and vertical intercept of your graph, including units. The units of your squeezes can just be called 'squeeze units'.
Starting in the next line give a brief sample calculation of one of your pressures.
Your answer (start in the next line):
1/10 6cm
5/10 31cm
8/10 46cm
588 Pa, 3038 Pa, 4508 Pa
1/10 588Pa
5/10 3038Pa
8/10 4508Pa
m = 612.5 P/squeeze units y-intercept = -24.5 P
P = rho*g*y = (1000kg/m^3)(9.8m/s/s)(0.06m) = 588 Pa
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Now report and graph your results from the 1b experiment:
In the first line give the level of squeeze necessary to shorten the air column by 10%, as you determined it in the 1b experiment.
In the next three lines give the percent change in air column length vs. the level of squeeze, for squeezes of 2, 5 and 8 as you perceived them.
Graph percent change in air column length vs. level of squeeze (you will have three or perhaps four percents and four levels of squeeze), fit the best straight line you can to your data, and in the fifth line give the slope and vertical intercept of your line.
Your answer (start in the next line):
9 squeeze units, 1.9cm
2 units 0.35cm
5 units 1cm
8 units 1.5cm
m = 0.217 y-intercept = -0.085
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You didn't find the percent change in air-column length. Assuming about a 25-cm air column, which is fairly typical, your percents would run from about 1.4% to about 8%.
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According to the appropriate graph:
To what height would water rise in an open vertical tube, given a level-10 squeeze? Answer and explain the basis for your answer.
Your answer (start in the next line):
y = 0.217x - 0.085 for x=10
y = 0.217(10) - 0.085
y = 2.085 cm
The equation for the line of best fit for the height vs. squeeze data can be used to find the expected height for a given squeeze when squeeze units are substituted in for x. The equation is solved for y, height. This equation suggests that a squeeze of 10/10 would raise the water level to approximately 2.1cm.
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This is what you would expect for the pressure tube.
According to your previous data for the vertical tube, a squeeze of 10 might be expected to raise the water about 60 cm.
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What would be the additional pressure achieved by a level-10 squeeze, as indicated by the results from the open vertical tube?
Answer and explain the basis for your answer.
Your answer (start in the next line):
P = rho*g*h
P = (1000kg/m^3)(9.8m/s/s)(2.085cm)
P = 20433 Pa
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kg / m^3 * m/s^2 * cm does not give you units of Pascals.
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In any case the water in the vertical tube rose much more than 2 cm.
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Substituting in the projected height upon a level-10 squeeze suggests that it would result in an additional pressure 20433 Pa.
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What would be percent increase in the pressure in the bottle due to a level-10 squeeze?
Answer and explain the basis for your answer.
Your answer (start in the next line):
Since the bottle had been open to atmospheric pressure before it was capped, it follows that the pressure inside the bottle would be approximately 101300Pa, so:
(20433/101300)*100 = 20.2%
The increase in pressure is 20.2% of the initial pressure, so an increase of 20433Pa in a system starting at atmospheric pressure is a 20.2% increase.
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Based on the amount of increase in pressure, and on the percent increase, determine the original pressure in the bottle.
Answer and explain the basis for your answer.
Your answer (start in the next line):
As above, I would have expected/assumed that the pressure in the bottle at the start of the experiment was atmospheric pressure. Is this not the case???? If not, how would I go about finding percent increase without an initial pressure to compare it to????
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The original pressure was atmospheric pressure.
This experiment measures atmospheric pressure, so you cannot assume a value for it.
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Based on the length and changes in length of the pressure tube you can determine the percent change in atmospheric pressure.
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(optional for Phy 122 students) What is the meaning of the slope of the graph of pressure vs. squeeze, and what is the meaning of the slope of the graph of percent increase in pressure vs. squeeze?
Answer and explain the basis for your answer.
Your answer (start in the next line):
The slope of a graph of pressure vs. squeeze gives the amount of pressure per squeeze unit.
The slope of a graph of percent increase vs. squeeze gives the change in pressure per squeeze unit. If human perception was perfectly linear, each squeeze would result in exactly the same pressure change with each successively larger squeeze, which would mean that the percent increase in pressure vs. squeeze line would have a slope of 0.
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A slope of 0 would correspond to no change in pressure as the squeeze increases.
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(optional for Phy 122 students) Based on the slopes of the two graphs from the preceding question, what is the original pressure in the bottle?
Answer and explain the basis for your answer.
Your answer (start in the next line):
Since the pressures graphed in both graphs were pressure increases, the pressures in the graphs were gauge pressures, or pressure that was additional to the initial pressure, which is essentially 0 Pa on the graphs in question. The first graph y-intercept suggests that the initial pressure was -24.5Pa, and the second graph y-intercept suggests an initial pressure of -0.085Pa.
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The initial pressure was 1 atmosphere, not zero.
Gauge pressure might have been zero, but gas laws don't apply to gauge pressure.
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It is possible with the caps having multiple tubes to observed the height to which water rises in the open vertical tube, and the changing length of the air column, simultaneously. This eliminates the uncertainty that results from 'level-of-squeeze' estimates.
The remainder of this experiment will be concerned with obtaining this data. Part 2 of the experiment, which is part of a subsequent assignment, will analyze this data.
Be very sure to keep a copy of your data for this experiment, since you will probably be using it in Part 2. It is recommended that you keep a word-processing document open, copy this document into that document, and copy any data you put into these boxes in the appropriate place in that document. That document can then be used as a course from which you can easily access the necessary data to copy into the data processing program.
Your setup for the preceding experiment Raising Water in a Vertical Tube included a vertical tube, with terminating caps on the other two tubes. You will use the vertical tube again in this experiment.
Your kit included two bottlecaps connected by a long tube. The long tube is to be used as a vertical tube, as in the previous experiment.
Each bottlecap has three tubes. One is a short tube; its intended use is to release pressure in the system when and if this becomes necessary. The third is fairly long. This tube is to be used as a 'pressure tube'.
First fill the 'pressure tube' with water. You can do this in any way you wish. One way:
The easiest way to do this is to temporarily disconnect the vertical tube and replace it with the new tube, so that when you squeeze the container you can fill the new tube. Add water to the container until it is nearly full, then screw on the bottlecap.
Hold the open end of the pressure-indicating tube a little higher than the top of the container, near the point where you just connected it, and squeeze the bottle so that water fills the tube. Since the water level in the container is higher than in the preceding experiment, and since the end of the new tube isn't much higher than the water level, this shouldn't require a very hard squeeze.
When the tube is full, maintain the squeeze so the water doesn't return to the container and disconnect it. You will have a tube full of water.
Now empty about half the water from the 'pressure tube'. Cap it and connect it to the system, and replace the vertical tube. You can do this in any way you wish, but one way is described below:
Just raise one end of the full 'pressure tube' and/or lower the other, and some water will flow out.
Once the tube is about half full, place a terminating cap on one end of this tube. This will hold the water in the 'pressure tube'.
You should at this point have:
The vertical tube, extended down into the water and out of the top of the container
The extended pressure-measuring tube, open on one end (through its connection to the newly opened tube in the stopper) to the air inside the container, half full of water, and capped at the other end.
A third tube short through the bottlecap, still closed off at the 'top end'.
In the picture below you see:
the short capped tube (the 'third' tube) sticking out of the top of the stopper,
the 'vertical' tube not yet in a vertical position but extending forward and to the right into a graduated cylinder, and
the pressure-indicating tube half full of caramel-colored liquid (the liquid is cola) and draped over a second graduated cylinder toward the back left. The pressure-indicating tube is capped at its end (hanging down near the tabletop), and the last 25 cm segment of the tube contains no liquid.
The picture below shows how the liquid in the tube comes to a point just below the 'peak' of the tube. This leaves an air column about 25 cm long in the capped end of the tube.
In the new picture the pressure-indicating tube is simply lying on the tabletop so the air column at the capped end is clearly visible.
The figure below shows a sketch of a tube which rises out of the bottle at left, then bends to form a U, then to the right of the U again levels off. The tube continues a ways to the right and is sealed at its right end. Liquid occupies the U up to almost the point of leveling, so that an increase in the pressure of the container will cause the liquid to move into the level region. As is the case in our experiment, the tube is assumed thin enough that the plane of the meniscus remains parallel to the cross-section of the tube (i.e., the meniscus doesn't 'level off' when it moves into a horizontal section of tube).
You should manipulate the pressure tube until its configuration resembles the one shown. The length and depth of the U can vary from that depicted, but the air column at the end of the tube should be at least 15 (actual) cm long. The liquid levels at the left and right ends of U should be very nearly equal.
The basic idea is that as you squeeze the system to raise water in the vertical tube, as in your previous experiment, the pressure in the system increases and compresses the 'air column' in thepressure tube. By measuring the lengths of this 'air column' you can determine relative pressures, and by measuring the heights of the water column in the 'vertical tube' you can determine theactual pressure differences required to support those columns.
Support the end of the vertical tube so that it is more or less vertical, as it was in the previous experiment.
The bottle should be pretty full, but not so full that it covers the open end of the tube to which the pressure tube is connected; the left end of the pressure tube should have an 'open path' to the gas inside the bottle, so that the pressure on the left-hand side of the water column in that tube is essentially equal to the pressure in the bottle.
If you squeeze the container a little, water will rise a little way in the vertical tube and the water in the pressure tube will also move is such a way as to slightly shorten the air column. The harder you squeeze the higher water will rise in the vertical tube and the shorter the air column will become.
Go ahead and observe this phenomenon. There is no need to measure anything yet, just get the 'feel' of the system.
Indicate below how the system behaves (what changes when you do what, how the system's reactions to your actions appear to be related to one another) and how it 'feels'.
----->>>>> behavior
Your answer (start in the next line):
It takes quite a bit of pressure to raise the water almost all the way to the top of the open vertical tube; the air column in the pressure indicator tube shortens somewhat but not by much.
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Using a measuring device you will measure the relative positions of the meniscus as you vary your squeeze:
One of the ruler copies used in the previous experiment on the distortion of paper rulers should be used here; a reduced copy should be used for greater precision. You may choose the level of reduction at which you think you will achieve the greatest level of precision. Only relative measurements will be important here; it will not be necessary to convert your units to actual millimeters or centimeters.
Indicate below the level of reduction you have chosen, and your reasons for this choice.
----->>>>> level of reduction and reasons
Your answer (start in the next line):
I used the triply-reduced ruler for best precision. It was easiest to see an accurate measurement on this finer ruler.
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In the units of the measuring device you have chosen, write down in your lab notebook the readings you used to indicate length of the air column, from the meniscus to the barrier at the capped end. No conversion of the units of your device to standard units (e.g., millimeters or centimeters) is required. Your information should include the marking at one end of the measuring device, and the marking at the other. If necessary two or more copies of paper rulers may be carefully taped together.
Indicate in the first line below the length of the air column in the units of your measuring device.
In the second line explain how you obtained your result, including the readings at the two ends and how you used those readings to indicate the length.
----->>>>> air column length, how obtained incl readings and how used
Your answer (start in the next line):
42.9 units (centimeters)
Two measuring devices taped together - one spanning from 13.0 to 49.6 units, and the other spanning from 37.0 to where the end of the tube was measured at 30.7 units. I subtracted 13.0 from 49.6, then added that result to the difference between 37.0 and 30.7, which then gave me 42.9 units total.
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Now place the same measuring device along the tube, positioned so you can observe as accurately as possible the relative positions of the meniscus in the pressure tube.
It is recommended that the initial position of the meniscus be in the vicinity of the center of the measuring device, so that position changes in both directions can be observed.
It is not necessary for the measuring device to extend the entire length of the air column, as long as you know the reading on the measuring device that corresponds to the initial position of the meniscus. From this information and from subsequent readings it will be easy to determine the varying lengths of the air column.
Take whatever precautions are necessary to make sure neither the measuring device nor the pressure tube can move until you have completed the necessary trials.
Mark positions along the vertical tube at 10-cm intervals (actual 10-cm intervals as indicated by a full-sized ruler) above the surface of the water in the bottle.
If the bottle is pretty full, as described before, it might be possible to make the first mark on the vertical tube at 10 or 15 cm above the water surface.
Marks may be made using an actual marker, or pieces of tape, or anything else that happens to be convenient.
Write your information in your lab notebook:
Write down the position of the first mark on the vertical tube with respect to the water surface (e.g., 10 cm or 15 cm).
Write down the position of the meniscus in the pressure tube. This position will simply be the reading on your measuring device. For example if the meniscus is at marking 17.35, that is what you write down.
As in all labs, you directly record what you read. Never do any arithmetic between making an observation and recording it.
You will now conduct 5 trials, raising water to the first mark on your vertical tube and reading the position of the meniscus before the squeeze and while water is at the given level.
Squeeze the bottle until water reaches the first mark in the vertical tube, and carefully read the position of the meniscus in the pressure tube. Release the bottle and immediately write down that position.
Repeat, being sure to again write down the position of the meniscus before squeezing the bottle (this position might or might not be the same as before) and the position of the meniscus when the water is at the first mark in the vertical tube.
Repeat three more times, so that you have a total of five trials in which the water was raised to the first mark in the vertical tube. With each repetition you will write down two more numbers.
Record your information below:
Indicate on the first line the vertical position of the first mark on the vertical tube, relative to the water surface, giving a single number in the first line.
On the second line give the length of the air column, as measured in units of the device you used to measure it.
On the third line, give the position of the meniscus before the first squeeze then the position of the meniscus when the water in the vertical tube was at the first mark. Give this information as two numbers, delimited by commas.
On lines four through seven, give the same information for the second through the fifth trials.
Starting in the eighth line give a brief synopsis of the meaning of the information you have given and how you obtained it.
----->>>>> vert pos mark vert tube, air column lgth, meniscus pos 1st trial, same 2d, same 3d, same 4th, same 5th trial, meaning
Your answer (start in the next line):
20cm
42.9 units
28.0, 27.5
28.0, 27.6
28.1, 27.6
28.0, 27.7
28.0, 27.7
I had a measuring device taped alongside the air column in the pressure measuring tube. Keeping the tube in a stable position throughout the experiment, I squeezed the bottle to raise the water up the tube to the desired heights and, maintaining the same squeeze as best as I could, I read the new position of the start of the air column. In all 5 trials, the air column shortens by approximately 0.3-0.5 units. The first mark in the vertical tube was 20cm above the surface of the water in the bottle (by some fortunate coincidence, I had filled the water bottle to 10 cm underneath the bottlecap, which made this experiment rather simpler!
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Now repeat the 5-trial process, this time raising water to the second mark. Write down everything as before.
In the space below report your results, using the same format as before:
----->>>>> same for 2d vert pos
Your answer (start in the next line):
30.0
42.9 units
28.0, 27.4
28.0, 27.4
28.1, 27.4
28.0, 27.3
28.0, 27.3
I did the same as I explained for the previous question, except that instead of raising the water to the first mark (20cm) I raised it to the second (30 cm). In all trials, the air column shortened upon squeezing the bottle.
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Repeat again, raising water to the highest mark you can manage with normal effort. Remember that this isn't supposed to be a test of strength.
In the space below report your results, using the same format as before:
----->>>>> same for 3d pos
Your answer (start in the next line):
120.0 cm
42.9 units
28.0, 25.0
27.8, 25.0
27.7, 25.0
27.8, 25.0
27.8, 25.1
The same as above, except that I raised the water to as high as I could without letting it spill out (120 cm above water surface in the bottle). In all cases, the air column decreased in length upon squeezing the bottle.
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If the highest mark you can easily manage is the third mark, then you may stop. If you have raised the water to a mark higher than the third, then do one more series of 5 trials, this time choosing a mark about halfway between the second and the highest mark.
In the space below report your results, using the same format as before. If you were not able to raise the water higher than your third mark, simply leave these lines empty.
Then report the approximate percent change in the length of the water column for each of the three vertical heights. Report in a single line separated by commas, and in the last line indicate how you got these results, including a sample calculation for the second set of trials.
----->>>>> same for 4th pos if possible
Your answer (start in the next line):
70cm
43.3 units
28.4, 26.5
28.2, 26.7
28.3, 26.4
28.0, 26.5
28.1, 26.5
I calculated the difference between 30 and 120cm, divided by two, then added the result to 30 in order to find the midpoint between 30 and 120cm. The midpoint was actually 75cm but since I did not have the tube labeled in increments of 5cm, I rounded down to 70cm. Again, as before, the air column decreased in the pressure measuring tube for all 5 trials.
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Make your estimate of atmospheric pressure:
What was the maximum height to which the water column was raised?
How much pressure was required to support the column?
Based on the behavior of the air column in the pressure tube, what percent do you think this is of atmospheric pressure?
What therefore do you conclude is atmospheric pressure?
----->>>>> max ht, pressure to support column, percent of atm pressure, conclusion atm pressure
Your answer (start in the next line):
120cm
The average decrease in length of the air column for these trials was 2.8 units, which is 6.5% of the total air column length. Since the air column was initially at atmospheric pressure before the squeeze, it follows that a 6.5% decrease in its length suggests a pressure that is 6.5% lower than atmospheric pressure.
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A decrease in the length of the air column in the pressure tube implies a decrease in the volume of the gas, which would be associated with an increase in pressure, not a decrease.
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P = pgh = (1000kg/m^3)(9.8m/s/s)(1.2m) = 11760 Pa
The above equation shows that, all other variables constant, the pressure required to raise this water 1.2m is 11760 Pa.
11760/P_atm = 0.065
P_atm = 180923 Pa
If the required 11760 Pa was indeed the amount of pressure utilized in this experiment and that led to a 6.5% decrease in air column length, then atmospheric pressure would be 180923 Pa. Of course we know this is not true, so an error (or errors) were incurred during the duration of the experiment.
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This conclusion is valid.
And obviously it doesn't agree all that well with accepted measurements of atmospheric pressure.
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Interestingly, your results for 'squeeze units' appear to give a low result for atmospheric pressure. This is based on an assumption I made about tube length (and about the ruler used to measure changes in the air column for that part of the experiment).
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I do want you to address the questions I've posed. Hopefully in the process you'll see either how to reconcile the two results, or to just let them both stand with the explanation that the perception of a squeeze isn't likely to produce a compelling result.
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&#Please see my notes and submit a copy of this document with revisions, comments 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. &#