#$&*
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 **
2.5
** **
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):
0.5cm, 1
3cm, 4
5cm, 8
49Pa, 294Pa, 490Pa
1 49Pa
4 294Pa
8 490Pa
m=62.243Pa per squeeze unit b=7.9459 squeeze units
P = `rho g y = 1000 kg / m^3 * 9.8 m/s^2 * .03 meters = 49 Pa
#$&*
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):
6
2 0%
5 5%
8 20%
m=.0333percent change/squeeze unit b=-.0833 percent change
#$&*
According to the appropriate graph:
To what height would water rise in an open vertical tube, given a level-10
squeeze? Answer and explain the basis for your answer.
Your answer (start in the next line):
6.4cm
I made a graph for the height of the water level vs squeeze level, then graphed
a line of best fit. I plugged in 10 as my x-value and found the corresponding
y-value using my best fit equation for the graph.
#$&*
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):
630.38Pa
I plugged in 10 as my x-value and found the corresponding y-value using my betst
fit line for pressure vs squeeze.
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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):
25%
I plugged in 10 as my x-value and found the corresponding y-value using my best
fit equation for percent chang vs squeeze
#$&*
Based on the amount of increase in pressure, and on the percent increase,
determine the original pressure in the bottle.
Answer and explain the basis for your answer.
Your answer (start in the next line):
.25x+x=630.38Pa
1.25x=630.38Pa
x=504.304Pa
The 25% incerase of pressure is added to the original. So, the 25% increase
should equal the original pressure (x) plus 25% of the original pressure.
#$&*
(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 the graph of pressure vs. squeeze shows the amount of pressure
caused by a level of squeeze.
The slope of the graph of percent increase in pressure vs. squeeze shows the
percent of pressure that increases due to a certain level of squeeze.
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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):
????I do not understand how I can only use both slopes from the two graphs to
find the original pressure. I could use the line of best fit to estimate but I'm
not clear on how to only use the slope of each graph with out using the entire
slope line equation?????
#$&*
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 the pressure 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 the actual
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):
Water moves up the vertical tube towards the stopper placed in the pressure tube
when you squeeze the bottle. The pressure tube will respond every time you
squeeze the bottle.
#$&*
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):
9.5-11cm
I measured from the capped end to the meniscus.
#$&*
In the units of the measuring device you have chosen, write down in your lab
notebook the readings you used to indicate length of the air column, from the
meniscus to the barrier at the capped end. No conversion of the units of your
device to standard units (e.g., millimeters or centimeters) is required. Your
information should include the marking at one end of the measuring device, and
the marking at the other. If necessary two or more copies of paper rulers may
be carefully taped together.
Indicate in the first line below the length of the air column in the units of
your measuring device.
In the second line explain how you obtained your result, including the readings
at the two ends and how you used those readings to indicate the length.
----->>>>> air column length, how obtained incl readings and how used
Your answer (start in the next line):
9.5-11cm pressure tube 9.5-43 vertical tube
I measured from the capped end to the meniscus.
#$&*
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):
10cm
33cm
33, 35.2
33.5, 35.1
33.2, 35.2
33.6, 35.4
33.3, 35.4
The same pressure being acting on the bottle is acting on the tube as well.
#$&*
Now repeat the 5-trial process, this time raising water to the second mark.
Write down everything as before.
In the space below report your results, using the same format as before:
----->>>>> same for 2d vert pos
Your answer (start in the next line):
20cm
33.5cm
33.5, 37.4
33.6, 37.3
33.7, 37.5
33.8, 37.8
33.9, 37.7
#$&*
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):
30cm
33.6cm
33.6, 38.5
33.7, 39.1
33.8, 39.2
33.8, 39.5
34.0, 39.6
#$&*
If the highest mark you can easily manage is the third mark, then you may stop.
If you have raised the water to a mark higher than the third, then do one more
series of 5 trials, this time choosing a mark about halfway between the second
and the highest mark.
In the space below report your results, using the same format as before. If you
were not able to raise the water higher than your third mark, simply leave these
lines empty.
Then report the approximate percent change in the length of the water column for
each of the three vertical heights. Report in a single line separated by
commas, and in the last line indicate how you got these results, including a
sample calculation for the second set of trials.
----->>>>> same for 4th pos if possible
Your answer (start in the next line):
35.4-35.2/35.2 * 100 = .6%
37.7-37.4/37.4 * 100 = .8%
39.6-38.5/38.5 * 100 = 2.9%
I subtracted the starting distance of each trial from the end distance of each
trial and then divided the answer by the starting distance and multiplied by 100
to find the % change of the column.
#$&*
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):
39.6cm
P = `rho g y = 1000 kg / m^3 * 9.8 m/s^2 * .396m = 3880.8 Pa
10%
.1x+x=3880.8
1.1x=3880.8
x=3528Pa
@&
If 3880 Pa is 10% of atmospheric pressure, then atmospheric pressure is not 3528 Pa. 10% of 3528 Pa is only 352.8 Pa, not 3880 Pa.
If 3880 Pa is 10% of atmospheric pressure, then atmospheric pressure is 38 800 Pa.
*@
#$&*
&#Good responses on this lab exercise. See my notes and let me know if you have questions.
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