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PHY 232
Your 'bottle thermometer' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Your optional message or comment: **
I couldn't understand the relationship between altitude change and Pressure or Temperature or Volume. However I will resubmit after your comment.
** #$&* What happens when you pull water up into the vertical tube then remove the tube from your mouth? **
1 hour
** #$&* What happens when you remove the pressure-release cap? **
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.
The water rushes back toward the bottle but stops about 20 cm outside the
bottle up the tube. I expected this to happen but not nearly as much. When I
brought the water out of the tube it ran almost the entire length of the tube,
so it reduced to about 20% of the original length.
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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.
Air will enter the system because it has been taken out. This will equalize the
pressure between the air inside and the air outside the system.
When I removed the cap the water column dropped back into the bottle.
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Now replace the cap on the pressure-valve tube and, while keeping an eye on the
air column in the pressure-indicating tube, blow just a little air through the
vertical tube, making some bubbles in the water inside the tube. Blow enough
that the air column in the pressure-indicating tube moves a little, but not more
than half a centimeter or so. Then remove the tube from your mouth, keeping an
eye on the pressure-indicating tube and also on the vertical tube.
What happens?
Why did the length of the air column in the pressure-indicating tube change
length when you blew air into the system? Did the air column move back to its
original position when you removed the tube from your mouth? Did it move at all
when you did so?
What happened in the vertical tube?
Why did all these things happen? Which would would you have anticipated, and
which would you not have anticipated?
When I blew into the vertical tube, the water in the pressure indicating tube
moved toward the plug, away from the bottle by less than a cm. Once the pressure
was removed, the column moved back.
The water in the vertical tube went into the bottle when pressure was applied
and air followed it. Once pressure was released the water went back up through
the tube.
The reason these things happened is because the pressure and volume are
inversely proportional and are trying to balance out.
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Place the thermometer that came with your kit near the bottle, with the bulb not
touching any surface so that it is sure to measure the air temperature in the
vicinity of the bottle and leave it alone until you need to read it.
Now you will blow enough air into the bottle to raise water in the vertical tube
to a position a little ways above the top of the bottle.
Use the pressure-valve tube to equalize the pressure once more with atmospheric
(i.e., take the cap off). Measure the length of the air column in the
pressure-indicating tube, and as you did before place a measuring device in the
vicinity of the meniscus in this tube.
Replace the cap on the pressure-valve tube and again blow a little bit of air
into the bottle through the vertical tube. Remove the tube from your mouth and
see how far the water column rises. Blow in a little more air and remove the
tube from your mouth. Repeat until water has reached a level about 10 cm above
the top of the bottle.
Place the bottle in a pan, a bowl or a basin to catch the water you will soon
pour over it.
Secure the vertical tube in a vertical or nearly-vertical position.
The water column is now supported by excess pressure in the bottle. This excess
pressure is between a few hundredths and a tenth of an atmosphere.
The pressure in the bottle is probably in the range from 103 kPa to 110 kPa,
depending on your altitude above sea level and on how high you chose to make the
water column. You are going to make a few estimates, using 100 kPa as the
approximate round-number pressure in the bottle, and 300 K as the approximate
round-number air temperature. Using these ball-park figures:
If gas pressure in the bottle changed by 1%, by how many N/m^2 would it change?
What would be the corresponding change in the height of the supported air
column?
By what percent would air temperature have to change to result in this change in
pressure, assuming that the container volume remains constant?
Report your numbers in the first three lines below, one number to a line, then
starting in the fourth line explain how you made your estimates:
1N/m^2
It would change in the other direction by 1%
I'm not sure how the height would change.
1%
1% of 100kPa = 1kPa = 1N/m^2
1 kPa = 1000 Pa, not 1 Pa
'dP = 1/'dT
P is proportional to T, not 1 / T.
In any case if T increases by 1%, or .01, 1/T does not change by a factor 1 / .01 = 100. The new T would be 1.01 times the old value, and the factor would be 1 / 1.01, not 1 / .01.
Since P is proportional to T, though, a 1% increase in T would imply a 1% increase in P.
The height would increase or decrease depending on the pressure acting on the
bottle increasing or decreasing respectively.
The temperature and pressure are directly proportional. So if the pressure
increased by 1% then the temperature would increase by 1%.
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Continuing the above assumptions:
How many degrees of temperature change would correspond to a 1% change in
temperature?
How much pressure change would correspond to a 1 degree change in temperature?
By how much would the vertical position of the water column change with a 1
degree change in temperature?
Report your three numerical estimates in the first three lines below, one number
to a line, then starting in the fourth line explain how you made your estimates:
3K
1kPa
1mm
your first two results look good; however you don't indicate how you got 1 mm, and that result isn't correct
Since the Temperature and Pressure changes are direclty proportional both would
just be a 1% change.
I'm not sure how to determine the height change. I think it has something to do
with volume which is inversely proportional to the temperature and the pressure.
Height change governs one of the quantities in Bernoulli's equation.
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How much temperature change would correspond to a 1 cm difference in the height
of the column?
How much temperature change would correspond to a 1 mm difference in the height
of the column?
Report your two numerical estimates in the first two lines below, one number to
a line, then starting in the third line explain how you made your estimates:
10% change
1% change
Those were guesses.
Go back to Bernoulli's Equation and see if you can figure this out.
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A change in temperature of 1 Kelvin or Celsius degree in the gas inside the
container should correspond to a little more than a 3 cm change in the height of
the water column. A change of 1 Fahrenheit degree should correspond to a little
less than a 2 cm change in the height of the water column. Your results should
be consistent with these figures; if not, keep the correct figures in mind as
you make your observations.
The temperature in your room is not likely to be completely steady. You will
first see whether this system reveals any temperature fluctuations:
Make a mark, or fasten a small piece of clear tape, at the position of the water
column.
Observe, at 30-second intervals, the temperature on your alcohol thermometer and
the height of the water column relative to the mark or tape (above the tape is
positive, below the tape is negative).
Try to estimate the temperatures on the alcohol thermometer to the nearest .1
degree, though you won't be completely accurate at this level of precision.
Make these observations for 10 minutes.
Report in units of Celsius vs. cm your 20 water column position vs. temperature
data, in the form of a comma-delimited table below.
21C, 0cm
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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):
22.1C, 0.7cm
The water in the tube increased by 0.7cm and the temperature went up 1.1C
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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.
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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:
The column barely moved, maybe about a fraction of 1mm.
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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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Repeat the experiment with your warm hands near the bottle. Report below what
you observe:
It changed by about 2mm this time.
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When in the first bottle experiment you squeezed water into a horizontal section
of the tube, how much additional pressure was required to move water along the
horizontal section?
By how much do you think the pressure in the bottle changed as the water moved
along the horizontal tube?
If the water moved 10 cm along the horizontal tube, whose inner diameter is
about 3 millimeters, by how much would the volume of air inside the system
change?
By what percent would the volume of the air inside the container therefore
change?
Assuming constant pressure, how much change in temperature would be required to
achieve this change in volume?
If the air temperature inside the bottle was 600 K rather than about 300 K, how
would your answer to the preceding question change?
Give your answers, one to a line, in the first 5 lines below. Starting in the
sixth line, explain how you reasoned out these results:
The pressure has changed significantly.
Not sure.
10%
10%
The same
Using the ideal gas law I used the proportionality between the variables to
figure out the changes in the variables.
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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 air was negligible.
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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 6 cm in
the vertical direction.
By how much would the pressure of the gas have to change to increase the
altitude of the water by 6 cm?
Assuming a temperature in the neighborhood of 300 K, how much temperature change
would be required, at constant volume, to achieve this pressure increase?
The volume of the gas would change by the additional volume occupied by the
water in the tube, in this case about .7 cm^3. Assuming that there are 3 liters
of gas in the container, how much temperature change would be necessary to
increase the gas volume by .7 cm^3?
Report your three numerical answers, one to a line. Then starting on the fourth
line, explain how you obtained your results. Also make note of the relative
magnitudes of the temperature changes required to increase the altitude of the
water column, and to increase the volume of the gas.
Gravity would affect the estimate of temperature difference because it would
pull the liquid down.
1600%
Since V = A*h then PAh = nRT
This reduces to a constant A, n, R, and T leaving P and h to be inversely
proportional. Since h changes by 0.06m then it changes by 0.06. This leaves an
inverse proportion to be 16.7
I am not getting the relationship between height and pressure and temperature. I
will need some clarification before I can completely finish this.
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Continue to assume a temperature near 300 K and a volume near 3 liters:
If the tube was in the completely vertical position, by how much would the
position of the meniscus change as a result of a 1 degree temperature increase?
What would be the change if the tube at the position of the meniscus was
perfectly horizontal? You may use the fact that the inside volume of a 10 cm
length tube is .7 cm^3.
A what slope do you think the change in the position of the meniscus would be
half as much as your last result?
Report your three numerical answers, one to a line. Then starting on the fourth
line, explain how you obtained your results. Also indicate what this
illustrates about the importance in the last part of the experiment of having
the tube in a truly horizontal position.
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As you state, you are missing the relationship between the water column height and the pressure.
`d(rho g y) + `d(1/2 rho v^2) + `dP = 0.
v is constant.
This gives you the relationship between change in height and change in pressure.