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Phy 122
Your 'bottle thermometer' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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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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What happened was that the water in the vertical tube went back in to the bottle and increased the pressure. I know this because I observed the air column decrease. I wasnt sure if the water would go back in to the bottle, but it did. The pressure in the bottle increased because we trapped the air in the bottle once we had water in the tube. Once the vacuum from my mouth was gone the weight of the water pushed it back into the bottle and then compressed the air in the bottle. This caused the air column to decrease.
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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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If I remove the cap from the pressure valve, air will escape the bottle and neutralize to atmospheric pressure. Air does escape. I believe this is so because I observed the pressure go down from the air column.
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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 decreases and then increases about half of what it decreased.
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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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The air column decreased when I blew into the tube because I was adding air to a confined space and this increased the pressure. The air column moved about half way back to its original positon.
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What happened in the vertical tube?
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Some water went up the vertical tube when my mouth released. The pressure induced into the bottle by blowing pushed some water up the vertical tube when it was taken out of my mouth.
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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 air column decreased because the gas pressure in the bottle was increased by blowing in the bottle. The air coumn returned to about half of what it decreased when my mouth was released. This was because the pressure in the bottle was enough to push some water up the vertical tube. I anticipated the pressure becoming greater in the bottle. I did not anticipate the pressure going back down some when my mouth released.
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What happened to the quantities P, V, n and T during various phases of this process?
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P increased when blowing in the bottle and then decreased slightly when mouth released.
V stayed constant when blowing into bottle. When mouth released V increased slightly from where it was as some water was pushed up the vertical tube.
I think T went up slightly as the gas compressed.
I believe n went up when I blew into the bottle.
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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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1 kPa = 1000 N/m^2. 100kPa * .01 = 1 kPa. A 1% change in pressure is 1000 N/m^2.
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What would be the corresponding change in the height of the supported air column?
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P=rho * g * h, P/(rho * g)=h. 1000N/m^2/(1000kg/m^3 * 9.8m/s^2) =.102m or 10.2cm
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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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PV=NKT. 300K/100kPa=3. 3*101kPa = 303K. 3K/300K=.01.
T needs to rise 1%.
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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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+3K or about 5 degree F.
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How much pressure change would correspond to a 1 degree change in temperature?
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1K/300K = .0033. 1.0033*100kPa=100.33kPa.
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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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dP=rho*g*dh, dP/(rho*g)=dh, .33kPa= 330N/m^2.
(330N/m^2)/(1000kg/m^3 * 9.8m/s^2)=.0337m or 3.37cm
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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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1cm=.01m
P=rho*g*h= (1000kg/m^3)*(9.8m/s^2)*(.01m)=98 N/m^2
(P1V1)/T1 = (P2V2)/T2, V is constant so it cancels to P1/T1=P2/T2.
We are given 100kPa in the bottle = 100,000N/m^2
We are given 300K in the bottle is approx 80.33 degree F.
(100,000N/m^2)/(300K)=(100,098N/m^2)/T2
Solve for T2=300.294 K is approx 80.86 degree F or approx a .53 degree F increase.
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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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1mm=.001m
P=rho*g*h= (1000kg/m^3)*(9.8m/s^2)*(.001m)=9.8 N/m^2
(P1V1)/T1 = (P2V2)/T2, V is constant so it cancels to P1/T1=P2/T2.
We are given 100kPa in the bottle = 100,000N/m^2
We are given 300K in the bottle is approx 80.33 degree F.
(100,000N/m^2)/(300K)=(100,009.8N/m^2)/T2
Solve for T2=300.0294 K is approx 80.38 degree F or approx a .05 degree F increase.
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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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19.9C, 10.2cm
20.0C, 10.25cm
20.0C, 10.25cm
20.0C, 10.25cm
19.9C, 10.25cm
20.0C, 10.25cm
19.9C, 10.25cm
19.9C, 10.25cm
20.0C, 10.25cm
20.0C, 10.25cm
19.9C, 10.25cm
19.9C, 10.25cm
19.9C, 10.20cm
19.8C, 10.20cm
19.8C, 10.20cm
19.8C, 10.20cm
19.9C, 10.20cm
19.8C, 10.20cm
19.9C, 10.20cm
19.C, 10.20cm
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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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Not much trend to talk about. Perhaps 10.25cm to 10.20cm deviation of .05cm. Temp deviation 19.8C to 20.0C or .2C deviation.
From calcs above we said that 1mm was about equal to .05 degree F. The thermometer range was about .2C or approx 1.64 degree F.
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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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10.0, 10.3, 9.9, 9.8, 9.7, 9.8, 9.9, 9.95, 10.0, 10.0, 10.0, 10.0, 10.05, 10.1, 10.1, 10.1, 10.1, 10.0
The air column actually increased in length. I expected it to decrease in length.
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This isn't what would be expected. If somehow the pressure tube was also warmed this could account for the observation.
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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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My hands near the bottle reduced the air column by approx 2mm. From above I said that 1mm was about .05 degree F. Then the air temp rose by .1 degree F.
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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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18.8C, 1mm
18.9C, 4mm
19.0, 15mm
19.1C, 15mm
19.5C, 19mm
19.5C, 22mm
19.6C, 24mm
19.8C, 25mm
19.9C, 25mm
19.9C, 26mm
Cool night. Started a fire in the wood burning stove and that was driving the air temp up.
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Repeat the experiment with your warm hands near the bottle. Report below what you observe:
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19.0C, 0mm
19.0C, 30mm
19.5C, 40mm
19.5C, 45mm
19.5C, 54mm
19.5C, 63mm
19.5C, 69mm
19.5C, 75mm
19.5c, 82mm
19.5C, 89mm
Between the first and second reading I touched the bottle and that caused the 30mm jump.
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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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The water in the horizontal tube moved a total of 89mm
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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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The above info says that a slug of water 10cm long in a 3mmdiameter tube moved up into the tube to move the water we are observing move. So, 10cm=.1m and 3mm=.003m. Then cross sectional area is (pie*.003m^2)/4=.00000707m^2.
.00000707m^2 * .1m = .00000071m^3.
The volume of air increased the amount of water up the tube or .00000071m^3
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By what percent would the volume of the air inside the container therefore change?
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Not knowing the volume in the bottle, Im going to take a guess here. Its a 2 litre bottle about half full. So there is 1 liter of air and 1 liter of water. So there is initially approx .001m^3 of air.
(.00000071m^3)/(.001m^3)=.00071 or .071% increase in air 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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Assuming we start at 300K. We have P constant so it cancels out. Then V1/T1=V2/T2. Plug in the known values and solve for T2=302.13K. To get this change in volume the temperature rose from 300K to 302.13K.
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I suspect a factor-of-10 error here.
.071% of 300 K is about .2 degrees, not 2 degrees.
Otherwise good.
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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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V1/T1=V2/T2. Plug in the known values and solve for T2=604.26K. to get the change in volume required a temperature change from 600K to 604.26K.
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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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That tube was open end to atmosphere. In that case the air is not compressing, just moving around. The volume of air on the open end of the tube doesnt have any effect on in bottle.
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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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P=rho*g*h
dP=(1000kg/m^3)*(9.8m/s^2)*.06m=588N/m^2 or approx .085psi
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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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P1V1/T1=P2V2/T2
V constant cancels. P1/T1=P2/T2. Ill assume P1 is the 100kPa or 100,000N/m^2
(100,000N/m^2)/(300K)=(100,588N/m^2)/T2.
Solve for T2=301.75K
We need a temperature change from 300K to 301.75K
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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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If P is constant, V1/T1=V2/T2.
.003m^3/300K=.0030007m^3/T2
Solve for T2=300.07K
Temperature needs to go from 300K to 300.07K.
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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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P1=100,000N/m^2, V2=.003m^3, T1=300K, V2=.003m^3, T2=301K.
(100,000N/m^2 * .003m^3)/300k = (P2*.003m^3)/301K
Solve for P2 = 100,333N/m^2
How far will additional 333N/m^2 push the water slug?
P=rho*g*h
P/(rho*g)=h
(333N/m^2)/(1000kg/m^3 * 9.8m/s^2)=.034m
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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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Im going to guess I use P=rho*g*h and remove the g. I dont think thats correct.
Using same data as above:
P/rho=h
(333N/m^2)/(1000kg/m^3)=.333m
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Figure this based on a volume expansion at constant pressure.
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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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Im not sure how to approach this.
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This can be done using the techniques of calculus, more specifically differentials.
However that exceeds the scope of this course.
So consider instead that a 1% increase in temperature, with water in the vertical tube, would raise the water about 10 cm. A 1% increase in volume would be 30cm^3, which would correspond to over 400 cm of tube volume. This is a ratio of 40-to-1.
We might then conjecture that 40 units of horizontal displacement would be equivalent to 1 cm of vertical displacement.
Your current results are that 1 degree raises the water 3.4 cm in a vertical tube, or increases the volume by 10 cm^3, which would correspond to a 140 cm change in meniscus position. The ratio is still basically 40-to-1.
Thus a slope of 1/40 might be expected to 'split' the effects between the pressure increase that raises the water and the expansion that displaces water in the horizontal direction.
This conjecture could be validated or refuted by using PV = n R T.
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Very well done. That last question is pretty challenging. See my note but otherwise don't worry about it.
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