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: **
** #$&* What happens when you pull water up into the vertical tube then remove the tube from your mouth? **
2 hours
** #$&* What happens when you remove the pressure-release cap? **
Describe below what happens and what you expected to happen. Also indicate why you think this happens.
When you remove the tube from your mouth, the water should fall back out of the tube and into the container because air enters back into the system through the vertical tube and forces the water back down. Furthermore, there is no change in the pressure indicating tube because when the pressure valve tube is uncapped, air is entering to maintain the atmospheric pressure and when it is capped air is entering through the vertical tube. After removing the tube, my observations matched my expectations
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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.
There will be no change when the cap is removed because although air will escape through this tube, it will also enter through the tube at the same rate thus maintaining the atmospheric pressure. When the cap was removed, nothing occurred.
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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?
The length of the air column in the pressure indicating tube decreased because the pressure inside the bottle was increased by blowing air through the vertical tube. However, when I stopped blowing, the length of the air column stayed the same and the water level rose in the vertical tube. This occurred because after being blown into the bottle, the air was unable to escape through the vertical tube. Therefore, the increased pressure forced the water up through the vertical tube. I anticipated the shortening of the air column in the pressure indicating tube. However, I initially did not anticipate the rise of the water in the vertical tube.
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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:
1000
10 cm
1%
1 N/m^2 is equal to 1 Pa. Therefore, a 1% change in a pressure of 100kPa would equal 1000 N/m^2. According to Bernoullis equation, this pressure increase would have to be matched by the same increase in rho g h. Therefore, solving this equation results in a height of approximately 10 cm. Finally, at constant volume and number of moles, temperature is proportional to pressure. Therefore, a 1% increase in pressure would require a 1% increase in temperature.
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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:
3 degrees Celsius
333 N/m^2
3 cm
The first value was found by multiplying 300 K by .01. The next two values were then found simply by dividing the height and pressure values found in the previous question by 3 because these values corresponded to a 3 degree Celsius temperature increase rather than a 1 degree increase.
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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:
.3 degrees Celsius
.03 degrees Celsius
1 cm is .1 of the height increase corresponding to a 3 degree Celsius temperature increase. Therefore, 3 was multiplied by .1 to find the first value. The second value was found by multiplying 3 by .01.
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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.
22.1, .5
22.3, 1.1
22.5, 1.1
22.5, 1.2
22.5, 1.3
22.7, 1.4
22.5, 1.5
22.5, 2.0
22.5, 2.0
22.8, 2.2
23.0, 2.2
22.8, 2.1
23.0, 2.2
23.0, 2.2
23.0, 2.2
23.0, 2.3
23.0, 2.3
23.0, 2.4
23.0, 2.4
22.9, 2.4
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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):
Over time, both the alcohol thermometer and bottle thermometer appeared to slowly increase. For the bottle thermometer, the maximum deviation from the starting position was 2.4 cm. For the alcohol thermometer, the starting temperature was not known. Therefore, the maximum deviation from the first measurement was .9 degrees Celsius.
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Now you will change the temperature of the gas in the system by a few degrees and observe the response of the vertical water column:
Read the alcohol thermometer once more and note the reading.
Pour a single cup of warm tap water over the sides of the bottle and note the water-column altitude relative to your tape, noting altitudes at 15-second intervals.
Continue until you are reasonably sure that the temperature of the system has returned to room temperature and any fluctuations in the column height are again just the result of fluctuations in room temperature. However don't take data on this part for more than 10 minutes.
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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:
After warming my hands and holding them near the bottle, the water column rose by a maximum of 4 cm.
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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:
23.0, 0
23.0, 0
23.0, .2
22.8, 0
22.7, -.1
23.0, .1
23.0, 0
23.2, .3
23.2, .2
23.0, .1
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Repeat the experiment with your warm hands near the bottle. Report below what you observe:
The water column rose again. However, it did not rise enough to reach the horizontal section.
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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 did not change as the water moved along the horizontal section.
The volume of air increased by 10 cm * (.15 cm)^2 * pi = .225 pi cm^3
Assuming the air began at 1000 cm^3, this would be an increase of .07%.
This change in volume would require a temperature change of approximately .21 degrees Celsius.
The required temperature change would double.
These results were found simply by using the gas law T1/V1=T2/V2.
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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 was not accounted for in this situation because the main mechanism by which the water column was moving was the increasing pressure of the gas due to the increasing temperature.
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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.
588 Pascals
1.76 K
.07 K
These values were found by using Bernoullis Equation as well as the Ideal Gas Law. As shown above, the amount of thermal energy required to increase pressure is far greater than that required to increase the gas volume.
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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.
3 cm
143 cm
45 degrees
Like previous answers, these answers were obtained from Bernoullis equation and the Ideal Gas Law. The first two values show that with an equal amount of thermal energy put into a system, a fluid can be moved a far greater distance horizontally than vertically.
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&#This looks very good. Let me know if you have any questions. &#