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course Phy 202
1/18 2
Myra Gentry Assignment 1Jan 14, 2014
This assignment consists of five problems based on today's class, four brief lab-related activities using the bottlecap and tube, a reading assignment in the text, and a single additional problem for you to view.
You will find that the most extensive assignments will be given on Wednesdays. On Mondays we will generally go over any difficulties you've had with the assignments, and the assignment between Monday and Wednesday will be fairly light.
Links to the four parts of this assignment:
• Five_problems
• Four_Brief_Experiments_with_the_Bottlecap_and_Tube
• Reading_in_Text_
• Introductory_Problem_Set_Problem
Monday we will try to reconcile any difficulties you have with this assignment.
Five problems
At a depth of 1 cm, water pressure would be .0098 Newtons / cm^2 or 980 dynes / cm^2. For intuitive approximations we can use .01 N / cm^2 and 1000 dynes / cm^2. Hopefully you recall how to get these results, but in any case be sure you know them for future reference.
We also determined that at depth 1 meter, water pressure is 9800 N / m^2, or for intuitive approximation, 10 000 N / m^2. You should also know this, and hopefully how it was determined.
Based on these results, see how you can do with the following. Bring your work with you on Monday. If you have trouble or questions, email me and I’ll be glad to answer.
1. Based on the 10 000 N / m^2 pressure at depth 1 meter, what should be the pressure at a depth of 10 cm?
1,000 N/m^2
2. Based on the 10 000 N / m^2 pressure at depth 1 meter, what should be the pressure at a depth of 1 cm?
100 N/m^2
3. The pressure at a depth of 1 cm is about .01 N / cm^2. How many cm^2 are there in one m^2? How many N of force would we therefore have on one m^2, if the pressure on one cm^2 was .01 N? (To be sure you’re on the right track, think about whether there would be more force on one cm^2 or on one m^2).
10 N/m^2
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There were several questions here, and you probably should have answered all of them rather than giving one final answer (which you might note is inconsistent with that given to the preceding question). I can't tell what you did so I can't tell where the inconsistency came from.
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4. As you’ve seen, heating the air in the bottle can raise a water column. Suppose we heated the air in a bottle, originally at absolute temperature 300 Kelvin (this is 27 Celsius, a fairly warm room temperature) enough to raise a water column 1.2 meters high.
· How much extra pressure does it take to accomplish this? (note that you’ve been told how much pressure corresponds to a 1 meter depth)
12,000 extra pressure
· What percent is this of atmospheric pressure? (atmospheric pressure is about 100 000 N / m^2)
88%
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12,000 is 12% of 100,000, not 88%.
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· If the temperature change necessary to increase the pressure is the same percent of absolute temperature, by how many degrees would the temperature have changed? What would be the new temperature in Celsius?
291 C ? I feel like this is really off. I did 273 plus 88 % and got 564 K and then converted to C.
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It's very good that you see the unreasonableness of this answer.
The explanation is simple enough. See my previous note. You should have added 12% of the original temperature (which, incidentally, was 300 K rather than 273 K).
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5. By making some simple measurements (as we will do in class Monday) we find that when the temperature of the gas in a bottle increases from 0 Celsius to 20 Celsius, the pressure in the bottle increases from 1 atmosphere to 1.08 atmosphere. This gives us two points on a graph of pressure vs. temperature, the points being (0 Celsius, 1 atmosphere) and (20 Celsius, 1.08 atmosphere).
Sketch these points on a graph, and sketch the straight line through these points. Extend the line until it intersects the horizontal axis.
· What is the temperature corresponding to this point?
0 C
Note that you can do this graphically and get a reasonable result. You can also calculate where a line through (0, 1) and (20, 1.08) would intersect the horizontal axis. Either method, or both, will be fine.
· How does this temperature correspond to the accepted value of absolute zero, which is -273 Celsius?
-273 = 0 K
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A line through (0, 1) and (20, 1.08) doesn't intercept the horizontal axis at -273 Celsius. It's fairly close, but it's not at -273.
You need to have sketched the graph, drawn the line and estimated the coordinate without thinking of the ideal answer.
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Four Brief Experiments with the Bottlecap and Tube
If you click on a link it might or might not work, but ‘might not’ is fairly likely. This is an anomaly of Blackboard and there isn’t much we can do about it.
To get the link to work, copy it and paste it into the address box of your Internet browser. If this doesn’t work, email me a copy of the contents of the address box and a description of what you’re seeing.
The document at
http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/ph1/classes_fall_2010/brief_expts/pictures/Physics_II_Initial_Bottlecap-and-tube_Experiments.htm
demonstrates a number of possible uses of bottle-and-tube systems. It includes at the beginning a number of short videos. You’re welcome to watch them all, but concentrate on the four entitled as follows:
• raising_water_by_squeezing
• raising_water_by_changing_temperature
• siphoning_water_into_tube
• pressure_tube
Then read through the rest of the document. Lots of pictures, not that many words.
This should prepare you to do the following experiments. I suggest you print each one out and pen in your results. Then copy the instructions into a text document, insert your results (which are brief and shouldn’t take you long to transcribe), and email me a copy.
Don’t let yourself get too hung up on the instructions. If there’s something you don’t understand, email me a question. If we can’t resolve it fairly easily, I can demonstrate it in class Monday.
· http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/ph1/classes_fall_2010/brief_expts/individual_experiments/brief_bottle_experiment_1a.htm
Comparing the state of the bottle before and after you squeeze:
Does the amount of air in the bottle increase or decrease?
decrease and then increases
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It would increase if you released the bottle after squeezing. If you just squeezed it would be strictly a decrease.
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Does the volume of air enclosed in the bottle increase or decrease?
decrease and the increase
Does the pressure in the bottle increase or decrease?
increase and then decrease
Does the temperature of the air in the bottle increase or decrease?
remains the same
Be sure you have explained all your answers.
Comparing the state of the bottle before and after you squeeze:
Does the amount of air in the system increase or decrease?
remains the same
Does the volume of air enclosed in the system increase or decrease?
remains the same
Does the pressure in the system increase or decrease?
increase and then decrease back down to normal
Does the temperature of the air in the system increase or decrease?
with increase in pressure, temp does rise some
· http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/ph1/classes_fall_2010/brief_expts/individual_experiments/brief_bottle_experiment_1b.htm
Does the air column get longer or shorter? By what percent do you estimate the length of the column changes?
shorter, by like 10 %
Does the volume of the air column increase or decrease? By what percent do you estimate the volume of the column changes?
decrease by 10%
Does the number of molecules in the air column increase, decrease or remain the same? By what percent do you estimate the number of molecules changes?
remains the same
Does the mass of the air in the air column increase or decrease? By what percent do you estimate the mass of the air in the column changes?
decrease 10 %
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Air can't get into or out of the air column, so there would be no change in the mass of the air. It would become 10% denser, though.
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Does the pressure in the air column increase, decrease or remain the same? By what percent do you conjecture the pressure in the column changes?
increase by 10 %
Does the pressure in the bottle increase, decrease or remain the same? By what percent do you conjecture the pressure in the bottle changes?
increase by 5 %
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The water 'plug' would move if the forces on its two ends were different, which would be the case in this situation if the pressures were different.
So the pressures are the same. If one increases by 10%, so does the other.
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When you hold the bottle in the squeezed position, with the water plug stationary, the pressure in the bottle results in a force on the plug which pushes it toward the capped end, while the pressure in the air column results in a force that pushes the plug away from that end. Which force do you think is the greater, or are they equal?
equal
Which do you think is greater, the pressure in the bottle or the pressure in the air column?
pressure in the air column
Measure the length of the air column.
What is the length of the air column?
18cm
How far would the water plug have to move to make the air column 10% shorter?
my original was around 20-21 and ended up being 18 cm, so that is 10 %
Squeeze the bottle so the air column becomes 10% shorter. It's up to you to figure out how to tell when it's 10% shorter. If you can't squeeze hard enough to achieve the 10% difference, then figure out what percent you can manage and note the percent in your answer.
10 % of 20 didn't take much of a squeeze but from 18 to 16 is a bigger difference
On a 1-10 scale, with 10 the hardest squeeze of which you are capable without risking injury, how hard did you have to squeeze the bottle and what percent change did you achieve in the length of the air column?
about 3 the first time and 4 the second from 18-16
Now, using the same 1-10 scale, give the bottle squeezes of 2, 5 and 8. Estimate the percent changes in the length of the air column.
What were your percent changes in air column length?
starting at 20 cm. 2- 19 5- 13 8-10
Now by heating and/or cooling the bottle, what extremes in air column length can you achieve? Careful not to melt the bottle. It won't handle boiling water, and you shouldn't mess with water hot enough to scald you or cold enough to injure you (e.g., don't use dry ice, which in any case is too cold for the bottle, and certainly don't use liquid nitrogen).
Report your results:
I tried heating it up and got nearly no results.
· http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/ph1/classes_fall_2010/brief_expts/individual_experiments/brief_bottle_experiment_1c.htm
Starting with the cap in place on an empty bottle, siphon water from an adjacent full bottle. Allow the siphon to run a few minutes until the water levels in the two bottles stabilize.
Estimate the percent change in the volume of the air in the capped bottle.
5-10% experiment did not go so well
Estimate the percent change in the number of molecules in the air within the capped bottle.
5-10 %
Estimate the percent change in the volume of the water in the open bottle.
maybe 10-15
What do you think is the percent change in the air pressure in the capped bottle?
10-15%
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These are good estimates.
As it turns out the percent change in air pressure is close to the percent change in volume.
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What is the difference in the two fluid levels?
I got about 20 ml in the other bottle
What is the percent change in the number of air molecules in the capped bottle?
2.5-5%
Raise the open bottle as high as possible without disturbing the capped bottle. Allow time for the water levels in the two bottles to stabilize.
What percent of the volume of the capped bottle do you now estimate is occupied by water?
about 40 ml about 50 % more than before
Estimate the percent change in the number of molecules in the air within the capped bottle.
about 10 %
By what percent do you estimate the pressure in the capped bottle exceeds the original pressure (i.e., the pressure when the bottle was first capped)?
probably about 10% more pressure that before
What percent of the uncapped bottle do you estimate is now occupied by air?
about 80%
What is the difference in the two water levels?
473 -40 is 433 and 40 ml is in the capped bottle
Return the uncapped bottle to the tabletop. What happens?
What is now the difference in the two water levels?
I messed up my hose, so nothing happened
What do you think is the pressure in the uncapped bottle as a percent of its original pressure (before the bottle was capped)?
about 5 to 10 % higher
· http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/ph1/classes_fall_2010/brief_expts/individual_experiments/brief_bottle_experiment_1d.htm (ignore the part about the extension on the tube; the tube I gave you is long enough)
Add the extension to the tube, so that by squeezing you can force water from the bottle into the tube. Squeeze hard enough to raise the water to as high as possible into the tube. Evaluate how hard you had to squeeze, on the 1-10 scale you used in part 1b. Measure how far you were able to raise water in the tube above the level of the water in the bottle.
How high did you raise the water, and how hard did you have to squeeze (using the 1-10 scale)?
to the top of the hose, about 80 cm, about 2-3
Give the bottle a squeeze corresponding to 1 on the 1-10 scale, and observe how high water rises. Then give it another squeeze, halfway between 1 and the squeeze you used to raise water to the top of the tube. Do this blind. Don't look at the tube, just feel the squeeze. Then look at the tube and see where the water is.
Report a table of water column height vs. squeeze.
1
About ˝ way up the tube (80ml)
2
So close to the top (40ml)
3
Out of the tube
4
Out
5
Out
6
Out
7
Out
8
Out
9
couldn’t squeeze this hard
10
I have no grip
**Remember to bring the caps, tubes and bottles (if you took one) back to class on Monday.
Reading in Text
Read through Chapter 10, Sections 1-6. Some of this will be familiar from our discussion in class, some will be related to it but might not be completely familiar.
Jot down questions about what you are reading and bring them to class on Monday.
Introductory Problem Set Problem
At the site
http://vhmthphy.vhcc.edu/ph2introsets/default.htm#Set 5: Fluids and Thermodynamics
you will see a menu of problems. Click on Problem #1, read it and see if you can solve it. Then look at the given solution and see how you did. The look at the Generalized Solution and see what you do and do not understand.
This problem is one of about 150 problems, entitled Introductory Problem Set Problems. These problems introduce most of the main ideas in your course. If you can master them, you will do well on the tests.
P=pgh
P=1,000 kg/m^3 (9.8m/s^2)(6.5 m)
P=637000 N/m^2
This is as far as I got so I have to look in the solution to times it by the area. Now I will know for next time."
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Overall this looks good. Check out my notes, and thanks for submitting it.
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