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course Phy 242
measuring atmospheric pressure part 2** **
Overview of the analysis:
The relative lengths of the air column in the pressure tube before and after 'squeezing' give you direct information about the pressure in atmospheres.
The height of the water column in the vertical tube give you the increase in pressure, in units of N / m^2, that reult from your 'squeeze'.
In this analysis you will
use the meniscus positions in the pressure tube to determine pressure ratios corresponding to the various marks on the vertical tube,
use the density of water and the heights of the water columns in the vertical tube to determine the corresponding pressure differences,
obtain two expressions for the various pressure ratios, which can in each case be set equal to one another and solved for the atmospheric pressure
graph pressure difference vs. pressure ratio and determine the meaning of the slope associated with that graph.
Preliminary analysis:
Using your data from the preceding experiment Measure Atmospheric Pressure (Part 1), what is the greatest height to which you raised water in the vertical tube?
How much pressure, in N / m^2 or Pa, is necessary to support the water column?
How long was the air column in the 'pressure tube' at the beginning when the system was at atmospheric pressure, and how long was it when the water was raised to its maximum height?
By what percent did the air pressure therefore exceed atmospheric pressure when the water was at its maximum height?
Answer these two questions below:
50 cm
1000*.5*9.8 = 4900 N/m^2
40 cm (singly reduced ruler)
decreased to 38.1 cm; 1.9 cm / 40 cm * 100 = 4.75 %.
----->>>>> max ht from before, pressure to support in N/m^2, air column lgths, % atm pressure exceeded
The original pressure of the system was 1 atmosphere. Additional pressure was required to support the water column. You have just calculated this excess pressure in N / m^2, or Pa, and you have also calculated it as a percent of the original atmospheric pressure. What percent of atmospheric pressure corresponds to how many N / m^2? Explain.
What therefore must be atmospheric pressure?
----->>>>> % of atm pressure to how many N/m^2; concluded atm pressure
Your answer (start in the next line):
1 atm = 101,325 Pascals or N/m^2.
4.75% of 101,325 N/m^2 is 4,813 N/m^2.
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Your preliminary analysis gave you a result based on the data for one of your trials. The analysis you do here will take into account all the points. Ultimately you will make a graph of pressure in N / m^2, or Pa, vs. pressure in atmospheres, and you will use this graph to determine atmospheric pressure.
Throughout this analysis we will assume that the pressure tube is of uniform cross-sectional area. This isn't really the case; the end cap creates a small segment where the cross-sectional area is a greater than that of the rest of the tube, but we're going to assume here that the effect on our overall results is not significant.
Making this assumption:
If the air column was 30 cm long in one part of the experiment, and its length decreased by 10%, what would be its new length in cm?
If the air column was 30 cm long in one part of the experiment, and the volume of the air column decreased by 10%, what would be its new length in cm?
Give your two answers, separated by a comma, below:
----->>>>> new lgth if lgth decr by 10%, new vol if vol decr 10%
Your answer (start in the next line):
30 cm - 30 cm * .1 = 27 cm.
27 cm because the length of the air column is directly proportional to its volume. Therefor 10% volume decrease causes a 10% length decrease.
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Your answers should have been the same. Since the cross-sectional area is the same in both cases, and since the volume is the product of the cross-sectional area and the length, the ratio of the two volumes will be the same as the ratio of the two lengths.
Explain why, provided cross-sectional area remains the same, the volume ratio must be identical to the length ratio.
----->>>>> why vol ratio identical to lgth ratio
Your answer (start in the next line):
The volume is multiplied by the cross sectional area, which is a constant so therefor the only other variable is the height and so volume and length are proportional.
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In what follows, atmospheric pressure is to be regarded as an unknown. When and if you refer to atmospheric pressure, use P_atm to represent this quantity. Do not use any numerical value, just the symbol P_atm.
When the system was pressurized, water rose in the vertical tube to an observed height above the water surface in the container.
Let point A be on the meniscus of the water column in the vertical tube, and let B be a point on the surface of the water in the container.
It is possible to trace a path from point A to point B, such that at every point the path is completely surrounded by water. Thus, the water is continuous along this path from A to B, and Bernoulli's Equation therefore holds.
Bernoulli's equation relates the density, velocity, pressure and altitude at different points in a continuous fluid.
The density of water is constant in this situation (it is very difficult to significantly compress liquid water, requiring very high pressures).
Of the three remaining variables (velocity, pressure and altitude, subsequently referred to as v, P and y):
Which variable(s), if any, do not change between point A and point B?
Which variable(s), if any, increase as you move from A to B?
Which variable(s), if any, decrease as you move from A to B?
Which variable(s), if any, are zero at both points?
Answer below. Be sure you indicate clearly which increase, which decrease, etc..
----->>>>> A top vert tube B water surf unchanging vbls, vbls incr A to B, decr A to B, same at both
Your answer (start in the next line):
Between A and B, only V does not change because you can be assumed that at both points the water is stationary and has no velocity. Y increases from A to B since water is raised to a greater height. P is forced to decrease from A to B to due to increase in y and v is zero at both points.
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Using the following symbols:
rho for density,
v_A and v_B for velocities at A and B,
P_A and P_B for pressures at A and B, and
y_A and y_B for altitudes at A and B,
and leaving out any terms of the equation that happen to be 0, write down Bernoulli's equation for this situation and report your equation below.
If you left out a term because it was 0, explain which term is was and why that term is 0.
----->>>>> Bernoulli's eqn, expl of term(s) omitted because 0
Your answer (start in the next line):
Rho*g*y_A + P_A = rho*g*y_B + P_B
.5*rho*v_A^2 and .5*rho*v_B^2 are equal to zero because the water column isn't moving.
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For the trial where the water in the vertical tube was at the first mark, give the following, one number or symbolic expression to each line. Give the numerical value if it is known, in units of m/s for velocity or m for altitude. If, and only if, a quantity is not known for this system and has not been assigned a specific symbol, report using just the appropriate symbol P_A, P_B, y_A, y_B, v_A or v_b.
The pressure at point A:
The altitude at point A:
The velocity at point A:
The pressure at point B:
The altitude at point B:
The velocity at point B:
----->>>>> at 1st mark P, y, v at A, P, y, v at B numbers or symbolic as appropriate
Your answer (start in the next line):
P_atm
.1 m
0 m/s
P_atm + 980 N/m^2
0 m
0 m/s
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The velocities at both A and B are zero; neither water surface should have been moving with any substantial velocity during any surveillance of the location of the meniscus in the pressure tube.
The point A is touching the air in the tube, which is open to the atmosphere and is thus at atmospheric pressure, so the pressure at that point is P_atm; remember that we are treating this pressure as an unknown here and are thus representing it as P_atm.
There is the ability to measure the vertical positions at A and B from any point you wish, butonly if you use the same relative altitude point; as long as your value of y_A is greater than y_B by the distance from the water surface to the mark, your values are reproducible.
You might have chosen to regard the water surface as point A, in which case y_A would be zero and y_B would be the distance from the water surface to the mark. Since both velocities are zero neither of the .5*rho*v^2 terms are relevant to this equation.
Check the equation you gave against these notes, and if you need to correct your equation, do it here; don't go back and correct your previous answer.
----->>>>> correct eqn if necessary
Your answer (start in the next line):
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Solve Bernoulli's equation as you wrote it for P_B, by algebraically rearranging it so that P_B is on the left-hand side.
On the first line below give the right-hand side of the rearranged equation.
On the second line substitute into this expression the accepted value rho = 1000 kg / m^3 for rho and simplify. Leave P_atm in symbolic form.
----->>>>> solve eqn for P_B give rt side; substitute for rho and simplify
Your answer (start in the next line):
Patm + .1*rho*g
Patm + 980
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The pressure of the gas changes very little from one part of the bottle to the other. Since point B is on the water surface and hence in contact with both water and gas, the gas pressure in the bottle is equal to the pressure at point B.
Write the ratio of pressure in the bottle to atmospheric pressure. The ratio would be
ratio = (pressure inside) / P_atm,
where (pressure inside) is the expression you obtained for P_B.
----->>>>> expression P inside / P_atm (at 1st mark)
Your answer (start in the next line):
(Patm + 980) / Patm
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Following the same sequence of steps, find the symbolic expression for the ratio of gas pressure inside the bottle to atmospheric pressure for the second mark on the vertical tube.
Then do the same for the highest mark.
If you did set of trials for a fourth mark, find that expression also.
Report all four of your ratios below, in order from the lowest to the highest mark. If you did not make the fourth observation, you may leave the fourth line blank.
----->>>>> pressure ratio expressions for water at all 3 or 4 marks
Your answer (start in the next line):
(P_atm + 980) / P_atm
(P_atm + 1960) / P_atm
(P_atm + 2940) / P_atm
(P_atm + 3920) /P_atm
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At this point you should have a symbolic expression for the pressure ratio in each sequence of trials.
Now you will obtain numerical results for these ratios, based on your observations. Once you have these results, you can set each of your symbolic expressions equal to the corresponding number and solve each for P_atm.
Open the data analysis program.
Copy into the data window the 5 rows of your data corresponding to the beginning and ending positions of the meniscus (i.e., the meniscus positions corresponding to atmospheric pressure, and to the pressure when the water was at the first mark).
Also copy the same five lines into the space given below:
----->>>>> for 1st vert pos copy of 5 rows copied into data analysis program
Your answer (start in the next line):
20.00, 19.40
20.00, 19.40
20.10, 19.50
20.10, 19.60
20.10, 19.55
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Find the mean and standard deviation of your unpressurized and pressurized meniscus positions:
Click on the Mean and Standard Deviation button and the program will report the mean and standard deviation of the numbers in the first column. These results will represent the mean and standard deviation of the 'unpressurized' positions of the meniscus.
If the Select Column button is on the form, click it to select column 2. Column 2 will appear in the data window. Otherwise you will need to manually remove the first column, including the comma, so that the data window shows only the numbers from the second column.
Find the mean and standard deviation of that column. These results will represent the mean and standard deviation of the 'unpressurized' positions of the meniscus.
Make a sketch representing these results:
Sketch on a number line (in your lab notebook) the interval corresponding to mean +- standard deviation for the no-pressure position, and for the pressurized position of the meniscus (i.e., sketch for each the interval from mean - standard deviation to mean + standard deviation).
Make note of the left- and right-hand endpoints of each interval.
Report your results as follows:
In the first line, the height from the water surface to the mark.
In the second line, the mean and standard deviation of the 'unpressurized' positions of the meniscus (the first column), reported in that order in comma-delimited format.
In the third line, the mean and standard deviation of the 'pressurized' ('squeezed') positions of the meniscus (originally in the second column), in the same format used for the second line.
In the fourth line, the difference of the two means.
Starting in the fifth line, an explanation of what the difference of the two means represents, and a discussion of the uncertainty in your measurement of this quantity.
In the sixth line, a short discussion of the uncertainty in your measurement of this quantity.
Starting in the seventh line briefly describe how you calculated your results and what they mean.
----->>>>> vert pos of water column, mean sdev unpressurized meniscus, same pressurized, difference of means, meaning and uncertainty, discuss uncertainty, how results calculated
Your answer (start in the next line):
10 cm
20.04, .05477
19.49, .08944
.55
Fourth line is average length the air column decreased when the water column was raised to 10 cm. This value has more uncertainty than the two mean values because the individual uncertainties of the means are increased through this calculation. But it is minimal due to the small standard deviations.
The uncertainty in the value can be assumed to be minimal.
Values are the means and standard deviations of the unpressurized and pressurized data for the water column at 10 cm calculated using data program.
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Repeat for the second set of 5 trials, where water was raised to the second mark
Copy your 5 lines of data here:
----->>>>> 5 lines water at 2d mark
Your answer (start in the next line):
20.00, 19.10
20.00, 18.95
20.00, 19.00
20.00, 19.05
20.00, 18.95
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Calculate and report your results for that data, using the same format as before:
In the first line report the height from the water surface to the mark.
In the second line, the mean and standard deviation of the 'unpressurized' positions of the meniscus (the first column), reported in that order in comma-delimited format.
In the third line, the mean and standard deviation of the 'pressurized' ('squeezed') positions of the meniscus (originally in the second column), in the same format used for the second line.
In the fourth line, the difference of the two means.
Starting in the fifth line, an explanation of what the difference of the two means represents.
In the sixth line, a short discussion of the uncertainty in your measurement of this quantity.
Starting in the seventh line briefly describe how you calculated your results and what they mean.
----->>>>> vert pos of water column, mean sdev unpressurized meniscus, same pressurized, difference of means, meaning and uncertainty, discuss uncertainty, how results calculated
Your answer (start in the next line):
20 cm
20.00, 0
19.01, .06519
.99
Fourth line is mean length the air column decreased when the water column was raised to 20 cm. Since the standard deviation of the unpressurized data is 0, the standard deviation of the difference continues to be .06519.
Values are the averages and standard deviations of the unpressurized and pressurized data for the water column at 20 cm calculated using the data program.
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Repeat for the third set of 5 trials, where water was raised to the highest mark you could easily manage:
Copy your 5 lines of data here:
----->>>>> 5 lines 3d mark
Your answer (start in the next line):
20.00, 18.90
20.15, 18.70
20.10, 18.80
20.10, 18.65
20.00, 18.50
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Calculate and report your results for that data, using the same six-line format as before:
----->>>>> vert pos of water column, mean sdev unpressurized meniscus, same pressurized, difference of means, meaning and uncertainty, discuss uncertainty, how results calculated
Your answer (start in the next line):
30 cm
20.07, .06708
18.71, .1517
1.36
The fourth line is the average length the air column decreased when the water column was altered to 30 cm. This value has increased uncertainty due to the fact that the individual uncertainties of the means are increased through this calculation. The standard deviation of the difference is .1659.
Values are the averages and standard deviations of the unpressurized and pressurized data for the water column at 30 cm calculated using the data program.
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Repeat for the fourth set of 5 trials, where water was raised to an intermediate mark (if you did not have data for the fourth set of trials this answer and the next may be left blank):
Copy your 5 lines of data here:
----->>>>> 5 lines 4th mark
Your answer (start in the next line):
20.00, 18.50
20.10, 18.50
20.05, 18.40
20.00, 18.30
20.10, 18.30
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Calculate and report your results for that data, using the same six-line format as before:
----->>>>> vert pos of water column, mean sdev unpressurized meniscus, same pressurized, difference of means, meaning and uncertainty, discuss uncertainty, how results calculated
Your answer (start in the next line):
40 cm
20.05, .05
18.4, .09999
1.65
The fourth line is the mean length the air column shrunk when the water column was raised to 40 cm. This has more uncertainty than the two mean values due to the fact that the individual uncertainties of the means are increased through calculation. The standard deviation of the difference is .1118.
Values are the means and standard deviations of the unpressurized and pressurized data for the water column at 40 cm calculated using the data program.
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Summarize below the heights of the vertical water columns and the difference between the mean unpressurized position and the mean pressurized position of the meniscus in the pressure tube. Report these quantities in order of the water column heights from least to greatest.
Each line should consist of the water column height and the difference in mean meniscus positions, both in units of centimeters, using comma-delimited format.
Your data should consist of four lines; if you did not to the fourth set of trials, enter 0, 0 in that line.
----->>>>> vert ht diff in mean meniscus pos in 4 lines
Your answer (start in the next line):
10, .55
20, .99
30, 1.36
40, 1.65
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For the first set of trials, where the water column was at the first mark, write down the difference you just reported in the meniscus positions, and then write down the original length of the air column in the pressure tube. Using these numbers, answer the following:
By what percent of the original length of the air column did the that length appear to change when the system was pressurized?
What therefore was the ratio of volumes of the air columns (the ratio would be (volume after compression / volume before compression)?
Report your results below, giving in each line the percent change in the length of the air column and the volume ratio, separated by commas. Report in order from the lowest mark to the highest mark you observed. If you did not complete the fourth set of trials, enter 0, 1 as your fourth line.
----->>>>> 4 lines % change in air column length , ratio of volumes
Your answer (start in the next line):
1.38%, .98625
2.48%, .97525
3.4%, .966
4.13%.95875
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We know that for a confined ideal gas, PV = n R T. The air column in the pressure tube consists of confined air, and air acts very nearly as an ideal gas.
The temperature T can be regarded as constant, for the reasons indicated in the following paragraph:
When you squeezed the container and thereby increased the air pressure inside, that increased pressure was transmitted to the water column in the pressure tube, and in turn to the air column in the pressure tube. The result was that the air column was compressed to a lesser volume. In the process positive work was done on the gas in this column, increasing its internal energy, and as a result its temperature also increased. However the gas being in contact with the walls of the tube, this excess thermal energy was quickly absorbed into the walls of the tube and eventually dissipated into the surrounding air. So any change in T was small and short-lived. So T can be regarded as being very nearly the same before and after your reading of the position of the meniscus.
So n and T, and of course R, remain constant. The right-hand side n R T therefore remains constant, and it follows that the product PV must also remain constant.
If PV remains constant then P1 V1 = P2 V2, where P1 and V1 are the pressure and volume before compression and P2 and V2 after. This is the same as saying that
P2 / P1 = V1 / V2, or
P2 / P1 = 1 / (V2 / V1).
That is, the pressure ratio is the reciprocal of the volume ratio.
In your own words explain below why the pressure ratio must be equal to the reciprocal of the volume ratio.
----->>>>> why pressure ratio reciprocal of vol ratio
Your answer (start in the next line):
Ideal Gas Law, when n, R, and T are constant the PV product must remain constant as well. P1 V1 = P2 V2 or P1 / P2 = V2 / V1.
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Figure out the pressure ratio corresponding to each of the volume ratios you reported above. Each pressure ratio will the the reciprocal of the corresponding volume ratio.
In the same order as before, from the lowest mark to the highest, report the pressure ratios in the air column, giving the ratio of the pressure after squeezing to the pressure before squeezing.
In the next available line give a sample calculation.
----->>>>> pressure ratios
Your answer (start in the next line):
1.013
1.025
1.035
1.043
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You reported earlier the symbolic expressions for the pressure ratios, in terms of the symbol P_atm.
Set the symbolic expression for the first mark equal to the number you just reported for the first mark. This gives you an equation. Report the equation below:
----->>>>> equation for 1st mark symbolic expr equal to just-reported pressure ratio
Your answer (start in the next line):
1.013 = (P_atm + 980) / P_atm
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Solve this equation for P_atm and give your solution in the first line below. Starting on the second line, show how you solved the equation, step by step.
----->>>>> solve for P_atm, steps of soln
Your answer (start in the next line):
75,385 N/m^2
1.013 = (Patm + 980) / Patm
1.013 * Patm = 1 Patm +980
.013 Patm = 980
Patm = 980/.013 = 75385
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Do the same for the remaining marks.
Report in the first line your equation for the second mark, then your solution to that equation, separated by commas.
In the second line report your equation and solution for the next-highest mark (this will be the highest mark if you did only three 5-trial sets).
If you did a fourth trial, then report in the third line your equation and solution for the highest mark.
----->>>>> each mark eqn then soln
Your answer (start in the next line):
1.025 = (Patm + 1960) / Patm, 78400
1.035 = (Patm + 2940) / Patm, 84000
1.043 = (Patm + 3920) / Patm, 91163
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You will now use rho g h to find the pressure added to the system by each water column:
Using h = y_A - y_B for each mark, find the additional pressure rho g h indicated by the water column.
For the first mark, report in the first line below your values of the pressure ratio and rho g h, in comma-delimited format.
In the same format report in the second line your values for the second mark.
In the third line report your values for the highest mark.
In the fourth line report your values for fourth mark you observed, if you observed one. Otherwise just put 1, 0 into the fourth line.
In the fifth line indicate the units of rho g h, and explain why it has these units.
Starting in the sixth line give a brief explanation of how you obtained your results and what they mean, Include a sample calculation.
----->>>>> rho g h each vert pos, units of rho g h, explanation
Your answer (start in the next line):
1.013, 980
1.025, 1960
1.035, 2940
1.043, 3920
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Sketch and analyze a graph of rho g h vs. pressure ratio:
Using the table you have just constructed, sketch a graph of rho g h vs. pressure ratio.
Sketch the straight line you think comes as close as possible, on the average, to your points. Unless your points lie perfectly on a straight line, none of the points should actually lie on the line.
Give below the coordinates of the leftmost and rightmost points on your straight line (these should be points on the line, and should not coincide with data points).
Put the horizontal and vertical coordinates of the first point in the first line, in comma-delimited format, and the coordinates of the second point in the second line.
----->>>>> coord of two points on graph rho g h vs pressure ratio
Your answer (start in the next line):
1.0125, 850
1.0425, 3760
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Report the slope of your straight line:
Give in the first line the slope of your straight line, as calculated from the two points you have just reported.
In the second line give the units of this slope (units of rise divided by units of run).
In the third line explain how you calculated the slope from the coordinates you reported in the previous box.
----->>>>> slope, units of slope, explanation
Your answer (start in the next line):
97000
N/m^2
(Y2-y1)/(x2-x1)
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Interpret the slope of your straight line:
Explain in terms of the system you observed what the meaning of the slope might be:
----->>>>> meaning of slope
Your answer (start in the next line):
The slope is positive meaning that if you apply more pressure there is a positive correlation that the air column will be longer and has more pressure pushing against it.
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&#Your work on this lab exercise is good. Let me know if you have questions. &#