course Phy 122
2/25 8
measuring atmospheric pressure part 2Phy 122
Your 'measuring atmospheric pressure part 2' 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: **
** Length if 30 cm long tube with uniform cross-sectional area has length decreased by 10%, and if volume decreased by 10%. **
27 cm, 27 cm
** Why, provided cross-sectional area remains the same, is volume ratio identical to the length ratio? **
Volume equals the cross-sectional area times the length. If area remains the same, then as you increase either volume or length, it will increase the other and therefore the ratios will simplify to an equal number.
If volume is increased, then the length must increase, when area remains unchanged. If the length is increased, then volume must increase, if area remains unchanged.
** Which variable(s) do not change, which increase, which decrease as you move from A (top of water column in the vertical tube) to B (water surface in the container)? **
Velocity doesn't change and should equal zero. Pressure increases and altitude decreases.
** Bernoulli's equation for this situation. **
P_A = P_B(rho)(g)('d'y_A - y_B)
Velocity was left out because it was 0 when the system reached point A and stayed there.
** Your report of pressure, altitude velocity at point A, then at point B: **
P_atm
15cm
0
P_B
0cm
0
** Bernoulli's equation for this situation (possibly corrected from before). **
** Bernoulli's equation for points A and B, solution for P_B in terms of P_atm. **
P_A - rho(g)('d' y_A - y_B)
P_atm - 1000kg/m^3 (9.8m/s^2)(15cm)
** Symbolic expression for the ratio of gas pressure inside the bottle to atmospheric pressure for the first mark on the vertical tube. **
ratio = [P_atm - 1000kg/m^3 (9.8m/s^2)(15cm)] / P_atm
** Symbolic expression for the ratios of gas pressure inside the bottle to atmospheric pressure, lowest mark to highest. **
ratio = [P_atm - 1000kg/m^3 (9.8m/s^2)(15cm)] / P_atm
ratio = [P_atm - 1000kg/m^3(9.8m/s^2)(25cm)] / P_atm
ratio = [P_atm - 1000kg/m^3(9.8m/s^2)(35cm)] / P_atm
** Your report of meniscus positions for corresponding to the first mark: **
10, 10.8
10, 10.7
10, 10.8
10, 10.6
10, 10.8
** For first mark, height of vertical column, mean and standard deviation of meniscus position for unpressurized system, then for pressurized system, difference of the two means. **
15cm
0, 0
0.07200, 0.08944
0.07200
It represents the difference between the mean and standard deviation.
My measurements are not truly accurate because it was difficult to maintain the pressure at the exact level while reading the measurement accurately.
I calculated the first number by using a ruler to measure from the surface of the water to the point marked on the bottle. The rest of the numbers were calculated using the data program and represent the mean and standard deviation of the unpressurized position of the menisus, the pressurized position, and the difference of the two means, respectively.
** Your report of meniscus positions for corresponding to the second mark: **
10, 11.1
10, 11.3
10, 11.3
10, 11.2
10, 11.1
** For second mark, height of vertical column, mean and standard deviation of meniscus position for unpressurized system, then for pressurized system, difference of the two means. **
10, 11.55
10, 11.45
10, 11.55
10, 11.40
10, 11.40
** Your report of meniscus positions for corresponding to the second mark: **
** For highest mark, height of vertical column, mean and standard deviation of meniscus position for unpressurized system, then for pressurized system, difference of the two means. **
25cm
0, 0
0.08000, .1000
0.08000
It represents the difference between the mean and standard deviation.
My measurements are not truly accurate because it was difficult to maintain the pressure at the exact level while reading the measurement accurately.
I calculated the first number by using a ruler to measure from the surface of the water to the point marked on the bottle. The rest of the numbers were calculated using the data program and represent the mean and standard deviation of the unpressurized position of the menisus, the pressurized position, and the difference of the two means, respectively.
** Your report of meniscus positions for corresponding to the second mark: **
35cm
47.05
0, 0
0.06400, 0.07583
0.06400
It represents the difference between the mean and standard deviation.
My measurements are not truly accurate because it was difficult to maintain the pressure at the exact level while reading the measurement accurately.
I calculated the first number by using a ruler to measure from the surface of the water to the point marked on the bottle. The rest of the numbers were calculated using the data program and represent the mean and standard deviation of the unpressurized position of the menisus, the pressurized position, and the difference of the two means, respectively.
** For fourth set of trials, height of vertical column, mean and standard deviation of meniscus position for unpressurized system, then for pressurized system, difference of the two means. **
** Your report of the difference in the mean meniscus positions vs. the height of the vertical water column. **
15cm, 0.07200cm
25cm, 0.08000cm
35cm, 0.06400cm
0cm, 0cm
** Percent changes in air column length and corresponding volume ratios (pressurized to unpressurized): **
0.15%, 47.122/47.05
0.17%, 47.130/47.05
0.14%, 47.114/47.05
0, 1
** Why is the pressure ratio must be equal to the reciprocal of the volume ratio. **
If the beginning product of pressure and volume are directly proportional to the final product of pressure and volume, then as we rearrange each into ratios by using simple algebra, we must keep those proportions.
** Pressure ratios, lowest mark to highest: **
47.122/47.05
47.130/47.05
47.114/47.05
0
If meniscus positions were 10.8, 11.2, 11.5, all in cm, and the length of the air column was about 50 cm, then the change between 10.8 cm and 11.2 cm would be .4 cm, which is .4 cm / 50 cm = .008 or .8%, and the pressure ratio would be about 1.008. The change between 11.2 and 11.5 would be .006 or .6%
** Your symbolic equation for the first mark: **
[P_atm - 1000kg/m^3 (9.8m/s^2)(0.15m)] / P_atm = 47.122/47.05
If there is a difference of, say, .4 cm then you might have a ratio of 47.05 / 46.75 = 1.008.
&&&& The pressure ratio is the same as the inverse volume ratio. I'm really lost going back and looking at all of these numbers and trying to correspond them with the original lab because they're in different formats. The pressure ratio is the pressure inside / P_atm. I was apparently using the measurements from the reduced ruler, which don't even have units, to obtain the 47.122 and 47.05 for the P_atm. I think I got mixed up because of using different measuring methods. What is the correct pressure ratio supposed to be? &&&&
If you measured the air column with the same ruler, then the units would divide out when you find the ratio.
Based on the length of the air column (which has a fairly small percent change) and the changes in meniscus position, you can find the percent change in the length of the air column. Just divide the change in meniscus position by the length of the air column (it wouldn't much matter if you used 47 cm or 48 cm or 46 cm, the difference in the quotient would be insignificant).
** Your solution for P_atm. **
????
[P_atm - 1470] / P_atm = 1.00153
???? I'm getting mixed up I think. I don't know what to do next and I don't think this is even correct.
** Equations and solutions for the remaining marks: **
????
** Your values of rho g h vs. pressure ratio. **
1.00153, 1470
1.00170, 2450
1.00136, 3430
1,0
(kg/m^3)(m/s^2)(m), which is density, gravity, and height, which equals Pascals.
The left numbers are the final pressure divided by the original pressure and the right numbers are equal to the change in pressure found by using rho x g x h.
** Coordinates of two points on the approximate best-fit line. **
1.0013, 2000
1.0017, 4000
** Slope of your best-fit line. **
2000/1.0013 = 1997, 4000/1.0017 = 3993
slope isn't y coordinate / x cooordinate, it's change in y coordinate / change in x coordinate.
The pressure changes from 2000 pa to 4000 pa, which would be a change of 2000 pa.
The number of atmosphere, according to your calculations (which I believe need to be modified per preceding notes) changes by 1.0017 - 1.0013 = .0004. So the slope would be 2000 Pa / .0004 = 5 000 000 Pa.
Pa/reduced ruler measurements
&&&& Ok, so slope is change in y / change in x. I can't really figure out from this document what numbers I need to use and/or change. &&&&
Your graph would be based on your table
1.00153, 1470
1.00170, 2450
1.00136, 3430
1,0
The first number in each pair is the pressure ratio, which doesn't appear to be correct (per previous notes) but which can easily be recalculated based on the approximate length of the air column and the changes in meniscus position.
You would use the methods of the 'fitting a straight line' exercise to get the slope: graph the points, using a scale the spreads the points out both vertically and horizontally, and fit a reasonable straight line to your points. Identify two points on that line and determine the slope.
** Meaning of the slope. **
????no idea????
** Optional additional comments and/or questions: **
3hrs +
** **
I think I need some help, as you can see from my numbers.
You've done most of the experiment quite well. However I'm not sure about your pressure ratios (e.g., 1.00153). See my notes and see if you can modify accordingly; of course you're welcome to ask questions.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end). &#
It's worth another 30 minutes of your time to make another revision, which would reinforce a few ideas and procedures that will be important in later labs. However if it takes you longer than that, it's not worth it at this point. You've done a good job and your grade will be OK either way.