My readings are only 1.5% to 2.5% low, as opposed to the 30% that you mention. If I had to venture a guess as to why other people's readings are lower, I would speculate that turning away to look at the horizontal column and get the reading distracts them from keeping a steady grip on the bottle." "
@&
I'll have to repeat the experiment myself and see if I have a similar tendency.
*@
Vertical column pressure in excess of atmospheric pressure (hereafter ""vertical column pressure"") = density:fluid * gravity * height:fluid = 1,000 (kg/[m^3]) * 9.81 (m/[s^2]) * height = height * (9,810 kg/[m^3 s^2])
For constant #MolesGas, type of gas or gas mixture, and temperature, P * V = constant; therefore, new pressure * new volume = old pressure * old volume; new pressure / old pressure = old volume / new volume.
- Note that because the horizontal column was sealed at atmospheric pressure, the ""new pressure / old pressure"" (= ""old volume / new volume"") ratio above represents the new pressure expressed in multiples of atmospheric pressure, i.e., in atmospheres.
Therefore:
- (Atmospheric pressure + vertical column pressure) / atmospheric pressure = old volume / new volume
- (Atmospheric pressure + vertical column pressure) = atmospheric pressure (old volume / new volume)
- Vertical column pressure = atmospheric pressure (old volume / new volume) - atmospheric pressure = atmospheric pressure ([old volume / new volume] - 1)
- Atmospheric pressure = vertical column pressure / ([old volume / new volume] - 1)
And because for two cylinders with equal radii, volume varies directly with height (if vertical) or length (if horizontal), ...
... Atmospheric pressure = vertical column pressure / ([old horizontal column length / new horizontal column length] - 1).
I got the following four data points before I accidentally squeezed the bottle too hard and pushed some water out of the vertical tube. Because this changed the volume of the water in the bottle and the initial water level, I worried that because the bottle isn't fully elastic and won't perfectly contract to maintain the pressure of the sealed air, there might be a significant upward change in sealed-air volume and a corresponding significant drop in sealed-air pressure, so I had to reset the system by unsealing the bottle, unplugging the horizontal column, and letting each reach a new equilibrium level before resealing them.
TABLE 1: Water level = -4.0 cm; starting horizontal column length = 20.0 cm for all
(Vertical column height [m], vertical column pressure in excess of atmospheric [Pa], initial horizontal column length (m), final horizontal column length (m), value for atmospheric pressure [Pa])
0.340, 3335.400, 0.200, 0.195, 130080.600
0.590, 5787.900, 0.200, 0.190, 109970.100
0.990, 9711.900, 0.200, 0.185, 119780.100
1.140, 11183.400, 0.200, 0.180, 100650.600
Analysis (per data program): Mean = 115,100 Pa; median = 114,871.1 Pa; StdDev (although less meaningful for this few values): 12,670 Pa = 11.0% of mean or roughly same (slightly higher but still rounds to 11.0%) of median
After resetting the system, I did another trial but soon noticed that my horizontal column was getting shorter each time, indicating that the seal was leaking. Although I had to disregard these data and reset the system yet again, I've listed the bad data at the bottom in case you're interested.
On the ""redo of the redo,"" I was able to get a more full data set, although I stayed away from the upper ends of the vertical-column-height range in order to avoid disrupting the system yet again. Data are as follows:
TABLE 2.1: Water level = -3.5 cm (i.e., 3.5 cm below cap level); measurement intervals are 10-cm increments above cap level; starting horizontal column length range = 16.7 cm to 16.35 cm (variation suggests continued nonzero leaking over time; chronological order of measurements was [vertical column length, horizontal ' ']: [0.435 m, 0.166 m], [0.635 m, 0.167 m], [0.735 m, 0.1655 m], [0.135 m, 0.1635 m], [0.235 m, 0.1635 m], [0.335 m, 0.1655 m], and [0.535 m, 0.1635 m])
(Vertical column height [m], vertical column pressure in excess of atmospheric [Pa], initial horizontal column length (m), final horizontal column length (m), value for atmospheric pressure [Pa])
0.135, 1324.350, 0.1635, 0.162, 143029.800*
0.235, 2305.350, 0.1635, 0.1595, 91925.831
0.335, 3286.350, 0.1655, 0.157, 60700.818*
0.435, 4267.350, 0.166, 0.159, 96929.807
0.535, 5248.350, 0.1635, 0.155, 95705.206
0.635, 6229.350, 0.167, 0.157, 94190.281
0.735, 7210.350, 0.1655, 0.1535, 92232.393
Analysis (per data program): Mean = 96,390 Pa; median = 94,947.7835 Pa; StdDev = 24,120 Pa = 25.0% of mean or 25.4% of median
Analysis (per data program) with high and low values (*) discarded: Mean = 94.190 Pa; median = [same]; StdDev = 2166 Pa = 2.3% (rounded from 2.2996%) of mean or 2.3% (rounded from 2.2813%) of median
Analysis of overall data set:
- Average of averages:
- - ""Straight-up"" comparison without weighting for difference in number of data points: 105,745 Pa if high and low values in second set kept; 104,645 Pa if high and low values in second set rejected
- - Weighted for number of data points in each set: (4/9)115,100 + (5/9)96,390 = 103,193.636 Pa if high and low values in second set kept; 103,483+(1/3) Pa if high and low values in second set rejected (note that the decrease of the average value within the second subset is more than offset by the decrease in its weighting relative to the smaller subset with higher average value)
- Comparison vs. generally accepted actual values:
- - (101,325 Pa - [103,193.636 or 103,483.3repeating]) / 101,325 Pa = approx. -1.84% with outliers in second subset or -2.13% without them
- Discussion of relationship between excess pressure in Pa and pressure in atmospheres:
- - As discussed above, the ratio of old volume to new volume is equal to the ratio of new pressure to atmospheric pressure, i.e., the value of the new pressure as expressed in atmospheres.
- - Therefore, a plot with vertical column pressure in excess of atmospheric pressure (in Pa) on the x-axis and the ""old volume / new volume"" ratio (= new pressure in atmospheres) on the y-axis will show how much each additional increment of pressure as expressed in Pa changes the expression of that pressure in atmospheres.
- - Therefore, the first derivative of the best-fit curve for this function (equal to the slope of the line of best fit when, as here, the relationship is linear, i.e., does not change with the value of pressure in Pa) shows a) how many Pa it takes to raise pressure by one atmosphere, b) how many Pa are in one atmosphere, and c) what the pressure in Pa is of our atmosphere under the conditions measured (here, i] 23.5 to 24 degC and ii] above sea level by a negligible amount: Our apartment thermostat is inconveniently placed in the coldest part of the first floor, with the effect that it's always too hot everywhere else, especially upstairs; Portsmouth's elevation is listed in Wikipedia as 20 feet, and adding another 10 for the second floor of my building still puts it at sea level for practical purposes.).
----Bad data from first attempt at redo, in case you're really interested---
TABLE 2.0: Water level = -4.0 cm; starting horizontal column length was erroneously assumed to be 15.7 cm each time and not rechecked; chronological order of measurements was (1.040 m, 0.240 m, 0.340 m, 0.440 m); drop in pressure suggests leak in seal (likely shaped so that outflow exceeds any backflow)
(Vertical column height [m], vertical column pressure in excess of atmospheric [Pa], initial horizontal column length (m), final horizontal column length (m), value for atmospheric pressure [Pa])
0.240, 2354.400, 0.157, 0.1535, 103257.257
0.340, 3335.400, 0.157[?], 0.151, 83940.900
0.440, 4316.400, 0.157[??], 0.148, 70980.800
1.040, 10202.400, 0.157[???], 0.1415, 123279.000
@&
Very well done.
A best-fit line for your data, without omitting any of the points, indicates a slope of 96 500 Pa / atmosphere.
The first point isn't really an outlier; on a graph of pressure in Pascals vs. pressure in atmospheres it fits the pattern of the other points fine. A calculation of atmospheric pressure based on that point will be subject to a great deal of uncertainty, due to the small differences being observed.
For this reason we wouldn't base our measurement of atmospheric pressure on the values calculated for the individual points, but rather on the graph.
The third point does appear to be an outlier. When it is omitted the best-fit line remains quite close to the other points, but the slope decreases to about 90 000 Pa / atmosphere.
Even at that your results are a good bit closer than my best attempts.
*@