Click here for a clean data sheet.
DATA SHEET:
Offset = | 14.7 – P1 |; add the offset to all
pressures if P1 was below 14.7; otherwise, subtract the offset from
all pressure values.
|
offset = 14.70-14.25 = 0.45
|
error in Tn = LC/4 = 1/4 =0.25 = 0.3 (note correction here.) |
error in P with offset = (1 + 1.41...)*LC/4 = (2.41...)*)0.5/4 = 0.301..= 0.30 |
|
|
T1 =
23.00 |
P1 = 14.25 |
P1 with offset = 14.70 |
|
|
T2 =
40.00 |
P2 = 15.00 |
P2 with offset = 15.45
|
|
|
T3 =
60.00 |
P3 = 16.00 |
P3 with |
|
|
T4 =
80.00 |
P4 = 17.00 |
P4 with |
|
|
T5 =
100.00 |
P5 = 18.00 |
P5 with offset=18.45 |
|
mbest = (18.35 -
14.65)/(100.0 - 20.0)= 0.04625
psi/ oC (see attached graphs); this computation may be discussed
further in
the future. You will receive an email alert indicating if the
above pressure's 1/100 place precision is called into question. You might ask,
"The graphing
precision appears to be better than the precision inherent in the measuring
process, which only recognizes 1/10 precision above---thus, should you really write
(18.35 - 14.65)/(100.0 - 20.0)= 0.04625
psi/ oC? " It turns out you can keep the "extra level"
of precision associated with high graphing paper resolution since it will be
washed out in computations using real data of lower precision--see
the error in T computation below. Furthermore, if you computed your
experimental absolute zero value (near -273 0C) to a very high
degree of graphing precision, you would be comparing your answer to an expected
value of limited precision. The accepted value ( -273.15 0C)
would limit the significant figures you could effectively retain in your
experimental value (near -273 0C) . More on this later.
Δm = (mmax - mmin)/2
(see attached graphs) . Attached graphs. Use same method to compute the
maximum and
minimum slope as you did with the "best" value of m above.
TBest = -P1/mbest + T1. Compute this using correct significant figures. Use mbest , P1 and T1 in the data table above.
Example computation: Using the numbers above from
this overall example problem---P1 with offset = 14.70.
Reason for the 1/10 precision is because the above mentioned
pressure error with offset should be written as 0.3 after
rounding to one
sig. fig. ) Also, mbest = 0.04625
numerically. Note: T1 = 23.00 if the above mentioned
temperature error (error in Tn ) is rounded to one sig. fig.
With these provisos, here is the computation, using consistency in sig. figs.
TBest = -P1/mbest +
T1. = -(14.70/0.04625)
+ 23.00 = -317.83
+ 23.00 = -294.8378. See below
for the computation of the percent error which you will use to compare with the
error in T.
Error in T = delta T1 + (P1best/mbest)*(delta
m/mbest + delta P1/P1best ) =
0.3 + (14.70/0.04625)*(delta m/mbest + delta P1/P1best
).
Now, delta m can be obtained by taking the difference between the maximum and
minimum slopes, then dividing by 2. Looking at the sample graphs handed out in
classes, we get:
mmax = (18.70 - 14.275)/(100.0 - 20.0) = 0.0553125.
mmin = (18.075 - 14.95)/(100.0 -
20.0) = 0.0390625.
delta m = Δm = (mmax - mmin)/2
= (0.0553125 - 0.0390625)/2 = 0.01625/2 =
0.008125.
Thus: Error in T = 0.3 + (14.70/0.04625)*(0.008125/0 .04625
+ 0.3/14.70) =
0.3 + (317.8378)*( 0.175675 + 0.020408) = 0.3
+ 62.3215 = 62.6225 = 63 degrees Celsius
Percent error = | Tbest - (-273) |*100%/273 =
| (-294.8378 + 273.000....)*100%/273 |
(21.8378)*100 %/273.000... = 7.999 % = 8.0 %.
Does the the accepted value (-273 oC) fall within
the range of final T values given by
TBest + (error in T ) and TBest - (error in T )?
Check: | Tbest - (-273) |< error in T ?
|-294.8378 + 273.00000..| < 63 ?
21.8378 < 63?
Yes !
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