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  offset = 16.45
T4   =   80.00		P4  = 17.00	P4 with  offset = 17.45
T5   =  100.00		P5  = 18.00	P5 with offset=18.45

 

m_best    =  (18.35 - 14.65)/(100.0 - 20.0)=  0.04625    psi/ ^oC (see attached graphs)

Δm  = (m_max - m_min)/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. 

T_Best =  -P₁/m_best + T₁. Compute this using correct significant figures. Use m_best , 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,  m_best = 0.04625  numerically.  Note: T₁ = 23.00 if the above mentioned temperature error (error in T_n ) is rounded to one sig. fig.  With these provisos, here is the computation, using consistency in sig. figs.

T_Best =  -P₁/m_best + 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 + (P_1best/m_best)*(delta m/m_best    + delta P₁/P_1best ) = 0.3 + (14.70/0.04625)*(delta m/m_best    + delta P₁/P_1best  ).
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:
m_max = (18.70 - 14.275)/(100.0 - 20.0) = 0.0553125.

m_min   =  (18.075 - 14.95)/(100.0 - 20.0) = 0.0390625.

delta m = Δm  = (m_max - m_min)/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  = | T_best - (-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 
T_Best + (error in T ) and T_Best - (error in T )?

Check: | T_best - (-273) |< error in T ?
|-294.8378 + 273.00000..| < 63 ?
21.8378 < 63?
Yes !
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 Show work  here for above computations