The predicted normal boiling point of the particular organic chemical should be determined.
Answer to Problem 1.29P
Explanation of Solution
Given:
Laboratory Data for the particular organic chemical.
T (C) | T (K) | Psat |
-18.5 | 254.65 | 3.18 |
-9.5 | 263.65 | 5.48 |
0.2 | 273.35 | 9.45 |
11.8 | 284.95 | 16.9 |
23.1 | 296.25 | 28.2 |
32.7 | 305.85 | 41.9 |
44.4 | 317.55 | 66.6 |
52.1 | 325.25 | 89.5 |
63.3 | 336.45 | 129 |
75.5 | 348.65 | 187 |
Below is the Antoine equation:
Here Psatis the function of A, B, C and T
Now differentiating (1) w.r.t to A
Now differentiating (1) w.r.t to B
Now differentiating (1) w.r.t to B
As the data given:
T (C) | T (K) | Psat |
-18.5 | 254.65 | 3.18 |
-9.5 | 263.65 | 5.48 |
0.2 | 273.35 | 9.45 |
11.8 | 284.95 | 16.9 |
23.1 | 296.25 | 28.2 |
32.7 | 305.85 | 41.9 |
44.4 | 317.55 | 66.6 |
52.1 | 325.25 | 89.5 |
63.3 | 336.45 | 129 |
75.5 | 348.65 | 187 |
Now drawing the graph between Psat vs T as follows:
Using the MADCAD genfit function to fit the above data as follows:
Here a0 = A, a1 = B and a2 = C
Take the initial guess as
Now using the genfit function as follows:
The solution is
Thus, the fitted curve is
Calculating the values of function F as follows:
T (C) | T (K) | Psat | (A-B)/(T+C) | F (A, B, C, T) |
-18.5 | 254.65 | 3.18 | 1.081638 | 2.949506685 |
-9.5 | 263.65 | 5.48 | 1.652271 | 5.218820016 |
0.2 | 273.35 | 9.45 | 2.210994 | 9.124778606 |
11.8 | 284.95 | 16.9 | 2.813238 | 16.6637818 |
23.1 | 296.25 | 28.2 | 3.340775 | 28.24099506 |
32.7 | 305.85 | 41.9 | 3.749392 | 42.49523245 |
44.4 | 317.55 | 66.6 | 4.204708 | 67.00104723 |
52.1 | 325.25 | 89.5 | 4.481671 | 88.38219805 |
63.3 | 336.45 | 129 | 4.856051 | 128.5156723 |
75.5 | 348.65 | 187 | 5.229728 | 186.742037 |
Hence the above graph is fitted with the data graph and thus the correct values of A, B an C are
Now calculating the normal boiling point by substituting Psat = 1 atm
Therefore, the normal boiling point is
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Chapter 1 Solutions
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