Chapter 1, Problem 1.29P

Interpretation Introduction

The predicted normal boiling point of the particular organic chemical should be determined.

Answer to Problem 1.29P

$\boxed{\text{328}\text{.98 K}}$.

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:

$\begin{array}{l}{\text{ln P}}^{\text{sat}}=A-\frac{B}{T+C}\\ P^{sat}=\exp \left(A-\frac{B}{T+C}\right)\end{array}$

Here P^satis the function of A, B, C and T

$F(A,B,C,T)=\exp \left(A-\frac{B}{T+C}\right)\mathrm{.........}\text{(1)}$

Now differentiating (1) w.r.t to A

$\begin{array}{l}\frac{\partial }{\partial A}F=\frac{\partial }{\partial A}\exp \left(A-\frac{B}{T+C}\right)\\ \frac{\partial F}{\partial A}=\exp \left(A-\frac{B}{T+C}\right)\left(1-0\right)\\ \frac{\partial F}{\partial A}=\exp \left(A-\frac{B}{T+C}\right)\mathrm{...........}\text{ (2)}\end{array}$

Now differentiating (1) w.r.t to B

$\begin{array}{l}\frac{\partial }{\partial B}F=\frac{\partial }{\partial B}\exp \left(A-\frac{B}{T+C}\right)\\ \frac{\partial F}{\partial B}=\exp \left(A-\frac{B}{T+C}\right)\left(0-\frac{1}{T+C}\right)\\ \frac{\partial F}{\partial B}=-\frac{1}{T+C}\exp \left(A-\frac{B}{T+C}\right)\mathrm{...........}\text{ (3)}\end{array}$d

Now differentiating (1) w.r.t to B

$\begin{array}{l}\frac{\partial }{\partial C}F=\frac{\partial }{\partial C}\exp \left(A-\frac{B}{T+C}\right)\\ \frac{\partial F}{\partial C}=\exp \left(A-\frac{B}{T+C}\right)\left(0+\frac{B}{{\left(T+C\right)}^2}\right)\\ \frac{\partial F}{\partial C}=\frac{B}{{\left(T+C\right)}^2}\exp \left(A-\frac{B}{T+C}\right)\mathrm{...........}\text{ (4)}\end{array}$

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 P^sat vs T as follows:

Using the MADCAD genfit function to fit the above data as follows:

$F(T,a)=\left[\begin{array}{l}\exp \left(a_0-\frac{a_1}{T+a_2}\right)\\ \exp \left(a_0-\frac{a_1}{T+a_2}\right)\\ -\frac{1}{T+C}\exp \left(a_0-\frac{a_1}{T+a_2}\right)\\ \frac{a_1}{{\left(T+a_2\right)}^2}\exp \left(a_0-\frac{a_1}{T+a_2}\right)\end{array}\right]$

Here a₀ = A, a₁ = B and a₂ = C

Take the initial guess as $=\left(\begin{array}{l}15\\ 3000\\ -50\end{array}\right)$

Now using the genfit function as follows:

$\left(\begin{array}{l}A\\ B\\ C\end{array}\right)=genfit\left(T,P_{sat},guess,F\right)$

The solution is $\left(\begin{array}{l}A\\ B\\ C\end{array}\right)=\left(\begin{array}{l}13.42\\ 2290\\ -69.05\end{array}\right)$

Thus, the fitted curve is $F(A,B,C,T)=\exp \left(13.42-\frac{2290}{T-69.05}\right)$

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 $\boxed{13.42}$, $\boxed{2290}$and $\boxed{-69.05}$.

Now calculating the normal boiling point by substituting P_sat = 1 atm

$\begin{array}{l}{T}_{\text{Boiling Point}}=\frac{B}{A-\ln (P_{sat})}-C\\ \text{ = }\frac{2290}{13.42-\ln (101.32)}-(-69.05)\\ \text{ = }\boxed{328.98K}\end{array}$

Therefore, the normal boiling point is $\boxed{328.98K}$.

Conclusion

$\boxed{\text{328}\text{.98 K}}$.