Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
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Chapter 1, Problem 1.29P
Interpretation Introduction

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

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Answer to Problem 1.29P

328.98 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:

ln Psat= ABT+C   Psat=exp(ABT+C)

Here Psatis the function of A, B, C and T

F(A,B,C,T)=exp(ABT+C).........(1)

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

AF=Aexp(ABT+C)FA=exp(ABT+C)(10)FA=exp(ABT+C) ........... (2)

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

BF=Bexp(ABT+C)FB=exp(ABT+C)(01T+C)FB=1T+Cexp(ABT+C) ........... (3)d

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

CF=Cexp(ABT+C)FC=exp(ABT+C)(0+B ( T+C) 2)FC=B( T+C)2exp(ABT+C) ........... (4)

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:

Introduction to Chemical Engineering Thermodynamics, Chapter 1, Problem 1.29P , additional homework tip  1

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

F(T,a)=[  exp(a0 a1 T+a2 )  exp(a0 a1 T+a2 )1T+Cexp(a0 a1 T+a2 )a1 ( T+a2 )2exp(a0 a1 T+a2 )]

Here a0 = A, a1 = B and a2 = C

Take the initial guess as =(  15300050)

Now using the genfit function as follows:

(ABC)=genfit(T,Psat,guess,F)

The solution is (ABC)=(  13.42  229069.05)

Thus, the fitted curve is F(A,B,C,T)=exp(13.422290T69.05)

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

Introduction to Chemical Engineering Thermodynamics, Chapter 1, Problem 1.29P , additional homework tip  2

Hence the above graph is fitted with the data graph and thus the correct values of A, B an C are 13.42, 2290and 69.05.

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

TBoiling Point=BAln(Psat)C               = 229013.42ln(101.32)(69.05)               = 328.98 K

Therefore, the normal boiling point is 328.98 K.

Conclusion

328.98 K.

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