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Solution Manual For Introduction to Chemical Engineering Thermodynamics 9e J.M. Smith, Mark Swihart Hendrick C. Van Ness, Michael Abbott

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This document provides a complete and well-structured Solution Manual for Introduction to Chemical Engineering Thermodynamics, 9th Edition by J.M. Smith, Mark Swihart, Hendrick C. Van Ness, and Michael Abbott. It includes accurate, step-by-step solutions designed to help students understand core thermodynamics concepts such as phase equilibria, energy balances, equations of state, and chemical processes. The content is organized chapter-by-chapter, making it easy to follow complex calculations and apply theoretical principles to real-world engineering problems. This resource is ideal for assignments, exam preparation, and in-depth revision. Perfect for students in chemical engineering seeking reliable academic support aligned with the latest edition.

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Solution 1.1


Problem Statement



Electric current is the fundamental SI electrical dimension, with the ampere (A) as its unit. Determine units for the

following quantities as combinations of fundamental SI units.


(a) Electric power

(b) Electric charge

(c) Electric potential difference

(d) Electric resistance

(e) Electric capacitance



Solution



(a) Power is power, whether it is electrical, mechanical, or otherwise. Thus, electric power has the usual units of

power:

energy J N m kg m 2
power    
time s s s3

(b) Electric current is by definition the time rate of transfer of electrical charge. Thus

charge
current 
time

or charge  current*time  A s

(you probably recall that the Coulomb is the usual derived unit of charge, defined as 1 A s)



Solution continued on next page…




Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill Education.

,(c) Because power is given by the product of current and electric potential,

energy
power   current*electric potential
time

energy kg m2
or electrical potential  
current*time A s3

(you probably recall that this is defined as the volt)



(d) Because (by Ohm’s Law) current is electric potential divided by resistance,

electrical potential kg m2
or resistance   2 3
current A s

(this is defined as the ohm)



(e) Because electric potential is electric charge divided by electric capacitance,

charge
electrical potential 
electrical capacitance

charge As A 2 s4
or electrical capacitance   
electrical potential kg m 2 kg m2
3
As




Solution 1.2


Problem Statement

Liquid/vapor saturation pressure Psat is often represented as a function of temperature by the Antoine equation, which

can be written in the form:

b
log10 P sat / (torr )  a 
t / C  c




Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill Education.

,Here, parameters a, b, and c are substance-specific constants. Suppose this equation is to be rewritten in the equivalent

form:

B
ln P sat / kPa  A 
T / K C

Show how the parameters in the two equations are related.




Solution


We must convert both the units and the logarithm (between base 10 and natural logarithm). We know that t in

degrees Celsius is equal to T in Kelvins minus 273.15. Also 1 kPa is equal to 7.50 torr (we might have to look up this

conversion factor). So, we have

a b 
 t /C c   B 
P sat / torr  10  7.5 • P sat / kPa  7.5exp  A  
 T / K  C 
Next, we might recognize that 10 can be written as exp(ln(10)) or exp(2.303). That is how we convert from base 10 log
to natural log in general. So,
  b   B 
exp2.303a    7.5exp A  
  T / K  273.15  c   T / K  C 

Here, I have also substituted T 273.15 for t. Taking the natural log of both sides gives

 b  B
2.303a    ln7.5  A 
 T / K  273.15  c  T /K C

For the two functions to be equal for all values of T, each part of the functions must be the same, so we must have

A = 2.303a ln(7.5) or A = ln(10)*a ln(7.5)

B = 2.303b or B = ln(10)*b

C=c 273.15




Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill Education.

, Solution 1.3


Problem Statement



Table B.2 in Appendix B provides parameters for computing the vapor pressure of many substances by the Antoine
sat
equation (see Prob. 1.2). For one of these substances, prepare two plots of P versus T over the range of temperature

sat sat
for which the parameters are valid. One plot should present P on a linear scale and the other should present P on

a log scale.



Problem 1.2


Liquid/vapor saturation pressure Psat is often represented as a function of temperature by the Antoine equation, which

can be written in the form:

b
log10 P sat / (torr )  a 
t / C  c

Here, parameters a, b, and c are substance-specific constants. Suppose this equation is to be rewritten in the equivalent

form:

B
ln P sat / kPa  A 
T / K C

Show how the parameters in the two equations are related.



Solution


The point of this problem is just for you to practice evaluating and plotting a simple function. You will do many

problems over the course of the semester (and many more over the course of your career) in which the results are best

presented in graphical form. The only thing to be careful of in plotting the Antoine equation is to pay attention to the
sat
units of T and P and to whether the constants are given for use with the base 10 logarithm or the natural logarithm.


Solution continued on next page…



Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill Education.

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