Solution Manual — Fundamentals of Chemical Engineering Thermodynamics, 1st Edition (Dahm & Visco, 2014), Chapters 1-15 | All Chapters Covered

SOLUTIONS MANUAL
Fundamentals of Chemical Engineering
Thermodynamics
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Kevin D. Dahm and Donald P. Visco, Jr.
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1st Edition
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, TABLE OF CONTENTS
Solution Manual: Fundamentals of Chemical Engineering Thermodynamics, 1st
Edition
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By Kevin D. Dahm and Donald P. Visco
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Chapter Title

Chapter 1 Introduction

Chapter 2 The Physical Properties of Pure Compounds
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Chapter 3 Material and Energy Balances

Chapter 4 Entropy
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Chapter 5 Thermodynamic Processes and Cycles

Chapter 6 Thermodynamic Models of Real, Pure Compounds
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Chapter 7 Equations of State

Chapter 8 Modeling Phase Equilibrium for Pure Compounds
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Chapter 9 An Introduction to Mixtures

Chapter 10 Vapor-Liquid Equilibrium
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Chapter 11 Theories and Models for Vapor–Liquid Equilibrium of Mixtures: Modified
Raoult’s Law Approaches

Chapter 12 Theories and Models for Vapor–Liquid Equilibrium of Mixtures: Using
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Equations of State

Chapter 13 Liquid-Liquid, Vapor-Liquid-Liquid, and Solid-Liquid Equilibrium

Chapter 14 Fundamentals of Chemical Reaction Equilibrium
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Chapter 15 Synthesis of Thermodynamic Principles




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, Chapter 1: Introduction
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1-16) The value g = 9.81 m/s2 is specific to the force of gravity on the surface of the
earth. The universal formula for the force of gravitational attraction is:
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�1�2
�= G
�2
Where m1 and m2 are the masses of the two objects, r is the distance between the
centers of the two objects, and G is the universal gravitation constant,
G = 6.674 × 10-11 N(m/kg)2.
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A) Research the diameters and masses of the Earth and Jupiter.
B) Demonstrate that F = m(9.81 m/s2) is a valid relationship on the surface of the
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earth.
C) Determine the force of gravity acting on a 1000 kg satellite that is 2000 miles
above the surface of the Earth.
D) One of the authors of this book has a mass of 200 lbm. If he was on the surface
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of Jupiter, what gravitational force in lbf would be acting on him?

Solution:
A) Measurements obtained from different sources will vary slightly.
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DEarth~ 12,742 km DJupiter~ 142,000 km
Massearth= 5.97 × 1024 kg Massjupiter= 1.90 × 1027 kg
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B) Massearth= 5.97 × 1024 kg RadiusEarth= 6.371 × 106 meters
� � Nm2 (5.97×1024kg) kg m
( )
�= G 1 2
 � = mobject (6.674 × 10−11 ) ( sec2
)
�2 kg2 (6.37×106 m)2 (1 N)
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m
 F = m(9. 81 )
sec2



C) 2000 miles = 3218.68 km = 3218680 m
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2
� = 1000kg (6.674 × 10−11 Nm (5.97×1024kg)
) (6371000�+3218680� )2 = 4348 N
kg2




1

, Chapter 1: Introduction


D) RadiusJupiter= 66854000 m MassJupiter= 1.898 × 1027 kg 200lbm = 90.7 kg


Nm2 (1.90 × 1027kg) ( kg m )
sec2
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−11
� = 90.7 kg (6.674 × 10 ) ( ) = 2281 N
kg2 (7.10 × 107 m)2 (1 N)
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1-17) A gas at T=300 K and P=1 bar is contained in a rigid, rectangular vessel that is 2
meters long, 1 meter wide and 1 meter deep. How much force does the gas exert on
the walls of the container?

Solution:
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1 Bar = 100,000 Pa

AreaSurface = (2 × W × H) + (2 × W × L) + (2 × H × L)
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Force = Pressure × Area
N
( 2)
Force = (100000Pa)(2 × 1m × 1m + 2 × 1m × 2m + 2 × 1m × 2m) ( m )
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Pa
Force = 1 × 106N
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1-18) A car weighs 3000 lbm, and is travelling 60 mph when it has to make an emergency
stop. The car comes to a stop 5 seconds after the brakes are applied.
A) Assuming the rate of deceleration is constant, what force is required?
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B) Assuming the rate of deceleration is constant, how much distance is covered
before the car comes to a stop?

Solution:
5280ft ft
A) Force = mass × acceleration 60mph ( 1hr
)( ) = 88
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3600sec 1mile sec

velocityfinal − velocityinitial
Acceleration =
time
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ft
88 sec − 0
a=
5 sec
ft
a = 17.6
sec2




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