introd to chem eng thermodynamics

1.1 THE SCOPE OF THERMODYNAMICS
The word thermodynamics means heat power, or power developed from heat,
rellecting its origin in the analysis of steam engines. As a fully developed modem
science, thermodynamics deals with transformations of energy of all kinds from
one form to another. The general restrictions within which all such transformations
are observed to occur are known as the first and second laws of thermodynamics.
These laws cannot be proved in the mathematical sense. Rather, their validity
rests upon experience.
Given mathematical expression, these laws lead to a network of equations
from which a wide range of practical results and conclusions can be deduced.
The universal applicability of this science is shown by the fact that it is employed
alike by physicists, chemists, and engineers. The basic principles are always the
same, but the applications differ. The chemical engineer must be able to cope
with a wide variety of problems. Among the most important are the determination
of heat and work requirements for physical and chemical processes, and the
determination of equilibrium conditions for chemical reactions and for the
transfer of chemical species between phases.
TheQl odynamic considerations by themselves are not sufficient to allow
calculation of the rates of chemical or physical processes. Rates depend on both
driving force and resistance. Although driving forces are thermodynamic variables,
resistances are not. Neither can thermodynamics, a macroscopic-property
formulation, reveal the microscopic (molecular) mechanisms of physical or
chemical processes. On the other hand, knowledge of the microscopic behavior
of matter can be useful in the calculation of thermodynamic properties. Such
property values are essential to the practical application of thermodynamics;
numerical results of thermodynamic analysis are accurate only to the extent that
the required data are accurate. The chemical engineer must deal with many
chemical species and their mixtures, and experimental data are often unavailable.
Thus one must make effective use of correlations developed from a limited data
base, but generalized to provide estimates in the absence of data.
The application of thermodynamics to any real problem starts with the
identification of a particular body of matter as the focus of attention. This quantity
of matter is called the system, and its thermodynamic state is defined by a few
measurable macrosCopic properties. These depend on the fundamental
dimensions of science, of which length, time, mass, temperature, and amount of
substance are of interest here.
1.2 DIMENSIONS AND UNITS
The fundamental dimensions are primitives, recognized through our sensory
perceptions and not definable in terms of anything simpler. Their use, however,
requires the definition of arbitrary scales of measure, divided into specific units
of size. Primary units have been set by international agreement, and are codified
as the International System of Units (abbreviated SI, for Systeme International).
The second, symbol s, is the SI unit of time, defined as the duration of
9,192,631,770 cycles of radiation associated with a specified transition of the
cesium atom. The meter, symbol m, is the fundamental unit of length, defined
as the distance light travels in a vacuum during 1/299,792,458 of a second. The
kilogram, symbol kg, is the mass of a platinum/iridium cylinder kept at the
International Bureau of Weights and Measures at Sevres, France. The unit of
temperature is the kelvin, symbol K, equal to 1/273.16 of the thermodynamic
temperature of the triple point of water. A more detailed discussion of temperature,
the characteristic dimension of thermodynamics, is given in Sec. 1.4.
The measure of the amount of substance is the mole, symbol mol, defined as the
amount of substance represented by as many elementary entities (e.g., molecules)

1.3 FORCE
The SI unit of force is the newton, symbol N, derived from Newton s second law,
which expresses force F as the product of mass m and acceleration a:
F=ma
The newton is defined as the force which when applied to a mass of I kg produce,
an acceleration of I m s -2; thus the newton is a derived unit representin~
I kgms-2.
In the English engineering system of units, force is treated as an additional
independent dimension along with length, time, and mass. The pound force (Ib,:
is defined as that force which accelerates I pound mass 32.1740 feet per second
per second. Newton s law must here include a dimensional proportionalit)
constant if it is to be reconciled with this definition. Thus, we write
whencet
and
I
F=-ma
go
1(lb,} =.!. x 1(lbm} x 32.1740(ft)(s}-2
go
go = 32.1740(lbm)(ft)(lb,}- (s}-2
The pound force is equivalent to 4.4482216 N.
Since force and mass are different concepts, a pound force and a pound ma"
are different quantities, and their units cannot be cancelled against one another
When an equation contains both units, (Ib,) and (Ibm), the dimensional constanl
go must-also appear in the equation to make it dimensionally correct.
Weight properly refers to the force of gravity on a body, and is therefon
correctly expressed in newtons or in pounds force. Unfortunately, standards
mass are often called "weights," and the use of a balance to compare masses is
called "weighing." Thus, one must discern from the context whether force or
mass is meant when the word "weight" is used in a casual or informal way.
Example 1.1 An astrQnaut weighs 730 N in Houston, Texas, where the local acceleration
of gravity is 9 = 9.792 m S-2. What is the mass of the astronaut, and what does
he weigh on the moon, where 9 = 1.67 m s-21
SOLUTION Letting a = g, we write Newton s law as
F=mg
whence
F 730N
m = - = = 74.55 N m-I S
9 9.792ms
Since the newton N has the units kgms-2, this result simplifies to
m = 74.55 kg
This mass of the astronaut is independent of loc~tion, but his weight depends on the
local acceleration of gravity. Thus on the moon his weight is
Fmoon = mOmoon = 74.55 kg x 1.67 m S-2
or
Fmoon = 124.5 kg m s- = 124.5 N
To work this problem in the English epgineering system of units, we convert the
astronaut s weight to (lb,) and the values of 9 to (ft)(s)- . Since 1 N is equivalent to
O.2248090b,) and 1 m to 3.28084(ft), we have:
Weight of astronaut in Houston = 164.1 (lb,)
gHn~ nn = 32.13 and gmoon = 5.48(ft)(s)-
Newton s law here gives
Fg,
m=-
9
or
164.1(lb,) x 32.174O(lbm )(ft)(lb,)- (s)-
32.13(ft)(s) ,
m = 164.3(lbm )
Thus the astronaufs m!1ss in (Ibm) and weight in (lbf ) in Houston
almost the same, but on the moon this is not the case:
F = mgmoon = (164.3)(5.48) 28 O(lb)
moon g, 32.1740 .,