Modern Engineering Thermodynamics

Summary 51
change in the mass of X due to the reaction”: m GX = m PX = δm X . Then, for a chemical reaction in a closed system, the mass
balance for each of the chemical species present can be written as
m GA = m PA = δm A
m GB = m PB = δm B
m GC = m PC = δm C
m GD = m PD = δm D
If we add these equations together, we get
δm A + δm B + δm C + δm D = m PA + m PB + m PC + m PD
Now, since total mass must be conserved, it follows that
∑m P = m PA + m PB + m PC + m PD = 0
then the previous equation can be written as
δm A + δm B = −δm C − δm D
If we now convert this equation into a molar equation by dividing each mass term by its corresponding species molecular
mass, then this equation becomes the stoichiometric chemical reaction equation n a A + n b B ! n c C + n d D, where
n a = δm A
;
M A
n c = − δm C
M C
n b = δm B
; n d = − δm D
M B M D
and M A , M B , M C ,andM D are the molecular masses of chemical species A, B, C, andD. Consequently, all stoichiometric
chemical reaction equations are just molar mass balances that utilize the conservation of mass law.
SUMMARY 7
In this chapter, we provide the definitions of many of the concepts necessary in the development of thermodynamics.
First, we review the phases of matter and define thermodynamic equilibrium. Then we see that one of the fundamental
elements underlying modern technology is the thermodynamic processes that systems undergo. Next we reexamine and
expand the pressure and temperature scales information given in Chapter 1. The zeroth law of thermodynamics is found
to provide an axiomatic way to develop temperature measurement technology, and the continuum hypothesis is found
to be a useful and valid analysis tool so long as the system’s dimensions are very large compared to those of the
molecules it contains. The balance and conservation concepts round out the chapter with an important discussion of
the form of all the equations used to represent the basic thermodynamic laws in this book.
Some of the more important equations introduced in this chapter are as follows.
1. The equations for the conversion of temperature units, Eqs. (2.1)–(2.4):
Tð°FÞ = 9 Tð°CÞ + 32 = TðRÞ − 459:67
5
Tð°CÞ = 5 ½Tð°FÞ − 32Š = TðKÞ − 273:15
9
TðRÞ = 9 TðKÞ = 1:8TðKÞ = Tð°FÞ + 459:67
5
TðKÞ = 5 TðRÞ
TðRÞ = = Tð°CÞ + 273:15
9 1:8
2. The equation for absolute to gauge pressure conversion:
Absolute pressure = Gauge pressure + Local atmospheric pressure
3. The general balance equations, (2.11) and (2.12), are
7 Well, what do you think of this course so far? Isn’t this a fascinating subject? I know, I know, it is a little vague right now, but it gets better.