Modern Engineering Thermodynamics

622 CHAPTER 15: Chemical Thermodynamics
and dividing through by the fuel molar flow rate _n fuel gives _S P
r /_n fuel = ðS P Þ r
/n fuel = ðs P Þ r
, where ðs P Þ r
is the specific
entropy production per unit mole of fuel consumed:
ðs P
Þ r
= ∑ðn i /n fuel Þs i −∑ðn i /n fuel Þs i − q r /T b (15.20)
P
In the case of a closed system with an isothermal boundary, the entropy balance equation gives the total entropy
production of the reaction as (see Eq. (7.77))
ðS P Þ r
= ðS 2 − S 1 Þ r
− 1 ðQ r Þ 2 /T b
R
= n P s P − n R s R − ðQ r Þ/T b
where the reactants and products are assumed to be mixed at the beginning and end of the reaction, respectively.
Again assuming the reactants and the products to be ideal gases and dividing through by the number of
moles of fuel present in the reactants gives
ðS P Þ r
/n fuel = ðs P
Þ r
= ∑ðn i /n fuel Þs i −∑ðn i /n fuel Þs i − q r /T b
P
which is identical to Eq. (15.20). This is as it should be, since the specific entropy production per unit mole of
fuel consumed should depend on only the reaction itself and not the analysis frame (i.e., open or closed) used
to determine it.
As in the case of enthalpy discussed earlier, we need a common zero point reference state from which to measure
the entropies of all of the components in the reaction. Enthalpy and internal energy have no physically
well-defined absolute zero values. Even at absolute zero temperature, it can be shown that the enthalpy and
internal energy are not generally zero. Entropy, on the other hand, does have an absolute zero point dictated by
a state of absolutely perfect molecular order. This is the postulate called the third law of thermodynamics.
The primary value of the third law of thermodynamics as far as we are concerned is that it gives us a reference
state from which we can construct an absolute entropy scale. This means that we now have three thermodynamic
properties with well-defined absolute zero value states: pressure, temperature, and entropy. We assume
that all the substances we deal with have ordered crystalline, rather than amorphous solid, phases at absolute
zero temperature. Therefore, we can compute the absolute molar specific entropy of an incompressible substance
at any pressure and temperature from Eq. (7.32) in Chapter 7 as
sp, ð TÞabs
incompressible =
substance
R
Z T
°
cdT/T ð Þ (15.21)
where c is the molar specific heat and T is in absolute temperature units (K or R). Similarly, the absolute molar
specific entropy in SI units of an ideal gas at any pressure and temperature can be determined from Eq. (7.35) in
Chapter 7 as
Z
T
sp, ð TÞabs
ideal gas = c P ðdT/TÞ− R ln ðp/0:1 MPaÞ (15.22)
°
where T is in K and p is in MPa. Note that both c and c p ! 0asT ! 0 therefore, Eqs. (15.21) and (15.22) cannot
be integrated by assuming constant specific heats in the temperature range from 0 to T.
WHAT IS THE THIRD LAW OF THERMODYNAMICS?
Unlike the first two laws of thermodynamics, the third law is not a statement about conservation or production. It was
developed from quantum statistical mechanics theories in 1906 by Walther Hermann Nernst (1864–1941), for which he
won the 1920 Nobel Prize in Chemistry. Basically, it states that the entropy of a perfect crystalline substance vanishes at
absolute zero temperature and is independent of the pressure at that point. That is,
lim ðSÞ perfect = ∂S
T
= 0
T!0
crystal ∂p =0
Therefore, if we choose absolute zero temperature and any convenient pressure as a reference state for an entropy scale, we
have produced an absolute entropy scale (i.e., one with an absolute zero point). A pressure of 0.1 MPa (about 1 atm) is
usually chosen for the reference state pressure. Consequently, we construct an absolute entropy scale from the point S = 0
at 0.1 MPa and 0 K.