Page 113 - Physical chemistry eng

90 CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics

You will show in the end-of-chapter problems that V and V can be eliminated from
c
d
the set of Equations (5.5) to yield
V b
w cycle =-nR(T hot - T cold ) ln 6 0 (5.7)
V a
Because ¢U a:b = 0 , the heat withdrawn from the hot reservoir is
V b
q ab =-w ab = nRT hot ln (5.8)
V a
and the efficiency of the reversible Carnot heat engine with an ideal gas as the working
substance can be expressed solely in terms of the reservoir temperatures.
ƒ w cycle ƒ T - T T
e = = hot cold = 1 - cold 6 1 (5.9)
q ab T hot T hot
The efficiency of this reversible heat engine can approach one only as T hot : q
or T cold : 0 , neither of which can be accomplished in practice. Therefore, heat can
never be totally converted to work in a reversible cyclic process. Because w cycle for an
engine operating in an irreversible cycle is less than the work attainable in a reversible
cycle, e irreversible 6 e reversible 6 . 1

EXAMPLE PROBLEM 5.1
Calculate the maximum work that can be done by a reversible heat engine operating
between 500. and 200. K if 1000. J is absorbed at 500. K.

Solution
The fraction of the heat that can be converted to work is the same as the fractional fall
in the absolute temperature. This is a convenient way to link the efficiency of an
engine with the properties of the absolute temperature.
T cold 200. K
e = 1 - = 1 - = 0.600
T hot 500. K
w = eq ab = 0.600 * 1000. J = 600. J

In this section, only the most important features of heat engines have been discussed.
It can also be shown that the efficiency of a reversible heat engine is independent of the
working substance. For a more in-depth discussion of heat engines, the interested reader
is referred to Heat and Thermodynamics, seventh edition, by M. W. Zemansky and
R. H. Dittman (McGraw-Hill, 1997). We will return to a discussion of the efficiency of
engines when we discuss refrigerators, heat pumps, and real engines in Section 5.11.

5.3 Introducing Entropy

Equating the two formulas for the efficiency of the reversible heat engine given in
Equations (5.4) and (5.9),
T hot - T cold q ab + q cd or q ab q cd
= + = 0 (5.10)
T hot q ab T hot T cold

The last expression in Equation (5.10) is the sum of the quantity q reversible > T around the
Carnot cycle. This result can be generalized to any reversible cycle made up of any
number of segments to give the important result stated in Equation (5.11):
dq reversible
= 0 (5.11)
T
C