Page 119 - Physical chemistry eng

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

Using Entropy to Calculate the Natural
Direction of a Process in an Isolated
5.5 System

To show that ¢S is useful in predicting the direction of spontaneous change, we now
return to the two processes introduced in Section 5.1. The first process concerns the
q P
natural direction of change in a metal rod with a temperature gradient. Will the gradient
become larger or smaller as the system approaches its equilibrium state? To model this
T 1 T 2
process, consider the isolated composite system shown in Figure 5.5. Two systems, in
the form of metal rods with uniform, but different, temperatures T > T , are brought
1
2
FIGURE 5.5
into thermal contact.
Two systems at constant P, each consisting
In the following discussion, heat is withdrawn from the left rod. (The same reason-
of a metal rod, are placed in thermal con-
ing would hold if the direction of heat flow were reversed.) To calculate ¢S for this
tact. The temperatures of the two rods differ
by ¢T . The composite system is contained irreversible process using the heat flow, one must imagine a reversible process in which
in a rigid adiabatic enclosure (not shown) the initial and final states are the same as for the irreversible process. In the imaginary
and is, therefore, an isolated system. reversible process, the rod is coupled to a reservoir whose temperature is lowered very
slowly. The temperatures of the rod and the reservoir differ only infinitesimally
throughout the process in which an amount of heat, q , is withdrawn from the rod. The
P
total change in temperature of the rod, ¢T , is related to q by
P
1 q P
dq = C dT or ¢T = dq = (5.25)
P
P
P
P C P
C L
It has been assumed that ¢T = T - T 1 is small enough that C P is constant over the
2
T ,V i interval.
i
1
T , / V i Because the path is defined (constant pressure), dq P is independent of how rapidly
2
i
1
the heat is withdrawn (the path); it depends only on C P and ¢T . More formally, because
q =¢H and because H is a state function, q P is independent of the path between the
P
Initial state Final state
initial and final states. Therefore, q = q reversible if the temperature increment ¢T is
P
Irreversible process
identical for the reversible and irreversible processes.
(a)
Using this result, the entropy change for this irreversible process in which heat
flows from one rod to the other is calculated. Because the composite system is isolated,
q + q = 0, and q =-q = q P . The entropy change of the composite system is the
1
2
2
1
sum of the entropy changes in each rod
q reversible,1 q reversible,2 q 1 q 2 1 1
¢S = + = + = q a - b (5.26)
P
T 1 T 2 T 1 T 2 T 1 T 2
Because T 1 > T 2 , the quantity in parentheses is negative. This process has two possible
Piston
directions:
Piston • If heat flows from the hotter to the colder rod, the temperature gradient will become
T ,V i
i
1
T , / V i smaller. In this case, q P < 0 and ¢S > 0.
2
i
• If heat flows from the colder to the hotter rod, the temperature gradient will become
larger. In this case, q P > 0 and ¢S < 0.
Initial state Final state
Note that ¢S has the same magnitude, but a different sign, for the two directions of
Reversible process
change. ¢S appears to be a useful function for measuring the direction of natural
(b)
change in an isolated system. Experience tells us that the temperature gradient will
FIGURE 5.6 become less with time. It can be concluded that the process in which S increases is the
(a) An irreversible process is shown in direction of natural change in an isolated system.
which an ideal gas confined in a container Next, consider the second process introduced in Section 5.1 in which an ideal
with rigid adiabatic walls is spontaneously gas spontaneously collapses to half its initial volume without a force acting on it.
reduced to half its initial volume. (b) A This process and its reversible analog are shown in Figure 5.6. Recall that U is
reversible isothermal compression is shown
between the same initial and final states as independent of V for an ideal gas. Because U does not change as V increases, and U
for the irreversible process. Reversibility is is a function of T only for an ideal gas, the temperature remains constant in the irre-
achieved by adjusting the rate at which the versible process. Therefore, the spontaneous irreversible process shown in Figure 5.6a
beaker on top of the piston is filled with is both adiabatic and isothermal and is described by , T : 1/2VV i i i , . The imagi-
T i
water relative to the evaporation rate. nary reversible process that we use to carry out the calculation of ¢S is shown in