Page 117 - Physical Chemistry

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Chapter 3
The Second Law of Thermodynamics System
Partition proceeds
removed to
Figure 3.12 equilibrium
Irreversible mixing of perfect
gases at constant T and P. 1 2

what really determines the equilibrium position of an isolated thermodynamic system?
To answer this, consider a simple example, the mixing at constant temperature and
pressure of equal volumes of two different inert perfect gases d and e in an isolated sys-
tem (Fig. 3.12). The motion of the gas molecules is completely random, and the mole-
cules do not interact with one another. What then makes 2 in Fig. 3.12 the equilibrium
state and 1 a nonequilibrium state? Why is the passage from the unmixed state 1 to the
mixed state 2 irreversible? (From 2, an isolated system will never go back to 1.)
Clearly the answer is probability. If the molecules move at random, any d mole-
cule has a 50% chance of being in the left half of the container. The probability that
all the d molecules will be in the left half and all the e molecules in the right half
(state 1) is extremely small. The most probable distribution has d and e molecules each
equally distributed between the two halves of the container (state 2). An analogy to the
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spatial distribution of 1 mole of d molecules would be tossing a coin 6 10 times.
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The chance of getting 6 10 heads is extremely tiny. The most probable outcome
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is 3 10 heads and 3 10 tails, and only outcomes with a very nearly equal ratio
of heads to tails have significant probabilities. The probability maximum is extremely
sharply peaked at 50% heads. (For example, Fig. 3.13 shows the probabilities for ob-
taining various numbers of heads for 10 tosses of a coin and for 100 tosses. As the
number of tosses increases, the probability of significant deviations from 50% heads
diminishes.) Similarly, any spatial distribution of the d molecules that differs signifi-
cantly from 50% d in each container has an extremely small probability because of the
large number of d molecules; similarly for the e molecules.
It seems clear that the equilibrium thermodynamic state of an isolated system is
the most probable state. The increase in S as an isolated system proceeds toward equi-
librium is directly related to the system’s going from a state of low probability to one
of high probability. We therefore postulate that the entropy S of a system is a function
of the probability p of the system’s thermodynamic state:
S f 1p2 (3.47)
Amazingly, use of the single fact that entropy is an extensive state function allows
us to find the function f in our postulate (3.47). To do this, we consider a system com-
posed of two independent, noninteracting parts, 1 and 2, separated by a rigid, imper-
meable, adiabatic wall that prevents flow of heat, work, and matter between them.
Entropy is an extensive property, so the entropy of the composite system 1 2 is
S 1 2 S S , where S and S are the entropies of parts 1 and 2. Substitution of
1
2
1
2
(3.47) into this equation gives
h1p 1 2 2 f1p 2 g1p 2 (3.48)
1
2
where f, g, and h are three functions. Since systems 1, 2, and 1 2 are not identical,
the functions f, g, and h are not necessarily identical. What is the relation between the
Figure 3.13 probability p 1 2 of the composite system’s thermodynamic state and the probabilities
p and p of the states of parts 1 and 2? The probability that two independent events
2
1
Probabilities for various numbers
of heads when a coin is tossed will both happen is shown in probability theory to be the product of the probabilities
10 times and 100 times. for each event. For example, the probability of getting two heads when two coins are