Page 118 - Physical Chemistry

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1
1
1
tossed is . Since parts 1 and 2 behave independently of each other, we have Section 3.7
2
2
4
p p p . Equation (3.48) becomes What Is Entropy?
1 2 1 2
h1p p 2 f1p 2 g1p 2 (3.49)
1
2
1 2
Our task is to find the functions that satisfy
h1xy2 f1x2 g1y2 (3.50)
Before reading ahead, you might try to guess a solution for h.
It isn’t hard to prove that the only way to satisfy (3.50) is with logarithmic func-
tions. Problem 14.54 shows that the functions in (3.50) must be
f1x2 k ln x a, g1y2 k ln y b, h1xy2 k ln 1xy2 c (3.51)
where k is a constant and a, b, and c are constants such that c a b. The constant
k must be the same for all systems [otherwise, (3.50) would not be satisfied], but the
additive constant (a, b, or c) differs for different systems.
Since we postulated S f(p) in Eq. (3.47), we have from (3.51) that
S k ln p a (3.52)
where k and a are constants and p is the probability of the system’s thermodynamic
state. Since the second law allows us to calculate only changes in entropy, we cannot
use thermodynamics to find a. We can, however, evaluate k as follows.
Consider again the spontaneous mixing of equal volumes of two different perfect
gases (Fig. 3.12). State 1 is the unmixed state of the middle drawing of Fig. 3.12, and
state 2 is the mixed state. Equation (3.52) gives for the process 1 → 2:
¢S S S k ln p a k ln p a
2
1
2
1
S S k ln 1p >p 2 (3.53)
2
1
1
2
(Don’t confuse the probabilities p and p with pressures.) We want p /p . The proba-
1
2
1
2
1
bility that any particular d molecule is in the left half of the container is . Since the
2
perfect-gas molecules move independently of one another, the probability that every d
molecule is in the left half of the container is the product of the independent proba-
bilities for each d molecule, namely, 1 2 , where N is the number of d molecules.
1 N d
2
d
Likewise, the probability that all the e molecules are in the right half of the container
is 1 2 . Since d and e molecules move independently, the simultaneous probability
1 N e
2
that all d molecules are in the left half of the box and all e molecules are in the right
half is the product of the two separate probabilities, namely,
p 1 2 1 2 1 2 1 2 (3.54)
1 N d N e
1 2N d
1 N d 1 N e
1
2
2
2
2
since N N . (We took equal volumes of d and e at the same T and P.)
e
d
State 2 is the thermodynamic state in which, to within the limits of macroscopic
measurement, the gases d and e are uniformly distributed through the container. As
noted, the probability of any departure from a uniform distribution that is large enough
to be directly detectable is vanishingly small because of the large number of molecules
composing the system. Hence the probability of the final state 2 is only “infinitesi-
mally” less than one and can be taken as one: p 1. Therefore, for the mixing, (3.53)
2
and (3.54) give
S S k ln 11>p 2 k ln 2 2N d 2N k ln 2 (3.55)
d
1
2
1
However, in Sec. 3.4 we used thermodynamics to calculate S for the constant-T-
and-P irreversible mixing of two perfect gases; Eq. (3.33) with mole fractions set
equal to one-half gives
S S 2n R ln 2 (3.56)
2
1
d