Physical Chemistry Third Edition

1062 25 Equilibrium Statistical Mechanics. I. The Probability Distribution for Molecular States

The probability distribution of Eq. (25.3-23) is the same as theBoltzmann probability

distributionof Eq. (22.5-2). It has the following general characteristics: (1) At a finite

temperature the equilibrium probability of a molecular state decreases exponentially

as a function of the molecular energy. States with energy much larger thankBTare

not significantly populated. (2) In the limit of zero temperature, only the lowest-energy

level is occupied. (3) If the temperature is increased, the probabilities of states of high

energy increase. (4) In the limit of infinite temperature, all molecular states are equally

populated.

EXAMPLE25.5

Consider two states with an energy difference of 1.00 eV. Find the ratio of their probabilities

at (a) 298.15 K; (b) 1000.0 K; and (c) 10000.0 K.

Solution

a.ratioexp

(

−

(1.00 eV)(1. 6022 × 10 −^19 JEv−^1 )

(1. 3807 × 10 −^23 JK−^1 )(298.15 K)

)

 1. 25 × 10 −^17

b.ratioexp

(

−

(1.00 eV)(1. 6022 × 10 −^19 JEv−^1 )

(1. 3807 × 10 −^23 JK−^1 )(1000.0K)

)

 9. 13 × 10 −^6

c.ratioexp

(

−

(1.00 eV)(1. 6022 × 10 −^19 JEv−^1 )

(1. 3807 × 10 −^23 JK−^1 )(10000.0K)

)

 0. 313

Our probability distribution is valid for equilibrium macroscopic states. Nonequi-

librium system states can occur in which a molecular state of high energy has a higher

population than a state of low energy. This situation is achieved in lasers. A distribution

with a larger population for a state of higher energy is said to correspond to a negative

temperature. Smaller values of the reciprocal temperature correspond to larger values of

the temperature. As a positive temperature approaches very large values, the reciprocal

temperature approaches zero. If the reciprocal temperature can pass through zero and

become negative, this must correspond to being “hotter” than any positive reciprocal

temperature. Negative temperatures pertain to nonequilibrium systems and cannot be

identified with the thermodynamic temperature, which is an equilibrium concept and

must be positive.

Exercise 25.15

In a certain laser, the population of a state of energy 2.08 eV is 50% larger than the population

of a state of energy zero. Find the negative temperature corresponding to this situation.

We can rewrite the expression for the thermodynamic energy shown in Eq. (25.3-7)

in terms ofT, using the chain rule (see Appendix B):

U−N

(

∂ln(z)

∂T

)

V

(

∂T

∂β

)

NkBT^2

(

∂ln(z)

∂T

)

V

(any dilute gas) (25.3-26)