Physical Chemistry Third Edition

(C. Jardin) #1

27.1 The Canonical Ensemble 1125


and


k

pk


k

nk

n

 1 A


k

e−βEk (27.1-16)

which is equivalent to

1
A




k

e−βEkZ (definition ofZ) (27.1-17)

Equation (27.1-17) defines the sumZ, which is called thecanonical partition function.
In German, a partition function is called aZustandsumme(sum over states), a name
that better represents the sum and is our motivation for usingZfor its symbol. The
canonical probability distribution can now be written as

pk

1

Z

e−βEk (27.1-18)

The ensemble average energy can be written

〈E〉U

∑∞

k 1

Ekpk

1

Z


k

Eke−βEk (27.1-19)

where we use the symbol〈···〉for an ensemble average value. The sum is over all of
the infinitely many system microstates, but we will omit the limits on our sums. This
equation can be rewritten using a mathematical trick that we applied in Chapter 25:

〈E〉−

1

Z


k

(

∂e−βEk
∂β

)

V,N



1

Z

(


∂β


k

e−βEk

)

V,N

−

1

Z

(

∂Z

∂β

)

V,N

−

(

∂ln(Z)
∂β

)

V,N

(27.1-20)

where we have interchanged the order of summing and differentiating. This is permis-
sible if the sum is uniformly convergent, which we assume to be the case. A sum is
convergent if it approaches more and more closely to a finite value as more and more
terms are taken. It is uniformly convergent if it converges with at least a certain rate
for all cases. We must keepVandNconstant in the differentiation in order to keep the
energy eigenvalues fixed, because the energy eigenvalues depend onVandN.
We can write an equation for the pressure, using some information from thermody-
namics. LetPkbe the pressure of the system if it is in system microstatek. The statistical
entropy was defined in Eq. (26.1-1). If the system is known to be in microstate num-
berk,Ωequals 1 and the statistical entropy is equal to a constant. We assert that if
the statistical entropy is constant we can use Eq. (26.1-8), which applies to constant
thermodynamic entropy:

Pk−

(

∂U

∂V

)

S,N

−

(

∂Ek
∂V

)

N

(system in microstatek) (27.1-21)

where we assume that the energy eigenvalueEkdepends only onVandN. The ensemble
average pressure becomes

〈P〉


k

Pkpk−

1

Z


k

(

∂Ek
∂V

)

N

e−βEk

1

β

(

∂ln(Z)
∂V

)

β,N

(27.1-22)

where we used the mathematical trick applied in Eq. (27.1-20).
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