1 (447)

Prévia do material em texto

SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 439
Next, consider the di�erential of term B
∂∑ j [(N j + 1
2 ) lnN j]
∂N i
=∑
j
[(
∂(N j + 1
2 )
∂N i
) lnN j + (N j + 1
2 )(
∂ lnN j
∂N i
)]
=∑
j
[(
∂(N j + 1
2 )
∂N i
) lnN j + (N j + 1
2 )
1
N j
(
∂N j
∂N i
)]
Only the derivatives with j = i are non-zero to give
= lnN i +
N i + 1
2
N i
Bringing the terms together gives the expression for the derivative
∂ lnW
∂N i
= lnN +
N + 1
2
N
− lnN i −
N i + 1
2
N i
For a macroscopic sample N ≫ 1 and N i ≫ 1, so N + 1
2 ≈ N and N i +
1
2 ≈ N i ;
with these approximations
∂ lnW
∂N i
= lnN + 1 − (lnN i + 1)
�is is identical to the expression found inHow is that done? 13A.3 on page 534,
and so it is concluded that for macroscopic samples the use of the full version
of Stirling’s approximation has no e�ect (with the possible exception of states
with very low populations).
13B Partition functions
Answer to discussion questions
D13B.2 It is possible for there to be di�erent wavefunctions which have the same en-
ergy: such wavefunctions are said to be degenerate. If this is the case, for a
given ‘energy level’, that is a given value of the energy, there are several ‘states’
each of which is distinct but has the same energy.
�e partition function is computed as a sum over the states. However, because
degenerate states have the same energy, the summaybe computed as a sumover
energy levels, as long as the degeneracy g i of each level is taken into account.
q = ∑
states i
e−βε i = ∑
levels i
g ie−βε i