Concise Physical Chemistry

c08 JWBS043-Rogers September 13, 2010 11:25 Printer Name: Yet to Come

THE PARTITION FUNCTION OF A SIMPLE SYSTEM 115

equally spaced. Because of this, the sum, written without degeneracy,

Qvib=

∑

i

e−εi/kBT, i= 0 , 1 , 2 ,...

can be represented by

Qvib= 1 +e−ε/kBT+e−^2 ε/kBT+e−^3 ε/kBT+···

whereεis the size of equal steps up an energy ladder. Multiplying both sides of this

equation bye−ε/kBT, we get

e−ε/kBTQvib=e−ε/kBT+e−^2 ε/kBT+e−^3 ε/kBT+···

but this sum is the same as the one above it except that it lacks the first term. The

difference between the two sums is just unity:

Qvib−e−ε/kBTQvib= 1

so

Qvib=

1

1 −e−ε/kBT

An ideal harmonic oscillator has only one vibrational spectral line at frequencyν,

which gives its energy through the Planck equationε=hν, wherehis Planck’s

constant 6. 626 × 10 −^34 J s. Knowing the energyεfor the oscillator, one knowsQvib.

Spectroscopists give the frequency of vibration as ̃νin units of reciprocal cm, cm−^1.

Converting this unit toν,wehaveν=cν ̃wherecis the speed of electromagnetic

radiation, 3. 00 × 108 ms−^1. The formula forQvibin spectrocopist’s terms is

Qvib=

1

1 −e−ε/kBT

=

1

1 −e−hcνβ ̃

=

1

1 −e−a

whereβ= 1 /kBTand constants have been gathered for convenience into

a=

1 .439 ̃ν

T

The diatomic molecule Na 2 (g) at 1000 K, for example, has a strong vibrational

resonance at 159.2 cm−^1. This leads toa= 1 .439(159.2)/ 1000 = 0 .229 and

Qvib=

1

1 −e−^0.^229

= 4. 88