Physical Chemistry Third Edition

26.2 Working Equations for the Thermodynamic Functions of a Dilute Gas 1099

Solution

The vibrational contribution is independent of the pressure. Letxihνi/kBT, so that at

373.15 K,x 1  14 .079,x 2  6 .1498, andx 3  14 .482.

Uvib

∑^3

i 1

Nhνi

ehνi/kBT− 1

Nh

∑^3

i 1

νi

exi− 1

(1.000 mol)(6. 022 × 1023 mol−^1 )(6. 6261 × 10 −^34 Js)

×

[

1. 0947 × 1014 s−^1

e^14.^079 − 1

+

4. 7817 × 1013 s−^1

e^6.^1498 − 1

+

1. 1260 × 1014 s−^1

e^14.^482 − 1

]

 40 .86 J

AvibNkBTln

(

zvib

)

nRTln

[(

1

1 −e−^14.^079

)(

1

1 −e−^6.^1498

)(

1

1 −e−^14.^482

)]

(1.000 mol)(8.3145 J K−^1 mol−^1 )(373.15 K)(2. 137 × 10 −^3 )

 6 .632 J

Svib

U−A

T



40 .86 J− 6 .63 J

373 .15 K

 0 .0917 J K−^1

In this example, vibrational frequencies have been used. Most tables of values give

frequencies divided by the speed of light, usually expressed in cm−^1 and denoted

byν ̃. Multiply these values by the speed of light to get the frequencies.

The Chemical Potential of a Dilute Gas

The four contributions to the chemical potential are given by the formulas

μtr−kBTln(ztr/N) (26.2-17a)

μrot−kBTln(zrot) (26.2-17b)

μvib−kBTln(zvib) (26.2-17c)

μel−kBTln(zel) (26.2-17d)

As usual, the divisorNis placed with the translational contribution. The formulas for

the Gibbs energy can be obtained by multiplying those for the chemical potential by

N, the number of molecules in the system.