Physical Chemistry , 1st ed.

0.08205

m

L

o

a

l

t



m

K

6.022  1023 /mol k

k

The mole units cancel on both sides of the fraction; for kwe get:

k1.363  10 ^25 La

K

tm

which is a perfectly good value for kif the units are appropriate otherwise.

We have seen that statistical thermodynamics gives the same translational

(that is, internal) energies and pressures that we find from other, phenomeno-

logical perspectives. But values for Aand Gdepend on the entropy of our

gaseous sample. It remains to be seen how (or rather, if!) statistical thermody-

namics’ predictions for Sagree with phenomenological values of entropy.

Using equation 17.43:

SNk T





ln

T

q



V

ln 

N

q

 (^1)
we can make some similar substitutions to get an expression for S. Without go-
ing through the math (which is left to the end-of-chapter exercises), we get
SNk ln 

2 

h

m

2

kT



3/2



k

p

T

 

5

2

 (17.61)

Equation 17.61 is one form of what is called the Sackur-Tetrode equation.It

provides what is probably the best example of how well statistical thermo-

dynamics applies to gaseous systems, because we can measure absolute en-

tropies. The following example illustrates.

Example 17.6

What is the absolute entropy of 1 mole of He at 25.0°C and 1.000 atm pres-

sure? Compare this with the tabulated value of 126.04 J/(molK). Don’t for-

get that proper units are necessary.

Solution

The “proper units” warning was to remind us that masses should be expressed

in units of kg, and that volumes should be expressed in m^3. For He, the mass

m6.65  10 ^27 kg (see Example 17.4), and 1 m^3 1000 L. From equa-

tion 17.61, we have

S(6.022  1023 /mol)1.381  10 ^23 

K

J



ln


3/2



1

1

00

m

0

3

L



5

2



1.381 ^10

 (^23) 
K

J

)(298 K



(1.000 atm)

1

L

0



1

a

.3

tm

2J



2 (6.65  10 ^27 kg)(1.381  10 ^23 

K

J

)(298 K)



(6.626  10 ^34 Js)^2

0.08205

m

L

o

a

l

t



m

K





6.022  1023 /mol

610 CHAPTER 17 Statistical Thermodynamics: Introduction