1096 26 Equilibrium Statistical Mechanics. II. Statistical Thermodynamics
whereMis the molar mass of the gas. The symbolP◦represents thestandard-state
pressure, which is defined to equal exactly 100,000 Pa1.000 bar. Substituting in
the values of the constants we obtain theSackur–Tetrode equationfor the translational
contribution to the entropy of any dilute gas:SmSm,trR[
9. 2100 +
3
2
ln(
M
1kg mol−^1)
+
5
2
ln(
T
1 K
)
−ln(
P
P◦
)]
(26.2-15)
where the divisors 1 kg mol−^1 and 1 K are inserted to make the arguments of the log-
arithms dimensionless. For a monatomic gas without significant electronic excitation,
the translational contribution is the only contribution to the entropy.EXAMPLE26.7
Calculate the molar entropy of helium gas at 1.000 bar pressure and 298.15 K.
Solution
At this temperature there is no significant electronic excitation and the ground electronic
state is a^1 S 0 term, sozel1 andzztr. The molar mass of helium is equal toM
0 .0040026 kg mol−^1 , so that from the Sackur–Tetrode equationSm◦,tr(8.3145 J K−^1 mol−^1 )(
9. 2100 +
3
2ln(0.0040026)+
5
2ln(298.15)+ln(1)) 126 .23 J K−^1 mol−^1The third-law entropy value of the standard-state entropy of He gas is 126.152 J K−^1 mol−^1.
This degree of agreement lends credence to our identification of the statistical entropy with
the thermodynamic entropy.Exercise 26.9
The value of the energy in the ground state has been assigned the value zero in the calculation
of Example 26.7. Show that the value of the entropy of a dilute gas is unchanged if the zero of
energy is changed.EXAMPLE26.8
Calculate the partition function and the molar energy, enthalpy, and entropy and constant-
pressure heat capacity 1.000 mol of^35 Cl 2 at 298.15 K and the standard-state pressureP◦.
Assume that the uncorrected harmonic oscillator–rigid rotor energy levels can be used.Data: ̃νe 559 .7cm−^1
B ̃e 0 .2440 cm−^1
M 0 .0699375 kg mol−^1
m 1. 16134 × 10 −^25 kg
V 0 .02479 m^3 (the molar volume of an ideal gas atP◦)