Page 88 - Essentials of physical chemistry

50 Essentials of Physical Chemistry

Example
Given the viscosity of CH 4 is 11.2 mPa s (112 micropoise), calculate the effective diameter d, the
mean free path l, and the collision numbers Z 1 and Z 11 at 1 atm and 278C.

s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
r ffiffiffiffiffiffiffiffiffi

8RT 1 mhvi mhvi
; M ¼ 12:0108 þ 4(1:00785) ¼ 16:0422g=mol,
ffiffiffi ffiffiffi
hvi¼ ; h ¼ p ; d ¼ p
pM 2 2pd 2 2 2ph
23
6:022 10 =mol 19 3
¼ 2:4463 10 molecules=cm ,
n* ¼ 3 300 K
22414 cm =mol
273:15 K

s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
7
8(8:314 10 erg= K mol)(300 K) 4

¼ 6:2922 10 cm=s,
p(16:0422 g=mol)
hvi¼
v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

u
16:0422 g=mol 4
u (6:2922 10 cm=s)
23
6:022 10 =mol
t 8 8
¼ 4:104 10 cm ¼ 4:104 A,
ffiffiffi
d ¼ p
2 2p(1:12 10 4 g=cm s)
1 1 6 8
¼ 5:463 10 cm ¼ 546:3 A,
ffiffiffi ffiffiffi
l ¼ p 2 ¼ p 8 2 19 3
2pd n* 2p(4:104 10 cm) (2:4463 10 =cm )
4
6:2922 10 cm=s
¼ 1:1518 10 collisions=s,
hvi 10
l 5:463 10 cm
Z 1 ¼ ¼ 6
19
3
Z 1 n* (1:1518 10 10 collisions=s)(2:4463 10 =cm ) 29 3
¼ 1:4089 10 collisions=cm s:
Z 11 ¼ ¼
2 2
We have rounded the numbers to four significant figures because 8.314 J=K mol is only given to
four places. Several results are especially worth emphasizing. First, the mean free path is more than
100 times the effective size of the molecule at 1 atm pressure; there is a lot of empty space in a gas at
1 atm. This is typical of conditions at 1 atm. Second, the effective diameter of a little more than
4.1 Å is an effective sphere equivalent shape of the methane molecule. Third, while the Z 1 is high,
3
the Z 11 value of over 10 29 collisions=cm s is truly amazing. Is it any wonder that gas-phase
reactions are fast? One question, students often ask is how do you calculate n*? Here, we have
used the molar volume of 22.414 L ¼ 22,414 mL at 0 C and 1 atm and corrected it with Charles law

for the temperature ratio of 300:273.15.
Finally, a comment is in order regarding the factor of 1=2 in the formula for h. A more complete
treatment by Pease [3] integrates over a hemispherical region of the upper and lower layers that
produces a factor of (5p=32) ¼ 0:49087, which is very close to the factor of 1=2 ¼ 0:5 used here.
Thus, we see that we are using a very good approximation.
SUMMARY
This chapter has involved a lot of numerical work as well as some new integral formulas to burrow
into the details of KMTG and the use of the Boltzmann principle. Now, we have a more detailed
understanding of gases somewhere between the idea of a flowing macroscopic fluid and the inner
electronic structure of molecules assuming the atoms=molecules are approximately small spheres.
We have learned that there is a lot of action going on inside a gas but there is still a lot of empty space.
Perhaps, the most important result of this consideration of the Boltzmann KMTG and gas viscosity is
a rough approximate determination of the size of atoms and molecules. In fact, the cgs angstrom unit
of 10 8 cm is well suited to the size of atoms and molecules and it is common in older texts, but the
modern SI unit is the nanometer, easily remembered as ‘‘ten-to-the-minus-nine-meter’’ unit. Since
the nanometer is in meters and angstroms are in centimeters, 1 nm ¼ 10 Å. Therefore, a useful