9.6 Effusion and Wall Collisions
417
Therefore
⎛
⎝
total number
striking area
A in time τ
⎞
⎠ NA τ
k B T
2πm
1/2
1
4
NA τ
8k B T
πm
1/2
1
4
NA τv (9.6-5)
The number of particles that strike the area A in time τ is proportional to the area A ,
proportional to the length of time τ, proportional to N , the number of particles per unit
volume, and proportional to the mean speed of the particles. The number of molecules
striking unit area per unit time is denoted by ν:
ν
1
4
N v
N
4
8k B T
πm
1/2
(9.6-6)
This equation for the rate of wall collisions also gives the rate of effusion per unit area
in the case of a small hole in the wall. The effusion rate predicted by Eq. (9.6-6) is
inversely proportional to the square root of the mass of the particles, and thus to the
square root of the density, in agreement with Graham’s law of effusion.
E X A M P L E 9.13
Estimate the number of molecules of air striking a person’s eardrum in 1.00 s at 298 K and
1.00 atm. Assume air to be 79 mol% nitrogen and 21 mol% oxygen by moles, and assume
that the area of the eardrum is 0.50 cm 2 .
Solution
v(O 2 ) 444 m s −1 from Example 9.4 and v(N 2 ) 475 m s −1 by a similar calculation.
N (N 2 ) N(N 2 )/V P(N 2 )/k B T
101325 N m −2 atm −1
(0.79 atm)
1.3807 × 10 −23 J K −1
(298 K)
1.9 × 10 25 m −3
where P(N 2 ) is the partial pressure of N 2 , N(N 2 ) is the number of N 2 molecules and
N (N 2 ) is the number density of N 2 molecules.
N (O 2 ) N(O 2 )/V 5.2 × 10 24 m −3 by a similar calculation
ν(N 2 )
1
4
(1.9 × 10 25 m −3 )(475 m s −1 ) 2.3 × 10 27 m −2 s −1
ν(O 2 ) 5.7 × 10 26 m −2 s −1 by a similar calculation
(number per second) ν(N 2 ) + ν(O 2 )(5.0 × 10 −5 m 2 ) 1.4 × 10 23 s −1
Exercise 9.19
a. Estimate the number of air molecules striking the palm of your hand in 24 hours.
b. A certain solid catalyst has a surface area of 55 m 2 per gram of catalyst. A mixture of gases
with carbon monoxide mole fraction equal to 0.0050 passes over the catalyst at 350 K and
1.00 atm. Find the amount of CO in moles striking 1 gram of the catalyst per second.
417
Therefore
⎛
⎝
total number
striking area
A in time τ
⎞
⎠ NA τ
k B T
2πm
1/2
1
4
NA τ
8k B T
πm
1/2
1
4
NA τv (9.6-5)
The number of particles that strike the area A in time τ is proportional to the area A ,
proportional to the length of time τ, proportional to N , the number of particles per unit
volume, and proportional to the mean speed of the particles. The number of molecules
striking unit area per unit time is denoted by ν:
ν
1
4
N v
N
4
8k B T
πm
1/2
(9.6-6)
This equation for the rate of wall collisions also gives the rate of effusion per unit area
in the case of a small hole in the wall. The effusion rate predicted by Eq. (9.6-6) is
inversely proportional to the square root of the mass of the particles, and thus to the
square root of the density, in agreement with Graham’s law of effusion.
E X A M P L E 9.13
Estimate the number of molecules of air striking a person’s eardrum in 1.00 s at 298 K and
1.00 atm. Assume air to be 79 mol% nitrogen and 21 mol% oxygen by moles, and assume
that the area of the eardrum is 0.50 cm 2 .
Solution
v(O 2 ) 444 m s −1 from Example 9.4 and v(N 2 ) 475 m s −1 by a similar calculation.
N (N 2 ) N(N 2 )/V P(N 2 )/k B T
101325 N m −2 atm −1
(0.79 atm)
1.3807 × 10 −23 J K −1
(298 K)
1.9 × 10 25 m −3
where P(N 2 ) is the partial pressure of N 2 , N(N 2 ) is the number of N 2 molecules and
N (N 2 ) is the number density of N 2 molecules.
N (O 2 ) N(O 2 )/V 5.2 × 10 24 m −3 by a similar calculation
ν(N 2 )
1
4
(1.9 × 10 25 m −3 )(475 m s −1 ) 2.3 × 10 27 m −2 s −1
ν(O 2 ) 5.7 × 10 26 m −2 s −1 by a similar calculation
(number per second) ν(N 2 ) + ν(O 2 )(5.0 × 10 −5 m 2 ) 1.4 × 10 23 s −1
Exercise 9.19
a. Estimate the number of air molecules striking the palm of your hand in 24 hours.
b. A certain solid catalyst has a surface area of 55 m 2 per gram of catalyst. A mixture of gases
with carbon monoxide mole fraction equal to 0.0050 passes over the catalyst at 350 K and
1.00 atm. Find the amount of CO in moles striking 1 gram of the catalyst per second.