Question

Many of the concepts developed in this chapter can be applied to understanding the atmosphere. Because atmospheric air is comprised primarily of N₂ (roughly 78% by volume), approximate the atmosphere as consisting only of N₂ in answering the following questions:

A) What is the single-particle collisional frequency at sea level, with T= 349 K and P= 1.0 atm. The corresponding single-particle collisional frequency is reported as 10¹⁰s⁻¹ in the CRC Handbook of Chemistry and Physics (62nd ed., p. F−171). z₁₁ = ?

B) At the tropopause (11 km), the collisional frequency decreases to 3.16×10⁹s⁻¹, primarily due to a reduction in temperature and barometric pressure (i.e., fewer particles). The temperature at the tropopause is 220 K. What is the pressure of N₂ at this altitude? P = ?

C) At the tropopause, what is the mean free path for N₂? λ = ?

Answer 1

given

T= 349 K

and

P= 1.0 atm = 101325 Pa

a )

using equation for

the single-particle colissional frequency Z₁₁ = ( P₁ N_A / RT ) ( 8 R T / M₁ )^1/2  

= ( 101325 x 6.023 x 10²³ / 8.314 x 349 ) x 0.43 x 10^-18 x ( 8 x 8.314 x 349 / 3.14 x 0.028 )^1/2   

Z₁₁ = 6.571 x 10⁹ /sec

b )

here T = 220 K

Z₁₁ = 3.16 x 10⁹ s⁻¹

using equation for

the single-particle colissional frequency Z₁₁ = ( P₁ N_A / RT ) ( 8 R T / M₁ )^1/2  

3.16 x 10⁹ = ( P₁ x 6.023 x 10²³ / 8.314 x 349 ) x0.43x10^-18x(8x8.314x349/3.14x 0.028 )^1/2   

3.16 x 10⁹ = x P₁ x 3.3 x 10²⁰ x 0.43 x 10^-18 x 407.8

P₁ = 3.16 x 10⁹ / 81836.04

P₁ = 38613.8 kg/m.sec²  

= 38613.8 / 101325

P₁ = 0.381 atm

c )

= ( RT / P₁ N_A ) x 1/

= ( 8.314 x 220 / 0.381 x 6.023 x 10²³ ) ( 1/ x 0.43 x 10^-18 )

= 1.29 x 10^-7 m