Question

How long will it take for an initially clean surface to become 1.0% covered if it is bathed by an atmosphere of nitrogen at T-77 K and pressure P-1 bar? Assume that every nitrogen molecule that strikes the surface sticks and that a nitrogen molecule covers an area of 1.1 x 105 pm?

Answer 1

From the kinetic theory you can find the number of N₂-molecules hitting the surface per unit area and time:
A = (1/4)∙(N/V)∙v_av
N/V is the number of molecules per unit volume, which can be calculated from ideal gas law
p∙V = N∙k_b∙T
=>
N/V = p/(k_b∙T)
v_av is the average speed of the molecules.

Because speed is Maxwell-Boltzmann distributed, these average speed is given by:

v_av = √( 8∙k_b∙T /(π∙m))
Hence:
A = (1/4)∙ [p/(k_b∙T)] ∙ √( 8∙k_b∙T /(π∙m))
= (1/2)∙ p ∙ √(2 / (π∙m∙k_b∙T))
For the given conditions
A = (1/2)∙ 1.1 * 10⁵ Pm² ∙ √(2 / (π ∙ 28∙1.6605 * 10^-27kg ∙ 1.3807 * 10^-23J/K ∙ 77K))
= 5.6744 * 10²⁷m⁻²s⁻¹

Assume that every portion of the surface, can only be covered by a single nitrogen molecule. Let x be the fraction of the surface not occupied by N₂. Then the rate which N₂-molecules hitting "free places" per unit surface is A∙x. Moreover assume that all molecules hitting a "free place" at the surface get absorbed,. If multiply the surface specific rate above by you by the area occupied of a single nitrogen molecule (a) are you get the rate at which fraction of "free area" decreases:
That means
- dx/dt = a∙A∙x
or
dx/dt = - k∙x
with
k = 1.1 * 10⁵ Pm² ∙ 5.6744 * 10²⁷m⁻²s⁻¹
= 1 * 10^{-19 m²} ∙ 5.6744 * 10²⁷m⁻²s⁻¹
= 1 * 10^{-19 m²} ∙ 5.6744 * 10²⁷m⁻²s⁻¹
= 6.242 * 10⁹s⁻¹

Solve this differential equation by separation of variables:
1/x dx = - k dt
<=>
∫ 1/x dx = - ∫ k dt
<=>
ln(x) = - k∙t + c

Apply initial condition to find constant c:
x(t=0) = 0
<=>
ln(1) = - k∙0 + c
=>
c = 0

The the fraction of free area is given by
ln(x) = - k∙t

So the time elapsed until 1% of the area, , i.e x=0.99, is occupied is
t = -ln(x)/k
= -ln(0.99) / 6.242 * 10⁹ s⁻¹
= 1.61 * 10^-12 s