4.E: UHV and Effects of Gas Pressure (Exercises)

This section provides a limited number of examples of the application of the formulae given in the previous section to determine the:

• Density of Molecules in the Gas Phase
• Mean Free Path of Molecules in the Gas Phase
• Flux of Molecules incident upon a Surface
• Rate of Adsorption of Molecules and Surface Coverages

If you have not already been through Section 4.2 then I would suggest that you stop now and return to this page only after you have done so !

Within any one of the following sub-sections, it will be assumed that you have already done the previous questions and may make use of the answers from these questions - you are therefore advised to work through the questions in the order they are presented.

A. Molecular Gas Densities

Example

Calculate the molecular gas density for an ideal gas at 300 K, under the following conditions (giving your answer in molecules m^-3 ): At a pressure of 10-6 Torr

Solution

The molecular gas density at this pressure is 3.2 x 10¹⁶ molecules m^-3

Rationale

The gas density is easily obtained from the ideal gas law :

\[n = \dfrac{N}{V} = \dfrac{P}{k\,T} \nonumber \]   [ molecules. m^-3 ]

where :	P - pressure [ N m^-2 ]
	k - Boltzmann's constant ( = 1.38 x 10^-23 J K^-1 )
	T - temperature [ K ]

Hence,

   P = ( 1 x 10^-6 / 760 ) x 101325 = 1.333 x 10^-4 N m^-2

   n = 1.333 x 10^-4 / (1.38x10^-23 x 300 ) = 3.2 x 10¹⁶ molecules m^-3

Example

Calculate the molecular gas density for an ideal gas at 300 K, under the following conditions (giving your answer in molecules m^-3 ): At a pressure of 10-9 Torr

Solution

The molecular gas density at this pressure is 3.2 x 10¹³ molecules m^-3

Rationale

The gas density is directly proportional to the pressure , hence when the pressure of the gas is reduced by 3 decades from 10^-6 Torr (as in the previous question) to 10^-9 Torr, the gas density will also be lowered by a factor of 1000.

i.e. n = 3.2 x 10¹⁶ / 1000 = 3.2 x 10¹³ molecules m^-3

B. Mean Free Path of Molecules in the Gas Phase

Example

Calculate the mean free path of CO molecules in a vessel at the indicated pressure and temperature, using a value for the collision cross section of CO of 0.42 nm².

P = 10-4 Torr, at 300 K

Solution

 The mean free path of CO under these conditions is 0.52 m

Rationale

The mean free path, , is given by the equation :

\(\lambda=\frac{k T}{1.414 P \sigma}\)[ m ]

where, in this instance,

  P = (1 x 10^-4 / 760) x 101325 = 1.333 x 10^-2 N m^-2

and

  σ = 0.42 nm² = 4.2 x 10^-19 m²

Substitution gives

  λ = ( 1.38x10^-23 x 300 ) / (1.414 x 1.33x10^-2 x 4.2x10^-19 )

⇒     λ = 0.52 m

Example

Calculate the mean free path of CO molecules in a vessel at the indicated pressure and temperature, using a value for the collision cross section of CO of 0.42 nm².

P = 10-9 Torr, at 300 K

Solution

The mean free path of CO under these new conditions is 52000 m ( = 5.2 x 10⁴ m)

Rationale

The mean free path of molecules in the gas phase is inversely proportional to the pressure , hence when the pressure of the gas is reduced by 5 decades from 10^-4 Torr to 10^-9 Torr, the mean free path will increase by a factor of 100000.

i.e.   λ = 0.52 x 100000 = 5.2 x 10⁴ m

C. Fluxes of Molecules Incident upon a Surface

Example

Calculate the flux of molecules incident upon a solid surface under the following conditions:
[Note - 1 u = 1.66 x 10^-27 kg: atomic masses ; m(O) =16.0 u, m(H) = 1.0 u]

• Oxygen gas ( P = 1 Torr ) at 300 K

Solution

  The flux of oxygen molecules under these conditions is 3.58 x 10²⁴ molecules m^-2 s^-1

Rationale

The incident flux is given by the following equation (with all quantities expressed in SI units).

\[F=\frac{P}{\sqrt{2 \pi m k T}} \nonumber \][ molecules m^-2 s^-1 ]

In this instance ,

  P = (1/760) x 101325 = 133.3 N m^-2

and

  m = 32 x 1.66x10^-27 = 5.32x10^-26 kg

so

  F = 133.3 / (2 x 3.1416 x 5.32x10^-26 x 1.38x10^-23 x 300 )^1/2

i.e.  F = 3.58 x 10²⁴ molecules m^-2 s^-1

Example

Calculate the flux of molecules incident upon a solid surface under the following conditions:
[Note - 1 u = 1.66 x 10^-27 kg: atomic masses ; m(O) =16.0 u, m(H) = 1.0 u]

• Oxygen gas ( P = 10-6Torr ) at 300 K

Solution

 The flux of oxygen molecules under these conditions is 3.58 x 10¹⁸ molecules m^-2 s^-1

Rationale

The flux is directly proportional to the pressure : consequently a decrease in pressure by 6 orders of magnitude (all other conditions remaining the same) leads to a corresponding decrease in the flux i.e.

F = 3.58x10²⁴ x ( 10^-6 / 1 )

⇒  F = 3.58 x 10¹⁸ molecules m^-2 s^-1

Example

Calculate the flux of molecules incident upon a solid surface under the following conditions:
[Note - 1 u = 1.66 x 10^-27 kg: atomic masses ; m(O) =16.0 u, m(H) = 1.0 u]

• Hydrogen gas ( P = 10-6Torr ) at 300 K

Solution

The flux of hydrogen molecules under these conditions is 1.43 x 10¹⁹ molecules m^-2 s^-1

Rationale

The flux is inversely proportional to the square root of the mass of the incident gas species - hence, using the answer obtained for oxygen at this pressure as a starting point

F = 3.58x10¹⁸ x ( 32 / 2 )^1/2

⇒  F = 1.43 x 10¹⁹ molecules m^-2 s^-1

Example

Calculate the flux of molecules incident upon a solid surface under the following conditions:
[Note - 1 u = 1.66 x 10^-27 kg: atomic masses ; m(O) =16.0 u, m(H) = 1.0 u]

• Hydrogen gas ( P = 10-6Torr ) at 1000 K

Solution

The flux of hydrogen molecules under these conditions is 7.83 x 10¹⁸ molecules m^-2 s^-1

Rationale

The flux is inversely proportional to the square root of the gas phase temperature - hence , by comparison with the previous result

F = 1.43 x 10¹⁹ x ( 300 / 1000 )^1/2

⇒  F = 7.83 x 10¹⁸ molecules m^-2 s^-1

D. The Kinetically Limited Uptake of Molecules onto a Surface

The rate of adsorption of molecules onto a surface can be determined from the flux of molecules incident on the surface and the sticking probability pertaining at that instant in time (note that in general the sticking probability itself will be dependent upon a number of factors including the existing coverage of adsorbed species).

In the following examples we will assume that the surface is initially clean (i.e. the initial coverage is zero), and that there is no desorption of the molecules once they have adsorbed. You should determine coverages as the ratio of the adsorbate concentration to the density of surface substrate atoms (which you may assume to be 10¹⁹ m^-2 ). In the first two questions we will assume that the sticking probability is constant over the coverage range concerned.

Example

Calculate the surface coverage obtained after exposure to a pressure of 10-8 Torr of CO for 20 s at 300 K - you may take the sticking probability of CO on this surface to have a constant value of 0.9 up to the coverage concerned.

Solution

 The surface coverage of CO molecules obtained is θ = 0.069

Rationale

Firstly we need to calculate the incident flux of CO using the Hertz-Knudsen formula : this gives

   F = 3.83 x 10¹⁶ molecules m^-2 s^-1

Since the sticking probability is constant, the number of molecules per m² is simply given by

Coverage	= F x S x t
	= 3.83 x 10^16 x 0.9 x 20
	= 6.90 x 10^17 molecules m^-2

which, when ratioed to the surface density of substrate atoms, gives

   θ = 6.90x10¹⁷ / 1.0x10¹⁹  ⇒  θ = 0.069

Example

Calculate the surface coverage of atomic nitrogen obtained by dissociative adsorption after exposure to a pressure of 10-8 Torr of nitrogen gas for 20 s at 300 K - you may take the dissociative sticking probability of molecular nitrogen on this surface to be constant and equal to 0.1

Solution

The surface coverage of N atoms obtained is θ = 0.015

Rationale

We can use the answer to the previous question to save some work here since the flux of incident molecules will be the same - the differences are that

1. each nitrogen molecule that sticks gives rise to two adsorbed N atoms,
2. the sticking probability is lower by a factor of 9

Hence,

Coverage , θ = 0.069 x (2/1) x (0.1/0.9)  ⇒  θ = 0.015

In general, the sticking probability varies with coverage - most obviously, the sticking probability must tend to zero as the coverage approaches its saturation value. These calculations are not quite so easy !

Example

Calculate the surface coverage obtained after exposure to a pressure of 10-8 Torr of CO for 200 s at 300 K - the sticking probability of CO in this case should be taken to vary linearly with coverage between a value of unity at zero coverage and a value of zero at saturation coverage (which you should take to be 6.5 x 1018 molecules m-2 ).

Solution

The surface coverage of CO molecules obtained is 4.5 x 10¹⁸ molecules m^-2 or θ = 0.45

Rationale

We have calculated the incident flux, F , of CO under these conditions in a previous question; it is equal 3.83 x 10¹⁶ molecules m^-2 s^-1. We are also told that the maximum surface coverage, N_sat , is 6.5 x 10¹⁸ molecules m^-2.

Let N(t) be the surface coverage in molecules m^-2 at any time, t.

Then	\(\frac{d N}{d t}=F \cdot S(N) \Rightarrow \int \frac{d N}{S(N)}=F \int d t\)
where	S(N) = ( 1 - ( N / Nsat ) )

 
If we use the Langmuir definition for the surface coverage, i.e.   θ = ( N / N_sat )  ,

Then  dθ = dN / N_sat  and

\[N_{s a t} \frac{d \theta}{d t}=F . S(\theta) \quad \Rightarrow \quad \int \frac{d \theta}{S(\theta)}=\frac{F}{N_{s a t}} \int d t \nonumber \]

where S(θ) = ( 1 - θ ) .

To answer this question we need to integrate the differential equation from t = 0 to t = 200 s which gives

\[[-\ln (1-\theta)]_{\theta=0}^{\theta \operatorname{final}}=\frac{F}{N_{\text {sat }}}[t]_0^{200} \nonumber \]

  ⇒   θ_final = 0.692 (where θ is still defined in the Langmuir manner)

So, the absolute coverage of CO molecules is given by

  ⇒   Coverage = 0.692 x 6.5 x 10¹⁸ molecules m^-2 = 4.5 x 10¹⁸ molecules m^-2.

and the coverage defined in the standard surface science manner is obtained by ratioing with the atom density of substrate atoms in the first layer

  ⇒   θ = ( 4.5 x 10¹⁸ / 1.0 x 10¹⁹ ) = 0.450