Physical Chemistry Third Edition

(C. Jardin) #1
416 9 Gas Kinetic Theory: The Molecular Theory of Dilute Gases at Equilibrium

9.6 Effusion and Wall Collisions

Effusion is a process by which molecules of a gas pass through a small hole into a
vacuum. The hole must be small enough so that the gas does not flow through the hole
as a fluid, but passes as individual molecules. According toGraham’s law of effusion,
at a given temperature and a given pressure the rate of effusion of a gas is inversely
proportional to the square root of its density.

Graham’s law is named for Thomas
Graham, 1805–1869, a British chemist
who not only studied diffusion and
effusion, but also determined the
chemical formulas of the various
species formed from phosphoric acid
in aqueous solution.


Our analysis of effusion will be similar to that of the pressure in Section 9.5 except
that we will now compute the number of molecules striking an areaAinstead of the
force exerted on this area. If this area is a hole in the wall, we obtain the rate of effusion.
If the area is a section of the wall, we obtain the rate of wall collisions. The number of
molecules with velocities in the velocity intervaldvxdvydvzthat will strike the areaA
in timeτis given by Eq. (9.5-9):

(number strikingAin timeτ)NAvyτg(v)dvxdvydvz (9.6-1)

The total number of molecules striking the areaAin the time intervalτis


total number
striking area
Ain timeτ


⎠NAτ

∫∞

−∞

∫∞

0

∫∞

−∞

vyg(v)dvxdvydvz

NA

(

m
2 πkBT

) 3 / 2

τ

∫∞

−∞

∫∞

0

∫∞

−∞

vye−mv

(^2) / 2 kBT
dvxdvydvz (9.6-2)
This triple integral can be factored:
∫∞
−∞


∫∞

0

∫∞

−∞

vye−mv

(^2) / 2 kBT
dvxdvydvz




∫∞

−∞

e−mv

(^2) x/ 2 kBT
dvx


∫∞

0

vye−mv
y^2 /^2 kBT
dvy

∫∞

−∞

e−mv

(^2) / 2 kBT
dvz
The integrations overvxandvzgive factors of


(

2 πkBT
m

) 1 / 2

so that



total number
striking area
Ain timeτ


⎠NA

(

m
2 πkBT

) 1 / 2

τ

∫∞

0

vye−mv
y^2 /^2 kBT
dvy (9.6-3)

The integral in this equation can be performed by the method of substitution. We let
wv^2 y, so that the integral becomes

∫∞

0

vye−mv
y^2 /^2 kBT
dvy

1

2

∫∞

0

e−mw/^2 kBTdw



1

2

2 kBT
m

(

−e−mw/^2 kBT

)∣∣




0



kBT
m

(9.6-4)
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