Membranes - Introduction to Carbon Capture and Sequestration - page 335

0 the Helmholtz free energy has only entropic contribu-
tions, and hence the Henry coeffi cient has only entropic contributions. An
adsorbed molecule loses entropy because the volume in which the par-
ticle can move ( V c + V w ) is smaller than the corresponding volume of the
gas phase ( V ). The density of adsorbed particles is given by:

If U c = U w =

( )

P
kT

VV VV + +

U = 0

cw cw gas

ρ = = ρ

V V

B

If we want a higher density in our pores, we therefore need to compen-
sate for this entropy effect with an energetic component: the interactions
with the wall, U c < 0.
To simplify our calculation we assume that the energy of a molecule
in the cavity is lower compared to a molecule in a narrow window
( U c < U w ), which gives for the Henry coeffi cient:

1 V V 1 V

− UkT / − UkT / − UkT /

c w c

H = e + e ≈ e

cB wB cB

kT V V kT V

B B

Because of the exponent in the energy we see that we only need a small
energy difference for the Henry coeffi cient to be dominated by the lowest
energy, which we assume is the cavity (see also Question 7.6.1 ).

Diffusion

We can also make an estimate of the diffusion coeffi cient using these free
energy calculations. The Henry coeffi cient is related to the (excess)
chemical potential, which is for a pure component equal to the (Gibbs)
free energy ( F ) per particle. The trick here is to use the same random
insertion formula, but now to compute our free energy as a function of
the position of the gas molecule in the channel of the membrane. This
can be done by inserting a molecule at a random position and computing
the test particle's energy as a function of the position in the z direction:

()

− UkT /

( )

exp − Fz kT / = e δ z' z − ,

B

B random

where the delta function expresses that only those particles that are ran-
domly inserted at the position z '

= z (but can differ in x and y coordinates)