Membrane Gas Separation

11.2
Sorption Theory for Multiple Gas Components 
11.2.1
Rubbery Polymeric Membranes 
Rubbery membranes operate above the glass transition temperature, therefore polymer 
chains are able to rearrange on a meaningful timescale and are usually in thermodynamic 
equilibrium. Examples of rubbery membranes are polydimethylsiloxane, polyethylene 
glycol and silicone rubber. These membranes generally achieve their selectivity by dif-
ferences in solubility [12] . Hence, they often allow larger gas molecules to permeate at 
a greater rate than smaller gases. A good example of this is hydrocarbon recovery from 
natural gas [13] . This is because the liquid - like polymer matrix has a poor ability to sieve 
molecules based on size because of chain motion. This is of considerable interest in syngas 
separation (pre - combustion capture), where it is desirable to keep hydrogen at high pres-
sure and therefore allow the larger gases, such as CO 
2
, to transport through the membrane. 
This ensures hydrogen is at the pressure required for combustion as part of an integrated 
gasifi cation and combined cycle process (IGCC). 
For rubbery polymeric membranes, the process of mixing gas A into a polymer P can 
be described by Flory – Huggins regular solution theory for mixing components within 
polymer solutions [14] :
μ
μ
φ
φ
χ φ
A
A
A
A
A
A
P
P
A
P
−
=
= ⎛⎝⎜
⎞
⎠⎟ =
+ −
⎛
⎝⎜
⎞
⎠⎟
+
0
0
0
2
1
Δ
G
RT
f
f
V
V
ln
ln
(11.3)
Where f
A
and f
A0
are the fugacity of A and the saturated fugacity of A respectively,    
μ
   
A
and 
µ
A
0
are the corresponding chemical potentials,  
φ
 
A
and  
φ
 
P
are the volume fraction of gas A 
and the polymer P respectively, V
A
and V
P
are the relevant molar volumes and  
χ
 
A
the 
Flory – Huggins parameter of gas A and the polymer P. 
The true concentration of the penetrant A expressed as the volume of gas (cm 
3
at STP) 
dissolving in 1 cm 
3
of polymer, can be related to the volume fraction in the above expres-
sion by [15] :
C
V
V
A
A
G
A
= φ
(11.4)
where V
G
is the molar volume of a gas at STP, which is 22400 cm 
3
/mol. 
Recognising that:
ln
ln
ln
ln
ln
ln
S
C
f
C
f
f
f
A
A
A
A
A
A0
A0
=
( )
−
( )
=
( )
−
−
(11.5)
Substitution of Equations (11.4) and (11.5) into Equation (11.3) gives:
ln
ln
ln
ln
S
V
V
V
V
f
A
A
G
A
A
A
P
P
A
P
A0
( )
=
−
− −
⎛
⎝⎜
⎞
⎠⎟
−
−
( )
φ
φ
φ
χ φ
1
2
(11.6)
Given  
φ
 
P
= 1 –  
φ
 
A
and assuming that both V
A
<< V
P
and
φ
A
2
is vanishingly small; Equation 
(11.6) becomes:

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