Michael J. Suess
World Health Organization Regionol Office for Europe Copenhagen, Denmark
1 Reverse OSIIIOS~S I
A thermodynamic analysis a n d exercise
The application of the reverse-osmosis process in modern science and technology has widely spread in recent years. Today, this process is used for the production of potable water through desalination of hrakish and sea water, for the advanced treatment of municipal and industrial waste effluents to achieve further purification before disposal to the natural environment, for the concentration of natural juices and other products in the food industry, for the concentration of chemical and biological solutions in medicine and the related life sciences, and more. The review of the basic thermodynamic principles of reverse-osmosis is, therefore, considered useful. Since reverse-osmosis is directly related to the natural and well-known osmosis process, the latter will be discussed first. The Osmosis Process
If we assume a membrane permeable to a solvent but impermeable to a solute, and if on the one side of the membrane we have a pure solvent, while on the other side a solution, the solvent molecules will pass from the pure solvent side through the membrane into the solution. This phenomenon takes place due to a diierence in the thermodynamic potential-the Gibbs free energy of the pure solvent being higher than that of the solution. Because of the tendency of any system to reach equilibrium, a state at which the free energy of both the solvent and the solution are equal, the initial difference in the free energy of the system can be considered as the driving force of the above described process called osmosis. The volume of the solvent, which is transferred through the membrane, if capable of performing work against a pressure a, denoted the osmotic pressure. If the Gibbs free energy is defined as G=H-TS
(1)
and the enthalpy is defined as H=EfPV
(2)
G=E+PV-TS
(3)
then
E = q - w and dE - dp dG
-
dE
+ PdV + VdP - TdS - SdT
For a reversible process, and osmosis is one, dq = TdS. If all the work done is expansion, then, dw = PdV. From the first law of thermodynamics it is recalled that
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190 / Journol of Chemical Education
0. Thus, eqn. (4)
=
VdP
- SdT
(5)
AG = VAP
(6)
At equilibrium the diierence in pressure AP is the pressure difference between solution (Pa) and solvent (PI). Thus AP,, = P, - p, = , (7) Remembering that the equation describing t,he free energy change of a dilute solution is AG = RT in (XA), XA being the mole fraction of the solvent in t,he solution, then, for an equilibrium state to be established, the increase in free energy of the solution (change of concentration of solute) must be balanced by the increase in pressure; and IV
+ RT in ( X A )
=
0
(8)
After a number of intermediate derivations (see also later in the section on mathematical derivation) the final form can be written down as rVs
"RTns
(9)
where n~ is the number of moles of solute in solution, and Vs is the volume of the solution. The Reverse-Osmosis Process
In the light of the above equations, the reversible, isotbermic work for the reverse-osmosis process can be described by The negative sign refers to work done on the system to cause the process to take place, and to transfer a unit volume of the solvent from the solution to the pure solvent. For such a process, (Pz - PI) is greater than 7. If work overcomes only the osmotic pressure of the solution, (P2 - PI) would equal 8, and -W
(4)
=
For a process taking place in isothermal conditions SdT = 0, and the final equation of reversible and isothermal work becomes simply dG = VdP, or for small incremental changes
For the change in the free energy, eqn. (3) becomes dG
+ dw
can be rewritten in the form
=
UVS = R T ~ B
(11)
This is the work required to keep the system continuously a t its initial volumetric balance against the osmotic forces. However, to achieve the transfer of solvent from the solution to the pure solvent, i.e., a reverse-osmosis process I-W'I
and
> I-WI
(12)
where A P ' + A P , and stands for any pressure in addition to s. The removal of the solvent from the solution will change the concentration of the solution but not the number of moles of solute, n,, in solution. However, the volume, Vs, of the solution will change. To fulfill the condition of reversihility, the solvent removed must at all times be in equilibrium with the solution, thus calling for a differential treatment of eqn. (13) -dW' = r'dVs Vsdr' (14)
+
If the pressure on the system is kept constant, namely, V,daf = 0, then the final equation for the change of required work with change of volume becomes and the total work required to transfer a volume of solvent equals
To transfer a volume of water-the solvent-from the solution to the pure solvent, it is necessary to overcome first the osmotic pressure rr and then add an additional pressure A P , to overcome the increase in the "osmotic potential." XA and XB are defined as the mole fractions of the solvent and the solute, respectively, both being smaller than unity; n~ and ns are defined as the moles of the solvent and the solute, respectively. Also, by definition XA = 1 XR and XB = nR/(nA nB). For very dilute solutions XB
