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From: chooft@ruunsa.fys.ruu.nl (Rob Hooft)
Newsgroups: sci.bio,sci.chem,sci.physics,sci.misc
Subject: Re: Osmosis - the cause at the molecular level.
Message-ID: <1726@ruunsa.fys.ruu.nl>
Date: 2 Nov 90 08:21:56 GMT
References: <1990Oct28.115303.7221@newcastle.ac.uk> <4396@pkmab.se> <29046@boulder.Colorado.EDU>
Followup-To: sci.bio
Organization: University of Utrecht, Dept. of Physics
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In <29046@boulder.Colorado.EDU> eddy@boulder.Colorado.EDU (Sean Eddy) writes:

>Indeed, that's the explanation I was taught, and in turn taught to
>undergrads here at Boulder. It was intellectually satisfying to
>me until a couple of weeks ago, when my complacency was thrashed by
>a fellow grad student. We started arguing, and reading, and arguing
>some more, and pretty soon the debate was spreading through the
>department... but we're just a bunch o' biologists, so I'll
>relate the problem to you guys. 

>Osmosis is a colligative property. That is, osmotic pressure is
>dependent on the *number* of particles in solution -- *not*
>their size, mass, etc. 
.
.
.
>What is the detailed rationale for osmotic pressure being
>dependent upon only the number of solute molecules, not
>their size, shape, and mass?

Entropy is. And the ideal gas law p.V=n.R.T is derived from entropy too.
If water flows from the pure-water to the mixture compartment the entropy of the
system rises, because there are more possible realisations of the locations of
the solute. This is exactly the same reason that some amount of gas will always
take all the volume it can get. You need to input energy (compressing the gas,
applying gravity) to stop the expansion process. From the analogy it can be
derived that the ideal gas law also holds for osmosis, such that p=n.R.T/V
is the osmotic pressure.

some theory
-----------
Equilibrium in the system means that the thermodynamic potential (mu) of the
components that can diffuse must be equal in both compartments. In the 
compartment with the solute present, the potential is raised by pressure,
and lowered by the presence of the solute, such that the equilibrium can be
stated as follows:

    mu(pure water)
        +(derivative of mu to the pressure)*(osmotic pressure)
        -R*T*(molefraction solute)     =  mu(pure water)

and so:
        (derivative of mu to the pressure)*(osmotic pressure)
        =R*T*(molefraction solute)

the derivative of mu to the pressure is the volume of one mole of water 
in the solution. This can be replaced by the total volume of a
"unit of solution" in case of dilute solutions:

        ("unit solution" Volume)*(osmotic pressure)
        =R*T*(molefraction solute)

Multiply both sides by the number of moles present:

         (total volume)*(osmotic pressure)
         =R*T*(number of moles of solute)

or:
                             R*T*(number of moles of solute)
         osmotic pressure = ---------------------------------
                                     Total volume

Hope this helps again.

Reference in Dutch:

Korte inleiding in de chemische Thermodynamica,
J.M. Bijvoet, A.F. Peerdeman, A.Schuijff, E.H. Wiebenga.
Utrecht, 1984.
-- 
Rob Hooft, Chemistry department University of Utrecht.
hooft@hutruu54.bitnet hooft@chem.ruu.nl chooft@fys.ruu.nl