🚀 go-pugleaf

None of the diagrams is in error, and I disagree with your assessment of which
diagram best illustrates the phenomenon, as well.

In article <1990Nov11.132436.2836@newcastle.ac.uk: w.p.coyne@newcastle.ac.uk writes:
:    1.'Beaker'-with semi-permeable        2. 'U' shaped tube with membrane
:       membrane down middle.                  at the bottom of the bend.
:  |     *     |    |^^^^^*     |          |  |        |  |  |^^|        |  |
:  |^^^^^*^^^^^|    |     *     |          |^^|        |^^|  |  |        |  |
:  | high*  lo |    |     *^^^^^|          |  |        |  |  |  |        |^^|
:  |     *     |    |     *     |          |hi'----*---'lo|  |  '----*---'  |
:  '-----------'    '-----------'          '-------*------'   -------*------
:     BEFORE            AFTER                   BEFORE            AFTER
: 
:   3. Enclosed cylinder with semi-permeable membrane which can slide, so
:  .---------------------.    .---------------------.
:  | high     *  lo      |    |                 *   | as to alter the
:  |          *          |    |                 *   | volumes on either side.
:  '---------------------'    '---------------------'
:         BEFORE                      AFTER
: 
:With both fig 1 and fig 2 mention is made of the levels ceasing to change
:when the difference in height becomes great enough.
:BUT
:     I believe Fig 1 is false as no mention made of the changing membrane area
:     available to each side.

The area of the membrane is totally irrelevant, and does not appear in the
equations describing the pheonomenon.  This area will affect only the rate at
which osmotic equilibrium is achieved, not the position of the equilibrium.
The position of equilibrium is calculated by equating the chemical
potential of the solvent (actually, the chem. potential of all permeable
substances) on both sides of the membrane.  This comes from the fact that if 
there is a phase equilibrium at constant T, any ingredient common to both 
phases must have the same chem. potential in both.  Any part of the membrane 
wet by one phase but not the other does not come into the equation, since at
such a part there is no phase equilibrium;  only the part wet by both phases 
enters into the calculation.  Again, the area of this part does not 
enter into the calculation.

:comments of the 3 diagrams:-
:Fig 1 is a very poor way of illustrating osmosis in that the effects of gravity
:      and air pressure on both sides and the area of membrane exposed each side
:      all change as one side rises and one falls
:Fig 2 much better as the area of membrane for each side is constant. But if
:      the lo side falls enough it becomes the same as Fig 1.

I believe that two is clearer, but 1 is also correct.

:Fig 3 is the best in that it shows the result of osmosis without the added
:      complications of the two others, unlike 1 and 2 the membrane should
:      continue to move until both sides have same conc. or (in the case when
:      the lo conc is actually pure water) all the water ends up on one side.

Fig 3 is also correct, but does not give a feeling for osmostic *pressure*.
Apparatus 3 is useless for calculating the molecular weight of a macromolecule,
for example.  Apparatus 2 (or 1) could in principle be so used.

Note that in Figure 1 and Figure 2, after equilibrium has been achieved, the
high-pressure side still has a higher concentration than the low-pressure
side, but not by as much as at the beginning of the experiment.  In Figure 3,
at equilibrium, there is neither a pressure nor a concentration difference
between the two sides.

	-P.
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Peter S. Shenkin, Department of Chemistry, Barnard College, New York, NY  10027
(212)854-1418  shenkin@cunixc.cc.columbia.edu(Internet)  shenkin@cunixc(Bitnet)
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