The water potential in a dilute aqueous solution depends on the pressure P and t
ID: 216112 • Letter: T
Question
The water potential in a dilute aqueous solution depends on the pressure P and the solute concentration xsolute as follows: uH2O=u*H2O-RTxsolute+(P+P0)Vo where u*H2O is the chemical potential of pure water at the reference pressure P0 and Vo =18 mL/mol is the volume per mole of water. Here, the chemical potential is given per mole of H2O rather than per molecule.
A) Show that H2O in a salt solution can never be at chemical equilibrium with pure water at the same temperature and pressure. In which liquid wil teh H2O molecules accumulate?
B) Show that a salt solution can be at chemical equilibrium with pure water at the same temperature if the two liquids are held at different pressures, and find the required pressure difference (osmotic pressue) as a function of solute concentration.
C) Cytoplasm has a concentration of xsolute = 0.005. Find the osmotic pressure of cytoplasm relative to pure water at 300K.
Explanation / Answer
A) the chemical potential of pure water = u*H2O
For a salt solution with concetration Y, the chemical potential is uH2O = u*H2O - RT Y + (P + P0) V0
as we can see from the equation, at a salt concentration of Y with the same temperature and pressure a as pure water, the chemical potential will be u*H2O - Y.
Hence at no salt concetration will the chemical potential be equal to that of pure water.
As pure water always has higher chemical potential, water will flow from pure water (higner potential) to water contatining solutes (lower potential). Hence water accumilates in solute containing water.
B) uH2O can be equal to u*H2O only if P changes.
Lets calculate the P at which uH2O + u*H2O
From the formula we know, uH2O = u*H2O - RT X + ( P + Po) Vo
we know uH2O = u*H2O and Vo = 18, so putting in the formula, we get
u*H2O = u*H2O - RT X + (P + Po) 18
therefore,
Rt X = (P + Po) 18,
P + Po = RT X/18
Therefore, P = (RT X)/18 - Po
Hence the required pressure is P = (RT X)/18 - Po
C) xsolute = 0.005 and RT = 300k
THerefore uH2O = u*H2O - 300 * 0.005 + (P -Po) * 18
Hence osmotic pressure is = u*H2O -1.5 + 18(P + Po)
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