Crossing the membrane: osmotic pressure

Classical Mechanics Level 4

Ideal gases exert a pressure on the walls of their container given by $nRT/V$ . One way to interpret this pressure is as the minimum pressure the container must withstand so that the gas does not alter the shape of its enclosure. Though most containers used for ideal gas examples are rigid, and never come anywhere near this limit, it is a valid interpretation.

Molecules dissolved in water establish a similar sort of pressure. Each molecule of solute introduced to pure water lowers the chemical potential of the solution relative to pure water. If the solution is allowed to interact with a less concentrated solution through a membrane permeable to water, water will flow until the concentration of the solute is equal on both sides.

This is an example of an entropic force: the two solutions have the highest number of molecular arrangements available to them when the concentration is equal on both sides. That this combinatorics generates a physical force is quite amazing when you think about it.

Anyways, just like ideal gases, ideal solutions obey a law of exactly the same form, Morse's law. This law states that the osmotic pressure felt by a solution is given by the ideal gas constant times the absolute temperature times the concentration of the solute.

$\displaystyle\Delta p_{osm} = MRT$

where $R$ is the ideal gas constant.

Suppose $\rho_{suc}$ from the last problem is 0.4 M (moles per liter), the xylem contains pure water, and the tree is at 25 $^{\circ}$ C. What osmotic pressure $\Delta p_{osm}$ (in Pascals ) is felt by the phloem from the xylem?

Note : This is the opposite perspective of the one usually taken by chemists, that the sap in the phloem exerts a negative osmotic pressure on the xylem. It is a matter of convention but ours is the more physical interpretation, in line with the usual understanding of hydrostatic fluid, and gas pressures.

The answer is 991085.078.

1 solution

Chew-Seong Cheong
Jul 31, 2014

The osmotic pressure $\Delta p_{osm}$ is given by: $\Delta p_{osm} = MRT$

And it is given that:

$M=0.4$ $mol/l$ , but since $\Delta p_{osm}$ is to be in SI unit $Pa$ , $M$ used should be $400$ $mol\space m^{-3}$
$R = 8.31447$ $J\space mol^{-1}\space K^{-1}$
$T=25^\circ C = (25+273)K = 298\space K$

Therefore, $\Delta p_{osm} = MRT = 400 \times 8.31447 \times 298 =\boxed {991084.82}\space Pa$

Crossing the membrane: osmotic pressure

The answer is 991085.078.

1 solution

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