Entropy as the driving force for osmosis

Question

How is entropy responsible for osmosis and is movement of solvent possible from its higher concentration to lower concentration?

By concentration, I will be referring to the concentration of solvent molecules below.

I understand that the entropy of the universe tends to increase. But I do not understand how this principle applies to osmosis. Basically, in osmosis the solvent moves from its higher concentration to lower concentration. Here, it appears that the entropy of the lower concentration side is increasing and the entropy of the higher concentration of solvent side is decreasing. How can we claim that the net change is increase in entropy?

Also, is the reverse process i.e. movement of solvent molecules from lower concentration side to their higher concentration side ever possible (= thermodynamically favorable) without the presence of external driving forces?

Answer 1

(a) The tendency for the solute undergoing diffusion to occupy as large a volume as possible is similar to that of a gas filling the volume available to it and in each case the driving cause is the increase in entropy. In any distribution of particles there are more ways of distributing them in a larger volume than in a smaller one.

Recall that as the solvent enters the concentrated solution the solution volume increases and this continues until the chemical potential on either side of the semi-permeable membrane is equalised. The pressure is also increased as solution is pushed up into a column, the difference in pressure is the osmotic pressure.

(b) From the statements above you can see that reverse osmosis cannot occur without applying external energy.

Notes: The change in entropy of a gas with volume is $\displaystyle \frac{\partial S}{\partial V}=\frac{R}{V}$. In a dilute solution the mole fraction $x_s \sim 1/V$ where $V$ is the volume of the solvent (or solution) with one mole of solute. As $G-G^0=-T S = RT\ln(x_s)$ then $\displaystyle \frac{\partial S}{\partial V}=\frac{R}{V}$. Thus the change in entropy is the same for the dilution of a gas as for the solute in a dilute solution.

Answer 2

Consider two decks of 52 playing cards. One deck has blue backs and one has yellow backs. A group plays poker with the yellow deck. If blue cards slowly get mixed into the yellow deck, the possibilities of types of five card hands increases. The faces in five card hands will have increased variations (a player could get two cards with the same face, like two aces of spades). Additionally, the variations of card back colors adds to the possible unique types of five card hands (four yellow-backed and one blue-backed). If the yellow-backed cards is the solute and the blue-backed cards is the solvent (and only the blue-backed cards can enter or exit the game) then the blue-backed cards are getting diluted from their higher "concentration" to a lower "concentration". A pure solvent which can make five hydrogen bonds with neighboring solvent molecules will have less entropy than a solvent that has the opportunity to make any combination of five ion-dipole attractions and hydrogen bonds. The solvent is effectively getting diluted and entropy increases.