Is Gibbs energy change during osmosis positive? If so then why is it spontaneous?

Question

For ideal solution, μ_A=μ_A^* +RT lnx_A

ln function is an increasing function, means that the higher the x_A the higher the μ_A......(1)

Now consider osmosis. From Atkins' Physical Chemistry Ninth Edition, page 173,

osmotic pressure, π, is the pressure that must be applied to the solution to stop the influx of solvent (into solution across a semi-permeable membrane).

Here is my interpretation and thought process:

Had you not exerted π on solution, the μ(solution)< μ(solvent) based on (1) because lnx_A is negative for x_A<1, so by Second Law, the osmosis is spontaneous and solvent should move from pure solvent (high μ) to solution (low μ). Sounds right.

However, during this process, μ(pure solvent) unchanged (μ is intensive, not affected by amount) but μ(solution) increases due to dilution. Based on (1), for the solution, since x_A increases μ(solution) must also increase. Although the process stops when μ(solution)=μ(pure solvent), the effect is that μ increases. μ increases would in turn imply ΔG>0 which means non-spontaneous, but as far as I know osmosis is spontaneous, so where is my misconception?

Answer 1

Let the subscript A refer to the solvent and B refer to the solute, and let the subscript 1 refer to the pure solvent container and 2 refer to the mixture container. Then the free energy of the combined system is given by: $$G=n_{A1}\mu_A^0+n_{A2}\left(\mu_A^0+RT\ln{\left[\frac{n_{A2}}{n_{A2}+n_{B_2}}\right]}\right)+n_{B2}\left(\mu_B^0+RT\ln{\left[\frac{n_{B2}}{n_{A2}+n_{B_2}}\right]}\right)$$If the number of moles of solvent in container 2 increases by $dn_{A2}$ and the number of moles of solvent in container 1 decreases by $dn_{A2}$, the change in the free energy of the combined system changes by dG. If we do the math correctly, we obtain: $$dG=RT\ln{x_{A2}dn_{A2}}\tag{1}$$where $x_{A2}$ is the mole fraction of solvent in the mixture container at any point: $$x_{A2}=\frac{n_{A2}}{n_{A2}+n_{B_2}}$$Eqn. 1 indicates that the change in free energy is negative (as expected).

Answer 2

To evaluate whether the process is spontaneous you have to take the difference of the chemical potentials $\mu$. Solvent molecules go from the pure solvent (reactant) to the solution (product):

$$\Delta_r G = \mu_\text{product} - \mu_\text{reactant} =$$ $$ = (\mu_\text{solvent}^\circ + R T \ln x_A) - \mu_\text{solvent}^\circ < 0$$ Throughout the process, $\mu$(solvent with solute)< $\mu$(pure solvent). As the process goes on, the solution becomes more and more diluted, so $\mu$(solvent with solute) will continuously increase, but never reach $\mu$(pure solvent). So the transport will go on, but the osmotic pressure will decrease as you get closer to equilibrium

In a different example, if one solution is at 1 mM and the other at 2 mM, solvent will go from the lower solute concentration to the higher solute concentration (or you could say from higher solvent concentration to lower solvent concentration) until the solute concentrations are equal and the osmotic pressure difference is zero.