Initially suppose we have a mixture with mole fractions $x_1$ and $x_2$. Let $x_1$ be the solute and $x_2$ the water and assume for simplicity that deviations from Raoult’s law behaviour are insignificant. In this case the heat of mixing is zero and entropy alone drives the mixing process.
In this case the free energy of mixing per mole at constant temperature is
$\Delta G =RTx_1\ln(x_1)+RTx_2\ln(x_2)$
and in as there are two components $x_2=1-x_1$ (and changing $x_1 \rightarrow x$ for simplicity ) gives
$$\Delta G =RTx\ln(x)+RT(1-x)\ln(1-x)$$
This energy is negative over the whole range, except at $x=0,1$ where it is zero. The minimum value at $x=1/2$ is $-RT\ln(2)$
After the pure water is added in the left hand compartment and equilibrium is established again, the initial mole fraction can only drop to half its value by mixing equal volumes of water which is a mixing energy of
$$\Delta G\left(\frac{x}{2}\right) = RT\frac{x}{2}\ln\left(\frac{x}{2}\right)+RT(1-\frac{x}{2})\ln\left(1-\frac{x}{2}\right)$$
Again this quantity is negative over the whole range, thus the solution will continue to be diluted if nothing prevents this.
Dilution of the mixture could, in principle, be prevented by the gravitational energy $mgh$ where m is the (molar) mass and h the height the liquid has to be raised to. To prevent further dilution
$mgh - \Delta G(x_f) \gt 0 $
where $x_f$ is the final mole fraction. If $mgh \rightarrow 0$ then $x_f=x/2$.
Using values for water gives $mgh = 0.176h$ Joule for height h whereas thermal energy $RT=2494$ J. If, for example the height is $h=9800 $ m then $mgh \approx RTln(2)$ which is the minimum value of $\Delta G$ then no dilution can take place as $\Delta G \gt 0 $ for all x.
At smaller heights whether or not dilution occurs depends on the mole fraction of the initial mixture but at a height of $1000$ m as long as the initial mole fraction is greater than $\approx 0.035$ then complete dilution will occur. Below $1000 $ m the final mole fraction is effectively half the initial value, $x/2$.
Of course the heights involved in effecting any change are quite ridiculous, but this is just because gravity is a weak force; you may not find it so if you ever go hill running :)
The figure shows different values of $\Delta G/RT$ per mole.
(a) shows the normal mixing free energy $\Delta G/RT$ vs. mole fraction solute x (b) the change upon no dilution and a potential of $1000$ m .(c) maximal dilution (doubling volume) at $1000$ m, thus dilution can occur for all initial values of molec fraction above $\approx 0.035$.(d) $\Delta G/RT$ plus $mgh = RT\ln(2)$ for water $h \approx 9800$ m. No transfer of water can occur in this case as $\Delta G \gt 0$
