Finding the equilibrium vapor pressure of two solutions(of different concentrations) enclosed in a vessel

Question

The saturated vapour pressure above an aqueous solution of sugar is known to be lower than that above pure water, where it is equal to $p_{sat}$, by $\Delta p = 0.05p_{sat}c$. where c is the molar concentration of the solution. A cylindrical vessel filled to height $h_1 =10$cm with a sugar solution of concentration $c_1 = 2\times 10^{-3}$ is placed under a wide bell jar. The same solution of concentration $c_2 = 10^{-3}$ is poured under the bell to a level $h_2$ << $h_1$ (Shown in the given figure).Determine the level h of the solution in the cylinder after the equilibrium has been set in. The temperature is maintained constant and equal to 20 °C. The vapour above the surface of the solution contains only water molecules, and the molar mass of water vapour is $\mu = 18\times 10^{-3}$ kg/mol.

This is how solved:
It is clear from the above data that as the equilibrium is achieved concentration of solution in cylinder decreases whereas concentration of the solution outside the cylinder increases.Let the height of the water in the cylinder after the equilibrium be $h$.As the water evaporates from cylinder the number of molecules in the cylinder remains the same. $$h_1c_1=hc$$(where c=conc. of solution in cylinder after equilibrium)
The conc. of solution outside of cylinder will also be $c$.
So vapour pressure in the bell jar will be $p=p_{sat}-0.05p_{sat}c$
Using ideal gas law
$$p\mu =dRT (where\enspace d=density\enspace of\enspace vapour)$$ $$\implies (p_{sat}-0.05p_{sat}c) \mu=dRT$$ $$\implies (p_{sat}(1-0.05\frac{h_1c_1}{h})) \mu =dRT$$ Now here is where I am stuck.As it is clear from the above equation that we need to know the density of the vapour to find $h$.So how do we find it.I thought a lot about it but literally have no idea.
The author has provided with a solution which contradicts my logic "conc. of solution in the cylinder decreases whereas that of outside the cylinder increases."Also the rest of authors solution doesn't seem to be logical to me. Following is his solution:

The concentration of the solution of sugar poured above a horizontal surface practically remains unchanged.
After the equilibrium sets in, the concentration of the solution in the vessel will be $c=\frac{c_1h_1}{h}$.
The concentration changes as a result of evaporation of water molecules from the surface (concentration increases) or as a result of condensation of vapour molecules into the vessel (concentration decreases). The saturated vapour pressure above the solution in the cylindrical vessel is lower than that above the solution at the bottom by $\Delta p = 0.05p_{sat}(c — c_2)$. This difference in pressure is balanced by the pressure of the vapour column of height h: $$\rho_{v}gh=0.05p_{sat}(c — c_2)$$ Hence we obtain:
$$h=\frac{0.05p_{sat}(c — c_2)}{\rho_{v}gh}$$
The density pv of vapour at a temperature T =293 K can be found from the equation of state for an ideal gas: $$\rho_{v}=\frac{p_{sat}\mu }{RT}$$ Thus, the height h of vapour column satisfies the quadratic equation $$h=\frac{0.05 c_2 RT}{\mu g}\frac{2h_1}{h-1}$$ Substituting the numerical values and solving the quadratic equation, we obtain $h\simeq 16.4cm$.

As it can be easily seen from the author's solution that he has assumed different value of vapour pressure above cylinder and at the bottom even though they are in the same closed bell jar which is very counter intuitive . I would be glad if someone logically and conceptually justifies the author's method.

Answer 1

Since the key question is where matter will move, this can be broadly framed as a chemical potential problem. With only the water gas phase of real interest (i.e., no phase changes—the equilibrium vapor pressure tells us everything we need to know about the liquid—and no chemical reactions), we can use the water gas pressure as a surrogate.

The description of "wide bell jar" and the fact that $h_2$ isn't given alerts us to think of the surrounding solution as a large reservoir of low sugar concentration (and high water chemical potential). This reservoir surrounds a cylinder of high sugar concentration (and low water chemical potential). Thus, we would expect water to move from the reservoir to the cylinder.

Wait—is the solution trivial? Since the concentration difference is a factor of two and the vapor pressure dependence is linear, the height in the cylinder must double to attain the concentration of the reservoir! Thus, $h=20\,\mathrm{cm}$! Well, wait a second. The cylinder is narrow and solution height within it high, and we know that altering the pressure (here, hydrostatic pressure) alters the chemical potential. Maybe we should incorporate that.

Let's move within the bell jar, always considering equilibrium conditions.

Start by writing the water vapor pressure at the surface of the reservoir:

$$p_\mathrm{sat}(1-0.05c_2).$$

Now move up to the top of the solution in the cylinder, accounting for hydrostatic pressure changes in the water gas. The barometric pressure formula gives a difference of $\Delta p=p_\mathrm{sat}(1-e^{-\mu gh/RT})$, which is approximately $p_\mathrm{sat}\mu gh/RT$ for small $h$. (It looks like the given solution uses this same approximation, which is equivalent to assuming a constant gas density over this small height.) Thus, the pressure over the cylinder solution (as controlled by the vapor pressure of the reservoir solution) is

$$p_\mathrm{sat}(1-0.05c_2)-\frac{p_\mathrm{sat}\mu gh}{RT}.$$

At equilibrium, this is equal to the saturation vapor pressure above the cylinder solution, which we'll assume to be well mixed, so $c=h_1c_1/h=2h_1c_2/h$:

$$p_\mathrm{sat}(1-0.05c_2)-\frac{p_\mathrm{sat}\mu gh}{RT}=p_\mathrm{sat}\left(1-2\frac{0.05h_1c_2}{h}\right);$$

$$\frac{\mu gh^2}{RT}=0.05c_2(2h_1-h).$$

Solve for $h$ to obtain 16.2 cm. Does this make sense?