# Adkins, Equilibrium Thermodynamics, problem 11.4

I'm stuck on problem 11.4 in Adkins Equilibrium Thermodynamics. The problem is as follows.

An inverted U-tube of cross-sectional area 1 cm$$^{2}$$ has 100mg of sodium chloride at the bottom of one arm and 5ml of water in the other, the remaining space being filled with water vapour. It is kept at a constant temperature of 20 deg C and the water vapour condenses on the salt side to form a salt solution. What is the difference in the levels of the liquids in the arms when the system reaches equilibrium? [Molecular weight of sodium chloride=58.4 . Vapour pressure of water at 20 deg C=17.5mm Hg.]

Here is my attempt at solving the problem. There are three phases. The phase in the left arm denoted by a superscript "L", the vapour denoted by superscript "V", the phase in the right arm denoted by superscript "R". The L phase is initially solid sodium chloride and becomes a solution of salt at equilibrium. The R phase is initially liquid water and becomes a salt solution at equilibrium. The V phase is initially pure water vapour and becomes a mixture of water vapour and sodium chloride vapour at equilibrium. However, the vapour pressure of sodium chloride is so low at 20 deg C that this is, presumably, negligible. There are two components denoted by subscripts "w" for water and "s" for sodium chloride. The equilibrium conditions are that the chemical potentials match up as follows. $$$$\mu_{w}^{L}=\mu_{w}^{V}=\mu_{w}^{R} \\ \mu_{s}^{L}=\mu_{s}^{V}=\mu_{s}^{R}$$$$ Assuming an ideal salt solution on the left, the chemical potential is, $$$$\mu_{w}^{L}(P,T,x_{w}^{L})=\mu_{w}^{L}(P,T,x_{w}^{L}=1)+RT\ln{x_{w}^{L}}$$$$ Here $$\mu_{w}^{L}(P,T,x_{w}^{L}=1)$$ is the chemical potential of pure liquid water and $$x_{w}^{L}$$ is the mole fraction of the water in the salt solution. We can write a similar equation for $$\mu_{w}^{R}$$ and the equilibrium condition $$\mu_{w}^{L}=\mu_{w}^{R}$$ implies the mole fractions in each arm are equal at equilibrium $$x_{w}^{L}=x_{w}^{R}$$. This is intuitively obvious; at equilibrium there are salt solutions of the same composition in each arm.

Now I tried to get some equations for how many moles of water and sodium chloride are transferred between the arms. Let the number of moles of water on the right at equilibrium be $$N_{w}^{R}(0)+\Delta N_{w}^{R}$$ where $$N_{w}^{R}(0)$$ is the initial number of moles of water in the right arm. Similarly, the number of moles of sodium chloride on the left at equilibrium is $$N_{s}^{L}(0)+\Delta N_{s}^{L}$$ . Now write the equality of the mole fractions $$x_{w}^{L}=x_{w}^{R}$$ at equilibrium. $$$$\frac{\Delta N_{w}^{L}}{\Delta N_{w}^{L}+N_{s}^{L}(0)+\Delta N_{s}^{L}}= \frac{N_{w}^{R}(0)+\Delta N_{w}^{R}}{N_{w}^{R}(0)+\Delta N_{w}^{R}+\Delta N_{s}^{R}}$$$$ There are two many variables in this equation, but the total moles of water and salt distributed across the phases is constant so that, $$$$\Delta N_{w}^{L}+\Delta N_{w}^{V}+\Delta N_{w}^{R}=0\\ \Delta N_{s}^{L}+\Delta N_{s}^{V}+\Delta N_{s}^{R}=0$$$$ I think the moles of vapour can be neglected compared to the moles in solution, so that, $$$$\Delta N_{w}^{L}\sim-\Delta N_{w}^{R}\\ \Delta N_{s}^{L}\sim-\Delta N_{s}^{R}$$$$ With this simplification the equilibrium condition becomes, $$$$\frac{-\Delta N_{w}^{R}}{-\Delta N_{w}^{R}+N_{s}^{L}(0)+\Delta N_{s}^{L}}= \frac{N_{w}^{R}(0)+\Delta N_{w}^{R}}{N_{w}^{R}(0)+\Delta N_{w}^{R}-\Delta N_{s}^{L}}$$$$ This simplifies to, $$$$N_{w}^{R}(0)N_{s}^{L}(0)+N_{w}^{R}(0)\Delta N_{s}^{L}+\Delta N_{w}^{R}N_{s}^{L}(0)=0$$$$ In order to calculate the changes in the levels in arms I need to solve for the two variables $$\Delta N_{w}^{R},\Delta N_{s}^{L}$$ so there must be another equation. Since the problem gives the vapour pressure of liquid water at 20 deg C, I feel the extra condition should bring in the vapour pressure. However, if one uses the chemical potential of the vapour $$\mu_{w}^{V}$$ this just gives Raoult's law which doesn't seem to help.

My final thought is that, since the vapour pressure of sodium chloride at 20 deg C is sensibly zero, there is no sodium chloride in the vapour phase and so no sodium chloride can move from the left to the right arm. Consequently $$\Delta N_{s}^{L}=0$$ and the last equation then gives $$\Delta N_{w}^{R}=-N_{w}^{R}(0)$$ so that the water in the right arm evaporates completely. However, I'm not convinced this argument is correct because I haven't made any use of the vapour pressure of water given in the question.

If all salt in left arm gets dissolved in the volume $$V$$ of water, the corresponding osmotic pressure will be equal to (assuming ideal solution): $$P_{osm}=2\frac{\nu_{\text{NaCl}} R T}{V},\tag{1}$$ where the factor of 2 appears because NaCl dissociates into positive and negative ions. The osmotic pressure must be equal to the hydrodynamic pressure difference in the arms: $$P_{osm}=\rho g \Delta h,\tag{2}$$ where $$\rho=1\text{g/cm}^3$$ is the density of water (we assume the density of salt water is the same) and $$\Delta h$$ is the water level height difference between the arms.
Lets assume that not only the salt will suck in all the water from the right arm, but also that the salt will condense additional water from the vapors in the tube. Assuming that the right arm is empty of water, the volume $$V=\Delta h A$$ ($$A=1 \text{cm}^2$$ is the area of the tube). We obtain from equations (1) and (2): $$\rho g \Delta h = 2\frac{\nu_{\text{NaCl}} R T}{\Delta h A}.\tag{3}$$ Solving for $$\Delta h$$ yields: $$\Delta h = \sqrt{2\frac{\nu_{\text{NaCl}} R T}{\rho g A}}\approx 2.9\text{m}.\tag{4}$$ This answer is, of course, an upper limit, since, as water evaporates completely from the right arm, the vapor pressure in the tube is going start dropping, until reaching equilibrium with the salt solution in the left arm.