# How can you explain the energy change in this phenomenon?

Tagged is the diagram of the apparatus used in this thought experiment. Pure water is taken in one side of the beaker and a solution of common salt in water in the other. After a very long time, the pure water should get transferred to the NaCl solution side because according to Raoult's law, the NaCl solution should have lesser vapor pressure than pure water, thus causing the water-vapour equilibrium in the pure water side to move forward, resulting in the condensation of the vapour in the other side (the solution side) . But how has the potential energy (=mgh) of the pure water increased during the process?

• Get in a car and drive all the way to the top of hill. Your potential energy (yes, that mgh) will increase, too. How? It happened at the cost of the chemical energy. Same thing here. – Ivan Neretin Apr 13 '16 at 14:04
• Energy of the salt solution and pure water. True, it is far less effective than gasoline, but still enough to do the trick. See, now that they are mixed, you can't have them separated again. Or rather, you can, but it will cost you some energy. – Ivan Neretin Apr 13 '16 at 14:12
• With infinite height, the setup will not work. In fact, a finite height of a few miles would be enough to stop the pure water from creeping into the solution. – Ivan Neretin Apr 13 '16 at 14:31
• That's complicated. We might have to look up the data on enthalpy and entropy of dissolution. In short, the temperature might change either way (very slightly, though) or not change at all. – Ivan Neretin Apr 13 '16 at 14:47
• Sorry, but I think a textbook on chemical thermodynamics would not fit in a comment. – Ivan Neretin Apr 13 '16 at 15:30

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 $\Delta G \gt 0$

It is important to realize that you deal with a lot of particles and that you can apply the usual approach and approximations in statistical thermodynamics.

It makes more sense to ask for the potential energy of individual water molecules. Because there are so many molecules, we assume the Maxwell-Boltzmann distribution to be continuous. This means, that there will be particles with enough kinetic energy to overcome the potential energy wall (including non gravitational forces like dipole-dipole interaction).

If you had two solutions of pure water at $T > 0 K$, this would just mean that they exchange some of their water molecules over time.

If you have two different solutions as in your example there are more possible microstates for the same macrostate in the salt solution: "Putting the water into the right pot is entropically favoured, because solving salt in water is entropically favoured." The exchange of water molecules is happening even in equilibrium.

As with any thermodynamical answer this ignores kinetic effects and other addends of the chemical potential like surface energy.