Intuitively, why does removing solutes cost $k_B T$ of free energy per molecule?

Question

I can calculate that if you want to, for example, desalinate water, you will have to pay a free energy cost of $k_B T$ for each ion you remove. In other words, removing an ion from a volume of water requires $\log_2 e$ bits of information. Is there an intuitive reason why?

Below is the calculation:

Imagine a tank of water from which we are removing ions, which currently have concentration $c$. We move the ions to a reservoir of concentration $c_0$.

Each ion can be considered to occupy a volume $1/c$, so the entropy change in moving an ion from our tank is $k_B\ln(c/c_0)$. As we empty the tank starting from $c = c_0$ to $c = 0$, the average entropy change per ion is

$$\mathrm{d}\bar{S} = \frac{k_B}{c_0}\int_{c_0}^0 \ln(c/c_0)\mathrm{d}c =- k_B$$

Because the internal energy doesn't change, the free energy cost is $$N k_BT$$ to completely remove $N$ solutes.

Answer 1

I am not sure if this answer will give you an intuitive understanding of the result, but I think it may be useful as it shows the assumptions behind it.

What your result means is that in an idealized situation when the volume gas or solute occupies is shrunk slightly by $\delta << V$ while its energy remains the same (let's say, isothermal compression of ideal gas), the entropy decreases by $k_B N'$, where $N'$ is the number of molecules in the lost volume before the shrinking occurs.

This can be shown as follows. Let the entropy of the gas be defined as log of volume of phase space available to it. Initially, the gas of $N$ molecules has volume $V$, in the end, $V-\delta$ and $N'$ molecules have moved to the remaining space. The change of entropy is

$$ \Delta \sigma = \log (V-\delta)^N - \log V^N $$

$$ \Delta \sigma = N\log \frac{V-\delta}{ V} $$

Since $\delta << V$, approximately $$ \Delta \sigma = - N \frac{\delta}{ V}. $$

Quantity $N/V$ is concentration $c_0$ of the gas, so the change of entropy is

$$ \Delta \sigma = - c_0 \delta = -N'. $$

Thus entropy decreases by 1 for each molecule removed from the lost space.

This is similar to change in thermodynamic entropy of the system. Since we have isothermal compression of ideal gas, all energy transferred through work eventually gets transferred to environment as heat. Change in the Clausius entropy is

$$ \Delta S = Q'/T $$

where $Q'$ is total received heat (actually the system gives off heat, so $Q'<0$.)

From 1st law of thermodynamics, $Q' = \Delta U - \Delta W$ and since internal energy does not change, it is equal in value to minus work done on the system. Since $\delta << V$, approximately $$ Q' = - P \delta. $$

Putting this into the formula for $\Delta S$, we obtain

$$ \Delta S = Q'/T = - \frac{P}{T} \delta $$

and using the equation of state $P = c_0 k_B T$, we arrive at

$$ \Delta S = - N' k_B. $$

This result is similar to $\delta \sigma$, except for the additional $k_B$ that we need in thermodynamic entropy.

For ideal gas, this result is perhaps not that surprising. There is work $k_B T$ per molecule to overcome the pressure of the gas itself, since it obeys the ideal gas equation of state.

For a solute in solvent, it seems more interesting: it means that after we strip solvent out of $N'$ solute molecules and dump them in the rest of the solution containing many times higher number of molecules, the system will have released heat $k_BTN'$. Entropy of the system decreases, so necessarily it must give off heat.

The calculation was exact for ideal gas, but solutions are more complicated. We neglected the solvent and its interaction with the solute which will probably not allow the process to proceed along constant internal energy (it could be easily made isothermal by making it slow enough, but energy is no longer function of $T$ for general solution). Thus the entropy cannot be expressed as function of configuration space volume only; momentum phase space part will become important as well and the calculation will be more difficult.

The heat produced in the filtering process thus may be lower or higher than the ideal value $k_BT$ per molecule.

Answer 2

Assuming a constant temperature, every particle has on average 1/2 kB T of "kinetic" thermal energy in each degree of freedom.

When a piston compresses a gas with such properties, it needs to do work on the gas by opposing the pressure, which is in kPa = J/L = J/dm^2 * dm. The piston has an area and it is moving along an axis, which is dimensionally equivalent to volume. We can call this "PV work" and simply accept it.

Nonetheless, I think we can also take it a step further apart. What are we doing? We're reflecting an individual particle in one dimension, which has a kinetic energy of 1/2 kB in that dimension. And we can't just stop it dead in its tracks or we'd have free refrigeration to absolute zero; we have to send it back the way it came with another 1/2 kB of work done to it. This is the same as a juggler doing work on a ball by catching it and tossing it back up again.

Now a solute in solution is very much like a gas, in that it is made of mostly non-interacting particles with the same thermal energy per degree of freedom. Ideally, osmotic pressure P = cRT = (n/V)RT, whether we're speaking of liquid or gas. (Even non-ideal behavior is considered comparably in many ways)

So at the moving osmosis membrane, or the surface of the boiling or freezing solvent, the same thing is always happening - a conceptual "piston" is forcing the solute into an ever-tighter space, with predictable effect on entropy and free energy.

At the moving boundary, the cost is kB T; but when considering a change in volume, it is kB T ln(V1/V2). The reason for this is that the pressure is the inverse of the volume, which is also the inverse of the linear axis on which the work is being done, and so we're integrating 1/x to ln x. The pressure increases all the while we compress the solute, which implies that some of the particles we bounce back once will be bounced back again before we finish changing the volume. For every factor-of-e change in volume that occurs, another kB T of energy is required; this is analogous to the mean-life of a radioisotope being the time until 1/e of it is left.