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

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.

• What is $e$ in the second sentence? – Ján Lalinský May 11 '14 at 6:19
• "Each ion can be considered to occupy a volume 1/c," This does not seem right. Each ion is free to move all around the tank, so the configuration part of the phase volume entropy is $k_B N\ln V$, where $V$ is volume of the whole tank. If the reservoir has 1000x larger volume than the tank, moving ion from the tank to the reservoir will decrease the former's entropy by $k_B\ln V$ and increase the latter's entropy by $k_B\ln (1000V)$... – Ján Lalinský May 11 '14 at 6:25
• I think we should agree on some concrete process by which this ion removal is done, because that may also have impact on the total entropy change (the above takes only configurations into account, not momenta). – Ján Lalinský May 11 '14 at 6:33
• $e$ is the base of the natural logarithm. My expression for the entropy is obvious from extensivity. The problem with what you have is that the ions are indistinguisable, so you're missing a factor 1/N!. You can confirm that modifying your expression in this way is equivalent to what I said. – Mark Eichenlaub May 11 '14 at 6:37
• momentum is unimportant because it doesn't change between the two reservoirs. I'm expecting the reader to use some discretion. This lets me avoid unnecessary lists of caveats and assumptions. – Mark Eichenlaub May 11 '14 at 6:39

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.