Consider a concentrated dye solution, say 10µM of Rhodamine 6G in water, which is rapidly diluted down to 10nM in an open container. I'm trying to estimate the (presumably minute) temperature change, assuming both solutions (dye and water) are initially at the same room temperature. I believe dilution increases entropy ever so slightly, but I'm struggling to turn this statement into a quantitative temperature change. Specifically, how do I calculate the change in entropy? Once I know this, the change in temperature should be straight-forward.

  • $\begingroup$ You need to measure the heat of mixing of the solute and solvent. $\endgroup$ – Chet Miller Dec 26 '17 at 2:52
  • $\begingroup$ @ChesterMiller sure, I could (try to) measure the temperature change, but I'm trying to estimate it theoretically. $\endgroup$ – baptiste Dec 26 '17 at 4:52
  • $\begingroup$ Why are you concerned with entropy but not with enthalpy contribution? $\endgroup$ – Alchimista Dec 26 '17 at 13:41
  • $\begingroup$ Also 10µM is particularly concentrated. You could get like 10 mM or more depending on counterion. $\endgroup$ – Mithoron Dec 26 '17 at 15:13
  • $\begingroup$ @Alchimista if enthalpy is more relevant please provide an explanation. I'm after the temperature change, and I assumed it would occur because of a change in entropy but maybe there's another contribution, I'm quite rusty on all this. $\endgroup$ – baptiste Dec 26 '17 at 20:24

At such low concentrations, can’t you just assume an ideal solution? Then the change in entropy from mixing would be $\Delta S = -R(x_A\ln x_A + x_B \ln x_B)$. But because the solution is so dilute, you can see that both terms are going to be essentially zero. (For a 10 nM solution, the contribution from entropy of mixing is on the order of $10^{-8}$ J K$^{-1}$ mol$^{-1}$.)

The dilution process would thus lower the total entropy, since there’s less uncertainty in the system. (You’d get the value you compute at 10 nM, the final state, minus the value at 10 µM, the initial state.)

  • $\begingroup$ sorry what's $x_A$? I do expect the answer to be small, but is it $10^{-10}K$ small or more like $10^{-2}K$ $\endgroup$ – baptiste Dec 26 '17 at 4:51
  • $\begingroup$ I'd expect something in between, like $10^{-5}$ or $10^{-6}$. $\endgroup$ – Ivan Neretin Dec 26 '17 at 6:27
  • $\begingroup$ The x’s are mole fractions of solute and solvent. $\endgroup$ – aeismail Dec 26 '17 at 6:32
  • $\begingroup$ I’ve included an order estimate in my answer. $\endgroup$ – aeismail Dec 26 '17 at 6:46
  • $\begingroup$ A solution typically does not approach ideal solution behavior at low concentrations (particularly for an aqueous ionic solution). This would only be the case if you were mixing molecules that have very similar molecular characteristics. The activity coefficient at infinite dilution in a liquid solution usually deviates maximally from unity. $\endgroup$ – Chet Miller Dec 26 '17 at 13:05

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