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Why does entropy increase when we mix two ideal liquids? In my opinion entropy shouldn't change.

Say we have two beakers of equal volume filled with $\ce{H2O(l)}$. I mark them red and blue (the marking doesn't change the chemical or physical configuration or composition). Now I pour half of the water from each container into an empty container so that there is 50% red $\ce{H2O(l)}$ and 50% blue $\ce{H2O(l)}$ in the new one.

Did the entropy increase?

Neither thermal nor configurational entropy could have been affected since they are the exact same molecule, i.e no intermolecular force has been affected (some sources mention configurational entropy). Also, the entropy in question is after the mixture settles after mixing. Commenting about entropy would be hard as we don't know how exactly the molecules will randomize.

$\ce{H2O}$ is an appropriate example because intermolecular force of attraction between the two molecules (red and blue) will be same to the force between 2 red/blue $\ce{H2O}$, cohesive forces = adhesive forces ($F_{RR}=F_{RB}=F_{BB}$).

Edit 1:

Summarizing, why exactly does the entropy increase for ideal solutions? If considering the example of $\ce{H2O}$ is incorrect, consider another example, wherein we mix $\ce{HCl}$ and $\ce{H2O}$. The entropy should decrease (in my opinion) as intermolecular forces between the two molecules have increased (hydrogen bonding between the molecules of $\ce{HCl}$ and $\ce{H2O}$, $\ce{H2O}$ will surround the $\ce{HCl}$ molecule), but the entropy increases for some reason. What exactly is the reason for that?

These questions aren't different as the reason why entropy increases for ideal solutions (theoretically ideal, and not near ideal) should be the reason why entropy increases in case of mixing of $\ce{HCl}$ and $\ce{H2O}$.

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Chemistry Meta, or in Chemistry Chat. Comments continuing discussion may be removed. $\endgroup$
    – andselisk
    Commented May 30 at 11:36
  • $\begingroup$ Mark the water molecules make half H2O18; mix them then try to separate them. To have an entropy change there must be a difference. Take 2 glasses of water pour into a container and refill the glasses with the exact same amount of mixed water Same state, all state functions unchanged. the entropy increase to accomplish the feat is in the environment. $\endgroup$
    – jimchmst
    Commented yesterday

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Entropy is a measure of the number of ways (configurations) in which a system (the liquids before or after mixing) can be arranged subject to constraints such as the pressure and temperature. At the molecular level computing the entropy is a combinatorial problem, which is why the branch devoted to the subject is called statistical thermodynamics.

For a pure substance entropy is a linear additive property. Additivity follows from application (implicitly) of logarithmic algebra to the number of configurations ($\Omega$): $$\log(\Omega_\textrm{final})=\log(\Omega_\textrm{1}\times\Omega_\textrm{2})=\log(\Omega_\textrm{1}) + \log(\Omega_\textrm{2}) \\ \implies S_\textrm{final} = S_\textrm{1}+S_\textrm{2} = S_\textrm{initial} $$

To compute the number of configurations of two independent volumes of the same substance (at the same T, p) you don't sum together the number of configurations for each volume, you multiply them, because for each configuration of volume 1 there are $\Omega_2$ possible configurations of volume 2.

For a combined volume of the substances the same applies because for each configuration of volume 1 the same original configurations remain possible for that volume, and again for each configuration of volume 1 there are $\Omega_2$ possible configurations of volume 2.

When the substances are different, however, the final number of possible configurations will be larger than in the original solutions, because the particles are not indistinguishable. Therefore after mixing the number of possible configurations for the particles in volume 1 will not be $\Omega_1$, it will be $\Omega_1'>\Omega_1$. Strangely, however, once mixed the same rules apply as for a pure substance, and splitting and combining will not change the total entropy.


Thermal entropy couldn't have been affected nor could configurational entropy be affected since they are the exact same molecule, i.e no intermolecular force has been affected (before asking this question, some sources mentioned otherwise about the configurational entropy).

Thermal entropy presumably refers to a change in the entropy of the surroundings or the system due to heat released during mixing. During mixing of ideal substances no heat is released (mixing enthalpy is zero), so this contribution to the entropy will be zero.

The configurational entropy has many sources. Configurations refers to both where molecule are in space and how they are positioned, oriented and moving relative to other molecules, and to their internal structure (conformations) and motions. Mixing two volumes of a homogeneous substance would not alter the possible configurations for the whole even if it does for individual molecules.

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  • $\begingroup$ How exactly is Ω related to degree of freedom (the one which tells us the internal energy of a thermodynamic system, U = nRfT/2, are they the same thing? $\endgroup$
    – Gauransh
    Commented May 27 at 9:39
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    $\begingroup$ They aren't the same, but they are related. Degrees of freedom can refer to various things. In this context it tells you how many coordinates can be changed, for instance translational degrees of freedom. To compute $\Omega$ you would integrate (sum) over all independent degrees of freedom to obtain all possible configurations. $\endgroup$
    – Buck Thorn
    Commented May 27 at 9:50
  • $\begingroup$ Configurations refer to both where molecule are in space and how they are positioned and oriented relative to other molecules, not just their structure. Isn't saying the above statement and then considering only translation degree of freedom contradictory? We'd have to count rotational and whatever other types of degree of freedom that exist as well right? $\endgroup$
    – Gauransh
    Commented May 27 at 11:23
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    $\begingroup$ You should consider all degrees of freedom. $\endgroup$
    – Buck Thorn
    Commented May 27 at 11:24
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    $\begingroup$ See also en.wikipedia.org/wiki/Gibbs_paradox#The_mixing_paradox $\endgroup$
    – arkeet
    Commented Jun 3 at 7:47
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If we have $1$ mol each of two different types of ideal gases ($\ce{He}$ and $\ce{Ne}$ for example) at the same pressure and in two equal volumes separated by a partition if you remove the partition the gases will diffuse into one another and you can do experiments to determine that mixing has happened. The entropy increase is $\Delta S=-R(1\ln(0.5)+1\ln(0.5))=2R\ln(2)$ where the mole fraction is $0.5$. The same result is true for any two different inert gases no matter how alike or not they may be. The change in entropy is only due to the fact that the volume of each gas is increased, not the mixing per se.

Suppose that after a while the partition is replaced then it will be clear by experiment that the two parts are distinguishable from those initially. For example the two volumes could be weighed and found to be identical but different from their initial weights.

If the two gases are as before in two partitions and of equal molarity and pressure but now they are identical, removing the partition causes no measurable diffusion. If after some time the partition is replaced the situation is just as if it had never been removed and so any increase in entropy does not make sense. There is no way that any experiment on the gas can be made that will show that any change has happened, by weighting for example as was done with different gases.

This is often called 'Gibbs Paradox' and arises from using classical arguments to calculate thermodynamic properties (partition function) rather than quantum mechanics where the essential indistinguishability of particles is explicitly accounted for.

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  • $\begingroup$ Indeed this is essentially the argument made by Jaynes: that entropy can be "subjective" depending on whether you have the capacity to distinguish the two types of particle or not: damtp.cam.ac.uk/user/tong/statphys/jaynes.pdf $\endgroup$
    – issy
    Commented May 29 at 12:37
  • $\begingroup$ @issy I'm not sure that subjective is quite the right word. The important point is that in a mixture of different types of molecules work need to be done to restore the original thermodynamic state with molecules separated into two boxes, but with identical molecules no work is needed to do this. $\endgroup$
    – porphyrin
    Commented Jun 7 at 7:53
  • $\begingroup$ I think the point Jaynes is making is that if you have an ability to (or care about) one set of macroscopic observables and I have a different set, then we may say that the entropy change is different for the same process; e.g. if I can only measure T, p but you can distinguish Ne from He as well, then you will say DeltaS > 0 and I will say DeltaS = 0. $\endgroup$
    – issy
    Commented Jun 8 at 14:24
  • $\begingroup$ Not withstanding the Jaynes argument, the 'paradox' arises from allowing the properties of two types of molecules to approach each other gradually, and then it will not be possible to tell at which point they become so similar that the entropy no longer changes on mixing them. This classical model is false. In quantum theory we know that properties change discontinuously, i.e. molecules are identical or they are different and the paradox no longer exists. $\endgroup$
    – porphyrin
    Commented Jun 15 at 9:04

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