Why is the change in entropy is different for different interpretations?

Question

Forgive me if the title is a bit vague. Read on and I'll clarify.

So, today in class, my teacher was explaining entropy. He gave various examples explaining how the degree of randomness changes. One example in particular stuck out to me. Here it goes:

Imagine there is a container with a hypothetical boundary in the middle. Neon gas is on the left of the boundary, and Argon on the right. As long as the boundary exists, the gases can't cross it to intermix. Now the boundary is removed, and the gases mix. This increases the randomness in the system and thus entropy increases. We get $\Delta S \gt 0$. Thus the process is spontaneous.

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My Question:

How can we definitively say that entropy has increased (or that randomness has increased)?

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My Reasoning:

So my teacher says that entropy increases. These following cases show that change in randomness depends on your interpretation.

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Interpretation 1:

Before mixing (boundary exists):

If we take some volume of gas, say $1 \text{ cm}^3$, at a random position in the container, we might get:

1. Pure $\ce{Ne}$
2. Pure $\ce{Ar}$
3. A mixture of the two gases

The point is that we aren't sure what composition we get, because the distribution of gases isn't homogeneous.

After Mixing (boundary removed):

The gases diffuse and intermix to form a homogeneous mixture.

Now if we take $1 \text{ cm} ^3$ from a random position, we can be $100\%$ sure of the composition of the gases, because the mixture is homogeneous.

This means there is more uniformity now. Or, we can say that there is less randomness. This tells us that $\Delta S \lt 0$.

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Interpretation 2:

Using a new approach now.

Before Mixing (boundary exists):

Take any single atom from either side of the boundary at random. My chances of correctly predicting what atom I get (either $\ce{Ne}$ or $\ce{Ar}$) are $50\%.$

After mixing (boundary removed):

I again pick any atom at random from a random point in the container. My chances of correctly predicting what atom I get is again $50\%.$

This tells us that since I still have the same chance of correctly predicting, the randomness of the system has not changed.

This gives us $\Delta S = 0$.

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Same physical process, but we get 3 different values of degree of randomness ($\Delta S$) depending on our interpretation. We know that diffusion does take place, so the process is spontaneous, and hence $\Delta S \gt 0$.

My Questions:

• How to accurately know whether the degree of randomness increases or not?
• How is the exact value of $\Delta S$ measured in actual experiments?

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Note:

• Assume the gases to be ideal.
• Assume the container to be sufficiently large.

Answer 1

From the perspective of Statistical Mechanics:

Multiplicities do depend on the external parameters that specify the energy level of the system.

For ideal gas of $N$ indistinguishable molecules, the multiplicity is given by $$\Omega= \frac{1}{N!}\frac{V^N}{h^{3N}}\,\frac{2\pi^{3N/2}}{(3N/2- 1)!}(\sqrt{2mE})^{3N-1}$$

So, indeed the multiplicity increases when the position space increases.

When the external parameter like volume of the container is increased, the energy levels come closer and then the particles can have more microstates for the same $E$.

Prior to the volume increase, the microstates of the system are $\Omega_{N,E,V_i}\;_;$ when the volume is increased, not will be there the microstates $\Omega_{N,E,V_i}$ still be available after the removal of the constraint, but also, there be addition of microstates. In general, $$\Omega_{V_f}\geq \Omega_{V_i}\;.$$

So, as the volume increases, the multiplicity increases and so does the entropy.

As requested by OP:

From the perspective of Thermodynamics:

Suppose the two dissimilar indistinguishable gases $A$ and $B$ having the same number of particles, energy, volume are partitioned from each other.

Now, when the partition is removed, the energy of the gas $A$ confined to a localised area gets dispersed over a greater volume.

Thus the entropy of gas $A$ increases and that is given by \begin{align}\Delta S_A &= Nk\,\ln \frac{V_f}{V_i}\\ &= Nk\,\ln \frac{2V}{V}\\ &= Nk\,\ln 2\;.\end{align}

Same happens for gas $B$ also.

Therefore, the change in entropy of the system is given by \begin{align}\Delta S_\textrm{system}&= \Delta S_A + \Delta S_B\\ &= 2Nk\ln 2 \gt 0\end{align}

Therefore, entropy increases when the partition is removed.

Answer 2

A non-mathematical look at entropy

Now while this doesn't directly answer the OP's questions, it provides another way of looking at the concept of entropy, something that's not the dreaded ‘disorder’[1]. An addendum to @user36790's answer perhaps.

Rather than disorder or chaos entropy is probably better described as the number of ways I can do something at the individual particle level and still get the same overall result on the large scale.

Entropy is a measure of the number of micro-states[2] a system can potentialy occupy and still give the same macro-state. A micro-state is the way the particles (in your case Ar and Ne atoms) are arranged and the energy of the system is partitioned between these particles. A macro-state is a particular arrangement of a group of particles characterised by temperature, volume (sometimes pressure), number of particles etc. it is a collection of microstates. (See [1] for a some technical definitions)

Entropy compares the number of microstates that make up different macrostates. So given one macrostate we can look at the number of ways we can make that state, and then look at the number of ways we can make a different macrostate. That number we are comparing is analogous to the entropy. If one of the macrostates has more ways it can be made than the other – it has a higher entropy.

Examining the OP's System

Let's start with just looking at one gas, neon. (Why Ne? All the others are gone ;)

Before Mixing (boundary exists):

The N particles of neon occupies a volume V at temperature T. There are a number of different ways you can arrange the particles and partition the energy to give you that N, V and T.

After mixing (boundary removed):

Now lets remove the partition and let the particles equilibrate. It doesn't really matter if the partition was halfway, one third, all that matters is that there is now more volume for the particles to occupy. There are more ways to arrange the neon atoms in your container (it's bigger).[3]

You need to be careful. The macrostate your microstates can be combined to give before the partition is removed is different to the one after the partition is removed. What entropy compares is macrostates. It does this my enumerating microstates.

You can do the same for argon. The process and thinking are the same, increase in space give an increase in the number of ways to arrange the particles to give the same (new) outcome.

An analogy: You have 10 books and one shelf in your study. There are a certain number of ways you can arrange the books on your shelf (microstates) to obtain the same end result, books on the shelf (macrostate).

Now, you buy a second shelf. There is now more space (and ways) to arrange your books and obtain a macrostate.

Be careful though, this is a different macrostate to before, your shelf space (volume), and pressure (roughly how close the books are packed) have changed. There are still more ways to arrange the books on the shelf in the second state than the first though, there's more space, so the second state is considered to have a higher entropy.

($S$ is larger in the second state, going from state one [one shelf] to state two [two shelves] results in an increase in entropy of the system $\Delta S \gt 0$)

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[1] ‘Disorder—A Cracked Crutch for Supporting Entropy Discussions’ Frank L. Lambert, J. Chem. Educ., 2002, 79 (2), p. 187, DOI: 10.1021/ed079p187 (Behind a paywall I'm afraid.)

Quote from the above paper:

A macrostate is measured by its temperature, volume, and number of molecules; a group of molecules in microstates (“molecular configurations”, a microcanonical ensemble) by their energy, volume, and number of molecules. In a microcanonical ensemble the entropy is found simply by counting: one counts the number $W$ of microstates that correspond to the given macrostate and computes the entropy of that macrostate by Boltzmann’s relationship, $S = k_B \ln{W}$, where $k_B$ is Boltzmann’s constant.

[2] When I see microstates I think ‘ways the system can be arranged’, every time I see microstates I just substitute in ‘ways the system can be arranged’. You have to be careful though, the way a system can be arranged includes both the positions of the particles and how the energy could be distributed through out the system. For an preliminary understanding though you can just look at the positions.

[3] You can actually look at the the box before and after, as one box, it just gets bigger. Removing the partition is just a way of visualising transitioning from one volume to another, the mixing has very little to do with it. The gasses don't interact anyway, and they occupy negligible space compared to the volume of the box, they're assumed to be ideal.