(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

My question is more global than a specific exercise, but I will illustrate my question with two examples.

My question is, when determining entropy for systems of say, 1000 molecules, 200 cards, and other small amounts (in contrast to the 6.022 E23 molecules in one mole), is it okay to give our result according to the well-known formulas (cited in2.)?

Let me put two examples:

a)You have 5 identical sets of cards, each with 24 different cards. Then you shuffle the five sets, calculate the entropy.

b)You have 1000 molecules of N2 and 250 molecules of O2 in two different containers until you mix them together. Imagine that they reach equilibrium and the process let no temperature change due to the large-enough volume.

2. Relevant equations

Fora)I would just use:

[tex]S = k_B lnW[/tex]

Where W are the possible combinations, so:

[tex]S = k_B ln \frac{n!}{(n-r)!r!}[/tex]

Now forb), similarly:

[tex]S = k_B ln\Omega[/tex]

And [tex]\Omega[/tex] is the different kind of arranging possibles, i.e.:

[tex]S = k_B ln \frac{N!}{N_1!N_2!}[/tex]

Where [tex]N = N_1 + N_2[/tex]

3. The attempt at a solution

The calculations are really straight forward at this point.

For a) I would use n = 24 and r = 5, and at the end, it gives me S = 10.65 * k_B = 1.47 E-22

And b) I use N1=1000 and N2=250, being N=1250. And after plugging the numbers in the computer (since the number of different arrangements is rather high). It gives me S= 621.93*k_B = 8.58 E-21.

As you can see both results are really small, I don't know if I'm missing something, or if in these statistical entropy cases there are other ways to deal with it.

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# Homework Help: Determining entropy

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