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We have that

$$ S = \frac{U}{T} + k\ln Q $$

Using that for indistinguishable particles in the canonical ensemble

$$ \ln Q_{\mathrm{indis}} = N\ln \frac{q_{\mathrm{tr}}e}{N} + N\ln q_{\mathrm{rot}} + N\ln q_{\mathrm{vib}} + N\ln q_{\mathrm{el}} $$

Similarly for distinguishable particles we have that

$$ \ln Q_{\mathrm{dis}} = N\ln q_{\mathrm{tr}} + N\ln q_{\mathrm{rot}} + N\ln q_{\mathrm{vib}} + N\ln q_{\mathrm{el}} $$

Then we assume the indistinguishable particles suddenly become distinguishable. By how much does the entropy of the system change?

$$ \Delta S = S_{\mathrm{dis}} - S_{\mathrm{indis}} = k(\ln Q_\mathrm{dis} - \ln Q_\mathrm{indis}) = Nk[\ln(N)-1] $$

In the special case that the system contains only one particle ($N=1$), the entropy changes by $-k$.

Intuitively, it does not matter whether one particle becomes distinguishable from the rest if there is only one particle; Assigning a label to just one particle does not really modify the information within the system.

So how can this be understood?

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    $\begingroup$ How can an indistinguishable particle suddenly become distinguishable without any energy? $\endgroup$ – DHMO Sep 7 '16 at 11:45
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    $\begingroup$ Whether the particles are indistinguishable or distinguishable does not affect the energy. $\endgroup$ – orthocresol Sep 7 '16 at 14:44
  • $\begingroup$ Just out of curiosity does this describe a physical process? It's very interesting regardless, but just wondering. $\endgroup$ – jheindel Sep 8 '16 at 19:59
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Very interesting question. The issue is that your formula for $\ln Q_\mathrm{indis}$ does not hold for $N = 1$.

Since the rotational, vibrational and electronic degrees of freedom do not come into play I will just ignore them. The way you derived the term $\ln (q_\mathrm{tr}e/N)$ comes from the use of Stirling's approximation

$$\begin{align} Q_\mathrm{tr,indis} &= \frac{q_\mathrm{tr}^N}{N!} \\ \ln Q_\mathrm{tr,indis} &= N \ln q_\mathrm{tr} - \ln(N!) \\ &\approx N \ln q_\mathrm{tr} - N \ln N + N \\ &= N\ln\left(\frac{q_\mathrm{tr}e}{N}\right) \end{align}$$

Plugging $N = 1$ into Stirling's approximation however tells you

$$\ln (1!) \approx 1\ln 1 - 1$$

or $0 \approx -1$, which is obviously a problem. Stirling's approximation is really meant to be used for large $N$ only!

Now, you could try working it out without using the approximation: plug $N = 1$ into the line before the approximation $\ln Q_\mathrm{tr,indis} = N \ln q_\mathrm{tr} - \ln(N!)$ and you will get the result

$$\ln Q_\mathrm{tr,indis} = \ln q_\mathrm{tr}$$

which matches the expression for the distinguishable case, so all your terms will cancel out and you will get the desired answer of $\Delta S = 0$.

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