I'm having trouble with the following problem:

Boltzmann defined entropy as a measure of the number of possible microscopic states of a system, $W$, as:

${S = k_BlnW}$

Here, $k_B$ is the Boltzmann constant. Hereafter, it is assumed that neither energy nor particles are exchanged between the system of interest and the surrounding environment, and that the system is in equilibrium at a constant temperature. In addition, the gas atoms and molecules (henceforth referred to as ‘molecules’) are assumed to behave as an ideal gas.

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Figure 1. Example of a microscopic state (4,6) in which ten molecules are distributed in two chambers separated by a wall.

Removing the boundary wall that separates the two chambers in Figure 1 allows the molecules to move throughout the entire container. Assume that it is possible to measure the arrangement of the molecules instantly after the removal of the wall. If this measurement is performed a sufficiently high number of times, the probability that a state with an arrangement (n,m) will be found can be calculated.

1) Calculate the arrangement ($n^*$, $N$$n^*$) with the highest probability of occurrence for $N = 10$ and $N = 100$, as well as the corresponding probability of occurrence for each state.

2) Calculate the probability of observing a state in which $n^*$ is in the range $n^* – 0.02 N ≤ n^* ≤ n^* + 0.02 N$ for $N = 10$ and $N = 100$. Here both $N$ and $n^*$ are integers.

I was able to calculate the first part and got the following results using a spreadsheet.

For $N = 10$:
$W(5,5) = 252$ and $P(5,5) = 0.226$

For $N = 100$:
$\ce{W(50,50) = 1.0*10^{29}}$ and $P(50,50) = 0.0788$

On the other hand, I don't understand what the mentioned range in the second part of the problem means and I'm therefore unable to solve it:

$n^*$ is in the range $n^* – 0.02 N ≤ n^* ≤ n^* + 0.02 N$


1 Answer 1


Some notes I had easily at hand may help. You have already calculated the probability of obtaining the chance with a number $k$ of type of ball (or molecule) out of a total on $n$ is

$$\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n} \tag{25c}$$

This distribution is a maximum when $k=n/2$. This can be seen with a straightforward argument. The factorial terms are symmetric, $k!(n-k)!=(n-k)!k!$ and always positive. When $k=0$ or $k=n$ the probability is very small tending to zero; i.e $1/2^n$ is small so there must be a maximum somewhere in the range $0\to n$. The symmetric nature ensures that this will be at $k = n/2$. This can also be determined by differentiation. It is not possible to differentiate a factorial, as $n$ is discrete, but replacing $n!$ with the Sterling approximation $(n!=n^ne^n)$ and setting the derivative in $n$ to zero allows the maximum to be found.

The second feature of this probability distribution is that it becomes extremely narrow as $n$ increases. This is shown in the figure which shows the normalised probability $p/p_{max}$ vs $k/n$ which is a fraction between $0$ to $1$. At large $n$ the distribution becomes so narrow, or peaked at $n/2$, that almost all the probability is described by that at $n/2$, i.e. this value is so great compared to others that $p_{max}$ is in effect greater than all others combined.

binomial dist

The normalised probability function $\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n}$ vs $k/n$ for different $n$. The the data only exists for integer values as shown as circles for $n=20$, at larger $n$ a line is drawn as the points become too congested. The curve for $n=10000$ is in red. At large $n$ the distribution becomes very narrow and the maximum value at $n/2$ is greater than all others combined and the distribution can be approximated by its maximum value.

You next need to choose $k/n$ corresponding to 2% either side of $k/n=0.5$ and calculate $p/p_{max}$.

  • 1
    $\begingroup$ This problem is the 5th Problem taken from the list of the 30 Preparatory Problems published by the Japanese organizers of the 53th International Chemistry Olympiad in July 2021. It is used for training the candidates of this Olympiad. As these 30 problems may be used in different countries for selecting their national candidates, I don't think it is advisable to give an answer or a solution to this problem 5. $\endgroup$
    – Maurice
    May 7, 2021 at 19:05
  • 2
    $\begingroup$ This answer was part of a document I wrote some years ago and has nothing to do with any competition. Besides I firmly believe that on this site we should not censor ourselves because someone might or might not be in a competition. $\endgroup$
    – porphyrin
    May 8, 2021 at 7:07
  • $\begingroup$ @Maurice that seems a problem for the organisers rather than SE. One can argue that every question here is written by someone sitting in a class or at home but during assignments. Writing such a neat question won't be feasible during a competition, anyway. $\endgroup$
    – Alchimista
    May 10, 2021 at 11:04

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