A box contains N identical gas molecules equally divided...

A box contains $N$ identical gas molecules equally divided between its two halves. For $N=50,$ what are (a) the multiplicity $W$ of the central configuration, (b) the total number of microstates, and (c) the percentage of the time the system spends in the central configuration? For $N=100,$ what are (d) $W$ of the central configuration, (e) the total number of microstates, and (f) the percentage of the time the system spends in the central configuration? For $N=200,$ what are $(\mathrm{g}) W$ of the central configuration, $(\mathrm{h})$ the total number of microstates, and (i) the percentage of the time the system spends in the central configuration?(j) Does the time spent in the central configuration increase or decrease with an increase in $N ?$

Key Concepts

Multiplicity

Multiplicity in statistical mechanics is the count of how many microscopic arrangements (microstates) correspond to a given macroscopic state (macrostate) of a system. For example, when gas molecules are divided between two halves of a container, the central configuration (equal split) has a multiplicity determined by the number of ways to choose half the molecules to be in one half of the box. This concept links microscopic randomness with macroscopic observables like pressure or temperature.

Total Microstates

The total microstates of a system are the complete set of distinct configurations available to all the particles. In systems where each particle can independently be on one side or the other (such as in a two?compartment box), the overall number of microstates is given by 2^N. This huge number underlies the probabilistic description in equilibrium statistical mechanics.

Binomial Coefficient

The binomial coefficient arises in counting problems where one divides a set of items into two groups. In the context of gas molecules in a box, it is used to calculate the multiplicity of a configuration where exactly N/2 molecules lie in one half. The coefficient, often written as C(N, N/2), quantitatively expresses the number of ways to choose half of the molecules from the total N.

Probability of a Configuration

The probability of observing a specific configuration is the ratio of its multiplicity to the total number of microstates. In practice, for a system with equally likely microstates, this fraction indicates how frequently the system will be found in that specific macroscopic arrangement over time.

Thermodynamic Limit Effects

When the number of particles N increases, the distribution of microstates becomes sharply peaked around the most probable configuration, even though the absolute number of microstates grows exponentially. This means that in the thermodynamic limit, the system spends an overwhelming amount of time near the configuration with the highest multiplicity (the central configuration in this case), and relative fluctuations around it become negligibly small.

Transcript

00:01 For part a, we can denote the figuration with n heads out of n trials as n colon, or it's semicolon n, where n would be number of trials, and n would be number of heads.

00:24 And we can then use equation 2020 and say that the work for 25 heads of 50 trials would be equal to 50 factorial.

00:38 Divided by 25 factorial, multiplied by 50 minus 25 quantity factorial.

00:47 And this is equaling 1 .26 times 10 to the 14th jules.

00:57 So this would be our answer for part a.

01:00 For part b, we know that here there are two possible choices for each molecule.

01:05 It can either be inside one or side two of the box.

01:08 That means that for a total of an independent molecule, molecules, we can say that the total number of available states of the n particle system.

01:32 So however many particles there are denoted by n, n total would be equal to 2 times 2 times 2.

01:44 And you can get the point.

01:46 The general formula would be 2 to the n power, essentially.

01:51 And for n equals 50, we can then say the total n total, total number of states would be 2 to the 50th power.

02:06 This is equaling 1 .13 times 10 to the 15th power.

02:15 For part c, this would be our answer.

02:20 For part c, we want to find the percentage of time in question.

02:25 This is going to be equal to the probability for the system to be in.

02:29 The central configuration.

02:31 So here we can say that the probability of 2550 would be equal to the work of 2550 divided by 2 to the 50th power the number of possible states.

02:47 So this would be 1 .26 times 10 to the 14th divided by 1 .13 times 10 to the 15th and and here it'll be 11 .1%.

03:11 This would be our answer for part c.

03:18 And then for part d, we know that we have n equals 100 now.

03:25 So we can then say that work of n over 2 comma n would be simply equal to n factorial divided by n over 2 factorial to the second power.

03:46 And this is equaling 1 .01 times 10 to the 29th, and this is for n equals 100.

04:00 So we're essentially doing all these calculations except for now n equals 100...

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