Question

20.14 Imagine dividing a box into two equal parts, part \(A\) on the left and part \(B\) on the right. Four identical gas atoms, numbered 1 through 4 , are placed in the box. What are most probable and second most probable distributions (for example, 3 atoms in \(\mathrm{A}, 1\) atom in \(\mathrm{B}\) ) of gas atoms in the box? Calculate the entropy, \(S\), for these two distributions. Note that the configuration with 3 atoms in \(\mathrm{A}\) and 1 atom in \(\mathrm{B}\) and that with 1 atom in A and three atoms in B count as different configurations.

Short answer

Answer: The most probable distribution is AABB, with an entropy value of 1.792. The second most probable distributions are AAAB and BBBA, both with entropy values of 1.386.

Step-by-step solution

Determine all possible configurations

We need to find all possible ways to distribute the gas atoms in parts A and B. The configurations are as follows: 1. All atoms in A and none in B (AAAA, B is empty) 2. All atoms in B and none in A (BBBB, A is empty) 3. Three atoms in A and one in B: (AAAB, B has one atom) 4. Three atoms in B and one in A: (BBBA, A has one atom) 5. Two atoms in each part: (AABB)

Calculate the multiplicity for the configurations

Multiplicity (denoted as W) is the number of ways a particular configuration can be arranged. We will calculate the multiplicity for each configuration. 1. (AAAA) - only one possible arrangement, so \(W_1=1\) 2. (BBBB) - only one possible arrangement, so \(W_2=1\) 3. (AAAB) - 4 different possible arrangements, so \(W_3=4\) (since any one of the 4 atoms can be in part B) 4. (BBBA) - 4 different possible arrangements, so \(W_4=4\) (since any one of the 4 atoms can be in part A) 5. (AABB) - 6 different possible arrangements, so \(W_5=6\) (as there are \({4 \choose 2} = 6\) ways to choose which atoms are in part A)

Calculate the entropy for each configuration

Entropy is given by \(S=k \ln W\), where k is the Boltzmann constant. We can ignore the constant k since we just need to compare the values to find the most and second most probable distributions. So, we only need to calculate \(\ln W\) for each configuration. 1. \(\ln W_1 = \ln(1) = 0\) 2. \(\ln W_2 = \ln(1) = 0\) 3. \(\ln W_3 = \ln(4) = 1.386\) 4. \(\ln W_4 = \ln(4) = 1.386\) 5. \(\ln W_5 = \ln(6) = 1.792\)

Determine the most and second most probable distributions

Comparing the entropy values for all the configurations, we can determine the most probable and second most probable distributions. Most probable: AABB, with entropy value of 1.792 Second most probable: configurations AAAB and BBBA, each with entropy values of 1.386

Key concepts

Entropy

Entropy is a central concept in statistical mechanics, representing the measure of disorder or randomness in a system. In our gas atom box example, entropy helps to understand how the atoms are distributed between parts A and B.
To calculate entropy (\( S \)), we use the formula:

• \( S = k \ln W \)

Here, \( k \) is the Boltzmann constant, and \( W \) is the multiplicity of the configuration. Multiplicity is merely the number of ways a specific distribution can occur. Higher multiplicity usually means higher entropy because there are more possible ways for the system to achieve that configuration.
In our scenario, the entropy calculation identifies the most probable distribution of atoms, which is the one with the highest entropy.

Probability Distribution

Probability distribution plays a crucial role in understanding the likely configurations of atoms or particles in a system. In statistical mechanics, it helps to foresee how particles are likely to be distributed over different states or regions.
For the given problem, we approach probability distributions with the combinations of gas atoms in parts A and B. Each possible arrangement of atoms corresponds to a different configuration, and the probability distribution is used to determine the likelihood of these configurations occurring in nature.
In simple terms, the probability of each configuration is proportional to its multiplicity. This means the configuration with the highest multiplicity is the most likely, explaining why the AABB configuration is the most probable in our example.

Multiplicity

Multiplicity, denoted as \( W \), is a measure of the number of different ways a configuration can be realized. It deeply influences which atom arrangement is most likely in a statistical mechanical system.
Consider our box with gas atoms; each allocation of atoms into parts A and B has its own multiplicity:

• AAAA and BBBB configurations have multiplicities of 1, as there is only one way to have all atoms in one part.
• AAAB and BBBA have multiplicities of 4, allowing any one of the 4 atoms to be in the opposite part.
• AABB has the highest multiplicity of 6, given by the binomial coefficient \({4 \choose 2}\), which highlights more arrangement possibilities.

The higher the multiplicity, the greater the disorder and consequentially the entropy, indicating a more probable configuration.

Boltzmann Constant

The Boltzmann constant (\( k = 1.38 \times 10^{-23} \, \text{J/K} \)) bridges the macroscopic and microscopic worlds, vital in statistical mechanics. It's a fundamental constant that links the average kinetic energy of particles in a gas with the temperature of the gas.
In our entropy formula (\( S = k \ln W \)), the Boltzmann constant makes the macroscopic quantity of entropy from the microscopic count of microstates, \( W \).
Even though in comparing configurations, we often ignore \( k\) for relative comparisons, it plays a crucial role when absolute values are required. It helps us understand how microscopic randomness contributes to macroscopic thermodynamic properties, enhancing our grasp of nature's probabilistic framework.