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Chapter 20: Problem 14

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

Expert verified
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

01

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)
02

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)
03

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. \(\ln W_5 = \ln(6) = 1.792\)
04

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

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Most popular questions from this chapter

A volume of \(6.00 \mathrm{~L}\) of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergoes the following processes, all done reversibly: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the entropy change for each process.

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