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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A heat engine cycle often used in refrigeration, is the Brayton cycle, which involves an adiabatic compression. followed by an isobaric expansion, an adiabatic expansion and finally an isobaric compression. The system begins at temperature \(T_{1}\) and transitions to temperatures \(T_{2}, T_{3},\) and \(T_{4}\) after respective parts of the cycle. a) Sketch this cycle on a \(p V\) -diagram. b) Show that the efficiency of the overall cycle is given by \(\epsilon=1-\left(T_{A}-T_{1}\right) /\left(T_{3}-T_{2}\right)\)

A refrigerator with a coefficient of performance of 3.80 is used to \(\operatorname{cool} 2.00 \mathrm{~L}\) of mineral water from room temperature \(\left(25.0^{\circ} \mathrm{C}\right)\) to \(4.00^{\circ} \mathrm{C} .\) If the refrigerator uses \(480 . \mathrm{W}\) how long will it take the water to reach \(4.00^{\circ} \mathrm{C}\) ? Recall that the heat capacity of water is \(4.19 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K}),\) and the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\). Assume the other contents of the refrig. erator are already at \(4.00^{\circ} \mathrm{C}\).

An ideal gas undergoes an isothermal expansion. What will happen to its entropy? a) It will increase. c) It's impossible to determine. b) It will decrease. d) It will remain unchanged.

Suppose a person metabolizes \(2000 .\) kcal/day. a) With a core body temperature of \(37.0^{\circ} \mathrm{C}\) and an ambient temperature of \(20.0^{\circ} \mathrm{C}\), what is the maximum (Carnot) efficiency with which the person can perform work? b) If the person could work with that efficiency, at what rate, in watts, would they have to shed waste heat to the surroundings? c) With a skin area of \(1.50 \mathrm{~m}^{2}\), a skin temperature of \(27.0^{\circ} \mathrm{C}\) and an effective emissivity of \(e=0.600,\) at what net rate does this person radiate heat to the \(20.0^{\circ} \mathrm{C}\) surroundings? d) The rest of the waste heat must be removed by evaporating water, either as perspiration or from the lungs At body temperature, the latent heat of vaporization of water is \(575 \mathrm{cal} / \mathrm{g}\). At what rate, in grams per hour, does this person lose water? e) Estimate the rate at which the person gains entropy. Assume that all the required evaporation of water takes place in the lungs, at the core body temperature of \(37.0^{\circ} \mathrm{C}\).

A Carnot engine operates between a warmer reservoir at a temperature \(T_{1}\) and a cooler reservoir at a temperature \(T_{2}\). It is found that increasing the temperature of the warmer reservoir by a factor of 2 while keeping the same temperature for the cooler reservoir increases the efficiency of the Carnot engine by a factor of 2 as well. Find the efficiency of the engine and the ratio of the temperatures of the two reservoirs in their original form.

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