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On page 822 the terms states and microstates were introduced. Consider a system that has four states (i.e., energy levels), with energy \(\varepsilon=0,1,2,\) and 3 energy units, and three particles labeled \(A, B,\) and \(C .\) The total energy of the system, in energy units, is 3 . How many microstates can be generated?

Short Answer

Expert verified
Based on the enumerated combinations, the system can generate 10 different microstates.

Step by step solution

01

Enumeration

Start by enumerating all possible combinations of energy levels for the three particles \(A, B, C\) that provide a total of 3 energy units. The combinations are as follows: {3,0,0}, {0,3,0}, {0,0,3}, {2,1,0}, {2,0,1}, {0,2,1}, {1,2,0}, {1,0,2}, {0,1,2}, {1,1,1}. Each set of curly braces, {}, represents a unique configuration of the energies of the three particles.
02

Count

Count the number of combinations obtained in step 1. These combinations represent the different microstates for the system. The goal is to find the total number of unique combinations, which represents the number of possible microstates that can exist given the conditions set for the system.

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Key Concepts

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

Statistical Mechanics
When diving into the fascinating world of statistical mechanics, one discovers that it bridges the gap between macroscopic observations and microscopic behaviors. This branch of physics uses probability theory to predict and explain the thermodynamic properties of systems composed of a large number of particles. Microstates are a fundamental concept in statistical mechanics, as they represent the specific ways in which a system’s particles can be arranged while still conforming to a given energy level.

For example, let’s say you have a system with three distinct particles, labeled A, B, and C. If these particles can each have an energy of 0, 1, 2, or 3 energy units, statistical mechanics allows us to calculate all possible ways these energy units can be distributed among the particles. The sum of the energy levels must equal the system's total energy. In practice, this means listing every possible combination of energies that each particle could have and ensuring that the total energy adds up correctly – just like the problem from the textbook.
Energy Levels in Particles
Delving into energy levels in particles provides a clearer understanding of microstates. In quantum mechanics, energy levels are the distinct states that particles like electrons or photons can occupy. Each level corresponds to a specific amount of energy.

When applied to a system of particles like in our exercise, we can see that the particles A, B, and C can each occupy one of four energy levels. The possible arrangements of these energy levels, given the total system energy, constitute the set of microstates.

Visualizing Energy Levels

Imagine each particle as a runner on a staircase with four steps. Each step represents a different energy level. The sum of the steps on which the three runners stand must equal three to satisfy the system's energy. This analogy helps visualize the possible combinations and their respective energies. Through this lens, the textbook exercise reviews the foundational concept of quantized energy in the context of thermal systems.
Entropy and Probability
Tying together entropy and probability unveils why certain states of matter are more prevalent than others. Entropy, a measure of disorder in a system, is closely related to the number of microstates: the greater the number of microstates, the higher the entropy. Probability comes into play because systems tend to evolve naturally towards states with higher probabilities, which, in turn, are states with higher entropy.

In our exercise, we count each combination of energy levels for the three particles, and this count reflects the system's entropy. High entropy is synonymous with high probability. So, a system with many microstates means there's a higher chance the system will be found in one of those microstates, making such a state more 'likely' from a statistical point of view.

The Link to Thermodynamics

In thermodynamics, this concept is crucial for understanding why systems tend towards equilibrium – a state of maximum entropy. The textbook exercise offers a snapshot into this broader principle, where students can see firsthand the stochastic nature of thermodynamic systems and how it underpins the second law of thermodynamics.

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Most popular questions from this chapter

Which substance in each of the following pairs would have the greater entropy? Explain. (a) at \(75^{\circ} \mathrm{C}\) and 1 atm: \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(1)\) or \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) at \(5^{\circ} \mathrm{C}\) and 1 atm: \(50.0 \mathrm{g} \mathrm{Fe}(\mathrm{s})\) or \(0.80 \mathrm{mol} \mathrm{Fe}(\mathrm{s})\) (c) 1 mol \(\mathrm{Br}_{2}\left(1,1 \text { atm }, 8^{\circ} \mathrm{C}\right)\) or \(1 \mathrm{mol} \mathrm{Br}_{2}(\mathrm{s}, 1 \mathrm{atm},\) \(\left.-8^{\circ} \mathrm{C}\right)\) (d) \(0.312 \mathrm{mol} \mathrm{SO}_{2}\left(\mathrm{g}, 0.110 \mathrm{atm}, 32.5^{\circ} \mathrm{C}\right)\) or \(0.284 \mathrm{mol}\) \(\mathrm{O}_{2}\left(\mathrm{g}, 15.0 \mathrm{atm}, 22.3^{\circ} \mathrm{C}\right)\)

What must be the temperature if the following reaction has \(\Delta G^{\circ}=-45.5 \mathrm{kJ}, \Delta H^{\circ}=-24.8 \mathrm{kJ},\) and \(\Delta S^{\circ}=15.2 \mathrm{JK}^{-1} ?\) $$\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})$$

\(\mathrm{H}_{2}(\mathrm{g})\) can be prepared by passing steam over hot iron: \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) (a) Write an expression for the thermodynamic equilibrium constant for this reaction. (b) Explain why the partial pressure of \(\mathrm{H}_{2}(\mathrm{g})\) is independent of the amounts of \(\mathrm{Fe}(\mathrm{s})\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})\) present. (c) Can we conclude that the production of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) could be accomplished regardless of the proportions of \(\mathrm{Fe}(\mathrm{s})\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})\) present? Explain.

Indicate whether the entropy of the system would increase or decrease in each of the following reactions. If you cannot be certain simply by inspecting the equation, explain why. (a) \(\mathrm{CCl}_{4}(1) \longrightarrow \mathrm{CCl}_{4}(\mathrm{g})\) (b) \(\mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow\) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) (c) \(\mathrm{SO}_{3}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{SO}_{2}(\mathrm{g})\) (not balanced)

Explain the important distinctions between each of the following pairs: (a) spontaneous and nonspontaneous processes; (b) the second and third laws of thermodynamics; (c) \(\Delta G\) and \(\Delta G^{\circ}\).

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