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Consider the following energy levels, each capable of holding two particles: $$\begin{array}{l}{E=2 \mathrm{kJ}} \\ {E=1 \mathrm{kJ}} \\ {E=0 \quad \underline{X X}}\end{array}$$ Draw all the possible arrangements of the two identical particles (represented by X) in the three energy levels. What total energy is most likely, that is, occurs the greatest number of times? Assume that the particles are indistinguishable from each other.

Short Answer

Expert verified
We have six possible arrangements of the two identical particles in the three energy levels: \(XX \ \_ \ \_\), \(\_ \ XX \ \_\), \(\_ \ \_ \ XX\), \(X \ X \ \_\), \(X \ \_ \ X\), and \(\_ \ X \ X\). The corresponding total energies are 0 kJ, 2 kJ, 4 kJ, 1 kJ, 2 kJ, and 3 kJ. The most likely total energies are 1 kJ and 2 kJ, as they occur two times each.

Step by step solution

01

Understanding the problem statement

We are given three energy levels, each capable of holding two particles, and we have two distinguishable particles represented by X. The goal is to find all the possible arrangements of the two particles in the three energy levels.
02

Enumerate all possible arrangements

Let's start with the possible arrangements for placing the two particles in the energy levels. Remember, each energy level can hold up to two particles, and the particles are indistinguishable. 1. Both particles in the 0 kJ energy level: \(XX \ \_ \ \_\) 2. Both particles in the 1 kJ energy level: \(\_ \ XX \ \_\) 3. Both particles in the 2 kJ energy level \(\_ \ \_ \ XX\) 4. One particle in 0 kJ and another in 1 kJ energy level: \(X \ X \ \_\) 5. One particle in 0 kJ and another in 2 kJ energy level: \(X \ \_ \ X\) 6. One particle in 1 kJ and another in 2 kJ energy level: \(\_ \ X \ X\)
03

Calculate the total energy for each arrangement

Now let's calculate the total energy for each arrangement: 1. Both particles in the 0 kJ energy level: Total energy = 0 kJ 2. Both particles in the 1 kJ energy level: Total energy = 1 kJ + 1 kJ = 2 kJ 3. Both particles in the 2 kJ energy level Total energy = 2 kJ + 2 kJ = 4 kJ 4. One particle in 0 kJ and another in 1 kJ energy level: Total energy = 0 kJ + 1 kJ = 1 kJ 5. One particle in 0 kJ and another in 2 kJ energy level: Total energy = 0 kJ + 2 kJ = 2 kJ 6. One particle in 1 kJ and another in 2 kJ energy level: Total energy = 1 kJ + 2 kJ = 3 kJ
04

Find the total energy that occurs the most

Now that we have the total energy for each arrangement, let's count how many times each energy value appears: - 0 kJ occurs 1 time - 1 kJ occurs 2 times - 2 kJ occurs 2 times - 3 kJ occurs 1 time - 4 kJ occurs 1 time Since the total energy of 1 kJ and 2 kJ occur the greatest number of times (2 times each), they are the most likely total energy values.

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Key Concepts

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

Indistinguishable Particles
When we talk about indistinguishable particles, we mean that no matter how we arrange them, they look identical. This concept is crucial in many areas of physics, especially in quantum mechanics. In our problem, the particles are represented by X, and distinguishing between different Xs is impossible. Whether one X is placed in one energy level and the other in another, or vice versa, these setups are the same.

Understanding indistinguishable particles helps in simplifying complex systems. It allows us to count only unique configurations rather than each possible permutation, making calculations more straightforward. This distinction is essential for accurately determining energy and statistical properties.
Energy Calculations
In analyzing particle arrangements, we need to calculate the energy levels associated with each arrangement. These calculations are straightforward once you know where each particle is placed. Each particle contributes an energy corresponding to its level, and we simply add these values.

Here's how it's done:
  • If both particles are in the 0 kJ energy level, the total energy is 0 kJ.
  • If they're both in the 1 kJ level, it's 2 kJ (1 kJ per particle).
  • If both are in the 2 kJ level, it's 4 kJ (2 kJ per particle).
  • When particles are in different levels, like one in 0 kJ and the other in 1 kJ, the total energy is their sum, 1 kJ.

These calculations help evaluate which energy states are most probable.
Energy Level Diagrams
Energy level diagrams provide a visual representation of where particles are located across different energy levels. This diagram is a valuable tool for conceptualizing the distribution of particles. In our scenario, we have three energy levels: 0 kJ, 1 kJ, and 2 kJ, each capable of holding two particles.

An energy level diagram helps identify possible configurations. For instance, the arrangement \(XX \ \_ \ \_\) indicates both particles at 0 kJ while \(_ \ XX \ \_\) shows both at 1 kJ. This visualization aids in understanding how particles distribute themselves in various states and allows us to compute the total energy for each configuration effortlessly.
Probability in Energy Distribution
Probability in energy distribution helps determine which energy states are more likely to occur. It's about finding how often a particular total energy appears among all possible arrangements.

In our example, 1 kJ and 2 kJ appeared twice, making them the most probable energy levels. To determine these probabilities, count the number of configurations leading to each possible total energy.

For instance:
  • 0 kJ occurs once.
  • 1 kJ appears in two setups.
  • Similarly, 2 kJ appears twice too.
  • 3 kJ and 4 kJ each occur once.

Understanding this probability provides insights into the behavior of particle systems, which is vital for predicting system behavior in various physical conditions.

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Most popular questions from this chapter

The third law of thermodynamics states that the entropy of a perfect crystal at 0 \(\mathrm{K}\) is zero. In Appendix \(4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)\) and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

Some nonelectrolyte solute (molar mass \(=142 \mathrm{g} / \mathrm{mol} )\) was dissolved in \(150 . \mathrm{mL}\) of a solvent (density \(=0.879 \mathrm{g} / \mathrm{cm}^{3} )\) The elevated boiling point of the solution was 355.4 \(\mathrm{K}\) . What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is 33.90 \(\mathrm{kJ} / \mathrm{mol}\) , the entropy of vaporization is 95.95 \(\mathrm{J} / \mathrm{K} \cdot\) mol, and the boiling- point elevation constant is 2.5 \(\mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}\) .

Consider the following equilibrium constant versus temperature data for some reaction: $$\begin{array}{ll} {\boldsymbol{T}\left(^{\circ} \mathbf{C}\right)} & \quad {\text { K }} \\ \hline {109} & {2.54 \times 10^{4}} \\ {225} & {5.04 \times 10^{2}} \\ {303} & {6.33 \times 10^{1}} \\ {412} & {2.25 \times 10^{-1}} \\\ {539} & {3.03 \times 10^{-3}}\end{array}$$ Predict the signs for \(\Delta G^{\circ}, \Delta H^{\circ},\) and \(\Delta S^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Consider a weak acid, HX. If a \(0.10-M\) solution of \(\mathrm{HX}\) has a pH of 5.83 at \(25^{\circ} \mathrm{C},\) what is \(\Delta G^{\circ}\) for the acid's dissociation reaction at \(25^{\circ} \mathrm{C} ?\)

For the reaction $$\mathrm{CS}_{2}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g)$$ \(\Delta S^{\circ}\) is equal to \(-143 \mathrm{JK}\) . Use this value and data from Appendix 4 to calculate the value of \(S^{\circ}\) for \(\mathrm{CS}_{2}(g) .\)

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