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Consider a system of two Einstein solids, with \(N_{A}=300, N_{B}=\) \(200,\) and \(q_{\text {total }}=100\) (as discussed in Section 2.3). Compute the entropy of the most likely macrostate and of the least likely macrostate. Also compute the entropy over long time scales, assuming that all microstates are accessible. (Neglect the factor of Boltzmann's constant in the definition of entropy; for systems this small it is best to think of entropy as a pure number.)

Short Answer

Expert verified
The entropy of the most likely macrostate is largest, while for the least likely it is zero. Average entropy considers all states.

Step by step solution

01

Determine Total Number of Microstates

For two Einstein solids, the total number of microstates is represented by the combination function \( \Omega(N, q) = \frac{(q + N - 1)!}{q!(N - 1)!} \). For the entire system, we are given \(q_{\text{total}} = q_A + q_B = 100\). We need to find configurations that maximize and minimize the number of microstates for \(q_A\) and \(q_B\), given that \(N_A = 300\) and \(N_B = 200\).
02

Find Most Likely Macrostate

Assume equal distribution of energy, so there are many possible configurations with this distribution. Calculate microstates when energies are equally distributed, typically when \(q_A \approx \frac{300}{500} \times 100\) and \(q_B \approx \frac{200}{500} \times 100\). Calculate \( \Omega \) for \(q_A = 60\) and \(q_B = 40\).
03

Calculate Entropy for Most Likely Macrostate

The entropy, \( S \), for the most likely macrostate is the logarithm of the number of microstates: \( S_{\text{most likely}} = \ln(\Omega_{A}(60) \times \Omega_{B}(40)) \).
04

Find Least Likely Macrostate

This macrostate occurs when one solid has all the energy. Calculate \( \Omega \) for \(q_A = 0\) and \(q_B = 100\) and vice versa. Calculate the entropy using \( \Omega_{A} = 1 \) for \((q_A=0, q_B=100)\) or \( \Omega_{B} = 1 \) for \((q_A=100, q_B=0)\).
05

Calculate Entropy for Least Likely Macrostate

The entropy for the least likely macrostate is: \( S = \ln(1) = 0 \), because the number of microstates, \( \Omega \), is 1 when all energy is in one solid.
06

Average Entropy Over Long Time Scales

Over long time scales, we consider the average entropy. All microstates are considered equally probable, so calculate the sum of entropies for all configurations and divide by the total number of configurations.

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Key Concepts

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

Einstein Solid
An Einstein solid is a simple model used to understand the vibrational behavior in solid matter. In an Einstein solid, each atom is considered as a quantum harmonic oscillator. This means each atom can behave like it is vibrating in a fixed position. To simplify, think of atoms in a lattice structure, each with its own unique energy levels.

  • Each atom can store energy in discrete amounts known as quanta.
  • The energy configuration across these atoms determines the system's microstates.
  • The total energy of the solid is distributed among the available atoms.
In the context of two solids, as discussed in this exercise, the idea is to see how energy can be distributed. This distribution of energy impacts the number of ways, or microstates, in which this can happen. Understanding the behavior of Einstein solids helps us explore complex thermodynamic properties, like entropy and heat capacity, in a straightforward manner.
Entropy
Entropy is a measure of disorder or randomness within a system. In statistical mechanics, entropy helps determine the number of ways energy can be distributed across states in a system. Here, it's understood as a numerical description of uncertainty or disorder across different configurations.

  • Entropy, denoted as \( S \), is calculated as the natural logarithm of the number of microstates \( \Omega \).
  • For the most likely configuration, where energy is evenly distributed, entropy reaches a peak.
  • For the least likely configuration, where all energy is concentrated, entropy is minimized (usually zero).
  • When all microstates become accessible over long periods, the entropy is averaged out across these states.
In this exercise, you were tasked with calculating the entropy for both the most and least likely macrostates, as well as over long time scales. The concept is crucial for understanding how systems evolve towards equilibrium and predict how energy is naturally spread out.
Macrostate
A macrostate is defined by the macroscopic features of a system, such as total energy, volume, and particle number. In the case of an Einstein solid, a macrostate is the state of the system with specified energy amounts distributed among the solids.

  • Macrostates are characterized by macroscopic quantities without detailing individual positions or energy levels.
  • Typically, the macrostate with the highest entropy is the most likely to be observed because it corresponds to the greatest number of microstates.
  • This exercise specifically asked you to identify the most likely macrostate (energy shared equally) versus the least likely one (all energy on one side).
Understanding macrostates lets us predict how tangible properties (like temperature) behave as they correspond to numerous microscopic configurations (microstates) within a system.
Microstate
In statistical mechanics, a microstate is a specific arrangement of atoms and energy within a system. Microstates are detailed configurations that correspond to a particular macrostate but describe the system at a much finer level.

  • Each microstate represents a possible way the energy of the system can be arranged.
  • The number of microstates corresponding to a macrostate is crucial for calculating entropy.
  • A greater number of microstates means higher entropy and a greater tendency for the system to exist in that macrostate.
  • This concept is essential for understanding statistical mechanics as it highlights the link between molecular dynamics and thermodynamic properties.
In this exercise, computing microstates for different distributions of energy between two Einstein solids illuminated how entropy and the likelihood of macrostates are closely connected.

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Most popular questions from this chapter

Consider a two-state paramagnet with \(10^{23}\) elementary dipoles, with the total energy fixed at zero so that exactly half the dipoles point up and half point down. (a) How many microstates are "accessible" to this system? (b) Suppose that the microstate of this system changes a billion times per second. How many microstates will it explore in ten billion years (the age of the universe)? (c) Is it correct to say that, if you wait long enough, a system will eventually be found in every "accessible" microstate? Explain your answer, and discuss the meaning of the word "accessible."

Use a computer to produce a table and graph, like those in this section, for two interacting two-state paramagnets, each containing 100 elementary magnetic dipoles. Take a "unit" of energy to be the amount needed to flip a single dipole from the "up" state (parallel to the external field) to the "down" state (antiparallel). Suppose that the total number of units of energy, relative to the state with all dipoles pointing up, is \(80 ;\) this energy can be shared in any way between the two paramagnets. What is the most probable macrostate, and what is its probability? What is the least probable macrostate, and what is its probability?

Fun with logarithms. (a) Simplify the expression \(e^{a \ln b} .\) (That is, write it in a way that doesn't involve logarithms.) (b) Assuming that \(b \ll a,\) prove that \(\ln (a+b) \approx(\ln a)+(b / a) .\) (Hint: Factor out the \(a\) from the argument of the logarithm, so that you can apply the approximation of part (d) of the previous problem.)

According to the Sackur-Tetrode equation, the entropy of a monatomic ideal gas can become negative when its temperature (and hence its energy) is sufficiently low. Of course this is absurd, so the Sackur-Tetrode equation must be invalid at very low temperatures. Suppose you start with a sample of helium at room temperature and atmospheric pressure, then lower the temperature holding the density fixed. Pretend that the helium remains a gas and does not liquefy. Below what temperature would the Sackur-Tetrode equation predict that \(S\) is negative? (The behavior of gases at very low temperatures is the main subject of Chapter \(7 .)\)

Suppose you flip 50 fair coins. (a) How many possible outcomes (microstates) are there? (b) How many ways are there of getting exactly 25 heads and 25 tails? (c) What is the probability of getting exactly 25 heads and 25 tails? (d) What is the probability of getting exactly 30 heads and 20 tails? (e) What is the probability of getting exactly 40 heads and 10 tails? (f) What is the probability of getting 50 heads and no tails? (g) Plot a graph of the probability of getting \(n\) heads, as a function of \(n\).

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