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Chapter 9: Problem 3

Given an arbitrary thermodynanic systemn, which is larger. the number of possible macrostates, or the number of possible microstates, or is it impossible to say? Explain your answer. (For most systems, both are infinite, but it is still possible to answer the question)

Short Answer

Expert verified
The number of possible microstates is usually larger than the number of macrostates. Even though the quantities of both may be potentially infinite, each observable macrostate is a result of many different arrangements of its underlying microstates.

Step by step solution

01

Define Macrostates and Microstates

Macrostates correspond to various properties or characteristics that can be macroscopically observed like volume, pressure and temperature. Microstates, however, mean the specific arrangements of individual particles (each particle's position and velocity) that gives rise to a macrostate. For instance, for a flipped coin, heads is a macrostate but the rotational trajectory the coin experienced while flipping is the microstate.
02

Relation Between Microstates and Macrostates

A single macrostate can include multiple microstates. For instance, if coins are flipped, the combination of heads and tails (such as HT, TH, HH, TT) are considered as the microstates that result in the macrostate. Another example would be in gases, where the macrostate includes the pressure, volume and temperature, while the microstates are linked to the position and velocities of the individual molecules.
03

Number of Microstates vs Macrostates

In practicality, the number of microstates is larger than the number of macrostates. This is due to the fact that each macrostate can be produced by many different arrangements or permutations of microstates. Thus, there can exist many more specific arrangements (microstates) than the observed characteristics (macrostates).
04

Summary

Therefore, even if the quantities of both states may be infinite, the number of possible microstates is usually larger because there are many ways for particles to arrange in order to bring about the same macroscopic observation.

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Key Concepts

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

Macrostates
When we talk about macrostates in thermodynamics, we're referring to the macroscopic characteristics of a thermodynamic system that are observable and measurable. These include properties such as volume, pressure, and temperature. Essentially, a macrostate is what you can observe without knowing every little detail about the system. Imagine looking at a container of gas and observing its pressure and temperature. These parameters together describe a macrostate because they tell you the general condition of the system.

One key thing about macrostates is that they're less detailed than microstates. Often, many different arrangements of particles and their energies (microstates) can lead to the same macrostate. This means macrostates provide a broader or more general picture of what's happening in a system.
Microstates
Microstates are like the little details or the zoomed-in view of a thermodynamic system. They describe the specific arrangements of particles, including their individual positions and velocities, that can exist within a given macrostate. To understand this, imagine flipping a coin: while the outcome (heads or tails) is a macrostate, the path the coin took to get there鈥攊ts flips and spins鈥攊s the microstate.

Each macrostate is associated with a vast number of microstates. This is because there are many ways the particles can be arranged and countless possible combinations of their energies within a single macrostate. For example, in a gas, even with the same pressure and temperature (macrostate), the molecules can be moving in countless different ways, creating numerous microstates.
Thermodynamic Systems
A thermodynamic system is a specific portion of the physical universe that is chosen for analysis. Everything outside this system is known as the surroundings. These systems can be as simple as a block of ice or as complex as a living organism.

Thermodynamic systems are classified into three types based on their interactions with the surroundings:
  • Open Systems: Exchange both energy and matter with their surroundings. An example is a boiling pot of water where steam and heat are exchanged.
  • Closed Systems: Exchange only energy with their surroundings, not matter. A sealed beaker is a good example.
  • Isolated Systems: Neither energy nor matter is exchanged with the surroundings. A perfectly sealed and insulated flask approaches this ideal.
Understanding thermodynamic systems is crucial because it helps us communicate what part of the universe we are studying and how it interacts with the environment. This understanding paves the way to exploring how macrostates and microstates interplay within these systems.

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Most popular questions from this chapter

The fact that a laser's resonant cavity so effectively sharpens the wavelength can lead to the output of several closely spaced laser wavelengths, called longitudinal modes. Here we see how. Suppose the spontaneous emission serving as the seed for stimulated emission is of wavelength \(633 \mathrm{~nm}\), but somewhat fuzzy, with a line width of roughly \(0.001 \mathrm{~nm}\) either side of the central value. The resonant cavity is exactly \(60 \mathrm{~cm}\) kng. (a) How many wavelengths fit the standing-wave condition? (b) If only u single wavelength were desired. would changing the length of the cavity help? Explain.

Suppose we have a system of identical particles moving in just one dimension and for which the energy quantization relationship is \(E=b n^{2 \Omega}\), where \(b\) is a constantand \(n\) an integer quantum number. Discuss whether the density of states should be independent of E. an increasing function of \(E,\) or a decreasing function of \(E\).

There are more permutations of particle labels when of particles have energy 0 and two have energy 1 than when three particles have energy 0 and one has energy 2\. (The total energies are the same.) From this observation alone argue that the Boltzmann distribution should be lower than the Bose-Einstein at the lowest energy level.

When a star has nearly bumed up its intemal fuel, it may become a white dwarf. It is crushed under its own enonnous gravitational forces to the point at which the exclusion principle for the electrons becomes a factor. A smaller size would decrease the gravitational potential energy, but assuming the electrons to be packed into the lowest energy states consistent with the exclusion principle. "squeezing" the potential well necessarily increases the ener gies of all the electrons (by shortening their wavelengths). If gravitation and the electron exclusion principle are the only factors, there is a mini. mum total energy and corresponding equilibrium radius. (a) Treat the electrons in a white dwarf as a quantum gas. The minimum energy allowed by the exclusion principle (see Exercise 67 ) is $$ U_{\text {elecimns }}=\frac{3}{10}\left(\frac{3 \pi^{2} \hbar^{3}}{m_{e}^{3 / 2} V}\right)^{2 / 3} N^{5 / 3} $$ Note that as the volume \(V\) is decreased, the energy does increase. For a neutral star. the number of electrons, \(N\), equals the number of protons. Assuming that protons account for half of the white dwarf's mass \(M\) (neutrons accounting for the other half). show that the minimum electron energy may be written $$ U_{\text {electrons }}=\frac{9 \hbar^{2}}{80 m_{e}}\left(\frac{3 \pi^{2} M^{5}}{m_{\mathrm{p}}^{5}}\right)^{1 / 3} \frac{1}{R^{2}} $$ where \(R\) is the star's radius. (b) The gravitational potential energy of a sphere of mass \(M\) and radius \(R\) is given by $$ U_{\operatorname{mav}}=-\frac{3}{5} \frac{G M^{2}}{R} $$ Taking both factors into account, show that the minimum total energy occurs when $$ R=\frac{3 h^{2}}{8 G}\left(\frac{3 \pi^{2}}{m^{3} m_{p}^{5} M}\right)^{1 / 3} $$ (c) Evaluate this radius for a star whose mass is equal to that of our Sun, \(\sim 2 \times 10^{30} \mathrm{~kg}\). (d) White dwarfs are comparable to the size of Eath. Does the value in part (c) agree?

What is special about a metastable state, and why is it so useful in a laser? Why wouldn't a nonmetastable state at the same energy work?
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