/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Verify that the quantity \((k / ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Verify that the quantity \((k / \mathcal{N}) \ln \Gamma\), where $$ \Gamma(\mathcal{N}, U)=\sum_{\left\langle n_{r}\right\rangle}^{\prime} W\left\\{n_{r}\right\\}, $$ is equal to the (mean) entropy of the given system. Show that this leads to essentially the same result for \(\ln \Gamma\) if we take, in the foregoing summation, only the largest term of the sum, namely the term \(W\left\\{n_{r}^{*}\right\\}\) that corresponds to the most probable distribution set. [Surprised? Well, note the following example: For all \(N\), the summation over the binomial coefficients \({ }^{N} C_{r}=N ! /[r !(N-r !)]\) gives $$ \sum_{r=0}^{N}{ }^{N} C_{r}=2^{N_{N}} $$ therefore, $$ \ln \left\\{\sum_{r=0}^{N}{ }^{N} C_{r}\right\\}=N \ln 2 . $$ Now, the largest term in this sum corresponds to \(r \simeq N / 2\); so, for large \(N\), the logarithm of the largest term is very nearly equal to $$ \begin{aligned} & \ln \\{N !\\}-2 \ln \\{(N / 2) !\\} \\ \approx & N \ln N-2 \frac{N}{2} \ln \frac{N}{2}=N \ln 2 \end{aligned} $$

Short Answer

Expert verified
Yes, the quantity \((k / \mathcal{N}) \ln \Gamma\) equals the mean entropy of the system, and taking the most probable distribution in summation gives a very similar result to \(\ln \Gamma\), which is \(N \ln 2\), using approximation for large N.

Step by step solution

01

Identify the Given

Identify the given values and equations. Here we have the expression \((k / \mathcal{N}) \ln \Gamma\), where \(\Gamma\) can be calculated using the given formula \(\Gamma(\mathcal{N}, U)=\sum_{\left\langle n_{r}\right\rangle}^{\prime} W\left\{n_{r}\right\}\). We are asked to prove this expression equals the mean entropy of the given system.
02

Understand the Entropy Formula

It is known in statistical mechanics that entropy \(S\), is given by the Boltzmann equation \(S = k \ln \Omega\), where \(\Omega\) is the multiplicity (or the number of microstates). Equating this with the given, \(S = (k / \mathcal{N}) \ln \Gamma\). which implies \(\Gamma = \Omega^\mathcal{N}\). Thus, the entropy formula holds.
03

Evaluate Taking the Most Probable Distribution Set

Express \(\ln \Gamma\) taking the largest term. In terms of most probable distribution set, we can represent \(\Gamma(\mathcal{N}, U)\) as \(W\left\{n_{r}^{*}\right\}\). So \(\ln \Gamma = \ln W\left\{n_{r}^{*}\right\}\), where \(W\left\{n_{r}^{*}\right\}\) is the most probable distribution.
04

Apply the Logarithm Properties

Apply logarithm properties to find the equivalent expression. The given example, \(\ln \{N !\}-2 \ln \{(N / 2) !\}\), can be approximated or navigated using Stirling's approximation for large N: \(n! \approx n \ln n - n\). Thus, resulting into \(N \ln 2\).
05

Relate to the Initial Entropy Expression

Relating this back to \(\ln \Gamma\), it can be inferred that taking the most probable distribution in summation will lead to the same result as the initial entropy expression which proves the last part of the problem statement.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Boltzmann Equation
In statistical mechanics, the Boltzmann Equation is essential for understanding entropy. Entropy measures the disorder in a system, describing how microstates—different arrangements of particles—relate to the macroscopic properties of a system. According to the Boltzmann Equation, entropy \( S \) is defined as \( S = k \ln \Omega \), where \( k \) is the Boltzmann constant and \( \Omega \) is the number of microstates representing the system.
By associating the entropy with \( (k / \mathcal{N}) \ln \Gamma \) in the exercise, where \( \Gamma \) represents a sum over weighted microstate distributions, we equate it to the classical definition through \( \Omega \). This illustrates the consistency of entropy measurement through both expressions, verifying their equivalence.
Understanding this equivalence helps us see how probabilities and microstate counts combine elegantly using Boltzmann's insight, forming a foundational pillar of statistical mechanics.
Most Probable Distribution
The Most Probable Distribution is a crucial concept in statistical mechanics, signifying the configuration of particle distributions that is most likely to occur. Typically, this is where the distribution of energy among particles reaches a state of equilibrium, or maximum entropy. In practice, when dealing with large ensembles in systems, a specific arrangement or configuration, marked as \( W\{n_{r}^{*}\} \), stands out because it is statistically more frequent compared to others.
In the exercise, \( W\{n_{r}^{*}\} \) represents the sum's largest term, bringing reinforced clarity to why complex calculations sometimes simplify to their most probable terms. This reflects on how minute deviations in configurations become negligible as systems tend towards equilibrium.
  • Most probable distribution simplifies complex sums by focusing on dominant terms.
  • This aligns closely with thermodynamic tendencies towards equilibrium.
  • Understanding this helps simplify complex system behaviors into comprehensible terms.
Stirling's Approximation
Stirling's Approximation is a mathematical tool that aids in simplifying factorial expressions, particularly for large numbers. In the realm of statistical mechanics, systems often involve huge particle counts, leading to the computation of large factorials. Stirling's Approximation provides a means to approximate these expressions by: \( n! \approx n \ln n - n \).
This approximation becomes especially useful when evaluating terms in logarithmic expressions. For instance, in the exercise, it simplifies the evaluations by allowing comparisons between \( \ln \{N!\} \) and other factorial terms without overwhelming calculations. This approximate value gives insight for handling probability distributions in extensive systems.
By using Stirling's approximation, the task of calculating factorial terms in entropy or distribution equations becomes manageable, saving computational resources, and making theoretical assessments easily investigable for realistic and large-scale systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A system of \(N\) spins at a negative temperature \((E>0)\) is brought into contact with an ideal-gas thermometer consisting of \(N^{\prime}\) molecules. What will the nature of their state of mutual equilibrium be? Will their common temperature be negative or positive, and in what manner will it be affected by the ratio \(N^{\prime} / N\) ?

Prove that, quite generally, $$ C_{P}-C_{V}=-k \frac{\left[\frac{\partial}{\partial T}\left\\{T\left(\frac{\partial \ln Q}{\partial V}\right)_{T}\right\\}\right]_{V}^{2}}{\left(\frac{\partial^{2} \ln Q}{\partial V^{2}}\right)_{T}}>0 . $$ Verify that the value of this quantity for a classical ideal classical gas is \(N k\).

(a) The volume of a sample of helium gas is increased by withdrawing the piston of the containing cylinder. The final pressure \(P_{f}\) is found to be equal to the initial pressure \(P_{i}\) times \(\left(V_{i} / V_{f}\right)^{1.2}, V_{i}\) and \(V_{f}\) being the initial and final volumes. Assuming that the product \(P V\) is always equal to \(\frac{2}{3} U\), will (i) the energy and (ii) the entropy of the gas increase, remain constant, or decrease during the process? (b) If the process were reversible, how much work would be done and how much heat would be added in doubling the volume of the gas? Take \(P_{i}=1 \mathrm{~atm}\) and \(V_{i}=1 \mathrm{~m}^{3}\).

(a) Consider a gaseous system of \(N\) noninteracting, diatomic molecules, each having an electric dipole moment \(\mu\), placed in an external electric field of strength \(E\). The energy of such a molecule will be given by the kinetic energy of rotation as well as translation plus the potential energy of orientation in the applied field: where \(I\) is the moment of inertia of the molecule. Study the thermodynamics of this system, including the electric polarization and the dielectric constant. Assume that (i) the system is a classical one and (ii) \(|\mu E| \ll k T^{16}\) (b) The molecule \(\mathrm{H}_{2} \mathrm{O}\) has an electric dipole moment of \(1.85 \times 10^{-18}\) e.s.u. Calculate, on the basis of the preceding theory, the dielectric constant of steam at \(100^{\circ} \mathrm{C}\) and at atmospheric pressure.

Making use of the fact that the Helmholtz free energy \(A(N, V, T)\) of a thermodynamic system is an extensive property of the system, show that $$ N\left(\frac{\partial A}{\partial N}\right)_{V, T}+V\left(\frac{\partial A}{\partial V}\right)_{N, T}=A . $$ [Note that this result implies the well-known relationship: \(N \mu=A+P V(\equiv G)\).]

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.