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At high temperature, the average energy of a classical one-diniensional oscillator is \(k_{\mathrm{h}} T,\) and tor an atom in \(\mathrm{u}\) monatomic ideal gas. it is \(\frac{1}{2} k_{B} T\), Explain the difference. using the equipartition theorem.

Short Answer

Expert verified
The difference in average energies arises due to the different number of degrees of freedom for a classical one-dimensional oscillator and an atom in a monatomic ideal gas. A one-dimensional oscillator has two degrees of freedom (kinetic and potential) contributing to the total energy, leading to average energy of \(k_{B}T\). Contrastingly, an atom in a monatomic ideal gas, having only one degree of freedom (kinetic), contributes \(\frac{1}{2}k_{B}T\) to the total energy.

Step by step solution

01

Understanding thermal energy of a one-dimensional oscillator and an atom in monatomic ideal gas

The average thermal energy of a classical one dimensional oscillator at high temperature is given by the formula \(E=k_{B}T\), where \(k_{B}\) is Boltzmann's constant and T is the absolute temperature. For an atom in a monatomic ideal gas, the average thermal energy is given by \(\frac{1}{2}k_{B}T\). These formulas pertain to the kinetic theory of gases wherein the total energy of the system is assumed to be divided equally among all available degrees of freedom.
02

Discussing the equipartition theorem

The equipartition theorem states that for a system in equilibrium at temperature T, each degree of freedom contributes an average energy of \(\frac{1}{2}k_{B}T\) to the total energy of the system. A degree of freedom refers to each independent way in which the energy can be changed without altering the total momentum.
03

Explaining the difference

For a one-dimensional oscillator, there are two degrees of freedom - one for the kinetic energy due to motion and one for the potential energy due to position. Therefore, according to the equipartition theorem, the total energy is \(2*(\frac{1}{2}k_{B}T)=k_{B}T\). However, for an atom in a monatomic ideal gas, there is only one degree of freedom – the kinetic energy as the potential energy through position in a gas is zero due to lack of restrictive forces. Therefore, average energy is \(\frac{1}{2}k_{B}T\) according to the equipartition theorem.

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Key Concepts

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

Classical Oscillator
A classical oscillator can be imagined as a system that behaves like a mass attached to a spring. It oscillates back and forth in a predictable manner. When it comes to understanding the energy of a classical one-dimensional oscillator, especially at high temperatures, the equipartition theorem plays a crucial role.
The equipartition theorem helps in determining how energy is distributed among various degrees of freedom. For the classical oscillator, these degrees of freedom refer to kinetic energy and potential energy. Each component contributes equally to the system's total energy. Thus, for the classical oscillator, the average energy is broken into two parts:
  • Kinetic energy, which is due to motion, amounts to \(\frac{1}{2}k_{B}T\).
  • Potential energy, due to its position, also amounts to \(\frac{1}{2}k_{B}T\).
Combining these contributions, the total energy for a one-dimensional classical oscillator at high temperature is \(k_{B}T\). This highlights the importance of each individual form of energy adding to the system's ability to store energy under normal oscillating conditions.
Monatomic Ideal Gas
A monatomic ideal gas consists of individual atoms that do not interact with each other through chemical bonds. Think of them as tiny, independent particles moving around freely in a container.
For a monatomic ideal gas, the energy analysis is a bit simpler than for a classical oscillator. In this scenario, each atom represents a single point with only one degree of freedom contributing its energy through kinetic motion. Potential energy is not considered because, in an ideal gas, there are no forces acting between atoms over a distance to create potential energy.
In this case, only the kinetic energy part is considered, contributing an average energy of:
  • \(\frac{1}{2}k_{B}T\).
The energy calculation doesn't account for any potential energy as the atoms are free to move and there are no restrictive positions in the system. Consequently, the average energy per atom in a monatomic ideal gas is simpler and entirely kinetic, shaped primarily by the absence of potential interactions.
Degrees of Freedom
Degrees of freedom in physics describe the ways that a system can independently change or vary without breaking any conservation rules. This is tied closely to how energy gets apportioned in systems like our classical oscillator and the monatomic ideal gas. Understanding this concept can help demystify why energy is distributed the way it is.
In a classical one-dimensional oscillator, there are two degrees of freedom. One degree of freedom accounts for the kinetic energy (movement), while the other accounts for the potential energy (position). According to the equipartition theorem, each degree contributes \(\frac{1}{2}k_{B}T\) to the total average energy.
On the other hand, in a monatomic ideal gas, each particle simply moves freely in space without the influence of forces binding them in potential wells. Thus, a monatomic ideal gas particle has only one degree of freedom – its kinetic energy itself. As per the equipartition theorem, this degree of freedom gives it the energy of \(\frac{1}{2}k_{B}T\).
Recognizing how degrees of freedom influence energy distribution is essential in thermodynamics, as they help us understand how energy gets shared across a system's components based on their ability to move and vibrate.

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Most popular questions from this chapter

We claim that the famous exponential decrease of probability with energy is natural, the vastly most probable and disordered state given the constraints on total energy and number of particles. It should be a state of maximum entropy! The proof involves mathematical techniques beyond the scope of the text, but finding support is good exercise and not difficult. Consider a system of 11 oscillators sharing a total energy of just \(5 h \omega_{0}\). In the symbols of Section \(9.3, N=11\) and \(M=5\). (a) Using equation \((9-9),\) calculate the probabilities of \(n\) being \(0.1,2 .\) and \(3 .\) (b) How many particles. \(N_{n}\), would be expected in each level? Round each other nearest integer. (Happily. the number is still 11 . and the energy still \(\operatorname{she}_{0}\) ) What you have is a distribution of the energy that is as close to expectations is possible, given that numbers at each level in a real case are integers. (c) Entropy is related to the number of microscopic ways the macrostate can be obtained. and the number of ways of permuting particle labels with \(N_{0}\). \(N_{1}, N_{2},\) an \(d N_{1}\) fixed and totaling 11 is \(11 ! /\left(N_{0} ! N_{1} !\right.\) \(\left.N_{2} ! N_{3}^{\prime}\right) .\) (See Appendix J for the proof.) Calculate the number of ways for your distribution. (d) Calculate the number of ways if there were 6 particles in \(n=0.5\) in \(n=1\), and none higher. Note that this also has the same total energy. (e) Find at least one other distribution in which the 11 oscillators share the same energy, and calculate the number of ways. (f) What do your findings suggest?

We based the exact probabilities of equation (9.9) on the claim that the number of ways of adding \(N\) distinct nonnegative integers/quuntum numbers to give a total of \(M\) is \(\\{M+N-1) ! /\left[M^{\prime}(N-1) !\right]\). Verify this claim (a) for the case \(N=2, M=5\) and (b) for the case \(N=5, M=2\)

A scientifically untrained but curious friend asks, "When I walk into a room. is there a chance that all the air will be on the other side?" How do you answer this question?

The exact probabilities of equation \((9-9)\) rest on the claim that the number of ways of adding \(N\) distinct nonnegative integers to give a total of \(M\) is \((M+N-1) !\) \([M !(N-1) !]\). One way to prove it involves the following trick. It represents two ways that \(N\) distinct integers can add to \(M-9\) and 5 , respectively, in this special case. $$ \begin{array}{c|cccccccccccccc} \hline \text { 1 } & \text { X } & \text { X } & \text { X } & \text { I } & \text { I } & \text { X } & \text { I } & \text { I } & \text { I } & \text { I } & \text { X } & \text { I } & \text { I } \\ \hline 2 & \text { I } & \text { X } & \text { X } & \text { I } & \text { I } & \text { I } & \text { I } & \text { X } & \text { I } & \text { I } & \text { I } & \text { X } & \text { X } \\ \hline \end{array} $$ The X's represent the total of the integers, \(M\) - each row has \(5 .\) The I's represent "dividers" between the distinct integers, of which there will of course be \(N-\) I -each row has 8 . The first row says that \(n_{1}\) is 3 (three \(X\) 's before the divider between it and \(n_{2}\) ). \(n_{2}\) is 0 (no \(X\) 's between its left divider with \(n_{1}\) and its right divider with \(\left.n_{3}\right), n_{3}\) is \(1, n_{4}\) through \(n_{6}\) are \(0, n_{2}\) is \(1,\) and \(n_{8}\) and \(n_{9}\) are 0\. The second row says that \(n_{2}\) is \(2, n_{6}\) is \(1, n_{9}\) is \(2,\) and all other \(n\) are 0 . Further rows could account for all possible ways that the integers can add to \(M\). Argue that. properly applied, the binomial coefficient (discussed in Appendix J) can be invoked to give the correct total number of ways for any \(N\) and \(M\).

At what wavelength does the human body emit the maximum electromagnetic radiation? Use Wien's law from Exercise 79 and assume a skin temperature of \(70^{\circ} \mathrm{F}\)

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