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For a systeFor a system of energy levels, \(\varepsilon_{m}=m^{2} \alpha,\) where \(\alpha\) is a constant with units of energy and \(m=0,1,2, \ldots, \infty,\) What is the internal energy and heat capacity of this system in the high-temperature limit?m of energy levels, \(\varepsilon_{m}=m^{2} \alpha,\) where \(\alpha\) is a constant with units of energy and \(m=0,1,2, \ldots, \infty,\) What is the internal energy and heat capacity of this system in the high-temperature limit?

Short Answer

Expert verified
In the high-temperature limit, the internal energy of the system is given by \(U = \frac{3}{2} k_B T\), and the heat capacity is constant and equals \(C_V = \frac{3}{2} k_B\).

Step by step solution

01

Boltzmann Distribution

To find the probabilities of different energy levels in the high-temperature limit, we will use the Boltzmann distribution, which states that the probability of finding the system in a state with energy level \(\varepsilon_m\) is proportional to \(e^{-\frac{\varepsilon_m}{k_BT}}\), where \(k_B\) is the Boltzmann constant and \(T\) is the temperature.
02

Normalization Factor

To find the normalization factor, we can sum over all probabilities to equal 1: \[Z = \sum_{m=0}^{\infty} e^{-\frac{m^2 \alpha}{k_BT}}\] As we are looking for the high-temperature limit (\(T \to \infty\)), the exponent becomes very small and the sum will approximate an integral: \[Z \approx \int_0^{\infty} e^{-\frac{m^2 \alpha}{k_BT}} dm\] This is a standard Gaussian integral, which has the value: \[\frac{1}{2} \sqrt{\frac{\pi k_B T}{\alpha}}\]
03

Probability distribution for m

Now, we can find the probability distribution for energy levels: \[P(m) = \frac{1}{Z} e^{-\frac{m^2 \alpha}{k_B T}} = \frac{2}{\sqrt{\pi k_B T / \alpha}} e^{-\frac{m^2 \alpha}{k_B T}}\]
04

Calculate internal energy

Using the probability distribution, we can find the internal energy of the system, which is the expected value of the energy: \[U = \sum_{m=0}^{\infty} \varepsilon_m P(m) = \frac{2}{\sqrt{\pi k_B T / \alpha}} \sum_{m=0}^{\infty} m^2 \alpha e^{-\frac{m^2 \alpha}{k_B T}}\] Again, let's approximate this sum by an integral: \[U \approx \frac{2}{\sqrt{\pi k_B T / \alpha}} \int_0^{\infty} m^2 \alpha e^{-\frac{m^2 \alpha}{k_B T}} dm\] To solve this integral, we can use the formula for the Gaussian integral: \[\int_0^{\infty} x^ne^{-ax^2}dx = \frac{1}{2} \sqrt{\frac{\pi}{a^{n+1}}} \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\] With \(n = 2\) and \(a = \frac{\alpha}{k_B T}\), we get the final result for internal energy: \[U = \frac{3}{2} k_B T\]
05

Calculate heat capacity

Now we can find the heat capacity at constant volume, which is defined as: \[C_V = \left( \frac{\partial U}{\partial T} \right)_V\] From the above internal energy expression, the heat capacity is: \[C_V = \frac{3}{2} k_B\] In the high-temperature limit, the heat capacity of the system is constant and equals \(\frac{3}{2} k_B\).

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Key Concepts

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

Boltzmann Distribution
Understanding the Boltzmann distribution is pivotal for studying statistical mechanics and thermodynamics. It describes the probability of a system to be in a particular energy state at thermal equilibrium. This distribution indicates that at higher temperatures, the particles are more likely to populate higher energy states, while at lower temperatures, lower energy states are more favored.

In practical terms, for a particle in a system with discrete energy levels \( \varepsilon_m \), the probability \( P(m) \) of finding it in a state \( m \) is given by the formula \( P(m) = \frac{e^{-\frac{\varepsilon_m}{k_BT}}}{Z} \) where \( k_B \) is the Boltzmann constant, \( T \) is the absolute temperature, and \( Z \) is the partition function. This relationship emphasizes the influence of temperature on energy distribution within a system.
Partition Function
The partition function, often symbolized by \( Z \) is a central concept in statistical thermodynamics. It plays a critical role by acting as a normalization factor in the Boltzmann distribution to ensure that the probabilities of all possible states sum up to 1.

The partition function is the sum of the Boltzmann factors over all possible energy states of a system and is mathematically expressed as \( Z = \sum_{m=0}^{\infty} e^{-\frac{\varepsilon_m}{k_BT}} \). In high-temperature conditions, many energy states are accessible, and the partition function helps us understand the multitude of possible configurations that a system can adopt. To calculate \( Z \) in these scenarios, one often uses integral approximations, as sums over large numbers of states can become intractable.
Internal Energy
Internal energy, often denoted as \( U \) in thermodynamics, is the total energy contained within a system. It encompasses all forms of energy, including kinetic and potential energies of the particles in the system.

The internal energy can be calculated by taking the expected value of the energy across all possible states, which is the sum of the energy levels weighted by their respective probabilities. For a system following the Boltzmann distribution, this amounts to \( U = \sum_{m=0}^{\infty} \varepsilon_m P(m) \), where \( P(m) \) is the probability of the system being in the state with energy \( \varepsilon_m \). At high temperatures, the internal energy approaches an approximation that simplifies the calculation, reflecting the energy spreading over an increasing number of accessible states.
Gaussian Integral
The Gaussian integral is a fundamental operation in mathematics, often encountered when dealing with normal distributions and statistical applications. It's utilized in various fields of physics and engineering to solve problems involving heat, diffusion, and quantum mechanics.

In a statistical mechanics context, particularly for continuous systems or those at high temperature limits, Gaussian integrals help approximate the sums over states. For example, the integral \( \int_0^{\infty} e^{-ax^2} dx \) calculates the area under the curve of a Gaussian function and has a closed-form solution related to the square root of \( \pi/a \). This analytical solution is a powerful tool when evaluating partition functions and internal energies, as it lets us bypass complex summations which are otherwise very challenging to compute.

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Most popular questions from this chapter

The molecule NO has a ground electronic level that is doubly degenerate, and a first excited level at \(121.1 \mathrm{cm}^{-1}\) that is also twofold degenerate. Determine the contribution of electronic degrees of freedom to the standard molar entropy of NO. Compare your result to \(R \ln (4)\). What is the significance of this comparison?

The three lowest energy levels for atomic carbon (C) have the following energies and\ degeneracies: $$\begin{array}{ccc} \text { Level }(\boldsymbol{n}) & \text { Energy }\left(\mathbf{c m}^{-1}\right) & \text {Degeneracy } \\ \hline 0 & 0 & 1 \\ 1 & 16.4 & 3 \\ 2 & 43.5 & 5 \end{array}$$ What is the contribution to the average molar energy from the electronic degrees of freedom for \(C\) when \(T=100 .\) K?

What is the contribution to the internal energy from translations for an ideal monatomic gas confined to move on a surface? What is the expected contribution from the equipartition theorem?

In "Direct Measurement of the Size of the Helium Dimer" by F. Luo, C. F. Geise, and W. R. Gentry (J. Chemical Physics \(104[1996]: 1151\) ), evidence for the helium dimer is presented. As one can imagine, the chemical bond in the dimer is extremely weak, with an estimated value of only \(8.3 \mathrm{mJ} \mathrm{mol}^{-1}\) a. An estimate for the bond length of \(65 \AA\) is presented in the paper. Using this information, determine the rotational constant for \(\mathrm{He}_{2}\). Using this value for the rotational constant, determine the location of the first rotational state. If correct, you will determine that the first excited rotational level is well beyond the dissociation energy of He \(_{2}\) b. Consider the following equilibrium between He \(_{2}\) and its atomic constituents: \(\mathrm{He}_{2}(g) \rightleftharpoons 2 \mathrm{He}(g) .\) If there are no rotational or vibrational states to consider, the equilibrium is determined exclusively by the translational degrees of freedom and the dissociation energy of He \(_{2}\). Using the dissociation energy provided earlier and \(V=1000 . \mathrm{cm}^{3}\) determine \(K_{P}\) assuming that \(T=10 .\) K. The experiments were actually performed at \(1 \mathrm{mK} ;\) why was such a low temperature employed?

Using the Helmholtz energy, demonstrate that the pressure for an ideal polyatomic gas is identical to that derived for an ideal monatomic gas in the text.

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