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The four energy levels for atomic vanadium (V) have the following energies and degeneracies: $$\begin{array}{ccc} \text { Level }(\boldsymbol{n}) & \text { Energy }\left(\mathrm{cm}^{-1}\right) & \text {Degeneracy } \\ \hline 0 & 0 & 4 \\ 1 & 137.38 & 6 \\ 2 & 323.46 & 8 \\ 3 & 552.96 & 10 \end{array}$$ What is the contribution to the average molar energy from electronic degrees of freedom for \(\mathrm{V}\) when \(T=298 \mathrm{K} ?\)

Short Answer

Expert verified
The contribution to the average molar energy from electronic degrees of freedom for atomic vanadium (V) when T = 298 K is approximately 291 J/mol.

Step by step solution

01

Calculate the Boltzmann factors for each energy level

We have been given the energies and degeneracies of the four energy levels. Let's compute the Boltzmann factors \(\frac{g_ne^{-E_n/k_BT}}{\sum_ng_ne^{-E_n/k_BT}}\), for each energy level, using the given temperature T = 298 K: 1. Level 0: \(\frac{4e^{-(0)/(k_B \cdot 298)}}{\sum_ng_ne^{-E_n/k_BT}}\) 2. Level 1: \(\frac{6e^{-(137.38 \cdot 100)/(k_B \cdot 298)}}{\sum_ng_ne^{-E_n/k_BT}}\) 3. Level 2: \(\frac{8e^{-(323.46 \cdot 100)/(k_B\cdot 298)}}{\sum_ng_ne^{-E_n/k_BT}}\) 4. Level 3: \(\frac{10e^{-(552.96 \cdot 100)/(k_B \cdot 298)}}{\sum_ng_ne^{-E_n/k_BT}}\)
02

Find the partition function Q

Now that we have the Boltzmann factors for each energy level, we need to find the partition function Q. We can do this by summing over the Boltzmann factors: Q = \(4e^{-(0)/(k_B \cdot 298)}\) + \(6e^{-(137.38 \cdot 100)/(k_B \cdot 298)}\) + \(8e^{-(323.46 \cdot 100)/(k_B\cdot 298)}\) + \(10e^{-(552.96 \cdot 100)/(k_B \cdot 298)}\) Calculate Q using the Boltzmann constant \(k_B = 1.380649\times10^{-23}\) JK^{-1}: Q ≈ 6.352
03

Differentiate the natural logarithm of Q with respect to temperature T

In order to find the average molar energy, we need to differentiate the natural logarithm of Q with respect to temperature T: \(\frac{d(\ln{Q})}{dT}\) = \(\frac{1}{Q} \cdot \frac{dQ}{dT}\) Using the chain rule, the derivative of Q with respect to T is: \(\frac{dQ}{dT}\) = \(\frac{d}{dT}\Big(4e^{-(0)/(k_B \cdot 298)} + 6e^{-(137.38 \cdot 100)/(k_B \cdot 298)} + 8e^{-(323.46 \cdot 100)/(k_B\cdot 298)} + 10e^{-(552.96 \cdot 100)/(k_B \cdot 298)}\Big)\) Now, calculate \(\frac{dQ}{dT}\) to get: \(\frac{dQ}{dT}\) ≈ -0.1796
04

Calculate the average molar energy

Now that we have Q and \(\frac{d(\ln{Q})}{dT}\), we can calculate the average molar energy using the formula: Average molar energy = \(k_BT^2\frac{d(\ln{Q})}{dT}\) Plug in the values for T, \(k_B\), and \(\frac{d(\ln{Q})}{dT}\) to get: Average molar energy ≈ (-1.380649 × 10^{-23} J/K × (298 K)² × (-0.1796)/6.352) × (6.022 × 10²³ atoms/mol) Average molar energy ≈ 291 J/mol The contribution to the average molar energy from electronic degrees of freedom for atomic vanadium (V) when T = 298 K is approximately 291 J/mol.

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Key Concepts

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

Boltzmann Factors
When studying statistical mechanics, one critical concept is the Boltzmann factor, named after the physicist Ludwig Boltzmann. These factors shed light on how energy is distributed among various atomic or molecular states at a given temperature, which significantly ties into how we understand thermodynamic properties and behaviors.

A Boltzmann factor for a state with energy level, represented by the symbol \texttt{E}, is calculated using the following expression: \( e^{-E/(k_BT)} \) where \( k_B \) denotes the Boltzmann constant and \( T \) is the absolute temperature in Kelvins. Each factor represents the probability that a system will occupy a particular energy level at thermal equilibrium.

Considering each energy level's degeneracy, which is the number of states with the same energy, we can adjust the Boltzmann factor to include this information, making it: \( g_ne^{-E_n/(k_BT)} \) where \( g_n \) is the degeneracy of level \texttt{n}. These corrected factors are ultimately essential for determining a system's partition function, as they directly influence the probabilities of each state.
Partition Function
The partition function, often denoted by \texttt{Q}, is an indispensable quantity in statistical thermodynamics. It provides a comprehensive sum of all possible states of a system, weighted by their Boltzmann factors, and plays a pivotal role in connecting statistical mechanics with thermodynamic quantities.

The partition function is defined as: \( Q = \sum g_ne^{-E_n/(k_BT)} \) with the summation extending over all states \texttt{n}. This function essentially captures the statistical distribution of a system over its possible quantum states, factoring in their energies and degeneracies.

Once we determine the partition function, we can derive many thermodynamic properties, such as internal energy, entropy, and heat capacity, by subsequent differentiation and arithmetic manipulation. The partition function serves as the foundation upon which the entire superstructure of equilibrium statistical mechanics is built.
Electronic Degrees of Freedom
In physical chemistry, the concept of degrees of freedom refers to independent ways in which a system can possess energy. Most notably, these include translational, rotational, vibrational, and electronic degrees of freedom.

Electronic degrees of freedom specifically describe the energy levels that electrons can occupy in an atom or molecule. Each of these levels has associated energies, and the electrons can transition between these levels by absorbing or emitting quanta of energy. The significance of electronic degrees of freedom arises when considering higher temperatures, where these transitions become more probable and hence more relevant in determining thermodynamic properties.

In the context of vanadium (V) with discrete energy levels, the calculation of the average molar energy includes contributions from these electronic states. Because different energy levels have various degeneracies, the probability of each state—as influenced by its Boltzmann factor—determines the impact it has on the system's overall energy.
Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In the realm of thermodynamics, we explore how energy is transferred within a system and between systems, and how this affects physical properties and processes.

At the center of thermodynamical studies are the laws of thermodynamics, which govern how systems react to changes in temperature, pressure, volume, or chemical composition. Together with concepts like entropy, enthalpy, and the Gibbs free energy, thermodynamics provides a framework within which the behavior of materials can be predicted and understood.

In the context of the given exercise on vanadium (V), thermodynamics allows us to calculate the contribution of electronic degrees of freedom to the average molar energy at a specific temperature. This involves using statistical mechanics - a discipline closely intertwined with thermodynamics - to evaluate how atomic or molecular states are populated and how these populations influence macroscopic properties such as temperature, pressure, and volume.

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Most popular questions from this chapter

The isotope exchange reaction for \(\mathrm{Cl}_{2}\) is as follows: \(^{35} \mathrm{Cl}^{35} \mathrm{Cl}+^{37} \mathrm{Cl}^{37} \mathrm{Cl} \rightleftharpoons 2^{37} \mathrm{Cl}^{35} \mathrm{Cl} .\) The equilibrium constant for this reaction is \(\sim 4 .\) Furthermore, the equilibrium constant for similar isotope-exchange reactions is also close to this value. Demonstrate why this would be so.

Determine the standard molar entropy for the hydroxyl radical, \(\mathrm{OH}\), for which \(\widetilde{\nu}=3735 \mathrm{cm}^{-1}\) and \(B=18.9 \mathrm{cm}^{-1},\) the ground electronic state is doubly degenerate, and \(P=1.00\) atm.

Consider an ensemble of units in which the first excited electronic state at energy \(\varepsilon_{1}\) is \(m_{1}\) -fold degenerate, and the energy of the ground state is \(m_{o}\) -fold degenerate with energy \(\varepsilon_{0}\) a. Demonstrate that if \(\varepsilon_{0}=0,\) the expression for the electronic partition function is \\[ q_{E}=m_{o}\left(1+\frac{m_{1}}{m_{o}} e^{-\varepsilon_{1} / k T}\right) \\]\ b. Determine the expression for the internal energy \(U\) of an ensemble of \(N\) such units. What is the limiting value of \(U\) as the temperature approaches zero and infinity?

For an ensemble consisting of a mole of particles having two energy levels separated by \(1000 . \mathrm{cm}^{-1},\) at what temperature will the internal energy equal \(3.00 \mathrm{kJ} ?\)

The equilibrium between hydrogen cyanide (HCN) and its isomer hydrogen isocyanide (HNC) is important in interstellar chemistry: \\[ \mathrm{HCN}(g) \rightleftharpoons \mathrm{HNC}(g) \\] A long-standing "puzzle" regarding this reaction is that in space \((\mathrm{T}=2.75 \mathrm{K})\) surprisingly large amounts of HNC are observed. For example, HNC/HCN ratios approaching \(20 \%\) have been observed in comets (Advances in Space Research 31 \((2003): 2577) .\) Using the spectroscopic information provided in the following table and knowledge that the potential- energy surface minimum of HNC lies roughly \(5200 \mathrm{cm}^{-1}\) higher in energy relative to \(\mathrm{HCN}\), calculate the theoretical value for \(K_{p}\) for this reaction in interstellar space. $$\begin{array}{cccccc} & {}{} {\boldsymbol{M}\left(\mathrm{g} \mathrm{mol}^{-1}\right) \boldsymbol{B}\left(\mathrm{cm}^{-1}\right)} & \boldsymbol{\nu}_{1}\left(\mathrm{cm}^{-1}\right) & \boldsymbol{\tilde { \nu }}_{2}\left(\mathrm{cm}^{-1}\right) & \widetilde{\boldsymbol{\nu}}_{3}\left(\mathrm{cm}^{-1}\right) \\ \hline \mathrm{HCN} & 27.03 & 1.477 & 2041 & 712 & 3669 \\ \mathrm{HNC} & 27.03 & 1.512 & 2024 & 464 & 3653 \end{array}$$

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