An ideal gas, in a box of volume \(V\), consists of a mixture of \(N_{r}\) "red" and \(N_{g}\) "green" atoms, both with mass \(m\). Red atoms are distinguishable from green atoms. The green atoms have an internal degree of freedom that allows the atom to exist in two energy states, \(E_{\mathrm{g}, 1}=p^{2} /(2 m)\) and \(E_{\mathrm{g} 2}=p^{2} /(2 m)+\Delta\). The red atoms have no internal degrees of freedom. Compute the chemical potential of the "green" atoms.

Short Answer

Expert verified
\( \mu = -k_B T \ln \left[ \frac{V}{N_{\text{g}}} \left( \frac{2 \pi m k_B T}{h^2} \right)^{3/2} \frac{1}{(1 + e^{-\frac{\Delta}{k_B T}})} \right] \)

Step by step solution

01

- Understand the Problem

We need to find the chemical potential of the 'green' atoms in an ideal gas mixture. The green atoms have two possible energy states: 1. State 1: \(E_{\text{g}, 1} = \frac{p^2}{2m}\)2. State 2: \(E_{\text{g}, 2} = \frac{p^2}{2m} + \Delta\)
02

- Chemical Potential Definition

The chemical potential \(\mu\) can be found from the partition function of the system. For an ideal gas, the chemical potential \(\mu\) is given by: \[\mu = -k_B T \ln \left[ \frac{V}{N_{\text{g}}} \left( \frac{2 \pi m k_B T}{h^2} \right)^{3/2} \frac{1}{Z_{\text{int}}} \right]\]
03

- Molecular Partition Function

The total partition function \(Z_{\text{g}}\) for the green atoms consists of the translational partition function \(Z_{\text{trans}}\) and the internal partition function \(Z_{\text{int}}\). The translational part is: \[Z_{\text{trans}} = \frac{V}{h^3} (2 \pi m k_B T)^{3/2} \] The internal partition function \(Z_{\text{int}}\) for the green atoms is given by: \[Z_{\text{int}} = \sum_{i} e^{-E_{\text{g}, i} / k_B T}\]
04

- Calculate Internal Partition Function

Since the green atoms have two possible energy states: \[Z_{\text{int}} = e^{-E_{\text{g}, 1} / k_B T} + e^{-E_{\text{g}, 2} / k_B T}\] Substitute the given energy states: \[Z_{\text{int}} = e^{-\frac{p^2}{2m k_B T}} + e^{-\frac{p^2}{2m k_B T} - \frac{\Delta}{k_B T}} \] Hence, \[Z_{\text{int}} = e^{-\frac{p^2}{2m k_B T}} + e^{-\frac{p^2}{2m k_B T}} e^{-\frac{\Delta}{k_B T}} \]
05

- Simplify Internal Partition Function

Factor out the common term: \[Z_{\text{int}} = e^{-\frac{p^2}{2m k_B T}} (1 + e^{-\frac{\Delta}{k_B T}}) \]
06

- Substitute in Chemical Potential Formula

Substitute the simplified internal partition function and the translational partition function into the formula for the chemical potential: \[\mu = -k_B T \ln \left[ \frac{V}{N_{\text{g}}} \left( \frac{2 \pi m k_B T}{h^2} \right)^{3/2} \frac{1}{e^{-\frac{p^2}{2m k_B T}} (1 + e^{-\frac{\Delta}{k_B T}})} \right]\]
07

- Finalize the Expression

Simplify the expression for \(\mu\): \[\mu = -k_B T \ln \left[ \frac{V}{N_{\text{g}}} \left( \frac{2 \pi m k_B T}{h^2} \right)^{3/2} \right] + k_B T \ln \(e^{-\frac{p^2}{2m k_B T}} (1 + e^{-\frac{\Delta}{k_B T}})\)\] The term \(\frac{p^2}{2m k_B T}\) appears in both the exponential and the logarithm, and they cancel each other out. So: \[\mu = -k_B T \ln \left[ \frac{V}{N_{\text{g}}} \left( \frac{2 \pi m k_B T}{h^2} \right)^{3/2} \frac{1}{(1 + e^{-\frac{\Delta}{k_B T}})} \right] \]

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Key Concepts

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

internal partition function
The internal partition function, also denoted as \(Z_{\text{int}}\), plays a crucial role in determining the thermodynamic properties of a system. For an ideal gas with atoms that have internal degrees of freedom, this function sums up the contributions of different internal energy states. In our particular problem, the green atoms have two distinct energy states: \(E_{\text{g}, 1} = \frac{p^2}{2m}\) and \(E_{\text{g}, 2} = \frac{p^2}{2m} + \Delta\). The internal partition function can be written as:
\[Z_{\text{int}} = \sum_{i} e^{-E_{\text{g}, i} / k_B T} \]
For this problem, we have two energy states, so:
\[Z_{\text{int}} = e^{-E_{\text{g}, 1} / k_B T} \ + \ e^{-E_{\text{g}, 2} / k_B T} \]
Substituting the given energy states:
\[Z_{\text{int}} = e^{-\frac{p^2}{2m k_B T}} + e^{-\frac{p^2}{2m k_B T} - \frac{\Delta}{k_B T}} \]
We can factor out the common term:
\[Z_{\text{int}} = e^{-\frac{p^2}{2m k_B T}} (1 + e^{-\frac{\Delta}{k_B T}}) \]
This function is essential in calculating the chemical potential of the gas.
translational partition function
The translational partition function, denoted as \(Z_{\text{trans}}\), is a significant part of the overall partition function for an ideal gas. It represents the contribution of the translational movements of the gas particles to the system's overall thermodynamics. For a monatomic ideal gas in a volume \(V\), the translational partition function is given by:
\[Z_{\text{trans}} = \frac{V}{h^3} (2 \pi m k_B T)^{3/2} \]
Here, \(h\) is Planck's constant, \(m\) is the mass of the gas particles, \(k_B\) is Boltzmann's constant, and \(T\) is the temperature. This function essentially counts the number of quantum states available for the translational motion within the given volume and temperature. Combining the translational partition function with the internal partition function gives us a complete picture of the partition function for the system, \(Z_{\text{g}}\).
energy states
Energy states are crucial in defining the behavior of the gas particles, especially when internal degrees of freedom are considered. For the green atoms in this problem, there are two energy states:
1. \ E_{\text{g}, 1} = \frac{p^2}{2m}\
2. \ E_{\text{g}, 2} = \frac{p^2}{2m} + \Delta = E_{\text{g}, 1} + \Delta \
State 1 represents the base energy level associated with the kinetic energy of the gas particle. State 2 is an excited state that includes an additional energy component, \(\Delta\), on top of the kinetic energy. These energy states contribute to the internal partition function, encapsulating the effect of internal degrees of freedom. Understanding these states is essential in calculating the internal partition function and, consequently, the chemical potential of the gas. In physical terms, these states and their statistical distribution determine how the energy is partitioned among the particles in the system.

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Most popular questions from this chapter

The vibrational frequency of the \(\mathrm{I}_{2}\) molecule is \(f=6.42 \times 10^{12} \mathrm{~s}^{-1}\). The vibrational temperature is \(\theta_{1_{2}}^{\text {vib }}=\frac{h f}{k_{B}}=308 \mathrm{~K}\). The rotational temperature is \(\theta_{1_{2}}^{\text {rot }}=0.0538 \mathrm{~K}\). Consider a gas of \(N\) \(\mathrm{I}_{2}\) molecules at temperature \(T=300 \mathrm{~K}\). (a) What fraction of the molecules is in the vibrational ground state and what fraction have one vibrational quantum of energy? (b) What percentage of the total internal energy of the gas is: (1) translational?; (2) vibrational?; (3) rotational?

Consider a solid surface to be a two-dimensional lattice with \(N_{\mathrm{s}}\) sites. Assume that \(N_{\mathrm{a}}\) atoms \(\left(N_{a} \ll N_{s}\right)\) are adsorbed on the surface, so that each lattice site has either zero or one adsorbed atom. An adsorbed atom has energy \(E=-\varepsilon\), where \(\varepsilon>0\). Assume the atoms on the surface do not interact with one another. If the surface is at temperature \(T\), compute the chemical potential of the adsorbed atoms as a function of \(T, \varepsilon\), and \(N_{\mathrm{a}} / N_{\mathrm{s}}\) (use the canonical ensemble).

The CO molecule has a rotational temperature \(\theta=\hbar^{2} /\left(2 I k_{\mathrm{B}}\right)=2.8 \mathrm{~K}\), where \(I\) is the moment of inertia of the CO molecule. The rotational partition function for one molecule is \(Z_{1}^{\mathrm{rot}}=\sum_{l=0}^{\infty}(2 l+1) \mathrm{e}^{-l l(1) \theta / T}\) (a) If one mole of CO molecules could freely rotate at temperature \(T=\) \(3.2 \mathrm{~K}\), what is their total rotational entropy? (b) What is the rotational entropy of one mole of \(\mathrm{CO}\) molecules at temperature \(T=320 \mathrm{~K}\) ? (Hint: At high temperature, where many angular momenta contribute \(Z_{1}^{\text {rot }} \approx \int_{0}^{\infty} \mathrm{d} l(2 l+1) \mathrm{e}^{-\| l+1) \theta / T}\).) (c) What is the translational entropy of one mole of \(\mathrm{CO}\) molecules in a box of volume \(V=1.0 \mathrm{~m}^{3}\) at temperature \(T=320 \mathrm{~K}\) ?

A one-dimensional lattice of spin-1/2 lattice sites can be decomposed into blocks of three spins each. Use renormalization theory to determine whether or not a phase transition can occur on this lattice. If a phase transition does occur, what are its critical exponents? Retain terms in the block Hamiltonian to order \((V)\), where \(V\) is the coupling between blocks.

A dilute gas, composed of a mixture of \(N_{1}\) iodine atoms 1 and \(N_{12}\) iodine molecules \(I_{2}\), is confined to a box of volume \(V=1.0 \mathrm{~m}^{3}\) at temperature \(T=300 \mathrm{~K}\). The rotational temperature of the iodine molecules is \(\theta_{\text {rot }}=0.0537 \mathrm{~K}\) (for simplicity we neglect vibrational modes). (a) Compute the chemical potentials, \(\mu_{1}\) and \(\mu_{12}\), of the iodine atoms and molecules, respectively. (b) The numbers of the iodine atoms and molecules can change via chemical reactions with one another. The condition for chemical equilibrium is \(\mu_{12}=2 \mu_{1}\). Use this condition to find the ratio \(N^{2}{ }_{1} / N_{12}\) when the gas is in equilibrium. (c) Does the inclusion of the rotational degree of freedom increase or decrease the number of \(\bar{I}_{2}\) molecules at chemical equilibrium.

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