# Internal Energy of a Mole of Particles each with 3 Energy Levels

• danyull
In summary, the single-particle partition function for a system consisting of indistinguishable particles is given by zN/(N!). However, when you take logs and then differentiate, the N! disappears.
danyull
Homework Statement
A particle has three energy levels, ##E = 0##, ##\Delta##, and ##4\Delta##, where ##\Delta## is a positive constant. The lowest energy level is nondegenerate, whereas the other two are both doubly degenerate. Find the internal energy ##U## of a system of consisting of a mole of such particles.
Relevant Equations
The canonical partition function is ##Z=\sum\limits_i g(E_i) e^{-\beta E_i}##, and the internal energy is related by ##U=-\frac{\partial}{\partial\beta} \ln Z##.
Hello, I'm doing some refreshers before going back to school. Stat mech is my shakiest and I'd appreciate some help on this problem.

I know that for a single particle, the partition function will be $$Z = 1 + 2e^{-\beta\Delta} + 2e^{-4\beta\Delta}$$ and so its internal energy is $$\frac{1}{Z} \left( 2\Delta e^{-\beta\Delta} + 8\Delta e^{-4\beta\Delta} \right).$$ My only concern is how does this extend to a mole of particles? Since each particle can be in one of 5 energy states, this means that a mole of particles will have ##5^{N_A}## energy states, with many degeneracies. In this case, is there a more explicit way to express the partition function than the general expression above?

I feel like I'm overthinking things, and it just comes down to something like tacking on an ##N_A## somewhere. If so, I'd appreciate any rough explanation why. Thank you.

In general, $$Z=z^N$$ is the partition function for a system consiting of ##N## distinguishable and noninteracting particles with single particle partition function ##z## (prove this from first principle, good exercise).

Thus, in your case where the single particle partition function is given by $$z = 1 + 2e^{-\beta\Delta} + 2e^{-4\beta\Delta}.$$ The partition function for a mole of distinguishable and noninteracting particles is therefore $$Z = \big(1 + 2e^{-\beta\Delta} + 2e^{-4\beta\Delta}\big)^{N_A}$$
and the internal energy becomes \begin{align*}U &= -\frac{\partial}{\partial\beta}\log Z \\ &= 2N_A\Delta\frac{e^{-\beta\Delta}+4e^{-4\beta\Delta}}{1+2e^{-\beta\Delta}+2e^{-4\beta\Delta}}.\end{align*}

danyull
William Crawford said:
In general, $$Z=z^N$$ is the partition function for a system consiting of ##N## distinguishable and noninteracting particles with single particle partition function ##z## (prove this from first principle, good exercise).
Ah I remember this now! This also reminds me that I should study up on distinguishable and indistinguishable particles too. Thanks a bunch!

If the particles are indistinguishable, which is probably the case here, Z = zN/(N!). But when you take logs and then differentiate, the N! disappears.

mjc123 said:
If the particles are indistinguishable, which is probably the case here, Z = zN/(N!). But when you take logs and then differentiate, the N! disappears.
No, this is generally not true for indistinguishable particles as the following minimal example will demonstrate.

Consider a system consisting of two particles, each of which can be in one out of two possible one-particle-states with the energies ##E_1 = 0## and ##E_2 = \epsilon## respectively.

Case 1 (distinguishable and noninteracting particles)
The single-particle partition function is in this case given by \begin{align}z = 1 + e^{-\beta\epsilon}\end{align} and the total partition function is \begin{align}Z_\text{dist} &= 1 + 2e^{-\beta\epsilon} + e^{-2\beta\epsilon} \nonumber\\ &= \big(1+e^{-\beta\epsilon}\big)^2 \\ &= z^2\nonumber\end{align} as expected.

Case 2 (indistinguishable bosons)
The single-particle partition function is still given by (1). However, this time, the total partition function given by \begin{align}Z_\text{bosons} &= 1 + e^{-\beta\epsilon} + e^{-2\beta\epsilon} \nonumber\\ &\neq \frac{z^2}{2} \\ &= \frac{1}{2} + e^{-\beta\epsilon} + \frac{1}{2}e^{-2\beta\epsilon}\nonumber\end{align} as otherwise claimed.

Case 3 (indistinguishable fermions)
Likewise is the single-particle partition function also given (1). However, this is the total partition function given as \begin{align}Z_\text{fermions} &= e^{-\beta\epsilon} \nonumber\\ &\neq \frac{z^2}{2} \\ &= \frac{1}{2} + e^{-\beta\epsilon} + \frac{1}{2}e^{-2\beta\epsilon}.\nonumber\end{align}

Conclusion: Neither ##Z_\text{bosons}## nor ##Z_\text{fermions}## satisfy ##z^2/2##.

EDIT: Three pictures illustrating the above three cases.

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## 1. What is the definition of internal energy?

Internal energy is the total energy stored within a system, including the kinetic energy of its molecules, potential energy due to the forces between molecules, and the energy associated with the molecular structure.

## 2. How is internal energy related to temperature?

The internal energy of a system is directly proportional to its temperature. As the temperature of a system increases, the molecules within it will have more kinetic energy, resulting in an overall increase in internal energy.

## 3. What is meant by "a mole of particles each with 3 energy levels"?

A mole is a unit of measurement that represents the number of particles in a given substance. In this case, it refers to a specific number of particles (Avogadro's number) each with 3 available energy levels.

## 4. How is the internal energy of a mole of particles with 3 energy levels calculated?

The internal energy of a mole of particles with 3 energy levels can be calculated by multiplying the number of particles by the average energy per particle. This average energy can be found by taking the sum of the energy levels and dividing by the total number of particles.

## 5. How does the internal energy of a mole of particles change with increasing temperature?

With increasing temperature, the internal energy of a mole of particles will also increase. This is because the average energy per particle will increase as a result of the molecules having more kinetic energy at higher temperatures.

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