Expressions for energy & entropy from free energy (discrete distribution)

In summary, the partition function is defined as the sum of the Boltzmann factors, and using this we can obtain an expression for the free energy. This can be used to solve for the average energy of the system, which is also known as the thermodynamic U. This relationship is standard in statistical mechanics and can be used to find the entropy and energy of a discrete system. Alternatively, the thermodynamic relation S = - F/T can be used to find the energy from the free energy expression.
  • #1
Homework Statement
(a) Find an expression for the free energy as a function of ##\tau## (temperature) of a system with two states, one at energy 0 and one at energy ##\epsilon##. (b) From the free energy, find expressions for energy and entropy of the system. The entropy is plotted in Figure 3.11 (see below).
Relevant Equations
$$ Z = \sum_i e^{ frac {-E_i} {k_BT} } $$
$$ F = -k_B T \ln{Z} $$
$$ F = U - TS $$
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So just by by using the definition of the partition function...
$$ Z = \sum_i e^{ \frac {-E_i} {k_BT} } = e^{ \frac {-0} {k_BT} } + e^{ \frac {-\epsilon} {k_BT} } = 1 + e^{ \frac {-\epsilon} {k_BT} } $$

And then, a result we obtained in class by using the Boltzmann H factor to solve for ##S## and finding an expression for ## U - TS## gave us ##F = -k_B T \ln{Z} ##. So applying that gives part (a)...

$$ F = -k_B T \ln{ 1 + e^{ \frac {-\epsilon} {k_BT} } } $$

Now for (b), I'm confused. I'm not sure if I have an equation to get S and U out of just this free energy expression. All I have is the free energy definition, ## F = U - TS ##, and if I try to solve using that and my above expression for free energy, I'm going to get something like ## U = U ## or ## S = S##. So I'm not sure which equation/principle to use. I have thought about using differential with constant volume but I don't think that would help and the question seems pretty explicit about using the free energy expression from (a) to obtain both U and S.
 
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  • #2
Think of ##U## as ##\langle E \rangle##. For the two-state system, how would you calculate ##\langle E \rangle##? Hint: What is the probability that the system has energy 0? What is the probability the system has energy ##\epsilon##?
 
  • #3
TSny said:
Think of ##U## as ##\langle E \rangle##. For the two-state system, how would you calculate ##\langle E \rangle##? Hint: What is the probability that the system has energy 0? What is the probability the system has energy ##\epsilon##?

Oh ok. So ...
$$ \langle E \rangle = \sum_{E = 0, E = \epsilon} E*P(E) = \frac {e^{\frac {0} {k_BT}}} {Z}*0 + \frac {e^{\frac {-\epsilon} {k_BT}}} {Z} *\epsilon $$

$$ \langle E \rangle = \frac {e^{\frac {-\epsilon} {k_BT} } } {1 + {e^{\frac {-\epsilon} {k_BT} } } *\epsilon $$

Is this the expression for "the energy of the system" that (b) is asking me for? So then I plug this into ## F = U - TS## to solve for entropy?
 
  • #4
Yes
 
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  • #5
That's interesting, I would have that by "energy of the system," they mean some sort of general expression, like energy as a function of the state. But it actually refers to an ensemble average of energy. Is there a reason for that? My guess is it has something to do with the system being discrete.
 
  • #6
baseballfan_ny said:
That's interesting, I would have that by "energy of the system," they mean some sort of general expression, like energy as a function of the state. But it actually refers to an ensemble average of energy. Is there a reason for that? My guess is it has something to do with the system being discrete.
I'm not sure I understand what you are asking here. Taking the thermodynamic ##U## to correspond to the statistical mechanics ##\langle E \rangle## is standard, I believe.

Another approach to the problem is to first find ##S## using the thermodynamic relation ##S = -\frac {\partial F}{\partial T}##. Then you can find ##U## from ##F = U - TS##.

Or, maybe you have seen the relation ##U= - \large \frac {\partial \ln Z}{\partial \beta}##, where ##\beta = \large \frac 1 {kT}##. This can be used to find ##U##.
 
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  • #7
TSny said:
Taking the thermodynamic U to correspond to the statistical mechanics ⟨E⟩ is standard, I believe.
That's exactly what I was asking :) Now that I think about, I think our instructor had mentioned it once briefly during class. Also got the same answer from trying the other two methods. Thanks for the help.
 
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1. What is the concept of free energy in thermodynamics?

Free energy is a measure of the amount of energy in a system that is available to do work. It is a combination of the system's internal energy and the energy that can be used to do work, such as mechanical or electrical work.

2. How is free energy related to energy and entropy?

In thermodynamics, free energy is related to energy and entropy through the equation: ΔG = ΔH - TΔS, where ΔG is the change in free energy, ΔH is the change in enthalpy, T is the temperature, and ΔS is the change in entropy. This equation shows that free energy is dependent on both the energy and the disorder of a system.

3. What is the significance of free energy in chemical reactions?

In chemical reactions, free energy is a measure of the spontaneity of the reaction. A negative ΔG indicates a spontaneous reaction, while a positive ΔG indicates a non-spontaneous reaction. Free energy also determines the equilibrium point of a reaction, where ΔG = 0.

4. How is free energy calculated for a discrete distribution?

For a discrete distribution, free energy can be calculated using the formula: F = -kT∑Pi ln(Pi), where F is the free energy, k is the Boltzmann constant, T is the temperature, and Pi is the probability of a specific state. This formula is based on the Boltzmann distribution, which describes the probability of a system being in a certain state at a given temperature.

5. What is the role of free energy in determining the stability of a system?

Free energy is a key factor in determining the stability of a system. A system with lower free energy is more stable, as it has a lower tendency to change or react. This is because a system with lower free energy has a lower energy state and is closer to its equilibrium point, making it less likely to undergo spontaneous changes.

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