Statistical mechanics - Helmholtz free energy and Z

In summary, the conversation is about finding the temperature of a certain star based on spectroscopic measurements of neutral carbon atoms. It is assumed that the atoms are in thermal equilibrium and that the population of higher levels is negligible. Using the fact that 10% of the atoms are in the excited level, a temperature of 5900 K is calculated. However, when plugging in the values into the equations P(s) = exp(-F/kT)/Z and F = -k*T*log(Z), there is a discrepancy and it is unclear where the error lies. Further analysis is needed to identify the mistake.
  • #1
Niles
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Homework Statement


The neutnral carbon atom has a 9-fold degeerate ground level and a 5-fold degenerate excited level at an energy 0.82 eV above the ground level. Spectroscopic measurements of a certain star show that 10% of the neutral carbon atoms are in the excited level, and that the population of higher levels is negligible. Assuming thermal equilibrium, find the temperature.

The Attempt at a Solution



Using the fact that P(s) = 0.1 = ..., I have found that T = 5900 K.

Now I want to verify F = -k*T*log(Z) and P(s) = exp(-F / kt)/Z by plugging in the same values, but it doesn't make sense. This is my F:

[tex]F = - k \cdot 5900K \cdot \ln \left( Z \right) = - k \cdot 5900K \cdot \ln \left( {9 + 5 \cdot e^{\frac{{ - 0.82}}{{k \cdot 5900K}}} } \right)[/tex]

I insert this is in P(s) = exp(-F/kT)/Z:

[tex]P(s) = \frac{1}{Z} \cdot \exp \left( {\frac{{ - F}}{{kT}}} \right) = \frac{1}{{9 + 5 \cdot e^{\frac{{ - 0.82}}{{k \cdot 5900K}}} }} \cdot \exp \left( {\ln \left( {9 + 5 \cdot e^{\frac{{ - 0.82}}{{k \cdot 5900K}}} } \right)} \right)[/tex]. I have NOT multiplied with 5 since we are using F.

I can see that it gives one, since e takes the logarithm, but still - it doesn't make sense, since it should equal 0.1. Can you see where my error is?
 
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  • #2
Is the problem that [tex]P(s) = \frac{1}{Z} \cdot \exp \left( {\frac{{ - F}}{{kT}}} \right)[/tex] only works when I write [tex]F=E-TS[/tex] and NOT [tex]F=-k\cdot T\cdot \ln(Z)[/tex]?

What I do know is that there is something fundementally wrong in the way I use this expression. It would be great if you could point out my error.
 
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Related to Statistical mechanics - Helmholtz free energy and Z

1. What is statistical mechanics and how does it relate to Helmholtz free energy and Z?

Statistical mechanics is a branch of physics that uses statistical methods to explain the behavior of a large number of particles. It relates to Helmholtz free energy and Z by providing a way to calculate the thermodynamic properties of a system, such as energy, temperature, and entropy, using statistical principles. Z, also known as the partition function, is a key concept in statistical mechanics that represents the sum of all possible states of a system.

2. How is Helmholtz free energy different from other thermodynamic potentials?

Helmholtz free energy, also known as the A potential, is a thermodynamic potential that is used to describe the energy available to do work in a system at a constant temperature and volume. It differs from other potentials, such as the internal energy and enthalpy, in that it takes into account the work that can be done by the system on its surroundings.

3. What is the significance of Helmholtz free energy in thermodynamics?

Helmholtz free energy is significant in thermodynamics because it represents the maximum amount of work that can be obtained from a system at a constant temperature and volume. It also allows for the calculation of other important thermodynamic quantities, such as entropy and temperature, which are essential in understanding the behavior of a system.

4. How does Helmholtz free energy relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. Helmholtz free energy is related to this law because it can be used to calculate the change in entropy of a system, and therefore determine whether a process is thermodynamically favorable or not.

5. Can Helmholtz free energy be used to predict the equilibrium state of a system?

Yes, Helmholtz free energy can be used to predict the equilibrium state of a system. At equilibrium, the free energy of a system is at its minimum value, meaning that there is no net change in the system. By calculating the free energy of a system at different conditions, we can determine which conditions will lead to equilibrium.

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