Free energy of a rotational system.

[Narcol2000]

If one has a diatomic molecule with energy levels

$$\epsilon_l = \frac{h^2 l(l+1)}{2I}$$

l = 0,1,2,3,4,5...

if the degneracy is given by $$g_l = (2l+1)$$

How does one show that the Helmholtz free energy at low temperature ($$h^2/Ikt$$ large)
is given by

$$F = -3kT e^{-h^2 / IkT} + ...$$

I got as far as getting the partition function to be

$$Z = \sum_{l=0}^{\inf} (2l+1)e^{-h^2 l(l+1)/2IkT}$$

 

[Count Iblis]

If one has a diatomic molecule with energy levels

$$\epsilon_l = \frac{h^2 l(l+1)}{2I}$$

l = 0,1,2,3,4,5...

if the degneracy is given by $$g_l = (2l+1)$$

How does one show that the Helmholtz free energy at low temperature ($$h^2/Ikt$$ large)
is given by

$$F = -3kT e^{-h^2 / IkT} + ...$$

I got as far as getting the partition function to be

$$Z = \sum_{l=0}^{\inf} (2l+1)e^{-h^2 l(l+1)/2IkT}$$

Take the first two terms of the summation and then use that
F = - k T Log(Z). The term Log(Z) is of the form

Log(1 + small term) = small term - small term^2/2 = approximately small term

 

[astrokreng]

I can provide an explanation for the given expression of Helmholtz free energy at low temperature. To start with, the Helmholtz free energy is a measure of the amount of work that a system can do while maintaining a constant temperature. In other words, it is the energy that is available to do work in a system at a given temperature.

In the case of a rotational system, the energy levels are given by \epsilon_l = \frac{h^2 l(l+1)}{2I}, where l is the rotational quantum number and I is the moment of inertia of the molecule. The degeneracy of each energy level is given by g_l = (2l+1), which represents the number of different ways a given energy level can be occupied by the molecule.

To calculate the Helmholtz free energy at low temperature, we need to consider the partition function, which is defined as the sum of all possible states of a system weighted by their respective energies. In this case, the partition function can be written as:

Z = \sum_{l=0}^{\infty} (2l+1)e^{-\epsilon_l / kT}

Substituting the expression for energy levels, we get:

Z = \sum_{l=0}^{\infty} (2l+1)e^{-h^2 l(l+1)/2IkT}

At low temperatures, the energy levels are highly populated and the degeneracy plays a significant role in the partition function. We can approximate the sum by considering only the first term, as the subsequent terms will contribute very little due to the exponential factor. This approximation is valid when h^2/IkT is large.

Therefore, we can write the partition function as:

Z \approx (2(0)+1)e^{-h^2 (0)(0+1)/2IkT} = 1

Using this value in the expression for Helmholtz free energy, we get:

F = -kT ln Z \approx -kT ln(1) = 0

This means that at low temperatures, the Helmholtz free energy is close to zero, indicating that there is a large amount of energy available for the system to do work. However, as the temperature increases, the higher energy levels become more populated and the degeneracy factor becomes less significant, leading to a decrease in the available free energy