Degeneracy (W) [ENDORSED]

[Alexia Joseph 2B]

Can someone quickly explain the concept of/meaning of degeneracy?

[Nisarg Shah 1C]

Degeneracy is the number of ways of achieving a given energy state, and is directly related to entropy using the equation Dr. Lavelle described in class with Boltzmann's constant. Gas has a higher degeneracy than a liquid or solid, because its molecules can exist in multiple states because the intermolecular interactions of a gas are much less, so its molecules are not as a rigid as in a liquid or solid. Therefore, a gas has higher entropy.

[Rachel Formaker 1E]

Degeneracy is the number of possible states that a system can have.

For example, if one particle can be in one of two sides of a flask (similar to the example in class) the degeneracy is W=2 because the system has two possible states (the particle on either side of the flask).

An equation for degeneracy of a 2-state system is W=2^N, where N is the number of objects in the system.

The base of the equation can be modified to fit the number of possible states.
For example, in a 3-state system (the object can be in one of 3 possible places), W=3^N.

So the base of the equation is equal to the number of states in the system, and N is equal to the number of objects in the system.

[Fatima_Iqbal_2E]

In lecture, Dr. Lavelle describes degeneracy as the number of ways that a given energy state could be achieved. You could find this number by using W=2^N if you had a 2-State system, 3^N if you had a 3-State system, etc.

[PeterTran1C]

Degeneracy, like previously stated, is the number of states a system can have. The more states, the higher its degeneracy. Moreover, it can be seen that the more complex a molecule is, the higher its degeneracy as well.

[Magdalena Palavecino 1A]

I have seen some problems in which W also sometimes is calculated using Avogadro's number. When would this be the case instead of doing 2^molecules in the system?

[Chem_Mod]

I have seen some problems in which W also sometimes is calculated using Avogadro's number. When would this be the case instead of doing 2^molecules in the system?

When you have n moles.
Using a similar example I did in class:

4 moles of CO would have a residual entropy of: S = k ln W where W = 2^{4 x Avogadro's number}

[Dang Lam]

I have seen some problems in which W also sometimes is calculated using Avogadro's number. When would this be the case instead of doing 2^molecules in the system?

When you have n moles.
Using a similar example I did in class:

4 moles of CO would have a residual entropy of: S = k ln W where W = 2^{4 x Avogadro's number}

So to clarify, when the question gives us a certain "# moles of molecules" we have to multiply by Avogadro's number to find the # of particles?

[Angela 1K]

Yes! As seen through stoichiometry.